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Stamp CD1 4/13 US ISM fornull Problem 1-1 Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column has a mass of 200 kg/m. a( ) Given: g 9.81 m s2 := wBC 300 kg m := LBC 3m:= wCA 400 kg m := FB 5kN:= LCA 1.2m:= FC 3kN:= Solution: +↑Σ Fy = 0; FA wBC g⋅( ) LBC⋅− wCA g⋅( ) LCA⋅− FB− 2FC− 0= FA wBC g⋅( ) LBC⋅ wCA g⋅( ) LCA⋅+ FB+ 2FC+:= FA 24.5 kN= Ans b( ) Given: g 9.81 m s2 := w 200 kg m := L 3m:= F1 6kN:= FB 8kN:= F2 4.5kN:= Solution: +↑Σ Fy = 0; FA w L⋅( ) g⋅− FB− 2F1− 2F2− 0= FA w L⋅( ) g⋅ FB+ 2F1+ 2F2+:= FA 34.89 kN= Ans Problem 1-2 Determine the resultant internal torque acting on the cross sections through points C and D of the shaft. The shaft is fixed at B. Given: TA 250N m⋅:= TCD 400N m⋅:= TDB 300N m⋅:= Solution: Equations of equilibrium: + TA TC− 0= TC TA:= TC 250 N m⋅= Ans + TA TCD− TD+ 0= TD TCD TA−:= TD 150 N m⋅= Ans Problem 1-3 Determine the resultant internal torque acting on the cross sections through points B and C. Given: TD 500N m⋅:= TBC 350N m⋅:= TAB 600N m⋅:= Solution: Equations of equilibrium: Σ Mx = 0; TB TBC TD−+ 0= TB TBC− TD+:= TB 150 N m⋅= Ans Σ Mx = 0; TC TD− 0= TC TD:= TC 500 N m⋅= Ans Problem 1-4 A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A. Given: P 80N:= θ 30deg:= φ 45deg:= a 0.3m:= b 0.1m:= Solution: Equations of equilibrium: + ΣFx'=0; NA P cos φ θ−( )⋅− 0= NA P cos φ θ−( )⋅:= NA 77.27 N= Ans + ΣFy'=0; VA P sin φ θ−( )⋅− 0= VA P sin φ θ−( )⋅:= VA 20.71 N= Ans + ΣΜA=0; MA P cos φ( )⋅ a⋅ cos θ( )+ P sin φ( )⋅ b a sin θ( )+( )⋅− 0= MA P− cos φ( )⋅ a⋅ cos θ( ) P sin φ( )⋅ b a sin θ( )+( )⋅+:= MA 0.555− N m⋅= Ans Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Problem 1-5 Determine the resultant internal loadings acting on the cross section through point D of member AB. Given: ME 70N m⋅:= a 0.05m:= b 0.3m:= Solution: Segment AB: Support Reactions + ΣΜA=0; −ME By 2 a⋅ b+( )⋅− 0= By ME− 2a b+:= By 175− N= +At B: Bx By 150 200 ⎛⎜⎝ ⎞ ⎠⋅:= Bx 131.25− N= Segment DB: NB Bx−:= VB By−:= + ΣFx=0; ND NB+ 0= ND NB−:= ND 131.25− N= Ans + ΣFy=0; VD VB+ 0= VD VB−:= VD 175− N= Ans + ΣΜD=0; MD− ME− By a b+( )⋅− 0= MD ME− By a b+( )⋅−:= MD 8.75− N m⋅= Ans Problem 1-6 The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal loadings acting on the cross section at point D. Given: P 5000N:= a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:= e 0.6m:= Solution: θ atan b d ⎛⎜⎝ ⎞ ⎠:= θ 36.87 deg= φ atan a b+ d ⎛⎜⎝ ⎞ ⎠ θ−:= φ 14.47 deg= Member AB: + ΣΜA=0; FBC sin φ( )⋅ a b+( )⋅ P b( )⋅− 0= FBC P b( )⋅ sin φ( ) a b+( )⋅:= FBC 12.01 kN= Segment BD: + ΣFx=0; ND− FBC cos φ( )⋅− P cos θ( )⋅− 0= ND FBC− cos φ( )⋅ P cos θ( )⋅−:= ND 15.63− kN= Ans + ΣFy=0; VD FBC sin φ( )⋅+ P sin θ( )⋅− 0= VD FBC− sin φ( )⋅ P sin θ( )⋅+:= VD 0 kN= Ans + ΣΜD=0; FBC sin φ( )⋅ P sin θ( )⋅−( ) d c−sin θ( )⋅ MD− 0= MD FBC sin φ( )⋅ P sin θ( )⋅−( ) d c−sin θ( )⋅:= MD 0 kN m⋅= Ans Note: Member AB is the two-force member. Therefore the shear force and moment are zero. Problem 1-7 Solve Prob. 1-6 for the resultant internal loadings acting at point E. Given: P 5000N:= a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:= e 0.6m:= Solution: θ atan b d ⎛⎜⎝ ⎞ ⎠:= θ 36.87 deg= φ atan a b+ d ⎛⎜⎝ ⎞ ⎠ θ−:= φ 14.47 deg= Member AB: + ΣΜA=0; FBC sin φ( )⋅ a b+( )⋅ P b( )⋅− 0= FBC P b( )⋅ sin φ( ) a b+( )⋅:= FBC 12.01 kN= Segment BE: + ΣFx=0; NE− FBC cos φ( )⋅− P cos θ( )⋅− 0= NE FBC− cos φ( )⋅ P cos θ( )⋅−:= NE 15.63− kN= Ans + ΣFy=0; VE FBC sin φ( )⋅+ P sin θ( )⋅− 0= VE FBC− sin φ( )⋅ P sin θ( )⋅+:= VE 0 kN= Ans + ΣΜE=0; FBC sin φ( )⋅ P sin θ( )⋅−( ) e⋅ ME− 0= ME FBC sin φ( )⋅ P sin θ( )⋅−( ) e⋅:= ME 0 kN m⋅= Ans Note: Member AB is the two-force member. Therefore the shear force and moment are zero. Problem 1-8 The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through points A, B, and C. Given: P 1500N:= w 750 N m := a 2.1m:= b 1.5m:= c 0.6m:= d 2.4m:= e 0.9m:= Solution: Equations of Equilibrium: For point A + ΣFx=0; NA 0:= Ans + VA w e⋅− P− 0=ΣFy=0; VA w e⋅ P+:= VA 2.17 kN= Ans + ΣΜA=0; MA− w e⋅( ) 0.5 e⋅( )⋅− P e( )⋅− 0= MA w e⋅( )− 0.5 e⋅( )⋅ P e( )⋅−:= MA 1.654− kN m⋅= Ans Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ΣFx=0; NB 0:= Ans + VB w d e+( )⋅− P− 0=ΣFy=0; VB w d e+( )⋅ P+:= VB 3.98 kN= Ans + ΣΜB=0; −MB w d e+( )⋅[ ] 0.5 d e+( )⋅[ ]⋅− P d e+( )⋅− 0= MB w d e+( )⋅[ ]− 0.5 d e+( )⋅[ ]⋅ P d e+( )⋅−:= MB 9.034− kN m⋅= Ans Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ΣFx=0; VC 0:= Ans + NC− w b c+ d+ e+( )⋅− P− 0=ΣFy=0; NC w− b c+ d+ e+( )⋅ P−:= NC 5.55− kN= Ans + ΣΜB=0; MC− w c d+ e+( )⋅[ ] 0.5 c d+ e+( )⋅[ ]⋅− P c d+ e+( )⋅− 0= MC w c d+ e+( )⋅[ ]− 0.5 c d+ e+( )⋅[ ]⋅ P c d+ e+( )⋅−:= MC 11.554− kN m⋅= Ans Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown on FBD. Problem 1-9 The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e., at the centroid point A of section a-a. Given: P 400N:= θ 30deg:= φ 45deg:= a 4mm:= b 5.75mm:= Solution: α φ θ−:= Equations of equilibrium: For section a -a + ΣFx'=0; VA P cos α( )⋅− 0= VA P cos α( )⋅:= VA 386.37 N= Ans + ΣFy'=0; NA sin α( )− 0= NA P sin α( )⋅:= NA 103.53 N= Ans + ΣΜA=0; MA− P sin α( )⋅ a⋅− P cos α( )⋅ b⋅+ 0= MA P− sin α( )⋅ a⋅ P cos α( )⋅ b⋅+:= MA 1.808 N m⋅= Ans Problem 1-10 The beam supports the distributed load shown. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical. Given: w1 4.5 kN m := w2 6.0 kN m := a 1.8m:= b 1.8m:= c 2.4m:= d 1.35m:= e 1.35m:= Solution: L1 a b+ c+:= L2 d e+:= Support Reactions: + ΣΜA=0; By L1⋅ w1 L1⋅( ) 0.5 L1⋅( )− 0.5w2 L2⋅( ) L1 L23+⎛⎜⎝ ⎞ ⎠⋅− 0= By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L23 L1⋅+ ⎛⎜⎝ ⎞ ⎠ ⋅+:= By 22.82 kN= + ΣFy=0; Ay By+ w1 L1⋅− 0.5w2 L2⋅− 0= Ay By− w1 L1⋅+ 0.5w2 L2⋅+:= Ay 12.29 kN= Equations of Equilibrium: For point C + ΣFx=0; NC 0:= Ans + ΣFy=0; Ay w1 a b+( )⋅−⎡⎣ ⎤⎦ VC− 0= VC Ay w1 a b+( )⋅−⎡⎣ ⎤⎦−:= VC 3.92 kN= Ans + ΣΜC=0; MC w1 a b+( )⋅⎡⎣ ⎤⎦ 0.5⋅ a b+( )⋅+ Ay a b+( )⋅− 0= MC w1 a b+( )⋅⎡⎣ ⎤⎦− 0.5⋅ a b+( )⋅ Ay a b+( )⋅+:= MC 15.07 kN m⋅= Ans Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD. Problem 1-11 The beam supports the distributed load shown. Determine the resultant internal loadings on the cross sections through points D and E. Assume the reactions at the supports A and B are vertical. Given: w1 4.5 kN m := w2 6.0 kN m := a 1.8m:= b 1.8m:= c 2.4m:= d 1.35m:= e 1.35m:= Solution: L1 a b+ c+:= L2 d e+:= Support Reactions: + ΣΜA=0; By L1⋅ w1 L1⋅( ) 0.5 L1⋅( )− 0.5w2 L2⋅( ) L1 L23+⎛⎜⎝ ⎞ ⎠⋅− 0= By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L23 L1⋅+ ⎛⎜⎝ ⎞ ⎠ ⋅+:= By 22.82 kN= + ΣFy=0; Ay By+ w1 L1⋅− 0.5w2 L2⋅− 0= Ay By− w1 L1⋅+ 0.5w2 L2⋅+:= Ay 12.29 kN= Equations of Equilibrium: For point D + ΣFx=0; ND 0:= Ans + ΣFy=0; Ay w1 a( )⋅−⎡⎣ ⎤⎦ VD− 0= VD Ay w1 a( )⋅−:= VD 4.18 kN= Ans + ΣΜD=0; MD w1 a( )⋅⎡⎣ ⎤⎦ 0.5⋅ a( )⋅+ Ay a( )⋅− 0= MD w1 a( )⋅⎡⎣ ⎤⎦− 0.5⋅ a( )⋅ Ay a( )⋅+:= MD 14.823 kN m⋅= Ans Equations of Equilibrium: For point E + ΣFx=0; NE 0:= Ans + ΣFy=0; VE 0.5w2 0.5 e⋅( )⋅− 0= VE 0.5w2 0.5 e⋅( )⋅:= VE 2.03 kN= Ans + ΣΜD=0; ME− 0.5w2 0.5 e⋅( )⋅⎡⎣ ⎤⎦ e3 ⎛⎜⎝ ⎞ ⎠⋅− 0= ME 0.5− w2 0.5 e⋅( )⋅⎡⎣ ⎤⎦ e3 ⎛⎜⎝ ⎞ ⎠⋅:= ME 0.911− kN m⋅= Ans Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD. Problem 1-12 Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section is located through the centroid, point C. Given: w 9 kN m := θ 45deg:= a 1.2m:= b 2.4m:= Solution: L a b+:= Support Reactions: + ΣΜA=0; Bx− L⋅ sin θ( )⋅ w L⋅( ) 0.5 L⋅( )+ 0= Bx w L⋅( ) 0.5 L⋅( )⋅ L sin θ( )⋅:= Bx 22.91 kN= + ΣFy=0; Ay w L⋅ sin θ( )⋅− 0= Ay w L⋅ sin θ( )⋅:= Ay 22.91 kN= + ΣFx=0; Bx w L⋅ cos θ( )⋅− Ax+ 0= Ax w L⋅ cos θ( )⋅ Bx−:= Ax 0− kN= (a) Equations of equilibrium: For Section a - a : + ΣFx=0; NC Ay sin θ( )⋅+ 0= NC Ay− sin θ( )⋅:= + ΣFy=0; VC Ay cos θ( )⋅+ w a⋅− 0= VC Ay− cos θ( )⋅ w a⋅+:= VC 5.4− kN= Ans + ΣΜA=0; MC− w a⋅( ) 0.5 a⋅( )⋅− Ay cos θ( ) a⋅+ 0= MC w a⋅( ) 0.5 a⋅( )⋅ Ay cos θ( ) a⋅−:= MC 12.96− kN m⋅= Ans (b) Equations of equilibrium: For Section b - b : + ΣFx=0; NC w a⋅ cos θ( )⋅+ 0= NC w− a⋅ cos θ( )⋅:= NC 7.64− kN= Ans + ΣFy=0; VC w a⋅ sin θ( )⋅− Ay+ 0= VC w a⋅ sin θ( )⋅ Ay−:= VC 15.27− kN= Ans + ΣΜA=0; MC− w a⋅( ) 0.5 a⋅( )⋅− Ay cos θ( ) a⋅+ 0= MC w a⋅( ) 0.5 a⋅( )⋅ Ay cos θ( ) a⋅−:= MC 12.96− kN m⋅= Ans NC 16.2− kN= Ans Problem 1-13 Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b) section b-b, each of which passes through point A. Take θ = 60 degree. The 650-N load is applied along the centroidal axis of the member. Given: P 650N:= θ 60deg:= (a) Equations of equilibrium: For Section a - a : + ΣFy=0; P Na_a− 0= Na_a P:= Na_a 650 N= Ans + ΣFx=0; Va_a 0:= Ans (b) Equations of equilibrium: For Section b - b : + ΣFy=0; Vb_b− P cos 90deg θ−( )⋅+ 0= Vb_b P cos 90deg θ−( )⋅:= Vb_b 562.92 N= Ans + ΣFx=0; Nb_b P sin 90deg θ−( )⋅− 0= Nb_b P sin 90deg θ−( )⋅:= Nb_b 325 N= Ans Problem 1-14 Determine the resultant internal normal and shear forces in the member at section b-b, each as a function of θ. Plot these results for 0o θ≤ 90o≤ . The 650-N load is applied along the centroidal axis of the member. Given: P 650N:= θ 0:= Equations of equilibrium: For Section b - b : + ΣFx0; Nb_b P cos θ( )⋅− 0= Nb_b P cos θ( )⋅:= Ans + ΣFy=0; Vb_b− P cos θ( )⋅+ 0= Vb_b P− cos θ( )⋅:= Ans Problem 1-15 The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N. Determine the resultant internal loadings acting on the cross section through point B in the beam. The beam has a weight of 600 N/m and is fixed to the wall at A. Given: W1 4000N:= w 600 N m := W2 450N:= a 1.2m:= b 1.2m:= c 0.9m:= d 0.9m:= e 1.2m:= f 0.45m:= r 0.075m:= Solution: Tension in rope: T W1 2 := T 2.00 kN= Equations of Equilibrium: For point B + ΣFx=0; NB− T−( ) 0= NB T−:= NB 2− kN= Ans + ΣFy=0; VB w e( )⋅− W1− 0= VB w e( )⋅ W1+:= VB 4.72 kN= Ans + ΣΜB=0; MB− w e( )⋅[ ] 0.5⋅ e( )⋅− W1 e r+( )⋅− T f( )⋅+ 0= MB w e( )⋅[ ]− 0.5⋅ e( )⋅ W1 e r+( )⋅− T f( )⋅+:= MB 4.632− kN m⋅= Ans Problem 1-16 Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob. 1-15. Given: W1 4000N:= w 600 N m := W2 450N:= a 1.2m:= b 1.2m:= c 0.9m:= d 0.9m:= e 1.2m:= f 0.45m:= r 0.075m:= Solution: Tension in rope: T W1 2 := T 2.00 kN= Equations of Equilibrium: For point C LC d e+:= + ΣFx=0; NC− T−( ) 0= NC T−:= NC 2− kN= Ans + ΣFy=0; VC w LC( )⋅− W1− 0= VC w LC( )⋅ W1+:= VC 5.26 kN= Ans + ΣΜC=0; MC− w LC( )⋅⎡⎣ ⎤⎦ 0.5⋅ LC( )⋅− W1 LC r+( )⋅− T f( )⋅+ 0= MC w LC( )⋅⎡⎣ ⎤⎦− 0.5⋅ LC( )⋅ W1 LC r+( )⋅− T f( )⋅+:= MC 9.123− kN m⋅= Ans Equations of Equilibrium: For point D LD b c+ d+ e+:= + ΣFx=0; ND 0:= ND 0 kN= Ans + ΣFy=0; VD w LD( )⋅− W1− W2− 0= VD w LD( )⋅ W1+ W2+:= VD 6.97 kN= Ans + ΣΜC=0; MD− w LD( )⋅⎡⎣ ⎤⎦ 0.5⋅ LD( )⋅− W1 LD r+( )⋅− W2 b( )⋅− 0= MD w LD( )⋅⎡⎣ ⎤⎦− 0.5⋅ LD( )⋅ W1 LD r+( )⋅− W2 b( )⋅−:= MD 22.932− kN m⋅= Ans Problem 1-17 Determine the resultant internal loadings acting on the cross section at point B. Given: w 900 kN m := a 1m:= b 4m:= Solution: L a b+:= Equations of Equilibrium: For point B + ΣFx=0; NB 0:= NB 0 kN= Ans + ΣFy=0; VB 0.5 w b L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ b( )⋅− 0= VB 0.5 w b L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ b( )⋅:= VB 1440 kN= Ans + ΣΜB=0; MB− 0.5 w b L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ b( )⋅ b 3 ⎛⎜⎝ ⎞ ⎠⋅− 0= MB 0.5− w b L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ b( )⋅ b 3 ⋅:= MB 1920− kN m⋅= Ans Problem 1-18 The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section through point C. Assume the reactions at the supports A and B are vertical. Given: w1 0.5 kN m := a 3m:= w2 1.5 kN m := Solution: L 3 a⋅:= w w2 w1−:= Support Reactions: + ΣΜA=0; By L⋅ w1 L⋅( ) 0.5 L⋅( )− 0.5 w( ) L⋅[ ] 2L3⎛⎜⎝ ⎞⎠⋅− 0= By w1 L⋅( ) 0.5( )⋅ 0.5 w( ) L⋅[ ] 23⎛⎜⎝ ⎞⎠⋅+:= By 5.25 kN= + ΣFy=0; Ay By+ w1 L⋅− 0.5 w( ) L⋅− 0= Ay By− w1 L⋅+ 0.5 w( ) L⋅+:= Ay 3.75 kN= Equations of Equilibrium: For point C + ΣFx=0; NC 0:= NC 0 kN= Ans + ΣFy=0; VC w1 a⋅+ 0.5 w a L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅+ Ay− 0= VC w1− a⋅ 0.5 w a L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅− Ay+:= VC 1.75 kN= Ans + ΣΜC=0; MC w1 a⋅( ) 0.5 a⋅( )+ 0.5 w aL⎛⎜⎝ ⎞⎠⋅⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅ a3⎛⎜⎝ ⎞⎠⋅+ Ay a⋅− 0= MC w1− a⋅( ) 0.5 a⋅( ) 0.5 w aL⎛⎜⎝ ⎞⎠⋅⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅ a3⎛⎜⎝ ⎞⎠⋅− Ay a⋅+:= MC 8.5 kN m⋅= Ans Problem 1-19 Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18. Given: w1 0.5 kN m := a 3m:= w2 1.5 kN m := Solution: L 3 a⋅:= w w2 w1−:= Support Reactions: + ΣΜA=0; By L⋅ w1 L⋅( ) 0.5 L⋅( )− 0.5 w( ) L⋅[ ] 2L3⎛⎜⎝ ⎞⎠⋅− 0= By w1 L⋅( ) 0.5( )⋅ 0.5 w( ) L⋅[ ] 23⎛⎜⎝ ⎞⎠⋅+:= By 5.25 kN= + ΣFy=0; Ay By+ w1 L⋅− 0.5 w( ) L⋅− 0= Ay By− w1 L⋅+ 0.5 w( ) L⋅+:= Ay 3.75 kN= Equations of Equilibrium: For point D + ΣFx=0; ND 0:= ND 0 kN= Ans + ΣFy=0; VD w1 2a( )⋅+ 0.5 w 2a L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ 2a( )⋅+ Ay− 0= VD w1− 2a( )⋅ 0.5 w 2 a⋅ L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ 2a( )⋅− Ay+:= VD 1.25− kN= Ans + ΣΜD=0; MD w1 2a( )⋅⎡⎣ ⎤⎦ a( )+ 0.5 w 2 a⋅L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ 2a( )⋅ 2a 3 ⎛⎜⎝ ⎞ ⎠⋅+ Ay 2a( )⋅− 0= MD w1− 2a( )⋅⎡⎣ ⎤⎦ a( ) 0.5 w 2 a⋅L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦⋅ 2a( )⋅ 2a 3 ⎛⎜⎝ ⎞ ⎠⋅− Ay 2a( )⋅+:= MD 9.5 kN m⋅= Ans Problem 1-20 The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kN on the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internal loadings at cross sections through points D, E, and F. Given: P 4kN:= a 1.2m:= b 1.8m:= Solution: Support Reactions: FBD (a) and (b). Given + ΣΜA=0; By a( )⋅ Bx 0.5 b⋅( )⋅+ P a( )⋅− 0= [1] + ΣΜC=0; Bx 0.5 b⋅( )⋅ P a( )⋅+ By a( )⋅− P a( )⋅− 0= [2] Solving [1] and [2]: Initial guess: Bx 1kN:= By 2kN:= Bx By ⎛⎜⎜⎝ ⎞ ⎠ Find Bx By,( ):= BxBy ⎛⎜⎜⎝ ⎞ ⎠ 2.67 2 ⎛⎜⎝ ⎞ ⎠ kN= From FBD (a): + ΣFx=0; Bx Ax− 0= Ax Bx:= Ax 2.67 kN= + ΣFy=0; Ay P− By− 0= Ay P By+:= Ay 6 kN= From FBD (b): + ΣFx=0; Cx Bx− 0= Cx Bx:= Cx 2.67 kN= + ΣFy=0; Cy By+ P− P− 0= Cy 2P By−:= Cy 6 kN= Equations of Equilibrium: For point D [FBD (c)]. + ΣFx=0; VD 0:= VD 0 kN= Ans + ΣFy=0; ND 0:= ND 0 kN= Ans + ΣΜD=0; MD 0:= MD 0 kN m⋅= Ans For point E [FBD (d)]. + ΣFx=0; VF Ax Cx−+ 0= VF Ax− Cx+:= VF 0 kN= Ans + ΣFy=0; NF Ay Cy−⋅− 0= NF Ay Cy+:= NF 12 kN= Ans + ΣΜF=0; MF Ax Cx+( ) 0.5 b⋅( )⋅− 0= MF Ax Cx+( ) 0.5 b⋅( )⋅:= MF 4.8 kN m⋅= Ans + ΣFx=0; Ax VE− 0= VE Ax:= + ΣFy=0; NE Ay− 0= NE Ay:= + ΣΜE=0; ME Ax 0.5 b⋅( )⋅− 0= ME Ax 0.5 b⋅( )⋅:= ME 2.4 kN m⋅= Ans For point F [FBD (e)]. VE 2.67 kN= Ans NE 6 kN= Ans Problem 1-21 The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. The gripping action on the drum chime is such that only horizontal and vertical forces are exerted on the drum at G and H. Determine the resultant internal loadings on the cross section through point I. Given: P 2.5 kN:= θ 60deg:= a 200mm:= b 125mm:= c 75mm:= d 125mm:= e 125mm:= f 50mm:= Solution: Equations of Equilibrium: Memeber Ac and BD are two-force members. ΣFy=0; P 2 F⋅ sin θ( )− 0= [1] F P 2 sin θ( )⋅:= [2] F 1.443 kN= Equations of Equilibrium: For point I. + ΣFx=0; VI F cos θ( )⋅− 0= VI F cos θ( )⋅:= VI 0.722 kN= Ans + ΣFy=0; NI− F sin θ( )⋅+ 0= NI F sin θ( )⋅:= NI 1.25 kN= Ans + ΣΜI=0; MI− F cos θ( )⋅ a( )⋅+ 0= MI F cos θ( )⋅ a( )⋅:= MI 0.144 kN m⋅= Ans Problem 1-22 Determine the resultant internal loadings on the cross sections through points K and J on the drum lifter in Prob. 1-21. Given: P 2.5 kN:= θ 60deg:= a 200mm:= b 125mm:= c 75mm:= d 125mm:= e 125mm:= f 50mm:= Solution: Equations of Equilibrium: Memeber Ac and BD are two-force members. ΣFy=0; P 2 F⋅ sin θ( )− 0= [1] F P 2 sin θ( )⋅:= [2] F 1.443 kN= Equations of Equilibrium: For point J. + ΣFy'=0; VI 0:= VI 0 kN= Ans + ΣFx'=0; NI F+ 0= NI F−:= NI 1.443− kN= Ans + ΣΜJ=0; MJ 0:= MJ 0 kN m⋅= Ans Note: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD. Support Reactions: For Member DFH : + ΣΜH=0; FEF c( )⋅ F cos θ( ) a b+ c+( )⋅− F sin θ( )⋅ f( )⋅+ 0= FEF F cos θ( )⋅ a b+ c+c⎛⎜⎝ ⎞ ⎠⋅ F sin θ( )⋅ f c ⎛⎜⎝ ⎞ ⎠⋅−:= FEF 3.016 kN= Equations of Equilibrium: For point K. + ΣFx=0; NK FEF+ 0= NK FEF:= NK 3.016 kN= Ans + ΣFy=0; VK 0:= VK 0 kN= Ans + ΣΜK=0; MK 0:= MK 0 kN m⋅= Ans Problem 1-23 The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section at B. Neglect the weight of the wrench CD. Given: g 9.81 m s2 := ρ 12 kg m := P 60N:= a 0.150m:= b 0.400m:= c 0.200m:= d 0.300m:= Solution: w ρ g⋅:= ΣFx=0; NBx 0N:= Ans ΣFy=0; VBy 0N:= Ans ΣFz=0; VBz P− P+ w b c+( )⋅− 0= VBz P P− w b c+( )⋅+:= VBz 70.6 N= Ans ΣΜx=0; TBx P b( )⋅+ P b( )⋅− w b⋅( ) 0.5 b⋅( )⋅− 0= TBx P− b( )⋅ P b( )⋅+ w b⋅( ) 0.5 b⋅( )⋅+:= TBx 9.42 N m⋅= Ans ΣΜy=0; MBy P 2a( )⋅− w b⋅( ) c( )⋅+ w c⋅( ) 0.5 c⋅( )⋅+ 0= MBy P 2 a⋅( )⋅ w b⋅( ) c( )⋅− w c⋅( ) 0.5 c⋅( )⋅−:= MBy 6.23 N m⋅= Ans ΣΜz=0; MBz 0N m⋅:= Ans Problem 1-24 The main beam AB supports the load on the wing of the airplane. The loads consist of the wheel reaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity at D, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A, determine the resultant internal loadings on the beam at this point. Assume that the wing does not transfer any of the loads to the fuselage, except through the beam. Given: PC 175kN:= PE 2kN:= PD 6kN:= a 1.8m:= b 1.2m:= e 0.3m:= c 0.6m:= d 0.45m:= Solution: ΣFx=0; VAx 0kN:= Ans ΣFy=0; NAy 0kN:= Ans ΣFz=0; VAz PD− PE− PC+ 0= VAz PD PE PC−+:= VAz 167− kN= Ans ΣΜx=0; MAx PD a( )⋅ PE a b+ c+( )⋅+ PC a b+( )⋅−:= MAx 507− kN m⋅= Ans ΣΜy=0; TAy PD d( )⋅+ PE e( )⋅− 0= TAy PD− d( )⋅ PE e( )⋅+:= TAy 2.1− kN m⋅= Ans ΣΜz=0; MAz 0kN m⋅:= Ans MAx PD a( )⋅− PE a b+ c+( )⋅− PC a b+( )⋅+ 0= Problem 1-25 Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of the sign. Given: a 4m:= d 2m:= p 50 N m2 := b 6m:= e 3m:= c 3m:= Solution: P p c( )⋅ d e+( )⋅:= ΣFx=0; VBx P− 0= VBx P:= VBx 750 N= Ans ΣFy=0; VBy 0N:= Ans ΣFz=0; NBz 0N:= Ans ΣΜx=0; MBx 0N m⋅:= Ans ΣΜy=0; MBy P b 0.5 c⋅+( )⋅− 0= MBy P b 0.5 c⋅+( )⋅:= MBy 5625 N m⋅= Ans ΣΜz=0; TBz P e 0.5 d e+( )⋅−[ ]⋅− 0= TBz P e 0.5 d e+( )⋅−[ ]⋅:= TBz 375 N m⋅= Ans Problem 1-26 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft. Given: P1z 400N:= P2y 200N:= P3y 80N:= a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:= Solution: L a b+ c+ d+:= Support Reactions: ΣΜz=0; 2P3y d( )⋅ 2P2y c d+( )⋅+ Ay L( )⋅− 0= Ay 2P3y d L ⋅ 2P2y c d+ L ⋅+:= Ay 245.71 N= ΣFy=0; Ay− By− 2 P2y⋅+ 2 P3y⋅+ 0= By Ay− 2 P2y⋅+ 2 P3y⋅+:= By 314.29 N= ΣΜy=0; 2P1z b c+ d+( )⋅ Az L( )⋅− 0= Az 2P1z b c+ d+ L ⋅:= Az 628.57 N= ΣFz=0; Bz Az+ 2 P1z⋅− 0= Bz Az− 2 P1z⋅+:= Bz 171.43 N= Equations of Equilibrium: For point D. ΣFx=0; NDx 0N:= Ans ΣFy=0; VDy By− 2 P3y⋅+ 0= VDy By 2 P3y⋅−:= VDy 154.3 N= Ans ΣFz=0; VDz Bz+ 0= VDz Bz−:= VDz 171.4− N= Ans ΣΜx=0; TDx 0N m⋅:= Ans ΣΜy=0; MDy Bz d 0.5 c⋅+( )⋅+ 0= MDy Bz d 0.5 c⋅+( )⋅⎡⎣ ⎤⎦−:= MDy 94.29− N m⋅= Ans ΣΜz=0; MDz By d 0.5 c⋅+( )⋅+ 2 P3y⋅ 0.5 c⋅( )⋅− 0= MDz By− d 0.5 c⋅+( )⋅ 2 P3y⋅ 0.5 c⋅( )⋅+:= MDz 148.86− N m⋅= Ans Problem 1-27 The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through point C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y direction. The journal bearings at A and B exert only y and z components of force on the shaft. Given: P1z 400N:= P2y 200N:= P3y 80N:= a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:= Solution: L a b+ c+ d+:= Support Reactions: ΣΜz=0; 2P3y d( )⋅ 2P2y c d+( )⋅+ Ay L( )⋅− 0= Ay 2P3y d L ⋅ 2P2y c d+ L ⋅+:= Ay 245.71 N= ΣFy=0; Ay− By− 2 P2y⋅+ 2 P3y⋅+ 0= By Ay− 2 P2y⋅+ 2 P3y⋅+:= By 314.29 N= ΣΜy=0; 2P1z b c+ d+( )⋅ Az L( )⋅− 0= Az 2P1z b c+ d+ L ⋅:= Az 628.57 N= ΣFz=0; Bz Az+ 2 P1z⋅− 0= Bz Az− 2 P1z⋅+:= Bz 171.43 N= Equations of Equilibrium: For point C. ΣFx=0; NCx 0N:= Ans ΣFy=0; VCy Ay− 0= VCy Ay:= VCy 245.7 N= Ans ΣFz=0; VCz Az+ 2P1z− 0= VCz Az− 2P1z+:= VCz 171.4 N= Ans ΣΜx=0; TCx 0N m⋅:= Ans ΣΜy=0; MCy Az a 0.5 b⋅+( )⋅− 2 P1z⋅ 0.5 b⋅( )⋅+ 0= MCy Az a 0.5 b⋅+( )⋅ 2 P1z⋅ 0.5 b⋅( )⋅−:= MCy 154.29 N m⋅= Ans ΣΜz=0; MCz Ay a 0.5 b⋅+( )⋅+ 0= MCz Ay− a 0.5 b⋅+( )⋅:= MCz 122.86− N m⋅= Ans Problem 1-28 Determine the resultant internal loadings acting on the cross section of the frame at points F and G. The contact at E is smooth. Given: a 1.2m:= b 1.5m:= c 0.9m:= P 400N:= d 0.9m:= e 1.2m:= θ 30deg:= Solution: L d2 e2+:= Member DEF : + ΣΜD=0; NE b( )⋅ P a b+( )⋅− 0= NE P a b+ b ⋅:= NE 720 N= Member BCE : + ΣΜB=0; FAC e L ⎛⎜⎝ ⎞ ⎠⋅ d( )⋅ NE sin θ( )⋅ c d+( )⋅− 0= FAC L e d⋅ ⎛⎜⎝ ⎞ ⎠ NE sin θ( )⋅ c d+( )⋅⎡⎣ ⎤⎦⋅:= FAC 900 N= + ΣFx=0; Bx FAC d L ⎛⎜⎝ ⎞ ⎠⋅+ NE cos θ( )⋅− 0= Bx FAC− d L ⎛⎜⎝ ⎞ ⎠⋅ NE cos θ( )⋅+:= Bx 83.54 N= + ΣFy=0; By− FAC e L ⎛⎜⎝ ⎞ ⎠⋅+ NE sin θ( )⋅− 0= By FAC e L ⎛⎜⎝ ⎞ ⎠⋅ NE sin θ( )⋅−:= By 360 N= Equations of Equilibrium: For point F. + ΣFy'=0; NF 0:= NF 0 N= Ans + ΣFx'=0; VF P− 0= VF P:= VF 400 N= Ans + ΣΜF=0; MF P 0.5 a⋅( )⋅− 0= MF P 0.5 a⋅( )⋅:= MF 240 N m⋅= Ans Equations of Equilibrium: For point G. + ΣFx=0; Bx NG− 0= NG Bx:= NG 83.54 N= Ans + ΣFy=0; VG By− 0= VG By:= VG 360 N= Ans + ΣΜG=0; MG− By 0.5 d⋅( )⋅+ 0= MG By 0.5 d⋅( )⋅:= MG 162 N m⋅= Ans Problem 1-29 The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting on the cross section at point C. Given: P 400N:= r 150mm:= θ 90deg:= Solution: Equations of Equilibrium: For segment AC. + ΣFx=0; NC P+ 0= NC P:= NC 400 N= Ans + ΣFy=0; VC 0:= VC 0 N= Ans + ΣΜG=0; MC P r( )⋅+ 0= MC P− r( )⋅:= MC 60− N m⋅= Ans Problem 1-30 The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings acting on the cross section through B. Given: P 750N:= MC 800N m⋅:= ρ 12 kg m := g 9.81 m s2 := a 1m:= b 2m:= c 2m:= Solution: Py 4 5 ⎛⎜⎝ ⎞ ⎠ P⋅:= Pz 3 5 − P⋅:= Equations of Equilibrium: For point B. ΣFx=0; VBx 0kip:= Ans ΣFy=0; NBy Py+ 0= NBy Py−:= Ans NBy 600− N= Ans ΣFz=0; VBz Pz+ ρ g⋅ c⋅− ρ g⋅ b⋅− 0= VBz Pz− ρ g⋅ c⋅+ ρ g⋅ b⋅+:= VBz 920.9 N= Ans ΣΜx=0; MBx Pz b( )⋅+ ρ g⋅ c⋅ b( )⋅− ρ g⋅ b⋅ 0.5 b⋅( )⋅− 0= MBx Pz− b( )⋅ ρ g⋅ c⋅ b( )⋅+ ρ g⋅ b⋅ 0.5 b⋅( )⋅+:= MBx 1606.3 N m⋅= Ans ΣΜy=0; TBy 0N m⋅:= Ans ΣΜz=0; MBy Mc+ 0= MBy MC−:= MBy 800− N m⋅= Ans Problem 1-31 The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings acting on the cross section through A which is located at an angle θ from the horizontal. Solution: P kN:= θ deg:= Equations of Equilibrium: For point A. + ΣFx=0; NA− P cos θ( )⋅+ 0= NA P cos θ( )⋅:= Ans + ΣFy=0; VA P sin θ( )⋅− 0= VA P sin θ( )⋅:= Ans + ΣΜA=0; MA P r⋅ 1 cos θ( )−( )⋅− 0= MA P r⋅ 1 cos θ( )−( )⋅:= r Ans Problem 1-32 The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine the resultant internal loadings acting on the cross section through point B. Hint: The distance from the centroid C of segment AB to point O is CO = 0.9745r. Given: θ 22.5deg:= r m:= a 0.9745r:= w kN m2 := Solution: Equations of Equilibrium: For point B. ΣFz=0; VB π 4 r⋅ w⋅− 0= VB 0.785 w⋅ r⋅:= Ans ΣFx=0; NB 0:= Ans ΣΜx=0; TB π 4 r⋅ w⋅ 0.09968r( )⋅− 0= TB 0.0783w r2⋅:= Ans ΣΜy=0; MB π 4 r⋅ w⋅ 0.37293r( )⋅+ 0= MB 0.293− w r2⋅:= Ans Problem 1-33 A differential element taken from a curved bar is shown in the figure. Show that dN/dθ = V, dV/dθ = -N, dM/dθ = -T, and dT/dθ = M, Solution: Problem 1-34 The column is subjected to an axial force of 8 kN, which is applied through the centroid of the cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution of stress acting over the area’s cross section. Given: P 8kN:= b 150mm:= d 140mm:= t 10mm:= Solution: A 2 b t⋅( ) d t⋅+:= A 4400.00 mm2= σ P A := σ 1.82 MPa= Ans Problem 1-35 The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine the average shear stress in the pin. Given: P 3.0kN:= d 6mm:= Solution: + ΣFy=0; 2 V⋅ P− 0= V 0.5P:= V 1500 N= A π d2⋅ 4 := A 28.2743 mm2= τavg V A := τavg 53.05 MPa= Ans Problem 1-36 While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times his weight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a. The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameter of 25 mm. Assume the fibula F does not support a load. Given: g 9.81 m s2 = M 75kg:= do 45mm:= di 25mm:= Solution: A π 4 do 2 di 2−⎛⎝ ⎞⎠⋅:= A 1099.5574 mm2= σ 5M g⋅ A := σ 3.345 MPa= Ans Problem 1-37 The thrust bearing is subjected to the loads shown. Determine the average normal stress developed on cross sections through points B, C, and D. Sketch the results on a differential volume element located at each section. Units Used: kPa 103Pa:= Given: P 500N:= Q 200N:= dB 65mm:= dC 140mm:= dD 100mm:= Solution: AB π dB2⋅ 4 := AB 3318.3 mm2= σB P AB := σB 150.7 kPa= Ans AC π dC2⋅ 4 := AC 15393.8 mm2= σC P AC := σC 32.5 kPa= Ans AD π dD2⋅ 4 := AD 7854.0 mm2= σD Q AD := σD 25.5 kPa= Ans Problem 1-38 The small block has a thickness of 5 mm. If the stress distribution at the support developed by the load varies as shown, determine the force F applied to the block, and the distance d to where it is applied. Given: a 60mm:= b 120mm:= t 5mm:= σ1 0MPa:= σ2 40MPa:= σ3 60MPa:= Solution: F Aσ⌠⎮⌡ d= F 0.5 σ2⋅ a t⋅( )⋅ σ2 b t⋅( )⋅+ 0.5 σ3 σ2−( )⋅ b t⋅( )⋅+:= F 36.00 kN= Ans Require: F d⋅ Ax σ⋅⌠⎮⌡ d= d 0.5 σ2⋅ a t⋅( )⋅⎡⎣ ⎤⎦ 2a3⋅ σ2 b t⋅( )⋅⎡⎣ ⎤⎦ a 0.5 b⋅+( )⋅+ 0.5 σ3 σ2−( )⋅ b t⋅( )⋅⎡⎣ ⎤⎦ a 2 b⋅3+⎛⎜⎝ ⎞⎠⋅+ F := d 110 mm= Ans Problem 1-39 The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a couple is applied to the lever, determine the average shear stress in the pin between the pin and lever. Given: a 250mm:= b 12mm:= d 6mm:= P 20N:= Solution: + ΣΜO=0; V b⋅ P 2a( )⋅− 0= V P 2a b ⎛⎜⎝ ⎞ ⎠⋅:= V 833.33 N= A π d2⋅ 4 := A 28.2743 mm2= τavg V A := τavg 29.47 MPa= Ans Problem 1-40 The cinder block has the dimensions shown. If the material fails when the average normal stress reaches 0.840 MPa, determine the largest centrally applied vertical load P it can support. Given: σallow 0.840MPa:= ao 150mm:= ai 100mm:= bo 2 1 2+ 3+( )⋅ 2+[ ] mm⋅:= bi 2 1 3+( )⋅[ ] mm⋅:= Solution: A ao bo⋅ ai bi⋅−:= A 1300 mm2= Pallow σallow A( )⋅:= Pallow 1.092 kN= Ans Problem 1-41 The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN, determine the average normal stress in the material. Show the result acting on a differential volume element of the material. Given: P 4kN:= ao 150mm:= ai 100mm:= bo 2 1 2+ 3+( )⋅ 2+[ ] mm⋅:= bi 2 1 3+( )⋅[ ] mm⋅:= Solution: A ao bo⋅ ai bi⋅−:= A 1300 mm2= σ P A := σ 3.08 MPa= Ans Problem 1-42 The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod is subjected to the greater average normal stress and compute its value. Take θ = 30°. The diameter of each rod is given in the figure. Given: W 250N:= θ 30deg:= φ 45deg:= dB 9mm:= dC 6mm:= dD 7.5mm:= Solution: Initial guess: FAC 1N:= FAD 1N:= Given + ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [1] + ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [2] Solving [1] and [2]: FAC FAD ⎛⎜⎜⎝ ⎞ ⎠ Find FAC FAD,( ):= FAC FAD ⎛⎜⎜⎝ ⎞ ⎠ 183.01 224.14 ⎛⎜⎝ ⎞ ⎠ N= Rod AB: AAB π dB2⋅ 4 := AAB 63.61725 mm2= σAB W AAB := σAB 3.93 MPa= Rod AD : AAD π dD2⋅ 4 := AAD 44.17865 mm2= σAD FAD AAD := σAD 5.074 MPa= Rod AC: AAC π dC2⋅ 4 := AAC 28.27433 mm2= σAC FAC AAC := σAC 6.473 MPa= Ans Problem 1-43 Solve Prob. 1-42 for θ = 45°. Given: W 250N:= θ 45deg:= φ 45deg:= dB 9mm:= dC 6mm:= dD 7.5mm:= Solution: Initial guess: FAC 1N:= FAD 1N:= Given + ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [1] + ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [2] Solving [1] and [2]: FAC FAD ⎛⎜⎜⎝ ⎞ ⎠ Find FAC FAD,( ):= FAC FAD ⎛⎜⎜⎝ ⎞ ⎠ 176.78 176.78 ⎛⎜⎝ ⎞ ⎠ N= Rod AB: AAB π dB2⋅ 4 := AAB 63.61725 mm2= σAB W AAB := σAB 3.93 MPa= Rod AD : AAD π dD2⋅ 4 := AAD 44.17865 mm2= σAD FAD AAD := σAD 4.001 MPa= Rod AC: AAC π dC2⋅ 4 := AAC 28.27433 mm2= σAC FAC AAC := σAC 6.252 MPa= Ans Problem 1-44 The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle of orientation θ of AC such that the average normal stress in rod AC is twice the average normal stress in rod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure. Given: W 250N:= φ 45deg:= dB 9mm:= dC 6mm:= dD 7.5mm:= Solution: Rod AB: AAB π dB2⋅ 4 := AAB 63.61725 mm2= Rod AD : AAD π dD2⋅ 4 := AAD 44.17865 mm2= Rod AC: AAC π dC2⋅ 4 := AAC 28.27433 mm2= Since σAC 2σAD= Therefore FAC AAC 2 FAD AAD = Initial guess: FAC 1N:= FAD 2N:= θ 30deg:= Given FAC AAC 2 FAD AAD = [1] + ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [2] + ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [3] Solving [1], [2] and [3]: FAC FAD θ ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find FAC FAD, θ,( ):= FACFAD ⎛⎜⎜⎝ ⎞ ⎠ 180.38 140.92 ⎛⎜⎝ ⎞ ⎠ N= θ 56.47 deg= σAB W AAB := σAB 3.93 MPa= Ans σAD FAD AAD := σAD 3.19 MPa= Ans σAC FAC AAC := σAC 6.38 MPa= Ans Problem 1-45 The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter hole in the fixed support A, determine the bearing stress acting on the collar C. Also, what is the average shear stress acting along the inside surface of the collar where it is fixed connected to the 52-mm diameter shaft? Given: P 30kN:= dhole 53mm:= dshaft 52mm:= dcollar 60mm:= hcollar 10mm:= Solution: Bearing Stress: Ab π 4 dcollar 2 dhole 2−⎛⎝ ⎞⎠⋅:= σb P Ab := σb 48.3 MPa= Ans Average Shear Stress: As π dshaft( )⋅ hcollar( )⋅:= τavg P As := τavg 18.4 MPa= Ans Problem 1-46 The two steel members are joined together using a 60° scarf weld. Determine the average normal and average shear stress resisted in the plane of the weld. Given: P 8kN:= θ 60deg:= b 25mm:= h 30mm:= Solution: Equations of Equilibrium: + ΣFx=0; N P sin θ( )⋅− 0= N P sin θ( )⋅:= N 6.928 kN= + ΣFy=0; V P cos θ( )⋅− 0= V P cos θ( )⋅:= V 4 kN= A h b⋅ sin θ( ):= σ N A := σ 8 MPa= Ans τavg V A := τavg 4.62 MPa= Ans Problem 1-47 The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine the average normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin. Given: P 775N:= a 40mm:= b 30mm:= d 8mm:= θ 20deg:= Solution: Support Reaction: ΣFA=0; P a( )⋅ FBC cos θ( )⋅ a b+( )⋅− 0= FBC P a⋅ a b+( ) cos θ( )⋅:= FBC 471.28 N= Average Normal Stress: ABC π d2⋅ 4 := σ FBC ABC := σ 9.38 MPa= Ans Problem 1-48 The board is subjected to a tensile force of 425 N. Determine the average normal and average shear stress developed in the wood fibers that are oriented along section a-a at 15° with the axis of the board. Given: P 425N:= θ 15deg:= b 25mm:= h 75mm:= Solution: Equations of Equilibrium: + ΣFx=0; V P cos θ( )⋅− 0= V P cos θ( )⋅:= V 410.518 N= + ΣFy=0; N P sin θ( )⋅− 0= N P sin θ( )⋅:= N 1 N= Average Normal Stress: A h b⋅ sin θ( ):= σ N A := σ 0.0152 MPa= Ans τavg V A := τavg 0.0567 MPa= Ans Problem 1-49 The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determine the average normal and average shear stress components that this loading creates on the face of the weld, section AB. Given: P 250kN:= θ 30deg:= b 150mm:= h 50mm:= Solution: Equations of Equilibrium: + ΣFx=0; V− P sin θ( )⋅+ 0= V P sin θ( )⋅:= V 125 kN= + ΣFy=0; N P cos θ( )⋅− 0= N P cos θ( )⋅:= N 216.506 kN= Average Normal and Shear Stress: A h b⋅ sin 2θ( ):= σ N A := σ 25 MPa= Ans τavg V A := τavg 14.434 MPa= Ans Problem 1-50 The specimen failed in a tension test at an angle of 52° when the axial load was 100 kN. If the diameter of the specimen is 12 mm, determine the average normal and average shear stress acting on the area of the inclined failure plane. Also, what is the average normal stress acting on the cross section when failure occurs? Given: P 100kN:= d 12mm:= θ 52deg:= Solution: Equations of Equilibrium: + ΣFx=0; V P cos θ( )⋅− 0= V P cos θ( )⋅:= V 61.566 kN= + ΣFy=0; N P sin θ( )⋅− 0= N P sin θ( )⋅:= N 78.801 kN= Inclined plane: A π 4 d2 sin θ( ) ⎛⎜⎝ ⎞ ⎠:= σ N A := σ 549.05 MPa= Ans τavg V A := τavg 428.96 MPa= Ans Cross section: A πd2 4 := σ P A := σ 884.19 MPa= Ans τavg 0:= τavg 0 MPa= Ans Problem 1-51 A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the maximum average shear stress in the specimen and indicate the orientation θ of a section on which it occurs. Solution: Equations of Equilibrium: + ΣFy=0; V P cos θ( )⋅− 0= V P cos θ( )⋅= Inclined plane: Aincl A sin θ( )= τ V Aincl = τ P cos θ( )⋅ sin θ( )⋅ A = τ P sin 2θ( )⋅ 2A = dτ dθ P cos 2θ( )⋅ A = dτ dθ 0= cos 2θ( ) 0= 2θ 90deg= θ 45deg:= Ans τmax P sin 90°( )⋅ 2A = τmax P 2A = Ans Problem 1-52 The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting on sections AB and BC. Assume the member is smooth and is 50-mm thick. Given: P 5kN:= θ 45deg:= φ 60deg:= dAB 40mm:= dBC 50mm:= t 50mm:= Solution: α 90deg φ−:= α 30.00 deg= AAB t dAB⋅:= ABC t dBC⋅:= + ΣFx=0; NAB cos α( )⋅ P cos θ( )⋅− 0= NAB P cos θ( )⋅ cos α( ):= NAB 4.082 kN= + ΣFy=0; NAB− sin α( )⋅ P sin θ( )⋅+ NBC− 0= NBC NAB− sin α( )⋅ P sin θ( )⋅+:= NBC 1.494 kN= σAB NAB AAB := σAB 2.041 MPa= Ans σBC NBC ABC := σBC 0.598 MPa= Ans Problem 1-53 The yoke is subjected to the force and couple moment. Determine the average shear stress in the bolt acting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couple moment is resisted by a set of couple forces developed in the shank of the bolt. Given: P 2.5kN:= M 120N m⋅:= ho 62mm:= hi 50mm:= d 6mm:= θ 60deg:= Solution: As a force on bolt shank is zero, then τA 0:= Ans Equations od Equilibrium: ΣFz=0; P 2Fz− 0= Fz 0.5P:= Fz 1.25 kN= ΣMz=0; M Fx hi( )⋅− 0= Fx M hi := Fx 2.4 kN= Average Shear Stress: A πd2 4 := The bolt shank subjected to a shear force of VB Fx 2 Fz 2+:= τB VB A := τB 95.71 MPa= Ans Problem 1-54 The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 2 kN. Given: P 4.0kN:= b 37.5mm:= hhalf 25m:= θ 30deg:= Solution: Equations of Equilibrium: + ΣFx=0; V− 0.5P cos θ( )⋅+ 0= V 0.5P cos θ( )⋅:= V 1.732 kN= + ΣFy=0; N 0.5P sin θ( )⋅− 0= N 0.5P sin θ( )⋅:= N 1 kN= Average Normal and Shear Stress: A hhalf( ) b⋅ sin θ( ):= σ N A := σ 533.33 Pa= Ans τavg V A := τavg 923.76 Pa= Ans Problem 1-55 The row of staples AB contained in the stapler is glued together so that the maximum shear stress the glue can withstand is τ max = 84 kPa. Determine the minimum force F that must be placed on the plunger in order to shear off a staple from its row and allow it to exit undeformed through the groove at C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mm Assume all the other parts are rigid and neglect friction. Given: τmax 0.084MPa:= a 12.5mm:= b 7.5mm:= t 1.25mm:= Solution: Average Shear Stress: A a b⋅ a 2t−( ) b t−( )⋅[ ]−:= τmax V A = V τmax( ) A⋅:= V 2.63 N= Fmin V:= Fmin 2.63 N= Ans Problem 1-56 Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to the ring at B, determine the average normal stress in each rod if θ = 60°. Given: W 8kN:= θ 60deg:= dA 4mm:= dC 6mm:= Solution: Rod AB: AAB π dA2⋅ 4 := Rod BC : ABC π dC2⋅ 4 := + ΣFy=0; FBC sin θ( )⋅ W− 0= FBC W sin θ( ):= FBC 9.238 kN= + ΣFx=0; FBC cos θ( )⋅ FAB− 0= FAB FBC cos θ( )⋅:= FAB 4.619 kN= σAB FAB AAB := σAB 367.6 MPa= Ans σBC FBC ABC := σBC 326.7 MPa= Ans Problem 1-57 Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN is applied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rod is equivalent. What is this stress? Given: W 8kN:= dA 4mm:= dC 6mm:= Solution: Rod AB: AAB π dA2⋅ 4 := Rod BC : ABC π dC2⋅ 4 := + ΣFy=0; FBC sin θ( )⋅ W− 0= + ΣFx=0; FBC cos θ( )⋅ FAB− 0= Since FAB σ AAB⋅= FBC σ ABC⋅= Initial guess: σ 100MPa:= θ 50deg:= Given σ ABC⋅ sin θ( )⋅ W− 0= [1] σ ABC⋅ cos θ( )⋅ σ AAB⋅− 0= [2] Solving [1] and [2]: σ θ ⎛⎜⎝ ⎞ ⎠ Find σ θ,( ):= θ 63.61 deg= Ans σ 315.85 MPa= Ans Problem 1-58 The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normal stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive. Given: P 40kN:= a 0.9m:= b 1.2m:= A 780mm2:= Solution: c a2 b2+:= c 1.5 m= h b c := v a c := Joint A: + ΣFy=0; v( ) FAB⋅ P− 0= FAB P v := FAB 66.667 kN= + ΣFx=0; h( ) FAB⋅ FAE− 0= FAE h( ) FAB⋅:= FAE 53.333 kN= σAB FAB A := σAB 85.47 MPa= (T) Ans σAE FAE A := σAE 68.376 MPa= (C) Ans Joint E: + ΣFy=0; FEB 0.75P− 0= FEB 0.75P:= FEB 30 kN= + ΣFx=0; FED FAE− 0= FED FAE:= FED 53.333 kN= σEB FEB A := σEB 38.462 MPa= (T) Ans σED FED A := σED 68.376 MPa= (C) Ans Joint B: + ΣFy=0; v( ) FBD⋅ v( ) FAB⋅− FEB− 0= FBD FAB FEB v ⎛⎜⎝ ⎞ ⎠+:= FBD 116.667 kN= + ΣFx=0; FBC h( )FAB− h( )FBD− 0= FBC h( )FAB h( )FBD+:= FBC 146.667 kN= σBC FBC A := σBC 188.034 MPa= (T) Ans σBD FBD A := σBD 149.573 MPa= (C) Ans Problem 1-59 The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normal stress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that can be applied to the truss. σallow 140MPa:=Given: a 0.9m:= b 1.2m:= A 780mm2:= Solution: c a2 b2+:= c 1.5 m= h b c := v a c := For comparison purpose, set P 1kN:= Joint A: + ΣFy=0; v( ) FAB⋅ P− 0= FAB P v := FAB 1.667 kN= + ΣFx=0; h( ) FAB⋅ FAE− 0= FAE h( ) FAB⋅:= FAE 1.333 kN= σAB FAB A := σAB 2.137 MPa= (T) σAE FAE A := σAE 1.709 MPa= (C) Joint E: + ΣFy=0; FEB 0.75P− 0= FEB 0.75P:= FEB 0.75 kN= + ΣFx=0; FED FAE− 0= FED FAE:= FED 1.333 kN= σEB FEB A := σEB 0.962 MPa= (T) σED FED A := σED 1.709 MPa= (C) Joint B: + ΣFy=0; v( ) FBD⋅ v( ) FAB⋅− FEB− 0= FBD FAB FEB v ⎛⎜⎝ ⎞ ⎠+:= FBD 2.917 kN= + ΣFx=0; FBC h( )FAB− h( )FBD− 0= FBC h( )FAB h( )FBD+:= FBC 3.667 kN= σBC FBC A := σBC 4.701 MPa= (T) σBD FBD A := σBD 3.739 MPa= (C) Since the cross-sectional areas are the same, the highest stress occurs in the member BC, which has the greatest force Fmax max FAB FAE, FEB, FED, FBD, FBC,( ):= Fmax 3.667 kN= Pallow P Fmax ⎛⎜⎝ ⎞ ⎠ σallow A⋅( )⋅:= Pallow 29.78 kN= Ans Problem 1-60 The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p = 650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed to hold the cap in place. Given: p 650Pa:= a 25mm:= di 35mm:= do 40mm:= Solution: Ap π di2⋅ 4 := As π do⋅( ) a( ):= P a( )⋅ FBC cos θ( )⋅ a b+( )⋅− 0= P p Ap( )⋅:= P 0.625 N= Average Shear Stress: τavg P As := τavg 199.1 Pa= Ans Problem 1-61 The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles, determine the average shear stress in the pin at A. The pin is subjected to double shear and has a diameter of 5 mm. Only a vertical force is exerted on the wire. Given: P 100N:= a 37.5mm:= b 50mm:= c 25mm:= d 125mm:= dpin 5mm:= Solution: From FBD (a): + ΣFx=0; Bx 0:= Bx 0 N= + ΣΜD=0; P d( )⋅ By c( )⋅− 0= By P d c ⋅:= By 500 N= From FBD (b): + ΣFx=0; Ax 0:= Ax 0 N= + ΣΜE=0; Ay a( )⋅ By a b+( )⋅− 0= Ay By a b+ a ⋅:= Ay 1166.67 N= Average Shear Stress: Apin π dpin2⋅ 4 := VA 0.5 Ay( )⋅:= VA 583.333 N= τavg VA Apin := τavg 29.709 MPa= Ans Problem 1-62 Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm. Given: a 37.5mm:= b 50mm:= c 25mm:= d 125mm:= dpin 5mm:= P 100N:= Solution: From FBD (a): + ΣFx=0; Bx 0:= Bx 0 N= + ΣΜD=0; P d( )⋅ By c( )⋅− 0= By P d c ⋅:= By 500 N= Average Shear Stress: Pin B is subjected to doule shear Apin π dpin2⋅ 4 := VB 0.5 By( )⋅:= VB 250 N= τavg VB Apin := τavg 12.732 MPa= Ans Problem 1-63 The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and the extension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed to support the lamp. Hint: The shear force in the pin is caused by the couple moment required for equilibrium at A. Given: w 8 N m := P 20N:= a 900mm:= h 32mm:= dpin 3mm:= Solution: From FBD (a): + ΣFx=0; Bx 0:= + ΣΜA=0; V h( )⋅ w a⋅( ) 0.5a( )⋅− P a( )⋅− 0= V w a⋅( ) 0.5 a h ⎛⎜⎝ ⎞ ⎠⋅ P a h ⋅+:= V 663.75 N= Average Shear Stress: Apin π dpin2⋅ 4 := τavg V Apin := τavg 93.901 MPa= Ans Problem 1-64 The two-member frame is subjected to the distributed loading shown. Determine the average normal stress and average shear stress acting at sections a-a and b-b. Member CB has a square cross section of 35 mm on each side. Take w = 8 kN/m. Given: w 8 kN m := a 3m:= b 4m:= A 0.0352( )m2:= Solution: c a2 b2+:= c 5 m= h a c := v b c := Member AB: ΣMA=0; By a( )⋅ w a⋅( ) 0.5a( )⋅− 0= By 0.5w a⋅:= By 12 kN= + ΣFy=0; v( ) FAB⋅ By− 0= FAB By v := FAB 15 kN= Section a-a: σa_a FAB A := σa_a 12.24 MPa= Ans τa_a 0:= τa_a 0 MPa= Ans Section b-b: + ΣFx=0; N FAB h( )⋅− 0= N FAB h( )⋅:= N 9 kN= + ΣFy=0; V FAB v( )⋅− 0= V FAB v( )⋅:= V 12 kN= Ab_b A h := σb_b N Ab_b := σb_b 4.41 MPa= Ans τb_b V Ab_b := τb_b 5.88 MPa= Ans Problem 1-65 Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determine the average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member B which has a thickness of 30 mm. Given: P 5kN:= θ 60deg:= φ 30deg:= drod 10mm:= h 40mm:= t 30mm:= Solution: AB t h⋅:= Arod π 4 drod 2⋅:= + ΣFx=0; P cos θ( )⋅ FB− 0= FB P cos θ( )⋅:= FB 2.5 kN= + ΣFy=0; Fc P sin θ( )⋅− 0= FC P sin θ( )⋅:= FC 4.33 kN= Average Normal Stress: σB FB AB := σB 2.083 MPa= Ans σC FC Arod := σC 55.133 MPa= Ans Problem 1-66 Consider the general problem of a bar made from m segments, each having a constant cross-sectional area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be used to determine the average normal stress at any specified location x. Show an application of the program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m, P2 = -1.5 kN, A2 = 625 mm2. Problem 1-67 The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a diameter of 18 mm. Given: P 15kN:= a 0.5m:= b 1m:= c 1.5m:= d 1.5m:= e 0.5m:= θ 30deg:= dpin 18mm:= Solution: L a b+ c+ d+ e+:= Support Reactions: ΣΜA=0; By− L( )⋅ P L a−( )⋅+ 4P c d+ e+( )⋅+ 4P d e+( )⋅+ 2P e( )⋅+ 0= By P L a− L ⋅ 4 P⋅ c d+ e+ L ⋅+ 4 P⋅ d e+ L ⋅+ 2 P⋅ e L ⋅+:= By 82.5 kN= + ΣFy=0; By− P+ 4 P⋅+ 4 P⋅+ 2 P⋅+ Ay− 0= Ay By− P+ 4 P⋅+ 4P+ 2 P⋅+:= Ay 82.5 kN= FBC By sin θ( ):= FBC 165 kN= Ax FBC cos θ( )⋅:= Ax 142.89 kN= Average Shear Stress: Apin π dpin2⋅ 4 := For Pins B and C: τB_and_C 0.5FBC Apin := τB_and_C 324.2 MPa= Ans For Pin A: FA Ax 2 Ay 2+:= FA 165 kN= τA 0.5FA Apin := τA 324.2 MPa= Ans Problem 1-68 The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are in double shear as shown, and each has a diameter of 18 mm. Given: τallow 80MPa:= a 0.5m:= b 1m:= c 1.5m:= d 1.5m:= e 0.5m:= θ 30deg:= dpin 18mm:= Solution: L a b+ c+ d+ e+:= For comparison purpose, set P 1kN:= Support Reactions: ΣΜA=0; By− L( )⋅ P L a−( )⋅+ 4P c d+ e+( )⋅+ 4P d e+( )⋅+ 2P e( )⋅+ 0= By P L a− L ⋅ 4 P⋅ c d+ e+ L ⋅+ 4 P⋅ d e+ L ⋅+ 2 P⋅ e L ⋅+:= By 5.5 kN= + ΣFy=0; By− P+ 4 P⋅+ 4 P⋅+ 2 P⋅+ Ay− 0= Ay By− P+ 4 P⋅+ 4P+ 2 P⋅+:= Ay 5.5 kN= FBC By sin θ( ):= FBC 11 kN= Ax FBC cos θ( )⋅:= Ax 9.53 kN= FA Ax 2 Ay 2+:= FA 11 kN= Require: Fmax max FBC FA,( ):= Fmax 11 kN= Apin π dpin2⋅ 4 := Pallow P Fmax ⎛⎜⎝ ⎞ ⎠ τallow 2Apin( )⋅⎡⎣ ⎤⎦⋅:= Pallow 3.70 kN= Ans Problem 1-69 The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as a function of the bar angle θ. Plot this function, 0 θ≤ 90o≤ , and indicate the values of θ for which this stress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear. Given: P 1kN:= dbolt 6mm:= a 0.6m:= b 0.45m:= c 0.15m:= Solution: Support Reactions: ΣΜC=0; FAB cos θ( )⋅ c( )⋅ FAB sin θ( )⋅ a( )⋅+ P a b+( )⋅− 0= FAB P a b+( )⋅ cos θ( ) c( )⋅ sin θ( ) a( )⋅+= Average Shear Stress: Pin B is subjected to doule shear τ FAB Abolt = Abolt π dbolt2⋅ 4 := τ 4P a b+( )⋅ cos θ( ) c( )⋅ sin θ( ) a( )⋅+⎡⎣ ⎤⎦ π dbolt2⋅⎛⎝ ⎞⎠⋅ = dτ dθ 4P a b+( )⋅ π dbolt2⋅ sin θ( ) c( )⋅ cos θ( ) a( )⋅− cos θ( ) c( )⋅ sin θ( ) a( )⋅+⎡⎣ ⎤⎦2 ⋅= dτ dθ 0= sin θ( ) c( )⋅ cos θ( ) a( )⋅− 0= tan θ( ) a c = θ atan a c ⎛⎜⎝ ⎞ ⎠:= θ 75.96 deg= Ans Problem 1-70 The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the beam, 1ft x≤ 12ft≤ . If the hoist is rated to support a maximum of 7.5 kN, determine the maximum average normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the 16-mm-diameter pin at B. Given: P 7.5kN:= xmax 3.6m:= a 3m:= θ 30deg:= drod 18mm:= dpin 16mm:= Solution: Support Reactions: ΣΜC=0; FBC sin θ( ) a( )⋅⋅ P x( )⋅− 0= FBC P x( )⋅ sin θ( ) a( )⋅= Maximum FBC occurs when x= xmax . Therefore, FBC P xmax( )⋅ sin θ( ) a( )⋅:= FBC 18.00 kN= Arod π drod2⋅ 4 := Apin π dpin2⋅ 4 := τpin 0.5 FBC⋅ Apin := τpin 44.762 MPa= Ans σrod FBC Arod := σrod 70.736 MPa= Ans Problem 1-71 The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal and average shear stresses acting over the shaded section, which is oriented at θ from the horizontal. Plot the variation of these stresses as a function of θ (0o θ≤ 90o≤ ). Solution: Equations of Equilibrium: + ΣFx=0; V P cos θ( )⋅− 0= V P cos θ( )⋅= + ΣFy=0; N P sin θ( )⋅− 0= N P sin θ( )⋅= Inclined plane: Aθ A sin θ( )= σ N A = σ P A sin θ( )2⋅= Ans τavg V A = τavg P 2A sin 2θ( )⋅= Ans Problem 1-72 The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable has a diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position θ for 0o θ≤ 90o≤ . Given: W 3kN:= a 1m:= do 15mm:= Solution: Angle B: φB 0.5 90deg θ+( )= φB 45deg 0.5θ+= Support Reactions: ΣΜA=0; FBC sin φB( )⋅ a( )⋅ W 0.5a( )⋅ cos θ( )− 0= FBC 0.5W cos θ( )⋅ sin 45deg 0.5θ+( )= Average Normal Stress: σBC FAB ABC = ABC π do2⋅ 4 := σBC 2W π do2⋅ ⎛⎜⎝ ⎞ ⎠ cos θ( ) sin 45deg 0.5θ+( )⋅= Ans Problem 1-73 The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads as shown, determine the average normal stress in the bar as a function of for 0 x< 0.5m≤ . Given: P1 3kN:= P2 6kN:= w 8 kN m := A 400 10 6−( )⋅ m2:= a 0.5m:= b 0.75m:= Solution: L a b+:= + ΣFx=0; N− P1+ P2+ w L x−( )⋅+ 0= N P1 P2+ w L x−( )⋅+= Average Normal Stress: σ N A = σ P1 P2+ w L x−( )⋅+ A = Ans Problem 1-74 The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed loading along its length and to two concentrated loads as shown, determine the average normal stress in the bar as a function of for 0.5m x< 1.25m≤ . Given: P1 3kN:= P2 6kN:= w 8 kN m := A 400 10 6−( )⋅ m2:= a 0.5m:= b 0.75m:= Solution: L a b+:= + ΣFx=0; N− P1+ w L x−( )⋅+ 0= N P1 w L x−( )⋅+= Average Normal Stress: σ N A = σ P1 w L x−( )⋅+ A = Ans Problem 1-75 The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axial compressive force of 15 kN. Determine the average normal stress in the column as a function of the distance z measured from its base. Note: The result will be useful only for finding the average normal stress at a section removed from the ends of the column, because of localized deformation at the ends. Given: P 3kN:= ρ 2.3 103( ) kg m3 := g 9.81 m s2 := r 180mm:= h 0.75m:= Solution: A π r2⋅:= w ρ g⋅ A⋅:= + ΣFz=0; N P− w h z−( )⋅− 0= N P w h z−( )⋅+= Average Normal Stress: σ N A = σ P w h z−( )⋅+ A = Ans Problem 1-76 The two-member frame is subjected to the distributed loading shown. Determine the largest intensity of the uniform loading that can be applied to the frame without causing either the average normal stress or the average shear stress at section b-b to exceed σ = 15 MPa, and τ = 16 MPa respectively. Member CB has a square cross-section of 30 mm on each side. Given: σallow 15MPa:= τallow 16MPa:= a 4m:= b 3m:= A 0.0302( )m2:= Solution: c a2 b2+:= c 5 m= v b c :=h a c := Set wo 1 kN m := Member AB: ΣMA=0; By− b( )⋅ wo b⋅( ) 0.5b( )⋅− 0= By 0.5wo b⋅:= By 1.5 kN= Section b-b: By FBC h( )⋅= FBC By h := FBC 1.88 kN= + ΣFx=0; FBC h( )⋅ Vb_b− 0= Vb_b FBC h( )⋅:= Vb_b 1.5 kN= + ΣFy=0; Nb_b− FBC v( )⋅+ 0= Nb_b FBC v( )⋅:= Nb_b 1.125 kN= Ab_b A v := σb_b Nb_b Ab_b := σb_b 0.75 MPa= τb_b Vb_b Ab_b := τb_b 1 MPa= Assume failure due to normal stress: wallow wo σallow σb_b ⎛⎜⎝ ⎞ ⎠ ⋅:= wallow 20.00 kN m = Assume failure due to shear stress: wallow wo τallow τb_b ⎛⎜⎝ ⎞ ⎠ ⋅:= wallow 16.00 kN m = Ans Controls ! Problem 1-77 The pedestal supports a load P at its center. If the material has a mass density ρ, determine the radial dimension r as a function of z so that the average normal stress in the pedestal remains constant. The cross section is circular. Solution: Require: σ P w1+ A = σ P w1+ dw+ A dA+= P dA⋅ w1 dA⋅+ A dw⋅= dw dA P w1+ A = dw dA σ= [1] dA π r dr+( )2 πr2−= dA 2πr dr⋅= dw πr2 ρ g⋅( )⋅ dz⋅= From Eq.