1. ISM for

2. Problem 1-1Determine the resultant internal normal force acting on the cross section through point A in eachcolumn. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the columnhas a mass of 200 kg/m.(a) Given: g 9.81ms2:= wBC 300kgm:=LBC := 3m wCA 400kgm:=FB := 5kN LCA := 1.2mFC := 3kNSolution:+ Fy = 0; FA wBC g ( )LBC (wCAg)LCA FB 2FC = 0FA := (wBCg)LBC + (wCAg)LCA + FB + 2FCFA = 24.5 kN Ans(b) Given: g 9.81ms2:= w 200kgm:=L := 3m F1 := 6kNFB := 8kN F2 := 4.5kNSolution:+ Fy = 0; FA w L ( )g FB 2F1 2F2 = 0FA := (wL)g + FB + 2F1 + 2F2FA = 34.89 kN Ans 3. Problem 1-2Determine the resultant internal torque acting on the cross sections through points C and D of theshaft. The shaft is fixed at B.Given: TA := 250NmTCD := 400NmTDB := 300NmSolution:Equations of equilibrium:+ TA TC = 0TC := TATC = 250Nm Ans+ TA TCD + TD = 0TD := TCD TATD = 150Nm Ans 4. Problem 1-3Determine the resultant internal torque acting on the cross sections through points B and C.Given: TD := 500NmTBC := 350NmTAB := 600NmSolution:Equations of equilibrium: Mx = 0; TB + TBC TD = 0TB := TBC + TDTB = 150Nm Ans Mx = 0; TC TD = 0TC := TDTC = 500Nm Ans 5. Problem 1-4A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings actingon the section through point A.Given: P := 80N := 30deg := 45dega := 0.3m b := 0.1mSolution:Equations of equilibrium:+ Fx'=0; NA Pcos( ) = 0NA := Pcos( )NA = 77.27N Ans+ Fy'=0; VA P sin( ) = 0VA := P sin( )VA = 20.71N Ans+ A=0; MA + Pcos()a cos() P sin() (b + a sin()) = 0MA := Pcos()a cos() + P sin() (b + a sin())MA = 0.555Nm AnsNote: Negative sign indicates that MA acts in the opposite direction to that shown on FBD. 6. Problem 1-5Determine the resultant internal loadings acting on the cross section through point D of member AB.Given: ME := 70Nma := 0.05m b := 0.3mSolution:Segment AB: Support Reactions+ A=0; ME By (2a + b) = 0ByME2a + b:= By = 175N150200At B: Bx := By Bx = 131.25 N+Segment DB: NB := Bx VB := By+ Fx=0; ND + NB = 0ND := NB ND = 131.25 N Ans+ Fy=0; VD + VB = 0VD := VB VD = 175N Ans+ D=0; MD ME By (a + b) = 0MD := ME By (a + b)MD = 8.75 Nm Ans 7. Problem 1-6The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internalloadings acting on the cross section at point D.Given: P := 5000Na := 0.8m b := 1.2m c := 0.6m d := 1.6me := 0.6mSolution: atanbd:= = 36.87 deg atana + bd:= = 14.47 degMember AB:+ A=0; FBC sin() (a + b) P (b) = 0FBCP (b)sin() (a + b):=FBC = 12.01 kNSegment BD:+ Fx=0; ND FBCcos() Pcos() = 0ND := FBCcos() Pcos()ND = 15.63 kN Ans+ Fy=0; VD + FBC sin() P sin() = 0VD := FBC sin() + P sin()VD = 0 kN Ans+ D=0; (FBC sin() P sin()) d csin() MD = 0MD (FBC sin() P sin()) d csin() := MD = 0 kNm AnsNote: Member AB is the two-force member. Therefore the shear force and moment are zero. 8. Problem 1-7Solve Prob. 1-6 for the resultant internal loadings acting at point E.Given: P := 5000Na := 0.8m b := 1.2m c := 0.6m d := 1.6me := 0.6mSolution: atanbd:= = 36.87 deg atana + bd:= = 14.47 degMember AB:+ A=0; FBC sin() (a + b) P (b) = 0FBCP (b)sin() (a + b):=FBC = 12.01 kNSegment BE:+ Fx=0; NE FBCcos() Pcos() = 0NE := FBCcos() Pcos()NE = 15.63 kN Ans+ Fy=0; VE + FBC sin() P sin() = 0VE := FBC sin() + P sin()VE = 0 kN Ans+ E=0; (FBC sin() P sin())e ME = 0ME := (FBC sin() P sin())eME = 0 kNm AnsNote: Member AB is the two-force member. Therefore the shear force and moment are zero. 9. Problem 1-8The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist andload weigh 1500 N, determine the resultant internal loadings in the crane on cross sections throughpoints A, B, and C.Given: P := 1500N w 750Nm:=a := 2.1m b := 1.5mc := 0.6m d := 2.4m e := 0.9mSolution:Equations of Equilibrium: For point A+ Fx=0; NA := 0 Ans+F y=0;VA we P = 0 VA := we + P VA = 2.17 kN Ans+ A=0; MA (we) (0.5e) P (e) = 0MA := (we) (0.5e) P (e) MA = 1.654 kNm AnsNote: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.Equations of Equilibrium: For point B+ Fx=0; NB := 0 Ans+F y=0;VB w (d + e) P = 0 VB := w (d + e) + P VB = 3.98 kN Ans+ B=0; MB [w (d + e)] [0.5 (d + e)] P (d + e) = 0MB := [w (d + e)] [0.5 (d + e)] P (d + e)MB = 9.034 kNm AnsNote: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.Equations of Equilibrium: For point C+ Fx=0; VC := 0 Ans+F y=0;NC w (b + c + d + e) P = 0 NC := w (b + c + d + e) PNC = 5.55 kN Ans+ B=0; MC [w (c + d + e)] [0.5 (c + d + e)] P (c + d + e) = 0MC := [w (c + d + e)] [0.5 (c + d + e)] P (c + d + e)MC = 11.554 kNm AnsNote: Negative sign indicates that NC and MC acts in the opposite direction to that shownon FBD. 10. Problem 1-9The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of thetooth, i.e., at the centroid point A of section a-a.Given: P := 400N := 30deg := 45dega := 4mm b := 5.75mmSolution: := Equations of equilibrium: For section a -a+ Fx'=0; VA Pcos() = 0VA := Pcos()VA = 386.37 N Ans+ Fy'=0; NA sin() = 0NA := P sin()NA = 103.53 N Ans+ A=0; MA P sin()a + Pcos()b = 0MA := P sin()a + Pcos()bMA = 1.808Nm Ans 11. Problem 1-10The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssection through point C. Assume the reactions at the supports A and B are vertical.Given: w1 4.5kNm:= w2 6.0kNm:=a := 1.8m b := 1.8m c := 2.4md := 1.35m e := 1.35mSolution: L1 := a + b + cL2 := d + eSupport Reactions:+ A=0; ByL1 (w1L1)(0.5L1) (0.5w2L2) L1L23+ = 0By (w1L1) (0.5) (0.5w2L2) 1L23L1+:= + By = 22.82 kN+ Fy=0; Ay + By w1L1 0.5w2L2 = 0Ay := By + w1L1 + 0.5w2L2 Ay = 12.29 kNEquations of Equilibrium: For point C+ Fx=0; NC := 0 Ans+ Fy=0; Ay w1 a b + ( ) VC = 0:= VC = 3.92 kN AnsVC Ay w1 a b + ( ) + C=0; MC w1 a b + ( ) + 0.5 (a + b) Ay (a + b) = 0MC w1 a b + ( ) := 0.5 (a + b) + Ay (a + b)MC = 15.07 kNm AnsNote: Negative sign indicates that VC acts in the opposite direction to that shown on FBD. 12. Problem 1-11The beam supports the distributed load shown. Determine the resultant internal loadings on the crosssections through points D and E. Assume the reactions at the supports A and B are vertical.Given: w1 4.5kNm:= w2 6.0kNm:=a := 1.8m b := 1.8m c := 2.4md := 1.35m e := 1.35mSolution: L1 := a + b + cL2 := d + eSupport Reactions:+ A=0; ByL1 (w1L1)(0.5L1) (0.5w2L2) L1L23+ = 0By (w1L1) (0.5) (0.5w2L2) 1L23L1+:= + By = 22.82 kN+ Fy=0; Ay + By w1L1 0.5w2L2 = 0Ay := By + w1L1 + 0.5w2L2 Ay = 12.29 kNEquations of Equilibrium: For point D+ Fx=0; ND := 0 Ans+ Fy=0; Ay w1 a ( ) VD = 0VD := Ay w1 (a) VD = 4.18 kN Ans+ D=0; MD w1 a ( ) + 0.5 (a) Ay (a) = 0:= 0.5 (a) + Ay (a) MD = 14.823 kNm AnsMD w1 a ( ) Equations of Equilibrium: For point E+ Fx=0; NE := 0 Ans+ Fy=0; VE 0.5w2 (0.5e) = 0VE := 0.5w2 (0.5e) VE = 2.03 kN Ans+ D=0; ME 0.5w2 0.5 e ( ) e3 = 0ME 0.5 w2 0.5 e ( ) e3:= ME = 0.911 kNm AnsNote: Negative sign indicates that ME acts in the opposite direction to that shown on FBD. 13. Problem 1-12Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section islocated through the centroid, point C.Given: w 9kNm:= := 45dega := 1.2m b := 2.4mSolution: L := a + bSupport Reactions:+ A=0; BxL sin() + (wL)(0.5L) = 0Bx(wL) (0.5L)L sin():= Bx = 22.91 kN+ Fy=0; Ay wL sin() = 0Ay := wL sin() Ay = 22.91 kN+ Fx=0; Bx wLcos() + Ax = 0Ax := wLcos() Bx Ax = 0 kN(a) Equations of equilibrium: For Section a - a :+ Fx=0; NC + Ay sin() = 0NC := Ay