Resolução do livro do Hibbeler - 7° Edição, Exercícios de Resistência dos materiais

Baixe Resolução do livro do Hibbeler - 7° Edição e outras Exercícios em PDF para Resistência dos materiais, somente na Docsity! OLICI MRILIVIN JOE 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 1-4 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 1-9 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 L2 3 + ⎛ ⎜ ⎝ ⎞ ⎠ ⋅− 0= By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L2 3 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 L2 3 + ⎛ ⎜ ⎝ ⎞ ⎠ ⋅− 0= By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L2 3 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⋅( )⋅⎡⎣ ⎤⎦ e 3 ⎛⎜ ⎝ ⎞ ⎠ ⋅− 0= ME 0.5− w2 0.5 e⋅( )⋅⎡⎣ ⎤⎦ e 3 ⎛⎜ ⎝ ⎞ ⎠ ⋅:= ME 0.911− kN m⋅= Ans Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD. 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 1-19 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⋅[ ] 2L 3 ⎛⎜ ⎝ ⎞ ⎠ ⋅− 0= By w1 L⋅( ) 0.5( )⋅ 0.5 w( ) L⋅[ ] 2 3 ⎛⎜ ⎝ ⎞ ⎠ ⋅+:= 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,( ):= Bx By ⎛⎜ ⎜⎝ ⎞ ⎠ 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 1-24 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 1-29 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( )−( )⋅:= Ans 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 1-39 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 1-44 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 dB 2⋅ 4 := AAB 63.61725 mm2= Rod AD : AAD π dD 2⋅ 4 := AAD 44.17865 mm2= Rod AC: AAC π dC 2⋅ 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,( ):= FAC FAD ⎛⎜ ⎜⎝ ⎞ ⎠ 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 1-49 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 1-54 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 dA 2⋅ 4 := Rod BC : ABC π dC 2⋅ 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 1-59 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 di 2⋅ 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 dpin 2⋅ 4 := VA 0.5 Ay( )⋅:= VA 583.333 N= τavg VA Apin := τavg 29.709 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 1-69 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 dbolt 2⋅ 4 := τ 4P a b+( )⋅ cos θ( ) c( )⋅ sin θ( ) a( )⋅+⎡⎣ ⎤⎦ π dbolt 2⋅⎛ ⎝ ⎞ ⎠⋅ = dτ dθ 4P a b+( )⋅ π dbolt 2⋅ 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 drod 2⋅ 4 := Apin π dpin 2⋅ 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 1-74 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 1-79 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 ⎛⎜ ⎝ ⎞ ⎠ d2⋅ Vbolt τallow = d 4 π Vbolt τallow ⎛ ⎜ ⎝ ⎞ ⎠ ⋅:= d 13.49 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( )⋅ 0.5P τallow = 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 ⎛⎜ ⎝ ⎞ ⎠ d2⋅ P σallow = 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⋅ P τallow = h 1 π dwasher( )⋅ P τallow ⎛ ⎜ ⎝ ⎞ ⎠ ⋅:= h 9.095 mm= Use d 10mm:= d 10 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,( ):= P1 FAC ⎛⎜ ⎜⎝ ⎞ ⎠ 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,( ):= FAB P2 ⎛⎜ ⎜⎝ ⎞ ⎠ 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 1-94 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 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 3a 3a ⎛⎜ ⎝ ⎞ ⎠ ⋅ P2 2a 3a ⎛⎜ ⎝ ⎞ ⎠ ⋅+ P3 a 3a ⎛⎜ ⎝ ⎞ ⎠ ⋅+ P b 3a ⎛⎜ ⎝ ⎞ ⎠ ⋅−= P P1 3a b ⎛⎜ ⎝ ⎞ ⎠ ⋅ P2 2a b ⎛⎜ ⎝ ⎞ ⎠ ⋅+ P3 a b ⎛⎜ ⎝ ⎞ ⎠ ⋅+ aA 2⎛ ⎝ ⎞ ⎠ σb_allow⋅ 3a b ⎛⎜ ⎝ ⎞ ⎠ ⋅−:= P 26.400 kN= Pcase_1 P:= For Plate B: By aB 2⎛ ⎝ ⎞ ⎠ σb_allow⋅:= aB 2⎛ ⎝ ⎞ ⎠ σb_allow⋅ P2 a 3a ⎛⎜ ⎝ ⎞ ⎠ ⋅ 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⋅ 3a L ⋅+:= 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