[1], πr2 ρ g⋅( )⋅ dz⋅ 2πr dr⋅ σ= r ρ g⋅( )⋅ dz⋅ 2dr σ= ρ g⋅ 2σ 0 z z1( ) ⌠⎮⌡ d r1 r r 1 r ⌠⎮ ⎮⌡ d= ρ g⋅ z⋅ 2σ ln r r1 ⎛⎜⎝ ⎞ ⎠= r r1 e ρ g⋅ 2σ ⎛⎜⎝ ⎞ ⎠ z⋅⋅= However, σ P π r12⋅ = Ans r r1 e π r12⋅ ρ⋅ g⋅ 2P ⎛⎜⎜⎝ ⎞ ⎠ z⋅⋅= Problem 1-78 The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the material has a density of 2.5 Mg/m3, determine the average normal stress at the support. Given: ro 0.5m:= h 3m:= g 9.81 m s2 = r ro e 0.08− y2m⋅= ρ 2.5 103( )⋅ kg m3 := yunit 1m:= Solution: dr π e 0.08− y 2⎛⎝ ⎞⎠ dy⋅= Ao πro2:= Ao 0.7854 m2= dV π r2( ) dy⋅= dV πro2 e 0.08− y 2⎛⎝ ⎞⎠ 2 dy⋅= V 0 3 yπro2 e 0.08− y 2⎛⎝ ⎞⎠ 2 yunit( )⋅⎡⎢⎣ ⎤⎥⎦ ⌠⎮ ⎮⌡ d:= V 1.584 m3= W ρ g⋅ V⋅:= W 38.835 kN= σ W Ao := σ 0.04945 MPa= Ans Problem 1-79 The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at its center. If it is rotating in the horizontal plane at a constant angular rate of ω, determine the average normal stress in the bar as a function of x. Solution: Equation of Motion : + ΣFx=MaN=Mω r2 ; N m L 2 x−⎛⎜⎝ ⎞ ⎠⋅ ω x 1 2 L 2 x−⎛⎜⎝ ⎞ ⎠+ ⎡⎢⎣ ⎤⎥⎦⋅= N m ω⋅ 8 L2 4 x2⋅−( )⋅= Average Normal Stress: σ N A = σ m ω⋅ 8A L2 4 x2⋅−( )⋅= Ans Problem 1-80 Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are 10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so that the average shear stress does not exceed τallow = 2.1 MPa. Given: P 4kN:= t 10mm:= τallow 2.1MPa:= a 300mm:= b 125mm:= Solution: c a2 b2+:= c 325 mm= h a c := v b c := V P v( )⋅:= V 1.54 kN= τallow V t h⋅= h V t τallow⋅ := h 73.26 mm= Use h 75mm:= h 75 mm= Ans Problem 1-81 The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure shear stress for the bolts is τfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5. Given: P 80kN:= τfail 350MPa:= γ 2.5:= Solution: τallow τfail γ:= τallow 140 MPa= Vbolt 0.5 P 2 ⎛⎜⎝ ⎞ ⎠⋅:= Vbolt 20 kN= Abolt Vbolt τallow = π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅ Vbolt τallow = d 4 π Vbolt τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 13.49 mm= Ans Problem 1-82 The rods AB and CD are made of steel having a failure tensile stress of σfail = 510 MPa. Using a factor of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load shown. The beam is assumed to be pin connected at A and C. Given: P1 4kN:= P2 6kN:= P3 5kN:= a 2m:= b 2m:= c 3m:= d 3m:= γ 1.75:= σfail 510MPa:= Solution: L a b+ c+ d+:= Support Reactions: ΣΜA=0; FCD L( )⋅ P1 a( )⋅− P2 a b+( )⋅− P3 a b+ c+( )⋅− 0= FCD P1 a L ⎛⎜⎝ ⎞ ⎠⋅ P2 a b+ L ⋅+ P3 a b+ c+ L ⋅+:= FCD 6.70 kN= ΣΜC=0; FAB− L( )⋅ P1 b c+ d+( )⋅+ P2 c d+( )⋅+ P3 d( )⋅+ 0= FAB P1 b c+ d+ L ⋅ P2 c d+ L ⋅+ P3 d L ⎛⎜⎝ ⎞ ⎠⋅+:= FAB 8.30 kN= Average Normal Stress: Design of rod sizes σallow σfail γ:= σallow 291.43 MPa= For Rod AB Abolt FAB σallow = π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅ FAB σallow = dAB 4 π FAB σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dAB 6.02 mm= Ans For Rod CD Abolt FCD σallow = π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅ FCD σallow = dCD 4 π FCD σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dCD 5.41 mm= Ans Problem 1-83 The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if the allowable shear stress for the key is τallow = 35 MPa. Given: P 200N:= τallow 35MPa:= L 500mm:= a 20mm:= b 25mm:= Solution: ΣΜA=0; Fa_a a( )⋅ P L( )⋅− 0= Fa_a P L a ⋅:= Fa_a 5000 N= For the key Aa_a Fa_a τallow = b d⋅ Fa_a τallow = d 1 b Fa_a τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 5.71 mm= Ans Problem 1-84 The fillet weld size a is determined by computing the average shear stress along the shaded plane, which has the smallest cross section. Determine the smallest size a of the two welds if the force applied to the plate is P = 100 kN. The allowable shear stress for the weld material is τallow = 100 MPa. Given: P 100kN:= τallow 100MPa:= L 100mm:= θ 45deg:= Solution: Shear Plane in the Weld: Aweld L a⋅ sin θ( )⋅= Aweld 0.5P τallow = L a⋅ sin θ( )⋅ 0.5Pτallow= a 1 L sin θ( )⋅ 0.5P τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= a 7.071 mm= Ans Problem 1-85 The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block along the shaded plane, which is the smallest cross section, determine the largest force P that can be applied to the plate. The allowable shear stress for the weld material is τallow = 100 MPa. Given: a 8mm:= L 100mm:= θ 45deg:= τallow 100MPa:= Solution: Shear Plane in the Weld: Aweld L a⋅ sin θ( )⋅= P τallow 2Aweld( )⋅= P τallow 2 L a⋅ sin θ( )⋅( )⎡⎣ ⎤⎦⋅:= P 113.14 kN= Ans Problem 1-86 The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of 5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washer will not penetrate or shear through it. The allowable normal stress for the bolt is σallow = 150 MPa and the allowable shear stress for the supporting material is τallow = 35 MPa. Given: P 25kN:= dwasher 25mm:= σallow 150MPa:= τallow 35MPa:= Solution: Allowable Normal Stress: Design of bolt size Abolt P σallow = π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅ Pσallow = d 4 π P σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 14.567 mm= Use d 15mm:= d 15 mm= Ans Allowable Shear Stress: Design of support thickness Asupport P τallow = π dwasher( )⋅ h⋅ Pτallow= h 1 π dwasher( )⋅ P τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= h 9.095 mm= Use d 10mm:= d 10 mm= Ans Problem 1-87 The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B if the allowable shear stress for the material is τallow = 42 MPa. Pin A is subjected to double shear, whereas pin B is subjected to single shear. Given: P 8kN:= τallow 42MPa:= a 1.5m:= b 1.5m:= c 1.5m:= d 0.6m:= Solution: θBC 45deg:= Support Reactions: From FBD (a), ΣΜD=0; FBC sin θBC( )⋅ c( )⋅ P c d+( )⋅− 0= FBC P c d+ sin θBC( ) c( )⋅⋅:= FBC 15.839 kN= From FBD (b), ΣΜA=0; Dy a b+( )⋅ P c d+( )⋅− 0= Dy P c d+ a b+⋅:= Dy 5.6 kN= + ΣFx=0; Ax P− 0= Ax P:= Ax 8 kN= + ΣFy=0; Dy Ay− 0= Ay Dy:= FA Ax 2 Ay 2+:= FA 9.77 kN= Apin 0.5FA τallow = π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅ 0.5FA τallow = d 4 π 0.5FA τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 12.166 mm= Ans For pin B: Pin A is subjected to single shear, and FB FBC:= Apin FB τallow = π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅ FB τallow = d 4 π FB τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 21.913 mm= Ans Problem 1-88 The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5 kN. Given: P 5kN:= σallow 200MPa:= a 4m:= b 3m:= θ 60deg:= Solution: c a2 b2+:= h a c := v b c := At joint A: Initial guess: FAB 1kN:= FAC 2kN:= Given + ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1] + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P− 0= [2] Solving [1] and [2]: FAB FAC ⎛⎜⎜⎝ ⎞ ⎠ Find FAB FAC,( ):= FAB FAC ⎛⎜⎜⎝ ⎞ ⎠ 4.3496 4.7086 ⎛⎜⎝ ⎞ ⎠ kN= For wire AB AAB FAB σallow = π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅ FAB σallow = dAB 4 π FAB σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dAB 5.26 mm= Ans For wire AC AAC FAC σallow = π 4 ⎛⎜⎝ ⎞ ⎠ dAC 2⋅ FAC σallow = dAC 4 π FAC σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dAC 5.48 mm= Ans Problem 1-89 The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile stress of σallow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm, determine the greatest force P that can be applied to the chain before one of the wires fails. Given: σallow 180MPa:= a 4m:= b 3m:= θ 60deg:= dAB 6mm:= dAC 4mm:= Solution: c a2 b2+:= h a c := v b c := Assume failure of AB: FAB AAB( ) σallow⋅= FAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅ σallow⋅:= FAB 5.09 kN= At joint A: Initial guess: P1 1kN:= FAC 2kN:= Given + ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1] + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P1− 0= [2] Solving [1] and [2]: P1 FAC ⎛⎜⎜⎝ ⎞ ⎠ Find P1 FAC,( ):= P1FAC ⎛⎜⎜⎝ ⎞ ⎠ 5.8503 5.5094 ⎛⎜⎝ ⎞ ⎠ kN= Assume failure of AC: FAC AAC( ) σallow⋅= FAC π 4 ⎛⎜⎝ ⎞ ⎠ dAC 2⋅ σallow⋅:= FAC 2.26 kN= At joint A: Initial guess: P2 1kN:= FAB 2kN:= Given + ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1] + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P2− 0= [2] Solving [1] and [2]: FAB P2 ⎛⎜⎜⎝ ⎞ ⎠ Find FAB P2,( ):= FABP2 ⎛⎜⎜⎝ ⎞ ⎠ 2.0895 2.4019 ⎛⎜⎝ ⎞ ⎠ kN= Chosoe the smallest value: P min P1 P2,( ):= P 2.40 kN= Ans Problem 1-90 The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress of σallow = 168 MPa. Determine the greatest load that can be supported without causing the cable to fail when θ = 30° and φ = 45°. Neglect the size of the winch. Given: σallow 168MPa:= do 6mm:= θ 30deg:= φ 45deg:= Solution: For the cable: Tcable Acable( ) σallow⋅= Tcable π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅ σallow⋅:= Tcable 4.7501 kN= At joint B: Initial guess: FAB 1 kN:= W 2 kN:= Given + ΣFx=0; Tcable− cos θ( ) FAB cos φ( )⋅+ 0= [1] + ΣFy=0; W− FAB sin φ( )⋅+ Tcable sin θ( )⋅− 0= [2] Solving [1] and [2]: FAB W ⎛⎜⎝ ⎞ ⎠ Find FAB W,( ):= FAB W ⎛⎜⎝ ⎞ ⎠ 5.818 1.739 ⎛⎜⎝ ⎞ ⎠ kN= Ans Problem 1-91 The boom is supported by the winch cable that has an allowable normal stress of σallow = 168 MPa. If it is required that it be able to slowly lift 25 kN, from θ = 20° to θ = 50°, determine the smallest diameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect the size of the winch. Set d = 3.6 m. Given: σallow 168MPa:= W 25 kN:= d 3.6m:= a 6m:= Solution: Maximum tension in canle occurs when θ 20deg:= sin θ( ) a sin ψ( ) d = ψ asin d a ⎛⎜⎝ ⎞ ⎠ sin θ( )⋅ ⎡⎢⎣ ⎤⎥⎦:= ψ 11.842 deg= At joint B: Initial guess: FAB 1 kN:= Tcable 2 kN:= Given φ θ ψ+:= + ΣFx=0; Tcable− cos θ( ) FAB cos φ( )⋅+ 0= [1] + ΣFy=0; W− FAB sin φ( )⋅+ Tcable sin θ( )⋅− 0= [2] Solving [1] and [2]: FAB Tcable ⎛⎜⎜⎝ ⎞ ⎠ Find FAB Tcable,( ):= FAB Tcable ⎛⎜⎜⎝ ⎞ ⎠ 114.478 103.491 ⎛⎜⎝ ⎞ ⎠ kN= For the cable: Acable P σallow = π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅ Tcable σallow = do 4 π Tcable σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= do 28.006 mm= Use do 30mm:= do 30 mm= Ans Problem 1-92 The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of the pins at A and B if the allowable shear stress for the material is τallow = 100 MPa. Both pins are subjected to double shear. Given: w 2 kN m := τallow 100MPa:= r 3m:= Solution: Member AB is atwo-force member θ 45deg:= Support Reactions: ΣΜA=0; FBC sin θ( )⋅ r( )⋅ w r⋅( ) 0.5r( )⋅− 0= FBC 0.5w r⋅ sin θ( ):= FBC 4.243 kN= + ΣFy=0; Ay FBC sin θ( )⋅+ w r⋅− 0= Ay FBC− sin θ( )⋅ w r⋅+:= Ay 3 kN= + ΣFx=0; Ax FBC cos θ( )⋅− 0= Ax FBC cos θ( )⋅:= Ax 3 kN= Average Shear Stress: Pin A and pin B are subjected to double shear FA Ax 2 Ay 2+:= FA 4.243 kN= FB FBC:= FB 4.243 kN= Since both subjected to the same shear force V = 0.5 FA and V 0.5FB:= Apin V τallow = π 4 ⎛⎜⎝ ⎞ ⎠ dpin 2⋅ Vτallow = dpin 4 π V τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dpin 5.20 mm= Ans Problem 1-93 Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required to support is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (σt)allow = 175 MPa, (σb)allow = 275 MPa, and τallow = 115 MPa. Given: P 150kN:= σt_allow 175MPa:= τallow 115MPa:= σb_allow 275MPa:= d2 30mm:= Solution: Allowable Normal Stress: Design of end cap outer diameter A P σt_allow = π 4 ⎛⎜⎝ ⎞ ⎠ d1 2 d2 2−⎛⎝ ⎞⎠⋅ Pσt_allow= d1 4 π P σt_allow ⎛⎜⎝ ⎞ ⎠ ⋅ d22+:= d1 44.62 mm= Ans Allowable Bearing Stress: Design of circular shaft diameter A P σb_allow = π 4 ⎛⎜⎝ ⎞ ⎠ d3 2⎛⎝ ⎞⎠⋅ Pσb_allow= d3 4 π P σb_allow ⎛⎜⎝ ⎞ ⎠ ⋅:= d3 26.35 mm= Ans Allowable Shear Stress: Design of end cap thickness A P τallow = π d3⋅( ) t⋅ Pτallow= t 1 π d3⋅ P τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= t 15.75 mm= Ans Problem 1-94 If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa, determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN. Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical. Given: σb_allow 2.8MPa:= P 7.5 kN:= P1 10 kN:= P2 10 kN:= P3 15 kN:= P4 10 kN:= a 1.5m:= b 2.5m:= Solution: L 3 a⋅ b+:= Support Reactions: ΣΜA=0; By 3a( ) P2 a( )⋅− P3 2a( )⋅− P4 3a( )⋅− P L( )⋅− 0= By P2 a 3a ⎛⎜⎝ ⎞ ⎠⋅ P3 2a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P4 3a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P L 3a ⎛⎜⎝ ⎞ ⎠⋅+:= By 35 kN= ΣΜB=0; Ay− 3 a⋅( )⋅ P1 3 a⋅( )⋅+ P2 2 a⋅( )⋅+ P3 a( )⋅+ P b( )⋅− 0= Ay P1 3a 3a ⎛⎜⎝ ⎞ ⎠⋅ P2 2a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P3 a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P b 3a ⎛⎜⎝ ⎞ ⎠⋅−:= Ay 17.5 kN= For Plate A: Aplate_A Ay σb_allow = aA 2 Ay σb_allow = aA Ay σb_allow := aA 79.057 mm= Use aA x aA plate: aA 80mm= Ans For Plate B Aplate_B By σb_allow = aB 2 By σb_allow = aB By σb_allow := aB 111.803 mm= Use aB x aB plate: aB 120mm= Ans ULoHS Text Box Rb = 35kNnullnull(b is in subscript) Problem 1-95 If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have square cross sections of 50mm x 50mm and 100mm x 100mm, respectively. Given: σb_allow 2.8MPa:= P1 10 kN:= P2 10 kN:= P3 15 kN:= P4 10 kN:= a 1.5m:= b 2.5m:= aA 50mm:= aB 100mm:= Solution: L 3 a⋅ b+:= Support Reactions: ΣΜA=0; By 3a( ) P2 a( )⋅− P3 2a( )⋅− P4 3 a⋅( )⋅− P L( )⋅− 0= By P2 a 3a ⎛⎜⎝ ⎞ ⎠⋅ P3 2 a 3a ⋅⎛⎜⎝ ⎞ ⎠⋅+ P4 3 a⋅ 3a ⎛⎜⎝ ⎞ ⎠⋅+ P L 3a ⎛⎜⎝ ⎞ ⎠⋅+= ΣΜB=0; Ay− 3 a⋅( )⋅ P1 3 a⋅( )⋅+ P2 2 a⋅( )⋅+ P3 a( )⋅+ P b( )⋅− 0= Ay P1 3a 3a ⎛⎜⎝ ⎞ ⎠⋅ P2 2a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P3 a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P b 3a ⎛⎜⎝ ⎞ ⎠⋅−= For Plate A: Ay aA 2⎛⎝ ⎞⎠ σb_allow⋅:= aA 2⎛⎝ ⎞⎠ σb_allow⋅ P1 3a3a ⎛⎜⎝ ⎞ ⎠⋅ P2 2a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P3 a 3a ⎛⎜⎝ ⎞ ⎠⋅+ P b 3a ⎛⎜⎝ ⎞ ⎠⋅−= P P1 3a b ⎛⎜⎝ ⎞ ⎠⋅ P2 2a b ⎛⎜⎝ ⎞ ⎠⋅+ P3 a b ⎛⎜⎝ ⎞ ⎠⋅+ aA 2⎛⎝ ⎞⎠ σb_allow⋅ 3ab ⎛⎜⎝ ⎞ ⎠⋅−:= P 26.400 kN= Pcase_1 P:= For Plate B: By aB 2⎛⎝ ⎞⎠ σb_allow⋅:= aB 2⎛⎝ ⎞⎠ σb_allow⋅ P2 a3a ⎛⎜⎝ ⎞ ⎠⋅ P3 2 a 3a ⋅⎛⎜⎝ ⎞ ⎠⋅+ P4 3 a⋅ 3a ⎛⎜⎝ ⎞ ⎠⋅+ P L 3a ⎛⎜⎝ ⎞ ⎠⋅+= P P2− a L ⎛⎜⎝ ⎞ ⎠⋅ P3 2 a L ⋅⎛⎜⎝ ⎞ ⎠⋅− P4 3 a⋅ L ⎛⎜⎝ ⎞ ⎠⋅− aB 2⎛⎝ ⎞⎠ σb_allow⋅ 3aL⋅+:= P 3.000 kN= Pcase_2 P:= Pallow min Pcase_1 Pcase_2,( ):= Pallow 3 kN= Ans Problem 1-96 Determine the required cross-sectional area of member BC and the diameter of the pins at A and B if the allowable normal stress is σallow = 21 MPa and the allowable shear stress is τallow = 28 MPa. Given: σallow 21MPa:= τallow 28MPa:= P 7.5kip:= θ 60deg:= a 0.6m:= b 1.2m:= c 0.6m:= Solution: L a b+ c+:= Support Reactions: ΣΜA=0; By L( )⋅ P a( )⋅− P a b+( )⋅− 0= By P a L ⋅ P a b+ L ⋅+:= FBC By sin θ( ):= FBC 38.523 kN=By 33.362 kN= Bx FBC cos θ( )⋅:= Bx 19.261 kN= + ΣFy=0; By− P+ P+ Ay− 0= Ay By− P+ P+:= Ay 33.362 kN= + ΣFx=0; Bx Ax− 0= Ax Bx:= Ax 19.261 kN= FA Ax 2 Ay 2+:= FA 38.523 kN= Member BC: ABC FBC σallow := ABC 1834.416 mm2= Ans Pin A: AA FA τallow = π 4 ⎛⎜⎝ ⎞ ⎠ dA 2⋅ FA τallow = dA 4 π FA τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dA 41.854 mm= Ans Pin B: AB 0.5FBC τallow = π 4 ⎛⎜⎝ ⎞ ⎠ dB 2⋅ 0.5FBC τallow = dB 4 π 0.5FBC τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dB 29.595 mm= Ans Problem 1-97 The assembly consists of three disks A, B, and C that are used to support the load of 140 kN. Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (τallow)b = 350 MPa and allowable shear stress is τallow = 125 MPa. Given: P 140kN:= τallow 125MPa:= σb_allow 350MPa:= hB 20mm:= hC 10mm:= Solution: Allowable Shear Stress: Assume shear failure dor disk C A P τallow = π d2⋅( ) hC⋅ Pτallow= d2 1 π hC⋅ P τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d2 35.65 mm= Ans Allowable Bearing Stress: Assume bearing failure dor disk C A P σb_allow = π 4 ⎛⎜⎝ ⎞ ⎠ d2 2 d3 2−⎛⎝ ⎞⎠⋅ Pσb_allow= d3 d2 2 4 π P σb_allow ⎛⎜⎝ ⎞ ⎠ ⋅−:= d3 27.60 mm= Ans Allowable Bearing Stress: Assume bearing failure dor disk B A P σb_allow = π 4 ⎛⎜⎝ ⎞ ⎠ d1 2⎛⎝ ⎞⎠⋅ Pσb_allow= d1 4 π P σb_allow ⎛⎜⎝ ⎞ ⎠ ⋅:= d1 22.57 mm= Since d3 > d1, disk B might fail due to shear. τ P A = τ Pπ d1⋅ hB⋅ := τ 98.73 MPa= < τallow (O.K.!) Therefore d1 22.57 mm= Ans Problem 1-98 Strips A and B are to be glued together using the two strips C and D. Determine the required thickness t of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that of strips C and D. Given: P 40N:= t 30mm:= bA 1.5m:= bB 1.5m:= bC 1m:= bD 1m:= Solution: Average Normal Stress: Requires, σA σB= σB σC= σC σD= N bA( ) t⋅ 0.5N bC( ) tC( )⋅= tC 0.5 bA( ) t⋅ bC := tC 22.5 mm= Ans Problem 1-99 If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa, determine the size of square bearing plates A' and B' required to support the loading. Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN. Given: σb_allow 2.8MPa:= P 7.5kN:= w 10 kN m := a 4.5m:= b 2.25m:= Solution: L a b+:= Support Reactions: ΣΜA=0; By a( ) w a( )⋅ 0.5 a⋅( )⋅− P L( )⋅− 0= By w 0.5 a⋅( )⋅ P L a ⎛⎜⎝ ⎞ ⎠⋅+:= By 33.75 kN= ΣΜB=0; Ay− a( )⋅ w a( )⋅ 0.5 a⋅( )⋅+ P L( )⋅− 0= Ay w 0.5 a⋅( )⋅ P b a ⎛⎜⎝ ⎞ ⎠⋅−:= Ay 18.75 kN= Allowable Bearing Stress: Design of bearing plates For Plate A: Area Ay σb_allow = aA 2 Ay σb_allow = aA Ay σb_allow := aA 81.832 mm= Use aA x aA plate: aA 90mm:= aA 90 mm= Ans For Plate B Area By σb_allow = aB 2 By σb_allow = aB By σb_allow := aB 109.789 mm= Use aB x aB plate: aB 110mm:= aB 110.00 mm= Ans Problem 1-100 If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa, determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have square cross sections of 50mm x 50mm and 100mm x 100mm, respectively. Given: σb_allow 2.8MPa:= w 10 kN m := a 4.5m:= b 2.25m:= aA 50mm:= aB 100mm:= Solution: L a b+:= Support Reactions: ΣΜA=0; By a( )⋅ w a( )⋅ 0.5 a⋅( )⋅− P L( )⋅− 0= P By a L ⎛⎜⎝ ⎞ ⎠⋅ w a L ⎛⎜⎝ ⎞ ⎠⋅ 0.5 a⋅( )⋅−= ΣΜB=0; Ay− a( )⋅ w a( )⋅ 0.5 a⋅( )⋅+ P b( )⋅− 0= P Ay− a b ⎛⎜⎝ ⎞ ⎠⋅ w a b ⎛⎜⎝ ⎞ ⎠⋅ 0.5 a⋅( )⋅+= Allowable Bearing Stress: Assume failure of material occurs under plate A. Ay aA 2⎛⎝ ⎞⎠ σb_allow⋅:= P aA 2⎛⎝ ⎞⎠ σb_allow⋅⎡⎣ ⎤⎦− ab ⎛⎜⎝ ⎞ ⎠⋅ w a⋅( ) 0.5a b ⋅+:= P 31 kN= Pcase_1 P:= Assume failure of material occurs under plate B. By aB 2⎛⎝ ⎞⎠ σb_allow⋅:= P By a L ⎛⎜⎝ ⎞ ⎠⋅ w a L ⎛⎜⎝ ⎞ ⎠⋅ 0.5 a⋅( )⋅−:= P 3.67 kN= Pcase_2 P:= Pallow min Pcase_1 Pcase_2,( ):= Pallow 3.67 kN= Ans Problem 1-101 The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the average shear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has a diameter of 12 mm. If the yield shear stress for the bolt is τy = 175 MPa, and the yield tensile stress for the rod is σy = 266 MPa, determine the factor of safety with respect to yielding in each case. Given: τy 175MPa:= w 12 kN m := σy 266MPa:= a 1.2m:= b 0.6m:= e 0.9m:= do 10mm:= drod 12mm:= Solution: c a2 e2+:= h a c := v e c := Support Reactions: L a b+:= ΣΜC=0; FAB v( )⋅⎡⎣ ⎤⎦ a( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0= FAB w L a v⋅ ⎛⎜⎝ ⎞ ⎠⋅ 0.5 L⋅( )⋅:= FAB 27 kN= For bolt A: Bolt A is subjected to double shear, and V 0.5FAB:= V 13.5 kN= A π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅:= τ V A := τ 171.89 MPa= Ans FS τy τ:= FS 1.02= Ans For rod AB: N FAB:= N 27 kN= A π 4 ⎛⎜⎝ ⎞ ⎠ drod 2⋅:= σ N A := σ 238.73 MPa= Ans FS σy σ:= FS 1.11= Ans Problem 1-102 Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of τallow = 95 MPa is not exceeded in the 10-mm-diameter bolts at A and B, and an allowable tensile stress of σallow = 155 MPa is not exceeded in the 12-mm-diameter rod AB. Given: τallow 95MPa:= σallow 155MPa:= a 1.2m:= b 0.6m:= e 0.9m:= do 10mm:= drod 12mm:= Solution: c a2 e2+:= h a c := v e c := Support Reactions: L a b+:= ΣΜC=0; FAB v( )⋅⎡⎣ ⎤⎦ a( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0= FAB w L a v⋅ ⎛⎜⎝ ⎞ ⎠⋅ 0.5 L⋅( )⋅= Assume failure of pin A or B: V 0.5FAB= V τallow A⋅= A π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅:= 0.5 w⋅ L a v⋅ ⎛⎜⎝ ⎞ ⎠⋅ 0.5 L⋅( )⋅ τallow π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅⎡⎢⎣ ⎤⎥⎦⋅= w a v⋅ 0.5L( )2 τallow⋅ π 4 ⎛⎜⎝ ⎞ ⎠ do 2⋅⎡⎢⎣ ⎤⎥⎦⋅:= w 6.632 kN m = (controls!) Ans Assuming failure of rod AB: N FAB= N σallow A⋅= A π 4 ⎛⎜⎝ ⎞ ⎠ drod 2⋅:= w L a v⋅ ⎛⎜⎝ ⎞ ⎠⋅ 0.5 L⋅( )⋅ σallow π 4 ⎛⎜⎝ ⎞ ⎠ drod 2⋅⎡⎢⎣ ⎤⎥⎦⋅= w a v⋅ 0.5L2 σallow⋅ π 4 ⎛⎜⎝ ⎞ ⎠ drod 2⋅⎡⎢⎣ ⎤⎥⎦⋅:= w 7.791 kN m = Problem 1-103 The bar is supported by the pin. If the allowable tensile stress for the bar is (σt)allow = 150 MPa, and the allowable shear stress for the pin is τallow = 85 MPa, determine the diameter of the pin for which the load P will be a maximum. What is this maximum load? Assume the hole in the bar has the same diameter d as the pin. Take t =6 mm and w = 50 mm. Given: τallow 85MPa:= σt_allow 150MPa:= t 6mm:= w 50mm:= Solution: Given Allowable Normal Stress: The effective cross-sectional area Ae for the bar must be considered here by taking into account the reduction in cross-sectional area introduced by the hole. Here, effective area Ae is equal to (w - d) t, and σallow equals to P /Ae . σt_allow P w d−( ) t⋅= [1] Allowable Shear Stress: The pin is subjected to double shear and therefore the allowable τ equals to 0.5P /Apin, and the area Apin is equal to ( π/4) d2. τallow 2 π ⎛⎜⎝ ⎞ ⎠ P d2 ⎛⎜⎝ ⎞ ⎠ ⋅= [2] Solving [1] and [2]: Initial guess: d 20mm:= P 10kN:= P d ⎛⎜⎝ ⎞ ⎠ Find P d,( ):= P 31.23 kN= Ans d 15.29 mm= Ans Problem 1-104 The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowable tensile stress for the bar is (σt)allow = 140 MPa, and the allowable bearing stress between the pin and the bar is (σb)allow =210 MPA, determine the dimensions w and t such that the gross area of the cross section is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the hole in the bar has the same diameter as the pin. Given: σt_allow 140 MPa:= σb_allow 210 MPa:= A 1250mm2:= d 25mm:= Solution: A w t⋅= Given Allowable Normal Stress: The effective cross-sectional area Ae for the bar must be considered here by taking into account the reduction in cross-sectional area introduced by the hole. Here, effective area Ae is equal to (w - d) t, that is (A- d t) and σallow equals to P /Ae . σt_allow P A d t⋅−= [1] Allowable Bearing Stress: The projected area Ab is equal to (d t), and σallow equals to P /Ab . σb_allow P d t⋅ ⎛⎜⎝ ⎞ ⎠= [2] Solving [1] and [2]: Initial guess: t 0.5in:= P 1kip:= P t ⎛⎜⎝ ⎞ ⎠ Find P t,( ):= P 105.00 kN= Ans t 20.00 mm= Ans And : w A t := w 62.50 mm= Ans Problem 1-105 The compound wooden beam is connected together by a bolt at B. Assuming that the connections at A, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at B and the required outer diameter of its washers if the allowable tensile stress for the bolt is (σt)allow = 150 MPa. and the allowable bearing stress for the wood is (σb)allow = 28 MPa. Assume that the hole in the washers has the same diameter as the bolt. Given: P1 3kN:= P2 1.5kN:= P3 2kN:= σt_allow 150MPa:= a 2m:= σb_allow 28MPa:= b 1.5m:= Solution: From FBD (a): Given ΣΜD=0; Cy− 4 b⋅( )⋅ By 3b( )⋅+ P2 2 b⋅( )⋅+ P3 b( )⋅+ 0= [1] From FBD (b): ΣΜA=0; By 2 a⋅ b+( )⋅ Cy 2 a⋅( )⋅−⎡⎣ ⎤⎦ P1 a( )⋅− 0= [2] Solving [1] and [2]: Initial guess: By 1kN:= Cy 2kN:= By Cy ⎛⎜⎜⎝ ⎞ ⎠ Find By Cy,( ):= ByCy ⎛⎜⎜⎝ ⎞ ⎠ 4.4 4.55 ⎛⎜⎝ ⎞ ⎠ kN= For bolt: Arae By σt_allow = π 4 ⎛⎜⎝ ⎞ ⎠ dB 2⎛⎝ ⎞⎠⋅ By σt_allow = dB 4 π By σt_allow ⎛⎜⎝ ⎞ ⎠ ⋅:= dB 6.11 mm= Ans For washer: Area By σb_allow = π 4 ⎛⎜⎝ ⎞ ⎠ dw 2 dB 2−⎛⎝ ⎞⎠⋅ By σb_allow = dw dB 2 4 π By σb_allow ⎛⎜⎝ ⎞ ⎠ ⋅+:= dw 15.41 mm= Ans Problem 1-106 The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore it involves double shear. The allowable shear stress for both pins is τallow = 150 MPa. If a uniform distributed load of w = 8 kN/m is placed on the bar, determine its minimum allowable position x from B. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar. Given: τallow 150MPa:= do 8mm:= a 2m:= w 8 kN m :=b 2m:= Solution: ΣΜA=0; By a( )⋅ w b x−( )⋅ a x+ 0.5 b x−( )⋅+[ ]⋅− 0= [1] ΣΜB=0; Ay a( )⋅ w b x−( )⋅ x 0.5 b x−( )⋅+[ ]⋅− 0= [2] Assume failure of pin A: Arae Ay τallow = π 4 ⎛⎜⎝ ⎞ ⎠ do 2⎛⎝ ⎞⎠⋅ Ay τallow = Ay π 4 do 2⎛⎝ ⎞⎠⋅ τallow( )⋅:= Ay 7.5398 kN= Substitute value of force A into Eq [2], Given Ay a( )⋅ w b x1−( )⋅ x1 0.5 b x1−( )⋅+⎡⎣ ⎤⎦⋅− 0= [2] Initial guess: x1 0.3m:= x1 Find x1( ):= x1 0.480 m= xcase_1 x1:= Assume failure of pin B: Arae 0.5By τallow = π 4 ⎛⎜⎝ ⎞ ⎠ do 2⎛⎝ ⎞⎠⋅ 0.5By τallow = By 2 π 4 ⎛⎜⎝ ⎞ ⎠⋅ do 2⎛⎝ ⎞⎠⋅ τallow( )⋅:= By 15.0796 kN= Substitute value of force A into Eq [1], Given By a( )⋅ w b x2−( )⋅ a x2+ 0.5 b x2−( )⋅+⎡⎣ ⎤⎦⋅− 0= [1] Initial guess: x2 0.3m:= x2 Find x2( ):= x2 0.909 m= xcase_2 x2:= Choose the larger x value: x max xcase_1 xcase_2,( ):= x 0.909 m= Ans Problem 1-107 The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore it involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m, determine the maximum distributed load w the bar will support. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar. Given: τallow 125MPa:= x 1m:= do 8mm:= a 2m:= b 2m:= Solution: wo kN m := Given ΣΜA=0; Bw a( )⋅ wo b x−( )⋅ a x+ 0.5 b x−( )⋅+[ ]⋅− 0= [1] ΣΜB=0; Aw a( )⋅ wo b x−( )⋅ x 0.5 b x−( )⋅+[ ]⋅− 0= [2] Initial guess: Bw 1kN:= Aw 1kN:= Solving [1] and [2]: Bw Aw ⎛⎜⎜⎝ ⎞ ⎠ Find Bw Aw,( ):= BwAw ⎛⎜⎜⎝ ⎞ ⎠ 1.75 0.75 ⎛⎜⎝ ⎞ ⎠ kN= For pin A: Ay w1 Aw wo ⎛⎜⎝ ⎞ ⎠ ⋅= Arae Ay τallow = π 4 ⎛⎜⎝ ⎞ ⎠ do 2⎛⎝ ⎞⎠⋅ w1 τallow ⎛⎜⎝ ⎞ ⎠ Aw wo ⎛⎜⎝ ⎞ ⎠ ⋅= w1 π 4 do 2⎛⎝ ⎞⎠⋅ τallow( )⋅ woAw ⎛⎜⎝ ⎞ ⎠ ⋅:= w1 8.378 kN m = For pin B By w Bw wo ⎛⎜⎝ ⎞ ⎠ ⋅= Arae 0.5By τallow = π 2 ⎛⎜⎝ ⎞ ⎠ do 2⎛⎝ ⎞⎠⋅ w2 τallow ⎛⎜⎝ ⎞ ⎠ Bw wo ⎛⎜⎝ ⎞ ⎠ ⋅= w2 π 2 do 2⎛⎝ ⎞⎠⋅ τallow( )⋅ woBw ⎛⎜⎝ ⎞ ⎠ ⋅:= w2 7.181 kN m = The smalleer w controls ! w min w1 w2,( ):= w 7.181 kN m = Ans Problem 1-108 The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a single leaf and therefore it involves single shear in the pin, and the support at B has a double leaf and therefore it involves double shear. The allowable shear stress for both pins is τallow = 125 MPa. If x = 1 m and w = 12 kN/m, determine the smallest required diameter of pins A and B. Neglect any axial force in the bar. Given: τallow 125MPa:= x 1m:= do 8mm:= a 2m:= b 2m:= Solution: w 12 kN m := Given ΣΜA=0; By a( )⋅ w b x−( )⋅ a x+ 0.5 b x−( )⋅+[ ]⋅− 0= [1] ΣΜB=0; Ay a( )⋅ w b x−( )⋅ x 0.5 b x−( )⋅+[ ]⋅− 0= [2] Initial guess: By 1kN:= Ay 1kN:= Solving [1] and [2]: By Ay ⎛⎜⎜⎝ ⎞ ⎠ Find By Ay,( ):= By Ay ⎛⎜⎜⎝ ⎞ ⎠ 21 9 ⎛⎜⎝ ⎞ ⎠ kN= For pin A: Arae Ay τallow = π 4 ⎛⎜⎝ ⎞ ⎠ dA 2⎛⎝ ⎞⎠⋅ Ay τallow ⎛⎜⎝ ⎞ ⎠ = dA 4 π Ay τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dA 9.57 mm= Ans For pin B Arae 0.5By τallow = π 2 ⎛⎜⎝ ⎞ ⎠ dB 2⎛⎝ ⎞⎠⋅ By τallow ⎛⎜⎝ ⎞ ⎠ = dB 2 π By τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dB 10.34 mm= Ans Problem 1-109 The pin is subjected to double shear since it is used to connect the three links together. Due to wear, the load is distributed over the top and bottom of the pin as shown on the free-body diagram. Determine the diameter d of the pin if the allowable shear stress is τallow = 70 MPa and the load P = 40 kN. Also, determine the load intensities w1 and w2 . Given: τallow 70 MPa:= P 40kN:= a 37.5mm:= b 25mm:= Solution: +Pin: ΣFy=0; P w1 a( )− 0= w1 P a := w1 1066.67 kN m = Ans +Link: ΣFy=0; P 2 0.5w2( ) b( )⋅− 0= w2 P b := w2 1600.00 kN m = Ans Shear Stress Area 0.5P τallow = π 2 ⎛⎜⎝ ⎞ ⎠ d 2⋅ Pτallow = d 2 π P τallow ⎛⎜⎝ ⎞ ⎠ ⋅:= d 19.073 mm= Ans Problem 1-110 The pin is subjected to double shear since it is used to connect the three links together. Due to wear, the load is distributed over the top and bottom of the pin as shown on the free-body diagram. Determine the maximum load P the connection can support if the allowable shear stress for the material is τallow = 56 MPa and the diameter of the pin is 12.5 mm. Also, determine the load intensities w1 and w2 . Given: τallow 56 MPa:= d 12.5mm:= a 37.5mm:= b 25mm:= Solution: Shear Stress Area 0.5P τallow = π 2 ⎛⎜⎝ ⎞ ⎠ d 2⋅ Pτallow = P π 2 ⎛⎜⎝ ⎞ ⎠ d 2⋅ τallow( )⋅:= P 13.7445 kN= Ans Pin: + ΣFy=0; P w1 a( )− 0= w1 P a := w1 366.52 kN m = Ans Link: + ΣFy=0; P 2 0.5w2( ) b( )⋅− 0= w2 P b := w2 549.78 kN m = Ans Problem 1-111 The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter and the smallest diameter d of the rods. All parts are made of steel for which the failure tensile stress is σfail = 500 MPa and the failure shear stress is τfail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 in tension and (F.S.)s = 1.75 in shear. Given: σfail 500MPa:= τfail 375MPa:= P 30kN:= d2 40mm:= h 10mm:= FSt 2.50:= FSs 1.75:= Solution: Allowable Normal Stress : Design of rod size σallow σfail FSt := σallow 200 MPa= Area P σallow = π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅ Pσallow ⎛⎜⎝ ⎞ ⎠ = d 4 π P σallow ⋅:= d 13.82 mm= Ans Allowable Shear Stress : Design of cotter size τallow τfail FSs := τallow 214.29 MPa= Area 0.5P τallow = h t⋅ 0.5Pτallow ⎛⎜⎝ ⎞ ⎠ = t 1 h 0.5P τallow ⋅:= t 7 mm= Ans Problem 1-112 The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determine the average normal stress in the shank, the average shear stress along the cylindrical area of the plate defined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical area defined by the section lines b-b. Given: P 8kN:= dshank 7mm:= da_a 18mm:= db_b 7mm:= hhead 8mm:= hplate 30mm:= Solution: Average Normal Stress: Ashank π 4 ⎛⎜⎝ ⎞ ⎠ dshank 2⎛⎝ ⎞⎠:= σt P Ashank := σt 207.9 MPa= Ans Average Shear Stresses: Aa_a π da_a( )hplate:= τa_avg P Aa_a := τa_avg 4.72 MPa= Ans Ab_b π db_b( )hhead:= τb_avg P Ab_b := τb_avg 45.47 MPa= Ans Problem 1-113 The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressive load of 6 kN. Determine the average normal and shear stress acting on the plane through section aa. Show the results on a ifferential volume element located on the plane. Given: P 6kN:= d 150mm:= θ 30deg:= Solution: φ 90deg θ−:= Equations of Equilibrium: + ΣFx=0; Va_a P cos φ( )⋅− 0= Va_a P cos φ( )⋅:= Va_a 3 kN= + ΣFy=0; Na_a P sin φ( )⋅− 0= Na_a P sin φ( )⋅:= Na_a 5.196 kN= At inclined plane: A d2 sin φ( ):= σa_a Na_a A := σa_a 0.2 MPa= Ans τa_a Va_a A := τa_a 0.115 MPa= Ans Problem 1-114 Determine the resultant internal loadings acting on the cross sections located through points D and E of the frame. Given: w 2.5 kN m := a 1.2m:= b 0.9m:= c 1.5m:= e 0.45m:= Solution: d a2 b2+:= d 1.5 m= v a d := h b d := Support Reactions: L b c+:= + ΣΜA=0; By b( )⋅ w L⋅( ) 0.5 L⋅( )− 0= By w L⋅( ) 0.5 L b ⎛⎜⎝ ⎞ ⎠⋅:= By 8 kN= + ΣFy=0; Ay By− w L⋅+ 0= Ay By w L⋅−:= Ay 2 kN= By FBC v( )⋅= FBC By v := FBC 10 kN= + ΣFx=0; FBC h( )⋅ Ax− 0= Ax FBC h( )⋅:= Ax 6 kN= Segment AD: + ΣFx=0; ND Ax− 0= ND Ax:= ND 6 kN= Ans + ΣFy=0; Ay w e( )⋅+ VD+ 0= VD Ay− w e( )⋅−:= VD 3.13− kN= Ans + ΣΜD=0; MD w e( )⋅[ ] 0.5 e⋅( )⋅+ Ay e( )⋅+ 0= MD w e( )⋅[ ]− 0.5 e⋅( )⋅ Ay e( )⋅−:= MD 1.153− kN m⋅= Ans Segment CE: + ΣFx=0; NE FBC+ 0= ND FBC−:= ND 10− kN= Ans + ΣFy=0; VD 0:= VD 0 kN= Ans + ΣΜE=0; MD 0:= MD 0 kN m⋅= Ans Problem 1-115 The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shear stress in the plate due to this loading. Given: P 2kN:= dpunch 4mm:= hplate 2mm:= Solution: Average Shear Stresses: Aa_a π dpunch( )hplate:= τa_avg P Aa_a := τa_avg 79.58 MPa= Ans Problem 1-116 The cable has a specific weight γ (weight/volume) and cross-sectional area A. If the sag s is small, so that its length is approximately L and its weight can be distributed uniformly along the horizontal axis, determine the average normal stress in the cable at its lowest point C. Solution: Equations of Equilibrium: ΣMA=0; T s⋅ γAL2 ⎛⎜⎝ ⎞ ⎠ L 4 ⋅− 0= γ γ A⋅ L⋅ 8 s⋅= Average Normal Stresses: σ T A = σ γ L⋅ 8 s⋅= Ans Problem 1-117 The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine the resultant internal loadings acting on cross sections located at points D and E. Neglect the thickness of both the beam and column in the calculation. Given: wb 2.0 kN m := L 3.6m:= d 1.8m:= wc 3.0 kN m := H 4.8m:= e 1.2m:= a 3.6m:= b 3.6m:= c 1.2m:= Solution: Beam AB: Lc L 2 c2+:= vb c Lc := hb L Lc := + ΣΜA=0; By L( )⋅ wAB L( )⋅ 0.5 L⋅( )− 0= By wb L( )⋅ 0.5 L L ⎛⎜⎝ ⎞ ⎠⋅:= By 7936.64 m s2 lb= + Ay By− wb L( )⋅+:= Ay 3.6 kN=ΣFy=0; Ay− By− wb L( )⋅+ 0= By FBC vb( )⋅= FBC Byvb:= FBC 11.38 kN=+ ΣFx=0; FBC− h( )⋅ Ax+ 0= Ax FBC hb( )⋅:= Ax 10.8 kN= Segment AD: + ΣFx=0; ND Ax+ 0= ND Ax−:= ND 10.8− kN= Ans + ΣFy=0; Ay− wb d( )⋅+ VD+ 0= VD Ay wb d( )⋅−:= VD 0 kN= Ans + ΣΜD=0; MD wb d( )⋅⎡⎣ ⎤⎦ 0.5 d⋅( )⋅+ Ay d( )⋅− 0= MD wb d( )⋅⎡⎣ ⎤⎦− 0.5 d⋅( )⋅ Ay d( )⋅+:= MD 3.24 kN m⋅= Ans Member CG: Hb H 2 b2+:= vc H Hb := hc b Hb := Column FC: + ΣΜC=0; Fx H( )⋅ Ax c⋅− 0= Fx Ax c H ⎛⎜⎝ ⎞ ⎠⋅:= Fx 2.7 kN= + ΣFx=0; FBC hb( )⋅ Ax− Fx+ FCG hc( )⋅− 0= + ΣFy=0; FCG FBC hb( )⋅ Ax− Fx+ hc := FCG 4.5 kN= Fy− By+ wc H( )⋅+ FBC vb( )⋅+ FCG vc( )⋅+ 0= Fy By wc H( )⋅+ FBC vb( )⋅+ FCG vc( )⋅+:= Fy 25.2 kN= Segment FE: + ΣFx=0; VE Fx− 0= VE Fx:= VE 2.7 kN= Ans + ΣFy=0; NE wc e( )⋅+ Fy− 0= NE wc− e( )⋅ Fy+:= NE 21.6 kN= Ans + ΣΜE=0; ME− Fy e( )⋅+ 0= ME Fy e( )⋅:= ME 30.24 kN m⋅= Ans Problem 1-118 The 3-Mg concrete pipe is suspended by the three wires. If BD and CD have a diameter of 10 mm and AD has a diameter of 7 mm, determine the average normal stress in each wire. Given: M 3000kg:= g 9.81 m s2 := h 2m:= r 1m:= α 120deg:= dBD 10mm:= dCD 10mm:= dAD 7mm:= Solution: W M g⋅:= W 29.43 kN= θ 0.5α:= θ 60 deg= L r2 h2+:= v h L := Equations of Equilibrium: ΣΜx=0; 2 F⋅ r cos θ( )⋅( )⋅ FAD r( )⋅− 0= FBD = FCD = F ΣΜy=0; FBD r sin θ( )⋅( )⋅ FCD r sin θ( )⋅( )⋅− 0= FAD = F ΣFz=0; 3 F v( )⋅[ ]⋅ W− 0= F W 3v := F 10.97 kN= Allowable Normal Stress : σBD = σCD = σ1 σ1 F Area = σ1 F π 4 dBD 2⎛⎝ ⎞⎠⋅ := σ1 139.65 MPa= Ans σAD = σ2 σ2 F Area = σ2 F π 4 dAD 2⎛⎝ ⎞⎠⋅ := σ2 285 MPa= Ans Problem 1-119 The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normal stress in each rod and the average shear stress in the pin A between the members. Given: P 5kN:= dpin 25mm:= d30 30mm:= d40 40mm:= Solution: Average Normal Stress: A40 π 4 ⎛⎜⎝ ⎞ ⎠ d40 2⎛⎝ ⎞⎠:= σ40 P A40 := σ40 3.979 MPa= Ans A30 π 4 ⎛⎜⎝ ⎞ ⎠ d30 2⎛⎝ ⎞⎠:= σ30 P A30 := σ30 7.074 MPa= Ans Average Shear Stresses: Apin π 4 ⎛⎜⎝ ⎞ ⎠ dpin 2⎛⎝ ⎞⎠:= τavg 0.5P Apin := τavg 5.093 MPa= Ans Problem 2-1 An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until the ball's diameter becomes 175 mm, determine the average normal strain in the rubber. Given: d0 150mm:= d 175mm:= Solution: ε πd πd0− πd0 := ε 0.1667 mm mm = Ans Problem 2-2 A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having an outer diameter of 125 mm, determine the average normal strain in the strip. Given: L0 375mm:= Solution: L π 125( )⋅ mm:= ε πL πL0− πL0 := ε 0.0472 mm mm = Ans Problem 2-3 The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD. Given: a 3m:= LCE 4m:= b 4m:= LBD 4m:= ∆LCE 10mm:= Solution: ∆LBD a a b+ ⎛⎜⎝ ⎞ ⎠ ∆LCE⋅:= ∆LBD 4.2857 mm= εCE ∆LCE LCE := εCE 0.00250 mm mm = Ans εBD ∆LBD LBD := εBD 0.00107 mm mm = Ans Problem 2-4 The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within it causes the balloon's diameter to become d = 125 mm, determine the average normal strain in the rubber. Given: d0 100mm:= d 125mm:= Solution: ε πd πd0− πd0 := ε 0.2500 mm mm = Ans Problem 2-5 The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace 10 mm downward, determine the normal strain developed in wires CE and BD. Given: a 3m:= b 2m:= c 2m:= LCE 4m:= LBD 3m:= ∆ tip 10mm:= Solution: ∆LBD a ∆LCE a b+= ∆LCE a b+ ∆ tip a b+ c+= ∆LCE a b+ a b+ c+ ⎛⎜⎝ ⎞ ⎠ ∆ tip⋅:= ∆LCE 7.1429 mm= ∆LBD a a b+ c+ ⎛⎜⎝ ⎞ ⎠ ∆ tip⋅:= ∆LBD 4.2857 mm= Average Normal Strain: εCE ∆LCE LCE := εCE 0.00179 mm mm = Ans εBD ∆LBD LBD := εBD 0.00143 mm mm = Ans Problem 2-6 The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normal strain in each wire is εmax = 0.002 mm/mm, determine the maximum vertical displacement of the load P. Given: a 3m:= b 2m:= c 2m:= LCE 4m:= LBD 3m:= εallow 0.002 mm mm := Solution: ∆LBD a ∆LCE a b+= ∆LCE a b+ ∆ tip a b+ c+= Average Elongation/Vertical Displacement: ∆LBD LBD εallow⋅:= ∆LBD 6.00 mm= ∆ tip a b+ c+ a ⎛⎜⎝ ⎞ ⎠ ∆LBD⋅:= ∆ tip 14.00 mm= ∆LCE LCE εallow⋅:= ∆LCE 8.00 mm= ∆ tip a b+ c+ a b+ ⎛⎜⎝ ⎞ ⎠ ∆LCE⋅:= ∆ tip 11.20 mm= (Controls !) Ans Problem 2-7 The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire. Given: a 300mm:= θ 30deg:= ∆A 2mm:= Solution: Consider the triangle CAA': φA 180deg θ−:= φA 150 deg= LCA' a 2 ∆A2+ 2 a⋅ ∆A( )⋅ cos φA( )⋅−:= LCA' 301.734 mm= ∆CA LCA' a− a := ∆CA 0.00578 mm mm = Ans Problem 2-8 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes it to rotate by θ = 0.3°, determine the normal strain in the cable. Originally the cable is unstretched. Given: a 400mm:= b 300mm:= c 300mm:= θ 0.3deg:= Solution: LAB a 2 b2+:= LAB 500 mm= Consider the triangle ACB': φC 90deg θ+:= φC 90.3 deg= LAB' a 2 b2+ 2 a⋅ b⋅ cos φC( )⋅−:= LAB' 501.255 mm= εAB LAB' LAB− LAB := εAB 0.00251 mm mm = Ans Problem 2-9 Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If a force is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mm determine the displacement of point D. Originally the cable is unstretched. Given: a 400mm:= b 300mm:= c 300mm:= εAB 0.0035 mm mm := Solution: LAB a 2 b2+:= LAB 500 mm= LAB' LAB 1 εAB+( )⋅:= LAB' 501.750 mm= Consider the triangle ACB': φC 90deg θ+= LAB' a 2 b2+ 2 a⋅ b⋅ cos φC( )⋅−= φC acos a2 b2+( ) LAB'2− 2 a⋅ b⋅ ⎡⎢⎣ ⎤⎥⎦:= φC 90.419 deg= θ φC 90deg−:= θ 0.41852 deg= θ 0.00730 rad= ∆D b c+( ) θ⋅:= ∆D 4.383 mm= Ans Problem 2-10 The wire AB is unstretched when θ = 45°. If a vertical load is applied to bar AC, which causes θ = 47°, determine the normal strain in the wire. Given: θ 45deg:= ∆θ 2deg:= Solution: LAB L 2 L2+= LAB 2 L= LCB 2L( ) 2 L2+= LCB 5 L= From the triangle CAB: φA 180deg θ−:= φA 135.00 deg= sin φB( ) L sin φA( ) LCB = φB asin L sin φA( )⋅ 5 L ⎛⎜⎝ ⎞ ⎠:= φB 18.435 deg= From the triangle CA'B: φ'B φB ∆θ+:= φ'B 20.435 deg= sin φ'B( ) L sin 180deg φA'−( ) LCB = φA' 180deg asin 5 L⋅ sin φ'B( )⋅ L ⎛⎜⎝ ⎞ ⎠−:= φA' 128.674 deg= φ'C 180deg φ'B− φA'−:= φ'C 30.891 deg= sin φ'B( ) L sin φ'C( ) LA'B = LA'B sin φ'C( ) sin φ'B( ) L⋅:= LAB 2 L:= εAB LA'B LAB− LAB := εAB 0.03977= Ans Problem 2-11 If a load applied to bar AC causes point A to be displaced to the left by an amount ∆L, determine the normal strain in wire AB. Originally, θ = 45°. Given: θ 45deg:= Solution: εAC ∆L L = LAB L 2 L2+= LAB 2 L= LCB 2L( ) 2 L2+= LCB 5 L= From the triangle A'AB: φA 180deg θ−:= φA 135.00 deg= LA'B ∆L2 LAB2+ 2 ∆L( )⋅ LAB( )⋅ cos φA( )⋅−= LA'B ∆L2 2 L2⋅+ 2 ∆L( )⋅ L⋅+= εAB LA'B LAB− LAB = εAB ∆L2 2 L2⋅+ 2 ∆L( )⋅ L⋅+ 2 L− 2 L = εAB 1 2 ∆L L ⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦ 2 1+ ∆L L ⎛⎜⎝ ⎞ ⎠+ 1−= Neglecting the higher-order terms, εAB 1 ∆L L +⎛⎜⎝ ⎞ ⎠ 0.5 1−= εAB 1 1 2 ∆L L ⎛⎜⎝ ⎞ ⎠⋅+ .....