sin()+ Fy=0; VC + Aycos() wa = 0NC = 16.2 kN AnsVC := Aycos() + wa VC = 5.4 kN Ans+ A=0; MC (wa) (0.5a) + Ay cos()a = 0MC := (wa) (0.5a) Ay cos()a MC = 12.96 kNm Ans(b) Equations of equilibrium: For Section b - b :+ Fx=0; NC + wacos() = 0NC := wacos() NC = 7.64 kN Ans+ Fy=0; VC wa sin() + Ay = 0VC := wa sin() Ay VC = 15.27 kN Ans+ A=0; MC (wa) (0.5a) + Ay cos()a = 0MC := (wa) (0.5a) Ay cos()a MC = 12.96 kNm Ans 14. Problem 1-13Determine 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 = 60 degree. The 650-N load is appliedalong the centroidal axis of the member.Given: P := 650N := 60deg(a) Equations of equilibrium: For Section a - a :+ Fy=0; P Na_a = 0Na_a := PNa_a = 650N Ans+ Fx=0; Va_a := 0 Ans(b) Equations of equilibrium: For Section b - b :+ Fy=0; Vb_b + Pcos(90deg ) = 0Vb_b := Pcos(90deg )Vb_b = 562.92N Ans+ Fx=0; Nb_b P sin(90deg ) = 0Nb_b := P sin(90deg )Nb_b = 325N Ans 15. Problem 1-14Determine the resultant internal normal and shear forces in the member at section b-b, each as afunction of . Plot these results for 0o 90o. The 650-N load is applied along the centroidal axisof the member.Given: P := 650N := 0Equations of equilibrium: For Section b - b :+ Fx0; Nb_b Pcos() = 0Nb_b := Pcos() Ans+ Fy=0; Vb_b + Pcos() = 0Vb_b := Pcos() Ans 16. Problem 1-15The 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. Thebeam has a weight of 600 N/m and is fixed to the wall at A.Given:W1 := 4000N w 600Nm:=W2 := 450Na := 1.2m b := 1.2m c := 0.9md := 0.9m e := 1.2mf := 0.45m r := 0.075mSolution:Tension in rope: TW12:=T = 2.00 kNEquations of Equilibrium: For point B+ Fx=0; (NB T) = 0NB := T NB = 2 kN Ans+ Fy=0; VB w (e) W1 = 0VB := w (e) + W1 VB = 4.72 kN Ans+ B=0; MB [w (e)]0.5 (e) W1 (e + r) + T (f) = 0MB := [w (e)]0.5 (e) W1 (e + r) + T (f)MB = 4.632 kNm Ans 17. Problem 1-16Determine the resultant internal loadings acting on the cross section through points C and D of thebeam in Prob. 1-15.Given: W1 := 4000Nw 600Nm:=W2 := 450Na := 1.2m b := 1.2m c := 0.9md := 0.9m e := 1.2mf := 0.45m r := 0.075mSolution:Tension in rope: TW12:= T = 2.00 kNEquations of Equilibrium: For point C LC := d + e+ Fx=0; (NC T) = 0NC := T NC = 2 kN Ans+ Fy=0; VC w (LC) W1 = 0VC := w (LC) + W1 VC = 5.26 kN Ans+ C=0; MC w LC ( ) 0.5 (LC) W1 (LC + r) + T (f) = 0MC w LC ( ) := 0.5 (LC) W1 (LC + r) + T (f)MC = 9.123 kNm AnsEquations of Equilibrium: For point D LD := b + c + d + e+ Fx=0; ND := 0 ND = 0 kN Ans+ Fy=0; VD w (LD) W1 W2 = 0VD := w (LD) + W1 + W2 VD = 6.97 kN Ans+ C=0; MD w LD ( ) 0.5 (LD) W1 (LD + r) W2 (b) = 0MD w LD ( ) := 0.5 (LD) W1 (LD + r) W2 (b)MD = 22.932 kNm Ans 18. Problem 1-17Determine the resultant internal loadings acting on the cross section at point B.Given: w 900kNm:=a := 1m b := 4mSolution: L := a + bEquations of Equilibrium: For point B+ Fx=0; NB := 0NB = 0 kN Ans+ Fy=0; VB 0.5 wbL (b) = 0VB 0.5 wbL := (b)VB = 1440 kN Ans+ B=0; MB 0.5 wbL (b)b3 = 0MB 0.5 wbL (b)b3:= MB = 1920 kNm Ans 19. Problem 1-18The beam supports the distributed load shown. Determine the resultant internal loadings acting on thecross section through point C. Assume the reactions at the supports A and B are vertical.Given: w1 0.5kNm:= a := 3mw2 1.5kNm:=Solution: L := 3a w := w2 w1Support Reactions:+ A=0; ByL (w1L)(0.5L) [0.5(w)L]2L3 = 0By (w1L) (0.5) [0.5(w)L]23:= + By = 5.25 kN+ Fy=0; Ay + By w1L 0.5(w)L = 0Ay := By + w1L + 0.5(w)LAy = 3.75 kNEquations of Equilibrium: For point C+ Fx=0; NC := 0 NC = 0 kN Ans+ Fy=0; VC + w1a 0.5 waL + (a) Ay = 0VC w1a 0.5 waL := (a) + AyVC = 1.75 kN Ans+ C=0; MC + (w1a)(0.5a) 0.5 waL (a)a3+ Aya = 0MC (w1a)(0.5a) 0.5 waL (a)a3:= + AyaMC = 8.5 kNm Ans 20. Problem 1-19Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.Given: w1 0.5kNm:= a := 3mw2 1.5kNm:=Solution: L := 3a w := w2 w1Support Reactions:+ A=0; ByL (w1L)(0.5L) [0.5(w)L]2L3 = 0By (w1L) (0.5) [0.5(w)L]23:= + By = 5.25 kN+ Fy=0; Ay + By w1L 0.5(w)L = 0Ay := By + w1L + 0.5(w)LAy = 3.75 kNEquations of Equilibrium: For point D+ Fx=0; ND := 0 ND = 0 kN Ans+ Fy=0; VD + w1 (2a) 0.5 w2aL + (2a) Ay = 0 VD w1 (2a) 0.5 w2aL:= (2a) + AyVD = 1.25 kN Ans+ D=0; MD w1 2a ( ) a ( ) + 0.5 w2aL (2a)2a3+ Ay (2a) = 0MD w1 2a ( ) (a) 0.5 w2aL (2a)2a3:= + Ay (2a)MD = 9.5 kNm Ans 21. Problem 1-20The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kNon the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internalloadings at cross sections through points D, E, and F.Given: P := 4kN a := 1.2m b := 1.8mSolution:Support Reactions: FBD (a) and (b).Given+ A=0; By (a) + Bx (0.5b) P (a) = 0 [1]+ C=0; Bx (0.5b) + P (a) By (a) P (a) = 0 [2]Solving [1] and [2]: Initial guess: Bx := 1kN By := 2kNBxBy:= Find(Bx , By)BxBy2.672= kNFrom FBD (a):+ Fx=0; Bx Ax = 0Ax := Bx Ax = 2.67 kN+ Fy=0; Ay P By = 0Ay := P + By Ay = 6 kNFrom FBD (b):+ Fx=0; Cx Bx = 0Cx := Bx Cx = 2.67 kN+ Fy=0; Cy + By P P = 0Cy := 2P By Cy = 6 kNEquations of Equilibrium: For point D [FBD (c)].+ Fx=0; VD := 0 VD = 0 kN Ans+ Fy=0; ND := 0 ND = 0 kN Ans+ D=0; MD := 0 MD = 0 kNm AnsFor point E [FBD (d)].+ Fx=0; Ax VE = 0 VE := AxVE = 2.67 kN Ans+ Fy=0; NE Ay = 0 NE := AyNE = 6 kN Ans+ E=0; ME Ax (0.5b) = 0 ME := Ax (0.5b) ME = 2.4 kNm AnsFor point F [FBD (e)].+ Fx=0; VF + Ax Cx = 0 VF := Ax + Cx VF = 0 kN Ans+ Fy=0; NF Ay Cy = 0 NF := Ay + Cy NF = 12 kN Ans+ F=0; MF (Ax + Cx) (0.5b) = 0 MF := (Ax + Cx) (0.5b) MF = 4.8 kNm Ans 22. Problem 1-21The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. Thegripping action on the drum chime is such that only horizontal and vertical forces are exerted on thedrum at G and H. Determine the resultant internal loadings on the cross section through point I.Given: P := 2.5 kN := 60dega := 200mm b := 125mm c := 75mmd := 125mm e := 125mm f := 50mmSolution:Equations of Equilibrium: Memeber Ac and BD aretwo-force members.Fy=0; P 2F sin() = 0 [1]FP:= [2]2 sin()F = 1.443 kNEquations of Equilibrium: For point I.+ Fx=0; VI Fcos() = 0 VI := Fcos() VI = 0.722 kN Ans+ Fy=0; NI + F sin() = 0 NI := F sin() NI = 1.25 kN Ans=0; MI + Fcos() (a) = 0+ IMI := Fcos() (a) MI = 0.144 kNm Ans 23. Problem 1-22Determine the resultant internal loadings on the cross sections through points K and J on the drum lifterin Prob. 1-21.Given: P := 2.5 kN := 60dega := 200mm b := 125mm c := 75mmd := 125mm e := 125mm f := 50mmSolution:Equations of Equilibrium: Memeber Ac and BD aretwo-force members.Fy=0; P 2F sin() = 0 [1]FP:= [2]2 sin()F = 1.443 kNEquations of Equilibrium: For point J.+ Fy'=0; VI := 0 VI = 0 kN Ans+ Fx'=0; NI + F = 0 NI := F NI = 1.443 kN Ans+ J=0; MJ := 0 MJ = 0 kNm AnsNote: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.Support Reactions: For Member DFH :+ H=0; FEF (c) F cos() (a + b + c) + F sin() (f) = 0FEF Fcos()a + b + cc F sin()fc:= FEF = 3.016 kNEquations of Equilibrium: For point K.