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ISM for 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 Rb = 35kN (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
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