+ ⎡⎢⎣ ⎤⎥⎦ 1−= (Binomial expansion) εAB 1 2 ∆L L ⎛⎜⎝ ⎞ ⎠⋅= Ans Alternatively, εAB LA'B LAB− LAB = εAB ∆L sin θ( )⋅ 2 L = εAB 1 2 ∆L L ⎛⎜⎝ ⎞ ⎠⋅= Ans Problem 2-12 The piece of plastic is originally rectangular. Determine the shear strain γxy at corners A and B if the plastic distorts as shown by the dashed lines. Given: a 400mm:= b 300mm:= ∆Ax 3mm:= ∆Ay 2mm:= ∆Cx 2mm:= ∆Cy 2mm:= ∆Bx 5mm:= ∆By 4mm:= Solution: Geometry : For small angles, α ∆Cx b ∆Cy+ := α 0.00662252 rad= β ∆By ∆Cy− a ∆Bx ∆Cx−( )+:= β 0.00496278 rad= ψ ∆Bx ∆Ax− b ∆By ∆Ay−( )+:= ψ 0.00662252 rad= θ ∆Ay a ∆Ax+ := θ 0.00496278 rad= Shear Strain : γxy_B β ψ+:= γxy_B 11.585 10 3−× rad= Ans γxy_A θ ψ+( )−:= γxy_A 11.585− 10 3−× rad= Ans Problem 2-13 The piece of plastic is originally rectangular. Determine the shear strain γxy at corners D and C if the plastic distorts as shown by the dashed lines. Given: a 400mm:= b 300mm:= ∆Ax 3mm:= ∆Ay 2mm:= ∆Cx 2mm:= ∆Cy 2mm:= ∆Bx 5mm:= ∆By 4mm:= Solution: Geometry : For small angles, α ∆Cx b ∆Cy+ := α 0.00662252 rad= β ∆By ∆Cy− a ∆Bx ∆Cx−( )+:= β 0.00496278 rad= ψ ∆Bx ∆Ax− b ∆By ∆Ay−( )+:= ψ 0.00662252 rad= θ ∆Ay a ∆Ax+ := θ 0.00496278 rad= Shear Strain : γxy_D α θ+:= γxy_D 11.585 10 3−× rad= Ans γxy_C α β+( )−:= γxy_C 11.585− 10 3−× rad= Ans Problem 2-14 The piece of plastic is originally rectangular. Determine the average normal strain that occurs along th diagonals AC and DB. Given: a 400mm:= b 300mm:= ∆Ax 3mm:= ∆Ay 2mm:= ∆Cx 2mm:= ∆Cy 2mm:= ∆Bx 5mm:= ∆By 4mm:= Solution: Geometry : LAC a 2 b2+:= LAC 500 mm= LDB a 2 b2+:= LDB 500 mm= LA'C' a ∆Ax+ ∆Cx−( )2 b ∆Cy+ ∆Ay−( )2+:= LA'C' 500.8 mm= LDB' a ∆Bx+( )2 b ∆By+( )2+:= LDB' 506.4 mm= Average Normal Strain : εAC LA'C' LAC− LAC := εAC 1.601 10 3−× mm mm = Ans εBD LDB' LDB− LDB := εBD 12.800 10 3−× mm mm = Ans Problem 2-15 The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columns of the frame tilt θ = 2°. Determine the approximate normal strain in the wire when the frame is in thi position. Assume the columns are rigid and rotate about their lower supports. Given: a 4m:= b 3m:= c 1m:= θ 2deg:= Solution: θ θ 180 ⎛⎜⎝ ⎞ ⎠ π⋅= θ 0.03490659 rad= Geometry : The vertical dosplacement is negligible. ∆Ax c θ⋅:= ∆Ax 34.907 mm= ∆Bx b c+( ) θ⋅:= ∆Bx 139.626 mm= LAB a 2 b2+:= LAB 5000 mm= LA'B' a ∆Bx+ ∆Ax−( )2 b2+:= LA'B' 5084.16 mm= Average Normal Strain : εAB LA'B' LAB− LAB := εAB 16.833 10 3−× mm mm = Ans Problem 2-16 The corners of the square plate are given the displacements indicated. Determine the shear strain along the edges of the plate at A and B. Given: ax 250mm:= ay 250mm:= ∆v 5mm:= ∆h 7.5mm:= Solution: At A : tan θ'A 2 ⎛⎜⎝ ⎞ ⎠ ax ∆h− ay ∆v+ = θ'A 2 atan ax ∆h− ay ∆v+ ⎛⎜⎝ ⎞ ⎠ ⋅:= θ'A 1.52056 rad= γnt_A π 2 ⎛⎜⎝ ⎞ ⎠ θ'A−:= γnt_A 0.05024 rad= Ans At B : tan φ'B 2 ⎛⎜⎝ ⎞ ⎠ ay ∆v+ ax ∆h− = φ'B 2 atan ay ∆v+ ax ∆h− ⎛⎜⎝ ⎞ ⎠ ⋅:= φ'B 1.62104 rad= γnt_B φ'B π 2 −:= γnt_B 0.05024 rad= Ans Problem 2-17 The corners of the square plate are given the displacements indicated. Determine the average normal strains along side AB and diagonals AC and DB. Given: ax 250mm:= ay 250mm:= ∆v 5mm:= ∆h 7.5mm:= Solution: For AB : LAB ax 2 ay 2+:= LA'B' ax ∆h−( )2 ay ∆v+( )2+:= εAB LA'B' LAB− LAB := LAB 353.55339 mm= LA'B' 351.89665 mm= εAB 4.686− 10 3−× mm mm = Ans For AC : LAC 2 ay( ):= LAC 500 mm= LA'C' 2 ay ∆v+( )⋅:= LA'C' 510 mm= εAC LA'C' LAC− LAC := εAC 20.000 10 3−× mm mm = Ans For DB : LDB 2 ax( ):= LDB 500 mm= LD'B' 2 ax ∆h−( )⋅:= LD'B' 485 mm= εDB LD'B' LDB− LDB := εDB 30.000− 10 3−× mm mm = Ans Problem 2-18 The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D'B' remains horizontal. Given: a 50mm:= b 50mm:= ∆Bx 3− mm:= ∆Cx 8mm:= ∆Cy 0mm:= θ'A 91.5deg:= LAD' 53mm:= Solution: For AB : ∆By LAD' cos θ'A 90deg−( )⋅ b−:= ∆By 2.9818 mm= LAB a 2 b2+:= LAB 70.7107 mm= LAB' a ∆Bx+( )2 b ∆By+( )2+:= LAB' 70.8243 mm= εAB LAB' LAB− LAB := εAB 1.606 10 3−× mm mm = Ans For CD : ∆Dy ∆By:= ∆Dy 2.9818 mm= LCD a 2 b2+:= LCD 70.7107 mm= LC'D' a ∆Cx+( )2 b ∆Dy+( )2+ 2 a ∆Cx+( )⋅ b ∆Dy+( )⋅ cos θ'A( )⋅−:= LC'D' 79.5736 mm= εCD LC'D' LCD− LCD := εCD 125.340 10 3−× mm mm = Ans Problem 2-19 The square deforms into the position shown by the dashed lines. Determine the shear strain at each of its corners, A, B, C, and D. Side D'B' remains horizontal. Given: a 50mm:= b 50mm:= ∆Bx 3− mm:= ∆Cx 8mm:= ∆Cy 0mm:= θ'A 91.5deg:= LAD' 53mm:= Solution: θ'A 1.597 rad= Geometry : ∆By LAD' cos θ'A 90deg−( )⋅ b−:= ∆By 2.9818 mm= ∆Dy ∆By:= ∆Dy 2.9818 mm= ∆Dx LAD'− sin θ'A 90deg−( )⋅:= ∆Dx 1.3874− mm= In triangle C'B'D' : LC'B' ∆Cx ∆Bx−( )2 b ∆By+( )2+:= LC'B' 54.1117 mm= LD'B' a ∆Bx+ ∆Dx−:= LD'B' 48.3874 mm= LC'D' a ∆Cx+ ∆Dx−( )2 b ∆Dy+( )2+:= LC'D' 79.586 mm= cos θB'( ) LC'B' 2 LD'B' 2+ LC'D'2− 2 LC'B'( )⋅ LD'B'( )⋅= θB' acos LC'B' 2 LD'B' 2+ LC'D'2− 2 LC'B'( )⋅ LD'B'( )⋅ ⎡⎢⎢⎣ ⎤⎥⎥⎦ := θB' 101.729 deg= θB' 1.7755 rad= θD' 180deg θ'A−:= θD' 88.500 deg= θD' 1.5446 rad= θC' 180deg θB'−:= θC' 78.271 deg= θC' 1.3661 rad= Shear Strain : γxy_A 0.5π θ'A( )−:= γxy_A 26.180− 10 3−× rad= Ans γxy_B 0.5π θB'( )−:= γxy_B 204.710− 10 3−× rad= Ans γxy_C 0.5π θC'( )−:= γxy_C 204.710 10 3−× rad= Ans γxy_D 0.5π θD'( )−:= γxy_D 26.180 10 3−× rad= Ans Problem 2-20 The block is deformed into the position shown by the dashed lines. Determine the average normal strain along line AB. Given: ∆xBA 70 30−( )mm:= ∆yBA 100mm:= ∆xB'A 55 30−( )mm:= ∆yB'A 1102 152−( )mm:= Solution: For AB : LAB ∆xBA2 ∆yBA2+:= LAB 107.7033 mm= LAB' ∆xB'A2 ∆yB'A2+:= LAB' 111.8034 mm= εAB LAB' LAB− LAB := εAB 38.068 10 3−× mm mm = Ans Problem 2-21 A thin wire, lying along the x axis, is strained such that each point on the wire is displaced ∆x = k x2 along the x axis. If k is constant, what is the normal strain at any point P along the wire? Given: ∆x k x2⋅= Solution: ε x ∆xd d = ε 2 k⋅ x⋅= Ans Problem 2-22 The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averag shear strain γxy of the plate. Given: a 150mm:= b 200mm:= ∆a 0mm:= ∆b 3− mm:= Solution: ∆θ atan ∆b a ⎛⎜⎝ ⎞ ⎠:= ∆θ 1.146− deg= ∆θ 19.9973− 10 3−× rad= Shear Strain : γxy ∆θ:= γxy 19.997− 10 3−× rad= Ans Problem 2-23 The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averag shear strain γxy of the plate. Given: a 200mm:= ∆a 3mm:= b 150mm:= ∆b 0mm:= Solution: ∆θ atan ∆a b ⎛⎜⎝ ⎞ ⎠:= ∆θ 1.146 deg= ∆θ 19.9973 10 3−× rad= Shear Strain : γxy ∆θ:= γxy 19.997 10 3−× rad= Ans Problem 2-24 The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the average normal strains along the diagonal AC and side AB. Given: a 200mm:= b 150mm:= ∆Ax 3− mm:= ∆Ay 0mm:= ∆Bx 0mm:= ∆By 0mm:= ∆Cx 0mm:= ∆Cy 0mm:= ∆Dx 3− mm:= ∆Dy 0mm:= Solution: Geometry : LAC a 2 b2+:= LAC 250 mm= LAB b:= LAB 150 mm= LA'C a ∆Cx+ ∆Ax−( )2 b ∆Cy+ ∆Ay−( )2+:= LA'C 252.41 mm= LA'B ∆Ax2 b2+:= LA'B 150.03 mm= Average Normal Strain : εAC LA'C LAC− LAC := εAC 9.626 10 3−× mm mm = Ans εAB LA'B LAB− LAB := εAB 199.980 10 6−× mm mm = Ans Problem 2-25 The piece of rubber is originally rectangular. Determine the average shear strain γxy if the corners B and D are subjected to the displacements that cause the rubber to distort as shown by the dashed lines Given: a 300mm:= b 400mm:= ∆Ax 0mm:= ∆Ay 0mm:= ∆Bx 0mm:= ∆By 2mm:= ∆Dx 3mm:= ∆Dy 0mm:= Solution: ∆θAB atan ∆By a ⎛⎜⎝ ⎞ ⎠:= ∆θAB 0.38197 deg= ∆θAB 6.6666 10 3−× rad= ∆θAD atan ∆Dx b ⎛⎜⎝ ⎞ ⎠:= ∆θAD 0.42971 deg= ∆θAD 7.4999 10 3−× rad= Shear Strain : γxy_A ∆θAB ∆θAD+:= γxy_A 14.166 10 3−× rad= Ans Problem 2-26 The piece of rubber is originally rectangular and subjected to the deformation shown by the dashed lines. Determine the average normal strain along the diagonal DB and side AD. Given: a 300mm:= b 400mm:= ∆Ax 0mm:= ∆Ay 0mm:= ∆Bx 0mm:= ∆By 2mm:= ∆Dx 3mm:= ∆Dy 0mm:= Solution: Geometry : LDB a 2 b2+:= LDB 500 mm= LAD b:= LAD 400 mm= LD'B' a ∆Bx+ ∆Dx−( )2 b ∆Dy+ ∆By−( )2+:= LD'B' 496.6 mm= LAD' ∆Dx2 b ∆Dy+( )2+:= LAD' 400.01 mm= Average Normal Strain : εBD LD'B' LDB− LDB := εBD 6.797− 10 3−× mm mm = Ans εAD LAD' LAD− LAD := εAD 28.125 10 6−× mm mm = Ans Problem 2-27 The material distorts into the dashed position shown. Determine (a) the average normal strains εx and εy , the shear strain γxy at A, and (b) the average normal strain along line BE. Given: a 80mm:= Bx 0mm:= Ex 80mm:= b 125mm:= By 100mm:= Ey 50mm:= ∆Ax 0mm:= ∆Cx 10mm:= ∆Dx 15mm:= ∆Ay 0mm:= ∆Cy 0mm:= ∆Dy 0mm:= Solution: ∆xAC ∆Cx ∆Ax−:= ∆xAC 10.00 mm= ∆yAC ∆Cy ∆Ay−:= ∆yAC 0.00 mm= ∆θAC atan ∆Cx b ⎛⎜⎝ ⎞ ⎠:= ∆θAC 79.8300 10 3−× rad= Since there is no deformation occuring along the y- and x-axis, εx_A ∆yAC:= εx_A 0= Ans εy_A ∆xAC2 b2+ b− b := εy_A 0.00319= Ans γxy_A ∆θAC:= γxy_A 79.830 10 3−× rad= Ans Geometry : ∆Bx ∆Cx By b = ∆Bx By b ⎛⎜⎝ ⎞ ⎠ ∆Cx⋅:= ∆Bx 8 mm= ∆By 0mm:= ∆Ex ∆Dx Ey b = ∆Ex Ey b ⎛⎜⎝ ⎞ ⎠ ∆Dx⋅:= ∆Ex 6 mm= ∆Ey 0mm:= LBE Ex Bx−( )2 Ey By−( )2+:= LBE 94.34 mm= LB'E' a ∆Ex+ ∆Bx−( )2 Ey ∆Ey+ ∆By−( )2+:= LB'E' 92.65 mm= εBE LB'E' LBE− LBE := εBE 17.913− 10 3−× mm mm = Ans Note: Negative sign indicates shortening of BE. Problem 2-28 The material distorts into the dashed position shown. Determine the average normal strain that occurs along the diagonals AD and CF. Given: a 80mm:= Bx 0mm:= Ex 80mm:= b 125mm:= By 100mm:= Ey 50mm:= ∆Ax 0mm:= ∆Cx 10mm:= ∆Dx 15mm:= ∆Ay 0mm:= ∆Cy 0mm:= ∆Dy 0mm:= Solution: ∆xAC ∆Cx ∆Ax−:= ∆xAC 10.00 mm= ∆yAC ∆Cy ∆Ay−:= ∆yAC 0.00 mm= ∆θAC atan ∆Cx b ⎛⎜⎝ ⎞ ⎠:= ∆θAC 79.8300 10 3−× rad= Geometry : LAD a 2 b2+:= LAD 148.41 mm= LCF a 2 b2+:= LCF 148.41 mm= LA'D' a ∆Dx+ ∆Ax−( )2 b ∆Dy+ ∆Ay−( )2+:= LA'D' 157.00 mm= LC'F a ∆Cx−( )2 b ∆Cy−( )2+:= LC'F 143.27 mm= Average Normal Strain : εAD LA'D' LAD− LAD := εAD 57.914 10 3−× mm mm = Ans εCF LC'F LCF− LCF := εCF 34.653− 10 3−× mm mm = Ans Problem 2-29 The block is deformed into the position shown by the dashed lines. Determine the shear strain at corners C and D. Given: a 100mm:= ∆Ax 15− mm:= b 100mm:= ∆Bx 15− mm:= LCA' 110mm:= Solution: Geometry : ∆θC asin ∆Ax LCA' ⎛⎜⎝ ⎞ ⎠ := ∆θC 7.84− deg= ∆θC 0.1368− rad= ∆θD asin ∆Bx LCA' ⎛⎜⎝ ⎞ ⎠ := ∆θD 7.84− deg= ∆θD 0.1368− rad= Shear Strain : γxy_C ∆θC:= γxy_C 136.790− 10 3−× rad= Ans γxy_D ∆θD−:= γxy_D 136.790 10 3−× rad= Ans Problem 2-30 The bar is originally 30 mm long when it is flat. If it is subjected to a shear strain defined by γxy = 0.0 x, where x is in millimeters, determine the displacement ∆y at the end of its bottom edge. It is distorte into the shape shown, where no elongation of the bar occurs in the x direction. Given: L 300mm:= γxy 0.02 x⋅= unit 1mm:= Solution: dy dx tan γxy( )= dy dx tan 0.02 x⋅( )= 0 ∆y y1 ⌠⎮⌡ d 0 L xtan 0.02 x⋅( ) unit( )⋅⌠⎮⌡ d= ∆y 0 30 xtan 0.02x( ) unit( )⋅⌠⎮⌡ d:= ∆y 9.60 mm= Ans Problem 2-31 The curved pipe has an original radius of 0.6 m. If it is heated nonuniformly, so that the normal strain along its length is ε = 0.05 cos θ, determine the increase in length of the pipe. Given: r 0.6m:= ε 0.05 cos θ( )⋅= Solution: ∆L Lε⌠⎮⌡ d= ∆L 0 90deg rθ( )0.05 cos θ( )⋅⌠⎮⌡ d= ∆L 0 90deg θ0.05 r⋅ cos θ( )⋅⌠⎮⌡ d:= ∆L 30.00 mm= Ans Problem 2-32 Solve Prob. 2-31 if ε = 0.08 sin θ . Given: r 0.6m:= ε 0.08 sin θ( )⋅= Solution: ∆L Lε⌠⎮⌡ d= ∆L 0 90deg rθ( )0.08 sin θ( )⋅⌠⎮⌡ d= ∆L 0 90deg θ0.08 r⋅ sin θ( )⋅⌠⎮⌡ d:= ∆L 0.0480 m= Ans Problem 2-33 A thin wire is wrapped along a surface having the form y = 0.02 x2, where x and y are in mm. Origina the end B is at x = 250 mm. If the wire undergoes a normal strain along its length of ε = 0.0002x, determine the change in length of the wire. Hint: For the curve, y = f (x), ds = 1 dy dx ⎛⎜⎝ ⎞ ⎠ 2 + dx⋅ . Given: Bx 250mm:= ε 0.0002 x⋅= unit 1mm:= Solution: y 0.02x2= dy dx 0.04x= ds 1 dy dx ⎛⎜⎝ ⎞ ⎠ 2 + dx⋅= ds 1 0.04x( )2+ dx⋅= ∆L sε⌠⎮⌡ d= ∆L 0 Bx x0.0002x( ) 1 0.04x( )2+⋅⌠⎮⌡ d= ∆L unit( ) 0 250 x0.0002x( ) 1 0.04x( )2+⋅⌠⎮⌡ d⋅:= ∆L 42.252 mm= Ans Problem 2-34 The fiber AB has a length L and orientation If its ends A and B undergo very small displacements uA and vB, respectively, determine the normal strain in the fiber when it is in position A'B'. Solution: Geometry: LA'B' L cos θ( )⋅ uA−( )2 L sin θ( )⋅ vB−( )2+= LA'B' L 2 uA 2+ vB2+ 2L vB sin θ( )⋅ uA cos θ( )⋅−( )⋅+= Average Normal Strain: εAB LA'B' L−⋅ L = εAB 1 uA 2 vB 2+ L2 + 2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅ L + 1−= Neglecting higher-order terms uA 2 and vB 2 , εAB 1 2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅ L + 1−= Using the binomial theorem: εAB 1 1 2 2 vB sin θ( )⋅ uA cos θ( )⋅−( )⋅ L ⎡⎢⎣ ⎤⎥⎦+ ..+ 1−= εAB vB sin θ( )⋅ L uA cos θ( )⋅ L −= Ans Problem 2-35 If the normal strain is defined in reference to the final length, that is, ε ' n = p'p ∆s' ∆s− ∆s' ⎛⎜⎝ ⎞ ⎠lim→ instead of in reference to the original length, Eq.2-2, show that the difference in these strains is represented as a second-order term, namely, εn - ε'n = εn ε'n. Solution: εn ∆S' ∆S− ∆S= εn ε'n− ∆S' ∆S− ∆S ⎛⎜⎝ ⎞ ⎠ ∆S' ∆S− ∆S' ⎛⎜⎝ ⎞ ⎠−= εn ε'n− ∆S'2 2 ∆S( )⋅ ∆S'( )⋅− ∆S2+ ∆S( ) ∆S'( )⋅= εn ε'n− ∆S' ∆S−( )2 ∆S( ) ∆S'( )⋅= εn ε'n− ∆S' ∆S− ∆S ⎛⎜⎝ ⎞ ⎠ ∆S' ∆S− ∆S' ⎛⎜⎝ ⎞ ⎠⋅= εn ε'n− εn( ) ε'n( )⋅= (Q.E.D.) 0 1 .10 4 2 .10 4 3 .10 4 4 .10 4 5 .10 4 6 .10 4 7 .10 4 2 4 6 8 10 12 14 16 σ Y x( ) ε x, δ 0.0000 0.0150 0.0300 0.0500 0.0650 0.0850 0.1000 0.1125 0.1250 0.1550 0.1750 0.1875 := Problem 3-1 A concrete cylinder having a diameter of 150 mm. and gauge length of 300 mm is tested in compression. The results of the test are reported in the table as load versus contraction. Draw the stress-strain diagram using scales of 10 mm = 2 MPa and 10 mm = 0.1(10-3) mm/mm. From the diagram, determine approximately the modulus of elasticity. Given: d 150:= L 300:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= * σ 103P A := ε δ L := σ=P/A (MPa) ε=δ/L (mm/mm) x min ε( ) 0.00005, max ε( )..:=Regression curve: Coeff loess ε σ, 1.5,( ):=σ 0.00 1.41 2.69 4.67 5.80 7.22 8.49 9.76 10.89 13.16 14.15 15.00 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = ε 0.000000 0.000050 0.000100 0.000167 0.000217 0.000283 0.000333 0.000375 0.000417 0.000517 0.000583 0.000625 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = Y x( ) interp Coeff ε, σ, x,( ):= Modulus of Elasticity: From the stress-strain diagram, ∆ε 0.0003 0−( ) mm mm := * ∆σ 8.0 0−( )MPa:= * Eapprox ∆σ ∆ε:= Eapprox 26.67 GPa= Ans Contraction (mm) P 0.0 25.0 47.5 82.5 102.5 127.5 150.0 172.5 192.5 232.5 250.0 265.0 := Load (kN) Problem 3-2 Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine the modulus of elasticity and the modulus of resilience. Unit used: MJ 106( )J:= σ 0.0 232.4 318.5 345.8 360.5 373.8 := ε 0.0000 0.0006 0.0010 0.0014 0.0018 0.0022 :=Solution: Regression curve: x min ε( ) 0.00005, max ε( )..:= Coeff loess ε σ, 0.9,( ):= Y x( ) interp Coeff ε, σ, x,( ):= 0 5 .10 4 0.001 0.0015 0.002 0.0025 40 80 120 160 200 240 280 320 360 400 σ Y x( ) ε x, Modulus of Elasticity: From the stress-strain diagram, ∆ε 0.0006 0−( ) mm mm := ∆σ 232.4 0−( )MPa:= Eapprox ∆σ ∆ε:= Eapprox 387.3 GPa= Ans Modulus of Resilience: The modulus of resilience is equal to the area under the initial linear portion of the curve. ∆ε 0.0006 0−( ) mm mm := ∆σ 232.4 0−( )103 kN m2 := ur 1 2 ∆ε⋅ ∆σ⋅:= ur 0.0697 MJ m3 = Ans σ=P/A (MPa) ε=δ/L (mm/mm) Problem 3-3 Data taken from a stress-strain test for a ceramic are given in the table. The curve is linear between the origin and the first point. Plot the diagram, and determine approximately the modulus of toughness. The rupture stress is σr = 373.8 MPa. Unit used: MJ 106( )J:= σ 0.0 232.4 318.5 345.8 360.5 373.8 := ε 0.0000 0.0006 0.0010 0.0014 0.0018 0.0022 := Solution: Regression curve: x min ε( ) 0.00005, max ε( )..:= Coeff loess ε σ, 0.9,( ):= Y x( ) interp Coeff ε, σ, x,( ):= 0 5 .10 4 0.001 0.0015 0.002 0.0025 40 80 120 160 200 240 280 320 360 400 σ Y x( ) ε x, Modulus of Resilience: The modulus of resilience is equal to the area under the curve. A1 1 2 232.4( )⋅ 0.0004 0.0010+( )⋅:= A2 318.5 0.0022 0.0010−( )⋅:= A3 1 2 373.8 318.5−( )⋅ 0.0022 0.0010−( )⋅:= A4 1 2 318.5 232.4−( )⋅ 0.0010 0.0006−( )⋅:= Atotal A1 A2+ A3+ A4+:= ut Atotal 10 6⋅ J m3 ⋅:= * ut 0.595 MJ m3 = * Ans σ=P/A (MPA) ε=δ/L (mm/mm) Problem 3-4 A tension test was performed on a steel specimen having an original diameter of 13 mm and gauge length of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determine approximately the modulus of elasticity, the yield stress, the ultimate stress, and the rupture stress. Use a scale of 10 mm = 209 MPa and 10 mm = 0.05 mm/mm. Redraw the elastic region, using the same stress scale but a strain scale of 10 mm = 0.001 mm/mm. Given: d 12.5:= * L 50:= * P 0.0 7.5 23.0 40.0 55.0 59.0 59.0 60.0 83.0 100.0 107.5 97.5 92.5 := δ 0.0000 0.0125 0.0375 0.0625 0.0875 0.1250 0.2000 0.5000 1.0000 2.5000 7.0000 10.0000 11.5000 := Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= * σ 103 P⋅ A := *ε δ L := * σ=P/A (MPa) ε=δ/L (mm/mm.) σ 0.00 61.12 187.42 325.95 448.18 480.78 480.78 488.92 676.34 814.87 875.99 794.50 753.76 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * ε 0.00000 0.00025 0.00075 0.00125 0.00175 0.00250 0.00400 0.01000 0.02000 0.05000 0.14000 0.20000 0.23000 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * Curve Fit: Use: F x( ) x x0.3 x0.6 ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ := Fit linfit ε σ, F,( ):= Fit 4189.696− 2285.571 596.538 ⎛⎜⎜⎝ ⎞ ⎠ = x min ε( ) 0.00005, max ε( )..:= Y x( ) F x( ) Fit⋅:= 0 0.001 0.002 0.003 0.004 0.005 200 400 600 800 1000 σ ε 0 0.05 0.1 0.15 0.2 0.25 200 400 600 800 1000 σ Y x( ) ε x, Modulus of Elasticity: From the stress-strain diagram, Load (kN) Elongation (mm) From the stress-strain diagram, σY 448MPa= Ans∆ε 0.00125 0−( ) mm mm := * σult 890MPa= Ans∆σ 326 0−( )MPa:= * σR 753.8MPa= Ans Eapprox ∆σ ∆ε:= * Eapprox 260.8 GPa= * Ans Problem 3-5 The stress-strain diagram for a steel alloy having an original diameter of 12mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of elasticity for the material, the load on the specimen that causes yielding, and the ultimate load the specimen will support. Given: d 12mm:= L 50mm:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Modulus of Elasticity: From the stress-strain diagram, ∆ε 0.001 0−( ) mm mm := ∆σ 290 0−( )MPa:= E ∆σ ∆ε:= * E 290 GPa= * Ans From the stress-strain diagram, σY 290MPa:= Yield Load: PY σY( ) A⋅:= PY 32.80 kN= Ans From th stress-strain diagram, σu 550MPa:= Ultimate Load: Pu σu( ) A⋅:= Pu 62.20 kN= Ans Problem 3-6 The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. If the specimen is loaded until it is stressed to 500MPa, determine the approximate amount of elastic recovery and the increase in the gauge length after it is unloaded. Given: d 12mm:= L 50mm:= σmax 500MPa:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Modulus of Elasticity: From the stress-strain diagram, ∆ε 0.001 0−( ) mm mm := ∆σ 290 0−( )MPa:= E ∆σ ∆ε:= * E 290 GPa= * Ans Elastic Recovery: rRe σmax E := rRe 0.00172 mm mm = AmountRe rRe L( )⋅:= AmountRe 0.08621 mm= Ans From the stress-strain diagram, εmax 0.08 mm mm := Permanent set: (Permanent elongation) rps εmax rRe−:= rps 0.07828 mm mm = Amountps rps L( )⋅:= Amountps 3.91379 mm= Ans Problem 3-7 The stress-strain diagram for a steel alloy having an original diameter of 12 mm and a gauge length of 50 mm is given in the figure. Determine approximately the modulus of resilience and the modulus of toughness for the material. Unit used: MJ 106( )J:= Given: d 12mm:= L 50mm:= Solution: Modulus of Resilience: The modulus of resilience is equal to the area under the initial linear portion of the curve. ∆ε 0.001 0−( ) mm mm := ∆σ 290 0−( )MPa:= ur 1 2 ∆ε⋅ ∆σ⋅:= ur 0.145 MPa= Ans Modulus of Toughness: The modulus of toughness is equal to the area under the curve, and could be approximated by counting the number of sqaures. the total number of squares is: n 33:= ∆εsq 0.04( ) mm mm := ∆σsq 100MPa:= ut n ∆εsq( )⋅ ∆σsq( )⋅:= ut 132 MPa= Ans Problem 3-8 The stress-strain diagram for a steel bar is shown in the figure. Determine approximately the modulus of elasticity, the proportional limit, the ultimate stress, and the modulus of resilience. If the bar is loaded until it is stressed to 450 MPa, determine the amount of elastic strain recovery and the permanent set or strain in the bar when it is unloaded. Unit used: kJ 103J:= Given: σmax 450MPa:= Solution: Modulus of Elasticity: From th stress-strain diagram, ∆ε 0.0015 0−( ) mm mm := ∆σ 325 0−( ) MPa⋅:= E ∆σ ∆ε:= E 216.67 GPa= Ans Modulus of Resilience: The modulus of resilience is equal to the area under the initial linear portion of the curve. ∆ε 0.0015 0−( ) mm mm := ∆σ 325 0−( ) MPa⋅:= ur 1 2 ∆ε⋅ ∆σ⋅:= ur 243.75 kJ m3 = Ans Elastic Recovery: rRe σmax E := rRe 0.00208 mm mm = Ans From th stress-strain diagram, εmax 0.0750 mm mm := Permanent set: (Permanent elongation) rps εmax rRe−:= rps 0.07292 mm mm = Ans Problem 3-9 The σ-ε diagram for elastic fibers that make up human skin and muscle is shown. Determine the modulus of elasticity of the fibers and estimate their modulus of toughness and modulus of resilience. Modulus of Elasticity: From the stress-strain diagram, ∆ε 2.00 0−( ) mm mm := ∆σ 77 0−( )MPa:= E ∆σ ∆ε:= E 38.5 MPa= Ans Modulus of Resilience: The modulus of resilience is equal to the area under the initial linear portion of the curve. ∆ε 2.00 0−( ) mm mm := ∆σ 77 0−( )MPa:= ur 1 2 ∆ε⋅ ∆σ⋅:= ur 77.00 MPa= Ans Modulus of Toughness: The modulus of toughness is equal to the area under the curve. ε0 0:= ε1 2.00:= ε2 2.25:= σ0 0:= σ1 77MPa:= σ2 385MPa:= A1 ur:= A2 0.5 σ1 σ2+( ) ε2 ε1−( )⋅:= ut A1 A2+:= ut 134.75 MPa= Ans Problem 3-10 An A-36 steel bar has a length of 1250 mm and cross-sectional area of 430 mm2. Determine the length of the bar if it is subjected to an axial tension of 25 kN. The material has linear-elastic behavior. Given: A 430mm2:= L0 1250mm:= P 25kN:= σY 250MPa:= Est 200GPa:= Solution: Normal Stress: σ P A := σ 58.140 MPa= [less than yield stress σy] Hence Hook's law is still valid. Normal Strain: ε σ Est := ε 290.6977 10 6−× mm mm = Thus, δL ε L0⋅:= δL 0.36337 mm= L L0 δL+:= L 1250.363 mm= Ans Problem 3-11 The stress-strain diagram for polyethylene, which is used to sheath coaxial cables, is determined from testing a specimen that has a gauge length of 250 mm. If a load P on the specimen develops a strain of ε = 0.024 mm/mm, determine the approximate length of the specimen, measured between the gauge points, when the load is removed. Assume the specimen recovers elastically. Given: L0 250mm:= Solution: Modulus of Elasticity: From th stress-strain diagram, ∆ε 0.004 0−( ) mm mm := ∆σ 14.0 0−( ) MPa⋅:= E ∆σ ∆ε:= E 3500.00 MPa= Elastic Recovery: From the stress-strain diagram, εmax 0.024 mm mm := σmax 26MPa:= rRe σmax E := rRe 0.00743 mm mm = Permanent set: rps εmax rRe−:= rps 0.01657 mm mm = Permanent elongation: ∆L rps L0⋅:= ∆L 4.14286 mm= L L0 ∆L+:= L 254.143 mm= Ans Problem 3-12 Fiberglass has a stress-strain diagram as shown. If a 50-mm-diameter bar of length 2 m made from this material is subjected to an axial tensile load of 60 kN, determine its elongation. Given: L0 2m:= d 50mm:= P 60kN:= unit 1Pa:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 30.558 106× Pa= σ' σ unit :=Given: σ' 300 106( ) ε0.5⋅= ε σ' 2 9 1016( ) ⎡⎢⎢⎣ ⎤⎥⎥⎦ := ε 0.010375 mm mm = ∆L ε L0( )⋅:= ∆L 20.75 mm= Ans Problem 3-13 The change in weight of an airplane is determined from reading the strain gauge A mounted in the plane's aluminum wheel strut. Before the plane is loaded, the strain gauge reading in a strut is ε1 = 0.00100 mm/mm, whereas after loading ε2 = 0.00243 mm/mm. Determine the change in the force on the strut if the cross-sectional area of the strut is 2200 mm2. Eal = 70 GPa. Given: A 2200mm2:= Eal 70 GPa⋅:= ε1 0.00100 mm mm := ε2 0.00243 mm mm := Solution: Stress-strain Relationship: Applying Hook's law σ = Eε . σ1 Eal ε1⋅:= σ1 70.00 MPa= σ2 Eal ε2⋅:= σ2 170.10 MPa= Normal Force: Applying equation σ = P / A . P1 A σ1⋅:= P1 154.00 kN= P2 A σ2⋅:= P2 374.22 kN= Thus, ∆P P2 P1−:= ∆P 220.22 kN= Ans Problem 3-14 A specimen is originally 300 mm long, has a diameter of 12 mm, and is subjected to a force of 2.5 kN. When the force is increased to 9 kN, the specimen elongates 22.5 mm. Determine the modulus of elasticity for the material if it remains elastic. Given: d 12mm:= L0 300mm:= P1 2.5kN:= P2 9kN:= ∆L 22.5mm:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Normal Force: Applying equation σ = P / A . σ1 P1 A := σ1 22.105 MPa= σ2 P2 A := σ2 79.577 MPa= Thus, ∆σ σ2 σ1−:= ∆σ 57.473 MPa= ∆ε ∆L L0 := E ∆σ ∆ε:= E 766.3 MPa= Ans Problem 3-15 A structural member in a nuclear reactor is made from a zirconium alloy. If an axial load of 20 kN is to be supported by the member, determine its required cross-sectional area. Use a factor of safety of 3 with respect to yielding. What is the load on the member if it is 1-m long and its elongation is 0.5 mm? Ezr = 100 GPa, σY = 400 MPa. The material has elastic behavior. Given: P 20kN:= L0 1m:= ∆L 0.5mm:= Ezr 100 GPa⋅:= σY 400 MPa⋅:= FoS 3:= Solution: σallow σY FoS := σallow 133.33 MPa= Areq P σallow := Areq 150 mm2= Ans A Areq:= ε ∆L L0 := ε 0.0005 mm mm = σ Ezr ε⋅:= σ 50 MPa= P σ A⋅:= P 7.5 kN= Ans Problem 3-16 The pole is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm, determine how much it stretches when a horizontal force of 15 kN acts on the pole. Given: P 15kN:= Est 200 GPa⋅:= a 1.2m:= b 1m:= d 5mm:= θ 30deg:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= L a b+:= Support Reactions: ΣΜB=0; Cx L( )⋅ P b( )⋅+ 0= Cx P− b L ⋅:= Cx 6.82− kN= + ΣFx=0; Cx P+ Bx+ 0= Bx Cx− P−:= FAB Bx− sin θ( ):= FAB 16.36 kN=Bx 8.18− kN= LAB L cos θ( ):= LAB 2.54 m= σAB FAB A := σAB 833.393 MPa= εAB σAB Est := εAB 0.0041670 mm mm = δAB εAB LAB⋅:= δAB 10.586 mm= Ans Problem 3-17 By adding plasticizers to polyvinyl chloride, it is possible to reduce its stiffness. The stress-strain diagrams for three types of this material showing this effect are given below. Specify the type that should be used in the manufacture of a rod having a length of 125 mm and a diameter of 50 mm, that is required to support at least an axial load of 100 kN and also be able to stretch at most 6 mm. Given: d 50mm:= L0 125mm:= P 100kN:= δL 6mm:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Normal Stress: σ P A := σ 50.930 MPa= Normal Strain: ε δL L0 := ε 0.048000 mm mm = From the stress-strain diagram,the copolymer will satisfy both stress and strain requirements. Ans Problem 3-18 The steel wires AB and AC support the 200-kg mass. If the allowable axial stress for the wires is σallow = 130 MPa, determine the required diameter of each wire. Also, what is the new length of wire AB after the load is applied? Take the unstretched length of AB to be 750 mm. Est = 200 GPa. Given: g 9.81 m s2 := m 200kg:= θ 60deg:= v 4 5 := h 3 5 := L0 750mm:= Est 200GPa:= σallow 130MPa:= Solution: W m g⋅:= Axial force in steel wires AB and AC: Initial guess: FAC 1N:= FAB 2N:= Given + ΣFx=0; FAB− cos θ( )⋅ FAC h( )⋅+ 0= [1] + ΣFy=0; FAB sin θ( )⋅ FAC v( )⋅+ W− 0= [2] Solving [1] and [2]: FAC FAB ⎛⎜⎜⎝ ⎞ ⎠ Find FAC FAB,( ):= FAC FAB ⎛⎜⎜⎝ ⎞ ⎠ 1066.75 1280.10 ⎛⎜⎝ ⎞ ⎠ N= Wire AB : FAB σallow π 4 ⎛⎜⎝ ⎞ ⎠⋅ dAB 2⋅= dAB 4 π FAB σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dAB 3.54 mm= Ans Wire AC: FAC σallow π 4 ⎛⎜⎝ ⎞ ⎠⋅ dAC 2⋅= dAC 4 π FAC σallow ⎛⎜⎝ ⎞ ⎠ ⋅:= dAC 3.23 mm= Ans Stress-strain Relationship: Applying Hook's law σ = Eε . εAB σallow Est := εAB 0.000650 mm mm = Thus, LAB L0 1 εAB+( )⋅:= LAB 750.487 mm= Ans Problem 3-19 The two bars are made of polystyrene, which has the stress-strain diagram shown. If the cross-sectional area of bar AB is 950 mm2 and BC is 2500 mm2, determine the largest force P that can be supported before any member ruptures. Assume that buckling does not occur. Ans Given: a 1.2m:= b 0.9m:= AAB 950mm 2:= ABC 2500mm2:= Solution: c a2 b2+:= h a c := v b c := + ΣFy=0; FAB v( )⋅ P− 0= [1] + ΣFx=0; FAB h( )⋅ FBC− 0= [2] Solving Eqs.[1] and [2]: FAB 5 3 P= [1a] FBC 4 3 P= [2a] Assume tension failure of BC: From the stress-strain diagram, σR_t 35MPa:= FBC ABC( ) σR_t( )⋅:= FBC 87.50 kN= From Eq.[2a], P 0.75 FBC⋅:= P 65.63 kN= Pcase_1 P:= Assume compression failure of AB: From the stress-strain diagram, σR_c 175MPa:= FAB AAB( ) σR_c( )⋅:= FAB 166.25 kN= From Eq.[1a], P 0.60 FAB⋅:= P 99.75 kN= Pcase_2 P:= Chosoe the smallest value: P min Pcase_1 Pcase_2,( ):= P 65.63 kN= Problem 3-20 The two bars are made of polystyrene, which has the stress-strain diagram shown. Determine the cross-sectional area of each bar so that the bars rupture simultaneously when the load P = 15 kN. Assume that buckling does not occur. Ans Given: a 1.2m:= b 0.9m:= P 15kN:= Solution: c a2 b2+:= h a c := v b c := + ΣFy=0; FAB v( )⋅ P− 0= [1] + ΣFx=0; FAB h( )⋅ FBC− 0= [2] Solving Eqs.[1] and [2]: FAB 5 3 P:= FAB 25.00 kN= FBC 4 3 P:= FBC 20.00 kN= For member BC: From the stress-strain diagram, σR_t 35MPa:= ABC FBC σR_t ⎛⎜⎝ ⎞ ⎠ := ABC 571.43 mm2= Ans For member AB: From the stress-strain diagram, σR_c 175MPa:= AAB FAB σR_c ⎛⎜⎝ ⎞ ⎠ := AAB 142.86 mm2= Problem 3-21 The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD, both made from this material, and subjected to a load of P = 80 kN, determine the angle of tilt of the beam when the load is applied. The diameter of the strut is 40 mm and the diameter of the post is 80 mm. Given: P 80kN:= LAB 2m:= LCD 0.5m:= LAC 1.5m:= dAB 40mm:= dCD 80mm:= Solution: Ay P 2 := Cy P 2 :=Support Reactions: FAB Ay:= FCD Cy:= AreaAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= AreaCD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= From the stress-strain diagram, E 32.2MPa 0.01 := E 3220.00 MPa= σAB FAB AreaAB := σAB 31.831 MPa= εAB σAB E := εAB 0.0098854 mm mm = δAB εAB LAB⋅:= δAB 19.771 mm= σCD FCD AreaCD := σCD 7.958 MPa= εCD σCD E := εCD 0.0024714 mm mm = δCD εCD LCD⋅:= δCD 1.236 mm= Angle of tilt α: tan α( ) δAB δCD− LAC := α atan δAB δCD− LAC ⎛⎜⎝ ⎞ ⎠ := α 0.708 deg= Ans Problem 3-22 The stress-strain diagram for a polyester resin is given in the figure. If the rigid beam is supported by a strut AB and post CD made from this material, determine the largest load P that can be applied to the beam before it ruptures. The diameter of the strut is 12 mm and the diameter of the post is 40 mm. Given: LAB 2m:= LCD 0.5m:= LAC 1.5m:= dAB 12mm:= dCD 40mm:= Solution: Support Reactions: Ay P 2 = Cy P 2 = FAB Ay= FCD Cy= AreaAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= AreaCD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= For rupture of strut AB: From the stress-strain diagram, σR_t 50.0MPa:= FAB AreaAB σR_t( )⋅= P 2AreaAB σR_t( )⋅:= P 11.31 kN= Ans (Controls!) For rupture of post CD: From the stress-strain diagram, σR_c 95.0MPa:= FCD AreaCD σR_c( )⋅= P 2AreaCD σR_c( )⋅:= P 238.76 kN= Problem 3-23 The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm, determine how much it stretches when a distributed load of w = 1.5 kN/m acts on the pipe. The material remains elastic. Given: L 3m:= w 1.5 kN m := dAB 5mm:= θ 30deg:= Est 200 GPa⋅:= Solution: Support Reactions: ΣΜC=0; FAB sin θ( )⋅ L( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0= FAB w L⋅ 2 sin θ( ):= AreaAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= LAB L cos θ( ):= σAB FAB AreaAB := σAB 229.183 MPa= εAB σAB Est := εAB 0.0011459 mm mm = δAB εAB LAB⋅:= δAB 3.970 mm= Ans Problem 3-24 The beam is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm, determine the distributed load w if the end B is displaced 18 mm downward. Given: L 3m:= δBy 18mm:= dAB 5mm:= θB 30deg:= Est 200 GPa⋅:= Solution: Consider triangle BB'C: L sin θC( )⋅ δBy= θC asin δBy L ⎛⎜⎝ ⎞ ⎠:= Consider triangle AB'C: θ'C 90deg θC+:= LAC L tan θB( )⋅:= L'AB LAC 2 L2+ 2 LAC( )⋅ L⋅ cos θ'C( )⋅−:= L'AB 3.4731 m= LAB L cos θB( ):=AreaAB π4⎛⎜⎝ ⎞⎠ dAB2⋅:= δAB L'AB LAB−:= δAB 8.9883 mm= εAB δAB LAB := εAB 0.0025947 mm mm = σAB εAB Est⋅:= σAB 518.942 MPa= FAB σAB AreaAB( )⋅:= FAB 10.19 kN= Support Reactions: ΣΜC=0; FAB sin θB( )⋅ L( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0= w 2 sin θB( ) L ⎛⎜⎝ ⎞ ⎠ FAB⋅:= w 3.40 kN m = Ans Problem 3-25 Direct tension indicators are sometimes used instead of torque wrenches to insure that a bolt has a prescribed tension when used for connections. If a nut on the bolt is tightened so that the six heads of the indicator that were originally 3 mm high are crushed 0.3 mm, leaving a contact area on each head of 1.5 mm2, determine the tension in the bolt shank. The material has the stressstrain diagram shown. Given: h 3mm:= δh 0.3mm:= Area 1.5mm2:= number 6:= unit 1MPa:= Solution: Stress-strain Relationship: ε δh h := ε 0.1000 mm mm = From the stress-strain diagram, σ 450− 600 450− ε 0.0015− 0.3 0.0015−= σ 450 600 450−( ) ε 0.0015− 0.3 0.0015− ⎛⎜⎝ ⎞ ⎠+ ⎡⎢⎣ ⎤⎥⎦ unit⋅:= σ 499.50 MPa= Axial Force: For each head P σ Area( )⋅:= P 0.749 kN= Thus, the tension in the bolt is T number( ) P⋅:= T 4.50 kN= Ans Problem 3-26 The acrylic plastic rod is 200 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Ep = 2.70 GPa, νp = 0.4. Given: P 300N:= L 200mm:= d 15mm:= EP 2.70 GPa⋅:= ν 0.4:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 1.698 MPa= εlong σ EP := εlong 0.0006288 mm mm = δ εlong L⋅:= δ 0.126 mm= Ans εlat ν− εlong⋅:= εlat 0.0002515− mm mm = ∆d εlat d( )⋅:= ∆d 0.003773− mm= Ans Problem 3-27 The block is made of titanium Ti-6A1-4V and is subjected to a compression of 1.5 mm along the y axis, and its shape is given a tilt of θ = 89.7°.Determine εy, εx, and γxy. Given: Lx 125mm:= Ly 100mm:= δy 1.5− mm:= θ 89.7deg:= ν 0.36:= Solution: Normal Strain: εy δy Ly := εy 0.01500− mm mm = Ans Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio. εx ν− εy( )⋅:= εx 0.00540 mmmm= Ans Shear Strain: β 180deg θ−:= β 90.30 deg= β 1.58 rad= Thus, γxy π 2 β−:= γxy 0.00524− rad= Ans Problem 3-28 A short cylindrical block of bronze C86100, having an original diameter of 38 mm and a length of 75 mm, is placed in a compression machine and squeezed until its length becomes 74.5 mm. Deterinme the new diameter of the block. Given: L 75mm:= L' 74.5mm:= d 38mm:= ν 0.34:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= εlong L' L− L := εlong 0.0066667− mm mm = εlat ν− εlong⋅:= εlat 0.0022667 mm mm = ∆d εlat d( )⋅:= ∆d 0.086133 mm= d' d ∆d+:= d' 38.0861 mm= Ans Problem 3-29 The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. When the applied load on the specimen is 50 kN, the diameter is 12.99265 mm. Determine Poisson's ratio for the material. Given: d 13mm:= d' 12.99265mm:= L 50mm:= P 50kN:= Solution: Normal Stress: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 376.698 MPa= Normal Strain: From the stress-strain diagram, the modulus of elasticity is E 400MPa 0.002 := E 200 GPa= Applying Hook's law σ = Eε . εlong σ E := εlong 0.0018835 mm mm = εlat d' d− d := εlat 0.0005654− mm mm = Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio. ν εlat− εlong := ν 0.30018= Ans Problem 3-30 The elastic portion of the stress-strain diagram for a steel alloy is shown in the figure. The specimen from which it was obtained had an original diameter of 13 mm and a gauge length of 50 mm. If a load of P = 20 kN is applied to the specimen, determine its diameter and gauge length. Take ν = 0.4. Given: L 50mm:= d 13mm:= P 20kN:= ν 0.4:= Solution: Normal Stress: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 150.679 MPa= Normal Strain: From the stress-strain diagram, the modulus of elasticity is E 400MPa 0.002 := E 200 GPa= Applying Hook's law σ = Eε . εlong σ E := εlong 0.0007534 mm mm = Thus, δL εlong L( )⋅:= δL 0.037670 mm= L' L δL+:= L' 50.0377 mm= Ans Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio. εlat ν− εlong( )⋅:= εlat 0.00030136− mmmm= ∆d εlat d( )⋅:= ∆d 0.003918− mm= d' d ∆d+:= d' 12.99608 mm= Ans Problem 3-31 The shear stress-strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 6 mm is made of this material and used in the lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take ν = 0.3. Given: d 6mm:= ν 0.3:= Solution: Modulus of Rigidity: From the stress-strain diagram, G 350MPa 0.004 := G 87500 MPa= Modulus of Elasticity: G E 2 1 ν+( )⋅= E 2G 1 ν+( )⋅:= E 227500 MPa= Ans Yielding Stress: From the stress-strain diagram, τY 350MPa:= A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= τY P A = P τY( ) A⋅:= P 9.896 kN= Ans Problem 3-32 The brake pads for a bicycle tire are made of rubber. If a frictional force of 50 N is applied to each side of the tires, determine the average shear strain in the rubber. Each pad has cross-sectional dimensions of 20 mm and 50 mm. Gr = 0.20 MPa. Given: a 20mm:= b 50mm:= V 50N:= G 0.20MPa:= Solution: A a b⋅:= A 1000.00 mm2= Average Shear Stress: The shear force is V. τ V A := τ 0.050 MPa= Shear Stress-strain Relationship: Applying Hooke's law for shear: τ G γ⋅= γ τ G := γ 0.250 rad= Ans Problem 3-33 The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to the top of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug be compressed downward in order to do this? The plug is made from a material for which E = 5 MPa, ν = 0.45. Given: d 30mm:= d' 32mm:= L 50mm:= E 5MPa:= ν 0.45:= Solution: εlat d' d− d := εlat 0.0666667 mm mm = ν εlat− εlong = εlong εlat− ν:= εlong 0.14815− mm mm = Applying Hook's law σ = Eε . p E εlong( )⋅:= p 0.741− MPa= Ans δL εlong( ) L⋅:= δL 7.41− mm= Ans Problem 3-34 The rubber block is subjected to an elongation of 0.75 mm. along the x axis, and its vertical faces are given a tilt so that θ = 89.3°. Deterinme the strains εx, εy and γxy. Take νr = 0.5. Given: Lx 100mm:= Ly 75mm:= δx 0.75mm:= θ 89.3deg:= ν 0.5:= Solution: Normal Strain: εx δx Lx := εx 0.00750 mm mm = Ans Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio. εy ν− εx⋅:= εy 0.00375− mm mm = Ans Shear Strain: θ 89.30 deg= θ 1.56 rad= Thus, γxy π 2 θ−:= γxy 0.01222 rad= Ans Problem 3-35 The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm. and a diameter of 12.5 mm. When the applied load is 45 kN, the new diameter of the specimen is 12.48375 mm. Compute shear modulus Gal for the aluinmum. Given: d 12.5mm:= d' 12.48375mm:= L 50mm:= P 45kN:= Solution: Normal Stress: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 366.693 MPa= Normal Strain: From the stress-strain diagram, the modulus of elasticity is E 500MPa 0.00614 := E 81433.22 MPa= Applying Hook's law σ = Eε . εlong σ E := εlong 0.0045030 mm mm = εlat d' d− d := εlat 0.0013000− mm mm = Poisson's Ratio: The lateral and longitudinal strain can be related using Poisson's ratio. ν εlat− εlong := ν 0.28870= G E 2 1 ν+( )⋅:= G 31.60 GPa= Ans Problem 3-36 The elastic portion of the tension stress-strain diagram for an aluinmum alloy is shown in the figure. The specimen used for the test has a gauge length of 50 mm and a diameter of 12.5 mm. If the applied load is 50 kN deterinme the new diameter of the specimen. The shear modulus is Gal = 28 GPa. Given: d 12.5mm:= L 50mm:= P 50kN:= G 28 GPa⋅:= Solution: Normal Stress: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 407.44 MPa= Normal Strain: From the stress-strain diagram, the modulus of elasticity is E 500MPa 0.00614 := E 81433.22 MPa= Applying Hook's law σ = Eε . εlong σ E := εlong 0.0050033 mm mm = Poisson's Ratio: G E 2 1 ν+( )⋅= Thus, ν E 2 G⋅ ⎛⎜⎝ ⎞ ⎠ 1−:= ν 0.454= The lateral and longitudinal strain can be related using Poisson's ratio. εlat ν− εlong( )⋅:= εlat 0.00227233− mmmm= ∆d εlat d( )⋅:= ∆d 0.028404− mm= d' d ∆d+:= d' 12.4716 mm= Ans Problem 3-37 The head H is connected to the cylinder of a compressor using six steel bolts. If the clamping force in each bolt is 4 kN, deterinme the normal strain in the bolts. Each bolt has a diameter of 5 mm. If σY = 280 MPa and Est = 200 GPa, what is the strain in each bolt when the nut is unscrewed so that the clamping force is released? Given: d 5mm:= σY 28MPa:= P 4kN:= E 200 GPa⋅:= Solution: Normal Stress: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σ P A := σ 203.7183 MPa= ( < σY = 280 MPa ) Normal Strain: Since σ < σY, Hook's law is still valid. ε σ E := ε 0.0010186 mm mm = Ans If the nut is unscrewed, the load is zero. Therefore, the strain ε = 0. Ans Problem 3-38 The rigid pipe is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm., determine how much it stretches when a load of P = 1.5 kN acts on the pipe. The material remains elastic. Given: P 1.5kN:= Est 200 GPa⋅:= LBC 2.4m:= d 5mm:= θ 60deg:= Solution: Support Reactions: ΣΜC=0; FAB− cos θ( )⋅ LAB( )⋅ P LAB( )⋅+ 0= FAB P cos θ( ):= FAB 3 kN= Normal Stress: Area π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= σAB FAB Area := σAB 152.789 MPa= Normal Strain: LAB LBC sin θ( ):= LAB 2.771 m= Applying Hook's law σ = Eε . εAB σAB Est := εAB 0.0007639 mm mm = Thus, δLAB εAB LAB( )⋅:= δLAB 2.1171 mm= Ans Problem 3-39 The rigid pipe is supported by a pin at C and an A-36 guy wire AB. If the wire has a diameter of 5 mm., determine the load P if the end B is displaced 2.5 mm. to the right. Est = 200 GPa. Given: L 2.4m:= d 5mm:= θ 60deg:= δBx 2.5mm:= Est 200 GPa⋅:= Solution: Consider triangle BB'C: L sin θC( )⋅ δBx= θC asin δBx L ⎛⎜⎝ ⎞ ⎠:= Consider triangle AB'C: θ'C 90deg θC+:= LAC L cot θ( )⋅:= L'AB LAC 2 L2+ 2 LAC( )⋅ L⋅ cos θ'C( )⋅−:= L'AB 2.7725 m= Normal Strain: LAB L sin θ( ):= εAB L'AB LAB− LAB := εAB 0.0004510 mm mm = Normal Stress: Area π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Applying Hook's law σ = Eε . σAB εAB Est⋅:= σAB 90.191 MPa= Thus, FAB σAB Area( )⋅:= FAB 1.771 kN= Support Reactions: ΣΜC=0; FAB cos θ( )⋅ L( )⋅ P L⋅− 0= P FAB cos θ( )⋅:= P 0.885 kN= Ans Problem 3-40 While undergoing a tension test, a copper-alloy specimen having a gauge length of 50 mm. is subjected to a strain of 0.40 mm./mm. when the stress is 490 MPa. If σY = 315 MPa when εY = 0.0025 mm./mm., determine the distance between the gauge points when the load is released. Given: L0 50mm:= ε1 0.40 mm mm := σ1 490 MPa⋅:= εY 0.0025 mm mm := σY 315MPa:= Solution: Modulus of Elasticity: E σY εY := E 126.00 GPa= Elastic Recovery: rRe σ1 E := rRe 0.0038889 mm mm = Permanent set: rps ε1 rRe−:= rps 0.39611 mm mm = Permanent elongation: ∆L rps L0⋅:= ∆L 19.806 mm= L L0 ∆L+:= L 69.806 mm= Ans Problem 3-41 The 8-mm-diameter bolt is made of an aluminum alloy. It fits through a magnesium sleeve that has an inner diameter of 12 mm and an outer diameter of 20 mm. If the original lengths of the bolt and sleeve are 80 mm and 50 mm, respectively, determine the strains in the sleeve and the bolt if the nut on the bolt is tightened so that the tension in the bolt is 8 kN. Assume the material at A is rigid. Eal = 70 GPa, Emg = 45 GPa. Given: Lb 80mm:= db 8mm:= Ls 30mm:= ds_o 20mm:= ds_i 12mm:= P 8kN:= Eal 70GPa:= Emg 45GPa:= Solution: Ab π 4 ⎛⎜⎝ ⎞ ⎠ db 2⋅:= As π 4 ⎛⎜⎝ ⎞ ⎠ ds_o 2 ds_i 2−⎛⎝ ⎞⎠⋅:= Normal Stress: σb P Ab := σb 159.15 MPa= σs P As := σs 39.79 MPa= Normal Strain: εb σb Eal := εb 0.002274 mm mm = Ans εs σs Emg := εs 0.000884 mm mm = Ans Problem 3-42 A tension test was performed on a steel specimen having an original diameter of 12.5mm and a gauge length of 50 mm. The data is listed in the table. Plot the stress-strain diagram and determine approximately the modulus of elasticity, the ultimate stress, and the rupture stress. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm/mm. Redraw the linear-elastic region, using the same stress scale but a strain scale of 20 mm = 0.001 mm/mm. Given: d 12.5:= * L 50:= * P 0.0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8 := δ 0.0000 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380 := Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= * σ P 103⋅ A := *ε δ L := * σ=P/A (MPa) ε=δ/L (mm/mm) Use F x( ) x x0.3 x0.6 ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ :=σ 0.0000 90.4509 259.9446 308.0221 333.2832 355.2848 435.1423 507.6661 525.5933 507.6661 479.1455 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * ε 0.000000 0.000350 0.001200 0.002040 0.003300 0.004980 0.020320 0.060960 0.127000 0.177800 0.238760 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * Fit 107.325 2123.113 2190.474− ⎛⎜⎜⎝ ⎞ ⎠ = x min ε( ) 0.005, max ε( )..:= Y x( ) F x( ) Fit⋅:= 0 0.001 0.002 0.003 0.004 0.005 50 100 150 200 250 300 350 400 450 500 550 σ ε 0 0.05 0.1 0.15 0.2 0.25 50 100 150 200 250 300 350 400 450 500 550 σ Y x( ) ε x, Load (kN) Elongation (mm) Fit linfit ε σ, F,( ):= Modulus of Elasticity: From th stress-strain diagram, From th stress-strain diagram, σult 530MPa= Ans ∆ε 0.0005 0−( ) mm mm := * σR 479MPa= Ans ∆σ 125 0−( )MPa:= * Eapprox ∆σ ∆ε:= * Eapprox 250 GPa= * Ans Problem 3-43 A tension test was performed on a steel specimen having an original diameter of 12.5 mm and a gauge length of 50 mm. Using the data listed in the table, plot the stress-strain diagram and determine approximately the modulus of toughness. Use a scale of 20 mm = 50 MPa and 20 mm = 0.05 mm/mm. Given: d 12.5:= * L 50:= * P 0.0 11.1 31.9 37.8 40.9 43.6 53.4 62.3 64.5 62.3 58.8 := δ 0.0000 0.0175 0.0600 0.1020 0.1650 0.2490 1.0160 3.0480 6.3500 8.8900 11.9380 := Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= * σ P 103⋅ A := *ε δ L := * σ=P/A (MPa) ε=δ/L (mm/mm) Use F x( ) x x0.3 x0.6 ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ :=σ 0.0000 90.4509 259.9446 308.0221 333.2832 355.2848 435.1423 507.6661 525.5933 507.6661 479.1455 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * ε 0.000000 0.000350 0.001200 0.002040 0.003300 0.004980 0.020320 0.060960 0.127000 0.177800 0.238760 ⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝ ⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟ ⎠ = * Fit linfit ε σ, F,( ):= Fit 107.325 2123.113 2190.474− ⎛⎜⎜⎝ ⎞ ⎠ = x min ε( ) 0.005, max ε( )..:= Y x( ) F x( ) Fit⋅:= 0 0.05 0.1 0.15 0.2 0.25 50 100 150 200 250 300 350 400 450 500 550 σ Y x( ) ε x, Modulus of Toughness: The modulus of toughness is equal to the area under the curve, and could be approximated by counting the number of sqaures. the total number of squares is: n 188.5:= ∆εsq 0.025( ) mm mm := ∆σsq 25MPa:= ut n ∆εsq( )⋅ ∆σsq( )⋅:= ut 117.81 MPa= Ans Elongation (mm) Load (kN) Problem 3-44 An 8-mm-diameter brass rod has a modulus of elasticity of Ebr = 100 GPa. If it is 3 m long and subjected to an axial load of 2 kN, determine its elongation. What is its elongation under the same load if its diameter is 6 mm? Given: P 2kN:= L 3m:= d1 8mm:= d2 6mm:= Ebr 100 GPa⋅:= Solution: Case 1: A1 π 4 ⎛⎜⎝ ⎞ ⎠ d1 2⋅:= σ1 P A1 := σ1 39.789 MPa= ε1 σ1 Ebr := ε1 0.0003979 mm mm = δ1 ε1 L⋅:= δ1 1.194 mm= Ans Case 2: A2 π 4 ⎛⎜⎝ ⎞ ⎠ d2 2⋅:= σ2 P A2 := σ2 70.736 MPa= ε2 σ2 Ebr := ε2 0.0007074 mm mm = δ2 ε2 L⋅:= δ2 2.122 mm= Ans Problem 4-1 The ship is pushed through the water using an A-36 steel propeller shaft that is 8 m long, measured from the propeller to the thrust bearing D at the engine. If it has an outer diameter of 400 mm and a wall thickness of 50 mm, determine the amount of axial contraction of the shaft when the propeller exerts a force on the shaft of 5 kN. The bearings at B and C are journal bearings. Given: F 5kN:= L 8m:= do 400mm:= t 50mm:= E 200 GPa⋅:= Solution: di do 2t−:= A π 4 ⎛⎜⎝ ⎞ ⎠ do 2 di 2−⎛⎝ ⎞⎠⋅:= Internal Force: As shown on FBD. P F−:= Displacement: δA P L⋅ A E⋅:= δA 0.00364− mm= Ans Note: Negative sign indicates that end A moves towards end D. Problem 4-2 The A-36 steel column is used to support the symmetric loads from the two floors of a building. Determine the vertical displacement of its top, A, if P1 = 200 kN, P2 = 310 kN, and the column has a cross-sectional area of 14625 mm2. Given: L 3.6m:= A 14625mm2:= P1 200kN:= P2 310kN:= E 200 GPa⋅:= Solution: Internal Force: As shown on FBD. Displacement: δAB 2− P1 L⋅ A E⋅:= δBC 2− P1 P2+( ) L⋅ A E⋅:= δA δAB δBC+:= δA 1.74769− mm= Ans Note: Negative sign indicates that end A moves towards end C. Problem 4-3 The A-36 steel column is used to support the symmetric loads from the two floors of a building. Determine the loads P1 and P2 if A moves downward 3 mm and B moves downward 2.25 mm when the loads are applied. The column has a cross-sectional area of 14625 mm2. Given: L 3.6m:= A 14625mm2:= δA 3− mm:= δB 2.25− mm:= E 200 GPa⋅:= Solution: Internal Force: As shown on FBD. Displacement: Initial guess: P1 1kN:= P2 2kN:= Given For AB: δA δB− 2− P1 L⋅ A E⋅= [1] For BC: δB 2− P1 P2+( ) L⋅ A E⋅= [2] Solving [1] and [2]: P1 P2 ⎛⎜⎜⎝ ⎞ ⎠ Find P1 P2,( ):= P1 P2 ⎛⎜⎜⎝ ⎞ ⎠ 304.69 609.38 ⎛⎜⎝ ⎞ ⎠ kN= Ans Problem 4-4 The copper shaft is subjected to the axial loads shown. Determine the displacement of end A with respect to end D if the diameters of each segment are dAB = 20 mm, dBC = 25 mm, and dCD = 12 mm. Take Ecu = 126 GPa. Given: LAB 2m:= dAB 20mm:= LBC 3.75m:= dBC 25mm:= LCD 2.5m:= dCD 12mm:= E 126 GPa⋅:= PA 40− kN:= PB 25kN:= PC 10kN:= PD 30− kN:= Solution: Internal Force: As shown on FBD. Displacement: AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= δAB PA LAB( )⋅ E AAB( )⋅:= ABC π 4 ⎛⎜⎝ ⎞ ⎠ dBC 2⋅:= δBC PA 2PB+( ) LBC( )⋅ E ABC( )⋅:= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= δCD PA 2PB+ 2PC+( ) LCD( )⋅ E ACD( )⋅:= δA_D δAB δBC+ δCD+:= δA_D 3.8483 mm= Ans Note: The positive sign indicates that end A moves away from end D. Problem 4-5 The A-36 steel rod is subjected to the loading shown. If the cross-sectional area of the rod is 60 mm2, Determine the displacement of B and A. Neglect the size of the couplings at B, C, and D. Given: LAB 0.50m:= LBC 1.50m:= LCD 0.75m:= PA 8kN:= PB 2kN:= PC 3.3kN:= θC 60deg:=hB 4 5 := vB 3 5 := A 60mm2:= E 200 GPa⋅:= Solution: Internal Force: As shown on FBD. Displacement: δAB PA LAB( )⋅ E A⋅:= δBC PA 2PB vB( )⋅+⎡⎣ ⎤⎦ LBC( )⋅ E A⋅:= δCD PA 2PB vB( )⋅+ 2PC sin θC( )⋅+⎡⎣ ⎤⎦ LCD( )⋅ E A⋅:= δB δBC δCD+:= δB 2.307 mm= Ans δA δAB δBC+ δCD+:= δA 2.641 mm= Ans Problem 4-6 The assembly consists of an A-36 steel rod CB and a 6061-T6 aluminum rod BA, each having a diameter of 25 mm. Determine the applied loads P1 and P2 if A is displaced 2 mm to the right and B is displaced 0.5 mm to the left when the loads are applied. The unstretched length of each segment is shown in the figure. Neglect the size of the connections at B and C, and assume that they are rigid. Given: LBA 1.2m:= LCB 0.6m:= δA 2mm:= δB 0.5− mm:= d 25mm:= ECB 200 GPa⋅:= EBA 68.9 GPa⋅:= Solution: Internal Force: As shown on FBD. Displacement: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Initial guess: P1 1kN:= P2 2kN:= Given For BA: δA δB− P1 LBA⋅ A EBA⋅ = [1] For CB: δB P1 P2−( ) LCB⋅ A ECB⋅ = [2] Solving [1] and [2]: P1 P2 ⎛⎜⎜⎝ ⎞ ⎠ Find P1 P2,( ):= P1P2 ⎛⎜⎜⎝ ⎞ ⎠ 70.46 152.27 ⎛⎜⎝ ⎞ ⎠ kN= Ans Problem 4-7 The 15-mm-diameter A-36 steel shaft AC is supported by a rigid collar, which is fixed to the shaft at B. If it is subjected to an axial load of 80 kN at its end, determine the uniform pressure distribution p on the collar required for equilibrium.Also, what is the elongation on segment BC and segment BA? Given: LAB 200mm:= LBC 500mm:= d 15mm:= PC 80kN:= E 200 GPa⋅:= rc 35mm:= Solution: Acollar π 4 ⎛⎜⎝ ⎞ ⎠ 2rc( )2 d2−⎡⎣ ⎤⎦⋅:= Equations of equilibrium: + ΣFy=0; p Acollar( )⋅ PC− 0= p PC Acollar := p 21.79 MPa= Ans Internal Force: As shown on FBD. Displacement: PBC PC:= PBA 0:=A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= δBC PBC LBC( )⋅ E A⋅:= δBC 1.132 mm= Ans δBA PBA LAB( )⋅ E A⋅:= δBA 0 mm= Ans Problem 4-8 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the vertical displacement of the 2.5-kN load if the members were horizontal when the load was originally applied. Each wire has a cross-sectional area of 16 mm2. Given: LDE 0.9m:= LCF 0.9m:=P 2.5kN:= LDH 0.3m:= LHC 0.6m:=A 16mm2:= LAH 0.54m:= LBG 1.5m:= E 193 GPa⋅:= LAI 0.9m:= LIB 0.3m:= Solution: LDC LDH LHC+:= LAB LAI LIB+:= Internal Forces in the wires : From FBD (b): ΣΜA=0; FBG LAB( )⋅ P LAI( )⋅− 0= + ΣFy=0; FAH FBG+ P− 0= From FBD (a): FBG P LAI LAB ⎛⎜⎝ ⎞ ⎠ ⋅:= FBG 1.875 kN= ΣΜD=0; FCF LDC( )⋅ FAH LDH( )⋅− 0= + FAH P FBG−:= FAH 0.625 kN= ΣFy=0; FCF FDE+ FAH− 0= FCF FAH LDH LDC ⎛⎜⎝ ⎞ ⎠ ⋅:= FCF 0.2083 kN= FDE FAH FCF−:= FDE 0.4167 kN= Displacement: δD FDE LDE( )⋅ E A⋅:= δD 0.12143782 mm= δC FCF LCF( )⋅ E A⋅:= δC 0.06071891 mm= δH δC LHC LDC ⎛⎜⎝ ⎞ ⎠ δD δC−( )⋅+:= δH 0.10119819 mm= δA δH FAH LAH( )⋅ E A⋅ ⎡⎢⎣ ⎤⎥⎦+:= δB FBG LBG( )⋅ E A⋅:= δI δA LAI LAB ⎛⎜⎝ ⎞ ⎠ δB δA−( )⋅+:= δI 0.736 mm= Ans LHC 0.6m:= LDE 0.9m:= LCF 0.9m:=Given: P 2.5kN:= LDH 0.3m:= Solution: A 16mm2:= LAH 0.54m:= LBG 1.5m:= E 193 GPa⋅:= LAI 0.9m:= LIB 0.3m:= Ans LDC LDH LHC+:= LAB LAI LIB+:= Internal Forces in the wires : From FBD (b): ΣΜA=0; FBG LAB( )⋅ P LAI( )⋅− 0= + ΣFy=0; FAH FBG+ P− 0= From FBD (a): FBG P LAI LAB ⎛⎜⎝ ⎞ ⎠ ⋅:= FBG 1.875 kN= ΣΜD=0; FCF LDC( )⋅ FAH LDH( )⋅− 0= + FAH P FBG−:= FAH 0.625 kN= ΣFy=0; FCF FDE+ FAH− 0= FCF FAH LDH LDC ⎛⎜⎝ ⎞ ⎠ ⋅:= FCF 0.2083 kN= FDE FAH FCF−:= FDE 0.4167 kN= Displacement: δD FDE LDE( )⋅ E A⋅:= δD 0.12143782 mm= δC FCF LCF( )⋅ E A⋅:= δC 0.06071891 mm= δH δC LHC LDC ⎛⎜⎝ ⎞ ⎠ δD δC−( )⋅+:= δH 0.10119819 mm= tan αDC( ) δD δC−LDC:= αDC atan δD δC− LDC ⎛⎜⎝ ⎞ ⎠ := αDC 0.0039 deg= Ans δA δH FAH LAH( )⋅ E A⋅ ⎡⎢⎣ ⎤⎥⎦+:= δA 0.21049223 mm= δB FBG LBG( )⋅ E A⋅:= δB 0.91078368 mm= tan βAB( ) δB δA−LAB:= βAB atan δB δA− LAB ⎛⎜⎝ ⎞ ⎠ := βAB 0.0334 deg= Problem 4-9 The load is supported by the four 304 stainless steel wires that are connected to the rigid members AB and DC. Determine the angle of tilt of each member after the 2.5-kN load is applied. The members were originally horizontal, and each wire has a cross-sectional area of 16 mm2. Problem 4-10 The bar has a cross-sectional area of 1800 mm2, and E = 250 GPa. Determine the displacement of its end A when it is subjected to the distributed loading. Given: L 1.5m:= E 250 GPa⋅:= A 1800mm2:= w 500 x 1 3⋅ N m = Solution: Internal Force: As shown on FBD. Px 0 x xw x⋅⌠⎮⌡ d= Px 0 x x500 x 1 3 ⎛⎜⎝ ⎞ ⎠⋅ x⋅ ⎡⎢⎣ ⎤⎥⎦ ⌠⎮ ⎮⌡ d= Px 1500 4 x 4 3 ⎛⎜⎝ ⎞ ⎠⋅= Displacement: unit 1N m⋅:= δA 0 L x Px E A⋅ ⌠⎮ ⎮⌡ d= L 1.50 m= δA unit A E⋅ 0 48 x 1500 4 x 4 3⋅ ⎛⎜⎜⎝ ⎞ ⎠ ⌠⎮ ⎮ ⎮⌡ d ⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ := δA 2.990 mm= Ans Problem 4-11 The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the horizontal displacement of point F. Given: LCD 1.2m:= LEC 0.6m:= LEF 0.3m:= LAB 1.8m:= LAE 0.3m:= P 30kN:= E 120 GPa⋅:= ACD 600mm 2:= AAB 900mm2:= AEF 1200mm 2:= Solution: LAC LAE LEC+:= Internal Forces in the rods : From FBD : FCD P LAE LAC ⎛⎜⎝ ⎞ ⎠ ⋅:=ΣΜA=0; FCD LAC( )⋅ P LAE( )⋅− 0= FCD 10.00 kN= + ΣFy=0; FCD FAB+ P− 0= FAB P FCD−:= FAB 20.00 kN= Displacement: δC FCD LCD( )⋅ E ACD⋅ := δC 0.1667 mm= δA FAB LAB( )⋅ E AAB⋅ := δA 0.3333 mm= δE δC LEC LAC ⎛⎜⎝ ⎞ ⎠ δA δC−( )⋅+:= δE 0.2778 mm= δF_E P LEF( )⋅ E AEF⋅ := δF_E 0.0625 mm= δF δE δF_E+:= δF 0.340278 mm= Ans Problem 4-12 The assembly consists of three titanium (Ti-6A1-4V) rods and a rigid bar AC.The cross-sectional area of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the angle of tilt of bar AC. Given: LCD 1.2m:= LEC 0.6m:= LEF 0.3m:= LAB 1.8m:= LAE 0.3m:= P 30kN:= E 120 GPa⋅:= ACD 600mm 2:= AAB 900mm2:= AEF 1200mm 2:= Solution: LAC LAE LEC+:= Internal Forces in the rods : From FBD : FCD P LAE LAC ⎛⎜⎝ ⎞ ⎠ ⋅:=ΣΜA=0; FCD LAC( )⋅ P LAE( )⋅− 0= FCD 10.00 kN= + ΣFy=0; FCD FAB+ P− 0= FAB P FCD−:= FAB 20.00 kN= Displacement: δC FCD LCD( )⋅ E ACD⋅ := δC 0.1667 mm= δA FAB LAB( )⋅ E AAB⋅ := δA 0.3333 mm= tan αAC( ) δA δC−LAC:= αAC atan δA δC− LAC ⎛⎜⎝ ⎞ ⎠ := αAC 0.01061 deg= Ans Problem 4-13 A spring-supported pipe hanger consists of two springs which are originally unstretched and have a stiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe and the fluid it carries have a total weight of 4 kN, determine the displacement of the pipe when it is attached to the support. Given: LAB 0.75m:= LCD 0.75m:= LEF 0.75m:= dAB 5mm:= dCD 5mm:= dEF 12mm:= LGE 0.25m:= LEH 0.25m:= k 60 kN m := P 4kN:= E 193 GPa⋅:= Solution: LGH LGE LEH+:= Internal Forces in the rods : From FBD (a) : ΣΜA=0; FCD LGH( )⋅ P LGE( )⋅− 0= + ΣFy=0; FCD FAB+ P− 0= FCD P LGE LGH ⎛⎜⎝ ⎞ ⎠ ⋅:= FCD 2.00 kN= FAB P FCD−:= FAB 2.00 kN= From FBD (b) : + ΣFy=0; FEF FCD FAB+( )−⋅ 0= FEF FCD FAB+:= FEF 4.00 kN= Displacement: AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= AEF π 4 ⎛⎜⎝ ⎞ ⎠ dEF 2⋅:= δE FEF LEF( )⋅ E AEF( )⋅:= δE 0.13744 mm= δD δE FCD k ⎛⎜⎝ ⎞ ⎠+:= δD 33.47077 mm= Due to symmetry, δB δD:= δC_D FCD LCD( )⋅ E ACD( )⋅:= δC_D 0.39583 mm= Due to symmetry, δB_A δC_D:= δtotal δD δC_D+:= δtotal 33.8666 mm= Ans Problem 4-14 A spring-supported pipe hanger consists of two springs, which are originally unstretched and have a stiffness of k = 60 kN/m, three 304 stainless steel rods, AB and CD, which have a diameter of 5 mm, and EF, which has a diameter of 12 mm, and a rigid beam GH. If the pipe is displaced 82 mm when it is filled with fluid, determine the weight of the fluid. Given: LAB 0.75m:= LCD 0.75m:= LEF 0.75m:= dAB 5mm:= dCD 5mm:= dEF 12mm:= LGE 0.25m:= LEH 0.25m:= k 60 kN m := δtotal 82mm:= E 193 GPa⋅:= Solution: LGH LGE LEH+:= Internal Forces in the rods : Iniatlly set: P 1kN:= From FBD (a) : ΣΜA=0; FCD LGH( )⋅ P LGE( )⋅− 0= + ΣFy=0; FCD FAB+ P− 0= FCD P LGE LGH ⎛⎜⎝ ⎞ ⎠ ⋅:= FCD 0.50 kN= FAB P FCD−:= FAB 0.50 kN= From FBD (b) : + ΣFy=0; FEF FCD FAB+( )−⋅ 0= FEF FCD FAB+:= FEF 1.00 kN= Displacement: AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= AEF π 4 ⎛⎜⎝ ⎞ ⎠ dEF 2⋅:= δE FEF LEF( )⋅ E AEF( )⋅:= δE 0.03436 mm= δD δE FCD k ⎛⎜⎝ ⎞ ⎠+:= δD 8.36769 mm= Due to symmetry, δB δD:= δC_D FCD LCD( )⋅ E ACD( )⋅:= δC_D 0.09896 mm= Due to symmetry, δB_A δC_D:= δ'total δD δC_D+:= δ'total 8.4666 mm= W P δtotal δ'total = W P δtotal δ'total ⎛⎜⎝ ⎞ ⎠ ⋅:= W 9.69 kN= Ans Problem 4-15 The assembly consists of three titanium rods and a rigid bar AC.The cross-sectional area of each rod is given in the figure. If a vertical force P = 20 kN is applied to the ring F, determine the vertical displacement of point F. Eti = 350 GPa. Given: LBA 2m:= LDC 2m:= LEF 1.5m:= LAE 0.5m:= LEC 0.75m:= ABA 60mm 2:= ADC 45mm2:= AEF 75mm2:= P 20kN:= E 350 GPa⋅:= Solution: LAC LAE LEC+:= Internal Forces in the rods : From FBD : FDC P LAE LAC ⎛⎜⎝ ⎞ ⎠ ⋅:=ΣΜA=0; FDC LAC( )⋅ P LAE( )⋅− 0= FDC 8.00 kN= + ΣFy=0; FDC FBA+ P− 0= FBA P FDC−:= FBA 12.00 kN= Displacement: δC FDC LDC( )⋅ E ADC⋅ := δC 1.0159 mm= δA FBA LBA( )⋅ E ABA⋅ := δA 1.1429 mm= δE δC LEC LAC ⎛⎜⎝ ⎞ ⎠ δA δC−( )⋅+:= δE 1.0921 mm= δF_E P LEF( )⋅ E AEF⋅ := δF_E 1.1429 mm= δF δE δF_E+:= δF 2.2349 mm= Ans Problem 4-16 The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 500 mm2. If a vertical force of P = 250 kN is applied to the end B of member AB, determine the vertical displacement of point B. Given: a 1.5m:= b 2m:= LAD a 2 b2+:= LAD 2.5 m= LAB 3m:= LAC a 2 b2+:= LAC 2.5 m= E 200 GPa⋅:= A 500mm2:= P 250kN:= Solution: c a2 b2+:= h a c := v b c := θ atan a b ⎛⎜⎝ ⎞ ⎠:= θ 36.869898 deg= Equation of equilibrium : +For AB : ΣFy=0; FAD FAC+( ) v⋅ P− 0= Since FAD FAC= FAC P 2v := FAC 156.25 kN= Displacement: δA_C FAC LAC( )⋅ E A⋅:= * δA_C 3.9063 mm= δB_A P LAB( )⋅ E A⋅:= δB_A 7.5 mm= Consider triangle AA'C: LA'C LAC δA_C+:= LAA' δA= θ'A 180deg θ−:= sin φA'( ) LAC sin θ'A( ) LA'C = φA' asin LAC sin θ'A( ) LA'C ⋅⎛⎜⎝ ⎞ ⎠ := φA' 36.80289 deg= θ'C 180deg θ'A− φA'−:= θ'C 0.0670094 deg= sin θ'C( ) δA sin θ'A( ) LA'C = δA LAC sin θ'C( ) sin θ'A( )⋅:= δA 4.8731 mm= δB δA δB_A+:= δB 12.37 mm= Ans Problem 4-17 The linkage is made of three pin-connected A-36 steel members, each having a cross-sectional area of 500 mm2. Determine the magnitude of the force P needed to displace point B 2.5 mm downward. Given: a 1.5m:= b 2m:= LAD a 2 b2+:= LAD 2.5 m= LAB 3m:= LAC a 2 b2+:= LAC 2.5 m= E 200 GPa⋅:= A 500mm2:= δB 2.5mm:= Solution: c a2 b2+:= h a c := v b c := θ atan a b ⎛⎜⎝ ⎞ ⎠:= θ 36.869898 deg= Equation of equilibrium : +For AB : ΣFy=0; FAD FAC+( ) v⋅ P− 0= Since FAD FAC= FAC P 2v = FAD P 2v = Displacement: δA_C FAC LAC( )⋅ E A⋅= δA_C P LAC( )⋅ 2v E A⋅( )= δB_A P LAB( )⋅ E A⋅= δB δA δB_A+= δA δB P LAB( )⋅ E A⋅−= Consider triangle AA'C: LA'C LAC δA_C+= LAA' δA= θ'A 180deg θ−:= θ'A 143.130102 deg= LA'C 2 δA2 LAC2+ 2 δA( ) LAC( )⋅ cos θ'A( )⋅−= Initial guess: P 1kN:= Given LAC P LAC( )⋅ 2v E A⋅( )+ ⎡⎢⎣ ⎤⎥⎦ 2 δB P LAB( )⋅ E A⋅− ⎡⎢⎣ ⎤⎥⎦ 2 LAC 2+ 2 δB P LAB( )⋅ E A⋅− ⎡⎢⎣ ⎤⎥⎦ LAC( )⋅ cos θ'A( )⋅−= Solving : P Find P( ):= P 50.47 kN= Ans Problem 4-18 Consider the general problem of a bar made from m segments, each having a constant cross-sectional area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be used to determine the displacement of the bar at any specified location x. Show an application of the program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m, P2 = -1.5 kN, A2 = 625 mm2. Problem 4-19 The rigid bar is supported by the pin-connected rod CB that has a cross-sectional area of 14 mm2 and is made from 6061-T6 aluminum. Determine the vertical deflection of the bar at D when the distributed load is applied. Given: L 4m:= w 300 N m := a 1.5m:= b 2m:= A 14mm2:= E 68.9 GPa⋅:= Solution: LBC a 2 b2+:= v a LBC := h b LBC := Support Reactions: ΣΜA=0; FBC v( )⋅ b( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0= FBC w L2⋅ 2 v⋅ b⋅:= FBC 2 kN= Displacement: δB_C FBC LBC( )⋅ E A⋅:= δB_C 5.183 mm= Consider triangle AB'C: LB'C LBC δB_C+:= LB'C 2.505183 m= LB'C 2 a2 b2+ 2 a( ) b( )⋅ cos θ'A( )⋅−= θ'A acos a2 b2+ LB'C2− 2 a⋅ b⋅ ⎛⎜⎝ ⎞ ⎠:= θ'A 90.24775 deg= φ θ'A 90deg−:= φ 0.24775 deg= φ 0.0043241 rad= δD φ L⋅:= δD 17.30 mm= Ans Problem 4-20 The rigid beam is supported at its ends by two A-36 steel tie rods. If the allowable stress for the steel is σallow = 115 MPa, the load w = 50 kN/m, and x = 1.2 m, determine the diameter of each rod so that the beam remains in the horizontal position when it is loaded. Given: LCD 1.8m:= LAC 2.4m:= LAB 1.8m:= x 1.2m:= w 50 kN m := σallow 115 MPa⋅:= Solution: Internal Forces in the rods : From FBD : FCD w x2⋅ 2 LAC⋅ :=ΣΜA=0; FCD LAC( )⋅ w x⋅( ) 0.5x( )⋅− 0= FCD 15.00 kN= + ΣFy=0; FCD FAB+ w x⋅− 0= FAB w x⋅ FCD−:= FAB 45.00 kN= Displacement: To maintain the rigid beam in the horizontal position, the elongation of both rods AB and CD must be the same. δA δC= AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅= δA FAB LAB( )⋅ E AAB⋅ = δC FCD LCD( )⋅ E ACD⋅ = Thus, FAB LAB( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅⎡⎢⎣ ⎤⎥⎦⋅ FCD LCD( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅⎡⎢⎣ ⎤⎥⎦⋅ = dCD dAB FCD LCD( )⋅ FAB LAB( )⋅= dAB 3 dCD⋅= Allowable Normal Stress: Assume failure of rods AB. σallow FAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅ = dAB 4 FAB⋅ π( ) σallow⋅:= dAB 22.321 mm= dCD dAB 3 := dCD 12.887 mm= Assume failure of rods CD. σallow FCD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅ = dCD 4 FCD⋅ π( ) σallow⋅:= dCD 12.887 mm= Ans dAB 3 dCD⋅:= dAB 22.321 mm= Ans Problem 4-21 The rigid beam is supported at its ends by two A-36 steel tie rods. The rods have diameters dAB = 12 mm and dCD = 7.5 mm. If the allowable stress for the steel is σallow = 115 MPa, determine the intensity of the distributed load w and its length x on the beam so that the beam remains in the horizontal position when it is loaded. Given: LCD 1.8m:= LAC 2.4m:= LAB 1.8m:= dAB 12mm:= dCD 7.5mm:= σallow 115 MPa⋅:= Solution: AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅= Allowable Normal Stress: Assume failure of rods AB. σallow FAB AAB = FAB σallow( ) π4⎛⎜⎝ ⎞⎠ dAB2⋅⎡⎢⎣ ⎤⎥⎦⋅:= FAB 13.006 kN= Displacement: To maintain the rigid beam in the horizontal position, the elongation of both rods AB and CD must be the same. δA δC= δA FAB LAB( )⋅ E AAB⋅ = δC FCD LCD( )⋅ E ACD⋅ = Thus, FAB LAB( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅⎡⎢⎣ ⎤⎥⎦⋅ FCD LCD( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅⎡⎢⎣ ⎤⎥⎦⋅ = FCD FAB dCD 2 dAB 2 = FCD dCD 2 dAB 2 ⎛⎜⎜⎝ ⎞ ⎠ FAB⋅:= FCD 5.081 kN= Internal Forces in the rods : Given From FBD : ΣΜA=0; FCD LAC( )⋅ w x⋅( ) 0.5x( )⋅− 0= + ΣFy=0; FCD FAB+ w x⋅− 0= Initial guess: w 1 kN m := x 1m:= Solving : w x ⎛⎜⎝ ⎞ ⎠ Find w x,( ):= w 13.41 kN m = Ans x 1.35 m= Allowable Normal Stress: Assume failure of rods CD. σallow FCD ACD = FCD σallow( ) π4⎛⎜⎝ ⎞⎠ dCD2⋅⎡⎢⎣ ⎤⎥⎦⋅:= FCD 5.081 kN= Displacement: To maintain the rigid beam in the horizontal position, the elongation of both rods AB and CD must be the same. δA δC= δA FAB LAB( )⋅ E AAB⋅ = δC FCD LCD( )⋅ E ACD⋅ = Thus, FAB LAB( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅⎡⎢⎣ ⎤⎥⎦⋅ FCD LCD( )⋅ E π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅⎡⎢⎣ ⎤⎥⎦⋅ = FCD FAB dCD 2 dAB 2 = FAB dAB 2 dCD 2 ⎛⎜⎜⎝ ⎞ ⎠ FCD⋅:= FAB 13.006 kN= The results are the same as assuming failure of rod AB. Therefore, rods AB and CD fail simultaneously. Ans Problem 4-22 The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is uniformly distributed along its sides of w = 4 kN/m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post. Given: L 2m:= d 60mm:= P 20kN:= w 4 kN m := E 13.1 GPa⋅:= Solution: Equation od Equilibrium : For entire post [FBD (a)] + ΣFy=0; P− w L⋅+ F+ 0= F P w L⋅−:= F 12 kN= Internal Force: FBD (b). + ΣFy=0; P− w y⋅+ Fy− 0= Fy P− w y⋅+= Displacement: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= unit 1kN:= δA_B 0 L y Fy A E⋅ ⌠⎮ ⎮⌡ d= δA_B unit A E⋅ 0 L yP− w y⋅+( ) 1 kN ⋅ ⌠⎮ ⎮⌡ d ⎡⎢⎢⎣ ⎤⎥⎥⎦ := δA_B 0.864− mm= Ans Note: Negative sign indicates that ens A moves towards end B. Problem 4-23 The post is made of Douglas fir and has a diameter of 60 mm. If it is subjected to the load of 20 kN and the soil provides a frictional resistance that is distributed along its length and varies linearly from w = 0 at y = 0 to w = 3 kN/m at y = 2 m, determine the force F at its bottom needed for equilibrium. Also, what is the displacement of the top of the post A with respect to its bottom B? Neglect the weight of the post. Given: L 2m:= d 60mm:= P 20kN:= wo 3 kN m := E 13.1 GPa⋅:= Solution: Equation od Equilibrium : For entire post [FBD (a)] + ΣFy=0; P− 0.5wo L⋅+ F+ 0= F P 0.5wo L⋅−:= F 17 kN= Internal Force: FBD (b). + ΣFy=0; P− 0.5 wo y L ⋅⎛⎜⎝ ⎞ ⎠⋅ y⋅+ Fy− 0= Fy P− wo 2 L⋅ ⎛⎜⎝ ⎞ ⎠ y 2⋅+= Displacement: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= unit 1kN:= δA_B 0 L y Fy A E⋅ ⌠⎮ ⎮⌡ d= δA_B unit A E⋅ 0 L yP− wo 2 L⋅ ⎛⎜⎝ ⎞ ⎠ y 2⋅+⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅ ⌠⎮ ⎮⌡ d ⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ := δA_B 1.026− mm= Ans Note: Negative sign indicates that ens A moves towards end B. Problem 4-24 The rod has a slight taper and length L. It is suspended from the ceiling and supports a load P at its end. Show that the displacement of its end due to this load is δ = PL/(πEr2r1). Neglect the weight of the material. The modulus of elasticity is E. Solution: Geometry : L xo+ r2 xo r1 = xo L r1⋅ r2 r1− = Thus, rx r1 r2 r1− L x⋅+= rx r1 L⋅ r2 r1−( ) x⋅+ L = Ax π rx2⎛⎝ ⎞⎠⋅= Ax π L2 r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2⋅= Displacement: δ 0 L x P Ax E⋅ ⌠⎮ ⎮⌡ d= δ P L 2⋅ π E⋅ 0 L x 1 r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2 ⌠⎮ ⎮ ⎮⌡ d ⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ ⋅= L 0 δ P L 2⋅ π E⋅ 1 r2 r1−( ) r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦⋅⋅= δ P− L 2⋅ π E⋅ r2 r1−( )⋅ 1 r1 L⋅ r2 r1−( ) L⋅+ 1 r1 L⋅ −⎡⎢⎣ ⎤⎥⎦ ⋅= δ P− L 2⋅ π E⋅ r2 r1−( )⋅ 1 r2 L⋅ 1 r1 L⋅ −⎛⎜⎝ ⎞ ⎠ ⋅= δ P− L 2⋅ π E⋅ r2 r1−( )⋅ r2 r1− r1 r2⋅ L⋅ ⎛⎜⎝ ⎞ ⎠ ⋅= δ P− L⋅π E⋅ r1⋅ r2⋅ = Ans Problem 4-25 Solve Prob. 4-24 by including both P and the weight of the material, considering its specific weight to be γ (weight per volume). Solution: Internal Force: FBD (b). + ΣFy=0; Px wx− P− 0= Px Wx P+= Geometry : L xo+ r2 xo r1 = xo L r1⋅ r2 r1− = Thus, rx r1 r2 r1− L x⋅+= rx r1 L⋅ r2 r1−( ) x⋅+ L = Ax π rx2⎛⎝ ⎞⎠⋅= Ax π L2 r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2⋅= Wx γ 3 ⎛⎜⎝ ⎞ ⎠ Ax⋅ x0 x+( )⋅ γ π⋅3⎛⎜⎝ ⎞⎠ r12⎛⎝ ⎞⎠⋅ x0( )⋅−= Wx γ π⋅ 3 L2⋅ ⎛⎜⎝ ⎞ ⎠ r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2⋅ L r1⋅r2 r1− x+ ⎛⎜⎝ ⎞ ⎠ ⋅ γ π⋅ 3 ⎛⎜⎝ ⎞ ⎠ r1 2⎛⎝ ⎞⎠⋅ L r1⋅ r2 r1− ⎛⎜⎝ ⎞ ⎠ ⋅−= Wx γ π⋅ 3 L2⋅ r2 r1−( )⋅ r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦ 3 r1 3 L3⋅⎛⎝ ⎞⎠−⎡⎣ ⎤⎦⋅= Displacement: δ1 γ 3 E⋅ r2 r1−( )⋅ 0 L x r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦3 r13 L3⋅⎛⎝ ⎞⎠− r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2 ⌠⎮ ⎮ ⎮⌡ d ⎡⎢⎢⎢⎢⎣ ⎤⎥⎥⎥⎥⎦ ⋅=δ1 0 L x Wx Ax E⋅ ⌠⎮ ⎮ ⎮⌡ d= δ1 γ 3 E⋅ r2 r1−( )⋅ 0 L xr1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦⌠⎮⌡ d r13 L3⋅⎛⎝ ⎞⎠ 0 L x 1 r1 L⋅ r2 r1−( ) x⋅+⎡⎣ ⎤⎦2 ⌠⎮ ⎮ ⎮⌡ d⋅− ⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ ⋅= Using the result of Prob.(4-24) for the 2nd Integral, we have δ1 γ 3 E⋅ r2 r1−( )⋅ r1 L2⋅ r2 r1−( ) L2 2 ⋅+⎡⎢⎣ ⎤⎥⎦ r1 3 L3⋅⎛⎝ ⎞⎠ 1L r1⋅ r2⋅ ⎛⎜⎝ ⎞ ⎠ ⋅−⎡⎢⎣ ⎤⎥⎦ ⋅= δ1 γ L2⋅ r2 r1+( )⋅ 6 E⋅ r2 r1−( )⋅ γ L2⋅ r12⋅ 3 E⋅ r2 r1−( )⋅ r2⋅−⋅= Using the result of Prob.(4-24) for the 2nd Integral, we have δ δ1 0 L x P Ax E⋅ ⌠⎮ ⎮⌡ d+= δ γ L2⋅ r2 r1+( )⋅ 6 E⋅ r2 r1−( )⋅ γ L2⋅ r12⋅ 3 E⋅ r2 r1−( )⋅ r2⋅−⋅ P L⋅ π E⋅ r1⋅ r2⋅ += Ans Problem 4-26 The support is made by cutting off the two opposite sides of a sphere that has a radius r0. If the original height of the support is r0 /2, determine how far it shortens when it supports a load P.The modulus of elasticity is E. Solution: Geometry : r ro cos θ( )⋅= y ro sin θ( )⋅= Ay π r2⋅= dy ro cos θ( )⋅ dθ⋅= Ay π ro2⋅ cos θ( )2⋅= h 0.25 ro⋅= h ro sin θo( )⋅=Displacement: 0.25 ro⋅ ro sin θo( )⋅= δ 2 0 h y P Ay E⋅ ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ ⋅= δ 0 θo θ 2P ro⋅ cos θ( )⋅ π ro2⋅ cos θ( )2⋅ E⋅ ⌠⎮ ⎮ ⎮⌡ d= θo asin 0.25( ):= θo 14.48 deg= δ 2Pπ ro⋅ E⋅ ⎛⎜⎝ ⎞ ⎠ 0 14.48o θ1 cos θ( ) ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ ⋅= o48.14 0δ 2 P⋅ π ro⋅ E⋅ ⎛⎜⎝ ⎞ ⎠ ln sec θ( ) tan θ( )+( )( )⋅= δ 0.511 P⋅π ro⋅ E⋅ = Ans Alternatively, Geometry : Ay π x2⋅= Ay π ro2 y2−⎛⎝ ⎞⎠⋅= Displacement: δ 0 0.25 ro⋅ θ2 P⋅ π ro2 y2−⎛⎝ ⎞⎠⋅ E⋅ ⌠⎮ ⎮ ⎮⌡ d=δ 2 0 h y P Ay E⋅ ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ ⋅= or25.0 0δ 2 P⋅ π E⋅ ⎛⎜⎝ ⎞ ⎠ 1 2 ro⋅ ⎛⎜⎝ ⎞ ⎠ ⋅ ln ro y+ ro y− ⎛⎜⎝ ⎞ ⎠ ⋅= δ 0.511 P⋅π ro⋅ E⋅ = Ans Problem 4-27 The ball is truncated at its ends and is used to support the bearing load P. If the modulus of elasticity for the material is E, determine the decrease in its height when the load is applied. Solution: Geometry : x2 r2 y2−= Ay π x2⋅= Ay π r2 y2−( )⋅= Displacement: δ 2 0 h y P Ay E⋅ ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ ⋅= r 2 ⎛⎜⎝ ⎞ ⎠ 2 r2 h2−= δ 0 0.866 r⋅ θ2 P⋅ π r2 y2−( )⋅ E⋅ ⌠⎮ ⎮⌡ d= h 3 2 r⋅= r866.0 0 h 0.866 r⋅= δ 2 P⋅π E⋅ ⎛⎜⎝ ⎞ ⎠ 1 2 r⋅ ⎛⎜⎝ ⎞ ⎠⋅ ln r y+ r y− ⎛⎜⎝ ⎞ ⎠⋅= δ 2.63 P⋅π ro⋅ E⋅ = Ans Problem 4-28 Determine the elongation of the aluminum strap when it is subjected to an axial force of 30 kN. Eal = 70 GPa. Given: L1 250mm:= L2 800mm:= d1 15mm:= d2 50mm:= t 6mm:= P 30kN:= E 70 GPa⋅:= Solution: Displacement: δ 2 P L1⋅ E t⋅ d2 d1−( )⋅⋅ ln d2 d1 ⎛⎜⎝ ⎞ ⎠ ⋅ P L2⋅ E t⋅ d2( )⋅+:= δ 2.371 mm= Ans Problem 4-29 The casting is made of a material that has a specific weight γ and modulus of elasticity E. If it is formed into a pyramid having the dimensions shown, determine how far its end is displaced due to gravity when it is suspended in the vertical position. Solution: Internal Force: FBD (b). + ΣFy=0; Pz 1 3 γ A⋅ z⋅− 0= Pz 1 3 γ A⋅ z⋅= Displacement: δ 0 L y Pz A E⋅ ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ = δ 0 L y γ A⋅ z⋅ 3A E⋅ ⌠⎮ ⎮⌡ d= δ γ 3 E⋅ 0 L yz ⌠⎮⌡ d ⎛⎜⎜⎝ ⎞ ⎠ = δ γ L 2⋅ 6 E⋅= Ans Problem 4-30 The pedestal is made in a shape that has a radius defined by the function r = 2/(2 + y1/2) m, where y is in meter. If the modulus of elasticity for the material is E = 100 MPa, determine the displacement of its top when it supports the 5-kN load. Given: H 4m:= do 1m:= d1 0.5m:= P 5kN:= E 100 106( )⋅ Pa:= r 2 2 y0.5+ = Solution: Ay 4 π⋅ 2 y0.5+( )2=Displacement: Ay π r2⋅= unit 1m:= H 4:= δ 0 H y P Ay E⋅ ⌠⎮ ⎮⌡ d= δ P E ⎛⎜⎝ ⎞ ⎠ 1 unit ⎛⎜⎝ ⎞ ⎠⋅ 0 H y 2 y0.5+( )2 4 π⋅ ⌠⎮ ⎮ ⎮⌡ d ⎡⎢⎢⎢⎣ ⎤⎥⎥⎥⎦ := δ 0.1804 mm= Ans Problem 4-31 The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 150 kN, determine the average normal stress in the concrete and in each rod. Each rod has a diameter of 20 mm. Given: L 1.2m:= rconc 100mm:= dst 20mm:= P 150kN:= Est 200 GPa⋅:= Econc 29 GPa⋅:= Solution: Ast 6 π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⎛⎝ ⎞⎠⋅:= Aconc π rconc2⎛⎝ ⎞⎠⋅ Ast−:= Compatibility: δst δconc= Given Pst( ) L⋅ Ast Est⋅ Pconc( ) L⋅ Aconc Econc⋅ = [1] Equations of equilibrium: +ΣFy=0; Pst Pconc+ P− 0= [2] Initial guess: Pconc 1kN:= Pst 2kN:= Solving [1] and [2]: Pconc Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pconc Pst,( ):= PconcPst ⎛⎜⎜⎝ ⎞ ⎠ 104.152 45.848 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal Stress: σst Pst Ast := σst 24.323 MPa= Ans σconc Pconc Aconc := σconc 3.527 MPa= Ans Problem 4-32 The column is constructed from high-strength concrete and six A-36 steel reinforcing rods. If it is subjected to an axial force of 150 kN, determine the required diameter of each rod so that one-fourth of the load is carried by the concrete and three-fourths by the steel. Given: L 1.2m:= rconc 100mm:= P 150kN:= Pst 0.75P:= Pconc 0.25P:= Est 200 GPa⋅:= Econc 29 GPa⋅:= Solution: Pst 112.50 kN= Pconc 37.50 kN= Compatibility: δst δconc= Pst( ) L⋅ Ast Est⋅ Pconc( ) L⋅ Aconc Econc⋅ = Ast 6 π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⎛⎝ ⎞⎠⋅= Aconc π rconc2⎛⎝ ⎞⎠⋅ Ast−= Thus, π rconc2⎛⎝ ⎞⎠⋅ Ast− Ast Est Pst Pconc Econc ⋅= π rconc2⎛⎝ ⎞⎠⋅ Ast Est Pst Pconc Econc ⋅ 1+⎛⎜⎝ ⎞ ⎠ = Ast π rconc2⎛⎝ ⎞⎠⋅ Est Pst Pconc Econc ⋅ 1+ := Ast 9523.2948 mm2= dst 2 2 3 π⋅ Ast= dst 2 3 π⋅ Ast:= dst 44.95 mm= Ans Problem 4-33 The A-36 steel pipe has a 6061-T6 aluminum core. It is subjected to a tensile force of 200 kN. Determine the average normal stress in the aluminum and the steel due to this loading. The pipe has an outer diameter of 80 mm and an inner diameter of 70 mm. Given: L 400mm:= do 80mm:= P 200kN:= di 70mm:= Est 200GPa:= Eal 68.9GPa:= Solution: Aal π 4 di 2⎛⎝ ⎞⎠⋅:= Ast π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Compatibility: δst δal= Given Pst( ) L⋅ Ast Est⋅ Pal( ) L⋅ Aal Eal⋅ = [1] Equations of equilibrium: + ΣFx=0; Pst Pal+ P− 0= [2] Initial guess: Pal 1kN:= Pst 2kN:= Solving [1] and [2]: Pal Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pal Pst,( ):= PalPst ⎛⎜⎜⎝ ⎞ ⎠ 105.899 94.101 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal Stress: σst Pst Ast := σst 79.88 MPa= Ans σal Pal Aal := σal 27.52 MPa= Ans Problem 4-34 The concrete column is reinforced using four steel reinforcing rods,each having a diameter of 18 mm. Determine the stress in the concrete and the steel if the column is subjected to an axial load of 800 kN. Est = 200 GPa, Ec = 25 GPa. Given: P 800kN:= bc 300mm:= dst 18mm:= Est 200 GPa⋅:= Ec 25 GPa⋅:= Solution: Ast 4 π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⎛⎝ ⎞⎠⋅:= Ac bc2⎛⎝ ⎞⎠ Ast−:= Set: L 1m:= Compatibility: δst δconc= Given Pst( ) L⋅ Ast Est⋅ Pc( ) L⋅ Ac Ec⋅ = [1] Equations of equilibrium: + ΣFy=0; Pst Pc+ P− 0= [2] Initial guess: Pc 1kN:= Pst 2kN:= Solving [1] and [2]: Pc Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pc Pst,( ):= PcPst ⎛⎜⎜⎝ ⎞ ⎠ 732.928 67.072 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal Stress: σst Pst Ast := σst 65.89 MPa= Ans σc Pc Ac := σc 8.24 MPa= Ans Problem 4-35 The column is constructed from high-strength concrete and four A-36 steel reinforcing rods. If it is subjected to an axial force of 800 kN, determine the required diameter of each rod so that one-fourth of the load is carried by the steel and three-fourths by the concrete. Est = 200 GPa, Ec = 25 GPa. Given: P 800kN:= Ec 25 GPa⋅:= bc 300mm:= Pc 0.75P:= Est 200 GPa⋅:= Pst 0.25P:= Solution: Pst 200 kN= Pc 600 kN= Compatibility: δst δc= Pst( ) L⋅ Ast Est⋅ Pc( ) L⋅ Ac Ec⋅ = Ast 4 π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⋅= Ac bc2⎛⎝ ⎞⎠ Ast−= Thus, bc 2 Ast− Ast Est Pst Pc Ec ⋅= bc 2 Ast Est Pst Pc Ec ⋅ 1+⎛⎜⎝ ⎞ ⎠ = Ast bc 2 Est Pst Pc Ec ⋅ 1+ := Ast 3600 mm2= dst 2 1 π Ast= dst 1 π Ast:= dst 33.85 mm= Ans Problem 4-36 The A-36 steel pipe has an outer radius of 20 mm and an inner radius of 15 mm. If it fits snugly between the fixed walls before it is loaded, determine the reaction at the walls when it is subjected to the load shown. Given: LAB 300mm:= ro 20mm:= LBC 700mm:= ri 15mm:= P 8kN:= Est 200GPa:= Solution: A π ro2 ri2−⎛⎝ ⎞⎠⋅:= L LAB LBC+:= By superposition : + ∆C− δC+ 0= 2P−( ) LAB⋅ A E⋅ FC( ) L⋅ A E⋅+ 0= FC 2 P⋅ LAB L ⎛⎜⎝ ⎞ ⎠⋅:= FC 4.80 kN= Ans Equations of equilibrium: + ΣFx=0; FA FC+ 2P− 0= FA 2P FC−:= FA 11.20 kN= Ans Problem 4-37 The 304 stainless steel post A has a diameter of d = 50 mm and is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 25 kN is applied to the rigid cap, determine the average normal stress developed in the post and the tube. Given: L 200mm:= dst 50mm:= ro 75mm:= t 12mm:= P 25kN:= Est 193 GPa⋅:= Ebr 101 GPa⋅:= Solution: ri ro t−:= Ast π 4 dst 2⎛⎝ ⎞⎠⋅:= Abr π ro2 ri2−⎛⎝ ⎞⎠⋅:= Compatibility: δst δbr= Given Pst( ) L⋅ Ast Est⋅ Pbr( ) L⋅ Abr Ebr⋅ = [1] Equations of equilibrium: + ΣFy=0; Pst Pbr+ P− 0= [2] Initial guess: Pbr 1kN:= Pst 2kN:= Solving [1] and [2]: Pbr Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pbr Pst,( ):= Pbr Pst ⎛⎜⎜⎝ ⎞ ⎠ 14.525 10.475 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal Stress: σst Pst Ast := σst 5.335 MPa= Ans σbr Pbr Abr := σbr 2.792 MPa= Ans Problem 4-38 The 304 stainless steel post A is surrounded by a red brass C83400 tube B. Both rest on the rigid surface. If a force of 25 kN is applied to the rigid cap, determine the required diameter d of the steel post so that the load is shared equally between the post and tube. Given: L 200mm:= ro 75mm:= t 12mm:= P 25kN:= Est 193 GPa⋅:= Ebr 101 GPa⋅:= Pst 0.5P:= Pbr 0.5P:= Solution: ri ro t−:= Pst 12.5 kN= Pbr 12.5 kN= Compatibility: δst δbr= Pst( ) L⋅ Ast Est⋅ Pbr( ) L⋅ Abr Ebr⋅ = Ast π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⋅= Abr π ro2 ri2−⎛⎝ ⎞⎠⋅:= Thus, π ro2 ri2−⎛⎝ ⎞⎠⋅ Ast Est Pst Pbr Ebr ⋅= Ast π ro2 ri2−⎛⎝ ⎞⎠⋅ Est Pst Pbr Ebr ⋅ := Ast 2722.54 mm2= dst 2 4 π Ast= dst 4 π Ast:= dst 58.88 mm= Ans Problem 4-39 The load of 7.5 kN is to be supported by the two vertical steel wires for which σY = 500 MPa. If, originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the force developed in each wire after the load is suspended. Each wire has a cross-sectional area of 12.5 mm2. Given: LAB 1250mm:= LAC 1252.5mm:= A 12.5mm2:= W 7.5kN:= E 200 GPa⋅:= EY 70500 MPa⋅:= Solution: ∆L LAC LAB−:= ∆L 2.50 mm= Compatibility: + δAC ∆L+ δAB= Given TAC( ) LAC⋅ A E⋅ ∆L+ TAB( ) LAB⋅ A E⋅= [1] Equations of equilibrium: + ΣFy=0; TAC TAB+ W− 0= [2] Initial guess: TAC 1kN:= TAB 2kN:= Solving [1] and [2]: TAC TAB ⎛⎜⎜⎝ ⎞ ⎠ Find TAC TAB,( ):= TACTAB ⎛⎜⎜⎝ ⎞ ⎠ 1.249 6.251 ⎛⎜⎝ ⎞ ⎠ kN= Ans Check Average Normal Stress: σAC TAC A := σAC 99.9 MPa= [ < σY = 500 MPa] σAB TAB A := σAB 500.1 MPa= [ < σY = 500 MPa] Problem 4-40 The load of 4 kN is to be supported by the two vertical steel wires for which σY = 560 MPa. If, originally, wire AB is 1250 mm long and wire AC is 1252.5 mm long, determine the cross-sectional area of AB if the load is to be shared equally between both wires. Wire AC has a cross-sectional area of 13 mm2. Given: W 4kN:= LAB 1250mm:= AAC 13mm2:= TAC 0.5W:= LAC 1252.5mm:= EY 560 MPa⋅:= TAB 0.5W:= E 200 GPa⋅:= Solution: ∆L LAC LAB−:= ∆L 2.50 mm= TAC 2 kN= TAB 2 kN= Compatibility: + δAC ∆L+ δAB= TAC( ) LAC⋅ AAC( ) E⋅ ∆L+ TAB( ) LAB⋅ AAB( ) E⋅= Thus, AAB TAB( ) LAB⋅ TAC( ) LAC⋅ AAC ∆L( ) E⋅+ := AAB 3.60911 mm 2= Ans Check Average Normal Stress: σAC TAC AAC := σAC 153.846 MPa= [ < σY = 560 MPa] σAB TAB AAB := σAB 554.15 MPa= [ < σY = 560 MPa] Problem 4-41 The support consists of a solid red brass C83400 post surrounded by a 304 stainless steel tube. Before the load is applied, the gap between these two parts is 1 mm. Given the dimensions shown, determine the greatest axial load that can be applied to the rigid cap A without causing yielding of any one of the materials. Given: Lbr 0.251m:= Ebr 101 GPa⋅:= σY_br 70 MPa⋅:= Lst 0.250m:= Est 193 GPa⋅:= dbr 60mm:= di 80mm:= t 10mm:= Solution: ∆L Lbr Lst−:= ∆L 1 mm= do di 2t+:= Abr π 4 dbr 2⎛⎝ ⎞⎠⋅:= Ast π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= +Compatibility: δbr ∆L+ δst= Fbr( ) Lbr⋅ Abr Ebr⋅ ∆L+ Fst( ) Lst⋅ Ast Est⋅ = [1] Equations of equilibrium: + ΣFy=0; Fst Fbr+ P− 0= [2] Assume brass yields, then Fbr σY_br( )Abr:= Fbr 197.92 kN= εbr σY_br Ebr := εbr 0.00069307 mm mm = δbr εbr( )Lbr:= δbr 0.17 mm= [ < ∆L = 1 mm ] Thus, only the brass is loaded. P Fbr:= P 197.92 kN= Ans Problem 4-42 Two A-36 steel wires are used to support the 3.25-kN (~325-kg) engine. Originally, AB is 800 mm long and A'B' is 800.2 mm long. Determine the force supported by each wire when the engine is suspended from them. Each wire has a cross-sectional area of 6.25 mm2. Given: LAB 800mm:= A 6.25mm2:= LA'B' 800.2mm:= W 3.25kN:= E 29 103( )⋅ ksi⋅:= Solution: ∆L LA'B' LAB−:= ∆L 0.200 mm= Compatibility: + δAB ∆L+ δA'B'= Given [1]TAB( ) LAB⋅ A E⋅ ∆L+ TA'B'( ) LA'B'⋅ A E⋅= Equations of equilibrium: + ΣFy=0; TAB TA'B'+ W− 0= [2] Initial guess: TAB 1kN:= TA'B' 2kN:= Solving [1] and [2]: TAB TA'B' ⎛⎜⎜⎝ ⎞ ⎠ Find TAB TA'B',( ):= TABTA'B' ⎛⎜⎜⎝ ⎞ ⎠ 1.469 1.781 ⎛⎜⎝ ⎞ ⎠ kN= Ans Problem 4-43 The bolt AB has a diameter of 20 mm and passes through a sleeve that has an inner diameter of 40 mm andan outer diameter of 50 mm. The bolt and sleeve are made of A-36 steel and are secured to the rigid brackets as shown. If the bolt length is 220 mm and the sleeve length is 200 mm, determine the tension in the bolt when a force of 50 kN is applied to the brackets. Given: Lb 220mm:= db 20mm:= do 50mm:= Ls 200mm:= P 25kN:= di 40mm:= E 200GPa:= Solution: Ab π 4 db 2⎛⎝ ⎞⎠⋅:= As π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Compatibility: δb δs= Given [1]Pb( ) Lb⋅ Ab E⋅ Ps( ) Ls⋅ As E⋅ = Equations of equilibrium: + ΣFx=0; Pb Ps+ 2P− 0= [2] Initial guess: Pb 1kN:= Ps 2kN:= Solving [1] and [2]: Pb Ps ⎛⎜⎜⎝ ⎞ ⎠ Find Pb Ps,( ):= PbPs ⎛⎜⎜⎝ ⎞ ⎠ 14.388 35.612 ⎛⎜⎝ ⎞ ⎠ kN= Problem 4-44 The specimen represents a filament-reinforced matrix system made from plastic (matrix) and glass (fiber). If there are n fibers, each having a cross-sectional area of Af and modulus of Ef , embedded in a matrix having a cross-sectional area of Am and modulus of Em, determine the stress in the matrix and each fiber when the force P is imposed on the specimen. Solution: Compatibility: δm δf= Pm( ) L⋅ Am Em⋅ Pf( ) L⋅ n Af⋅ Ef⋅ = Pm Am Em⋅ n Af⋅ Ef⋅ ⎛⎜⎝ ⎞ ⎠ Pf⋅= [1] Equations of equilibrium: + ΣFy=0; P Pm− Pf− 0= [2] Solving [1] and [2]: Pm Am Em⋅ n Af⋅ Ef⋅ Am Em⋅+ ⎛⎜⎝ ⎞ ⎠ P⋅= Pf n Af⋅ Ef⋅ n Af⋅ Ef⋅ Am Em⋅+ ⎛⎜⎝ ⎞ ⎠ P⋅= Average Normal Stress: σm Pm Am = σm Em n Af⋅ Ef⋅ Am Em⋅+ ⎛⎜⎝ ⎞ ⎠ P⋅= Ans σf Pf n Af⋅ = σf Ef n Af⋅ Ef⋅ Am Em⋅+ ⎛⎜⎝ ⎞ ⎠ P⋅= Ans Problem 4-45 The distributed loading is supported by the three suspender bars. AB and EF are made from aluminum andCD is made from steel. If each bar has a cross-sectional area of 450 mm2, determine the maximum intensity w of the distributed loading so that an allowable stress of (σallow)st = 180 MPa in the steel and (σallow)al = 94 MPa in the aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa. Given: L 2m:= b 1.5m:= A 450mm2:= Est 200 GPa⋅:= Eal 70 GPa⋅:= σal_allow 94MPa:= σst_allow 180MPa:= Solution: Compatibility: + δA δC= FAB( ) L⋅ Eal( ) A⋅ FCD( ) L⋅ Est( ) A⋅= FAB Eal Est ⎛⎜⎝ ⎞ ⎠ FCD⋅= [1] Equations of equilibrium: ΣΜC=0; FEF b( )⋅ FAB b( )⋅− 0= FAB FEF= + ΣFy=0; FAB FCD+ FEF+ w 2b( )⋅− 0= 2 FAB⋅ w 2b( )⋅ FCD−= [2] Assume failure of AB and EF: FAB σal_allow( ) A⋅:= FAB 42.30 kN= From [1]: FCD Est Eal ⎛⎜⎝ ⎞ ⎠ FAB⋅:= FCD 120.86 kN= From [2]: w FAB b FCD 2 b⋅+:= w 68.49 kN m = Assume failure of CD: FCD σst_allow( ) A⋅:= FCD 81.00 kN= From [1]: FAB Eal Est ⎛⎜⎝ ⎞ ⎠ FCD⋅:= FAB 28.35 kN= From [2]: w FAB b FCD 2 b⋅+:= w 45.90 kN m = [Controls !] Ans Problem 4-46 The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and cross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the average normal stress in the wire and the block. Est = 200 GPa, Eal = 70 GPa. Given: Lk 200mm:= Ak 22.5mm2:= eD 50mm:= AD 40mm2:= P 450N:= a 100mm:= b 150mm:= c 150mm:= Est 200GPa:= Eal 70GPa:= Solution: Compatibility: θBC θAD= δBC b δD c = Given [1]FBC( ) Lk⋅ Ak Est⋅ b⋅ FD( ) eD⋅ AD Eal⋅ c⋅ = Equations of equilibrium: ΣΜA=0; FBC− b( )⋅ FD c( )⋅− P a b+( )⋅+ 0= [2] Initial guess: FBC 1kN:= FD 2kN:= Solving [1] and [2]: FBC FD ⎛⎜⎜⎝ ⎞ ⎠ Find FBC FD,( ):= FBC FD ⎛⎜⎜⎝ ⎞ ⎠ 214.968 535.032 ⎛⎜⎝ ⎞ ⎠ N= Average Normal Stress: σBC FBC Ak := σBC 9.554 MPa= Ans σD FD AD := σD 13.376 MPa= Ans Problem 4-47 The rigid link is supported by a pin at A, a steel wire BC having an unstretched length of 200 mm and cross-sectional area of 22.5 mm2, and a short aluminum block having an unloaded length of 50 mm and cross-sectional area of 40 mm2. If the link is subjected to the vertical load shown, determine the rotation of the link about the pin A. Report the answer in radians. Est = 200 GPa, Eal = 70 GPa. Given: Lk 200mm:= Ak 22.5mm2:= eD 50mm:= AD 40mm2:= P 450N:= a 100mm:= b 150mm:= c 150mm:= Est 200GPa:= Eal 70GPa:= Solution: Compatibility: θBC θAD= δBC b δD c = Given [1]FBC( ) Lk⋅ Ak Est⋅ b⋅ FD( ) eD⋅ AD Eal⋅ c⋅ = Equations of equilibrium: ΣΜA=0; FBC− b( )⋅ FD c( )⋅− P a b+( )⋅+ 0= [2] Initial guess: FBC 1kN:= FD 2kN:= Solving [1] and [2]: FBC FD ⎛⎜⎜⎝ ⎞ ⎠ Find FBC FD,( ):= FBCFD ⎛⎜⎜⎝ ⎞ ⎠ 214.968 535.032 ⎛⎜⎝ ⎞ ⎠ N= Displacement: δD FD( ) eD⋅ AD Eal⋅ := δD 0.009554 mm= θAD δD c := θAD 63.69 10 6−× rad= Ans Problem 4-48 The three A-36 steel wires each have a diameter of 2 mm and unloaded lengths of LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the force in each wire after the 150-kg mass is suspended from the ring at A. Given: LAC 1.60m:= a 3:= d 2mm:= LAB 2.00m:= b 4:= M 150kg:= LAD 2.00m:= c 5:= g 9.81 m s2 = Solution: h a c := v b c := A π 4 ⎛⎜⎝ ⎞ ⎠d 2:= W M g⋅:= Compatibility: In triangle AA'B, since the displacement δA is very small, cos θA'( ) v= δAC cos θA'( )⋅ δAD= δAC v( )⋅ δAD= FAC( ) LAC⋅ A E⋅ v( )⋅ FAD( ) LAD⋅ A E⋅= [1] Equations of equilibrium: + ΣFx=0; FAB h( )⋅ FAD h( )⋅− 0= FAB FAD= + ΣFy=0; FAB v( )⋅ FAC+ FAD v( )⋅+ W− 0= Given 2 FAD⋅ v( )⋅ W FAC−= [2] From [1]: FAC LAC⋅ v( )⋅ FAD LAD⋅= [3] Initial guess: FAD 1N:= FAC 2N:= Solving [2] and [3]: FAD FAC ⎛⎜⎜⎝ ⎞ ⎠ Find FAD FAC,( ):= FADFAC ⎛⎜⎜⎝ ⎞ ⎠ 465.1 726.8 ⎛⎜⎝ ⎞ ⎠N= Ans FAB FAD= Problem 4-49 The A-36 steel wires AB and AD each have a diameter of 2 mm and the unloaded lengths of each wire are LAC = 1.60 m and LAB = LAD = 2.00 m. Determine the required diameter of wire AC so that each wire is subjected to the same force caused by the 150-kg mass suspended from the ring at A. Given: LAC 1.60m:= a 3:= dAB 2mm:= LAB 2.00m:= b 4:= dAD 2mm:= LAD 2.00m:= c 5:= g 9.81 m s2 = M 150kg:= Solution: h a c := v b c := W M g⋅:= AAB π 4 ⎛⎜⎝ ⎞ ⎠dAB 2:= AAD π 4 ⎛⎜⎝ ⎞ ⎠dAD 2:= Equations of equilibrium: Since each wire is required to carry the same amount of load. Hence, FAB F= FAC F= FAD F= Compatibility: In triangle AA'B, since the displacement δA is very small, cos θA'( ) v= δAC cos θA'( )⋅ δAD= δAC v( )⋅ δAD= FAC( ) LAC⋅ AAC E⋅ v( )⋅ FAD( ) LAD⋅ AAD E⋅ = AAC LAC LAD v( )⋅ AAD⋅:= AAC 2.010619 mm 2= AAC π 4 ⎛⎜⎝ ⎞ ⎠dAC 2= dAC 4 π AAC( ):= dAC 1.60 mm= Ans Problem 4-50 The three suspender bars are made of the same material and have equal cross-sectional areas A. Determinethe average normal stress in each bar if the rigid beam ACE is subjected to the force P. Solution: Compatibility: δC δE− d δA δE− 2d = 2δC δA δE+= 2 FCD( ) L⋅ A E⋅ FAB( ) L⋅ A E⋅ FEF( ) L⋅ A E⋅+= 2FCD FAB FEF+= [1] Equations of equilibrium: ΣΜA=0; FCD d( )⋅ FEF 2 d⋅( )⋅+ P d 2 ⎛⎜⎝ ⎞ ⎠⋅− 0= FCD 2FEF+ 0.5P= [2] + ΣFy=0; FAB FCD+ FEF+ P− 0= [3] Solving [1], [2] and [3]: FAB 7 12 P= FCD 1 3 P= FEF 1 12 P= σAB 7P 12A = σAB P 3A = σAB P 12A = Ans Problem 4-51 The assembly consists of an A-36 steel bolt and a C83400 red brass tube. If the nut is drawn up snug against the tube so that L = 75 mm, then turned an additional amount so that it advances 0.02 mm on the bolt, determine the force in the bolt and the tube. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2. Given: L 75mm:= Ebr 101 GPa⋅:= ∆L 0.02mm:= Est 200 GPa⋅:= dst 7mm:= Abr 100mm2:= Solution: Ast π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⋅:= Kst Est( ) Ast⋅:= Kbr Ebr( ) Abr⋅:= Equations of equilibrium: Since no external load is applie, the force acting on the tube and the bolt is the same. Compatibility: ∆L δst δbr+= ∆L P L⋅ Kst P L⋅ Kbr += ∆L P L⋅ Kst Kbr+ Kst Kbr⋅ ⎛⎜⎝ ⎞ ⎠ ⋅= P ∆L L Kst Kbr⋅ Kst Kbr+ ⎛⎜⎝ ⎞ ⎠ := P 1.165 kN= Ans Problem 4-52 The assembly consists of an A-36 steel bolt and a C83400 red brass tube. The nut is drawn up snug against the tube so that L = 75 mm. Determine the maximum additional amount of advance of the nut on the bolt so that none of the material will yield. The bolt has a diameter of 7 mm and the tube has a cross-sectional area of 100 mm2. Given: L 75mm:= Ebr 101 GPa⋅:= dst 7mm:= Est 200 GPa⋅:= Abr 100mm 2:= σY_st 250MPa:= σY_br 70MPa:= Solution: Ast π 4 ⎛⎜⎝ ⎞ ⎠ dst 2⋅:= Kst Est( ) Ast⋅:= Kbr Ebr( ) Abr⋅:= Allowable Normal Stress: Pst σY_st( ) Ast⋅:= Pst 9.621 kN= Pbr σY_br( ) Abr⋅:= Pbr 7.000 kN= Since Pst > Pbr , by comparison, the brass will yield first. P min Pst Pbr,( ):= P 7.00 kN= Compatibility: ∆L δst δbr+= ∆L P L⋅ Kst P L⋅ Kbr += ∆L P L⋅ Kst P L⋅ Kbr +:= ∆L 0.120 mm= Ans Problem 4-53 The 10-mm-diameter steel bolt is surrounded by a bronze sleeve. The outer diameter of this sleeve is 20 mm, and its inner diameter is 10 mm. If the bolt is subjected to a compressive force of P = 20 kN, determine the average normal stress in the steel and the bronze. Est = 200 GPa, Ebr = 100 GPa. Given: ds 10mm:= do 20mm:= Est 200GPa:= P 20kN:= di 10mm:= Ebr 100GPa:= Solution: Ast π 4 ds 2⎛⎝ ⎞⎠⋅:= Abr π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Compatibility: δb δs= Pbr( ) L⋅ Abr Ebr⋅ Pst( ) L⋅ Ast Est⋅ = Given Pbr Abr Ebr⋅ Pst Ast Est⋅ = [1] Equations of equilibrium: + ΣFy=0; Pbr Pst+ P− 0= [2] Initial guess: Pbr 1kN:= Pst 2kN:= Solving [1] and [2]: Pbr Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pbr Pst,( ):= PbrPst ⎛⎜⎜⎝ ⎞ ⎠ 12.000 8.000 ⎛⎜⎝ ⎞ ⎠ kN= σst Pst Ast := σst 101.86 MPa= Ans σbr Pbr Abr := σbr 50.93 MPa= Ans Problem 4-54 The 10-mm-diameter steel bolt is surrounded by a bronze sleeve.The outer diameter of this sleeve is 20 mm,and its inner diameter is 10 mm. If the yield stress for the steel is (σY)st = 640 MPa, and for the bronze (σY)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est = 200 GPa, Ebr = 100 GPa. Given: do 20mm:= Est 200GPa:= σY_st 640MPa:= di 10mm:= Ebr 100GPa:= σY_br 520MPa:= ds 10mm:= Solution: Ast π 4 ds 2⎛⎝ ⎞⎠⋅:= Abr π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Equations of equilibrium: + ΣFy=0; Pbr Pst+ P− 0= [1] Compatibility: δb δs= Pbr( ) L⋅ Abr Ebr⋅ Pst( ) L⋅ Ast Est⋅ = Pbr Abr Ebr⋅ Pst Ast Est⋅ = [2] Assume failure of bolt, then Pst σY_st( )Ast:= Pst 50.265 kN= From [2]: Pbr Abr Ebr⋅ Ast Est⋅ ⎛⎜⎝ ⎞ ⎠ Pst⋅:= Pbr 75.398 kN= From [1]: P Pbr Pst+:= P 125.66 kN= (Controls!): Ans Assume failure of sleeve, then Pbr σY_br Abr:= Pbr 122.522 kN= From [2]: Pst Ast Est⋅ Abr Ebr⋅ ⎛⎜⎝ ⎞ ⎠ Pbr⋅:= Pst 81.681 kN= From [1]: P Pbr Pst+:= P 204.20 kN= Problem 4-55 The rigid member is held in the position shown by three A-36 steel tie rods. Each rod has an unstretched length of 0.75 m and a cross-sectional area of 125 mm2. Determine the forces in the rods if a turnbuckle on rod EF undergoes one full turn. The lead of the screw is 1.5 mm. Neglect the size of the turnbuckle and assume that it is rigid. Note: The lead would cause the rod, when unloaded, to shorten 1.5 mm when the turnbuckle is rotated one revolution. Given: L 0.75m:= b 0.5m:= A 125mm2:= ∆L 1.5mm:= E 200 GPa⋅:= Solution: Equations of equilibrium: ΣΜE=0; TCD b( )⋅ TAB b( )⋅+ 0= TCD TAB= [1] + ΣFy=0; TAB TCD+ TEF− 0= [1] TEF 2TAB= [2] Compatibility: Rod EF shortens 1.5mm causing AB (and CD) to elongate. Thus,: + ∆L δA δC+= ∆L TAB( ) L⋅ E A⋅ TEF( ) L⋅ E A⋅+= ∆L TAB( ) L⋅ E A⋅ 2TAB( ) L⋅ E A⋅+= TAB E A⋅ 3L ∆L⋅:= TAB 16.667 kN= Ans TCD TAB:= TCD 16.667 kN= AnsFrom [1]: TEF 2TAB:= TEF 33.333 kN= AnsFrom [2]: Problem 4-56 The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and a modulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine the displacement of the end B when the force of 10 kN is applied. Given: LCD 0.6m:= LEF 0.3m:= d 25mm:= a 0.3m:= P 10kN:= Eal 70GPa:= Solution: A π 4 d2⋅:= Compatibility: δE 3 a⋅ δC a = Given FEF( ) LEF⋅ 3A Eal⋅ FCD( ) LCD⋅ A Eal⋅ = FEF LEF⋅ 3FCD LCD⋅= [1] Equations of equilibrium: ΣΜA=0; FEF− 3 a⋅( )⋅ FCD a( )⋅− P 2a( )⋅+ 0= 3− FEF FCD− 2P+ 0= [2] Initial guess: FEF 1kN:= FCD 2kN:= Solving [1] and [2]: FEF FCD ⎛⎜⎜⎝ ⎞ ⎠ Find FEF FCD,( ):= FEFFCD ⎛⎜⎜⎝ ⎞ ⎠ 6.31579 1.05263 ⎛⎜⎝ ⎞ ⎠ kN= Displacement: δE FEF( ) LEF⋅ A Eal⋅ := δE 0.055142 mm= δB 4 a⋅ δE 3a = δB 4 3 δE:= δB 0.073522 mm= Ans Problem 4-57 The bar is pinned at A and supported by two aluminum rods, each having a diameter of 25 mm and a modulus of elasticity Eal = 70 GPa. If the bar is assumed to be rigid and initially vertical, determine the force in each rod when the 10-kN load is applied. Given: LCD 0.6m:= LEF 0.3m:= d 25mm:= a 0.3m:= P 10kN:= Eal 70GPa:= Solution: A π 4 d2⋅:= Compatibility: δE 3 a⋅ δC a = Given FEF( ) LEF⋅ 3A Eal⋅ FCD( ) LCD⋅ A Eal⋅ = FEF LEF⋅ 3FCD LCD⋅= [1] Equations of equilibrium: ΣΜA=0; FEF− 3 a⋅( )⋅ FCD a( )⋅− P 2a( )⋅+ 0= 3− FEF FCD− 2P+ 0= [2] Initial guess: FEF 1kN:= FCD 2kN:= Solving [1] and [2]: FEF FCD ⎛⎜⎜⎝ ⎞ ⎠ Find FEF FCD,( ):= FEFFCD ⎛⎜⎜⎝ ⎞ ⎠ 6.316 1.053 ⎛⎜⎝ ⎞ ⎠ kN= Problem 4-58 The assembly consists of two posts made from material 1 having a modulus of elasticity of E1 and each a cross-sectional area A1 , and a material 2 having a modulus of elasticity E2 and cross-sectional area A2. If a central load P is applied to the rigid cap, determine the force in each material. Solution: FAB F1= FEF F2= Equations of equilibrium: ΣΜE=0; FCD d( )⋅ FAB d( )⋅− 0= FCD FAB= FCD F1= + ΣFy=0; FAB FCD+ FEF+ P− 0= 2F1 F2+ P= [1] Compatibility: δA δB= δB δC= F1( ) L⋅ A1 E1⋅ F2( ) L⋅ A2 E2⋅ = F1 A1 E1⋅ A2 E2⋅ ⎛⎜⎝ ⎞ ⎠ F2⋅= [2] Solving [1] and [2]: F1 A1 E1⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= Ans F2 A2 E2⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= Ans Problem 4-59 The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity E1 ofand each a cross-sectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a cross-sectional area A2. If posts AB and CD are to be replaced by those having a material 2, determine the required cross-sectional area of these new posts so that both assemblies deform the same amount when loaded. Solution: FAB F1= FEF F2= Equations of equilibrium: ΣΜE=0; FCD d( )⋅ FAB d( )⋅− 0= FCD FAB= FCD F1= + ΣFy=0; FAB FCD+ FEF+ P− 0= 2F1 F2+ P= [1] Compatibility: δA δB= δB δC= δA δ= F1( ) L⋅ A1 E1⋅ F2( ) L⋅ A2 E2⋅ = F1 A1 E1⋅ A2 E2⋅ ⎛⎜⎝ ⎞ ⎠ F2⋅= [2] Solving [1] and [2]: F1 A1 E1⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= F2 A2 E2⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= When material 1 has been replaced by material 2 for two side posts, then Equations of equilibrium: [1] becomes 2 F'1( ) F'2+ P= [1'] Compatibility: [2] becomes F'1 A'1 A2 ⎛⎜⎝ ⎞ ⎠ F'2⋅= [2'] Solving [1'] and [2']: F'1 A'1 2A'1 A2+ ⎛⎜⎝ ⎞ ⎠ P= F'2 A2 2A'1 A2+ ⎛⎜⎝ ⎞ ⎠ P= Requires, δB δ'B= F2( ) L⋅ A2 E2⋅ F'2( ) L⋅ A2 E2⋅ = A2 E2⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P A2 2A'1 A2+ ⎛⎜⎝ ⎞ ⎠ P= A'1 E1 E2 ⎛⎜⎝ ⎞ ⎠ A1⋅= Ans Problem 4-60 The assembly consists of two posts AB and CD made from material 1 having a modulus of elasticity E1 ofand each a cross-sectional area A1, and a central post EF made from material 2 having a modulus of elasticity E2 and a cross-sectional area A2. If post EF is to be replaced by one having a material 1, determine the required cross-sectional area of this new post so that both assemblies deform the same amount when loaded. Solution: FAB F1= FEF F2= Equations of equilibrium: ΣΜE=0; FCD d( )⋅ FAB d( )⋅− 0= FCD FAB= FCD F1= + ΣFy=0; FAB FCD+ FEF+ P− 0= 2F1 F2+ P= [1] Compatibility: δA δB= δB δC= δA δ= F1( ) L⋅ A1 E1⋅ F2( ) L⋅ A2 E2⋅ = F2 A2 E2⋅ A1 E1⋅ ⎛⎜⎝ ⎞ ⎠ F1⋅= [2] Solving [1] and [2]: F1 A1 E1⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= F2 A2 E2⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P= When material 2 has been replaced by material 1 for central post, then Equations of equilibrium: [1] becomes 2 F'1( ) F'2+ P= [1'] Compatibility: [2] becomes F'2 A'2 A1 ⎛⎜⎝ ⎞ ⎠ F'1⋅= [2'] Solving [1'] and [2']: F'1 A1 2A1 A'2+ ⎛⎜⎝ ⎞ ⎠ P= F'2 A'2 2A1 A'2+ ⎛⎜⎝ ⎞ ⎠ P= Requires, δA δ'A= F1( ) L⋅ A1 E1⋅ F'1( ) L⋅ A1 E1⋅ = A1 E1⋅ 2A1 E1⋅ A2 E2⋅+ ⎛⎜⎝ ⎞ ⎠ P A1 2A1 A'2+ ⎛⎜⎝ ⎞ ⎠ P= A'2 E2 E1 ⎛⎜⎝ ⎞ ⎠ A2⋅= Ans Problem 4-61 The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of 12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown, determine the force developed in each bolt. For the calculation, assume that the bolts carry no shear; rather, the vertical force of 4 kN is supported by the toe at A. Also, assume that the wall and bracket are rigid. A greatly exaggerated deformation of the bolts is shown. Given: a 12.5mm:= b 25mm:= c 50mm:= d 12.5mm:= e 50mm:= L 50mm:= P 4kN:= E 200GPa:= Solution: LAD a b+ c+:= LAC a b+:= A π 4 d2 ⋅:= LAB a:= Compatibility: δD LAD δC LAC = FD( ) L⋅ LAD( ) A⋅ E⋅ FC( ) L⋅ LAC( ) A⋅ E⋅= δB LAB δC LAC = FB( ) L⋅ LAB( ) A⋅ E⋅ FC( ) L⋅ LAC( ) A⋅ E⋅= Given FD LAD LAC ⎛⎜⎝ ⎞ ⎠ FC⋅= [1] FB LAB LAC ⎛⎜⎝ ⎞ ⎠ FC⋅= [2] Equations of equilibrium: ΣΜA=0; FD LAD( )⋅ FC LAC( )⋅+ FB LAB( )⋅+ P e( )⋅− 0= [3] Initial guess: FB 10kN:= FC 20kN:= FD 30kN:= Solving [1], [2] and [3]: FB FC FD ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find FB FC, FD,( ):= FB FC FD ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ 0.2712 0.8136 1.8983 ⎛⎜⎜⎝ ⎞ ⎠ kN= Ans Problem 4-62 The bracket is held to the wall using three A-36 steel bolts at B, C, and D. Each bolt has a diameter of 12.5 mm and an unstretched length of 50 mm. If a force of 4 kN is placed on the bracket as shown, determine how far, s, the top bracket at bolt D moves away from the wall. For the calculation, assume that the bolts carry no shear; rather, the vertical force of 4 kN is supported by the toe at A. Also, assume that the wall and bracket are rigid. A greatly exaggerated deformation of the bolts is shown. Given: a 12.5mm:= b 25mm:= c 50mm:= d 12.5mm:= e 50mm:= L 50mm:= P 4kN:= E 200GPa:= Solution: LAD a b+ c+:= LAC a b+:= LAB a:= A π 4 d2⋅:= Compatibility: δD LAD δC LAC = FD( ) L⋅ LAD( ) A⋅ E⋅ FC( ) L⋅ LAC( ) A⋅ E⋅= δB LAB δC LAC = FB( ) L⋅ LAB( ) A⋅ E⋅ FC( ) L⋅ LAC( ) A⋅ E⋅= Given FD LAD LAC ⎛⎜⎝ ⎞ ⎠ FC⋅= [1] FB LAB LAC ⎛⎜⎝ ⎞ ⎠ FC⋅= [2] Equations of equilibrium: ΣΜA=0; FD LAD( )⋅ FC LAC( )⋅+ FB LAB( )⋅+ P e( )⋅− 0= [3] Initial guess: FB 10kN:= FC 20kN:= FD 30kN:= Solving [1], [2] and [3]: FB FC FD ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find FB FC, FD,( ):= FB FC FD ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ 0.2712 0.8136 1.8983 ⎛⎜⎜⎝ ⎞ ⎠ kN= Displacement: δD FD( ) L⋅ A E⋅:= δD 0.003867 mm= Ans Problem 4-63 The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a stiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the force in each post after the load is applied to the bar. Given: L 1m:= Lk 1.02m:= b 2m:= w 50 kN m := k 2 103( )⋅ kN m ⋅:= A 600mm2:= E 9.65GPa:= Solution: ∆Lk Lk L−:= ∆Lk 0.02 m= Compatibility: + δA ∆Lk+ δk= [1] Equations of equilibrium: ΣΜC=0; FB b( )⋅ FA b( )⋅− 0= FA FB= + ΣFy=0; FA FB+ Fsp+ w b( )⋅− 0= Given 2FA Fsp+ w b( )⋅− 0= [2] [1']From [1]: FA( ) L⋅ E A⋅ ∆Lk+ Fsp k = Initial guess: FA 1kN:= Fsp 2kN:= Solving [2] and [1']: FA Fsp ⎛⎜⎜⎝ ⎞ ⎠ Find FA Fsp,( ):= FA Fsp ⎛⎜⎜⎝ ⎞ ⎠ 25.58 48.84 ⎛⎜⎝ ⎞ ⎠ kN= Ans FB FA:= FB 25.58 kN= Ans Problem 4-64 The rigid bar is supported by the two short white spruce wooden posts and a spring. If each of the posts has an unloaded length of 1 m and a cross-sectional area of 600 mm2, and the spring has a stiffness of k = 2 MN/m and an unstretched length of 1.02 m, determine the vertical displacement of A and B after the load is applied to the bar. Given: L 1m:= Lk 1.02m:= b 2m:= w 50 kN m := k 2 103( )⋅ kN m ⋅:= A 600mm2:= E 9.65GPa:= Solution: ∆Lk Lk L−:= ∆Lk 0.02 m= Compatibility: + δA ∆Lk+ δk= [1] Equations of equilibrium: ΣΜC=0; FB b( )⋅ FA b( )⋅− 0= FA FB= + ΣFy=0; FA FB+ Fsp+ w b( )⋅− 0= Given 2FA Fsp+ w b( )⋅− 0= [2] [1']From [1]: FA( ) L⋅ E A⋅ ∆Lk+ Fsp k = Initial guess: FA 1kN:= Fsp 2kN:= Solving [2] and [1']: FA Fsp ⎛⎜⎜⎝ ⎞ ⎠ Find FA Fsp,( ):= FAFsp ⎛⎜⎜⎝ ⎞ ⎠ 25.58 48.84 ⎛⎜⎝ ⎞ ⎠ kN= FB FA:= FB 25.58 kN= Displacement: δA FA( ) L⋅ A E⋅:= δA 4.418 mm= Ans δB FB( ) L⋅ A E⋅:= δB 4.418 mm= Ans Problem 4-65 The wheel is subjected to a force of 18 kN from the axle. Determine the force in each of the three spokes. Assume the rim is rigid and the spokes are made of the same material, and each has the same cross-sectional area. Given: L 0.4m:= θ 120deg:= P 18kN:= Solution: φ 180deg θ−:= Compatibility: δAB cos φ( )⋅ δAC= FAB( ) L⋅ E A⋅ cos φ( )⋅ FAC( ) L⋅ E A⋅= [1] Equations of equilibrium: At A : + ΣFx=0; FAC sin φ( )⋅ FAD sin b( )⋅− 0= FAC FAD= + ΣFy=0; FAB FAC cos φ( )⋅+ FAD cos φ( )⋅+ P− 0= Given FAB 2 FAC cos φ( )⋅( )+ P− 0= [2] From [1]: FAB cos φ( )⋅ FAC= [1'] Initial guess: FAB 1kN:= FAC 2kN:= Solving [2] and [1']: FAB FAC ⎛⎜⎜⎝ ⎞ ⎠ Find FAB FAC,( ):= FABFAC ⎛⎜⎜⎝ ⎞ ⎠ 12 6 ⎛⎜⎝ ⎞ ⎠ kN= Ans FAD FAC:= FAD 6 kN= Ans Problem 4-66 The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally uncompressed, determine the reactions at A and B when the force P = 40 kN is applied to the collar. Given: LCA 0.25m:= P 40kN:= d 50mm:= LBC 0.25m:= E 68.9GPa:= k 200 103( ) kNm:= Solution: A π 4 d2⋅:= kCA E A⋅ LCA := kBC E A⋅ LBC := Equations of equilibrium: + ΣFy=0; FA P− Fk+ FB+ 0= [1] Compatibility: Considera a combined force F (=FB+Fk) acted at free end B . ∆B δP δF−= ∆B 0= Given 0 P kBC FB Fk+ kCA FB Fk+ kBC k+ +⎛⎜⎝ ⎞ ⎠ −= [2] Also, ∆sp=∆BC Fk k FB Fk+ kBC k+ = [3] Initial guess: FB 1kN:= Fk 2kN:= Solving [1] and [2]: FB Fk ⎛⎜⎜⎝ ⎞ ⎠ Find FB Fk,( ):= FBFk ⎛⎜⎜⎝ ⎞ ⎠ 16.88 6.24 ⎛⎜⎝ ⎞ ⎠ kN= Ans From [1]: FA P Fk− FB−:= FA 16.88 kN= Ans Problem 4-67 The post is made from 6061-T6 aluminum and has a diameter of 50 mm. It is fixed supported at A and B, and at its center C there is a coiled spring attached to the rigid collar. If the spring is originally uncompressed, determine the compression in the spring when the load of P = 50 kN is applied to the collar. Given: LCA 0.25m:= P 50kN:= d 50mm:= LBC 0.25m:= E 68.9GPa:= k 200 103( ) kNm:= Solution: A π 4 d2⋅:= kCA E A⋅ LCA := kBC E A⋅ LBC := Equations of equilibrium: + ΣFy=0; FA P− Fk+ FB+ 0= [1] Compatibility: Considera a combined force F (=FB+Fk) acted at free end B . ∆B δP δF−= ∆B 0= Given 0 P kBC FB Fk+ kCA FB Fk+ kBC k+ +⎛⎜⎝ ⎞ ⎠ −= [2] Also, ∆sp=∆BC Fk k FB Fk+ kBC k+ = [3] Initial guess: FB 1kN:= Fk 2kN:= Solving [1] and [2]: FB Fk ⎛⎜⎜⎝ ⎞ ⎠ Find FB Fk,( ):= FBFk ⎛⎜⎜⎝ ⎞ ⎠ 21.101 7.799 ⎛⎜⎝ ⎞ ⎠ kN= Thus, ∆sp Fk k := ∆sp 0.03899 mm= Ans Problem 4-68 The rigid bar supports the uniform distributed load of 90 kN/m. Determine the force in each cable if each cable has a cross-sectional area of 36 mm2, and E = 200 GPa. Given: a 1m:= b 2m:= A 36mm2:= E 200GPa:= w 90 kN m := Solution: c a2 b2+:= v b c := h a c := LAC 4a 2 b2+:= L 3a:= LBC c:= LDC c:= Equations of equilibrium: ΣΜA=0; TBC v( )⋅ a⋅ TDC v( )⋅ L⋅+ w L⋅ 0.5L( )⋅− 0= [1] Compatibility: In triangle AB'C: LB'C 2 a2 LAC 2+ 2 a⋅ LAC( )⋅ cos θ'A( )⋅−= In triangle AD'C: LD'C 2 L2 LAC 2+ 2 L⋅ LAC( )⋅ cos θ'A( )⋅−= Thus, eliminating cos θ'A( ) : LD'C 2 L LB'C 2 a − L a− LAC2 1 L 1 a −⎛⎜⎝ ⎞ ⎠⋅+= LD'C 2 3LB'C 2− 6a2 2LAC2−= 3LB'C 2 LD'C 2− 2a2 2b2+= 3LB'C 2 LD'C 2− 2c2= But, LB'C LBC ∆BC+= LD'C LDC ∆DC+= LB'C c ∆BC+= LD'C c ∆DC+= Thus, 3 c ∆BC+( )2 c ∆DC+( )2− 2c2= Neglect squares of ∆'s since small strain occurs: 3 c2 2c∆BC+⎛⎝ ⎞⎠ c2 2c∆DC+⎛⎝ ⎞⎠− 2c2= 3 c2 2c∆BC+⎛⎝ ⎞⎠ c2 2c∆DC+⎛⎝ ⎞⎠− 2c2= 3∆BC ∆DC− 0= 3 TBC c( )⋅ E A⋅ TDC c( )⋅ E A⋅− 0= TDC 3TBC= [2] Substituting [2] into [1]: TBC v( )⋅ a⋅ 3TBC v( )⋅ L⋅+ w L⋅ 0.5L( )⋅− 0= TBC 9a c⋅ 20 b⋅ ⎛⎜⎝ ⎞ ⎠ w⋅:= TBC 45.2804 kN= Ans From [2] : TDC 3TBC:= TDC 135.8411 kN= Ans Problem 4-69 The rigid bar is originally horizontal and is supported by two cables each having a cross-sectional area of 36 mm2, and E = 200 GPa. Determine the slight rotation of the bar when the uniform load is applied. Given: a 1m:= b 2m:= A 36mm2:= E 200GPa:= w 90 kN m := Solution: c a2 b2+:= v b c := h a c := LAC 4a 2 b2+:= L 3a:= LBC c:= LDC c:= Equations of equilibrium: ΣΜA=0; TBC v( )⋅ a⋅ TDC v( )⋅ L⋅+ w L⋅ 0.5L( )⋅− 0= [1] Compatibility: See solution of Prob. 4-68. TDC 3TBC= [2] Solving [1] and [2]: TDC 27.1682 kN:= ∆DC TDC c( )⋅ E A⋅:= ∆DC 8.4374919 mm= Geometry: tan θA( ) b2a= θA atan b2a⎛⎜⎝ ⎞⎠:= θA 45.000 deg= In triangle AD'C: LD'C 2 L2 LAC 2+ 2 L⋅ LAC( )⋅ cos θ'A( )⋅−= LDC ∆DC+( )2 L2 LAC2+ 2 L⋅ LAC( )⋅ cos θ'A( )⋅−= θ'A acos L2 LAC 2+ LDC ∆DC+( )2− 2 L⋅ LAC( )⋅ ⎡⎢⎢⎣ ⎤⎥⎥⎦ := θ'A 45.180 deg= ∆θ θ'A θA−:= ∆θ 0.180 deg= Ans Problem 4-70 The electrical switch closes when the linkage rods CD and AB heat up, causing the rigid arm BDE both to translate and rotate until contact is made at F. Originally, BDE is vertical, and the temperature is 20°C. If AB is made of bronze C86100 and CD is made of aluminum 6061-T6, determine the gap s required so that the switch will close when the temperature becomes 110°C. Unit used: °C deg:= Given: LBD 400mm:= LDE 200mm:= L 300mm:= T1 20°C:= T2 110°C:= αcu 17.0 10 6−( )⋅ 1°C:= αst 24.0 10 6−( )⋅ 1°C:= Solution: ∆T T2 T1−:= ∆T 90 °C= Thermal Expansion: δAB αcu ∆T( )⋅ L⋅:= δAB 0.4590 mm= δCD αst ∆T( )⋅ L⋅:= δCD 0.6480 mm= Geomotry: θ δCD δAB− LBD := s δAB θ LBD LDE+( )⋅+:= s 0.7425 mm= Ans Problem 4-71 A steel surveyor's tape is to be used to measure the length of a line. The tape has a rectangular cross section of 1.25 mm by 5 mm and a length of 30 m when T1 = 20°C and the tension or pull on the tape is 100 N. Determine the true length of the line if the tape shows the reading to be 139 m when used with a pull of 175 N at T2 = 40°C. The ground on which it is placed is flat. αst = 17(10-6)/°C, Est = 200 GPa. Unit used: °C deg:= Given: a 5mm:= b 1.25mm:= T1 20°C:= P1 100N:= L1 30m:= T2 40°C:= P2 175N:= L2 139m:= αst 17 10 6−( )⋅ 1°C:= Est 200GPa:= Solution: A a b⋅:= ∆T T2 T1−:= ∆T 20 °C= Thermal Expansion: δT αst ∆T( )⋅ L2⋅:= δT 47.2600 mm= δT 0.0473 m= δP P2 P1−( ) L2⋅ A Est⋅ := δP 8.3400 mm= δP 0.0083 m= L' L2 δT+ δP+:= L' 139.056 m= Ans Problem 4-72 The assembly has the diameters and material make-up indicated. If it fits securely between its fixed supports when the temperature is T1 = 20°C, determine the average normal stress in each material when the temperature reaches T2 = 40°C. Unit used: °C deg:= Given: T1 20°C:= T2 40°C:= L1 1.2m:= d1 300mm:= L2 1.8m:= d2 200mm:= L3 0.9m:= d3 100mm:= α1 23 10 6−( )⋅ 1°C:= α2 17 10 6−( )⋅ 1°C:= α3 17 10 6−( )⋅ 1°C:= E1 73.1GPa:= E2 103GPa:= E3 193GPa:= Solution: ∆T T2 T1−:= ∆T 20 °C= A1 π 4 ⎛⎜⎝ ⎞ ⎠ d1 2⋅:= A2 π 4 ⎛⎜⎝ ⎞ ⎠ d2 2⋅:= A3 π 4 ⎛⎜⎝ ⎞ ⎠ d3 2⋅:= Equations of equilibrium: + ΣFx=0; FA FD− 0= FA FD= Let FA=F. Then, FA = FAB = FBC = FCD = FD = F Thermal Expansion: δT α1 ∆T( )⋅ L1⋅ α2 ∆T( )⋅ L2⋅+ α3 ∆T( )⋅ L3⋅+:= δT 1.470000 mm= Elongation: negative sign indicates shortening δF F L1⋅ A1 E1⋅ − F L2⋅ A2 E2⋅ − F L3⋅ A3 E3⋅ −= Compatibility: 0 δT δF+= Given 0 δT F L1⋅ A1 E1⋅ − F L2⋅ A2 E2⋅ − F L3⋅ A3 E3⋅ −= [1] Initial guess: F 1kN:= Solving [1] : F Find F( ):= F 1063.49 kN= Average Normal Stress: σal F A1 := σal 15.05 MPa= Ans σbr F A2 := σbr 33.85 MPa= Ans σst F A3 := σst 135.41 MPa= Ans Problem 4-73 A high-strength concrete driveway slab has a length of 6 m when its temperature is 10°C. If there is a gap of 3 mm on one side before it touches its fixed abutment, determine the temperature required to close the gap. What is the compressive stress in the concrete if the temperature becomes 60°C? Unit used: °C deg:= Given: α 11 10 6−( )⋅ 1°C:=L 6m:= T1 10°C:= ∆gap 3mm:= T2 60°C:= E 29GPa:= Solution: Require, ∆gap α T' T1−( )⋅ L⋅= T'1 ∆gap α L⋅ ⎛⎜⎝ ⎞ ⎠ T1+:= T'1 55.45 °C= Ans For ∆T T2 T1−:= ∆T 50 °C= ∆gap δT δF−= ∆gap α ∆T( )⋅ L⋅ F L⋅A E⋅−= F A α ∆T( )⋅ E⋅ ∆gap EL⎛⎜⎝ ⎞ ⎠⋅−= σ F A = σ α ∆T( )⋅ E⋅ ∆gap EL⎛⎜⎝ ⎞ ⎠⋅−:= σ 1.45 MPa= Ans Problem 4-74 A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbines A and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm.The connection was made at T1 = 20°C. If the turbines' points of attachment are assumed rigid, determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 135°C. Unit used: °C deg:= Given: L 1.8m:= T1 20°C:= T2 135°C:= σY 280MPa:= do 100mm:= t 6mm:= E 200GPa:= α 12 10 6−( )⋅ 1°C:= Solution: di do 2t−:= A π 4 do 2 di 2−⎛⎝ ⎞⎠⋅:= ∆T T2 T1−:= ∆T 115 °C= Compatibility: 0 δT δF+= 0 α ∆T( )⋅ L⋅ F L⋅ E A⋅−= F α ∆T( )⋅ E⋅ A⋅:= F 489.03 kN= Ans Check stress: σ F A := σ 276.00 MPa= (< σY = 280 MPa) ok. Problem 4-75 A 1.8-m-long steam pipe is made of steel with σY = 280 MPa. It is connected directly to two turbines A and B as shown.The pipe has an outer diameter of 100 mm and a wall thickness of 6 mm. The connection was made at T1 = 20°C. If the turbines' points of attachment are assumed to have a stiffness of k = 16 MN/m., determine the force the pipe exerts on the turbines when the steam and thus the pipe reach a temperature of T2 = 135°C. Unit used: °C deg:= Given: L 1.8m:= T1 20°C:= T2 135°C:= σY 280MPa:= do 100mm:= t 6mm:= E 200GPa:= α 12 10 6−( )⋅ 1°C:= k 16 103( )⋅ kNmm:= Solution: di do 2t−:= A π 4 do 2 di 2−⎛⎝ ⎞⎠⋅:= ∆T T2 T1−:= ∆T 115 °C= Compatibility: 2x δT δF+= 2x α ∆T( )⋅ L⋅ k x⋅( ) L⋅ E A⋅−= x α ∆T( )⋅ E⋅ A⋅ L⋅ k L⋅ 2E A⋅+:= x 0.029830 mm= F k x⋅:= F 477.29 kN= Ans Check stress: σ F A := σ 269.37 MPa= (< σY = 280 MPa) ok. Problem 4-76 The 12-m-long A-36 steel rails on a train track are laid with a small gap δ between them to allow forthermal expansion. Determine the required gap so that the rails just touch one another when the temperature is increased from T1 = -30°C to T2 = 30°C. Using this gap, what would be the axial force in the rails if the temperature were to rise to T3 = 40°C? The cross-sectional area of each rail is 3200 mm2. Unit used: °C deg:= Given: L 12m:= A 3200mm2:= T1 30− °C:= T2 30°C:= T3 40°C:= E 200GPa:= α 12 10 6−( )⋅ 1°C:= Solution: Require, ∆gap α T2 T1−( )⋅ L⋅:= ∆gap 8.640 mm= Ans For ∆T T3 T1−:= ∆T 70 °C= Compatibility: ∆gap δT δF+= ∆gap α ∆T( )⋅ L⋅ F L⋅E A⋅−= F α ∆T( )⋅ L⋅ ∆gap−⎡⎣ ⎤⎦ E A⋅L⋅:= F 76.80 kN= Ans Problem 4-77 The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid walls such that there is a gap of 0.2 mm between them when T1 = 15°C. What larger temperature T2 is required in order to just close the gap? Each rod has a diameter of 30 mm, αal = 24(10-6)/°C, Eal = 70 GPa, αcu = 17(10-6)/°C, Ecu = 126 GPa. Determine the average normal stress in each rod if T2 = 95°C. Unit used: °C deg:= Given: Lcu 100mm:= Lal 200mm:= d 30mm:= T1 15°C:= T2 95°C:= ∆gap 0.2mm:= Ecu 126GPa:= αcu 17 10 6−( )⋅ 1°C:= Eal 70GPa:= αal 24 10 6−( )⋅ 1°C:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= Require, ∆gap αcu T' T1−( )⋅ Lcu⋅ αal T' T1−( )⋅ Lal⋅+= T' T1 ∆gap αcu Lcu⋅ αal Lal⋅+ +:= T' 45.77 °C= Ans For ∆T T2 T1−:= ∆T 80 °C= Compatibility: δcu δ1T δ1F+= δal δ2T δ2F+= ∆gap δcu δal+= ∆gap αcu ∆T( )⋅ Lcu⋅ F Lcu⋅ Ecu A⋅ − αal ∆T( )⋅ Lal⋅+ F Lal⋅ Eal A⋅ −= F αcu ∆T( )⋅ Lcu⋅ αal ∆T( )⋅ Lal⋅+ ∆gap− Lcu Ecu A⋅ Lal Eal A⋅ + := F 61.958 kN= Ans Average Normal Stress: σ F A := σ 87.65 MPa= Ans Problem 4-78 The two circular rod segments, one of aluminum and the other of copper, are fixed to the rigid walls such that there is a gap of 0.2 mm between them when T1 = 15°C. Each rod has a diameter of 30 mm, αal = 24(10-6)/°C, Eal = 70 GPa, αcu = 17(10-6)/°C, Ecu = 126 GPa. Determine the average normal stress in each rod if T2 = 150°C, and also calculate the new length of the aluminum segment. Unit used: °C deg:= Given: Lcu 100mm:= Lal 200mm:= d 30mm:= T1 15°C:= T2 150°C:= ∆gap 0.2mm:= Ecu 126GPa:= αcu 17 10 6−( )⋅ 1°C:= Eal 70GPa:= αal 24 10 6−( )⋅ 1°C:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= For ∆T T2 T1−:= ∆T 135 °C= Compatibility: δcu δ1T δ1F+= δal δ2T δ2F+= ∆gap δcu δal+= ∆gap αcu ∆T( )⋅ Lcu⋅ F Lcu⋅ Ecu A⋅ − αal ∆T( )⋅ Lal⋅+ F Lal⋅ Eal A⋅ −= F αcu ∆T( )⋅ Lcu⋅ αal ∆T( )⋅ Lal⋅+ ∆gap− Lcu Ecu A⋅ Lal Eal A⋅ + := F 131.176 kN= Average Normal Stress: σ F A := σ 185.58 MPa= Ans Displacement: δal αal ∆T( )⋅ Lal⋅ F Lal⋅ Eal A⋅ −:= δal 0.117783 mm= L'al Lal δal+:= L'al 200.117783 mm= Ans Problem 4-79 Two bars, each made of a different material, are connected and placed between two walls when the temperature is T1 = 10°C. Determine the force exerted on the (rigid) supports when the temperature becomes T2 = 20°C. The material properties and cross-sectional area of each bar are given in the figure. Unit used: °C deg:= Given: T1 10°C:= T2 20°C:= L 300mm:= Ast 200mm 2:= Abr 450mm2:= αst 12 10 6−( )⋅ 1°C:= αbr 21 10 6−( )⋅ 1°C:= Est 200GPa:= Ebr 100GPa:= Solution: ∆T T2 T1−:= ∆T 10 °C= Equations of equilibrium: + ΣFx=0; FA FC− 0= FA FC= Let FA=F. Then, FA = FAB = FBC = FC = F Compatibility: 0 δT δF+= 0 αst ∆T( )⋅ L⋅ αbr ∆T( )⋅ L⋅+ F L⋅Ast Est⋅− F L⋅ Abr Ebr⋅ −= F αst αbr+( ) ∆T⋅ 1 Ast Est⋅ 1 Abr Ebr⋅ + := F 6.988 kN= Ans Problem 4-80 The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance heating. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of and 375 mm2. Est = 200 GPa, Eal = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C. Unit used: °C deg:= Given: T1 30°C:= T2 180°C:= ∆gap 0.7mm:= Ast 125mm 2:= Aal 375mm2:= αst 12 10 6−( )⋅ 1°C:= αal 23 10 6−( )⋅ 1°C:= Est 200GPa:= Eal 70GPa:= Lst 300mm:= Lal 240mm:= Solution: ∆T T2 T1−:= ∆T 150 °C= Equations of equilibrium: ΣΜC=0; FAB b( )⋅ FEF b( )⋅− 0= FAB FEF= Let FAB= Fst. Then, FAB = FEF = Fst + ΣFy=0; FAB FEF+ Fal− 0= Fal 2Fst= [1] Compatibility: δst δal ∆gap−= Fst Lst⋅ Est Ast⋅ αal ∆T( )⋅ Lal⋅ Fal Lal⋅ Eal Aal⋅ −⎡⎢⎣ ⎤⎥⎦ ∆gap−= [2] Substituting [1] into [2]: Fst αal ∆T( )⋅ Lal ∆gap− Lst Ast Est⋅ 2Lal Aal Eal⋅ + := Fst 4.226 kN= Ans From [1]: Fal 2Fst:= Fal 8.453 kN= Ans Problem 4-81 The center rod CD of the assembly is heated from T1 = 30°C to T2 = 180°C using electrical resistance heating. Also, the two end rods AB and EF are heated from T1 = 30°C to T2 = 50°C. At the lower temperature T1 the gap between C and the rigid bar is 0.7 mm. Determine the force in rods AB and EF caused by the increase in temperature. Rods AB and EF are made of steel, and each has a cross-sectional area of 125 mm2. CD is made of aluminum and has a cross-sectional area of and 375 mm2. Est = 200 GPa, Eal = 70 GPa, αst = 12(10-6)/°C, αal = 23(10-6)/°C. Unit used: °C deg:= Given: T1 30°C:= T2 180°C:= T'2 50°C:= Ast 125mm 2:= Aal 375mm2:= αst 12 10 6−( )⋅ 1°C:= αal 23 10 6−( )⋅ 1°C:= Est 200GPa:= Eal 70GPa:= ∆gap 0.7mm:= Lst 300mm:= Lal 240mm:= Solution: ∆T T2 T1−:= ∆T 150 °C= ∆T' T'2 T1−:= ∆T' 20 °C= Equations of equilibrium: ΣΜC=0; FAB b( )⋅ FEF b( )⋅− 0= FAB FEF= Let FAB= Fst. Then, FAB = FEF = Fst + ΣFy=0; FAB FEF+ Fal− 0= Fal 2Fst= [1] Compatibility: δst δal ∆gap−= αst ∆T'( )⋅ Lst⋅ Fst Lst⋅ Est Ast⋅ ⎛⎜⎝ ⎞ ⎠ + αal ∆T( )⋅ Lal⋅ Fal Lal⋅ Eal Aal⋅ −⎡⎢⎣ ⎤⎥⎦ ∆gap−= [2] Substituting [1] into [2]: Fst αal ∆T( )⋅ Lal ∆gap− αst ∆T'( )⋅ Lst⋅− Lst Ast Est⋅ 2Lal Aal Eal⋅ + := Fst 1.849 kN= Ans From [1]: Fal 2Fst:= Fal 3.698 kN= Ans Problem 4-82 The pipe is made of A-36 steel and is connected to the collars at A and B.When the temperature is 15° C, there is no axial load in the pipe. If hot gas traveling through the pipe causes its temperature to rise by ∆T = (20 + 30x)°C, where x is in meter, determine the average normal stress in the pipe. The inner diameter is 50 mm, the wall thickness is 4 mm. Unit used: °C deg:= Given: L 2.4m:= di 50m:= T1 15°C:= t 4mm:= ∆T 20 30x+( )°C= E 200GPa:= α 12 10 6−( )⋅ 1°C:= Solution: do di 2t+:= A π 4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Compatibility: 0 δT δF+= 0 0 L xα ∆T( )⋅⌠⎮⌡ d F L⋅ E A⋅−= unit 1m °C⋅:= Lx L m := F α E⋅ A⋅ L ⎛⎜⎝ ⎞ ⎠ unit( )⋅ 0 Lx x20 30x+( )⌠⎮⌡ d⋅:= F 84452.766 kN= Average Normal Stress: σ F A := σ 134.40 MPa= Ans Problem 4-83 The bronze 86100 pipe has an inner radius of 12.5 mm and a wall thickness of 5 mm. If the gas flowing through it changes the temperature of the pipe uniformly from TA = 60°C at A to TB = 15°C at B, determine the axial force it exerts on the walls. The pipe was fitted between the walls when T = 15°C. Unit used: °C deg:= Given: L 2.4m:= ri 12.5mm:= TA 60°C:= t 5mm:= TB 15°C:= To 15°C:= E 103GPa:= α 17 10 6−( )⋅ 1°C:= Solution: ro ri t+:= A π ro2 ri2−⎛⎝ ⎞⎠⋅:= Tx 60 TB TA− L x⋅+⎛⎜⎝ ⎞ ⎠°C= Tx 60 18.75x−( )°C= ∆T Tx To−( )= ∆T 45 18.75x−( )°C= Compatibility: 0 δT δF+= 0 0 L xα ∆T( )⋅⌠⎮⌡ d F L⋅ E A⋅−= unit 1m °C⋅:= Lx L m := F α E⋅ A⋅ L ⎛⎜⎝ ⎞ ⎠ unit( )⋅ 0 Lx x45 18.75x−( )⌠⎮⌡ d⋅:= F 18.566 kN= Ans Problem 4-84 The rigid block has a weight of 400 kN and is to be supported by posts A and B, which are made of A-36 steel, and the post C, which is made of C83400 red brass. If all the posts have the same original length before they are loaded, determine the average normal stress developed in each post when post C is heated so that its temperature is increased by 10°C. Each post has a cross-sectional area of 5000 mm2. Unit used: °C deg:= Given: ∆T 10°C:= A 5000mm2:= b 1m:= Est 200GPa:= Ebr 101GPa:= W 400kN:= αbr 18 10 6−( )⋅ 1°C:= Solution: Set: L 1m:= Equations of equilibrium: ΣΜC=0; FA b( )⋅ FB b( )⋅− 0= FA FB= Let FA= Fst. Then, FA = FB = Fst + ΣFy=0; FA FB+ Fbr+ W− 0= Given Fbr W 2Fst−= [1] Compatibility: δst = δbr Fst L⋅ Est A⋅ Fbr L⋅ Ebr A⋅ αbr ∆T( )⋅ L⋅−= [2] Initial guess: Fst 1kN:= Fbr 2kN:= Solving [1] and [2]: Fst Fbr ⎛⎜⎜⎝ ⎞ ⎠ Find Fst Fbr,( ):= FstFbr ⎛⎜⎜⎝ ⎞ ⎠ 123.393 153.214 ⎛⎜⎝ ⎞ ⎠ kN= Ans Average Normal Stress: σst Fst A := σst 24.68 MPa= Ans σbr Fbr A := σbr 30.64 MPa= Ans Problem 4-85 The bar has a cross-sectional area A, length L, modulus of elasticity E, and coefficient of thermal expansion α. The temperature of the bar changes uniformly along its length from TA at A to TB at B so that at any point x along the bar T = TA + x (TB - TA)/L. Determine the force the bar exerts on the rigid walls. Initially no axial force is in the bar. Solution: Compatibility: 0 δT δF+= [1] Tx TA TB TA− L x⋅+= ∆T Tx TA−( )= ∆T TB TA−( ) xL⋅= d ∆T( ) TB TA−L x⋅= However, d δT( ) α ∆T( )⋅ dx= d δT( ) α TB TA−( ) xL⋅⎡⎢⎣ ⎤⎥⎦⋅ dx= δT 0 L xα TB TA−( ) xL⋅⎡⎢⎣ ⎤⎥⎦⋅ ⌠⎮ ⎮⌡ d= δT α L⋅ 2 TB TA−( )⋅= From [1]: 0 α L⋅ 2 TB TA−( )⋅ F L⋅E A⋅−= F α E⋅ A⋅ 2 TB TA−( )⋅= Ans Problem 4-86 The rod is made of A-36 steel and has a diameter of 6 mm. If the springs are compressed 12 mm when the temperature of the rod is T = 10°C, determine the force in the rod when its temperature is T = 75°C. Unit used: °C deg:= Given: L 1.2m:= d 6mm:= k 200 N mm := T1 10°C:= T2 75°C:= xo 12mm:= E 200GPa:= α 12 10 6−( )⋅ 1°C:= Solution: A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= For ∆T T2 T1−:= ∆T 65 °C= F k x xo+( )⋅= Compatibility: 2x δT δF+= 2x α ∆T( )⋅ L⋅ F L⋅ E A⋅−= x α ∆T( )⋅ L 2 ⋅ k x xo+( )⋅ L⋅ 2E A⋅−= x α ∆T( )⋅ E⋅ A⋅ L⋅ k xo⋅ L⋅− 2E A⋅ k L⋅+:= x 0.20892 mm= F k x xo+( )⋅:= F 2.442 kN= Ans Problem 4-87 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN. Given: P 8kN:= d 20mm:= w 40mm:= h 20mm:= t 5mm:= r 10mm:= Solution: For the fillet: A h t⋅:= w h 2= r h 0.5= From Fig. 4-24, K 1.4:= σmax K σavg⋅= σmax K P A ⋅:= σmax 112 MPa= Ans For the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.25= From Fig. 4-25, Ko 2.375:= σmax K σavg⋅= σmax Ko P Ao ⋅:= σmax 190 MPa= Ans Problem 4-88 If the allowable normal stress for the bar is σallow = 120 MPa, determine the maximum axial force P that can be applied to the bar. Given: d 20mm:= t 5mm:= r 10mm:= w 40mm:= h 20mm:= σallow 120MPa:= Solution: Assume failure of the fillet: A h t⋅:= w h 2= r h 0.5= From Fig. 4-24, K 1.4:= σallow σmax= σmax K σavg⋅= σallow K P A ⋅= P σallow( ) A⋅ K := P 8.57 kN= Assume failure of the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.25= From Fig. 4-25, Ko 2.375:= σallow σmax= σmax K σavg⋅= σmax Ko P Ao ⋅:= P σallow( ) Ao⋅ Ko := P 5.05 kN= Ans (Controls!) Problem 4-89 The steel bar has the dimensions shown. Determine the maximum axial force P that can be applied so as not to exceed an allowable tensile stress of σallow =150 MPa. Given: d 24mm:= t 20mm:= r 15mm:= w 60mm:= h 30mm:= σallow 150MPa:= Solution: Assume failure occurs at the fillet: A h t⋅:= w h 2= r h 0.5= From Fig. 4-24, K 1.4:= σallow σmax= σmax K σavg⋅= σallow K P A ⋅= P σallow( ) A⋅ K := P 64.29 kN= Assume failure occrs at the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.2= From Fig. 4-25, Ko 2.45:= σallow σmax= σmax K σavg⋅= σmax Ko P Ao ⋅:= P σallow( ) Ao⋅ Ko := P 44.08 kN= Ans (Controls!) Problem 4-90 Determine the maximum axial force P that can be applied to the bar. The bar is made from steel and has an allowable stress of σallow = 147 MPa. Given: d 15mm:= t 4mm:= r 5mm:= w 37.5mm:= h 25mm:= σallow 147MPa:= Solution: Assume failure of the fillet: A h t⋅:= w h 1.5= r h 0.2= From Fig. 4-24, K 1.73:= σallow σmax= σmax K σavg⋅= σallow K P A ⋅= P σallow( ) A⋅ K := P 8.497 kN= Assume failure of the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.2= From Fig. 4-25, Ko 2.45:= σallow σmax= σmax K σavg⋅= σmax Ko P Ao ⋅:= P σallow( ) Ao⋅ Ko := P 5.4 kN= Ans (Controls!) Problem 4-91 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN. Given: d 15mm:= t 4mm:= r 5mm:= w 37.5mm:= h 25mm:= P 8kN:= Solution: For the fillet: A h t⋅:= w h 1.5= r h 0.2= From Fig. 4-24, K 1.73:= σmax K σavg⋅= σmax K P A ⋅:= σmax 138.4 MPa= For the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.20= From Fig. 4-25, Ko 2.45:= σmax K σavg⋅= σmax Ko P Ao ⋅:= σmax 217.78 MPa= Ans (Controls!) Problem 4-92 Determine the maximum normal stress developed in the bar when it is subjected to a tension of P = 8 kN. Given: P 8kN:= d 12mm:= w 60mm:= h 30mm:= t 5mm:= r 15mm:= Solution: At the fillet: A h t⋅:= w h 2= r h 0.5= From Fig. 4-24, K 1.4:= σmax K σavg⋅= σmax K P A ⋅:= σmax 74.67 MPa= At the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.1= From Fig. 4-25, Ko 2.65:= σmax K σavg⋅= σmax Ko P Ao ⋅:= σmax 88.33 MPa= Ans (Controls!) Problem 4-93 The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar. Also, what is the stress-concentration factor for this geometry? Given: h 80mm:= t 10mm:= u 5MPa:= v 20mm:= Solution: P Aσ⌠⎮⌡ d= P = Volume under curve Number of sqaures, n 19:= P n u v⋅( )⋅ t⋅:= P 19.00 kN= Ans σavg P h t⋅:= σavg 23.75 MPa= Ans From the Figure, σmax 30MPa:= K σmax σavg := K 1.26= Ans Problem 4-94 The resulting stress distribution along section AB for the bar is shown. From this distribution, determine the approximate resultant axial force P applied to the bar.Also, what is the stress-concentration factor for this geometry? Given: h 80mm:= t 20mm:= u 18MPa:= v 20mm:= P Aσ⌠⎮⌡ d= P = Volume under curve Number of sqaures, n 10:= P n u v⋅( )⋅ t⋅:= P 72.00 kN= Ans σavg P h t⋅:= σavg 45.00 MPa= Ans From the Figure, σmax 72MPa:= K σmax σavg := K 1.60= Ans Solution: Problem 4-95 The A-36 steel plate has a thickness of 12 mm. If there are shoulder fillets at B and C, and σallow = 150 MPa, determine the maximum axial load P that it can support. Compute its elongation neglecting the effect of the fillets. Given: t 12mm:= r 30mm:= w 120mm:= h 60mm:= L1 200mm:= L2 800mm:= σallow 150MPa:= E 200GPa:= Solution: A1 h t⋅:= A2 w t⋅:= Maximum Normal Stress at fillet: w h 2= r h 0.5= From Fig. 4-23, K 1.4:= σallow σmax= σmax K σavg⋅= σallow K P A1 ⋅= P σallow( ) A1⋅ K := P 77.14 kN= Ans Displacement: δ δ1 δ2+ δ1+= δ 2 P L1⋅ E A1⋅ P L2⋅ E A2⋅ +:= δ 0.429 mm= Ans Problem 4-96 The 1500-kN weight is slowly set on the top of a post made of 2014-T6 aluminum with an A-36 steel core. If both materials can be considered elastic perfectly plastic, determine the stress in each material. Given: ro 50mm:= Est 200 GPa⋅:= ri 25mm:= Eal 73.1 GPa⋅:= rs 25mm:= σY_st 250MPa:= P 1500kN:= σY_al 414MPa:= Solution: Ast π rs2⎛⎝ ⎞⎠⋅:= Aal π ro2 ri2−⎛⎝ ⎞⎠⋅:= Equations of equilibrium: +Given ΣFy=0; Pal Pst+ P− 0= [1] Compatibility: δst = δal Pal( ) L⋅ Aal Eal⋅ Pst( ) L⋅ Ast Est⋅ = Pal Aal Eal⋅ Pst Ast Est⋅ = [2] Initial guess: Pal 1kN:= Pst 2kN:= Solving [1] and [2]: Pal Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pal Pst,( ):= PalPst ⎛⎜⎜⎝ ⎞ ⎠ 784.5218 715.4782 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal stress: σal Pal Aal := σal 133.18 MPa= ( < σY_al = 414MPa ) o.k.! σst Pst Ast := σst 364.39 MPa= ( > σY_st = 250MPa ) Thererfore, the steel core yields and so the elastic analysis is invalid. Plastic Analysis: The stress in the steel is σst σY_st:= σst 250 MPa= Ans Pst σY_st( )Ast:= Pst 490.87 kN= From [1]: Pal P Pst−:= Pal 1009.13 kN= σal Pal Aal := σal 171.31 MPa= ( < σY_al = 414MPa ) o.k.! Ans Problem 4-97 The 10-mm-diameter shank of the steel bolt has a bronze sleeve bonded to it. The outer diameter of this sleeve is 20 mm. If the yield stress for the steel is (σY)st = 640 MPa, and for the bronze (σY)br = 520 MPa, determine the magnitude of the largest elastic load P that can be applied to the assembly. Est = 200 GPa, Ebr = 100 GPa. Given: do 20mm:= Est 200GPa:= σY_st 640MPa:= di 10mm:= Ebr 100GPa:= σY_br 520MPa:= ds 10mm:= Solution: Ast π 4 ds 2⎛⎝ ⎞⎠⋅:= Abr π4 do 2 di 2−⎛⎝ ⎞⎠⋅:= Equations of equilibrium: + ΣFy=0; Pbr Pst+ P− 0= [1] Compatibility: δb δs= Pbr( ) L⋅ Abr Ebr⋅ Pst( ) L⋅ Ast Est⋅ = Pbr Abr Ebr⋅ Pst Ast Est⋅ = [2] Assume yielding of bolt, then Pst σY_st( )Ast:= Pst 50.265 kN= From [2]: Pbr Abr Ebr⋅ Ast Est⋅ ⎛⎜⎝ ⎞ ⎠ Pst⋅:= Pbr 75.398 kN= From [1]: P Pbr Pst+:= P 125.66 kN= (Controls!): Ans Assume yielding of sleeve, then Pbr σY_br Abr:= Pbr 122.522 kN= From [2]: Pst Ast Est⋅ Abr Ebr⋅ ⎛⎜⎝ ⎞ ⎠ Pbr⋅:= Pst 81.681 kN= From [1]: P Pbr Pst+:= P 204.20 kN= Problem 4-98 The weight is suspended from steel and aluminum wires, each having the same initial length of 3 m and cross-sectional area of 4 mm2. If the materials can be assumed to be elastic perfectly plastic, with (σY)st = 120 MPa and (σY)al = 70 MPa, determine the force in each wire if the weight is (a) 600 N and (b) 720 N. Eal = 70 GPa, Est = 200 GPa. Given: L 3m:= Est 200GPa:= σY_st 120MPa:= A 4mm2:= Eal 70GPa:= σY_al 70MPa:= Solution: (a) W 600N:= Equations of equilibrium: +Given ΣFy=0; Pal Pst+ W− 0= [1] Compatibility: δst = δal Pal( ) L⋅ A Eal⋅ Pst( ) L⋅ A Est⋅ = Pal A Eal⋅ Pst A Est⋅ = [2] Initial guess: Pal 1N:= Pst 2N:= Solving [1] and [2]: Pal Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pal Pst,( ):= PalPst ⎛⎜⎜⎝ ⎞ ⎠ 155.56 444.44 ⎛⎜⎝ ⎞ ⎠ N= Ans Average Normal stress: σal Pal A := σal 38.89 MPa= ( < σY_al = 70MPa ) o.k.! σst Pst A := σst 111.11 MPa= ( < σY_st = 120MPa ) o.k.! The average normal stress for both wires do not exceed their respective yield stress. Thererfore, the elastic analysis is valid for both wires. Solution: (b) W 720N:= Equations of equilibrium: +Given ΣFy=0; Pal Pst+ W− 0= [1] Compatibility: δst = δal Pal( ) L⋅ A Eal⋅ Pst( ) L⋅ A Est⋅ = Pal A Eal⋅ Pst A Est⋅ = [2] Initial guess: Pal 1N:= Pst 2N:= Solving [1] and [2]: Pal Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pal Pst,( ):= PalPst ⎛⎜⎜⎝ ⎞ ⎠ 186.67 533.33 ⎛⎜⎝ ⎞ ⎠ N= Average Normal stress: σal Pal A := σal 46.67 MPa= ( < σY_al = 70MPa ) o.k.! σst Pst A := σst 133.33 MPa= ( > σY_st = 120MPa ) Thererfore, the steel wire yields and so the elastic analysis is invalid. Plastic Analysis: The stress in the steel is σst σY_st:= σst 120 MPa= Pst σY_st( )A:= Pst 480.00 N= Ans From [1]: Pal W Pst−:= Pal 240.00 N= Ans σal Pal A := σal 60 MPa= ( < σY_al = 70MPa ) o.k.! Problem 4-99 The bar has a cross-sectional area of 625 mm2. If a force of P = 225 kN is applied at B and then removed, determine the residual stress in sections AB and BC. σY = 210 MPa. Given: L 1m:= LAB 0.75L:= LBC 0.25L:= A 625mm2:= σY 210MPa:= P 225kN:= Solution: By superposition : + ∆C δC− 0= P( ) LAB⋅ A E⋅ FC( ) L⋅ A E⋅− 0= FC P LAB L ⎛⎜⎝ ⎞ ⎠⋅:= FC 168.75 kN= Equations of equilibrium: + ΣFy=0; FA FC+ P− 0= FA P FC−:= FA 56.25 kN= [1] Average Normal stress: σAB FA A := σAB 90 MPa= ( < σY = 210MPa ) o.k.! σBC FC A := σBC 270 MPa= ( > σY = 210MPa ) Thererfore, the segment BC yields and so the elastic analysis is invalid. Plastic Analysis: The stress in the BC is σBC σY:= σBC 210 MPa= FC σY( )A:= FC 131.25 kN= From [1]: FA P FC−:= FA 93.75 kN= σAB FA A := σAB 150 MPa= ( < σY = 210MPa ) o.k.! A reversal of force of 45kip applied results in a reversed FC=270 MPa and FA=90 MPa, which produce σ'AB 90 MPa:= (T) σ'BC 270 MPa:= (C) Hence, ∆σAB σAB σ'AB−:= ∆σAB 60 MPa= (T) Ans ∆σBC σBC σ'BC−( )−:= ∆σBC 60 MPa= (T) Ans Problem 4-100 The bar has a cross-sectional area of 300 mm2 and is made of a material that has a stressstrain diagram that can be approximated by the two line segments shown. Determine the elongation of the bar due to the applied loading. Given: LAB 1.5m:= LBC 0.6m:= A 300mm2:= PB 40kN:= PC 25kN:= σ1 140MPa:= σ2 280MPa:= ε2 0.021 mm mm := ε1 0.001 mm mm := Solution: PBC PC:= PAB PC PB+:= Average Normal stress and Strain: For segment BC σBC PBC A := σBC 83.33 MPa= (< 140MPa ) εBC σBC σ1 ⎛⎜⎝ ⎞ ⎠ ε1⋅:= εBC 0.00060 mm mm = Average Normal stress and Strain: For segment AB σAB PAB A := σAB 216.67 MPa= (> 140MPa ) εAB ε1 σAB σ1− σ2 σ1− ⎛⎜⎝ ⎞ ⎠ ε2 ε1−( )⋅+:= εAB 0.01195 mmmm= Elongation: δAB εAB LAB⋅:= δAB 17.92857 mm= δBC εBC LBC⋅:= δBC 0.35714 mm= δTotal δAB δBC+:= δTotal 18.286 mm= Ans Problem 4-101 The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine the intensity of the distributed load w that can be placed on the beam and will just cause wire EB to yield. What is the displacement of point G for this case? For the calculation, assume that the steel is elastic perfectly plastic. Given: a 400mm:= b 250mm:= c 150mm:= d 4mm:= Lwire 800mm:= Est 200GPa:= σY 530MPa:= Solution: L a b+ c+:= A π 4 d2⋅:= Plastic Analysis: Wire CD will yield first followed by wire BE. When both wores yield, FBE σY( )A:= FBE 6.660 kN= FCD σY( )A:= FCD 6.660 kN= Equations of equilibrium: ΣΜA=0; FBE a( )⋅ FCD a b+( )⋅+ w L⋅ 0.5L( )⋅− 0= w FBE 2a L2 ⎛⎜⎝ ⎞ ⎠ ⋅ FCD 2 a b+( ) L2 ⋅+:= w 21.85 kN m = Ans Displacement: When wire BE achieves yield stress, the corresponding yield strain is εY σY Est := εY 0.002650 mm mm = δBE εY Lwire( )⋅:= δBE 2.120 mm= Geometry: δBE a δG L = δG L a δBE( )⋅:= δG 4.24 mm= Ans Problem 4-102 The rigid bar is supported by a pin at A and two steel wires, each having a diameter of 4 mm. If the yield stress for the wires is σY = 530 MPa, and Est = 200 GPa, determine (a) the intensity of the distributed load w that can be placed on the beam that will cause only one of the wires to start to yield and (b) the smallest intensity of the distributed load that will cause both wires to yield. For the calculation, assume that the steel is elastic perfectly plastic. Given: a 400mm:= b 250mm:= c 150mm:= d 4mm:= Lwire 800mm:= Est 200GPa:= σY 530MPa:= Solution: L a b+ c+:= A π 4 d2⋅:= Equations of equilibrium: ΣΜA=0; FBE a( )⋅ FCD a b+( )⋅+ w L⋅ 0.5L( )⋅− 0= [1] a) By observation, wire CD will yield first. Then, FCD σY( )A:= FCD 6.660 kN= Geometry: δBE a δCD a b+= δBE a a b+ ⎛⎜⎝ ⎞ ⎠ δCD⋅= FBE Lwire⋅ Est A⋅ a a b+ FCD Lwire⋅ Est A⋅ ⎛⎜⎝ ⎞ ⎠ ⋅= FBE a a b+ ⎛⎜⎝ ⎞ ⎠ FCD⋅:= FBE 4.0986 kN= From [1]: w FBE 2a L2 ⎛⎜⎝ ⎞ ⎠ ⋅ FCD 2 a b+( ) L2 ⋅+:= w 18.65 kN m = Ans b) When both wires yield, FBE σY( )A:= FBE 6.660 kN= FCD σY( )A:= FCD 6.660 kN= From [1]: w FBE 2a L2 ⎛⎜⎝ ⎞ ⎠ ⋅ FCD 2 a b+( ) L2 ⋅+:= w 21.85 kN m = Ans Problem 4-103 The rigid beam is supported by the three posts A,B, and C of equal length. Posts A and C have a diameter of 75 mm and are made of aluminum, for which Eal = 70 GPa and (σY)al = 20 MPa. Post B has a diameter of 20 mm and is made of brass, for which Ebr = 100 GPa and (σY)br = 590 MPa. Determine the smallest magnitude of P so that (a) only rods A and C yield and (b) all the posts yield. Given: dA 75mm:= dB 20mm:= dC 75mm:= Eal 70GPa:= σY_al 20MPa:= L 4m:= Ebr 100GPa:= σY_br 590MPa:= Solution: AA π 4 dA 2⋅:= AB π 4 dB 2⋅:= AC π 4 dC 2⋅:= Equations of equilibrium: ΣΜB=0; FC L( )⋅ P 0.5L( )⋅− FA L( )⋅− P 0.5L( )⋅+ 0= FA FC= Let FA= Fal. Then, FA = FC = Fal + ΣFy=0; FA FC+ Fbr+ 2P− 0= 2Fal Fbr+ 2P− 0= [1] a) Post A and C will yield, Fal σY_al( )AA:= Fal 88.357 kN= εal σY_al Eal := εal 0.0002857 mm mm = Compatibility condition: δal δbr= Fal Lpost⋅ Eal AA⋅ Fbr Lpost⋅ Ebr AB⋅ = Fbr Ebr AB⋅ Eal AA⋅ ⎛⎜⎝ ⎞ ⎠ Fal⋅:= Fbr 8.976 kN= σbr Fbr AB := σbr 28.57 MPa= (< σY ) o.k.! From [1]: P Fal 0.5Fbr+:= P 92.85 kN= Ans b) All the posts yield. Then, Fal σY_al( )AA:= Fal 88.357 kN= Fbr σY_br( )AB:= Fbr 185.354 kN= From [1]: P Fal 0.5Fbr+:= P 181.0 kN= Ans Problem 4-104 The rigid beam is supported by the three posts A,B, and C. Posts A and C have a diameter of 60 mm and are made of aluminum, for which Eal = 70 GPa and (σY)al = 20 MPa. Post B is made of brass, for which Ebr = 100 GPa and (σY)br = 590 MPa. If P = 130 kN, determine the largest diameter of post B so that all the posts yield at the same time. Given: dA 60mm:= dC 60mm:= P 130kN:= Eal 70GPa:= σY_al 20MPa:= L 4m:= Ebr 100GPa:= σY_br 590MPa:= Solution: AA π 4 dA 2⋅:= AC π 4 dC 2⋅:= Equations of equilibrium: ΣΜB=0; FC L( )⋅ P 0.5L( )⋅− FA L( )⋅− P 0.5L( )⋅+ 0= FA FC= Let FA= Fal. Then, FA = FC = Fal + ΣFy=0; FA FC+ Fbr+ 2P− 0= 2Fal Fbr+ 2P− 0= [1] When all the posts yield, Fal σY_al( )AA:= Fal 56.549 kN= From [1]: Fbr 2P 2Fal−:= Fbr 146.90 kN= Also, Fbr σY_br( )AB= AB π4 dB2⋅= Fbr σY_br( ) π4 dB2⋅⎛⎜⎝ ⎞⎠= dB 4Fbr π σY_br⋅ := dB 17.805 mm= Ans Problem 4-105 The rigid beam is supported by three A-36 steel wires, each having a length of 1.2 m. The cross- sectional area of AB and EF is 10 mm2, and the cross-sectional area of CD is 4 mm2. Determine the largest distributed load w that can be supported by the beam before any of the wires begin to yield. If the steel is assumed to be elastic perfectly plastic, determine how far the beam is displaced downward just before all the wires begin to yield. Given: L 1.2m:= b 1.5m:= AAB 10mm2:= AEF 10mm 2:= ACD 4mm2:= E 200 GPa⋅:= σY 250MPa:= Solution: Compatibility: Beam remains horizontal after the displacement since the loading and the system are symmetrical. δAB δCD= FAB( ) L⋅ E AAB⋅ FCD( ) L⋅ E ACD⋅ = FCD ACD AAB ⎛⎜⎝ ⎞ ⎠ FAB⋅= [1] Equations of equilibrium: ΣΜC=0; FEF b( )⋅ FAB b( )⋅− 0= FAB FEF= + ΣFy=0; FAB FCD+ FEF+ w 2b( )⋅− 0= 2 FAB⋅ w 2b( )⋅ FCD−= [2] Plastic Analysis: Assume wires AB and EF yield. FAB σY( ) AAB⋅:= FAB 2.500 kN= From [1]: FCD ACD AAB ⎛⎜⎝ ⎞ ⎠ FAB⋅:= FCD 1.000 kN= From [2]: w FAB b FCD 2 b⋅+:= w 2.0000 kN m = Plastic Analysis: Assume failure of CD. FCD σY( ) ACD⋅:= FCD 1.000 kN= From [1]: FAB AAB ACD ⎛⎜⎝ ⎞ ⎠ FCD⋅:= FAB 2.500 kN= From [2]: w FAB b FCD 2 b⋅+:= w 2.0000 kN m = The three wires AB, CD and EF yield simultaneously. Hence, w 2.00 kN m = Ans Displacement: δ FCD L⋅ E ACD⋅ := δ 1.500 mm= Ans Problem 4-106 A material has a stressstrain diagram that can be described by the curve σ = cε 1/2. Determine the deflection δ of the end of a rod made from this material if it has a length L, cross-sectional area A, and a specific weight γ. Solution: Stress-strain relationship : σ c ε⋅= σ2 c2 ε⋅= However, σ P A = and ε dδ dx = Thus, P2 A2 c2 dδ dx ⋅= dδ dx P2 c2 A2 = Since P γ A⋅ x⋅= dδ dx γ2 c2 ⎛⎜⎜⎝ ⎞ ⎠ x2⋅= Displacement: δ 0 L x γ2 c2 ⎛⎜⎜⎝ ⎞ ⎠ x2⋅ ⌠⎮ ⎮ ⎮⌡ d= δ γ 2 c2 0 L xx2 ⌠⎮⌡ d ⎛⎜⎜⎝ ⎞ ⎠ ⋅= δ γ 2 L3⋅ 3 c2⋅ = Ans Problem 4-107 Solve Prob. 4-106 if the stressstrain diagram is defined by σ = cε 3/2. Solution: Stress-strain relationship : σ c ε3⋅= σ2 c2 ε3⋅= However, σ P A = and ε dδ dx = Thus, P2 A2 c2 dδ dx ⎛⎜⎝ ⎞ ⎠ 3 ⋅= dδ dx P c A⋅ ⎛⎜⎝ ⎞ ⎠ 2 3 = Since P γ A⋅ x⋅= dδ dx γ c ⎛⎜⎝ ⎞ ⎠ 2 3 x 2 3⋅= Displacement: δ 0 L x γ c ⎛⎜⎝ ⎞ ⎠ 2 3 x 2 3⋅ ⌠⎮ ⎮ ⎮ ⎮⌡ d= δ γ c ⎛⎜⎝ ⎞ ⎠ 2 3 0 L xx 2 3 ⌠⎮ ⎮⌡ d ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ ⋅= δ 3 5 γ c ⎛⎜⎝ ⎞ ⎠ 2 3 ⋅ L 5 3⋅= Ans Problem 4-108 The bar having a diameter of 50 mm is fixed connected at its ends and supports the axial load P. If the material is elastic perfectly plastic as shown by the stressstrain diagram, determine the smallest load P needed to cause segment AC to yield. If this load is released, determine the permanent displacement of point C. Given: LAC 0.6m:= LCB 0.9m:= d 50mm:= σY 140MPa:= εY 0.001 mm mm := Solution: A π 4 d2⋅⎛⎜⎝ ⎞ ⎠:= When P is increased, the segment AC will become plastic first, then CB will become plastic. Thus, FA σY( )A:= FA 274.889 kN= FB σY( )A:= FB 274.889 kN= Equations of equilibrium: + ΣFx=0; FA FB+ P− 0= [1] From the figure, E σY εY := E 140 103× MPa=P FA FB+:= P 549.78 kN= Ans The deflection of point C is, δC εY( ) LCB⋅:= δC 0.900 mm= Consider the reverse of P on the bar. F'A 1.5F'B= [2] F'A( ) LAC⋅ A E⋅ F'B( ) LCB⋅ A E⋅= Equations of equilibrium: + ΣFx=0; F'A F'B+ P− 0= [1'] Substituting [2] into [1']: F'A 0.6P:= F'A 329.87 kN= F'B 0.4P:= F'B 219.91 kN= The deflection of point C is, δ'C F'B( ) LCB⋅ A E⋅:= δ'C 0.72000 mm= Hence, ∆δ δC δ'C−:= ∆δ 0.180 mm= Ans Problem 4-109 Determine the elongation of the bar in Prob. 4-108 when both the load P and the supports are removed. Given: LAC 0.6m:= LCB 0.9m:= d 50mm:= σY 140MPa:= εY 0.001 mm mm := Solution: A π 4 d2⋅⎛⎜⎝ ⎞ ⎠:= L LAC LCB+:= When P is increased, the segment AC will become plastic first, then CB will become plastic. Thus, FA σY( )A:= FA 274.889 kN= FB σY( )A:= FB 274.889 kN= Equations of equilibrium: + ΣFx=0; FA FB+ P− 0= [1] From the figure, E σY εY := E 140 103× MPa=P FA FB+:= P 549.78 kN= Ans The deflection of point C is, δC εY( ) LCB⋅:= δC 0.900 mm= Consider the reverse of P on the bar. F'A 1.5F'B= [2] F'A( ) LAC⋅ A E⋅ F'B( ) LCB⋅ A E⋅= Equations of equilibrium: + ΣFx=0; F'A F'B+ P− 0= [1'] Substituting [2] into [1']: F'A 0.6P:= F'A 329.87 kN= F'B 0.4P:= F'B 219.91 kN= The resultant reactions are: F''A F'A FA−:= F''A 54.978 kN= F''B F'B FB−( )−:= F''B 54.978 kN= When the supports are removed, the elongation will be δ''C F''A L⋅ A E⋅:= δ''C 0.300 mm= Ans Problem 4-110 A 6-mm-diameter steel rivet having a temperature of 800°C is secured between two plates such that at this temperature it is 50 mm long and exerts a clamping force of 1.25 kN between the plates. Determine the approximate clamping force between the plates when the rivet cools to 5°C. For the calculation, assume that the heads of the rivet and the plates are rigid. Take αst = 14(10-6)/°C, Est = 200 GPa. Is the result a conservative estimate of the actual answer? Why or why not? Unit used: °C deg:= Given: T1 800°C:= T2 5°C:= P 1.250kN:= L 50mm:= d 6mm:= αst 14 10 6−( )⋅ 1°C:= Est 200GPa:= Solution: ∆T T1 T2−:= ∆T 795 °C= A π 4 ⎛⎜⎝ ⎞ ⎠ d 2⋅:= By superposition: + 0 δT δF+= 0 αst ∆T( )⋅ L⋅ FT( ) L⋅ A Est⋅ −= FT αst( ) ∆T⋅ A Est⋅( )⋅:= FT 62.939 kN= F P FT+:= F 64.189 kN= Ans Yes. Because as the rivet cools, the plates and the rivet head will also deform. Consequently, the force FT on the rivets will not be as great. Problem 4-111 Determine the maximum axial force P that can be applied to the steel plate. The allowable stress is σallow = 150 MPa. Given: d 24mm:= t 6mm:= r 6mm:= w 120mm:= h 60mm:= σallow 150MPa:= Solution: Assume failure of the fillet: A h t⋅:= w h 2= r h 0.1= From Fig. 4-24, K 2.4:= σallow σmax= σmax K σavg⋅= σallow K P A ⋅= P σallow( ) A⋅ K := P 22.5 kN= (Controls!) Ans Assume failure of the hole: Ao w d−( ) t⋅:= ro d 2 := ro w 0.1= From Fig. 4-25, Ko 2.65:= σallow σmax= σmax K σavg⋅= σmax Ko P Ao ⋅:= P σallow( ) Ao⋅ Ko := P 32.604 kN= Problem 4-112 The rigid link is supported by a pin at A and two A-36 steel wires, each having an unstretched length of 300 mm and cross-sectional area of 7.8 mm2. Determine the force developed in the wires when the link supports the vertical load of 1.75 kN. Given: L 300mm:= A 7.8mm2:= P 1.75kN:= a 150mm:= b 100mm:= c 125mm:= Solution: Compatibility: δB b δC b c+= FB( ) L⋅ b A E⋅( )⋅ FC( ) L⋅ b c+( ) A E⋅( )⋅= Given FB b FC b c+= [1] Equations of equilibrium: ΣΜA=0; FC− b c+( )⋅ FB b( )⋅− P a( )⋅+ 0= [2] Initial guess: FC 1kN:= FB 2kN:= Solving [1] and [2]: FC FB ⎛⎜⎜⎝ ⎞ ⎠ Find FC FB,( ):= FC FB ⎛⎜⎜⎝ ⎞ ⎠ 0.974 0.433 ⎛⎜⎝ ⎞ ⎠ kN= Ans Problem 4-113 The force P is applied to the bar, which is composed of an elastic perfectly plastic material. Construct a graph to show how the force in each section AB and BC (ordinate) varies as P (abscissa) is increased. The bar has cross-sectional areas of 625 mm2 in region AB and 2500 mm2 in region BC, and σY = 210 MPa. Given: LAB 150mm:= AAB 625mm2:= LBC 50mm:= ABC 2500mm2:= σY 210MPa:= Solution: Equations of equilibrium: + ΣFx=0; P FA− FC− 0= [1] Elastic behavior: + 0 ∆C δC−= 0 P( ) LAB⋅ E AAB⋅ FC( ) LBC⋅ E ABC⋅ FC( ) LAB⋅ E AAB⋅ +⎡⎢⎣ ⎤⎥⎦ −= 0 6P FC( ) 0.5 6+( )−= FC 1213 P= [2] Substituting [2] into [1]: FA 1 13 P= [3] By comparison, segment BC will yield first. Hence, FC σY( ) ABC⋅:= FC 525 kN= From [2]: P 13 12 FC⋅:= P 568.75 kN= From [3]: FA 1 13 P⋅:= FA 43.75 kN= When segment AB yields, FA σY( ) AAB⋅:= FA 131.25 kN= FC σY( ) ABC⋅:= FC 525 kN= From [1]: P FA FC+:= P 656.25 kN= Problem 4-114 The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A and B when T1 = 25°C. If the temperature becomes T2 = -20°C, and an axial force of P = 80 kN is applied to the rigid collar as shown, determine the reactions at A and B. Unit used: °C deg:= Given: LAC 125mm:= T1 25°C:= d 12mm:= LCB 200mm:= T2 20− °C:= P 80N:= E 73.1 GPa⋅:= α 23 10 6−( )⋅ 1°C:= Solution: ∆T T2 T1−:= ∆T 45− °C= L LAC LCB+:= A π 4 d2( )⋅:= By superposition : + 0 ∆B ∆T+ δB+= 0 P( ) LAC⋅ A E⋅ α ∆T( )⋅ L⋅+ FB( ) L⋅ A E⋅+= FB P− LAC L ⎛⎜⎝ ⎞ ⎠⋅ α ∆T( )⋅ A E⋅( )⋅−:= FB 8.526 kN= Ans Equations of equilibrium: + ΣFx=0; FA− P+ FB+ 0= FA P FB+:= FA 8.606 kN= Ans Problem 4-115 The 2014-T6 aluminum rod has a diameter of 12 mm and is lightly attached to the rigid supports at A and B when T1 = 40°C. Determine the force P that must be applied to the collar so that, when T = 0°C, the reaction at B is zero. Unit used: °C deg:= Given: LAC 125mm:= T1 40°C:= d 12mm:= LCB 200mm:= T2 0°C:= FB 0kN:= E 73.1 GPa⋅:= α 23 10 6−( )⋅ 1°C:= Solution: ∆T T2 T1−:= ∆T 40− °C= L LAC LCB+:= A π 4 d2( )⋅:= By superposition : + 0 ∆B ∆T+ δB+= 0 P( ) LAC⋅ A E⋅ α ∆T( )⋅ L⋅+ FB( ) L⋅ A E⋅+= P L LAC α− ∆T( )⋅ A E⋅( )⋅ FB−⎡⎣ ⎤⎦⋅:= P 19.776 kN= Ans Problem 4-116 The A-36 steel column, having a cross-sectional area of 11250 mm2, is encased in high-strength concrete as shown. If an axial force of 300 kN is applied to the column, determine the average compressive stress in the concrete and in the steel. How far does the column shorten? It has an original length of 2.4 m. Given: ac 225mm:= bc 400mm:= Ast 11250mm2:= L 2.4m:= P 300kN:= Est 200 GPa⋅:= Ec 29 GPa⋅:= Solution: Ac ac bc⋅( ) Ast−:= Compatibility: δst δconc= Given Pst( ) L⋅ Ast Est⋅ Pc( ) L⋅ Ac Ec⋅ = [1] Equations of equilibrium: + ΣFy=0; Pst Pc+ P− 0= [2] Initial guess: Pc 1kN:= Pst 2kN:= Solving [1] and [2]: Pc Pst ⎛⎜⎜⎝ ⎞ ⎠ Find Pc Pst,( ):= Pc Pst ⎛⎜⎜⎝ ⎞ ⎠ 151.117 148.883 ⎛⎜⎝ ⎞ ⎠ kN= Average Normal Stress: σst Pst Ast := σst 13.23 MPa= Ans σc Pc Ac := σc 1.92 MPa= Ans Displacement: Either the concrete or steel can be used for the calculation. δ Pst( ) L⋅ Ast Est⋅ := δ 0.15881 mm= Ans Problem 4-117 The A-36 steel column is encased in high-strength concrete as shown. If an axial force of 300 kN is applied to the column, determine the required area of the steel so that the force is shared equally between the steel and concrete. How far does the column shorten? It has an original length of 2.4 m. Given: ac 225mm:= bc 400mm:= L 2.4m:= P 300kN:= Est 200 GPa⋅:= Ec 29 GPa⋅:= Pc 0.5P:= Pst 0.5P:= Solution: Ac ac bc⋅( ) Ast−= Compatibility: δst δconc= Pst( ) L⋅ Ast Est⋅ Pc( ) L⋅ Ac Ec⋅ = Ast Ec Ec Est+ ac bc⋅( )⋅:= Ast 11397.38 mm 2= Ans Displacement: Either the concrete or steel can be used for the calculation. δ Pst( ) L⋅ Ast Est⋅ := δ 0.15793 mm= Ans Problem 4-118 The assembly consists of a 30-mm-diameter aluminum bar ABC with fixed collar at B and a 10-mm-diameter steel rod CD. Determine the displacement of point D when the assembly is loaded as shown. Neglect the size of the collar at B and the connection at C. Est = 200 GPa, Eal = 70 GPa. Given: LAB 300mm:= dAB 30mm:= LBC 500mm:= dBC 30mm:= Eal 70GPa:= LCD 700mm:= dCD 10mm:= Est 200GPa:= PB 8− kN:= PD 20kN:= Solution: Internal Force: As shown on FBD. Displacement: AAB π 4 ⎛⎜⎝ ⎞ ⎠ dAB 2⋅:= δAB PD PB+( ) LAB( )⋅ Eal AAB( )⋅:= ABC π 4 ⎛⎜⎝ ⎞ ⎠ dBC 2⋅:= δBC PD( ) LBC( )⋅ Eal ABC( )⋅:= ACD π 4 ⎛⎜⎝ ⎞ ⎠ dCD 2⋅:= δCD PD( ) LCD( )⋅ Est ACD( )⋅:= δD δAB δBC+ δCD+:= δD 1.166 mm= Ans Problem 4-119 The joint is made from three A-36 steel plates that are bonded together at their seams. Determine the displacement of end A with respect to end B when the joint is subjected to the axial loads shown. Each plate has a thickness of 5 mm. Given: b 100mm:= t 5mm:= LAB 600mm:= AAB t b⋅:= LBC 200mm:= ABC 3t b⋅:= LCD 800mm:= ACD 2t b⋅:= P 46kN:= E 200GPa:= Solution: Internal Force: As shown on FBD. δA_D δAB δBC+ δCD+= δA_D P LAB( )⋅ E AAB( )⋅ P LBC( )⋅ E ABC( )⋅+ P LCD( )⋅ E ACD( )⋅+:= δA_D 0.491 mm= Ans Problem 5-1 A shaft is made of a steel alloy having an allowable shear stress of τallow = 84 MPa. If the diameter of the shaft is 37.5 mm, determine the maximum torque T that can be transmitted. What would be the maximum torque T' if a 25-mm-diameter hole is bored through the shaft? Sketch the shear-stress distribution along a radial line in each case. Given: do 37.5mm:= di 25mm:= τallow 84MPa:= Solution: c do 2 := a) Allowable shear srtess : Aplying the torsion formula τallow T c⋅ J =J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= T τallow( ) J⋅ c := T 0.87 kN m⋅= Ans b) Allowable shear srtess : Aplying the torsion formula τallow T' c⋅ J' =J' π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= T' τallow( ) J⋅ c := T' 0.87 kN m⋅= Ans Shera stress at ρ 0.5 di⋅:= τρ T' ρ⋅ J' := τρ 69.78 MPa= Problem 5-2 The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of the shaft that resists one-half of the applied torque (T/2). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution. Problem 5-3 The solid shaft of radius r is subjected to a torque T. Determine the radius r' of the inner core of the shaft that resists one-quarter of the applied torque (T/4). Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution. Problem 5-4 The tube is subjected to a torque of 750 N·m. Determine the amount of this torque that is resisted by the gray shaded section. Solve the problem two ways: (a) by using the torsion formula, (b) by finding the resultant of the shear-stress distribution. Given: ro 100mm:= ri 25mm:= T 750N m⋅:= rc 75mm:= Solution: c ro:= a) Aplying the torsion formula: J π 2 ⎛⎜⎝ ⎞ ⎠ ro 4 ri 4−⎛⎝ ⎞⎠⋅:= J' π2 ⎛⎜⎝ ⎞ ⎠ ro 4 rc 4−⎛⎝ ⎞⎠⋅:= τmax T c⋅ J := τmax T' c⋅ J' = τmax 0.479 MPa= T' τmax J'⋅ c := T' 0.515 kN m⋅= Ans b) Integartion Method : dT' ρ τ⋅ dA⋅= dA 2π ρ dρ⋅= τ ρ c ⎛⎜⎝ ⎞ ⎠ τmax⋅= dT' ρ ρ c ⎛⎜⎝ ⎞ ⎠⋅ τmax⋅ 2π ρ dρ⋅( )⋅= dT' 2π c τmax⋅ ρ3 dρ⋅= T' 2π c τmax⋅ rc ro ρρ3⌠⎮⌡ d ⎛⎜⎜⎝ ⎞ ⎠ := T' 0.515 kN m⋅= Ans Problem 5-5 The solid 30-mm-diameter shaft is used to transmit the torques applied to the gears. Determine the absolute maximum shear stress on the shaft. Given: a 300mm:= TA 300− N m⋅:= b 400mm:= TC 500N m⋅:= c 500mm:= TD 200N m⋅:= do 30mm:= TB 400− N m⋅:= Solution: c' do 2 := Internal Torque : As shown in the torque diagram. Allowable shear srtess : Aplying the torsion formula From the torque diagram, Tmax TB:= J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax Tmax c'⋅ J := τmax 75.45 MPa= Ans x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= T1 x1( ) TA 1N m⋅⋅:= T2 x2( ) TA TC+( ) 1N m⋅⋅:= T3 x3( ) TA TC+ TD+( ) 1N m⋅⋅:= 0 0.2 0.4 0.6 0.8 1 500 0 500 Distance (m) To rq ue (N m ) T1 x1( ) T2 x2( ) T3 x3( ) x1 x2, x3, Problem 5-6 The solid 32-mm-diameter shaft is used to transmit the torques applied to the gears. If it is supported by smooth bearings at A and B, which do not resist torque, determine the shear stress developed in the shaft at points C and D. Indicate the shear stress on volume elements located at these points. Given: do 32mm:= T1 185N m⋅:= T2 260− N m⋅:= T3 75N m⋅:= Solution: c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= TC T1:= TD T1 T2+:= τC TC( ) c⋅ J := τC 28.75 MPa= Ans τD TD( ) c⋅ J := τD 11.66− MPa= Ans Problem 5-7 The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to the applied torques as shown, determine the absolute maximum shear stress developed in the shaft. The smooth bearings at A and B do not resist torque. Given: do 32mm:= di 25mm:= T1 185N m⋅:= T2 260− N m⋅:= T3 75N m⋅:= Solution: c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= Tmax T1:= τmax Tmax( ) c⋅ J := τmax 45.82 MPa= Ans Problem 5-8 The shaft has an outer diameter of 32 mm and an inner diameter of 25 mm. If it is subjected to the applied torques as shown, plot the shear-stress distribution acting along a radial line lying within region EA of the shaft. The smooth bearings at A and B do not resist torque. Given: do 32mm:= di 25mm:= T1 185N m⋅:= T2 260− N m⋅:= T3 75N m⋅:= Solution: c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= TEA T1:= τmax TEA( ) c⋅ J := τmax 45.82 MPa= Ans Shera stress at ρ 0.5 di⋅:= τρ TEA ρ⋅ J := τρ 35.80 MPa= Ans Problem 5-9 The assembly consists of two sections of galvanized steel pipe connected together using a reducing coupling at B. The smaller pipe has an outer diameter of 18.75 mm and an inner diameter of 17 mm, whereas the larger pipe has an outer diameter of 25 mm and an inner diameter of 21.5 mm. If the pipe is tightly secured to the wall at C, determine the maximum shear stress developed in each section of the pipe when the couple shown is applied to the handles of the wrench. Given: d1o 18.75mm:= d1i 17mm:= F 75N:= d2o 25mm:= d2i 21.5mm:= aL 150mm:= aR 200mm:= Solution: T F aL⋅ F aR⋅+:= T 26.25 N m⋅= Segment AB : c1 d1o 2 := J1 π 2 ⎛⎜⎝ ⎞ ⎠ d1o 2 ⎛⎜⎝ ⎞ ⎠ 4 d1i 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τAB T c1⋅ J1 := τAB 62.55 MPa= Ans Segment BC : c2 d2o 2 := J2 π 2 ⎛⎜⎝ ⎞ ⎠ d2o 2 ⎛⎜⎝ ⎞ ⎠ 4 d2i 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τBC T c2⋅ J2 := τBC 18.89 MPa= Ans Problem 5-10 The link acts as part of the elevator control for a small airplane. If the attached aluminum tube has an inner diameter of 25 mm and a wall thickness of 5 mm, determine the maximum shear stress in the tube when the cable force of 600 N is applied to the cables. Also, sketch the shear-stress distribution over the cross section. Given: di 25mm:= t 5mm:= P 600N:= a 75mm:= Solution: do di 2t+:= c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= T P 2a( )⋅:= T 90.00 N m⋅= τmax T c⋅ J := τmax 14.45 MPa= Ans Shera stress at ρ 0.5 di⋅:= τρ T ρ⋅ J := τρ 10.32 MPa= Ans Problem 5-11 The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T = 800 N·m is applied to the rigid disk fixed to its end, determin the maximum shear stress in the shaft. Given: ri1 20mm:= ro1 25mm:= ri2 26mm:= ro2 30mm:= ri3 32mm:= ro3 38mm:= T 800N m⋅:= L 2m:= Solution: cmax ro3:= J1 π 2 ⎛⎜⎝ ⎞ ⎠ ro1 4 ri1 4−⎛⎝ ⎞⎠⋅:= J2 π 2 ⎛⎜⎝ ⎞ ⎠ ro2 4 ri2 4−⎛⎝ ⎞⎠⋅:= J3 π 2 ⎛⎜⎝ ⎞ ⎠ ro3 4 ri3 4−⎛⎝ ⎞⎠⋅:= J J1 J2+ J3+:= τmax T cmax⋅ J := τmax 11.94 MPa= Ans Problem 5-12 The solid shaft is fixed to the support at C and subjected to the torsional loadings shown. Determine the shear stress at points A and B and sketch the shear stress on volume elements located at these points. Given: ro 35mm:= ρA 35mm:= ρB 20mm:= TD 800N m⋅:= TB 300− N m⋅:= Solution: J π 2 ro 4⋅:= TBA TD TB+:= TAB TD:= τA TBA ρA⋅ J := τB TAB ρB⋅ J := τA 7.42 MPa= τB 6.79 MPa= Ans Problem 5-13 A steel tube having an outer diameter of 62.5 mm is used to transmit 3 kW when turning at 27 rev/min. Determine the inner diameter d of the tube to the nearest multiples of 5mm if the allowable shear stress is τallow = 70 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 62.5mm:= ω 27 rpm⋅:= P 3kW:= τallow 70MPa:= Solution: ω 2.83 rad s = T P ω:= T 1061.03 N m⋅= Max. stress : c do 2 := τallow T c⋅ J = J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅= di 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 2 π ⎛⎜⎝ ⎞ ⎠ T c⋅ τallow ⎛⎜⎝ ⎞ ⎠ ⋅− ⎡⎢⎢⎣ ⎤⎥⎥⎦ 0.25 := di 56.83 mm= Use di 60mm= Ans Problem 5-14 The solid aluminum shaft has a diameter of 50 mm and an allowable shear stress of τallow = 6 MPa. Determine the largest torque T1 that can be applied to the shaft if it is also subjected to the other torsional loadings. It is required that T1 act in the direction shown. Also, determine the maximum she stress within regions CD and DE. Given: do 50mm:= τallow 6MPa:= TA 68N m⋅:= TC 49N m⋅:= TD 35N m⋅:= Solution: J π 32 do 4⋅:= c do 2 := Assume failure at region BC. TBC T1 TA+( )−= Aplying the torsion formula τallow TBC( ) c⋅ J = τallow T1 TA+( )− c⋅ J = T1 τallow J⋅ c − TA−:= T1 215.26− N m⋅= Ans Internal Torque : Maximum torque occurs ithin region BC as indicated on the torque diagram. Maximum shear srtesses at Other Regions : TCD TA T1+ TC+:= TCD 98.26− N m⋅= τmax.CD TCD( ) c⋅ J := τmax.CD 4.00− MPa= Ans TDE TCD TD+:= TDE 63.26− N m⋅= τmax.DE TDE( ) c⋅ J := τmax.DE 2.58− MPa= Ans Let a 1m:= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= x4 3a 1.01 3a( )⋅, 4a..:= T'1 x1( ) TA 1N m⋅⋅:= T'2 x2( ) TA T1+( ) 1N m⋅⋅:= T'3 x3( ) TA T1+ TC+( ) 1N m⋅⋅:= T'4 x3( ) TA T1+ TC+ TD+( ) 1N m⋅⋅:= 0 1 2 3 4 200 100 0 100 Distance (m) To rq ue (N m ) T'1 x1( ) T'2 x2( ) T'3 x3( ) T'4 x4( ) x1 x2, x3, x4, Problem 5-15 The solid aluminum shaft has a diameter of 50 mm. Determine the absolute maximum shear stress in the shaft and sketch the shear-stress distribution along a radial line of the shaft where the shear stress is maximum. Set T1 = 20 N·m. Given: do 50mm:= T1 20− N m⋅:= TA 68N m⋅:= TC 49N m⋅:= TD 35N m⋅:= Solution: J π 32 do 4⋅:= c do 2 := Maximum Torque : Maximum torque occurs ithin region DE as indicated on the torque diagram. TDE TA T1+ TC+ TD+:= TDE 132.00 N m⋅= Tmax TDE:= Aplying the torsion formula τmax Tmax( ) c⋅ J := τmax 5.38 MPa= Ans Let a 1m:= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= x4 3a 1.01 3a( )⋅, 4a..:= T'1 x1( ) TA 1N m⋅⋅:= T'2 x2( ) TA T1+( ) 1N m⋅⋅:= T'3 x3( ) TA T1+ TC+( ) 1N m⋅⋅:= T'4 x3( ) TA T1+ TC+ TD+( ) 1N m⋅⋅:= 0 1 2 3 4 0 50 100 150 Distance (m) To rq ue (N m ) T'1 x1( ) T'2 x2( ) T'3 x3( ) T'4 x4( ) x1 x2, x3, x4, Problem 5-16 The motor delivers a torque of 50 N·m to the shaft AB. This torque is transmitted to shaft CD using the gears at E and F. Determine the equilibrium torque T' on shaft CD and the maximum shear stress each shaft. The bearings B, C, and D allow free rotation of the shafts. Given: dAB 30mm:= RE 50mm:= TAB 50N m⋅:= dCD 35mm:= RF 125mm:= Solution: Equilibrium : ΣΜE=0; TAB F RE( )⋅− 0= F TABRE:= ΣΜF=0; T' F RF⋅− 0= T' F RF⋅:= T' 125 N m⋅= Ans Internal Torque : As shown in FBD. JAB π 32 dAB 4⋅:= cAB 0.5dAB:= JCD π 32 dCD 4⋅:= cCD 0.5dCD:= Maximum shear srtesses: τmax.AB TAB( ) cAB⋅ JAB := τmax.AB 9.43 MPa= Ans τmax.CD T' cCD⋅ JCD := τmax.CD 14.85 MPa= Ans Problem 5-17 If the applied torque on shaft CD is T' = 75 N·m, determine the absolute maximum shear stress in eac shaft. The bearings B, C, and D allow free rotation of the shafts, and the motor holds the shafts fixed from rotating. Given: dEA 30mm:= RE 50mm:= TAB 50N m⋅:= dCD 35mm:= RF 125mm:= T' 75N m⋅:= Solution: Equilibrium : ΣΜF=0; T' F RF⋅− 0= F T' RF := ΣΜE=0; TAB F RE( )⋅− TA− 0= TA TAB F RE⋅−:= TA 20 N m⋅= Ans Internal Torque : As shown in FBD. TEA TAB TA−:= TEA 30.00 N m⋅= JEA π 32 dEA 4⋅:= cEA 0.5dEA:= JCD π 32 dCD 4⋅:= cCD 0.5dCD:= Maximum shear srtesses: τmax.EA TEA( ) cEA⋅ JEA := τmax.EA 5.66 MPa= Ans τmax.CD T' cCD⋅ JCD := τmax.CD 8.91 MPa= Ans Problem 5-18 The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightly secured to the wall at C and a uniformly distributed torque is applied to it as shown, determine the shear stress developed at points A and B. These points lie on the pipe's outer surface. Sketch the shear stress on volume elements located at A and B. Given: do 62.5mm:= di 57.5mm:= q 625 N m⋅ m := LOA 300mm:= LAB 225mm:= LBC 100mm:= Solution: Internal Torque : As shown on FBD TA q LOA( )⋅:= TA 187.50 N m⋅= TB q LOA LAB+( )⋅:= TB 328.13 N m⋅= Max. shear stress : τ T c⋅ J = c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τA TA c⋅ J := τA 13.79 MPa= Ans τB TB c⋅ J := τB 24.14 MPa= Ans Problem 5-19 The copper pipe has an outer diameter of 62.5 mm and an inner diameter of 57.5 mm. If it is tightly secured to the wall at C and it is subjected to the uniformly distributed torque along its entire length, determine the absolute maximum shear stress in the pipe. Discuss the validity of this result. Given: do 62.5mm:= di 57.5mm:= q 625 N m⋅ m := LOA 300mm:= LAB 225mm:= LBC 100mm:= Solution: Internal Torque : The maximum torque occurs at the support C TC q LOA LAB+ LBC+( )⋅:= TC 390.63 N m⋅= Max. shear stress : τ T c⋅ J = c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τC TC c⋅ J := τC 28.73 MPa= Ans According to Saint-Venant's principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Problem 5-20 The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the shear stress at points A and B, and sketch the shear stress on volume elements located at these points. Given: L1 0.3m:= L2 0.4m:= T1 400N m⋅:= T2 600− N m⋅:= do 60mm:= q 2 kN m⋅ m := Solution: Internal Torque : As shown on FBD. TA T1:= TA 400.00 N m⋅= TB T1 T2+ q L2⋅+:= TB 600.00 N m⋅= Maximum shear stress : τ T c⋅ J = c do 2 := J π 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τA TA c⋅ J := τA 9.43 MPa= Ans τB TB c⋅ J := τB 14.15 MPa= Ans Problem 5-21 The 60-mm diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses in the shaft and specify their locations, measured from the fixed end. Given: L1 0.3m:= L2 0.4m:= T1 400N m⋅:= T2 600− N m⋅:= do 60mm:= q 2 kN m⋅ m := Solution: Internal Torque : TC T1 T2+ q 2L2( )⋅+:= TC 1400.00 N m⋅= The maximum torque occurs at the fixed support C. Tmax TC:= Tmax 1400.00 N m⋅= The minimum torque occurs in segment loaded with q. Tmin 0:= xo 2L2 TC q 2L2( )⋅= xo TC q := xo 0.700 m= Shear stress : τ T c⋅ J = c do 2 := J π 32 ⎛⎜⎝ ⎞ ⎠ do 4⋅:= τabs.min Tmin c⋅ J := τabs.min 0.00 MPa= Ans τabs.max Tmax c⋅ J := τabs.max 33.01 MPa= Ans According to Saint-Venant's principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, the absolute τmax is not valid. x1 0 0.01 2⋅ L2, 2L2( )..:= x2 2L2 1.01 2L2( )⋅, 2L2 2L1+( )..:= Ta x1( ) TC− q x1⋅+( ) 1N m⋅⋅:= Tb x2( ) TC− 2q L2⋅+ T2+( ) 1N m⋅⋅:= 0 0.2 0.4 0.6 0.8 1 1.2 1500 1000 500 0 Distance (m) To rq ue (N m ) Ta x1( ) Tb x2( ) x1 x2, Problem 5-22 The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is τallow = 175 MPa. Given: L1 0.3m:= L2 0.4m:= T1 400N m⋅:= T2 600− N m⋅:= τallow 175MPa:= q 2 kN m⋅ m := Solution: Internal Torque : TC T1 T2+ q 2L2( )⋅+:= TC 1400.00 N m⋅= The maximum torque occurs at the fixed support C. Tmax TC:= Tmax 1400.00 N m⋅= Allowable Shear stress : τ T c⋅ J = c do 2 = J π 32 ⎛⎜⎝ ⎞ ⎠ do 4⋅= τallow 16Tmax π do3⋅ = do 3 16Tmax π τallow⋅ := do 34.41 mm= Ans According to Saint-Venant's principle, application of the torsion formula should be at points sufficiently removed from the supports or points of concentrated loading. Therefore, the above analysis is not valid. x1 0 0.01 2⋅ L2, 2L2( )..:= x2 2L2 1.01 2L2( )⋅, 2L2 2L1+( )..:= Ta x1( ) TC− q x1⋅+( ) 1N m⋅⋅:= Tb x2( ) TC− 2q L2⋅+ T2+( ) 1N m⋅⋅:= 0 0.2 0.4 0.6 0.8 1 1.2 1500 1000 500 0 Distance (m) To rq ue (N m ) Ta x1( ) Tb x2( ) x1 x2, Problem 5-23 The steel shafts are connected together using a fillet weld as shown. Determine the average shear stress in the weld along section aa if the torque applied to the shafts is T = 60 N·m. Note: The critical section where the weld fails is along section aa. Given: do 50mm:= a 12mm:= T 60N m⋅:= θa 45deg:= Solution: The maen radius of weld is: rweld do a+ 2 := rweld 31.00 mm= Shear stress : V T rweld := Aweld 2π rweld⋅ a sin θa( )⋅( )⋅:= τavg V Aweld := τavg 1.17 MPa= Ans Problem 5-24 The rod has a diameter of 12 mm and a weight of 80 N/m. Determine the maximum torsional stress in the rod at a section located at A due to the rod's weight. Given: d 12mm:= w 80 N m := LA 0.3m:= Lx 0.9m:= Ly 0.9m:= Lz 0.3m:= Solution: Equilibrium : Σ Mx = 0; TA w Ly⋅ 0.5Ly( )⋅− w Lz⋅ Ly( )⋅− = 0 TA w Ly⋅ 0.5 Ly⋅( )⋅ w Lz⋅ Ly( )⋅+:= TA 54.00 N m⋅= Max. shear stress : τ T c⋅ J = c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τA TA c⋅ J := τA 159.15 MPa= Ans Problem 5-25 Solve Prob. 5-24 for the maximum torsional stress at B. Given: d 12mm:= w 80 N m := LA 0.3m:= Lx 0.9m:= Ly 0.9m:= Lz 0.3m:= Solution: Equilibrium : Σ Mx = 0; TB w Ly⋅ 0.5Ly( )⋅− w Lz⋅ Ly( )⋅− = 0 TB w Ly⋅ 0.5 Ly⋅( )⋅ w Lz⋅ Ly( )⋅+:= TB 54.00 N m⋅= Max. shear stress : τ T c⋅ J = c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τB TB c⋅ J := τB 159.15 MPa= Ans Problem 5-26 Consider the general problem of a circular shaft made from m segments each having a radius of cm. If there are n torques on the shaft as shown, write a computer program that can be used to determine the maximum shearing stress at any specified location x along the shaft. Show an application of the program using the values L1 = 0.6 m, c1 = 50 mm, L2 = 1.2 m, c2 = 25 mm, T1 = 1200 N·m, d1 = 0, T2 = -900 N·m, d2 = 1.5 m. Problem 5-27 The wooden post, which is half buried in the ground, is subjected to a torsional moment of 50 N·m that causes the post to rotate at constant angular velocity.This moment is resisted by a linear distribution of torque developed by soil friction, which varies from zero at the ground to t0 N·m/m at its base. Determine the equilibrium value for t0 , and then calculate the shear stress at points A and B, which lie on the outer surface of the post. Given: do 100mm:= L 0.75m:= LAB 0.5 m⋅:= T 50N m⋅:= Solution: Equilibrium : Σ Mz = 0; 0.5to( ) L⋅ T− = 0 to 2 T L := to 133.33 N m⋅ m = Internal Torque : As shown on FBD. TA T:= TA 50.00 N m⋅= TB T 0.5to LAB L ⎛⎜⎝ ⎞ ⎠⋅ LAB⋅−:= TB 27.78 N m⋅= Maximum Shear Stress : τ T c⋅ J = c do 2 := J π 32 do 4⋅:= τA TA c⋅ J := τA 0.255 MPa= Ans τB TB c⋅ J := τB 0.141 MPa= Ans Problem 5-28 A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the shaft, determine the maximum shear stress in the rubber. Solution: Shear stress : τ F A = F T r = A 2π r⋅ h⋅= τ T 2π r2⋅ h⋅ = Shear stress is maximm when r is the smallest, i.e. r = ri. Hence, τmax T 2π ri2⋅ h⋅ = Ans Problem 5-29 The shaft has a diameter of 80 mm and due to friction at its surface within the hole, it is subjected to a variable torque described by the function t = (25x ex2) N·m/m, where x is in meters. Determine the minimum torque T0 needed to overcome friction and cause it to twist. Also, determine the absolute maximum stress in the shaft. Given: do 80mm:= L 2m:= t 25x ex 2 ⋅ N m⋅ m = Solution: unit N m⋅:= Equilibrium : Σ Mz = 0; To unit 0 L m x25x ex 2 ⋅ ⌠⎮ ⎮⌡ d⋅:= To 669.98 N m⋅= Ans Maximum Shear Stress : τ T c⋅ J = c do 2 := J π 32 do 4⋅:= τabs.max To c⋅ J := τabs.max 6.664 MPa= Ans Problem 5-30 The solid shaft has a linear taper from rA at one end to rB at the other. Derive an equation that gives th maximum shear stress in the shaft at a location x along the shaft's axis. Problem 5-31 When drilling a well at constant angular velocity, the bottom end of the drill pipe encounters a torsion resistance TA. Also, soil along the sides of the pipe creates a distributed frictional torque along its length, varying uniformly from zero at the surface B to tA at A. Determine the minimum torque TB tha must be supplied by the drive unit to overcome the resisting torques, and compute the maximum shea stress in the pipe. The pipe has an outer radius ro and an inner radius ri . Problem 5-32 The drive shaft AB of an automobile is made of a steel having an allowable shear stress of τallow= 56 MPa. If the outer diameter of the shaft is 62.5 mm and the engine delivers 165 kW to the shaft when is turning at 1140 rev/min, determine the minimum required thickness of the shaft's wall. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 62.5mm:= ω 1140 rpm⋅:= P 165kW:= τallow 56MPa:= Solution: ω 119.38 rad s = T P ω:= T 1382.14 J= Max. shear stress : c do 2 := τallow T c⋅ J = J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅= di 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 2 π ⎛⎜⎝ ⎞ ⎠ T c⋅ τallow ⎛⎜⎝ ⎞ ⎠ ⋅− ⎡⎢⎢⎣ ⎤⎥⎥⎦ 0.25 := di 52.16 mm= t do di− 2 := t 5.17 mm= Ans Problem 5-33 The drive shaft AB of an automobile is to be designed as a thin-walled tube. The engine delivers 125 kW when the shaft is turning at 1500 rev/min. Determine the minimum thickness of the shaft's wall if the shaft's outer diameter is 62.5 mm.The material has an allowable shear stress of τallow = 50 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 62.5mm:= ω 1500 rpm⋅:= P 125kW:= τallow 50MPa:= Solution: ω 157.08 rad s = T P ω:= T 795.77 J= Max. shear stress : c do 2 := τallow T c⋅ J = J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅= di 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 2 π ⎛⎜⎝ ⎞ ⎠ T c⋅ τallow ⎛⎜⎝ ⎞ ⎠ ⋅− ⎡⎢⎢⎣ ⎤⎥⎥⎦ 0.25 := di 56.50 mm= t do di− 2 := t 2.998 mm= Ans Problem 5-34 The gear motor can develop 100 W when it turns at 300 rev/min. If the shaft has a diameter of 12 m, determine the maximum shear stress that will be developed in the shaft. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 12mm:= ω 300 rpm⋅:= P 100W:= Solution: ω 31.42 rad s = T P ω:= T 3.183 N m⋅= Max. shear stress : c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax T c⋅ J := τmax 9.382 MPa= Ans Problem 5-35 The gear motor can develop 100 W when it turns at 80 rev/min. If the allowable shear stress for the shaft is τallow = 28 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mm that can be used. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: P 100W:= ω 80 rpm⋅:= τallow 28MPa:= Solution: ω 8.38 rad s = T P ω:= T 11.937 N m⋅= Max. shear stress : c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow T c⋅ J = π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅ T d⋅ 2 τallow⋅ = d 16T π τallow⋅ ⎛⎜⎝ ⎞ ⎠ 1 3 := d 12.95 mm= Use d 15mm= Ans Problem 5-36 The drive shaft of a tractor is made of a steel tube having an allowable shear stress of τallow = 42 MPa If the outer diameter is 75 mm and the engine delivers 145 kW to the shaft when it is turning at 1250 rev/min., determine the minimum required thickness of the shaft's wall. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 75mm:= ω 1250 rpm⋅:= P 145kW:= τallow 42MPa:= Solution: ω 130.90 rad s = T P ω:= T 1107.718 N m⋅= Max. shear stress : c do 2 := τallow T c⋅ J = J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅= di 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 2 π ⎛⎜⎝ ⎞ ⎠ T c⋅ τallow ⎛⎜⎝ ⎞ ⎠ ⋅− ⎡⎢⎢⎣ ⎤⎥⎥⎦ 0.25 := di 68.15 mm= t do di− 2 := t 3.427 mm= Ans Problem 5-37 The 2.5-kW reducer motor can turn at 330 rev/min. If the shaft has a diameter of 20 mm, determine the maximum shear stress that will be developed in the shaft. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 20mm:= ω 330 rpm⋅:= P 2.5kW:= Solution: ω 34.56 rad s = T P ω:= T 72.343 N m⋅= Max. shear stress : c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax T c⋅ J := τmax 46.055 MPa= Ans Problem 5-38 The 2.5-kW reducer motor can turn at 330 rev/min. If the allowable shear stress for the shaft is τallow = 56 MPa, determine the smallest diameter of the shaft to the nearest multiples of 5mm that can be used. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 20mm:= ω 330 rpm⋅:= P 2.5kW:= τallow 56MPa:= Solution: ω 34.56 rad s = T P ω:= T 72.343 N m⋅= Max. shear stress : c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow T c⋅ J = π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅ T d⋅ 2 τallow⋅ = d 16T π τallow⋅ ⎛⎜⎝ ⎞ ⎠ 1 3 := d 18.74 mm= Use d 20mm= Ans Problem 5-39 The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C, which delivers 3 kW of power to the shaft while it is turning at 50 rev/s. If gears A and B remove 1 kW and 2 kW, respectively, determine the maximum shear stress developed i the shaft within regions AB and BC. The shaft is free to turn in its support bearings D and E. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 25mm:= ω 50 60( )⋅ rpm⋅:= PA 1− kW:= Po 3kW:= PB 2− kW:= Solution: ω 314.16 rad s = TC Po ω:= TC 9.549 N m⋅= TA PA Po TC⋅:= TA 3.183 N m⋅= Maximum Shear Stress : c do 2 := J π 2 do 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τAB.max TA c⋅ J := τAB.max 1.038 MPa= Ans τBC.max TC c⋅ J := τBC.max 3.113 MPa= Ans Problem 5-40 A ship has a propeller drive shaft that is turning at 1500 rev/min. while developing 1500 kW. If it is 2 m long and has a diameter of 100 mm, determine the maximum shear stress in the shaft caused by torsion. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 100mm:= ω 1500 rpm⋅:= P 1500kW:= L 2.4m:= Solution: ω 157.08 rad s = T P ω:= T 9549.297 N m⋅= Max. shear stress : c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax T c⋅ J := τmax 48.634 MPa= Ans Problem 5-41 The motor A develops a power of 300 W and turns its connected pulley at 90 rev/min. Determine the required diameters of the steel shafts on the pulleys at A and B if the allowable shear stress is τallow = 85 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: rA 60mm:= ωA 90 rpm⋅:= rB 150mm:= P 300W:= τallow 85MPa:= Solution: ωB ωA rA rB ⎛⎜⎝ ⎞ ⎠ ⋅:= ωB 3.77 rad s = TA P ωA := TA 31.831 N m⋅= TB P ωB := TB 79.577 N m⋅= Allowable Shear Stress : For shaft A: τallow TA c⋅ J = c dA 2 = J π 32 dA 4⋅= τallow 16TA π dA3⋅ = dA 3 16TA π τallow⋅ := dA 12.40 mm= Ans For shaft B: τallow TB c⋅ J = c dB 2 = J π 32 dB 4⋅= τallow 16TB π dB3⋅ = dB 3 16TB π τallow⋅ := dB 16.83 mm= Ans Problem 5-42 The motor delivers 400 kW to the steel shaft AB, which is tubular and has an outer diameter of 50 mm and an inner diameter of 46 mm. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for material is τallow = 175 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 50mm:= di 46mm:= P 400kW:= τallow 175MPa:= Solution: c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τallow T c⋅ J = T J τallow( )⋅ c := T 1218.13 N m⋅= ω P T := ω 328.37 rad s = ω 3135.714 rpm= Ans Problem 5-43 The motor delivers 40 kW while turning at a constant rate of 1350 rpm at A. Using the belt and pulley system this loading is delivered to the steel blower shaft BC. Determine to the nearest multiples of 5mm the smallest diameter of this shaft if the allowable shear stress for steel if τallow = 84 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: rA 100mm:= rB 200mm:= P 40kW:= ω 1350 rpm⋅:= τallow 84MPa:= Solution: ω 141.37 rad s = TA P ω:= TA 282.942 N m⋅= TA 2rA F' F−( )⋅= TB 2rB F F'−( )⋅= TB rB rA ⎛⎜⎝ ⎞ ⎠ TA⋅:= TB 565.88 N m⋅= Max. shear stress : c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow TB c⋅ J = π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅ TB d⋅ 2 τallow⋅ = d 16TB π τallow⋅ ⎛⎜⎝ ⎞ ⎠ 1 3 := d 32.49 mm= Use d 35mm= Ans Problem 5-44 The propellers of a ship are connected to a solid A-36 steel shaft that is 60 m long and has an outer diameter of 340 mm and inner diameter of 260 mm. If the power output is 4.5 MW when the shaft rotates at 20 rad/s, determine the maximum torsional stress in the shaft and its angle of twist. Given: do 340mm:= di 260mm:= L 60m:= P 4500kW:= G 75GPa:= ω 20 rad s := Solution: T P ω:= T 225 kN m⋅= Maximum Shear Stress : c do 2 := J π 32 do 4 di 4−⎛⎝ ⎞⎠⋅:= τmax T c⋅ J := τmax 44.31 MPa= Ans Angle of Twist : φ T L⋅ G J⋅:= φ 0.2085 rad= φ 11.946 deg= Ans Problem 5-45 A shaft is subjected to a torque T. Compare the effectiveness of using the tube shown in the figure with that of a solid section of radius c. To do this, compute the percent increase in torsional stress and angle of twist per unit length for the tube versus the solid section. Given: ro c= ri 0.5c= Solution: Maximum Shear Stress : For solid shaft:: τs.max T c⋅ J = c ro= Js π 2 ro 4⋅= τs.max 2T π ro3⋅ = For the tube:: τt.max T c⋅ J = c ro= Jt π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅= Jt 15π32 ro 4⋅= τt.max 32T 15π ro3⋅ = % increase in shear stress: τ% τt.max τs.max− τs.max 100⋅= τ% 32 15 2− 2 100⋅:= τ% 6.67= Ans Angle of Twist : For solid shaft:: φs T L⋅ G Js⋅ = For the tube:: φt T L⋅ G Jt⋅ = φ% 1 Js 1 Jt − 1 Js 100⋅=% increase in angle of twist: φ% φt τs.max− τs.max 100⋅= φ% π 2 15π 32 − 15π 32 100⋅:= φ% 6.67= Ans Problem 5-46 The tubular drive shaft for the propeller of a hover-craft is 6 m long. If the motor delivers 4 MW of power to the shaft when the propellers rotate at 25 rad/s, determine the required inner diameter of the shaft if the outer diameter is 250 mm.What is the angle of twist of the shaft when it is operating? Tak τallow = 90 MPa and G = 75 GPa. Given: do 250mm:= L 6m:= P 4000kW:= τallow 90MPa:= G 75GPa:= ω 25 rad s := Solution: T P ω:= T 160 kN m⋅= Maximum Shear Stress : τallow T c⋅ J = c do 2 := J π 32 do 4 di 4−⎛⎝ ⎞⎠⋅= τallow 16T do⋅ π do4 di4−⎛⎝ ⎞⎠⋅ = di 4 do 4 16T do⋅ π τallow⋅ −:= di 201.3 mm= Ans Angle of Twist : J π 32 do 4 di 4−⎛⎝ ⎞⎠⋅:= φ T L⋅ G J⋅:= φ 0.05760 rad= φ 3.30 deg= Ans Problem 5-47 The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques, determine the angle of twist of gear A relative to gear D. The tubes have an outer diameter of 30 mm and an inner diameter of 20 mm. The solid section has a diameter of 40 mm. Given: do 30mm:= di 20mm:= ds 40mm:= LAB 0.4m:= LBC 0.25m:= LCD 0.4m:= T 85N m⋅:= G 75GPa:= Solution: TAB T:= TBC T:= TCD T:= Angle of Twist : JAB π 32 do 4 di 4−⎛⎝ ⎞⎠⋅:= JBC π32 ds 4⋅:= JCD JAB:= φ 2 TAB LAB⋅ G JAB⋅ TBC LBC⋅ G JBC⋅ +:= φ 0.01534 rad= φ 0.879 deg= Ans Problem 5-48 The A-36 steel axle is made from tubes AB and CD and a solid section BC. It is supported on smooth bearings that allow it to rotate freely. If the gears, fixed to its ends, are subjected to 85-N·m torques, determine the angle of twist of the end B of the solid section relative to end C. The tubes have an oute diameter of 30 mm and an inner diameter of 20 mm.The solid section has a diameter of 40 mm. Given: do 30mm:= di 20mm:= ds 40mm:= LAB 0.4m:= LBC 0.25m:= LCD 0.4m:= T 85N m⋅:= G 75GPa:= Solution: TAB T:= TBC T:= TCD T:= Angle of Twist : JBC π 32 ds 4⋅:= φ TBC LBC⋅ G JBC⋅ := φ 0.001127 rad= φ 0.0646 deg= Ans Problem 5-49 The hydrofoil boat has an A-36 steel propeller shaft that is 30 m long. It is connected to an in-line diesel engine that delivers a maximum power of 2000 kW and causes the shaft to rotate at 1700 rpm. If the outer diameter of the shaft is 200 mm and the wall thickness is 10 mm, determine the maximum shear stress developed in the shaft. Also, what is the “wind up,” or angle of twist in the shaft at full power? Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 200mm:= t 10mm:= P 2000kW:= ω 1700rpm:= G 75GPa:= L 30m:= Solution: ω 178.02 rad s = T P ω:= T 11234.467 N m⋅= Max. shear stress : di do 2t−:= c do 2 := J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τmax T c⋅ J := τmax 20.797 MPa= Ans Angle of Twist : φ T L⋅ G J⋅:= φ 0.0832 rad= φ 4.766 deg= Ans Problem 5-50 The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of gear C with respect to gear D. The shaft has a diameter of 40 mm. Given: LAC 0.3m:= LCD 0.4m:= LDE 0.5m:= do 40mm:= G 75GPa:= TA 300− N m⋅:= TC 500N m⋅:= TD 200N m⋅:= TE 400− N m⋅:= Solution: TAC TA:= TCD TA TC+:= TDE TA TC+ TD+:= Angle of Twist : J π 32 do 4⋅:= φC_D TCD LCD⋅ G J⋅:= φC_D 0.004244 rad= φC_D 0.243 deg= Ans Problem 5-51 The 20-mm-diameter A-36 steel shaft is subjected to the torques shown. Determine the angle of twist of the end B. Given: LAD 0.2m:= LDC 0.6m:= LCB 0.8m:= TB 80N m⋅:= TC 20− N m⋅:= TD 30N m⋅:= do 20mm:= G 75GPa:= Solution: TCB TB:= TDC TB TC+:= TAD TB TC+ TD+:= Angle of Twist : J π 32 do 4⋅:= φ TCB LCB⋅ G J⋅ TDC LDC⋅ G J⋅+ TAD LAD⋅ G J⋅+:= φ 0.100162 rad= φ 5.739 deg= Ans Problem 5-52 The 8-mm-diameter A-36 bolt is screwed tightly into a block at A. Determine the couple forces F that should be applied to the wrench so that the maximum shear stress in the bolt becomes 18 MPa. Also, compute the corresponding displacement of each force F needed to cause this stress. Assume that the wrench is rigid. Given: do 8mm:= L 80mm:= a 150mm:= G 75GPa:= τallow 18MPa:= Solution: Allowable Shear Stress : c do 2 := J π 32 ⎛⎜⎝ ⎞ ⎠ do 4⋅:= τallow T c⋅ J = T τallow J⋅ c := T 1.8096 N m⋅= Equilibrium : T F 2a( )− 0= F T 2a := F 6.03 N= Ans Angle of Twist : φ T L⋅ G J⋅:= φ 0.00480 rad= Displacement : s' a φ⋅:= s' 0.720 mm= Ans Problem 5-53 The turbine develops 150 kW of power, which is transmitted to the gears such that C receives 70% and D receives 30%. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 800 rev/min., determine the absolute maximum shear stress in the shaft and the angle of twist of end E of the shaft relative to B. The journal bearing at E allows the shaft to turn freely about its axis. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 100mm:= ω 800 rpm⋅:= LBC 3m:= LCD 4m:= LDE 2m:= P 150kW:= G 75GPa:= TC 0.7T= TD 0.3T= Solution: ω 83.78 rad s = T P ω:= T 1.790 kN m⋅= TC 0.7T:= TC 1.253 kN m⋅= TD 0.3T:= TD 0.537 kN m⋅= TBC T:= TCD 0.3T:= TDE 0:= Maximum Shear Stress : Maximum torque occurs in region BC. c do 2 := J π 32 do 4⋅:= τmax T c⋅ J := τmax 9.119 MPa= Ans Angle of Twist : φE_B TBC LBC⋅ G J⋅ TCD LCD⋅ G J⋅+ TDE LDE⋅ G J⋅+:= φE_B 0.010213 rad= φE_B 0.5852 deg= Ans Problem 5-54 The turbine develops 150 kW of power, which is transmitted to the gears such that both C and D receive an equal amount. If the rotation of the 100-mm-diameter A-36 steel shaft is ω = 500 rev/min. determine the absolute maximum shear stress in the shaft and the rotation of end B of the shaft relativ to E. The journal bearing at C allows the shaft to turn freely about its axis. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 100mm:= ω 500 rpm⋅:= LBC 3m:= LCD 4m:= LDE 2m:= P 150kW:= G 75GPa:= TC 0.5T= TD 0.5T= Solution: ω 52.36 rad s = T P ω:= T 2.865 kN m⋅= TC 0.5T:= TC 1.432 kN m⋅= TD 0.5T:= TD 1.432 kN m⋅= TBC T:= TCD 0.5T:= TDE 0:= Maximum Shear Stress : Maximum torque occurs in region BC. c do 2 := J π 32 do 4⋅:= τmax T c⋅ J := τmax 14.59 MPa= Ans Angle of Twist : φB_E TBC LBC⋅ G J⋅ TCD LCD⋅ G J⋅+ TDE LDE⋅ G J⋅+:= φB_E 0.019454 rad= φB_E 1.1146 deg= Ans Problem 5-55 The motor delivers 33 kW to the 304 stainless steel shaft while it rotates at 20 Hz. The shaft is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 20 kW and 12 kW, respectively. Determine the diameter of the shaft to the nearest mm if the allowable shear stress is τallow = 56 MPa and the allowable angle of twist of C with respect to D is 0.20°. Given: LAC 250mm:= LCD 200mm:= LDB 150mm:= P 32kW:= PC 20− kW:= PD 12− kW:= f 20Hz:= G 75GPa:= φallow 0.20deg:= τallow 56MPa:= Solution: ω 2 π⋅ f⋅:= ω 125.6637 rad s = TAC P ω:= TAC 254.648 N m⋅= TCD P PC+ ω:= TCD 95.493 N m⋅= Tmax max TAC TCD,( ):= Tmax 254.6479 N m⋅= Max. shear stress : Assume failure due to shear stress c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow Tmax c⋅ J = π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅ Tmax d⋅ 2 τallow⋅ = d 16Tmax π τallow⋅ ⎛⎜⎝ ⎞ ⎠ 1 3 := d 28.5041 mm= Angle of Twist : Assume failure due to angle of twist limitation (occured between C and D) J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= φallow TCD LCD⋅ G J⋅= φallow 0.003491 rad= π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅ TCD LCD⋅ G φallow⋅ = d 2 2TCD LCD⋅ π G⋅ φallow⋅ ⎛⎜⎝ ⎞ ⎠ 0.25 := d 29.3601 mm= Use d 30mm= Ans Problem 5-56 The motor delivers 33 kW to the 304 stainless steel solid shaft while it rotates at 20 Hz. The shaft has a diameter of 37.5 mm and is supported on smooth bearings at A and B, which allow free rotation of the shaft. The gears C and D fixed to the shaft remove 20 kW and 12 kW, respectively. Determine the absolute maximum stress in the shaft and the angle of twist of gear C with respect to gear D. Given: LAC 250mm:= LCD 200mm:= LDB 150mm:= P 32kW:= PC 20− kW:= PD 12− kW:= f 20Hz:= G 75GPa:= d 37.5mm:= Solution: ω 2 π⋅ f⋅:= ω 125.66 rad s = TAC P ω:= TAC 254.648 N m⋅= TCD P PC+ ω:= TCD 95.493 N m⋅= Tmax max TAC TCD,( ):= Tmax 254.65 N m⋅= Max. shear stress : Maximum shear stress occured between A and C c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax Tmax c⋅ J := τmax 24.59 MPa= Ans Angle of Twist : φCD TCD LCD⋅ G J⋅:= φCD 0.001312 rad= φCD 0.075152 deg= Ans Problem 5-57 The motor produces a torque of T = 20 N·m on gear A. If gear C is suddenly locked so it does not turn, yet B can freely turn, determine the angle of twist of F with respect to E and F with respect to D of the L2-steel shaft, which has an inner diameter of 30 mm and an outer diameter of 50 mm. Also, calculate the absolute maximum shear stress in the shaft. The shaft is supported on journal bearings at G and H. Given: do 50mm:= di 30mm:= rA 30mm:= rF 100mm:= G 75GPa:= T 20N m⋅:= LEF 0.6m:= Solution: Equilibrium : T F rA⋅− 0= F T rA := F 666.67 N= T' F rF⋅− 0= T' F rF⋅:= T' 66.67 N m⋅= Angle of Twist : J π 32 do 4 di 4−⎛⎝ ⎞⎠⋅:= Since shaft is held fixed at C, the torque is only in region EF of the shaft. φF_E T' LEF⋅ G J⋅:= φF_E 9.986 10 4−× rad= Ans Since the torque in region ED is zero, φF_D φF_E:= φF_D 9.986 10 4−× rad= Ans Maximum Shear Stress : c do 2 := τmax T' c⋅ J := τmax 3.121 MPa= Ans Problem 5-58 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end B when the torques are applied to the assembly as shown. Given: LDH 250mm:= LHE 750mm:= LAG 200mm:= LGF 250mm:= LFB 300mm:= d 25mm:= rE 150mm:= rF 100mm:= TH 120N m⋅:= TG 60N m⋅:= G 75GPa:= Solution: Internal Torque : AS shown on FBD At F : TF=TG P TG rF := P 600 N= At E : TE P− rE( )⋅:= TE 90− N m⋅= Angle of Twist : J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= φE TH LDH⋅ G J⋅ TE LDH LHE+( )⋅ G J⋅+:= φE 0.020861− rad= φE rE( )⋅ φF− rF( )⋅= φF rErF− φE⋅:= φF 0.031291 rad= Since there is no torque applied between F and B, φB φF:= φB 0.031291 rad= Ans φB 1.793 deg= Problem 5-59 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by bearings at A, B, and C, which allow free rotation. If the support at D is fixed, determine the angle of twist of end A when the torques are applied to the assembly as shown. Given: LDH 250mm:= LHE 750mm:= LAG 200mm:= LGF 250mm:= LFB 300mm:= d 25mm:= rE 150mm:= rF 100mm:= TH 120N m⋅:= TG 60N m⋅:= G 75GPa:= Solution: Internal Torque : AS shown on FBD At F : TF=TG P TG rF := P 600 N= At E : TE P− rE( )⋅:= TE 90− N m⋅= Angle of Twist : J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= φE TH LDH⋅ G J⋅ TE LDH LHE+( )⋅ G J⋅+:= φE 0.020861− rad= φE rE( )⋅ φF− rF( )⋅= φF rErF− φE⋅:= φF 0.031291 rad= Since there is no torque applied between A and G, φA_F TG LGF⋅ G J⋅:= φA_F 0.005215 rad= φA φA_F φF+:= φA 0.036506 rad= φA 2.092 deg= Ans Problem 5-60 Consider the general problem of a circular shaft made from m segments, each having a radius of cm and shearing modulus Gm. If there are n torques on the shaft as shown, write a computer program that can be used to determine the angle of twist of its end A. Show an application of the program using the values L1 = 0.5 m, c1 = 0.02 m, G1 = 30 GPa, L2 = 1.5 m, c2 = 0.05 m, G2 = 15 GPa, T1 = -450 N·m, d1 = 0.25 m, T2 = 600 N·m , d2 = 0.8 m. Problem 5-61 The 30-mm-diameter shafts are made of L2 tool steel and are supported on journal bearings that allow the shaft to rotate freely. If the motor at A develops a torque of T = 45 N·m on the shaft AB, while the turbine at E is fixed from turning, determine the amount of rotation of gears B and C. Given: LAB 1.5m:= LDC 0.5m:= LCE 0.75m:= TA 45N m⋅:= rB 50mm:= rC 75mm:= do 30mm:= G 75GPa:= Solution: TAB TA:= Equilibrium : TAB F rB⋅− 0= F TAB rB := F 900.00 N= T' F rC⋅− 0= T' F rC⋅:= T' 67.50 N m⋅= TCE T':= Angle of Twist : J π 32 do 4⋅:= φB TAB LAB⋅ G J⋅:= φC TCE LCE⋅ G J⋅:= φB 0.011318 rad= φC 0.008488 rad= φB 0.648 deg= Ans φC 0.486 deg= Ans Problem 5-62 The 60-mm-diameter solid shaft is made of A-36 steel and is subjected to the distributed and concentrated torsional loadings shown. Determine the angle of twist at the free end A of the shaft due to these loadings. Given: LAC 0.6m:= LCB 0.8m:= do 60mm:= TA 400N m⋅:= TC 600− N m⋅:= G 75GPa:= q 2000 N m⋅ m := Solution: Internal Torque : As shown in the torque diagram. TAC TA:= TCB TA TC+ q x⋅+= Angle of Twist : J π 32 do 4⋅:= φA TAC LAC⋅ G J⋅ 1 G J⋅ 0 LCB xTCB ⌠⎮⌡ d⋅+= φA TAC LAC⋅ G J⋅ 1 G J⋅ 0 LCB xTA TC+ q x⋅+ ⌠⎮⌡ d⋅+:= φA 0.007545 rad= φA 0.432 deg= Ans Problem 5-63 When drilling a well, the deep end of the drill pipe is assumed to encounter a torsional resistance TA. Furthermore, soil friction along the sides of the pipe creates a linear distribution of torque per unit length, varying from zero at the surface B to t0 at A. Determine the necessary torque TB that must be supplied by the drive unit to turn the pipe. Also, what is the relative angle of twist of one end of the pipe with respect to the other end at the instant the pipe is about to turn? The pipe has an outer radius ro and an inner radius ri . The shear modulus is G. Problem 5-64 The assembly is made of A-36 steel and consists of a solid rod 15 mm in diameter connected to the inside of a tube using a rigid disk at B. Determine the angle of twist at A. The tube has an outer diameter of 30 mm and wall thickness of 3 mm. Given: do 30mm:= t 3mm:= dr 15mm:= LAB 0.3m:= LBC 0.3m:= G 75GPa:= TA 50N m⋅:= TB 30N m⋅:= Solution: di do 2t−:= TAB TA:= TBC TA TB+:= Angle of Twist : JAB π 32 dr 4⋅:= JBC π 32 do 4 di 4−⎛⎝ ⎞⎠⋅:= φA TAB LAB⋅ G JAB⋅ TBC LBC⋅ G JBC⋅ +:= φA 0.04706 rad= φA 2.696 deg= Ans Problem 5-65 The device serves as a compact torsional spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If a torque of T = 0.25 N·m. is applied to the shaft, determine the angle of twist at the end C and the maximum shear stress in the tube and shaft. Given: LBC 600mm:= LBA 300mm:= rti 18.75mm:= rto 25mm:= r 12.5mm:=TC 0.25kN m⋅:= G 75GPa:= Solution: Internal Torque : AS shown on FBD TBC TC:= Inner shaft : TBA TC:= Outer tube : Max. Shear Stress : Inner shaft : cBC r:= JBC π 2 r4⋅:= τBC TBC cBC( )⋅ JBC := τBC 81.49 MPa= Ans cBA rto:= JBA π 2 rto 4 rti 4−⎛⎝ ⎞⎠⋅:= Outer tube : τBA TBC cBA( )⋅ JBA := τBA 14.9 MPa= Ans Angle of Twist : φB TBA LBA⋅ G JBA⋅ := φB 0.002384 rad= φC_B TBC LBC⋅ G JBC⋅ := φC_B 0.052152 rad= φC φC_B φB+:= φC 0.054536 rad= φC 3.125 deg= Ans Problem 5-66 The device serves as a compact torsion spring. It is made of A-36 steel and consists of a solid inner shaft CB which is surrounded by and attached to a tube AB using a rigid ring at B. The ring at A can also be assumed rigid and is fixed from rotating. If the allowable shear stress for the material is τallow = 84 MPa and the angle of twist at C is limited to φallow = 3°, determine the maximum torque T that can be applied at the end C. Given: LBC 600mm:= LBA 300mm:= rti 18.75mm:= rto 25mm:= τallow 84MPa:= r 12.5mm:= φallow 3deg:= G 75GPa:= Solution: Internal Torque : AS shown on FBD Inner shaft : TBC=TC Outer tube : TBA=TC Allowable Shear Stress : Assume failure due to shear stress. Inner shaft : cBC r:= JBC π 2 r4⋅:= TBC τallow JBC( )⋅ cBC := TBC 257.71 N m⋅= cBA rto:= JBA π 2 rto 4 rti 4−⎛⎝ ⎞⎠⋅:= Outer tube : TBA τallow JBA( )⋅ cBA := TBA 1409.34 N m⋅= Angle of Twist : Assume failure due to angle of twist limitation (maximum occurred at C). φC φC_B φB+= φC_B TC LBC⋅ G JBC⋅ = φB TC LBA⋅ G JBA⋅ = Thus, φallow TC LBC⋅ G JBC⋅ TC LBA⋅ G JBA⋅ += TC G φallow⋅ LBC JBC LBA JBA + := TC 240.02 N m⋅= Ans (controls !) Problem 5-67 The shaft has a radius c and is subjected to a torque per unit length of t0 , which is distributed uniformly over the shaft's entire length L. If it is fixed at its far end A, determine the angle of twist φ o end B. The shear modulus is G. Solution: Internal Torque : As shown in the torque diagram. T x( ) to− x⋅= Angle of Twist : J π 2 c4⋅:= φB 1 G J⋅ 0 L xT x( ) ⌠⎮⌡ d⋅= φB to− G J⋅ 0 L xx ⌠⎮⌡ d⋅= φB to− L2⋅ 2G J⋅= φB to− L2⋅ πG c4⋅ = φB to L 2⋅ πG c4⋅ = Ans Problem 5-68 The A-36 bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (k x2) N·m /m, where x is in meters. If a torque of T = 50 N·m is applied to the bol head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm. Given: ro 4mm:= L 50mm:= TA 50N m⋅:= G 75GPa:= t k x2⋅ N m⋅ m = Solution: c ro:= Internal Torque : As shown in the torque diagram. t x( ) k x2⋅= Equilibrium : TA 0 L xt x( ) ⌠⎮⌡ d− 0= TA 0 L xk x2⋅⌠⎮⌡ d− 0= TA k L3⋅ 3 − 0= k 3TA L3 := k 1.200 MPa= Ans Hence, T x( ) TA xk x 2⋅ ⌠⎮ ⎮⌡ d−= T x( ) TA k x3⋅ 3 −= Angle of Twist : J π 2 c4⋅:= φ 1 G J⋅ 0 L xT x( ) ⌠⎮⌡ d⋅= φ 1 G J⋅ 0 L xTA k x3⋅ 3 −⎛⎜⎝ ⎞ ⎠ ⌠⎮ ⎮⌡ d⋅:= φ 0.06217 rad= φ 3.562 deg= Ans Problem 5-69 Solve Prob. 5-68 if the distributed torque is t = (k x2/3) N·m /m. Given: ro 4mm:= L 50mm:= TA 50N m⋅:= G 75GPa:= t k x 2 3⋅ N m⋅ m = Solution: c ro:= Internal Torque : As shown in the torque diagram. Equilibrium : TA 0 L xt x( ) ⌠⎮⌡ d− 0= TA 0 L xk x 2 3⋅ ⌠⎮ ⎮⌡ d− 0= Let unit m 4− 3:= TA 3k L 5 3 ⎛⎜⎝ ⎞ ⎠⋅ 5 − 0= k 5TA 3 ⎛⎜⎝ ⎞ ⎠ L 5− 3⋅ unit⋅:= k 0.01228 MPa= Ans Hence, T x( ) TA xk x 2 3⋅ ⌠⎮ ⎮ ⎮⌡ d−= T x( ) TA 3k x 5 3 ⎛⎜⎝ ⎞ ⎠⋅ 5 −= Angle of Twist : J π 2 c4⋅:= φ 1 G J⋅ 0 L xT x( ) ⌠⎮⌡ d⋅= φ 1 G J⋅ 0 L xTA 3k x 5 3 ⎛⎜⎝ ⎞ ⎠⋅ 5 unit⋅− ⌠⎮ ⎮ ⎮ ⎮⌡ d⋅:= φ 0.05181 rad= φ 2.968 deg= Ans Problem 5-70 The contour of the surface of the shaft is defined by the equation y = e ax, where a is a constant. If the shaft is subjected to a torque T at its ends, determine the angle of twist of end A with respect to end B. The shear modulus is G. Problem 5-71 The A-36 steel shaft has a diameter of 50 mm and is subjected to the distributed and concentrated loadings shown. Determine the absolute maximum shear stress in the shaft and plot a graph of the angle of twist of the shaft in radians versus x. Given: do 50mm:= L 0.5m:= TC 250− N m⋅:= G 75GPa:= q 200 N m⋅ m := Solution: Support Reaction: RA TC− q L⋅−:= RA 150.00 N m⋅= Internal Torque : As shown in the torque diagram. T x( ) RA q x⋅+= The maximum torque occurs at C. Tmax RA q L⋅+:= Tmax 250.00 N m⋅= Maximum Shear Stress : c 0.5 do⋅:= J π 2 c4⋅:= τmax Tmax c⋅ J := τmax 10.19 MPa= Ans Angle of Twist : φx 1 G J⋅ 0 x xT x( ) ⌠⎮⌡ d⋅= φx 1 G J⋅ 0 x xRA q x⋅+( )⌠⎮⌡ d⋅= φx 1 G J⋅ RA x⋅ q x2⋅ 2 +⎛⎜⎝ ⎞ ⎠⋅= At C, x = L φC 1 G J⋅ RA L⋅ q L2⋅ 2 +⎛⎜⎝ ⎞ ⎠⋅:= φC 0.00217 rad= φC 0.125 deg= Ans For L < x < 2L Since T(x) = 0, then, φx φC:= x1 0 0.01 L⋅, L..:= x2 L 1.01 L⋅, 2L..:= φ1 x1( ) 1G J⋅ RA x1⋅ q x1 2⋅ 2 + ⎛⎜⎝ ⎞ ⎠⋅ 1 N m⋅⋅:= φ2 x2( ) φC( ) 1N m⋅⋅:= 0 0.2 0.4 0.6 0.8 0 0.001 0.002 Distance (m) Tw is t ( ra d) φ1 x1( ) φ2 x2( ) x1 x2, Problem 5-72 A cylindrical spring consists of a rubber annulus bonded to a rigid ring and shaft. If the ring is held fixed and a torque T is applied to the rigid shaft, determine the angle of twist of the shaft. The shear modulus of the rubber is G. Hint: As shown in the figure, the deformation of the element at radius r can be determined from rdθ = drγ. Use this expression along with τ = T/(2π r2h) from Prob. 5-28, to obtain the result. Problem 5-73 The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB of the shaft. Given: do 50mm:= LAC 0.4m:= LBC 0.8m:= G 75GPa:= TC 300N m⋅:= Solution: c 0.5 do⋅:= J π 2 c4⋅:= Equilibrium : Given TA TB+ TC− 0= (1) Compatibility : φC/A = φC/B TA LAC⋅ G J⋅ TB LBC⋅ G J⋅= (2) Solving Eqs. (1) and (2): Guess TA 1N m⋅:= TB 1N m⋅:= TA TB ⎛⎜⎜⎝ ⎞ ⎠ Find TA TB,( ):= TATB ⎛⎜⎜⎝ ⎞ ⎠ 200 100 ⎛⎜⎝ ⎞ ⎠N m⋅= Maximum Shear Stress : τAC.max TA c⋅ J := τAC.max 8.15 MPa= Ans τBC.max TB c⋅ J := τBC.max 4.07 MPa= Ans Problem 5-74 The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling on it at C is being tightened using a wrench. If the torque developed at A is 16 N·m, determine the magnitude F of the couple forces. The pipe is fixed supported at end B. Given: LCB 200mm:= LCA 250mm:= Lw 300mm:= do 37.5mm:= t 3mm:= TA 16N m⋅:= G 38GPa:= Solution: ro 0.5do:= ri ro t−:= Compatibility : φC_B φC_A= TB LCB⋅ G J⋅ TB LCA⋅ G J⋅= TB LCB⋅ TA LCA⋅= TB LCA LCB ⎛⎜⎝ ⎞ ⎠ TA⋅:= TB 20.00 N m⋅= Equilibrium: F Lw( )⋅ TB− TA− 0= F TB TA+ Lw := F 120.00 N= Ans Problem 5-75 The bronze C86100 pipe has an outer diameter of 37.5 mm and a thickness of 3 mm. The coupling on it at C is being tightened using a wrench. If the applied force is F = 100 N, determine the maximum shear stress in the pipe. Given: LCB 200mm:= LCA 250mm:= Lw 300mm:= do 37.5mm:= t 3mm:= F 100N:= G 38GPa:= Solution: ro 0.5do:= ri ro t−:= Compatibility : φC_B φC_A= TB LCB⋅ G J⋅ TB LCA⋅ G J⋅= Given TB LCB⋅ TA LCA⋅= [1] Equilibrium: F Lw( )⋅ TB− TA− 0= [2] Initial guess: TA 1N m⋅:= TB 2N m⋅:= Solving [1] and [2]: TA TB ⎛⎜⎜⎝ ⎞ ⎠ Find TA TB,( ):= TATB ⎛⎜⎜⎝ ⎞ ⎠ 13.33 16.67 ⎛⎜⎝ ⎞ ⎠ N m⋅= Max. Shear Stress : c ro:= J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅:= τmax TB c( )⋅ J := τmax 3.21 MPa= Ans Problem 5-76 The steel shaft is made from two segments: AC has a diameter of 12 mm, and CB has a diameter of 25 mm. If it is fixed at its ends A and B and subjected to a torque of 750 N·m, determine the maximum shear stress in the shaft. Gst = 75 GPa. Given: LAC 125mm:= LCD 200mm:= LDB 300mm:= dAC 12mm:= dCD 25mm:= dDB 25mm:= TD 750N m⋅:= G 75GPa:= Solution: JAC π 2 dAC 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= JCD π 2 dCD 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= JDB π 2 dDB 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= Compatibility : φD_A φD_B= Given TA LAC⋅ G JAC⋅ TA LCD⋅ G JCD⋅ + TB LDB⋅ G JDB⋅ = [1] Equilibrium: TD TB− TA− 0= [2] Initial guess: TA 1N m⋅:= TB 2N m⋅:= Solving [1] and [2]: TA TB ⎛⎜⎜⎝ ⎞ ⎠ Find TA TB,( ):= TATB ⎛⎜⎜⎝ ⎞ ⎠ 78.82 671.18 ⎛⎜⎝ ⎞ ⎠N m⋅= Max. Shear Stress : cAC dAC 2 := τAC TA cAC( )⋅ JAC := τAC 232.3 MPa= cDB dDB 2 := τDB TB cDB( )⋅ JDB := τDB 218.77 MPa= τmax max τAC τDB,( ):= τmax 232.30 MPa= Ans Problem 5-77 The shaft is made of L2 tool steel, has a diameter of 40 mm, and is fixed at its ends A and B. If it is subjected to the couple, determine the maximum shear stress in regions AC and CB. Given: do 40mm:= LAC 0.4m:= LBC 0.6m:= G 75GPa:= PC 2kN:= rC 50mm:= Solution: c 0.5 do⋅:= J π 2 c4⋅:= Equilibrium : Given TA TB+ PC 2rC( )⋅− 0= (1) Compatibility : φC/A = φC/B TA LAC⋅ G J⋅ TB LBC⋅ G J⋅= (2) Solving Eqs. (1) and (2): Guess TA 1N m⋅:= TB 1N m⋅:= TA TB ⎛⎜⎜⎝ ⎞ ⎠ Find TA TB,( ):= TATB ⎛⎜⎜⎝ ⎞ ⎠ 120 80 ⎛⎜⎝ ⎞ ⎠N m⋅= Maximum Shear Stress : τAC.max TA c⋅ J := τAC.max 9.55 MPa= Ans τBC.max TB c⋅ J := τBC.max 6.37 MPa= Ans Problem 5-78 The composite shaft consists of a mid-section that includes the 20-mm-diameter solid shaft and a tube that is welded to the rigid flanges at A and B. Neglect the thickness of the flanges and determine the angle of twist of end C of the shaft relative to end D. The shaft is subjected to a torque of 800 N·m. The material is A-36 steel. Given: LCA 100mm:= LAB 150mm:= LBD 100mm:= ds 20mm:= dto 60mm:= t 5mm:= T 800N m⋅:= G 75GPa:= Solution: dti dto 2t−:= Js π 2 ds 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= Jt π 2 dto 2 ⎛⎜⎝ ⎞ ⎠ 4 dti 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= Compatibility : φs φt= Ts LAB⋅ G Js⋅ Tt LAB⋅ G Jt⋅ = Given Ts Js Tt Jt = [1] Equilibrium: T Ts− Tt− 0= [2] Initial guess: Ts 1N m⋅:= Tt 2N m⋅:= Solving [1] and [2]: Ts Tt ⎛⎜⎜⎝ ⎞ ⎠ Find Ts Tt,( ):= TsTt ⎛⎜⎜⎝ ⎞ ⎠ 18.63 781.37 ⎛⎜⎝ ⎞ ⎠ N m⋅= Angle of Twist: φC_D Ts LCA⋅ G Js⋅ Tt LAB⋅ G Jt⋅ + Ts LBD⋅ G Js⋅ +:= φC_D 0.005535 rad= φC_D 0.317 deg= Ans Problem 5-79 The shaft is made from a solid steel section AB and a tubular portion made of steel and having a brass core. If it is fixed to a rigid support at A, and a torque of T = 50 N·m is applied to it at C, determine the angle of twist that occurs at C and compute the maximum shear stress and maximum shear strain in the brass and steel. Take Gst = 80 GPa, Gbr = 40 GPa. Given: LAB 1.5m:= LBC 1m:= rs 20mm:= rto 20mm:= rti 10mm:= T 50N m⋅:= Gst 80GPa:= Gbr 40GPa:= Solution: Js π 2 rs 4⋅:= Jt π 2 rto 4 rti 4−⎛⎝ ⎞⎠⋅:= Jbr π 2 rti 4⋅:= Compatibility : φst φbr= Tst LBC⋅ Gst Jt⋅ Tbr LBC⋅ Gbr Jbr⋅ = Given Tst Gst Jt⋅ Tbr Gbr Jbr⋅ = [1] Equilibrium: T Tst− Tbr− 0= [2] Initial guess: Tst 1N m⋅:= Tbr 2N m⋅:= Solving [1] and [2]: Tst Tbr ⎛⎜⎜⎝ ⎞ ⎠ Find Tst Tbr,( ):= TstTbr ⎛⎜⎜⎝ ⎞ ⎠ 48.39 1.61 ⎛⎜⎝ ⎞ ⎠N m⋅= Angle of Twist: φC T LAB⋅ Gst Js⋅ Tbr LBC⋅ Gbr Jbr⋅ +:= φC 0.006297 rad= φC 0.361 deg= Ans Max. Shear Stress : cAB rs:= τAB T cAB( )⋅ Js := τAB 3.98 MPa= cBC rto:= τBC Tst cBC( )⋅ Jt := τBC 4.11 MPa= τst max τAB τBC,( ):= τst 4.11 MPa= Ans γst τst Gst := γst 51.34 10 6−× rad= Ans τbr Tbr rti( )⋅ Jbr := τbr 1.03 MPa= Ans γbr τbr Gbr := γbr 25.67 10 6−× rad= Ans Problem 5-80 The two 1-m-long shafts are made of 2014-T6 aluminum. Each has a diameter of 30 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 900 N·m is applied to the top gear as shown, determine the maximum shear stress in each shaft. Given: L 1000mm:= d 30mm:= rA 80mm:= rB 40mm:= T 900N m⋅:= G 27GPa:= Solution: J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= Compatibility : φE rA( )⋅ φF rB( )⋅= TA L⋅ G J⋅ rA( )⋅ TB L⋅ G J⋅ rB( )⋅= Given TA rA( )⋅ TB rB( )⋅= [1] Equilibrium: T TA− F rA( )⋅− 0= [2] TB F rB( )⋅− 0= [3] Initial guess: TA 1N m⋅:= TB 2N m⋅:= F 1N:= Solving [1], [2] and [3]: TA TB F ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find TA TB, F,( ):= TA TB ⎛⎜⎜⎝ ⎞ ⎠ 180 360 ⎛⎜⎝ ⎞ ⎠N m⋅= F 9000.00 N= Max. Shear Stress : c d 2 := τAC TA c( )⋅ J := τAC 33.95 MPa= Ans τBD TB c( )⋅ J := τBD 67.91 MPa= Ans Problem 5-81 The two shafts are made of A-36 steel. Each has a diameter of 25 mm and they are connected using the gears fixed to their ends. Their other ends are attached to fixed supports at A and B. They are also supported by journal bearings at C and D, which allow free rotation of the shafts along their axes. If a torque of 500 N·m is applied to the gear at E as shown, determine the reactions at A and B. Given: LAE 1.5m:= LBF 0.75m:= do 25mm:= rE 100mm:= rF 50mm:= TE 500N m⋅:= G 75GPa:= Solution: J π 32 do 4⋅:= Compatibility : φE rE( )⋅ φF rF( )⋅= TA LAE⋅ G J⋅ rE( )⋅ TB LBF⋅ G J⋅ rF( )⋅= Given TA LAE⋅ rE( )⋅ TB LBF⋅ rF( )⋅= [1] Equilibrium: TA F rE( )⋅+ TE− 0= [2] TB F rF( )⋅− 0= [3] Initial guess: TA 1N m⋅:= TB 2N m⋅:= F 1N:= Solving [1], [2] and [3]: TA TB F ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find TA TB, F,( ):= TA TB ⎛⎜⎜⎝ ⎞ ⎠ 55.56 222.22 ⎛⎜⎝ ⎞ ⎠N m⋅= A F 4444.44 N= Problem 5-82 Determine the rotation of the gear at E in Prob. 5-81. Given: LAE 1.5m:= LBF 0.75m:= do 25mm:= rE 100mm:= rF 50mm:= TE 500N m⋅:= G 75GPa:= Solution: J π 32 do 4⋅:= Compatibility : φE rE( )⋅ φF rF( )⋅= TA LAE⋅ G J⋅ rE( )⋅ TB LBF⋅ G J⋅ rF( )⋅= Given TA LAE⋅ rE( )⋅ TB LBF⋅ rF( )⋅= [1] Equilibrium: TA F rE( )⋅+ TE− 0= [2] TB F rF( )⋅− 0= [3] Initial guess: TA 1N m⋅:= TB 2N m⋅:= F 1N:= Solving [1], [2] and [3]: TA TB F ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find TA TB, F,( ):= TA TB ⎛⎜⎜⎝ ⎞ ⎠ 55.56 222.22 ⎛⎜⎝ ⎞ ⎠N m⋅= F 4444.44 N= Angle of Twist : φE TA LAE⋅ G J⋅:= φE 0.02897 rad= φE 1.660 deg= Ans Problem 5-83 The A-36 steel shaft is made from two segments: AC has a diameter of 10 mm and CB has a diameter of 20 mm. If the shaft is fixed at its ends A and B and subjected to a uniform distributed torque of 300 N·m/m along segment CB, determine the absolute maximum shear stress in the shaft. Given: LAC 0.1m:= LCB 0.4m:= dAC 10mm:= dCB 20mm:= G 75GPa:= q 300 N m⋅ m := Solution: JAC π 2 dAC 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= JCB π 2 dCB 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= Compatibility : φC_A φC_B= Given TA LAC⋅ G JAC⋅ TB LCB⋅ 0.5q LCB2⋅− G JCB⋅ = [1] Equilibrium: q LCB⋅ TA− TB− 0= [2] Initial guess: TA 1N m⋅:= TB 2N m⋅:= Solving [1] and [2]: TA TB ⎛⎜⎜⎝ ⎞ ⎠ Find TA TB,( ):= TATB ⎛⎜⎜⎝ ⎞ ⎠ 12 108 ⎛⎜⎝ ⎞ ⎠N m⋅= Max. Shear Stress : cAC dAC 2 := τAC TA cAC( )⋅ JAC := τAC 61.12 MPa= Ans cCB dCB 2 := τCB TB cCB( )⋅ JCB := τCB 68.75 MPa= Ans τmax max τAC τCB,( ):= τmax 68.75 MPa= Ans Problem 5-84 The tapered shaft is confined by the fixed supports at A and B. If a torque T is applied at its mid-poin determine the reactions at the supports. Problem 5-85 A portion of the A-36 steel shaft is subjected to a linearly distributed torsional loading. If the shaft has the dimensions shown, determine the reactions at the fixed supports A and C. Segment AB has a diameter of 30 mm and segment BC has a diameter of 15 mm. Given: LAB 1.2m:= LBC 0.96m:= dAB 30mm:= dBC 15mm:= G 75GPa:= q 1.5 kN m⋅ m := Solution: JAB π 2 dAB 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= JBC π 2 dBC 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= qx 1 x Lab −⎛⎜⎝ ⎞ ⎠ q⋅= TR qx( ) x⋅ 0.5 q qx−( ) x⋅+= TR 1 x Lab −⎛⎜⎝ ⎞ ⎠ q⋅ x⋅ 0.5 x Lab ⎛⎜⎝ ⎞ ⎠ q⋅ x⋅+= TR 1 0.5x Lab −⎛⎜⎝ ⎞ ⎠ q⋅ x⋅= β 0 LAB x1 0.5x LAB −⎛⎜⎝ ⎞ ⎠ x⋅⌠⎮⎮⌡ d:= β 0.48 m2= Compatibility : φB_A φB_C= Given TA LAB⋅ q β⋅− G JAB⋅ TC LBC⋅ G JBC⋅ = [1] Equilibrium: 0.5q LAB⋅ TA− TC− 0= [2] Initial guess: TA 1kN m⋅:= TC 2kN m⋅:= Solving [1] and [2]: TA TC ⎛⎜⎜⎝ ⎞ ⎠ Find TA TC,( ):= TATC ⎛⎜⎜⎝ ⎞ ⎠ 878.26 21.74 ⎛⎜⎝ ⎞ ⎠ N m⋅= Ans Problem 5-86 Determine the rotation of joint B and the absolute maximum shear stress in the shaft in Prob. 5-85. Given: LAB 1.2m:= LBC 0.96m:= dAB 30mm:= dBC 15mm:= G 75GPa:= q 1.5 kN m⋅ m := Solution: JAB π 2 dAB 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= JBC π 2 dBC 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= qx 1 x Lab −⎛⎜⎝ ⎞ ⎠ q⋅= TR qx( ) x⋅ 0.5 q qx−( ) x⋅+= TR 1 x Lab −⎛⎜⎝ ⎞ ⎠ q⋅ x⋅ 0.5 x Lab ⎛⎜⎝ ⎞ ⎠ q⋅ x⋅+= TR 1 0.5x Lab −⎛⎜⎝ ⎞ ⎠ q⋅ x⋅= β 0 LAB x1 0.5x LAB −⎛⎜⎝ ⎞ ⎠ x⋅⌠⎮⎮⌡ d:= β 0.48 m2= Compatibility : φB_A φB_C= Given TA LAB⋅ q β⋅− G JAB⋅ TC LBC⋅ G JBC⋅ = [1] Equilibrium: 0.5q LAB⋅ TA− TC− 0= [2] Initial guess: TA 1kN m⋅:= TC 2kN m⋅:= Solving [1] and [2]: TA TC ⎛⎜⎜⎝ ⎞ ⎠ Find TA TC,( ):= TATC ⎛⎜⎜⎝ ⎞ ⎠ 878.26 21.74 ⎛⎜⎝ ⎞ ⎠N m⋅= Angle of Twist: φB TC LBC⋅ G JBC⋅ := φB 0.055987 rad= φB 3.208 deg= Ans Max. Shear Stress : cAB dAB 2 := τAB TA cAB( )⋅ JAB := τAB 165.66 MPa= cBC dBC 2 := τBC TC cBC( )⋅ JBC := τBC 32.8 MPa= τmax max τAB τBC,( ):= τmax 165.66 MPa= Ans Problem 5-87 The shaft of radius c is subjected to a distributed torque t, measured as torque/length of shaft. Determine the reactions at the fixed supports A and B. Problem 5-88 Compare the values of the maximum elastic shear stress and the angle of twist developed in 304 stainless steel shafts having circular and square cross sections. Each shaft has the same cross-sectiona area of 5600 mm2, length of 900 mm, and is subjected to a torque of 500 N·m. Given: L 900mm:= A 5600mm2:= G 75GPa:= T 500N m⋅:= Solution: Max. Shear Stress : For circular shaft: A π r2⋅= r Aπ:= c r:= J π 2 r4⋅:= τC_max T c( )⋅ J := τC_max 4.23 MPa= Ans For squarr shaft: A a2= a A:= τS_max T 4.81( )⋅ a3 := τS_max 5.74 MPa= Ans Angle of Twist: For circular shaft: φC T L⋅ G J⋅:= φC 0.001202 rad= φC 0.0689 deg= Ans For sqaure shaft: φS 7.10( )T L⋅ G a4⋅ := φS 0.001358 rad= φS 0.0778 deg= Ans Note: The sqaure shaft has a greater maximum shear stress and angle of twist. Problem 5-89 The shaft is made of red brass C83400 and has an elliptical cross section. If it is subjected to the torsional loading shown, determine the maximum shear stress within regions AC and BC, and the ang of twist φ of end B relative to end A. Given: LAC 2m:= LCB 1.5m:= a 50mm:= b 20mm:= TA 50N m⋅:= TB 30N m⋅:= G 37GPa:= TC 20N m⋅:= Solution: Max. Shear Stress : τBC 2TB π a⋅ b2⋅ := τBC 0.955 MPa= Ans τAC 2TA π a⋅ b2⋅ := τAC 1.592 MPa= Ans Angle of Twist: φB_A n a2 b2+( )Tn Ln⋅ π a3⋅ b3⋅ G⋅∑= φB_A a2 b2+( ) TB− LCB⋅ TA LAC⋅−( ) π a3⋅ b3⋅ G⋅ := φB_A 0.003618− rad= φB_A 0.2073 deg= Ans Problem 5-90 Solve Prob. 5-89 for the maximum shear stress within regions AC and BC, and the angle of twist φ of end B relative to C. Given: LAC 2m:= LCB 1.5m:= a 50mm:= b 20mm:= TA 50N m⋅:= TB 30N m⋅:= G 37GPa:= TC 20N m⋅:= Solution: Max. Shear Stress : τBC 2TB π a⋅ b2⋅ := τBC 0.955 MPa= Ans τAC 2TA π a⋅ b2⋅ := τAC 1.592 MPa= Ans Angle of Twist: φB_C n a2 b2+( )Tn Ln⋅ π a3⋅ b3⋅ G⋅∑= φB_C a2 b2+( ) TB− LCB⋅( ) π a3⋅ b3⋅ G⋅ := φB_C 0.001123− rad= φB_C 0.0643 deg= Ans Problem 5-91 The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the largest couple forces F that can be applied to the shaft without causing the steel to yield. τY = 56 MPa. Given: L 300mm:= Lw 400mm:= a 25mm:= τY 56MPa:= Solution: Max. Shear Stress : τmax 4.81T a3 = T a3τY 4.81 := T 181.91 N m⋅= Equilibrium: F Lw( )⋅ T− 0= F TLw:= F 454.78 N= Ans Problem 5-92 The steel shaft is 300 mm long and is screwed into the wall using a wrench. Determine the maximum shear stress in the shaft and the amount of displacement that each couple force undergoes if the couple forces have a magnitude of F = 150 N. Gst = 75 GPa. Given: L 300mm:= Lw 400mm:= a 25mm:= F 150N:= Gst 75GPa:= Solution: Equilibrium: F Lw( )⋅ T− 0= T F Lw( )⋅:= T 60 N m⋅= Max. Shear Stress : τmax 4.81T a3 := τmax 18.47 MPa= Ans Angle of Twist: φ 7.10( )T L⋅ Gst a 4⋅ := φ 0.004362 rad= δF φ Lw 2 ⋅:= δF 0.872 mm= Ans Problem 5-93 The shaft is made of plastic and has an elliptical cross-section. If it is subjected to the torsional loadin shown, determine the shear stress at point A and show the shear stress on a volume element located at this point. Also, determine the angle of twist φ at the end B. Gp = 15 GPa. Given: LOC 2m:= LCB 1.5m:= a 50mm:= b 20mm:= TB 50N m⋅:= TC 40N m⋅:= G 15GPa:= Solution: TOC TB TC+:= Shear Stress : τA 2TOC π a⋅ b2⋅ := τA 2.865 MPa= Ans Angle of Twist: φB n a2 b2+( )Tn Ln⋅ π a3⋅ b3⋅ G⋅∑= φB a2 b2+( ) TB LCB⋅ TOC LOC⋅+( ) π a3⋅ b3⋅ G⋅ := φB 0.015693 rad= φB 0.8991 deg= Ans Problem 5-94 The square shaft is used at the end of a drive cable in order to register the rotation of the cable on a gauge. If it has the dimensions shown and is subjected to a torque of 8 N·m, determine the shear stres in the shaft at point A. Sketch the shear stress on a volume element located at this point. Given: a 5mm:= T 8N m⋅:= Solution: Maximum Shear Stress : τA.max 4.81T a3 := τA.max 307.8 MPa= Ans Problem 5-95 The brass wire has a triangular cross section, 2 mm on a side. If the yield stress for brass is τY = 205 MPa, determine the maximum torque T to which it can be subjected so that the wire will not yield. If this torque is applied to a segment 4 m long, determine the greatest angle of twist of one end of the wire relative to the other end that will not cause permanent damage to the wire. Gbr = 37 GPa. Given: a 2mm:= L 4m:= G 37GPa:= τY 205MPa:= Solution: Allowable Shear Stress : τallow 20T a3 = τallow τY:= T τY a3⋅ 20 := T 0.0820 N m⋅= Ans Angle of Twist : φ 46T L⋅ G a4⋅ := φ 25.49 rad= Ans Problem 5-96 It is intended to manufacture a circular bar to resist torque; however, the bar is made elliptical in the process of manufacturing, with one dimension smaller than the other by a factor k as shown. Determine the factor by which the maximum shear stress is increased. Given: a 0.5 do⋅= b 0.5 k⋅ do⋅= Solution: Maximum Shear Stress : For the circular shaft:: τc.max T c⋅ J = c 0.5 do⋅= J π 32 do 4⋅= τc.max 16T π do3⋅ = For the elliptical shaft: τe.max 2T π a⋅ b2⋅ = τe.max 2T π 0.5 do⋅( )⋅ 0.5 k⋅ do⋅( )2⋅= τt.max 16T π k2⋅ do3⋅ = Factor of increase in shear stress: Fτ τe.max τc.max = Fτ 16T π k2⋅ do3⋅ 16T π do3⋅ = Fτ 1 k2 = Ans Problem 5-97 The 2014-T6 aluminum strut is fixed between the two walls at A and B. If it has a 50 mm by 50 mm square cross section, and it is subjected to the torsional loading shown, determine the reactions at the fixed supports. Also, what is the angle of twist at C? Given: LAC 600mm:= LCD 600mm:= a 50mm:= LDB 600mm:= TC 60N m⋅:= TD 30N m⋅:= G 27GPa:= Solution: T TC TD+:= T 90 N m⋅= L LAC LCD+ LDB+:= Compatibility : φT φB− 0= 7.10 T LAC⋅( ) G a4⋅ 7.10 TD LCD⋅( ) G a4⋅ + 7.10 TB L⋅( ) G a4⋅ − 0= TB T LAC⋅ TD LCD⋅+ L := TB 40 N m⋅= Ans Equilibrium: TA TB+ T− 0= TA T TB−:= TA 50 N m⋅= Ans Angle of Twist: φC 7.10( )TA LAC⋅ G a4⋅ := φC 0.001262 rad= φC 0.0723 deg= Ans Problem 5-98 The 304 stainless steel tube has a thickness of 10 mm. If the allowable shear stress is τallow = 80 MPa, determine the maximum torque T that it can transmit. Also, what is the angle of twist of one end of th tube with respect to the other if the tube is 4 m long? Neglect the stress concentrations at the corners. The mean dimensions are shown. Given: a 70mm:= b 30mm:= L 4m:= τallow 80 MPa⋅:= t 10mm:= G 75GPa:= Solution: Section properties : S Σds= Am a b⋅:= Am 2100 mm2= S 2a 2 b⋅+:= S 200 mm= Average shear stress: τavg T 2t Am⋅ = τavg τallow:= T 2 t⋅ Am⋅ τallow⋅:= T 3.36 kN m⋅= Ans Angle of Twist : φ T L⋅ 4Am 2 G⋅ s 1 t ⌠⎮ ⎮⌡ d⋅= φ T L⋅ 4Am 2 G⋅ S t ⎛⎜⎝ ⎞ ⎠⋅:= φ 0.203175 rad= φ 11.641 deg= Ans Problem 5-99 The 304 stainless steel tube has a thickness of 10 mm. If the applied torque is T = 50 N·m, determine the average shear stress in the tube. Neglect the stress concentrations at the corners. The mean dimensions are shown. Given: a 70mm:= b 30mm:= t 10mm:= T 50N m⋅:= G 75GPa:= Solution: Section properties : Am a b⋅:= Am 2100 mm2= Average shear stress: τavg T 2t Am⋅ := τavg 1.19 MPa= Ans Problem 5-100 Determine the constant thickness of the rectangular tube if the average shear stress is not to exceed 84 MPa when a torque of T = 2.5 kN·m is applied to the tube. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown. Given: a 100mm:= b 50mm:= T 2.5kN m⋅:= τallow 84MPa:= Solution: Section properties : Am a b⋅:= Am 5000 mm2= Average shear stress: τavg T 2t Am⋅ = t T 2 Am⋅ τallow⋅ := t 2.98 mm= Ans Problem 5-101 Determine the torque T that can be applied to the rectangular tube if the average shear stress is not to exceed 84 MPa. Neglect stress concentrations at the corners. The mean dimensions of the tube are shown and the tube has a thickness of 3 mm. Given: a 100mm:= b 50mm:= t 3mm:= τallow 84MPa:= Solution: Section properties : Am a b⋅:= Am 5000 mm2= Average shear stress: τavg T 2t Am⋅ = T 2 t⋅ Am⋅ τallow⋅:= T 2.52 kN m⋅= Ans Problem 5-102 A tube having the dimensions shown is subjected to a torque of T = 50 N·m. Neglecting the stress concentrations at its corners, determine the average shear stress in the tube at points A and B. Show the shear stress on volume elements located at these points. Given: ao 50mm:= bo 50mm:= ta 3mm:= tb 5mm:= T 50N m⋅:= Solution: Section properties : a ao ta−:= b bo tb−:= Am a b⋅:= Am 2115 mm2= Average shear stress: τA.avg T 2ta Am⋅ := τA.avg 3.94 MPa= Ans τB.avg T 2tb Am⋅ := τB.avg 2.36 MPa= Ans Problem 5-103 The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the average shear stress at points A and B if it is subjected to the torque of T = 5 N·m. Show the shear stress on volume elements located at these points. Given: a 80mm:= b 110mm:= t 5mm:= c1 40mm:= c2 30mm:= T 5N m⋅:= Solution: Section properties : c c1 2 c2 2+:= c 50 mm= Am a b⋅ 1 2 a⋅ c2⋅+:= Am 10000 mm2= Average shear stress: τA.avg T 2t Am⋅ := τA.avg 0.05 MPa= Ans τB.avg T 2t Am⋅ := τB.avg 0.05 MPa= Ans Problem 5-104 The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t = 5 mm. If the allowable shear stress is τallow = 56 MPa, and the tube is to resist a torque of T = 375 N·m, determine the necessary dimension b. The mean area for the ellipse is Am = π b (0.5b). Given: T 375N m⋅:= t 5mm:= τallow 56MPa:= Solution: Section properties : Am π b⋅ 0.5 b⋅( )⋅= Average shear stress: τavg T 2t Am⋅ = τavg T 2t π b⋅ 0.5 b⋅( )⋅⎡⎣ ⎤⎦⋅= b T t π⋅ τallow⋅ := b 20.65 mm= Ans Problem 5-105 The tube is made of plastic, is 5 mm thick, and has the mean dimensions shown. Determine the average shear stress at points A and B if the tube is subjected to the torque of T = 500 N·m. Show the shear stress on volume elements located at these points. Neglect stress concentrations at the corners. Given: a 40mm:= b 100mm:= t 5mm:= c1 20mm:= c2 30mm:= T 500N m⋅:= Solution: Section properties : c c1 2 c2 2+:= c 36.06 mm= Am a b⋅ 2 1 2 a⋅ c2⋅⎛⎜⎝ ⎞ ⎠+:= Am 5200 mm 2= Average shear stress: τA.avg T 2t Am⋅ := τA.avg 9.62 MPa= Ans τB.avg T 2t Am⋅ := τB.avg 9.62 MPa= Ans Problem 5-106 The steel tube has an elliptical cross section of mean dimensions shown and a constant thickness of t = 5 mm. If the allowable shear stress is τallow = 56 MPa, determine the necessary dimension b needed to resist the torque shown.. The mean area for the ellipse is Am = π b (0.5b). Given: T1 75N m⋅:= τallow 56MPa:= T2 120− N m⋅:= t 5mm:= T3 450N m⋅:= Solution: Section properties : Am π b⋅ 0.5 b⋅( )⋅= Internal torque : Tmax T1 T2+ T3+:= Tmax 405 N m⋅= Average shear stress: τavg T 2t Am⋅ = τavg T 2t π b⋅ 0.5 b⋅( )⋅⎡⎣ ⎤⎦⋅= b Tmax t π⋅ τallow⋅ := b 21.46 mm= Ans Problem 5-107 The symmetric tube is made from a high-strength steel, having the mean dimensions shown and a thickness of 5 mm. If it is subjected to a torque of T = 40 N·m, determine the average shear stress developed at points A and B. Indicate the shear stress on volume elements located at these points. Given: a 40mm:= b 60mm:= t 5mm:= T 40N m⋅:= Solution: Section properties : Am a a⋅ 4 a b⋅( )+:= Am 11200 mm2= Average shear stress: τA.avg T 2t Am⋅ := τA.avg 0.357 MPa= Ans τB.avg T 2t Am⋅ := τB.avg 0.357 MPa= Ans Problem 5-108 Due to a fabrication error the inner circle of the tube is eccentric with respect to the outer circle. By what percentage is the torsional strength reduced when the eccentricity e is one-fourth of the difference in the radii? Given: e a b− 4 = Solution: For the aligned tube:: Section properties : t a b−= Am π a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅= Average shear stress: τA.avg T 2t Am⋅ = T τA.avg 2t Am⋅( )⋅= T τA.avg 2( )⋅ a b−( ) π⋅ a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅= For the eccentric tube: Section properties : A'm π a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅= t' a e 2 − e 2 b+⎛⎜⎝ ⎞ ⎠−= t' a e− b−= t' a a b− 4 − b−= t' 3 4 a b−( )= Average shear stress: τA.avg T' 2t' A'm⋅ = T' τA.avg 2t' A'm⋅( )⋅= T' τA.avg 2( )⋅ 3 4 a b−( ) π⋅ a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅= Factor of increase in shear stress: FT T' T = Fτ τA.avg 2( )⋅ 3 4 a b−( ) π⋅ a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅ τA.avg 2( )⋅ a b−( ) π⋅ a b+ 2 ⎛⎜⎝ ⎞ ⎠ 2 ⋅ = FT 3 4 := % reduction in strength : T% 1 FT−( ) 100⋅:= T% 25.00= Ans Problem 5-109 For a given average shear stress, determine the factor by which the torque-carrying capacity is increased if the half-circular sections are reversed from the dashed-line positions to the section shown. The tube is 2.5 mm thick. Given: a 30mm:= b 45mm:= r 15mm:= t 2.5mm:= Solution: Section properties : am a t−:= rm r 0.5t−:= A'm am( ) b⋅ 2 π2 rm2⋅⎛⎜⎝ ⎞⎠⋅−:= A'm 643.54 mm2= Am am( ) b⋅ 2 π2 rm2⋅⎛⎜⎝ ⎞⎠⋅+:= Am 1831.46 mm2= Average shear stress: τavg T 2t Am⋅ = T τavg 2t Am⋅( )⋅= T' τavg 2t A'm⋅( )⋅= Hence, the factor of increase is: α T T' = α Am A'm := α 2.85= Ans Problem 5-110 For a given maximum shear stress, determine the factor by which the torque carrying capacity is increased if the half-circular section is reversed from the dashed-line position to the section shown. The tube is 2.5 mm thick. Given: a 30mm:= b 45mm:= r 15mm:= t 2.5mm:= Solution: Section properties : am a t−:= bm b 0.5t−:= rm r 0.5t−:= A'm am( ) bm⋅ π2 rm2⋅⎛⎜⎝ ⎞⎠−:= A'm 906.15 mm2= Am am( ) bm⋅ π2 rm2⋅⎛⎜⎝ ⎞⎠+:= Am 1500.10 mm2= Average shear stress: τavg T 2t Am⋅ = T τavg 2t Am⋅( )⋅= T' τavg 2t A'm⋅( )⋅= Hence, the factor of increase is: α T T' = α Am A'm := α 1.66= Ans Problem 5-111 The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members are connected with a fillet weld of radius r = 4 mm, determine the maximum torque T that can be applied Given: D 50mm:= d 20mm:= r 4mm:= τallow 8MPa:= Solution: Stress Concentration Factor : D d 2.50= r d 0.20= From Fig. 5-36, K 1.25:= Allowable Shear Stress: c d 2 := J π 32 d4⋅:= τallow K 0.5T c⋅ J ⎛⎜⎝ ⎞ ⎠⋅= T τallow J⋅ 0.5K c⋅:= T 20.11 N m⋅= Ans Problem 5-112 The shaft is used to transmit 660 W while turning at 450 rpm. Determine the maximum shear stress in the shaft. The segments are connected together using a fillet weld having a radius of 1.875 mm. Unit used: rpm 2π 60 rad s := Given: D 25mm:= d 12.5mm:= r 1.875mm:= ω 450rpm:= P 660W:= Solution: T P ω:= T 14.01 N m⋅= Stress Concentration Factor : D d 2.00= r d 0.15= From Fig. 5-36, K 1.30:= Max. shear stress: c d 2 := J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅:= τmax K T c⋅ J ⎛⎜⎝ ⎞ ⎠⋅:= τmax 47.48 MPa= Ans Problem 5-113 The shaft is fixed to the wall at A and is subjected to the torques shown. Determine the maximum shear stress in the shaft. A fillet weld having a radius of 4.5 mm is used to connect the shafts at B. Given: D 60mm:= d 30mm:= r 4.5mm:= TC 250N m⋅:= TD 300− N m⋅:= TE 800N m⋅:= Solution: Internal Torque : As shown in the torque diagram. TCD TC:= TDB TC TD+:= TBE TC TD+:= TEA TC TD+ TE+:= Maximum Shear Stress : For segment CD: c d 2 := J π 32 d4⋅:= τCD TCD c( )⋅ J := τCD 47.16 MPa= Ans (Max.) For segment EA: c' D 2 := J' π 32 D4⋅:= τEA TEA c'( )⋅ J' := τEA 17.68 MPa= Ans For the fillet: Stress Concentration Factor : D d 2.00= r d 0.15= From Fig. 5-36, K 1.30:= Max. shear stress: τmax K TDB c⋅ J ⎛⎜⎝ ⎞ ⎠⋅:= τmax 12.26 MPa= Ans Problem 5-114 The built-up shaft is to be designed to rotate at 720 rpm while transmitting 30 kW of power. Is this possible? The allowable shear stress is τallow = 12 MPa. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 60mm:= D 75mm:= ω 720rpm:= P 30kW:= τallow 12MPa:= Solution: ω 75.40 rad s = T Pω:= T 397.89 N m⋅= Maximum Shear Stress : c d 2 := J π 32 d4⋅:= τallow K T c⋅ J ⋅= K τallow J⋅ T c⋅:= K 1.28= Stress Concentration Factor : D d 1.25= K 1.28= From Fig. 5-36, r d 0.133= r 0.133d:= r 7.98 mm= Check: D d− 2 7.50 mm= < r = 7.98mm No. It si impossible. Problem 5-115 The built-up shaft is designed to rotate at 540 rpm. If the radius of the fillet weld connecting the shaft is r = 7.20 mm, and the allowable shear stress for the material is τallow = 55 MPa, determine the maximum power the shaft can transmit. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: d 60mm:= D 75mm:= ω 540rpm:= r 7.20mm:= τallow 55MPa:= Solution: Stress Concentration Factor : D d 1.25= r d 0.12= From Fig. 5-36, K 1.30:= Maximum Shear Stress : c d 2 := J π 32 d4⋅:= τallow K T c⋅ J ⋅= T τallow J⋅ K c⋅:= T 1.7943 kN m⋅= Maximum Power : ω 56.55 rad s = P T ω⋅:= P 101.47 kW= Ans Problem 5-116 The steel used for the shaft has an allowable shear stress of τallow = 8 MPa. If the members are connected together with a fillet weld of radius r = 2.25 mm, determine the maximum torque T that can be applied. Given: d 15mm:= D 30mm:= r 2.25mm:= τallow 8MPa:= Solution: Stress Concentration Factor : D d 2.00= r d 0.15= From Fig. 5-36, K 1.30:= Maximum Shear Stress : c d 2 := J π 32 d4⋅:= τallow K 0.5T c⋅ J ⋅= T τallow J⋅ 0.5K c⋅:= T 8.156 N m⋅= Ans Problem 5-117 A solid shaft is subjected to the torque T, which causes the material to yield. If the material is elastic-plastic, show that the torque can be expressed in terms of the angle of twist φ of the shaft as T = 4/3 TY (1 - φY3/ 4φ3), where TY and φY are the torque and angle of twist when the material begins to yield. Problem 5-118 A solid shaft having a diameter of 50 mm is made of elastic-plastic material having a yield stress of τY = 112 MPa and shear modulus of G = 84 GPa. Determine the torque required to develop an elastic core in the shaft having a diameter of 25 mm. Also, what is the plastic torque? Given: d 50mm:= G 84GPa:= dY 25mm:= τY 112MPa:= Solution: c d 2 := ρY dY 2 := Elastic-plastic torque: Use Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅:= T 3.551 kN m⋅= Ans Plastic torque: Use Eq. 5-27 from the text : TP 2π τY( )⋅ 3 c3( )⋅:= TP 3.665 kN m⋅= Ans Problem 5-119 Determine the torque needed to twist a short 3-mm-diameter steel wire through several revolutions if it is made from steel assumed to be elastic-plastic and having a yield stress of τY = 80 MPa. Assume that the material becomes fully plastic. Given: d 3mm:= τY 80MPa:= Solution: c d 2 := Plastic torque: Use Eq. 5-27 from the text : TP 2π τY( )⋅ 3 c3( )⋅:= TP 0.565 N m⋅= Ans Problem 5-120 A solid shaft has a diameter of 40 mm and length of 1 m. It is made from an elastic-plastic material having a yield stress of τY = 100 MPa. Determine the maximum elastic torque TY and the corresponding angle of twist. What is the angle of twist if the torque is increased to T = 1.2TY? G = 80 GPa. Given: d 40mm:= G 80GPa:= L 1m:= τY 100MPa:= Solution: c d 2 := J π 32 d4⋅:= Maximum Elastic Torque: τY TY c⋅ J = TY τY( ) J⋅ c := TY 1.257 kN m⋅= Ans Angle of Twist: γY τY G := γY 0.00125 rad= φ γY( ) L⋅ c := φ 0.0625 rad= φ 3.581 deg= Ans Elastic-plastic torque: T 1.2TY= Use Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅= 1.2TY π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅= ρY 3 4c3 7.2 TY⋅ π τY( )⋅−:= ρY 14.74 mm= Angle of Twist: φ' γY( ) L⋅ ρY := φ' 0.084826 rad= φ' 4.86 deg= Ans Problem 5-121 The stepped shaft is subjected to a torque T that produces yielding on the surface of the larger diamet segment. Determine the radius of the elastic core produced in the smaller diameter segment. Neglect the stress concentration at the fillet. Given: do 60mm:= ds 55mm:= Solution: co do 2 := Jo π 32 do 4⋅:= cs ds 2 := Js π 32 ds 4⋅:= Set τY 1MPa:= Maximum Elastic Torque: For the larger diameter segment. τY TY c⋅ J = TY τY( ) Jo⋅ co := TY 42.41 N m⋅= Elastic-plastic torque: For the smaller diameter segment. TP 2π τY( )⋅ 3 cs 3⎛⎝ ⎞⎠⋅:= TP 43.56 N m⋅= > TY Applying Eq. 5-26 from the text : T π τY( )⋅ 6 4cs 3 ρY3−⎛⎝ ⎞⎠⋅= TY π τY( )⋅ 6 4cs 3 ρY3−⎛⎝ ⎞⎠⋅= ρY 3 4cs 3 6 TY⋅ π τY( )⋅−:= ρY 12.98 mm= Ans Problem 5-122 A bar having a circular cross section of 75 mm diameter is subjected to a torque of 12 kN·m. If the material is elastic-plastic, with τY = 120 MPa, determine the radius of the elastic core. Given: d 75mm:= T 12kN m⋅:= τY 120MPa:= Solution: c d 2 := Elastic-plastic torque: Use Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅= ρY 3 4c3 6T π τY( )⋅−:= ρY 27.12 mm= Ans Problem 5-123 A tubular shaft has an inner diameter of 20 mm, an outer diameter of 40 mm, and a length of 1 m. It i made of an elastic perfectly plastic material having a yield stress of τY = 100 MPa. Determine the maximum torque it can transmit.What is the angle of twist of one end with respect to the other end if the shear strain on the inner surface of the tube is about to yield? G = 80 GPa. Given: do 40mm:= G 80GPa:= L 1m:= di 20mm:= τY 100MPa:= Solution: ρY di 2 := Plastic Torque: co 0.5do:= ci 0.5di:= TP 2π ci co ρτY( ) ρ2⋅⌠⎮⎮⌡ d⋅:= TP 1.466 kN m⋅= Ans Angle of Twist: γY τY G := γY 0.00125 rad= φ γY( ) L⋅ ρY := φ 0.125 rad= φ 7.162 deg= Ans Problem 5-124 The 2-m-long tube is made from an elastic-plastic material as shown. Determine the applied torque T, which subjects the material of the tube's outer edge to a shearing strain of γmax = 0.008 rad. What would be the permanent angle of twist of the tube when the torque is removed? Sketch the residual stress distribution of the tube. Given: ro 45mm:= ri 40mm:= L 2m:= γY 0.003:= τY 240MPa:= γmax 0.008:= Solution: Determine if it is fully plastic : φmax γmax L⋅ ro := φmax 0.35556 rad= However, φmax γY L⋅ ρY = ρY γY L⋅ φmax := ρY 16.88 mm= < ri = 40mm Therefore, the tube is filly plastic. Also, at r = ri : γr ri ro γmax⋅:= γr 0.00711= > γY = 0.003γmax ro γr ri = Again, the tube is filly plastic. Plastic Torque : Tp 2π ri ro ρτY( ) ρ2⋅⌠⎮⎮⌡ d⋅:= Tp 13.63 kN m⋅= Ans Angle of Twist: When the torque is removed: The equal but opposite torque TP is applied. G τY γY := J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅:= φ' Tp L⋅ G J⋅:= φ' 0.14085 rad= Permanent angle of twist: φr φmax φ'−:= φr 0.21470 rad= φr 12.30 deg= Ans Shear Stresses : At r = ro : τ'po Tp ro⋅ J := τ'po 253.5 MPa= At r = ri : τ'pi Tp ri⋅ J := τ'pi 225.4 MPa= Problem 5-125 The tube has a length of 2 m and is made of an elastic-plastic material as shown. Determine the torqu needed to just cause the material to become fully plastic. What is the permanent angle of twist of the tube when this torque is removed? Given: ro 100mm:= ri 60mm:= L 2m:= γY 0.007:= τY 350MPa:= Solution: At Just Fully Plastic : ρY ri:= φp γY L⋅ ρY := φp 0.23333 rad= Plastic Torque : Tp 2π ri ro ρτY( ) ρ2⋅⌠⎮⎮⌡ d⋅:= Tp 574.70 kN m⋅= Ans Angle of Twist: When the torque is removed: The equal but opposite torque TP is applied. G τY γY := J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅:= φ' Tp L⋅ G J⋅:= φ' 0.16814 rad= Permanent angle of twist: φr φp φ'−:= φr 0.06520 rad= φr 3.74 deg= Ans Problem 5-126 The shaft is made from a strain-hardening material having a τ -γ diagram as shown. Determine the torque T that must be applied to the shaft in order to create an elastic core in the shaft having a radius of ρc = 12.5 mm. Given: r 15mm:= ρc 12.5mm:= γa 0.005:= τa 70MPa:= γb 0.010:= τb 105MPa:= Solution: c r:= ρa ρc:= For linear strain variations against ρ : γ ρ γa ρa = γ γa ρa ⎛⎜⎝ ⎞ ⎠ ρ⋅= Path 1: τ1 τa γa ⎛⎜⎝ ⎞ ⎠ γ⋅= τ1 τa ρa ⎛⎜⎝ ⎞ ⎠ ρ⋅= Path 2: τ2 τa− γ γa− τb τa− γb γa− = τ2 τb τa− γb γa− ⎛⎜⎝ ⎞ ⎠ γ γa−⋅( )⋅ τa+= τ2 τb τa− γb γa− ⎛⎜⎝ ⎞ ⎠ ρ ρa 1−⎛⎜⎝ ⎞ ⎠ ⋅ γa⋅ τa+= T 2π 0 c ρτ ρ2⋅⌠⎮⌡ d⋅= T 2π 0 ρa ρτ1 ρ2⋅ ⌠⎮⌡ d⋅ 2π ρa c ρτ2 ρ2⋅ ⌠⎮ ⎮⌡ d+= T 2π τa ρa ⎛⎜⎝ ⎞ ⎠ ⋅ 0 ρa ρρ3⌠⎮⌡ d⋅ 2π γa⋅ τb τa− γb γa− ⎛⎜⎝ ⎞ ⎠ ⋅ ρa c ρρρa 1−⎛⎜⎝ ⎞ ⎠ ρ2⋅ ⌠⎮ ⎮ ⎮⌡ d+ 2π τa⋅ ρa c ρρ2⌠⎮⌡ d⋅+:= T 434.27 N m⋅= Ans Problem 5-127 The 2-m-long tube is made of an elastic perfectly plastic material as shown. Determine the applied torque T that subjects the material at the tube's outer edge to a shear strain of γmax = 0.006 rad. What would be the permanent angle of twist of the tube when this torque is removed? Sketch the residual stress distribution in the tube. Given: ro 35mm:= ri 30mm:= L 2m:= γY 0.003:= τY 210MPa:= γmax 0.006:= Solution: At Fully Plastic : ρY ri:= φp γmax L⋅ ro := φp 0.34286 rad= Also, at r = ri : γr ri ro γmax⋅:= γr 0.00514=γmax ro γr ri = > γY = 0.003 Confirm that tube is filly plastic. Plastic Torque : Tp 2π ri ro ρτY( ) ρ2⋅⌠⎮⎮⌡ d⋅:= Tp 6.98 kN m⋅= Ans Angle of Twist: When the torque is removed: The equal but opposite torque TP is applied. G τY γY := J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅:= φ' Tp L⋅ G J⋅:= φ' 0.18389 rad= Permanent angle of twist: φr φp φ'−:= φr 0.15897 rad= φr 9.11 deg= Ans Residual Shear Stresses : At r = ro : τ'po Tp ro⋅ J := τ'po 225.3 MPa= τro τY− τ'po+:= τro 15.3 MPa= At r = ri : τ'pi Tp ri⋅ J := τ'pi 193.1 MPa= τri τY− τ'pi+:= τri 16.9− MPa= Problem 5-128 The shear stressstrain diagram for a solid 50-mm diameter shaft can be approximated as shown in the figure. Determine the torque required to cause a maximum shear stress in the shaft of 125 MPa. If the shaft is 3 m long, what is the corresponding angle of twist? Given: do 50mm:= L 3m:= γ1 0.0025:= τ1 50MPa:= γmax 0.010:= τmax 125MPa:= Solution: ro 0.5 do⋅:= τ-ρ Function : At r = ρ1 : γmax ro γ1 ρ1 = ρ1 γ1 γmax ro⋅:= ρ1 6.25 mm= For the region 0 < r < ρ1: k1 τ1 0− ρ1 0− := k1 8.00 MPa mm = k1 τ 0− ρ 0−= τ k1 ρ⋅= For the region ρ1 < r < r0: k2 τmax τ1− ro ρ1− := k2 4.00 MPa mm = k2 τ' τ1− ρ' ρ1− = τ' τ1 k2 ρ' ρ1−( )⋅+= Plastic Torque : Tp 2π 0 ρ1 ρ'τ'( ) ρ'2⋅⌠⎮⌡ d⋅ 2π ρ1 ro ρτ( ) ρ2⋅⌠⎮⌡ d⋅+= Tp 2π 0 ρ1 ρk1 ρ⋅( ) ρ2⋅⌠⎮⌡ d⋅ 2π ρ1 ro ρ'τ1 k2 ρ' ρ1−( )⋅+⎡⎣ ⎤⎦ ρ'2⋅⌠⎮⎮⌡ d⋅+:= Tp 3.27 kN m⋅= Ans Angle of Twist: φmax γmax L⋅ ro := φmax 1.20000 rad= φmax 68.75 deg= Ans Problem 5-129 The shaft consists of two sections that are rigidly connected. If the material is elastic perfectly plastic as shown, determine the largest torque T that can be applied to the shaft. Also, draw the shear-stress distribution over a radial line for each section. Neglect the effect of stress concentration. Given: r1 20mm:= r2 25mm:= γY 0.002:= τY 70MPa:= Solution: Plastic Torque : For the smaller-diameter segment c r1:= Tp 2π 0 c ρτY( ) ρ2⋅⌠⎮⌡ d⋅:= Tp 1.173 kN m⋅= Ans Max. Shear Stress : For the bigger-diameter segment c r2:= J π 2 r2 4⋅:= τmax Tp c( )⋅ J := τmax 47.79 MPa= Ans Problem 5-130 The shaft is made of an elastic-perfectly plastic material as shown. Plot the shear-stress distribution acting along a radial line if it is subjected to a torque of T = 2 kN·m. What is the residual stress distribution in the shaft when the torque is removed? Given: ro 20mm:= T 2kN m⋅:= γY 0.001875:= τY 150MPa:= Solution: c ro:= J π 2 ro 4⋅:= Maximum Elastic Torque: τY TY c⋅ J = TY τY( ) J⋅ c := TY 1.885 kN m⋅= < T = 2 kNm Plastic Torque: TP 2π τY⋅ 3 c3( )⋅:= TP 2.513 kN m⋅= > T = 2 kNm Therefore, it is elastic-plastic. Elastic-plastic Torque: Applying Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅= ρY 3 4c3 6 T⋅ π τY( )⋅−:= ρY 18.70 mm= Residual Shear Stresses : When the torque is removed: The equal but opposite torque T is applied. At r = ro : τ'po T ro⋅ J := τ'po 159.15 MPa= τro τY− τ'po+:= τro 9.15 MPa= At r = ρY : τ'pi T ρY⋅ J := τ'pi 148.78 MPa= τri τY− τ'pi+:= τri 1.22− MPa= Problem 5-131 A 40-mm-diameter shaft is made from an elasticplastic material as shown. Determine the radius of its elastic core if it is subjected to a torque of T = 300 N·m. If the shaft is 250 mm long, determine the angle of twist. Given: d 40mm:= L 250mm:= T 300N m⋅:= γY 0.006:= τY 21MPa:= Solution: c d 2 := Elastic-plastic torque: Use Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅= ρY 3 4c3 6T π τY( )⋅−:= ρY 16.77 mm= Ans Angle of Twist: φ γY( ) L⋅ ρY := φ 0.089445 rad= φ 5.1248 deg= Ans Problem 5-132 A torque is applied to the shaft having a radius of 100 mm. If the material obeys a shear stress-strain relation of τ = 20γ 1/3 MPa, determine the torque that must be applied to the shaft so that the maximum shear strain becomes 0.005 rad. Given: ro 100mm:= γmax 0.005:= τ 20 3 γ unit⋅= Solution: c ro:= τ-ρ Function : unit MPa:= γ ρ c γmax⋅= τ 20 3 γ unit⋅= τ 20 3 ρ c γmax⋅ unit⋅= Ultimate Torque : T 2π 0 c ρτ ρ2⋅⌠⎮⌡ d⋅= T 2π unit⋅ 0 c ρ20 3 ρ c γmax⋅ ρ2( )⋅ ⌠⎮ ⎮⌡ d⋅:= T 6.446 kN m⋅= Ans Problem 5-133 The shaft is made of an elastic-perfectly plastic material as shown. Determine the torque that the shaf can transmit if the allowable angle of twist is 0.375 rad. Also, determine the permanent angle of twist once the torque is removed. The shaft is 2-m-long. Given: ro 20mm:= φallow 0.375rad:= γY 0.001875:= τY 150MPa:= L 2m:= Solution: c ro:= Angle of Twist: γmax φallow c⋅ L := γmax 0.00375 rad= γmax c γY ρY = ρY γY γmax c⋅:= ρY 10.00 mm= Elastic-plastic Torque: Applying Eq. 5-26 from the text : T π τY( )⋅ 6 4c3 ρY3−⎛⎝ ⎞⎠⋅:= T 2.435 kN m⋅= Ans Permanent Angle of Twist: When the torque is removed: The equal but opposite torque T is applied. G τY γY := J π 2 ro 4⋅:= φ' T L⋅ G J⋅:= φ' 0.24219 rad= Permanent angle of twist: φr φallow φ'−:= φr 0.13281 rad= φr 7.61 deg= Ans Problem 5-134 Consider a thin-walled tube of mean radius r and thickness t. Show that the maximum shear stress in the tube due to an applied torque T approaches the average shear stress computed from Eq. 5-18 as r/t approaches infinity. Problem 5-135 The 304 stainless steel shaft is 3 m long and has an outer diameter of 60 mm. When it is rotating at 60 rad/s, it transmits 30 kW of power from the engine E to the generator G. Determine the smallest thickness of the shaft if the allowable shear stress is τallow = 150 MPa and the shaft is restricted not to twist more than 0.08 rad. Given: do 60mm:= L 3m:= ω 60 rad s ⋅:= P 30kW:= G 75GPa:= τallow 150MPa:= φallow 0.08rad:= Solution: T P ω:= T 500 N m⋅= Allowable Shear Stress : Assume failure due to shear strss. ro do 2 := c ro:= J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅= τallow T c⋅ J = J T c⋅ τallow = π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅ T ro⋅ τallow = ri 4 ro 4 2T ro⋅ π τallow⋅ −:= ri 29.392 mm= Angle of Twist : Assume failure due to angle of twist limitation. φ T L⋅ G J⋅= J' T L⋅ G φallow⋅ := π 2 ro 4 r'i 4−⎛⎝ ⎞⎠⋅ T L⋅G φallow⋅= r'i 4 ro 4 2T L⋅ π G⋅ φallow⋅ −:= r'i 28.403 mm= Choose the smallest value of ri : ri min ri r'i,( ):= ri 28.403 mm= t ro ri−:= t 1.597 mm= Ans Problem 5-136 The 304 stainless solid steel shaft is 3 m long and has a diameter of 50 mm. It is required to transmit 40 kW of power from the engine E to the generator G. Determine the smallest angular velocity the shaft can have if it is restricted not to twist more than 1.5°. Given: do 50mm:= L 3m:= P 40kW:= G 75GPa:= φallow 1.5deg:= Solution: Angle of Twist : ro do 2 := c ro:= J π 2 ro 4⋅:= φ T L⋅ G J⋅= T G J⋅ L φallow⋅:= T 401.60 N m⋅= Angular Velocity: T P ω= ω P T := ω 99.6 rad s = Ans Problem 5-137 The drilling pipe on an oil rig is made from steel pipe having an outside diameter of 112 mm and a thickness of 6 mm. If the pipe is turning at 650 rev/min while being powered by a 12-kW motor, determine the maximum shear stress in the pipe. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: do 112mm:= t 6mm:= ω 650 rpm⋅:= P 12kW:= τallow 70MPa:= Solution: ω 68.07 rad s = T P ω:= T 176.29 N m⋅= Max. stress : c do 2 := di do 2t−:= J π 2 ⎛⎜⎝ ⎞ ⎠ do 2 ⎛⎜⎝ ⎞ ⎠ 4 di 2 ⎛⎜⎝ ⎞ ⎠ 4 − ⎡⎢⎣ ⎤⎥⎦⋅:= τmax T c⋅ J := τmax 1.75 MPa= Ans Problem 5-138 The tapered shaft is made from 2014-T6 aluminum alloy, and has a radius which can be described by the function r = 0.02 (1 + x3/2) m, where x is in meters. Determine the angle of twist of its end A if it i subjected to a torque of 450 N·m. Given: T 450N m⋅:= L 4m:= G 27GPa:= r 0.02 1 x3+( ) unit⋅= Solution: unit m:= Angle of Twist : J π 2 r4⋅= r 0.02 1 x3+( ) unit⋅= φ 0 L x T G J⋅ ⌠⎮ ⎮⌡ d= φ unit 0 L m x 2T G π⋅ 0.02 1 x3+( ) unit⋅⎡⎣ ⎤⎦4⋅ ⌠⎮ ⎮ ⎮⌡ d⋅:= φ 0.02771 rad= φ 1.588 deg= Ans Problem 5-139 The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at 1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl shear stress is τallow = 56 MPa and the vibrations limit the angle of twist of the shaft to 0.05 rad. The shaft is 0.6 m long and made of L2 tool steel. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: L 600mm:= ω 1500 rpm⋅:= P 660kW:= τallow 56MPa:= G 75GPa:= φallow 0.05rad:= Solution: ω 157.08 rad s = T P ω:= T 4.202 kN m⋅= Allowable shear stress : Assume failure due to shear stress c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow T c⋅ J = Thus, d 2 2 π ⎛⎜⎝ ⎞ ⎠ T τallow ⎛⎜⎝ ⎞ ⎠ ⋅⎡⎢⎣ ⎤⎥⎦ 1 3 := d 72.57 mm= Angle of Twist : Assume failure due to angle of twist limitation J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= φ T L⋅ G J⋅= Thus, d 2 2 π ⎛⎜⎝ ⎞ ⎠ T L⋅ G φallow⋅ ⎛⎜⎝ ⎞ ⎠ ⋅⎡⎢⎣ ⎤⎥⎦ 0.25 := d 51.15 mm= Shear stress failure controls the design. Hence, Use d 75mm= Ans Problem 5-140 The engine of the helicopter is delivering 660 kW to the rotor shaft AB when the blade is rotating at 1500 rev/min. Determine to the nearest multiples of 5mm the diameter of the shaft AB if the allowabl shear stress is τallow = 75 MPa and the vibrations limit the angle of twist of the shaft to 0.03 rad. The shaft is 0.6 m long and made of L2 tool steel. Unit used: rpm 2π 60 ⎛⎜⎝ ⎞ ⎠ rad s := Given: L 600mm:= ω 1500 rpm⋅:= P 660kW:= τallow 75MPa:= G 75GPa:= φallow 0.03rad:= Solution: ω 157.08 rad s = T P ω:= T 4.202 kN m⋅= Allowable shear stress : Assume failure due to shear stress c d 2 = J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= τallow T c⋅ J = Thus, d 2 2 π ⎛⎜⎝ ⎞ ⎠ T τallow ⎛⎜⎝ ⎞ ⎠ ⋅⎡⎢⎣ ⎤⎥⎦ 1 3 := d 65.83 mm= Angle of Twist : Assume failure due to angle of twist limitation J π 2 d 2 ⎛⎜⎝ ⎞ ⎠ 4 ⋅= φ T L⋅ G J⋅= Thus, d 2 2 π ⎛⎜⎝ ⎞ ⎠ T L⋅ G φallow⋅ ⎛⎜⎝ ⎞ ⎠ ⋅⎡⎢⎣ ⎤⎥⎦ 0.25 := d 58.12 mm= Shear stress failure controls the design. Hence, Use d 70mm= Ans Problem 5-141 The material of which each of three shafts is made has a yield stress of τY and a shear modulus of G. Determine which shaft geometry will resist the largest torque without yielding. What percentage of th torque can be carried by the other two shafts? Assume that each shaft is made of the same amount of material and that it has the same cross-sectional area A. Given: AΟ A= Asq A= A∆ A= Solution: For circular shaft:: AΟ πc2= c A π= J π 2 c4⋅= J A 2 2π= τmax T c⋅ J = τmax 2 π⋅ T A A = T A A 2 π τY⋅= αΟ 1 2 π:= αΟ 0.2821= For square shaft: Asq a 2= a A= τmax 4.81T a3 = τmax 4.81T A A⋅= T A A 4.81 τY⋅= αsq 1 4.81 := αsq 0.2079= For triangular shaft: θ 60deg:= A∆ a 2 a⋅ sin θ( )⋅= a 2 A 4 3 = τmax 20T a3 = τmax 5 4 27⋅ T 2A A⋅= T 2A A 5 4 27⋅ τY⋅= α∆ 2 5 4 27⋅ := α∆ 0.1755= The circular shaft will carry the largest torque. Ans For square shaft: %sq αsq αΟ 100⋅:= %sq 73.7= Ans For triangular shaft: %∆ α∆ αΟ 100⋅:= %∆ 62.2= Ans Problem 5-142 The A-36 steel circular tube is subjected to a torque of 10 kN·m. Determine the shear stress at the mean radius ρ = 60 mm and compute the angle of twist of the tube if it is 4 m long and fixed at its far end. Solve the problem using Eqs. 5-7 and 5-15 and by using Eqs. 5-18 and 5-20. Given: ρ 60mm:= t 5mm:= L 4m:= T 10kN m⋅:= G 75GPa:= Solution: Section properties : ro ρ 0.5t+:= ri ρ 0.5t−:= Σds 2π ρ:= Σds 376.99 mm= Am π ρ2:= Am 11309.73 mm2= J π 2 ro 4 ri 4−⎛⎝ ⎞⎠⋅:= J 6797621.10 mm4= Shear Stress: Applying Eq. 5-7, c ρ:= τ T c⋅ J := τ 88.27 MPa= Ans Applying Eq. 5-18, τavg T 2 t⋅ Am := τavg 88.42 MPa= Ans Angle of Twist : Applying Eq. 5-20, φ' T L⋅ J G⋅:= φ' 0.07846 rad= φ' 4.495 deg= Ans Applying Eq. 5-20, φ T L⋅ 4Am 2 G⋅ s 1 t ⌠⎮ ⎮⌡ d⋅= φ T L⋅ 4Am 2 G⋅ Σds t ⎛⎜⎝ ⎞ ⎠⋅:= φ 0.07860 rad= φ 4.503 deg= Ans Problem 5-143 The aluminum tube has a thickness of 5 mm and the outer cross-sectional dimensions shown. Determine the maximum average shear stress in the tube. If the tube has a length of 5 m, determine th angle of twist. Gal = 28 GPa. Given: ao 150mm:= bo 100mm:= L 4m:= t 5mm:= TA 280N m⋅:= TB 135N m⋅:= LAB 2m:= LBC 3m:= G 28GPa:= Solution: Section properties : S Σds= a ao t−:= b bo t−:= Am a b⋅:= Am 13775 mm2= S 2a 2 b⋅+:= S 480 mm= Maximum Average shear stress: TAB TA:= TBC TA TB−:= Tmax max TAB TBC,( ):= Tmax 280 N m⋅= τavg_max Tmax 2t Am⋅ := τavg_max 2.03 MPa= Ans Angle of Twist : φ T L⋅ 4Am 2 G⋅ s 1 t ⌠⎮ ⎮⌡ d⋅= φ TAB LAB⋅ TBC LBC⋅+ 4Am 2 G⋅ S t ⎛⎜⎝ ⎞ ⎠⋅:= φ 0.004495 rad= φ 0.258 deg= Ans Problem 6-1 Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Given: a 250mm:= b 800mm:= F 24kN:= Solution: Given Equilibrium : + ΣFy=0; A B+ F− 0= ΣΜA=0; F− a⋅ B b⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 31.50 7.50− ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) F− 1kN⋅:= V2 x2( ) F− A+( ) 1kN⋅:= M1 x1( ) F− x1⋅kN m⋅:= M2 x2( ) F− x2( )⋅ A x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 0.5 1 40 20 0 20 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 0.5 1 5 0 Distane (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-2 The load binder is used to support a load. If the force applied to the handle is 250 N, determine the tensions T1 and T2 in each end of the chain and then draw the shear and moment diagrams for the arm ABC. Given: a 300mm:= b 75mm:= F 250N:= Solution: Given Equilibrium : + ΣFy=0; T1 T2− F− 0= ΣΜB=0; F− a⋅ T2 b⋅+ 0= Guess T1 1N:= T2 1N:= T1 T2 ⎛⎜⎜⎝ ⎞ ⎠ Find T1 T2,( ):= T1T2 ⎛⎜⎜⎝ ⎞ ⎠ 1.25 1.00 ⎛⎜⎝ ⎞ ⎠ kN= Ans x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) F− 1kN⋅:= V2 x2( ) F− T1+( ) 1kN⋅:= M1 x1( ) F− x1⋅N m⋅:= M2 x2( ) F− x2( )⋅ T1 x2 a−( )⋅+⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 0.2 1 0 1 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 0.2 100 50 0 Distane (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-3 Draw the shear and moment diagrams for the shaft. The bearings at A and D exert only vertical reactions on the shaft. The loading is applied to the pulleys at B and C and E. Given: a 350mm:= b 500mm:= c 375mm:= d 300mm:= B 400N:= C 550N:= E 175N:=Solution: Equilibrium : Given + ΣFy=0; A D+ B− C− E− 0= ΣΜD=0; A a b+ c+( )⋅ B b c+( )⋅− C c⋅− E d⋅+ 0= Guess A 1N:= D 1N:= A D ⎛⎜⎝ ⎞ ⎠ Find A D,( ):= A D ⎛⎜⎝ ⎞ ⎠ 411.22 713.78 ⎛⎜⎝ ⎞ ⎠ N= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= x4 a b+ c+ 1.01 a b+ c+( )⋅, a b+ c+ d+..:= V1 x1( ) A 1N⋅:= V2 x2( ) A B−( ) 1N⋅:= V3 x3( ) A B− C−( ) 1N⋅:= V4 x4( ) A B− C− D+( ) 1N⋅:= M1 x1( ) A x1⋅N m⋅:= M2 x2( ) A x2( )⋅ B x2 a−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= M3 x3( ) A x3( )⋅ B x3 a−( )⋅− C x3 a− b−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= M4 x4( ) A x4( )⋅ B x4 a−( )⋅− C x4 a− b−( )⋅− D x4 a− b− c−( )⋅+⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1000 500 0 500 Distance (m) Sh ea r ( N ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) x1 x2, x3, x4, 0 0.2 0.4 0.6 0.8 1 1.2 1.4 100 0 100 200 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) x1 x2, x3, x4, Problem 6-4 Draw the shear and moment diagrams for the beam. Given: a 1m:= F 10kN:= Solution: Equilibrium : Given + ΣFy=0; A 4F− B+ 0= ΣΜB=0; A 5a( )⋅ F 4a 3a+ 2a+ a+( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 20 20 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= x4 3a 1.01 3a( )⋅, 4a..:= x5 4a 1.01 4a( )⋅, 5a..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A F−( ) 1kN⋅:= V3 x3( ) A 2F−( ) 1kN⋅:= V4 x4( ) A 3F−( ) 1kN⋅:= V5 x5( ) A 4F−( ) 1kN⋅:= M1 x1( ) A x1⋅kN m⋅:= M2 x2( ) A x2( )⋅ F x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) A x3( )⋅ F x3 a−( )⋅− F x3 2a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M4 x4( ) A x4( )⋅ F x4 a−( )⋅− F x4 2a−( )⋅− F x4 3a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M5 x5( ) A x5( )⋅ F x5 a−( )⋅− F x5 2a−( )⋅− F x5 3a−( )⋅− F x5 4a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 5 20 0 20 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) V5 x5( ) x1 x2, x3, x4, x5, 0 1 2 3 4 5 0 10 20 30 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) M5 x5( ) x1 x2, x3, x4, x5, Problem 6-5 A reinforced concrete pier is used to support the stringers for a bridge deck. Draw the shear and moment diagrams for the pier when it is subjected to the stringer loads shown. Assume the columns a A and B exert only vertical reactions on the pier. Given: a 1m:= b 1.5m:= F1 60kN:= F2 35kN:= Solution: L 4a 2 b⋅+:= Equilibrium : Given + ΣFy=0; A B+ 2F1− 3F2− 0= ΣΜB=0; A 2a 2 b⋅+( )⋅ F1 L 2a−( )⋅− F2 3 b⋅ 3a+( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 112.5 112.5 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 2a b+( )..:= x4 2a b+( ) 1.01 2a b+( )⋅, 2a 2 b⋅+( )..:= x5 2a 2 b⋅+( ) 1.01 2a 2 b⋅+( )⋅, 3a 2 b⋅+( )..:= x6 3a 2 b⋅+( ) 1.01 3a 2 b⋅+( )⋅, 4a 2 b⋅+( )..:= V1 x1( ) F1− 1kN⋅:= V2 x2( ) A F1−( ) 1kN⋅:= V3 x3( ) A F1− F2−( ) 1kN⋅:= V4 x4( ) A F1− 2F2−( ) 1kN⋅:= V5 x5( ) A F1− 3F2−( ) 1kN⋅:= V6 x6( ) A F1− 3F2− B+( ) 1kN⋅:= M1 x1( ) F1− x1⋅kN m⋅:= M2 x2( ) A x2 a−( )⋅ F1 x2( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) A x3 a−( )⋅ F1 x3( )⋅− F2 x3 2a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M4 x4( ) A x4 a−( )⋅ F1 x4( )⋅− F2 2 x4 2 a⋅−( ) b−⎡⎣ ⎤⎦⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M5 x5( ) A x5 a−( )⋅ F1 x5( )⋅− F2 3 x5 2 a⋅−( ) 3 b⋅−⎡⎣ ⎤⎦⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M6 x6( ) A x6 a−( )⋅ F1 x6( )⋅− F2 3 x6 2 a⋅−( )⋅ 3 b⋅−⎡⎣ ⎤⎦⋅− B x6 L a−( )−⎡⎣ ⎤⎦⋅+⎡⎣ ⎤⎦ 1kN⋅⋅:= 0 1 2 3 4 5 6 7 50 0 50 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) V5 x5( ) V6 x6( ) x1 x2, x3, x4, x5, x6, 0 1 2 3 4 5 6 7 60 40 20 0 20 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) M5 x5( ) M6 x6( ) x1 x2, x3, x4, x5, x6, Problem 6-6 Draw the shear and moment diagrams for the shaft. The bearings at A and B exert only vertical reactions on the shaft. Also, express the shear and moment in the shaft as a function of x within the region 125 mm < x < 725 mm. Given: a 125mm:= b 600mm:= c 75mm:= F1 0.8kN:= F2 1.5kN:= Solution: L a b+ c+:= Equilibrium : Given + ΣFy=0; A F1− F2− B+ 0= ΣΜB=0; A L( )⋅ F1 b c+( )⋅− F2 c( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 0.8156 1.4844 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A F1−( ) 1kN⋅:= V3 x3( ) A F1− F2−( ) 1kN⋅:= M1 x1( ) A x1⋅N m⋅:= M2 x2( ) A x2( )⋅ F1 x2 a−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= M3 x3( ) A x3( )⋅ F1 x3 a−( )⋅− F2 x3 a− b−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 2 1 0 1 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 0.2 0.4 0.6 0 50 100 150 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-7 Draw the shear and moment diagrams for the shaft and determine the shear and moment throughout the shaft as a function of x. The bearings at A and B exert only vertical reactions on the shaft. Given: a 0.9m:= b 0.6m:= c 0.3m:= d 0.15m:= F1 4kN:= F2 2.5kN:= Solution: Equilibrium : Given + ΣFy=0; A F1− B+ F2− 0= ΣΜB=0; A a b+( )⋅ F1 b( )⋅− F2 c d+( )⋅+ 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 0.85 5.65 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A F1−( ) 1kN⋅:= V3 x3( ) A F1− B+( ) 1kN⋅:= M1 x1( ) A x1⋅N m⋅:= M2 x2( ) A x2( )⋅ F1 x2 a−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= M3 x3( ) A x3( )⋅ F1 x3 a−( )⋅− B x3 a− b−( )⋅+⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 0.5 1 1.5 4 2 0 2 4 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 0.5 1 1.5 1000 500 0 500 1000 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-8 Draw the shear and moment diagrams for the pipe. The end screw is subjected to a horizontal force of 5 kN. Hint: The reactions at the pin C must be replaced by equivalent loadings at point B on the axis the pipe. Given: a 400mm:= h 80mm:= F 5kN:= Solution: Given Equilibrium : + ΣFy=0; A C+ 0= ΣΜC=0; A a⋅ F h⋅+ 0= Guess A 1N:= C 1N:= A C ⎛⎜⎝ ⎞ ⎠ Find A C,( ):= A C ⎛⎜⎝ ⎞ ⎠ 1.00− 1.00 ⎛⎜⎝ ⎞ ⎠ kN= Ans x1 0 0.01 a⋅, a..:= V1 x1( ) A 1kN⋅:= M1 x1( ) A x1⋅N m⋅:= 0 0.2 0.4 2 1 0 1 Distance (m) Sh ea r ( kN ) V1 x1( ) x1 0 0.2 0 600 400 200 0 Distane (m) M om en t ( N -m ) M1 x1( ) x1 Problem 6-9 Draw the shear and moment diagrams for the beam. Hint: The 100-kN load must be replaced by equivalent loadings at point C on the axis of the beam. Given: a 1m:= b 1m:= c 1m:= d 0.25m:= F1 75kN:= F2 100kN:= Solution: Equilibrium : Given + ΣFy=0; A F1− B+ 0= ΣΜC=0; A a b+ c+( )⋅ F1 b c+( )⋅− F2 d( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 58.33 16.67 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A F1−( ) 1kN⋅:= V3 x3( ) A F1−( ) 1kN⋅:= M1 x1( ) A x1⋅kN m⋅:= M2 x2( ) A x2( )⋅ F1 x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) A x3( )⋅ F1 x3 a−( )⋅− F2 d⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 0.5 1 1.5 2 2.5 3 0 50 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 0.5 1 1.5 2 2.5 3 0 20 40 60 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-10 The engine crane is used to support the engine, which has a weight of 6 kN. Draw the shear and moment diagrams of the boom ABC when it is in the horizontal position shown. Given: a 0.9m:= b 1.5m:= c 1.2m:= W 6kN:= Solution: d a2 c2+:= v c d := h a d := Equilibrium : Given + ΣFy=0; Ay− B v⋅+ W− 0= ΣΜA=0; B− v⋅( ) a⋅ W a b+( )⋅+ 0= + ΣFx=0; Ax B h⋅− 0= Guess Ax 1kN:= Ay 1kN:= B 1kN:= Ax Ay B ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ Find Ax Ay, B,( ):= Ax Ay B ⎛⎜⎜⎜⎝ ⎞ ⎟ ⎠ 12 10 20 ⎛⎜⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) Ay− 1kN⋅:= V2 x2( ) Ay− B v⋅+( ) 1kN⋅:= M1 x1( ) Ay− x1⋅kN m⋅:= M2 x2( ) Ay− x2( )⋅ B v⋅( ) x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 10 0 10 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 1 2 10 5 0 Distane (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-11 Draw the shear and moment diagrams for the compound beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment and axial load. Set: a 1m:= P 1kN:= Solution: Equilibrium : Given + ΣFy=0; A P− C+ P− 0= ΣΜB=0; P 2a( )⋅ C a⋅− 0= Guess A 1kN:= C 1kN:= A C ⎛⎜⎝ ⎞ ⎠ Find A C,( ):= A C ⎛⎜⎝ ⎞ ⎠ 0 2 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= x4 3a 1.01 3a( )⋅, 4a..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A P−( ) 1kN⋅:= V3 x3( ) A P−( ) 1kN⋅:= V4 x4( ) A P− C+( ) 1kN⋅:= 0 0.5 1 1.5 2 2.5 3 3.5 4 2 1 0 1 2 Distance (m) Sh ea r ( P kN ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) x1 x2, x3, x4, MA C 3a( )⋅ P 4a( )⋅− P a( )⋅−:= MA 1.00 kN m⋅= M1 x1( ) MA A x1⋅+kN m⋅:= M2 x2( ) MA A x2( )⋅+ P x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) A P−( ) x3 2a−( )⋅⎡⎣ ⎤⎦ 1kN m⋅⋅:= M4 x4( ) A P−( ) x4 2a−( )⋅ C x4 3a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 1 0 1 Distance (m) M om en t ( P kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) x1 x2, x3, x4, Problem 6-12 Draw the shear and moment diagrams for the compound beam which is pin connected at B. Given: a 1m:= b 1.5m:= c 1m:= d 1m:= F1 30kN:= F2 40kN:= Solution: Equilibrium : Given + ΣFy=0; F1− A+ F2− C+ 0= ΣΜB=0; F1− a b+( )⋅ A b⋅+ 0= Guess A 1kN:= C 1kN:= A C ⎛⎜⎝ ⎞ ⎠ Find A C,( ):= A C ⎛⎜⎝ ⎞ ⎠ 50 20 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= x3 a b+ 1.01 a b+( )⋅, a b+ c+..:= x4 a b+ c+ 1.01 a b+ c+( )⋅, a b+ c+ d+..:= V1 x1( ) F1− 1kN⋅:= V2 x2( ) F1− A+( ) 1kN⋅:= V3 x3( ) F1− A+( ) 1kN⋅:= V4 x4( ) F1− A+ F2−( ) 1kN⋅:= 0 1 2 3 4 40 20 0 20 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) x1 x2, x3, x4, M1 x1( ) F1− x1⋅kN m⋅:= M2 x2( ) F1− x2( )⋅ A x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) F1− A+( ) x3 a− b−( )⋅⎡⎣ ⎤⎦ 1kN m⋅⋅:= M4 x4( ) F1− A+( ) x4 a− b−( )⋅ F2 x4 a− b− c−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 30 20 10 0 10 20 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) x1 x2, x3, x4, Problem 6-13 Draw the shear and moment diagrams for the beam. Set: a 1m:= Mo 1kN m⋅:= Solution: Equilibrium : Given + ΣFy=0; A B+ 0= ΣΜB=0; A 3a( )⋅ 2Mo+ Mo− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 0.33− 0.33 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A( ) 1kN⋅:= V3 x3( ) A( ) 1kN⋅:= M1 x1( ) Mo A x1⋅+kN m⋅:= M2 x2( ) 2 Mo⋅ A x2( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) 2 Mo⋅ A x3( )⋅+ Mo−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 0.5 0 0.5 Distance (m) Sh ea r ( M o/ a k N ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 1 2 0 1 2 Distance (m) M om en t ( M o kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-14 Consider the general problem of a simply supported beam subjected to n concentrated loads. Write a computer program that can be used to determine the internal shear and moment at any specified location x along the beam, and plot the shear and moment diagrams for the beam. Show an applicatio of the program using the values P1 = 2.5 kN, d1 = 1.5 m, P2 = 4 kN, d2 = 4.5 m, L1 = 3 m, L = 4.5 m. Problem 6-15 The beam is subjected to the uniformly distributed moment m (Moment/length). Draw the shear and moment diagrams for the beam. Set: L 1m:= mo 1 kN m⋅ m := Solution: Given Equilibrium : + ΣFy=0; A 0:= ΣΜA=0; MA mo L⋅:= x1 0 0.01 L⋅, L..:= V1 x1( ) A 1kN⋅:= M1 x1( ) MA A x1⋅+ mo x1⋅−( ) 1kN m⋅⋅:= 0 0.5 1 0 0.5 1 1.5 Distance (m) Sh ea r ( kN ) V1 x1( ) x1 0 0.5 1 0 0.5 1 Distane (m) M om en t m *L (k N -m ) M1 x1( ) x1 Problem 6-16 Draw the shear and moment diagrams for the beam. Given: a 2.5m:= w 10 kN m := b 2.5m:= Solution: Equilibrium : + A w a⋅ w a⋅−:=ΣFy=0; A 0 kN= ΣΜA=0; MA w a⋅( ) 0.5a( )⋅ w b⋅( ) a 0.5b+( )⋅−:= MA 62.50− kN m⋅= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) A w x1⋅−( ) 1kN⋅:= V2 x2( ) A w a⋅− w x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) MA− A x1⋅+ 0.5w x12⋅−⎛⎝ ⎞⎠ 1kN m⋅⋅:= M2 x2( ) MA− A x2⋅+ w a⋅( ) x2 0.5a−( )⋅− 0.5w x2 a−( )2⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 2 4 20 0 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 2 4 0 50 Distane (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-17 The 75-kg man sits in the center of the boat, which has a uniform width and a weight per linear foot of 50 N/m. Determine the maximum bending moment exerted on the boat. Assume that the water exerts a uniform distributed load upward on the bottom of the boat. Given: a 2.5m:= Mw 75kg:= b 2.5m:= w 50 N m := Solution: W Mw g⋅:= Equilibrium : + ΣFy=0; q W w 2a( )⋅+ 2a:= q 197.1 N m = x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) q w−( ) x1⋅⎡⎣ ⎤⎦ 1N⋅:= V2 x2( ) q w−( ) x2⋅ W−⎡⎣ ⎤⎦ 1N⋅:= M1 x1( ) 0.5 q w−( ) x12⋅⎡⎣ ⎤⎦ 1N m⋅⋅:= M2 x2( ) 0.5 q w−( ) x22⋅ W x2 a−( )⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 2 4 500 0 500 Distance (m) Sh ea r ( N ) V1 x1( ) V2 x2( ) x1 x2, 0 2 4 0 200 400 Distane (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-18 Draw the shear and moment diagrams for the beam. It is supported by a smooth plate at A which slides within the groove and so it cannot support a vertical force, although it can support a moment an axial load. Set: L 1m:= w 1 kN m := A 0:= Solution: Given Equilibrium : + ΣFy=0; A B+ w L⋅− 0= B w L⋅:= ΣΜA=0; MA A L⋅+ w L⋅( ) L 2 ⋅− 0= MA w L2⋅ 2 := x1 0 0.01 L⋅, L..:= V1 x1( ) A w x1⋅−( ) 1kN⋅:= M1 x1( ) MA A x1⋅+ w2 x12⋅−⎛⎜⎝ ⎞⎠ 1kN m⋅⋅:= 0 0.5 1 1 0 Distance (m) Sh ea r w *L ( kN ) V1 x1( ) x1 0 0.5 1 0 0.2 0.4 0.6 Distane (m) M om en t w *L *L (k N -m ) M1 x1( ) x1 Problem 6-19 Draw the shear and moment diagrams for the beam. Given: a 1.5m:= w 30 kN m := Mo 45kN m⋅:= Solution: Equilibrium : Given + ΣFy=0; w− a⋅ A+ B+ 0= ΣΜA=0; w a⋅( )− 0.5a( )⋅ Mo+ B 2a( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 41.25 3.75 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= V1 x1( ) w− x1⋅kN:= V2 x2( ) w− a⋅ A+( ) 1kN⋅:= V3 x3( ) w− a⋅ A+( ) 1kN⋅:= M1 x1( ) 0.5− w x1 2⋅ kN m⋅:= M2 x2( ) w a⋅( )− x2 0.5a−( )⋅ A x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) w a⋅( )− x3 0.5a−( )⋅ A x3 a−( )⋅+ Mo+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 40 20 0 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 1 2 3 4 40 20 0 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-20 Draw the shear and moment diagrams for the beam, and determine the shear and moment throughout the beam as functions of x. Given: a 2.4m:= b 1.2m:= P1 50kN:= w 30 kN m :=P2 40kN:= M2 60kN m⋅:= Solution: Equilibrium : + A w a⋅ P1+ P2+:= A 162 kN=ΣFy=0; MA w a⋅( ) 0.5a( )⋅ P1 a⋅+ P2 a b+( )⋅+ M2+:=ΣΜA=0; MA 410.40 kN m⋅= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+..:= V1 x1( ) A w x1⋅−( ) 1kN⋅:= V2 x2( ) A w a⋅− P1−( ) 1kN⋅:= M1 x1( ) MA− A x1⋅+ 0.5w x12⋅−⎛⎝ ⎞⎠ 1kN m⋅⋅:= M2 x2( ) MA− A x2⋅+ w a⋅( ) x2 0.5 a⋅−( )⋅− P1 x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 0 50 100 150 200 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 1 2 3 400 200 0 Distane (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-21 Draw the shear and moment diagrams for the beam and determine the shear and moment in the beam as functions of x, where 1.2 m < x < 3 m. Given: a 1.2m:= Mo 300N m⋅:= b 1.8m:= w 2.5 kN m :=c 1.2m:= Solution: Equilibrium : Given + ΣFy=0; w− b⋅ A+ B+ 0= ΣΜB=0; Mo− A b⋅+ w b⋅( ) 0.5b( )⋅− Mo+ 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 2.25 2.25 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a b+ c+( )..:= V1 x1( ) 0kN:= V2 x2( ) A w x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V3 x3( ) A w b⋅− B+( ) 1kN⋅:= M1 x1( ) Mo−N m⋅:= M2 x2( ) Mo− A x2 a−( )⋅+ 0.5w x2 a−( )2⋅−⎡⎣ ⎤⎦ 1N m⋅⋅:= M3 x3( ) Mo− A x3 a−( )⋅+ w b⋅( ) x3 a− 0.5 b⋅−( )⋅− B x3 a− b−( )⋅+⎡⎣ ⎤⎦ 1N m⋅⋅:= 0 1 2 3 4 2 0 2 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 1 2 3 4 200 0 200 400 600 800 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-22 Draw the shear and moment diagrams for the compound beam.The three segments are connected by pins at B and E. Given: a 2m:= b 1m:= F 3kN:= LBE a 2 b⋅+:= w 0.8 kN m :=LAB a b+:= Solution: L 3a 4 b⋅+:= Equilibrium : Consider segment AB: ΣΜB=0; A a b+( )⋅ F b( )⋅− 0= A ba b+ F⋅:= A 1.00 kN= + ΣFy=0; A B+ F− 0= B F A−:= B 2.00 kN= Consider segment BE: By symmetry, E B= D C= + ΣFy=0; 2C 2B− w a 2 b⋅+( )⋅− 0= C B w2 a 2 b⋅+( )⋅+:= C 3.60 kN= D C:= D 3.60 kN= E B:= E 2.00 kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a 2⋅+(..:= x4 a 2 b⋅+( ) 1.01 a 2 b⋅+( )⋅, 2a 2 b⋅+( )..:= x5 2a 2 b⋅+( ) 1.01 2a 2 b⋅+( )⋅, 2a 3 b⋅+( )..:= x6 2a 3 b⋅+( ) 1.01 2a 3 b⋅+( )⋅, 2a 4 b⋅+( )..:= x7 2a 4 b⋅+( ) 1.01 2a 4 b⋅+( )⋅, 3a 4 b⋅+( )..:= V1 x1( ) A 1kN⋅:= V2 x2( ) A F−( ) 1kN⋅:= V3 x3( ) A F− w x3 LAB−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V4 x4( ) A F− C+ w x4 LAB−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V5 x5( ) A F− C+ D+ w x5 LAB−( )⋅−⎡⎣ ⎤⎦⋅:= V6 x6( ) A F− C+ D+ w LBE( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V7 x7( ) A 2F− C+ D+ w LBE( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) A x1⋅kN m⋅:= M2 x2( ) A x2( )⋅ F x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) B− x3 LAB−( )⋅ 0.5w x3 LAB−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M4 x4( ) B− x4 LAB−( )⋅ 0.5w x4 LAB−( )2⋅− C x4 LAB− b−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M5 x5( ) B− x5 LAB−( )⋅ 0.5w x5 LAB−( )2⋅− C x5 LAB− b−( )⋅+ D x5 LAB− b− a−( )⋅+⎡⎣ ⎤⎦⋅:= M6 x6( ) E x6 LAB− LBE−( )⋅⎡⎣ ⎤⎦ 1kN m⋅⋅:= M7 x7( ) E x7 LAB− LBE−( )⋅ F x7 LAB− LBE− b−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 2 4 6 8 10 4 2 0 2 4 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) V5 x5( ) V6 x6( ) V7 x7( ) x1 x2, x3, x4, x5, x6, x7, 0 2 4 6 8 10 3 2 1 0 1 2 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) M5 x5( ) M6 x6( ) M7 x7( ) x1 x2, x3, x4, x5, x6, x7, Problem 6-23 Draw the shear and moment diagrams for the beam. Given: a 1.5m:= Mo 30kN m⋅:= w 30 kN m := Solution: Equilibrium : Given + ΣFy=0; w− a⋅ A+ w a⋅− B+ 0= ΣΜB=0; Mo w a⋅( ) 2.5a( )⋅− A 2a( )⋅+ w a⋅( ) 0.5a( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 57.50 32.50 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2⋅ a, 3a..:= V1 x1( ) w− x1⋅kN:= V2 x2( ) w a⋅( )− A+[ ] 1kN⋅:= V3 x3( ) w a⋅( )− A+ w x3 2a−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) Mo 0.5w x1 2⋅− kN m⋅:= M2 x2( ) Mo w a⋅( ) x2 0.5a−( )⋅− A x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) Mo w a⋅( ) x3 0.5a−( )⋅− A x3 a−( )⋅+ 0.5w x3 2a−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 40 20 0 20 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 1 2 3 4 10 0 10 20 30 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-24 The beam is bolted or pinned at A and rests on a bearing pad at B that exerts a uniform distributed loading on the beam over its 0.6-m length. Draw the shear and moment diagrams for the beam if it supports a uniform loading of 30 kN/m. Given: a 0.3m:= c 0.6m:= b 2.4m:= w 30 kN m := Solution: Equilibrium : Given + ΣFy=0; A w b⋅− qB( ) c⋅+ 0= ΣΜA=0; w b⋅( ) a 0.5b+( )⋅ qB c⋅( ) a b+ 0.5c+( )⋅− 0= Guess A 1kN:= qB 1 kN m := A qB ⎛⎜⎝ ⎞ ⎠ Find A qB,( ):= A 36.00 kN= qB 60.00 kN m = x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a b+ c+( )..:= V1 x1( ) AkN:= V2 x2( ) A w x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V3 x3( ) A w b⋅− qB x3 a− b−( )⋅+⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) A x1⋅kN m⋅:= M2 x2( ) A x2( )⋅ 0.5w x2 a−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) A x3( )⋅ w b⋅( ) x3 a− 0.5 b⋅−( )⋅− 0.5qB x3 a− b−( )2⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 0.5 1 1.5 2 2.5 3 40 20 0 20 40 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 0.5 1 1.5 2 2.5 3 0 10 20 30 40 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-25 Draw the shear and moment diagrams for the beam. The two segments are joined together at B. Given: a 0.9m:= P 40kN:= b 1.5m:= w 50 kN m :=c 2.4m:= Solution: Equilibrium : Given + ΣFy=0; A P− w c⋅− C+ 0= ΣΜB=0; w c⋅( ) 0.5c( )⋅ C c( )⋅− 0= Guess A 1kN:= C 1kN:= A C ⎛⎜⎝ ⎞ ⎠ Find A C,( ):= A C ⎛⎜⎝ ⎞ ⎠ 100 60 ⎛⎜⎝ ⎞ ⎠ kN= MA P a⋅ C w c⋅−( ) a b+( )⋅−:= MA 180 kN m⋅= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a b+ c+( )..:= V1 x1( ) AkN:= V2 x2( ) A P−( ) 1kN⋅:= V3 x3( ) A P− w x3 a− b−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) MA− A x1⋅+kN m⋅:= M2 x2( ) MA− A x2⋅+ P x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) MA− A x3⋅+ P x3 a−( )⋅− 0.5w x3 a− b−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 50 0 50 100 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 1 2 3 4 200 100 0 100 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-26 Consider the general problem of a cantilevered beam subjected to n concentrated loads and a constant distributed loading w. Write a computer program that can be used to determine the internal shear and moment at any specified location x along the beam, and plot the shear and moment diagrams for the beam. Show an application of the program using the values P1 = 4 kN, d1 = 2 m, w = 800 N/m, a1 = 2 m, a2 = 4 m, L = 4 m. Problem 6-27 Determine the placement distance a of the roller support so that the largest absolute value of the moment is a minimum. Draw the shear and moment diagrams for this condition. Problem 6-28 Draw the shear and moment diagrams for the rod. Only vertical reactions occur at its ends A and B. Given: a 900mm:= A 360N:= B 720N:= w' 2.4 kN m := Solution: + ΣFy=0; A 0.5w' xo a ⎛⎜⎝ ⎞ ⎠⋅ xo⋅− 0= xo A a⋅ 0.5w' := xo 519.62 mm= ΣΜ Mmax A xo⋅ 0.5w' xo a ⎛⎜⎝ ⎞ ⎠⋅ xo⋅ xo 3 ⎛⎜⎝ ⎞ ⎠⋅−:= Mmax 124.71 N m⋅= x 0 0.01 a⋅, a..:= V x( ) A 0.5w' x a ⎛⎜⎝ ⎞ ⎠⋅ x⋅− ⎡⎢⎣ ⎤⎥⎦ 1 N ⋅:= M x( ) A x⋅ 0.5w' x a ⎛⎜⎝ ⎞ ⎠⋅ x⋅ x 3 ⎛⎜⎝ ⎞ ⎠⋅− ⎡⎢⎣ ⎤⎥⎦ 1 N m⋅:= 0 0.2 0.4 0.6 0.8 500 0 Distance (m) Sh ea r ( N ) V x( ) x 0 0.2 0.4 0.6 0.8 0 50 100 Distance m) M om en t ( N -m ) M x( ) x Problem 6-29 Draw the shear and moment diagrams for the beam. Given: Set L 1m:= wo 1 kN m := a L 3 := Solution: Equilibrium : Given + ΣFy=0; A 2 0.5wo( ) a⋅− wo a⋅− B+ 0= ΣΜB=0; A 3 a⋅( )⋅ 0.5 wo⋅ a⋅( ) 2a a3+⎛⎜⎝ ⎞⎠⋅− wo a⋅( ) 1.5a( )⋅− 0.5 wo⋅ a⋅( ) 2a3⎛⎜⎝ ⎞⎠⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 0.33 0.33 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a( )..:= x3 2a( ) 1.01 2a( )⋅, 3a(..:= V1 x1( ) A wo2 x1 a ⎛⎜⎝ ⎞ ⎠⋅ x1⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= V2 x2( ) A 0.5 wo⋅ a⋅− wo x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V3 x3( ) A 0.5 wo⋅ a⋅− wo a⋅− wo x3 2a−( )⋅ 1 0.5 x3 2a−a⋅−⎛⎜⎝ ⎞ ⎠⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= 0 0.2 0.4 0.6 0.8 0.4 0.2 0 0.2 0.4 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, M1 x1( ) A x1⋅ wo2 x1 a ⎛⎜⎝ ⎞ ⎠⋅ x1⋅ x1 3 ⋅−⎡⎢⎣ ⎤⎥⎦ 1 N m⋅⋅:= M2 x2( ) A x2⋅ wo a⋅2 x2 2a3−⎛⎜⎝ ⎞⎠⋅− 0.5wo x2 a−( )2⋅− ⎡⎢⎣ ⎤⎥⎦ 1 N m⋅⋅:= M'3 x3( ) wo2 x3 2 a⋅−( )2⋅ 1 x3 2 a⋅− a ⎛⎜⎝ ⎞ ⎠ 1 3 ⋅−⎡⎢⎣ ⎤⎥⎦⋅:= M3 x3( ) A x3⋅ wo a⋅2 x3 2 a⋅3−⎛⎜⎝ ⎞⎠⋅− wo a⋅( ) x3 1.5 a⋅−( )⋅− M'3 x3( )− ⎡⎢⎣ ⎤⎥⎦ 1 N m⋅⋅:= 0 0.2 0.4 0.6 0.8 0 20 40 60 80 100 120 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-30 Draw the shear and moment diagrams for the beam. Set: L 1m:= wo 1 kN m := Solution: Equilibrium : Given + ΣFy=0; A B+ 0.5 wo⋅ L⋅− 0= ΣΜB=0; A 2 L⋅3⋅ wo 2 L⋅⎛⎜⎝ ⎞ ⎠ L 3 ⎛⎜⎝ ⎞ ⎠⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 0.25 0.25 ⎛⎜⎝ ⎞ ⎠ kN= Let a L 3 := x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 3a..:= V1 x1( ) wo2− x1 L ⋅ x1( )⋅⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= V2 x2( ) A wo2 x2 L ⋅ x2( )⋅−⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= M1 x1( ) wo2− x1 L ⋅ x1( )⋅ x13⎛⎜⎝ ⎞ ⎠⋅ ⎡⎢⎣ ⎤⎥⎦ 1 kN m⋅⋅:= M2 x2( ) A x2 a−( )⋅ wo2 x2 L ⋅ x2( )⋅ x23⎛⎜⎝ ⎞ ⎠⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN m⋅⋅:= 0 0.2 0.4 0.6 0.8 0.2 0 0.2 Distance (m) Sh ea r W o (k N ) V1 x1( ) V2 x2( ) x1 x2, 0 0.2 0.4 0.6 0.8 0 0.02 0.04 Distance (m) M om en t W o* L* L (k N -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-31 The T-beam is subjected to the loading shown. Draw the shear and moment diagrams. Given: a 2m:= P 10kN:= b 3m:= w 3 kN m :=c 3m:= Solution: Equilibrium : Given + ΣFy=0; P− A+ w c⋅− B+ 0= ΣΜB=0; P− a b+ c+( )⋅ A b c+( )⋅+ w c⋅( ) 0.5c( )⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 15.58 3.42 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a b+ c+( )..:= V1 x1( ) P−kN:= V2 x2( ) P− A+( ) 1kN⋅:= V3 x3( ) P− A+ w x3 a− b−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= M1 x1( ) P− x1⋅kN m⋅:= M2 x2( ) P− x2⋅ A x2 a−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) P− x3⋅ A x3 a−( )⋅+ 0.5 w⋅ x3 a− b−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 1 2 3 4 5 6 7 8 10 5 0 5 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 2 4 6 8 25 20 15 10 5 0 5 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-32 The ski supports the 900-N (~90-kg) weight of the man. If the snow loading on its bottom surface is trapezoidal as shown, determine the intensity w, and then draw the shear and moment diagrams for the ski. Given: a 0.5m:= P 900N:= Solution: Equilibrium : + ΣFy=0; P− w2 2a 4a+( )⋅+ 0= w P 3a := w 600 N m = x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a..:= x3 2a 1.01 2a( )⋅, 3a..:= x4 3a 1.01 3a( )⋅, 4a..:= V1 x1( ) w2 x1 a ⎛⎜⎝ ⎞ ⎠⋅ x1⋅ ⎡⎢⎣ ⎤⎥⎦ 1 N ⋅:= V2 x2( ) 0.5 w⋅ a⋅ w x2 a−( )⋅+⎡⎣ ⎤⎦ 1N⋅:= V3 x3( ) 0.5 w⋅ a⋅ w a⋅+ P− w x3 2a−( )⋅+⎡⎣ ⎤⎦ 1N⋅:= V4 x4( ) 0.5 w⋅ a⋅ w a⋅+ P− w a+ w x4 3a−( )⋅ 1 0.5 x4 3a−a⋅−⎛⎜⎝ ⎞ ⎠⋅+ ⎡⎢⎣ ⎤⎥⎦ 1 N ⋅:= 0 0.5 1 1.5 2 400 200 0 200 400 Distance (m) Sh ea r ( N ) V1 x1( ) V2 x2( ) V3 x3( ) V4 x4( ) x1 x2, x3, x4, M1 x1( ) w2 x1 a ⎛⎜⎝ ⎞ ⎠⋅ x1⋅ x1 3 ⋅⎡⎢⎣ ⎤⎥⎦ 1 N m⋅⋅:= M2 x2( ) w a⋅2 x2 2a3−⎛⎜⎝ ⎞⎠⋅ w2 x2 a−( )2⋅+⎡⎢⎣ ⎤⎥⎦ 1N m⋅⋅:= M3 x3( ) w a⋅2 x3 2a3−⎛⎜⎝ ⎞⎠⋅ w2 x3 a−( )2⋅+ P x3 2a−( )⋅−⎡⎢⎣ ⎤⎥⎦ 1N m⋅⋅:= M'4 x4( ) w2 x4 3 a⋅−( )2⋅ 1 x4 3 a⋅− a ⎛⎜⎝ ⎞ ⎠ 1 3 ⋅−⎡⎢⎣ ⎤⎥⎦⋅:= M4 x4( ) w a⋅2 x4 2 a⋅3−⎛⎜⎝ ⎞⎠⋅ 2w a⋅( ) x4 2 a⋅−( )⋅+ P x4 2a−( )⋅− M'4 x4( )+⎡⎢⎣ ⎤⎥⎦ 1N m⋅⋅:= 0 0.5 1 1.5 2 0 50 100 150 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) M3 x3( ) M4 x4( ) x1 x2, x3, x4, Problem 6-33 Draw the shear and moment diagrams for the beam. Given: L 9m:= wo 50 kN m :=a 0.5L:= Solution: Equilibrium : Given + ΣFy=0; A 2 0.5wo( ) a⋅− B+ 0= ΣΜB=0; A L⋅ 0.5 wo⋅ a⋅( ) a a3+⎛⎜⎝ ⎞⎠⋅− 0.5 wo⋅ a⋅( ) 2a3⎛⎜⎝ ⎞⎠⋅− 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 112.50 112.50 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, 2a( )..:= V1 x1( ) A wo x1⋅ 1 0.5 x1a⋅−⎛⎜⎝ ⎞ ⎠⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= V2 x2( ) A 0.5 wo⋅ a⋅− wo2 x2 a− a ⎛⎜⎝ ⎞ ⎠⋅ x2 a−( )⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN ⋅:= M1 x1( ) A x1⋅ wo2 x12⋅ 1 x1 a ⎛⎜⎝ ⎞ ⎠ 1 3 ⋅−⎡⎢⎣ ⎤⎥⎦⋅− ⎡⎢⎣ ⎤⎥⎦ 1 kN m⋅⋅:= M'2 x2( ) wo2 x2 a− a ⎛⎜⎝ ⎞ ⎠⋅ x2 a−( )⋅ x2 a− 3 ⎛⎜⎝ ⎞ ⎠⋅:= M2 x2( ) A x2⋅ wo a⋅2 x2 a3−⎛⎜⎝ ⎞⎠⋅− M'2 x2( )− ⎡⎢⎣ ⎤⎥⎦ 1 kN m⋅⋅:= 0 5 100 0 100 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) x1 x2, 0 5 0 100 200 Distance (m) M om en t ( N -m ) M1 x1( ) M2 x2( ) x1 x2, Problem 6-34 Draw the shear and moment diagrams for the wood beam, and determine the shear and moment throughout the beam as functions of x. Given: a 1m:= P 1kN:= b 1.5m:= w 2 kN m :=c 1m:= Solution: Equilibrium : Given + ΣFy=0; A w b⋅− B+ 2P− 0= ΣΜB=0; P− a b+( )⋅ A b⋅+ w b⋅( ) 0.5b( )⋅− P c⋅+ 0= Guess A 1kN:= B 1kN:= A B ⎛⎜⎝ ⎞ ⎠ Find A B,( ):= A B ⎛⎜⎝ ⎞ ⎠ 2.50 2.50 ⎛⎜⎝ ⎞ ⎠ kN= x1 0 0.01 a⋅, a..:= x2 a 1.01 a⋅, a b+( )..:= x3 a b+( ) 1.01 a b+( )⋅, a b+ c+( )..:= V1 x1( ) P−kN:= V2 x2( ) P− A+ w x2 a−( )⋅−⎡⎣ ⎤⎦ 1kN⋅:= V3 x3( ) P− A+ w b⋅− B+( ) 1kN⋅:= M1 x1( ) P− x1⋅kN m⋅:= M2 x2( ) P− x2⋅ A x2 a−( )⋅+ 0.5w x2 a−( )2⋅−⎡⎣ ⎤⎦ 1kN m⋅⋅:= M3 x3( ) P− x3⋅ A x3 a−( )⋅+ w b⋅( ) x3 a− 0.5 b⋅−( )⋅− B x3 a− b−( )⋅+⎡⎣ ⎤⎦ 1kN m⋅⋅:= 0 0.5 1 1.5 2 2.5 3 3.5 2 1 0 1 2 Distance (m) Sh ea r ( kN ) V1 x1( ) V2 x2( ) V3 x3( ) x1 x2, x3, 0 0.5 1 1.5 2 2.5 3 3.5 1 0.8 0.6 0.4 0.2 0 Distance (m) M om en t ( kN -m ) M1 x1( ) M2 x2( ) M3 x3( ) x1 x2, x3, Problem 6-35 The smooth pin is supported by two leaves A and B and subjected to a compressive load of 0.4 kN/m caused by bar C. Determine the intensity of the distributed load w0 of the leaves on the pin and draw the shear and moment diagrams for the pin. Given: L 100mm:= w 0.4 kN m := a 0.2L:= Solution: Equilibrium : + ΣFy=0; 2 0.5wo( ) a⋅ w 3a( )⋅− 0= wo 3w:= wo 1.20 kN m = x1 0 0.01 a⋅, a..:= x2 a
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