+ Fx=0; NK + FEF = 0 NK := FEF NK = 3.016 kN Ans+ Fy=0; VK := 0 VK = 0 kN Ans+ K=0; MK := 0 MK = 0 kNm Ans 24. Problem 1-23The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section at B. Neglect the weight of the wrench CD.Given: g 9.81ms2:= 12kgm:= P := 60Na := 0.150m b := 0.400mc := 0.200m d := 0.300mSolution: w := gFx=0; NBx := 0N AnsFy=0; VBy := 0N AnsFz=0; VBz P + P w (b + c) = 0VBz := P P + w (b + c)VBz = 70.6 N Ans x=0; TBx + P (b) P (b) (wb) (0.5b) = 0TBx := P (b) + P (b) + (wb) (0.5b) TBx = 9.42Nm Ans y=0; MBy P (2a) + (wb) (c) + (wc) (0.5c) = 0MBy := P (2a) (wb) (c) (wc) (0.5c) MBy = 6.23 Nm Ans=0; MBz := 0Nm Ans z 25. Problem 1-24The main beam AB supports the load on the wing of the airplane. The loads consist of the wheelreaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity atD, 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 nottransfer any of the loads to the fuselage, except through the beam.Given: PC := 175kN PE := 2kN PD := 6kNa := 1.8m b := 1.2m e := 0.3mc := 0.6m d := 0.45mSolution:Fx=0; VAx := 0kN AnsFy=0; NAy := 0kN AnsFz=0; VAz PD PE + PC = 0VAz := PD + PE PCVAz = 167 kN Ans x=0;MAx := PD (a) + PE (a + b + c) PC (a + b) MAx = 507 kNm Ans y=0; TAy + PD (d) PE (e) = 0TAy := PD (d) + PE (e) TAy = 2.1 kNm Ans=0; MAz := 0kNm Ans zMAx PD (a) PE (a + b + c) + PC (a + b) = 0 26. Problem 1-25Determine 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 thesign.Given: a := 4m d := 2m p 50 Nm2:=b := 6m e := 3mc := 3mSolution: P := p (c) (d + e)Fx=0; VBx P = 0VBx := PVBx = 750N AnsFy=0; VBy := 0N AnsFz=0; NBz := 0N Ans x=0; MBx := 0Nm Ans y=0; MBy P (b + 0.5c) = 0MBy := P (b + 0.5c)MBy = 5625Nm Ans z=0; TBz P [e 0.5 (d + e)] = 0TBz := P [e 0.5 (d + e)]TBz = 375Nm Ans 27. Problem 1-26The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.Given: P1z := 400N P2y := 200N P3y := 80Na := 0.3m b := 0.4m c := 0.3m d := 0.4mSolution: L := a + b + c + dSupport Reactions: z=0; 2P3y (d) + 2P2y (c + d) Ay (L) = 0Ay 2P3ydL 2P2yc + dL:= + Ay = 245.71NFy=0; Ay By + 2P2y + 2P3y = 0By := Ay + 2P2y + 2P3y By = 314.29 N y=0; 2P1z (b + c + d) Az (L) = 0Az 2P1zb + c + d:= Az = 628.57 NLFz=0; Bz + Az 2P1z = 0Bz := Az + 2P1z Bz = 171.43NEquations of Equilibrium: For point D.Fx=0; NDx := 0N AnsFy=0; VDy By + 2P3y = 0VDy := By 2P3yVDy = 154.3N AnsFz=0; VDz + Bz = 0VDz := BzVDz = 171.4N Ans x=0; TDx := 0Nm Ans y=0; MDy + Bz (d + 0.5c) = 0:= MDy = 94.29Nm AnsMDy Bz d 0.5 c + ( ) z=0; MDz + By (d + 0.5c) 2P3y (0.5c) = 0MDz := By (d + 0.5c) + 2P3y (0.5c) MDz = 148.86 Nm Ans 28. Problem 1-27The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to thepulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section throughpoint C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +ydirection. The journal bearings at A and B exert only y and z components of force on the shaft.Given: P1z := 400N P2y := 200N P3y := 80Na := 0.3m b := 0.4m c := 0.3m d := 0.4mSolution: L := a + b + c + dSupport Reactions: z=0; 2P3y (d) + 2P2y (c + d) Ay (L) = 0Ay 2P3ydL 2P2yc + dL:= + Ay = 245.71NFy=0; Ay By + 2P2y + 2P3y = 0By := Ay + 2P2y + 2P3y By = 314.29 N y=0; 2P1z (b + c + d) Az (L) = 0Az 2P1zb + c + d:= Az = 628.57 NLFz=0; Bz + Az 2P1z = 0Bz := Az + 2P1z Bz = 171.43NEquations of Equilibrium: For point C.Fx=0; NCx := 0N AnsFy=0; VCy Ay = 0VCy := AyVCy = 245.7N AnsFz=0; VCz + Az 2P1z = 0VCz := Az + 2P1zVCz = 171.4N Ans=0; TCx := 0Nm Ans x y=0; MCy Az (a + 0.5b) + 2P1z (0.5b) = 0MCy := Az (a + 0.5b) 2P1z (0.5b) MCy = 154.29 Nm Ans z=0; MCz + Ay (a + 0.5b) = 0MCz := Ay (a + 0.5b) MCz = 122.86 Nm Ans 29. Problem 1-28Determine 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 := 400Nd := 0.9m e := 1.2m := 30degSolution: L := d2 + e2Member DEF :+ D=0; NE (b) P (a + b) = 0NE Pa + bb:= NE = 720NMember BCE :+ B=0; FACeL (d) NE sin() (c + d) = 0FACLedNE sin ( ) c d + ( ) := FAC = 900N+ Fx=0; Bx FACdL+ NEcos() = 0Bx FACdL:= + NEcos()Bx = 83.54N+ Fy=0; By FACeL+ NE sin() = 0By FACeL:= NE sin()By = 360NEquations of Equilibrium: For point F.+ Fy'=0; NF := 0 NF = 0N Ans+ Fx'=0; VF P = 0 VF := P VF = 400N Ans+ F=0; MF P (0.5a) = 0 MF := P (0.5a) MF = 240Nm AnsEquations of Equilibrium: For point G.+ Fx=0; Bx NG = 0 NG := Bx NG = 83.54N Ans+ Fy=0; VG By = 0 VG := By VG = 360N Ans+ G=0; MG + By (0.5d) = 0 MG := By (0.5d) MG = 162Nm Ans 30. Problem 1-29The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting onthe cross section at point C.Given:P := 400Nr := 150mm := 90degSolution:Equations of Equilibrium: For segment AC.+ Fx=0; NC + P = 0 NC := P NC = 400N Ans+ Fy=0; VC := 0 VC = 0N Ans+ G=0; MC + P (r) = 0 MC := P (r) MC = 60Nm Ans 31. Problem 1-30The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadingsacting on the cross section through B.Given: P := 750N MC := 800Nm 12kgm:= g 9.81ms2:=a := 1m b := 2m c := 2mSolution:Py45:= P Pz35:= PEquations of Equilibrium: For point B.Fx=0; VBx := 0kip AnsFy=0; NBy + Py = 0NBy := Py AnsNBy = 600 N AnsFz=0; VBz + Pz gc gb = 0VBz := Pz + gc + gbVBz = 920.9N Ans x=0; MBx + Pz (b) gc (b) gb (0.5b) = 0MBx := Pz (b) + gc (b) + gb (0.5b)MBx = 1606.3 Nm Ans=0; TBy := 0Nm Ans y z=0; MBy + Mc = 0MBy := MCMBy = 800Nm Ans 32. Problem 1-31The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadingsacting on the cross section through A which is located at an angle from the horizontal.Solution: P := kN := degEquations of Equilibrium: For point A.+ Fx=0; NA + Pcos() = 0NA := Pcos() Ans+ Fy=0; VA P sin() = 0VA := P sin() Ans+ A=0; MA P r (1 cos()) = 0MA := P r (1 cos()) Ans 33. Problem 1-32The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determinethe resultant internal loadings acting on the cross section through point B. Hint: The distance from thecentroid C of segment AB to point O is CO = 0.9745r.Given: := 22.5deg r := m a := 0.9745r wkNm2:=Solution:Equations of Equilibrium: For point B.Fz=0; VB4 rw = 0 VB := 0.785w r AnsFx=0; NB := 0 Ans x=0; TB4 rw (0.09968r) = 0 TB := 0.0783w r2 Ans y=0; MB4+ rw (0.37293r) = 0 MB := 0.293w r2 Ans 34. Problem 1-33A differential element taken from a curved bar is shown in the figure. Show that dN/d = V,dV/d = -N, dM/d = -T, and dT/d = M,Solution: 35. Problem 1-34The column is subjected to an axial force of 8 kN, which is applied through the centroid of thecross-sectional area. Determine the average normal stress acting at section aa. Show this distributionof stress acting over the areas cross section.Given: P := 8kNb := 150mm d := 140mm t := 10mmSolution:A := 2(b t) + d t A = 4400.00mm2PA:= = 1.82MPa Ans 36. Problem 1-35The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine theaverage shear stress in the pin.Given: P := 3.0kN d := 6mmSolution:+ Fy=0; 2V P = 0V := 0.5PV = 1500NAd24:= A = 28.2743mm2avgVA:= avg = 53.05MPa Ans 37. Problem 1-36While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times hisweight. 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 diameterof 25 mm. Assume the fibula F does not support a load.Given: g 9.81ms2=M := 75kgdo := 45mm di := 25mmSolution:A4:= A = 1099.5574mm2do2 di2 5MgA:= = 3.345MPa Ans 38. Problem 1-37The thrust bearing is subjected to the loads shown. Determine the average normal stress developed oncross sections through points B, C, and D. Sketch the results on a differential volume element locatedat each section.Units Used: kPa := 103PaGiven: P := 500N Q := 200NdB := 65mm dC := 140mm dD := 100mmSolution:AB 24 dB:= AB = 3318.3mm2BPAB:= B = 150.7 kPa AnsAC 24 dC:= AC = 15393.8mm2CPAC:= C = 32.5 kPa AnsAD 24 dD:= AD = 7854.0mm2DQAD:= D = 25.5 kPa Ans 39. Problem 1-38The small block has a thickness of 5 mm. If the stress distribution at the support developed by the loadvaries 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 := 5mm1 := 0MPa 2 := 40MPa 3 := 60MPaSolution:= dF AF := 0.52 (a t) + 2 (b t) + 0.5 (3 2) (b t)F = 36.00 kN AnsRequire:= dFd x Ad0.5 2 a t ( ) 2a3a 0.5 b + ( ) + 0.5 3 2 ( ) b t ( ) 2 b t ( ) a2b3+ + F:=d = 110mm Ans 40. Problem 1-39The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If acouple is applied to the lever, determine the average shear stress in the pin between the pin and lever.Given: a := 250mm b := 12mmd := 6mm P := 20NSolution:+ O=0; Vb P (2a) = 0V P2ab:= V = 833.33 NAd24:= A = 28.2743mm2avgVA:= avg = 29.47MPa Ans 41. Problem 1-40The cinder block has the dimensions shown. If the material fails when the average normal stressreaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.Given: allow := 0.840MPaao := 150mm ai := 100mmbo := [2 (1 + 2 + 3) + 2]mmbi := [2 (1 + 3)]mmSolution:A := aobo aibi A = 1300mm2Pallow := allow (A)Pallow = 1.092 kN Ans 42. Problem 1-41The 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 volumeelement of the material.Given: P := 4kNao := 150mm ai := 100mmbo := [2 (1 + 2 + 3) + 2]mmbi := [2 (1 + 3)]mmSolution:A := aobo aibi A = 1300mm2PA:= = 3.08MPa Ans 43. Problem 1-42The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod issubjected to the greater average normal stress and compute its value. Take = 30. The diameter ofeach rod is given in the figure.Given: W := 250N := 30deg := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution: Initial guess: FAC := 1N FAD := 1NGiven+ Fx=0; FACcos() FADcos() = 0 [1]+ Fy=0; FAC sin() + FAD sin() W = 0 [2]Solving [1] and [2]:FACFAD:= Find(FAC, FAD)FACFAD183.01224.14= NRod AB:AAB 24 dB:= AAB = 63.61725mm2ABWAAB:= AB = 3.93MPaRod AD :AAD 24 dD:= AAD = 44.17865mm2ADFADAAD:= AD = 5.074MPaRod AC:AAC 24 dC:= AAC = 28.27433mm2ACFACAAC:= AC = 6.473MPa Ans 44. Problem 1-43Solve Prob. 1-42 for = 45.Given: W := 250N := 45deg := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution: Initial guess: FAC := 1N FAD := 1NGiven+ Fx=0; FACcos() FADcos() = 0 [1]+ Fy=0; FAC sin() + FAD sin() W = 0 [2]Solving [1] and [2]:FACFAD:= Find(FAC, FAD)FACFAD176.78176.78= NRod AB:AAB 24 dB:= AAB = 63.61725mm2ABWAAB:= AB = 3.93MPaRod AD :AAD 24 dD:= AAD = 44.17865mm2ADFADAAD:= AD = 4.001MPaRod AC:AAC 24 dC:= AAC = 28.27433mm2ACFACAAC:= AC = 6.252MPa Ans 45. Problem 1-44The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle oforientation of AC such that the average normal stress in rod AC is twice the average normal stress inrod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.Given: W := 250N := 45degdB := 9mm dC := 6mm dD := 7.5mmSolution:Rod AB: AAB 24 dB:= AAB = 63.61725mm2Rod AD : AAD 24 dD:= AAD = 44.17865mm2Rod AC: AAC 24 dC:= AAC = 28.27433mm2Since AC = 2AD ThereforeFACAAC2FADAAD=Initial guess: FAC := 1N FAD := 2N := 30degGiven FACAAC2FADAAD= [1]+ Fx=0; FACcos() FADcos() = 0 [2]+ Fy=0; FAC sin() + FAD sin() W = 0 [3]Solving [1], [2] and [3]:FACFAD:= Find(FAC, FAD, )FACFAD180.38140.92= N = 56.47 degABWAAB:= AB = 3.93MPa AnsADFADAAD:= AD = 3.19MPa AnsACFACAAC:= AC = 6.38MPa Ans 46. Problem 1-45The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter holein the fixed support A, determine the bearing stress acting on the collar C. Also, what is the averageshear stress acting along the inside surface of the collar where it is fixed connected tothe 52-mm diameter shaft?Given: P := 30kNdhole := 53mm dshaft := 52mmdcollar := 60mm hcollar := 10mmSolution:Bearing Stress:Ab4dcollar2 dhole2 := bPAb:= b = 48.3MPa AnsAverage Shear Stress:As := (dshaft) (hcollar)avgPAs:= avg = 18.4MPa Ans 47. Problem 1-46The two steel members are joined together using a 60 scarf weld. Determine the average normal andaverage shear stress resisted in the plane of the weld.Given:P := 8kN := 60degb := 25mm h := 30mmSolution:Equations of Equilibrium:+ Fx=0; N P sin() = 0N := P sin() N = 6.928 kN+ Fy=0; V Pcos() = 0V := Pcos() V = 4 kNAhbsin() :=NA:= = 8MPa AnsavgVA:= avg = 4.62MPa Ans 48. Problem 1-47The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine theaverage normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.Given: P := 775Na := 40mm b := 30mmd := 8mm := 20degSolution:Support Reaction:FA=0; P (a) FBCcos() (a + b) = 0FBCPa(a + b)cos():=FBC = 471.28NAverage Normal Stress:ABCd24:=FBCABC:= = 9.38MPa Ans 49. Problem 1-48The board is subjected to a tensile force of 425 N. Determine the average normal and average shearstress developed in the wood fibers that are oriented along section a-a at 15 with the axis of the board.Given: P := 425N := 15degb := 25mm h := 75mmSolution:Equations of Equilibrium:+ Fx=0; V Pcos() = 0V := Pcos() V = 410.518N+ Fy=0; N P sin() = 0N := P sin() N = 1NAverage Normal Stress:Ahbsin() :=NA:= = 0.0152MPa AnsavgVA:= avg = 0.0567MPa Ans 50. Problem 1-49The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determinethe average normal and average shear stress components that this loading creates on the face of theweld, section AB.Given: P := 250kN := 30degb := 150mm h := 50mmSolution:Equations of Equilibrium:+ Fx=0; V + P sin() = 0V := P sin() V = 125 kN+ Fy=0; N Pcos() = 0N := Pcos() N = 216.506 kNAverage Normal and Shear Stress:Ahbsin(2) :=NA:= = 25MPa AnsavgVA:= avg = 14.434MPa Ans 51. Problem 1-50The specimen failed in a tension test at an angle of 52 when the axial load was 100 kN. If the diameterof the specimen is 12 mm, determine the average normal and average shear stress acting on the area ofthe inclined failure plane. Also, what is the average normal stress acting on the cross section whenfailure occurs?Given: P := 100kNd := 12mm := 52degSolution:Equations of Equilibrium:+ Fx=0; V Pcos() = 0V := Pcos() V = 61.566 kN+ Fy=0; N P sin() = 0N := P sin() N = 78.801 kNInclined plane:A4d2sin():=NA:= = 549.05MPa AnsavgVA:= avg = 428.96MPa AnsCross section:Ad24:=PA:= = 884.19MPa Ansavg := 0 avg = 0MPa Ans 52. Problem 1-51A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine themaximum average shear stress in the specimen and indicate the orientation of a section on which itoccurs.Solution:Equations of Equilibrium:+ Fy=0; V Pcos() = 0V = Pcos()Inclined plane:AinclAsin() =VAincl= Pcos() sin()= AP sin(2)2A=ddPcos(2)A=dd= 0cos(2) = 02 = 90deg := 45deg AnsmaxP sin(90)= max2AP2A= Ans 53. Problem 1-52The joint is subjected to the axial member force of 5 kN. Determine the average normal stress actingon sections AB and BC. Assume the member is smooth and is 50-mm thick.Given: P := 5kN := 45deg := 60degdAB := 40mm dBC := 50mm t := 50mmSolution: := 90deg = 30.00 degAAB := tdABABC := tdBC+ Fx=0; NABcos() Pcos() = 0NABPcos()cos() :=NAB = 4.082 kN+ Fy=0; NAB sin() + P sin() NBC = 0NBC := NAB sin() + P sin()NBC = 1.494 kNABNABAAB:= AB = 2.041MPa AnsBCNBCABC:= BC = 0.598MPa Ans 54. Problem 1-53The yoke is subjected to the force and couple moment. Determine the average shear stress in the boltacting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couplemoment is resisted by a set of couple forces developed in the shank of the bolt.Given: P := 2.5kN M := 120Nmho := 62mm hi := 50mmd := 6mm := 60degSolution:As a force on bolt shank is zero, thenA := 0 AnsEquations od Equilibrium:Fz=0; P 2Fz = 0Fz := 0.5P Fz = 1.25 kNMz=0; M Fx (hi) = 0FxMhi:= Fx = 2.4 kNAverage Shear Stress:Ad24:=:= + 22 FzThe bolt shank subjected to a shear force of VB FxBVBA:= B = 95.71MPa Ans 55. Problem 1-54The two members used in the construction of an aircraft fuselage are joined together using a 30fish-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 := 4.0kNb := 37.5mm hhalf := 25m := 30degSolution:Equations of Equilibrium:+ Fx=0; V + 0.5Pcos() = 0V := 0.5Pcos() V = 1.732 kN+ Fy=0; N 0.5P sin() = 0N := 0.5P sin() N = 1 kNAverage Normal and Shear Stress:A(hhalf)bsin() :=NA:= = 533.33 Pa AnsavgVA:= avg = 923.76 Pa Ans 56. Problem 1-55The row of staples AB contained in the stapler is glued together so that the maximum shear stress theglue can withstand is max = 84 kPa. Determine the minimum force F that must be placed on theplunger in order to shear off a staple from its row and allow it to exit undeformed through the grooveat C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mmAssume all the other parts are rigid and neglect friction.Given: max := 0.084MPaa := 12.5mm b := 7.5mmt := 1.25mmSolution:Average Shear Stress:A := ab [(a 2t) (b t)]maxVA= V := (max)AV = 2.63 NFmin := VFmin = 2.63 N Ans 57. Problem 1-56Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to thering at B, determine the average normal stress in each rod if = 60.Given: W := 8kN := 60degdA := 4mm dC := 6mmSolution:Rod AB: AAB 24 dA:=Rod BC : ABC 24 dC:=+ Fy=0; FBC sin() W = 0FBCWsin() :=FBC = 9.238 kN+ Fx=0; FBCcos() FAB = 0FAB := FBCcos()FAB = 4.619 kNABFABAAB:= AB = 367.6MPa AnsBCFBCABC:= BC = 326.7MPa Ans 58. Problem 1-57Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN isapplied to the ring at B, determine the angle of rod BC so that the average normal stress in each rodis equivalent. What is this stress?Given: W := 8kNdA := 4mm dC := 6mmSolution:Rod AB: AAB 24 dA:=Rod BC : ABC 24 dC:=+ Fy=0; FBC sin() W = 0+ Fx=0; FBCcos() FAB = 0Since FAB = AABFBC = ABCInitial guess: := 100MPa := 50degGivenABC sin() W = 0 [1]ABCcos() AAB = 0 [2]Solving [1] and [2]::= Find( , ) = 63.61 deg Ans = 315.85MPa Ans 59. Problem 1-58The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normalstress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.Given: P := 40kNa := 0.9m b := 1.2m A := 780mm2Solution: c := a2 + b2 c = 1.5mhbc:= vac:=Joint A:+ Fy=0; (v)FAB P = 0 FABPv:= FAB = 66.667 kN+ Fx=0; (h)FAB FAE = 0 FAE := (h)FAB FAE = 53.333 kNABFABA:= AB = 85.47MPa (T) AnsAEFAEA:= AE = 68.376MPa (C) AnsJoint E:+ Fy=0; FEB 0.75P = 0 FEB := 0.75P FEB = 30 kN+ Fx=0; FED FAE = 0 FED := FAE FED = 53.333 kNEBFEBA:= EB = 38.462MPa (T) AnsEDFEDA:= ED = 68.376MPa (C) AnsJoint B:+ Fy=0; (v)FBD (v)FAB FEB = 0 FBD FABFEBv:= + FBD = 116.667 kN+ Fx=0; FBC (h)FAB (h)FBD = 0 FBC := (h)FAB + (h)FBD FBC = 146.667 kNBCFBCA:= BC = 188.034MPa (T) AnsBDFBDA:= BD = 149.573MPa (C) Ans 60. Problem 1-59The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normalstress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that canbe applied to the truss.Given: allow := 140MPaa := 0.9m b := 1.2m A := 780mm2Solution: c := a2 + b2 c = 1.5mhbc:= vac:=For comparison purpose, set P := 1kNJoint A:+ Fy=0; (v)FAB P = 0 FABPv:= FAB = 1.667 kN+ Fx=0; (h)FAB FAE = 0 FAE := (h)FAB FAE = 1.333 kNABFABA:= AB = 2.137MPa (T)AEFAEA:= AE = 1.709MPa (C)Joint E:+ Fy=0; FEB 0.75P = 0 FEB := 0.75P FEB = 0.75 kN+ Fx=0; FED FAE = 0 FED := FAE FED = 1.333 kNEBFEBA:= EB = 0.962MPa (T)EDFEDA:= ED = 1.709MPa (C)Joint B:+ Fy=0; (v)FBD (v)FAB FEB = 0 FBD FABFEBv:= + FBD = 2.917 kN+ Fx=0; FBC (h)FAB (h)FBD = 0 FBC := (h)FAB + (h)FBD FBC = 3.667 kNBCFBCA:= BC = 4.701MPa (T)BDFBDA:= BD = 3.739MPa (C)Since the cross-sectional areas are the same, the highest stress occurs in the member BC,which has the greatest forceFmax := max(FAB, FAE , FEB, FED, FBD, FBC) Fmax = 3.667 kNPallowPFmax:= (allowA) Pallow = 29.78 kN Ans 61. Problem 1-60The 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 tohold the cap in place.Given: p := 650Pa a := 25mmdi := 35mm do := 40mmSolution:Ap 24 di:= As := (do)(a)P (a) FBCcos() (a + b) = 0P p Ap:= ( ) P = 0.625 NAverage Shear Stress:avgPAs:=avg = 199.1 Pa Ans 62. Problem 1-61The 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 adiameter of 5 mm. Only a vertical force is exerted on the wire.Given: P := 100Na := 37.5mm b := 50mm c := 25mmd := 125mm dpin := 5mmSolution:From FBD (a):+ Fx=0; Bx := 0Bx = 0N+ D=0; P (d) By (c) = 0By Pdc:= By = 500NFrom FBD (b):+ Fx=0; Ax := 0Ax = 0N+ E=0; Ay (a) By (a + b) = 0Ay Bya + ba:= Ay = 1166.67NAverage Shear Stress:Apin 24 dpin:=VA := 0.5 (Ay) VA = 583.333NavgVAApin:= avg = 29.709MPa Ans 63. Problem 1-62Solve 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 := 25mmd := 125mm dpin := 5mm P := 100NSolution:From FBD (a):+ Fx=0; Bx := 0Bx = 0N+ D=0; P (d) By (c) = 0By Pdc:= By = 500NAverage Shear Stress: Pin B is subjected to doule shearApin 24 dpin:=VB := 0.5 (By) VB = 250NavgVBApin:= avg = 12.732MPa Ans 64. Problem 1-63The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and theextension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed tosupport the lamp. Hint: The shear force in the pin is caused by the couple moment required forequilibrium at A.Given: w 8Nm:= P := 20Na := 900mm h := 32mmdpin := 3mmSolution:From FBD (a):+ Fx=0; Bx := 0+ A=0; V (h) (wa) (0.5a) P (a) = 0V (wa) 0.5ah Pah:= + V = 663.75 NAverage Shear Stress:Apin 24 dpin:=avgVApin:= avg = 93.901MPa Ans 65. Problem 1-64The two-member frame is subjected to the distributed loading shown. Determine the average normalstress and average shear stress acting at sections a-a and b-b. Member CB has a square cross sectionof 35 mm on each side. Take w = 8 kN/m.Given: w 8kNm:=a := 3m b := 4m A := (0.0352)m2Solution: c := a2 + b2 c = 5mhac:= vbc:=Member AB:MA=0; By (a) (wa) (0.5a) = 0By := 0.5wa By = 12 kN+ Fy=0; (v)FAB By = 0FABByv:= FAB = 15 kNSection a-a:a_aFABA:= a_a = 12.24MPa Ansa_a := 0 a_a = 0MPa AnsSection b-b:+ Fx=0; N FAB (h) = 0 N := FAB (h) N = 9 kN+ Fy=0; V FAB (v) = 0 V := FAB (v) V = 12 kNAb_bAh:=b_bNAb_b:= b_b = 4.41MPa Ansb_bVAb_b:= b_b = 5.88MPa Ans 66. Problem 1-65Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determinethe average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member Bwhich has a thickness of 30 mm.Given: P := 5kN := 60deg := 30degdrod := 10mm h := 40mm t := 30mmSolution:AB := thArod4:= 2drod+ Fx=0; Pcos() FB = 0FB := Pcos() FB = 2.5 kN+ Fy=0; Fc P sin() = 0FC := P sin() FC = 4.33 kNAverage Normal Stress:BFBAB:= B = 2.083MPa AnsCFCArod:= C = 55.133MPa Ans 67. Problem 1-66Consider the general problem of a bar made from m segments, each having a constant cross-sectionalarea Am and length Lm. If there are n loads on the bar as shown, write a computer program that can beused to determine the average normal stress at any specified location x. Show an application of theprogram 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. 68. Problem 1-67The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shearstress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has adiameter of 18 mm.Given: P := 15kNa := 0.5m b := 1m c := 1.5md := 1.5m e := 0.5m := 30deg dpin := 18mmSolution: L := a + b + c + d + eSupport Reactions:A=0; By (L) + P (L a) + 4P (c + d + e) + 4P (d + e) + 2P (e) = 0By PL aL 4Pc + d + e+ 4PLd + eL+ 2PeL:= + By = 82.5 kN+ Fy=0; By + P + 4P + 4P + 2P Ay = 0Ay := By + P + 4P + 4P + 2PAy = 82.5 kNFBCBysin() := FBC = 165 kNAx := FBCcos() Ax = 142.89 kNAverage Shear Stress:Apin 24 dpin:=For Pins B and C:B_and_C0.5FBCApin:= B_and_C = 324.2MPa AnsFor Pin A::= + 2 FA = 165 kNFA Ax2 AyA0.5FAApin:= A = 324.2MPa Ans 69. Problem 1-68The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of theloads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins arein double shear as shown, and each has a diameter of 18 mm.Given: allow := 80MPaa := 0.5m b := 1m c := 1.5md := 1.5m e := 0.5m := 30deg dpin := 18mmSolution: L := a + b + c + d + eFor comparison purpose, set P := 1kNSupport Reactions:A=0; By (L) + P (L a) + 4P (c + d + e) + 4P (d + e) + 2P (e) = 0By PL aL 4Pc + d + e+ 4PLd + eL+ 2PeL:= + By = 5.5 kN+ Fy=0; By + P + 4P + 4P + 2P Ay = 0Ay := By + P + 4P + 4P + 2P Ay = 5.5 kNFBCBysin() := FBC = 11 kNAx := FBCcos() Ax = 9.53 kNFA := Ax2 + Ay2 FA = 11 kNRequire:Fmax := max(FBC, FA) Fmax = 11 kNApin 24 dpin:=PallowPFmaxallow 2Apin ( ) := Pallow = 3.70 kN Ans 70. Problem 1-69The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as afunction of the bar angle . Plot this function, 0 90o, and indicate the values of for which thisstress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.Given: P := 1kN dbolt := 6mma := 0.6m b := 0.45m c := 0.15mSolution:Support Reactions:C=0; FABcos() (c) + FAB sin() (a) P (a + b) = 0FABP (a + b)cos() (c) + sin() (a)=Average Shear Stress: Pin B is subjected to doule shearFABAbolt= Abolt 24 dbolt:=4P (a + b)2 cos() (c) + sin() (a) dbolt=dd4P (a + b) dbolt2sin() (c) cos() (a)cos() (c) + sin() (a)2= dd= 0 sin() (c) cos() (a) = 0 tan() ac= atanac:= = 75.96 deg Ans 71. Problem 1-70The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of thebeam, 1ft x 12ft. If the hoist is rated to support a maximum of 7.5 kN, determine the maximumaverage normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the16-mm-diameter pin at B.Given: P := 7.5kN xmax := 3.6ma := 3m := 30degdrod := 18mm dpin := 16mmSolution:Support Reactions:C=0; FBC sin() (a) P (x) = 0FBCP (x)sin() (a)=Maximum FBC occurs when x= xmax . Therefore,FBCP (xmax)sin() (a):= FBC = 18.00 kNArod 24 drod:= Apin 24 dpin:=pin0.5FBCApin:= pin = 44.762MPa AnsrodFBCArod:= rod = 70.736MPa Ans 72. Problem 1-71The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normaland average shear stresses acting over the shaded section, which is oriented at from the horizontal.Plot the variation of these stresses as a function of (0o 90o).Solution:Equations of Equilibrium:+ Fx=0; V Pcos() = 0V = Pcos()+ Fy=0; N P sin() = 0N = P sin()Inclined plane:AAsin() =NA= PA= sin()2 AnsavgVA= avgP2A= sin(2) Ans 73. Problem 1-72The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable hasa diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position for0o 90o.Given: W := 3kNa := 1mdo := 15mmSolution: Angle B: B = 0.5(90deg + )B = 45deg + 0.5Support Reactions:A=0; FBC sin(B) (a) W (0.5a) cos() = 0FBC0.5Wcos()sin(45deg + 0.5) =Average Normal Stress:BCFABABC= ABC 24 do:=BC2W do 2cos()sin(45deg + 0.5) = Ans 74. Problem 1-73The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0 < x 0.5m.Given: P1 := 3kN P2 := 6kN:= A := 400 (10 6)m2a := 0.5m b := 0.75mw 8kNmSolution: L := a + b+ Fx=0; N + P1 + P2 + w (L x) = 0N = P1 + P2 + w (L x)Average Normal Stress:NA=P1 + P2 + w (L x)= AnsA 75. Problem 1-74The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributedloading along its length and to two concentrated loads as shown, determine the average normal stress inthe bar as a function of for 0.5m < x 1.25m.Given: P1 := 3kN P2 := 6kN:= A := 400 (10 6)m2a := 0.5m b := 0.75mw 8kNmSolution: L := a + b+ Fx=0; N + P1 + w (L x) = 0N = P1 + w (L x)Average Normal Stress:NA=P1 + w (L x)= AnsA 76. Problem 1-75The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axialcompressive force of 15 kN. Determine the average normal stress in the column as a function of thedistance z measured from its base. Note: The result will be useful only for finding the average normalstress at a section removed from the ends of the column, because of localized deformation at the ends.Given: P := 3kN 2.3(103) kg:= g 9.81m3ms2:=r := 180mm h := 0.75mSolution:A := r2 w := gA+ Fz=0; N P w (h z) = 0N = P + w (h z)Average Normal Stress:NA=P + w (h z)= AnsA 77. Problem 1-76The two-member frame is subjected to the distributed loading shown. Determine the largestintensity of the uniform loading that can be applied to the frame without causing either the averagenormal stress or the average shear stress at section b-b to exceed = 15 MPa, and = 16 MParespectively. Member CB has a square cross-section of 30 mm on each side.Given: allow := 15MPa allow := 16MPaa := 4m b := 3m A := (0.0302)m2Solution: c := a2 + b2 c = 5mvbcach:=:= Set wo 1kNm:=Member AB:MA=0; By (b) (wob) (0.5b) = 0By := 0.5wobBy = 1.5 kNSection b-b:By = FBC (h) FBCByh:=FBC = 1.88 kN+ Fx=0; FBC (h) Vb_b = 0 Vb_b := FBC (h)Vb_b = 1.5 kN+ Fy=0; Nb_b + FBC (v) = 0 Nb_b := FBC (v)Nb_b = 1.125 kNAb_bAv:=b_bNb_bAb_b:= b_b = 0.75MPab_bVb_bAb_b:= b_b = 1MPaAssume failure due to normal stress: wallow woallowb_b:= wallow 20.00kNm=Assume failure due to shear stress: wallow woallowb_b:= wallow 16.00kNm= AnsControls ! 78. Problem 1-77The pedestal supports a load P at its center. If the material has a mass density , determine the radialdimension r as a function of z so that the average normal stress in the pedestal remains constant. Thecross section is circular.Solution:Require: P + w1A= P + w1 + dwA + dA=PdA + w1dA = AdwdwdAP + w1A=dwdA= [1]dA = (r + dr)2 r2dA = 2rdrdw = r2 (g)dzFrom Eq.[1],r2 (g)dz2rdr= r (g)dz2dr= z(1) zg2 0drr1r1r= dgz2lnrr1= r r1 eg2z= However,P r1 2=Ansr r1 e 2g2P r1z= 79. Problem 1-78The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the materialhas a density of 2.5 Mg/m3, determine the average normal stress at the support.Given:ro := 0.5m h := 3m g 9.81ms2== m 2.5 (103)r ro e 0.08y2kgm3:=yunit := 1mSolution:dr e 0.08y2 = dy:= 2 Ao = 0.7854 m2Ao rodV = (r2)dy2 e 0.08 y2 dV ro2= dyV3ro y02 e 0.08 y2 2 (yunit):= dV = 1.584m3W := gVW = 38.835 kNWAo:= = 0.04945MPa Ans 80. Problem 1-79The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at itscenter. If it is rotating in the horizontal plane at a constant angular rate of , determine the averagenormal stress in the bar as a function of x.Solution:Equation of Motion :+ Fx=MaN=M r2 ;N mL x 2 x12L x 2+ = N= (L2 4x2)m8Average Normal Stress:NA== (L2 4x2) Ansm8A 81. Problem 1-80Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so thatthe average shear stress does not exceed allow = 2.1 MPa.Given: P := 4kNt := 10mm allow := 2.1MPaa := 300mm b := 125mmSolution: c := a2 + b2 c = 325mmhac:= vbc:=V := P (v) V = 1.54 kNallowVth=hVt allow:= h = 73.26mmUse h := 75mm h = 75mm Ans 82. Problem 1-81The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failureshear stress for the bolts is fail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.Given: P := 80kN fail := 350MPa := 2.5Solution:allowfail:= allow = 140MPaVbolt 0.5P2:= Vbolt = 20 kNAboltVboltallow=4d2Vboltallow=d4Vboltallow:= d = 13.49mm Ans 83. Problem 1-82The rods AB and CD are made of steel having a failure tensile stress of fail = 510 MPa. Using a factorof safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the loadshown. The beam is assumed to be pin connected at A and C.Given: P1 := 4kN P2 := 6kN P3 := 5kNa := 2m b := 2m c := 3m d := 3m := 1.75 fail := 510MPaSolution: L := a + b + c + dSupport Reactions:A=0; FCD (L) P1 (a) P2 (a + b) P3 (a + b + c) = 0FCD P1aL:= P2+ FCD = 6.70 kNa + bL+ P3a + b + cLC=0; FAB (L) + P1 (b + c + d) + P2 (c + d) + P3 (d) = 0FAB P1b + c + d P2Lc + dL+ P3dL:= + FAB = 8.30 kNAverage Normal Stress: Design of rod sizesallowfail:= allow = 291.43MPaFor Rod ABAboltFABallow=4 2dABFABallow=dAB4FABallow:= dAB = 6.02mm AnsFor Rod CDAboltFCDallow=4 2dCDFCDallow=dCD4FCDallow:= dCD = 5.41mm Ans 84. Problem 1-83The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft isfixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d ifthe allowable shear stress for the key is allow = 35 MPa.Given: P := 200N allow := 35MPaL := 500mm a := 20mmb := 25mmSolution:A=0; Fa_a (a) P (L) = 0Fa_a PLa:= Fa_a = 5000NFor the keyAa_aFa_aallow= bdFa_aallow=d1bFa_aallow:= d = 5.71mm Ans 85. Problem 1-84The 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 forceapplied to the plate is P = 100 kN. The allowable shear stress for the weld material is allow = 100 MPa.Given: P := 100kN allow := 100MPaL := 100mm := 45degSolution:Shear Plane in the Weld: Aweld = La sin()Aweld0.5Pallow=La sin()0.5Pallow=a1L sin()0.5Pallow:= a = 7.071mm Ans 86. Problem 1-85The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block alongthe shaded plane, which is the smallest cross section, determine the largest force P that can be appliedto the plate. The allowable shear stress for the weld material is allow = 100 MPa.Given: a := 8mm L := 100mm := 45deg allow := 100MPaSolution:Shear Plane in the Weld: Aweld = La sin()P = allow (2Aweld)P := allow2(La sin())P = 113.14 kN Ans 87. Problem 1-86The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washerwill not penetrate or shear through it. The allowable normal stress for the bolt is allow = 150 MPa andthe allowable shear stress for the supporting material is allow = 35 MPa.Given: P := 25kN dwasher := 25mmallow := 150MPa allow := 35MPaSolution:Allowable Normal Stress: Design of bolt sizeAboltPallow=4d2Pallow=d4Pallow:= d = 14.567mmUse d := 15mm d = 15mm AnsAllowable Shear Stress: Design of support thicknessAsupportPallow= (dwasher)hPallow=h1 (dwasher)Pallow:= h = 9.095mmUse d := 10mm d = 10mm Ans 88. Problem 1-87The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B ifthe allowable shear stress for the material is allow = 42 MPa. Pin A is subjected to double shear,whereas pin B is subjected to single shear.Given: P := 8kN allow := 42MPaa := 1.5m b := 1.5m c := 1.5m d := 0.6mSolution: BC := 45degSupport Reactions: From FBD (a),D=0; FBC sin(BC) (c) P (c + d) = 0FBC Pc + dsin(BC) (c):= FBC = 15.839 kNFrom FBD (b),A=0; Dy (a + b) P (c + d) = 0Dy Pc + da + b:= Dy = 5.6 kN+ Fx=0; Ax P = 0 Ax := P Ax = 8 kN+ Fy=0; Dy Ay = 0 Ay := Dy:= + 2 FA = 9.77 kNFA Ax2 AyApin0.5FAallow=4d20.5FAallow=d40.5FAallow:= d = 12.166mm AnsFor pin B: Pin A is subjected to single shear, and FB := FBCApinFBallow=4d2FBallow=d4FBallow:= d = 21.913mm Ans 89. Problem 1-88The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of allow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5kN.Given: P := 5kN allow := 200MPaa := 4m b := 3m := 60degSolution:c := a2 + b2 hac:= vbc:=At joint A:Initial guess: FAB := 1kN FAC := 2kNGiven+ Fx=0; FAC (h) FAB sin() = 0 [1]+ Fy=0; FAC (v) + FABcos() P = 0 [2]Solving [1] and [2]:FABFAC:= Find(FAB, FAC)FABFAC4.34964.7086= kNFor wire ABAABFABallow=4 2dABFABallow=dAB4FABallow:= dAB = 5.26mm AnsFor wire ACAACFACallow=4 2dACFACallow=dAC4FACallow:= dAC = 5.48mm Ans 90. Problem 1-89The two steel wires AB and AC are used to support the load. If both wires have an allowable tensilestress of allow = 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: allow := 180MPaa := 4m b := 3m := 60degdAB := 6mm dAC := 4mmSolution:c := a2 + b2 hac:= vbc:=Assume failure of AB: FAB = (AAB)allowFAB4:= dAB2 allow FAB = 5.09 kNAt joint A:Initial guess: P1 := 1kN FAC := 2kNGiven+ Fx=0; FAC (h) FAB sin() = 0 [1]+ Fy=0; FAC (v) + FABcos() P1 = 0 [2]Solving [1] and [2]:P1FAC:= Find(P1 , FAC)P1FAC5.85035.5094= kNAssume failure of AC: FAC = (AAC)allowFAC4:= dAC2 allow FAC = 2.26 kNAt joint A:Initial guess: P2 := 1kN FAB := 2kNGiven+ Fx=0; FAC (h) FAB sin() = 0 [1]+ Fy=0; FAC (v) + FABcos() P2 = 0 [2]Solving [1] and [2]:FABP2:= Find(FAB, P2)FABP22.08952.4019= kNChosoe the smallest value: P := min(P1 , P2) P = 2.40 kN Ans 91. Problem 1-90The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stressof allow = 168 MPa. Determine the greatest load that can be supported without causing the cable to failwhen = 30 and = 45. Neglect the size of the winch.Given: allow := 168MPa do := 6mm := 30deg := 45degSolution:For the cable:Tcable = (Acable)allowTcable4:= do2 allowTcable = 4.7501 kNAt joint B:Initial guess: FAB := 1 kN W := 2 kNGiven+ Fx=0; Tcable cos() + FABcos() = 0 [1]+ Fy=0; W + FAB sin() Tcable sin() = 0 [2]Solving [1] and [2]:FABW:= Find(FAB,W)FABW5.8181.739= kN Ans 92. Problem 1-91The boom is supported by the winch cable that has an allowable normal stress of allow = 168 MPa. Ifit is required that it be able to slowly lift 25 kN, from = 20 to = 50, determine the smallestdiameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect thesize of the winch. Set d = 3.6 m.Given: allow := 168MPa W := 25 kNd := 3.6m a := 6mSolution:Maximum tension in canle occurs when := 20degsin()asin()d= asindasin ( ) := = 11.842 degAt joint B:Initial guess: FAB := 1 kN Tcable := 2 kNGiven := + + Fx=0; Tcable cos() + FABcos() = 0 [1]+ Fy=0; W + FAB sin() Tcable sin() = 0 [2]Solving [1] and [2]:FABTcable:= Find(FAB, Tcable)FABTcable114.478103.491= kNFor the cable:AcablePallow=4 2doTcableallow=do4Tcableallow:= do = 28.006mmUse do := 30mm do = 30mm Ans 93. Problem 1-92The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of thepins at A and B if the allowable shear stress for the material is allow = 100 MPa. Both pins aresubjected to double shear.Given: w 2kNm:= allow := 100MPar := 3mSolution: Member AB is atwo-force member := 45degSupport Reactions:A=0; FBC sin() (r) (w r) (0.5r) = 0FBC0.5w rsin() := FBC = 4.243 kN+ Fy=0; Ay + FBC sin() w r = 0Ay := FBC sin() + w r Ay = 3 kN+ Fx=0; Ax FBCcos() = 0Ax := FBCcos() Ax = 3 kNAverage Shear Stress: Pin A and pin B are subjected to double shearFA := Ax2 + Ay2 FA = 4.243 kNFB := FBC FB = 4.243 kNSince both subjected to the same shear force V = 0.5 FA and V := 0.5FBApinVallow=4 2dpinVallow=dpin4Vallow:= dpin = 5.20mm Ans 94. Problem 1-93Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required tosupport is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (t)allow = 175MPa, (b)allow = 275 MPa, and allow = 115 MPa.Given: P := 150kN t_allow := 175MPaallow := 115MPa b_allow := 275MPad2 := 30mmSolution:Allowable Normal Stress: Design of end cap outer diameterAPt_allow=4d12 d22 Pt_allow=d14Pt_allow:= + d22 d1 = 44.62mm AnsAllowable Bearing Stress: Design of circular shaft diameterAPb_allow=4d32 Pb_allow=d34Pb_allow:= d3 = 26.35mm AnsAllowable Shear Stress: Design of end cap thicknessAPallow= (d3) tPallow=t1d3Pallow:= t = 15.75mm Ans 95. Problem 1-94If the allowable bearing stress for the material under the supports at A and B is (b)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: b_allow := 2.8MPa P := 7.5 kNP1 := 10 kN P2 := 10 kNP3 := 15 kN P4 := 10 kNa := 1.5m b := 2.5mSolution: L := 3a + bSupport Reactions:A=0; By 3a ( ) P2 (a) P3 (2a) P4 (3a) P (L) = 0By P2a3a P32a3a+ P43a3a+ PL3a:= + By = 35 kNB=0; Ay (3a) + P1 (3a) + P2 (2a) + P3 (a) P (b) = 0Ay P13a3a P22a3a+ P3a3a+ Pb3a:= Ay = 17.5 kNFor Plate A:Aplate_AAyb_allow2 Ay= aAb_allow=aAAyb_allow:=aA = 79.057mmUse aA x aA plate: aA = 80mm AnsFor Plate BAplate_BByb_allow2 By= aBb_allow=aBByb_allow:=aB = 111.803mmUse aB x aB plate: aB = 120mm AnsRb = 35kN(b is in subscript) 96. Problem 1-95If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.Given: b_allow := 2.8MPaP1 := 10 kN P2 := 10 kNP3 := 15 kN P4 := 10 kNa := 1.5m b := 2.5maA := 50mm aB := 100mmSolution: L := 3a + bSupport Reactions:A=0; By 3a ( ) P2 (a) P3 (2a) P4 (3a) P (L) = 0By P2a3a P3 2a3a + P43a3a+ PL3a= + B=0; Ay (3a) + P1 (3a) + P2 (2a) + P3 (a) P (b) = 0Ay P13a3a P22a3a+ P3a3a+ Pb3a= 2 For Plate A: Ay aA:= b_allowaA2 b_allow P13a3a P22a3a+ P3a3a+ Pb3a= P P13ab P22ab+ P3ab2 + aAb_allow3ab:= P = 26.400 kNPcase_1 := P2 For Plate B: By aB:= b_allowaB2 b_allow P2a3a P3 2a3a + P43a3a+ PL3a= + P P2aL P3 2aL P43aL2 aBb_allow3aL:= + P = 3.000 kNPcase_2 := PPallow := min(Pcase_1 , Pcase_2) Pallow = 3 kN Ans 97. Problem 1-96Determine the required cross-sectional area of member BC and the diameter of the pins at A and B ifthe allowable normal stress is allow = 21 MPa and the allowable shear stress is allow = 28 MPa.Given: allow := 21MPa allow := 28MPaP := 7.5kip := 60dega := 0.6m b := 1.2m c := 0.6mSolution: L := a + b + cSupport Reactions:A=0; By (L) P (a) P (a + b) = 0By PaL Pa + bL:= + FBCBysin() := FBC = 38.523 kNBy = 33.362 kNBx := FBCcos() Bx = 19.261 kN+ Fy=0; By + P + P Ay = 0 Ay := By + P + P Ay = 33.362 kN+ Fx=0; Bx Ax = 0 Ax := Bx Ax = 19.261 kN:= + 2 FA = 38.523 kNFA Ax2 AyMember BC:ABCFBCallow:= ABC = 1834.416mm2 AnsPin A:AAFAallow=4 2dAFAallow=dA4FAallow:= dA = 41.854mm AnsPin B:AB0.5FBCallow=4 2dB0.5FBCallow=dB40.5FBCallow:= dB = 29.595mm Ans 98. Problem 1-97The 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 thediameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (allow)b =350 MPa and allowable shear stress is allow = 125 MPa.Given: P := 140kNallow := 125MPa b_allow := 350MPahB := 20mm hC := 10mmSolution:Allowable Shear Stress: Assume shear failure dor disk CAPallow= (d2)hCPallow=d21hCPallow:= d2 = 35.65mm AnsAllowable Bearing Stress: Assume bearing failure dor disk CAPb_allow=4d22 d32 Pb_allow=2 4d3 d2Pb_allow:= d3 = 27.60mm AnsAllowable Bearing Stress: Assume bearing failure dor disk BAPb_allow=4d12 Pb_allow=d14Pb_allow:= d1 = 22.57mmSince d3 > d1, disk B might fail due to shear.PA= P:= = 98.73MPa < allow (O.K.!)d1hBTherefore d1 = 22.57mm Ans 99. Problem 1-98Strips A and B are to be glued together using the two strips C and D. Determine the required thicknesst of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that ofstrips C and D.Given: P := 40N t := 30mmbA := 1.5m bB := 1.5mbC := 1m bD := 1mSolution:Average Normal Stress: Requires,A = B B = C C = DN(bA) t0.5N(bC) (tC)=tC0.5(bA) tbC:= tC = 22.5mm Ans 100. Problem 1-99If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the size of square bearing plates A' and B' required to support the loading. Dimension theplates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.Given: b_allow := 2.8MPa P := 7.5kNw 10kNm:= a := 4.5m b := 2.25mSolution: L := a + bSupport Reactions:A=0; By(a) w (a) (0.5a) P (L) = 0By w (0.5a) PLa:= + By = 33.75 kNB=0; Ay (a) + w (a) (0.5a) P (L) = 0Ay w (0.5a) Pba:= Ay = 18.75 kNAllowable Bearing Stress: Design of bearing platesFor Plate A:AreaAyb_allow2 Ay= aAb_allow=aAAyb_allow:=aA = 81.832mmUse aA x aA plate: aA := 90mm aA = 90mm AnsFor Plate BAreaByb_allow2 By= aBb_allow=aBByb_allow:=aB = 109.789mmUse aB x aB plate: aB := 110mm aB = 110.00mm Ans 101. Problem 1-100If the allowable bearing stress for the material under the supports at A and B is (b)allow = 2.8 MPa,determine the maximum load P that can be applied to the beam. The bearing plates A' and B' havesquare cross sections of 50mm x 50mm and 100mm x 100mm, respectively.Given: b_allow := 2.8MPaw 10kNm:= a := 4.5m b := 2.25maA := 50mm aB := 100mmSolution: L := a + bSupport Reactions:A=0;By (a) w (a) (0.5a) P (L) = 0P ByaL waL= (0.5a)B=0; Ay (a) + w (a) (0.5a) P (b) = 0P Ayab wab= + (0.5a)Allowable Bearing Stress:Assume failure of material occurs under plate A. Ay aA2 := b_allow2 b_allow P aAab (wa)0.5ab:= + P = 31 kNPcase_1 := P2 Assume failure of material occurs under plate B. By aB:= b_allowP ByaL waL:= (0.5a) P = 3.67 kNPcase_2 := PPallow := min(Pcase_1 , Pcase_2) Pallow = 3.67 kN Ans 102. Problem 1-101The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the averageshear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has adiameter of 12 mm. If the yield shear stress for the bolt is y = 175 MPa, and the yield tensile stress forthe rod is y = 266 MPa, determine the factor of safety with respect to yielding in each case.Given: y := 175MPa w 12kNm:=y := 266MPaa := 1.2m b := 0.6m e := 0.9mdo := 10mm drod := 12mmSolution:c := a2 + e2 hac:= vec:=Support Reactions: L := a + bC=0; FAB v ( ) (a) w (L) (0.5L) = 0FAB wLav:= (0.5L)FAB = 27 kNFor bolt A: Bolt A is subjected to double shear, and V := 0.5FAB V = 13.5 kNA4:= 2 doVA:= = 171.89MPa AnsFSy:= FS = 1.02 AnsFor rod AB: N := FAB N = 27 kNA4:= 2 drodNA:= = 238.73MPa AnsFSy:= FS = 1.11 Ans 103. Problem 1-102Determine the intensity w of the maximum distributed load that can be supported by the hangerassembly so that an allowable shear stress of allow = 95 MPa is not exceeded in the 10-mm-diameterbolts at A and B, and an allowable tensile stress of allow = 155 MPa is not exceeded in the12-mm-diameter rod AB.Given: allow := 95MPa allow := 155MPaa := 1.2m b := 0.6m e := 0.9mdo := 10mm drod := 12mmSolution: c := a2 + e2 hac:= vec:=Support Reactions: L := a + bC=0; FAB v ( ) (a) w (L) (0.5L) = 0FAB wLav= (0.5L)Assume failure of pin A or B:V = 0.5FAB V = allowA A4:= 2do0.5wLav (0.5L) allow 2 4do= wav(0.5L)2allow 2 4do:= w 6.632kNm= (controls!) AnsAssuming failure of rod AB:N = FAB N = allowA A4:= 2drodwLav (0.5L) allow 2 4drod= wav0.5L2allow 2 4drod:= w 7.791kNm= 104. Problem 1-103The bar is supported by the pin. If the allowable tensile stress for the bar is (t)allow = 150 MPa, andthe allowable shear stress for the pin is allow = 85 MPa, determine the diameter of the pin for whichthe load P will be a maximum. What is this maximum load? Assume the hole in the bar has the samediameter d as the pin. Take t =6 mm and w = 50 mm.Given: allow := 85MPat_allow := 150MPat := 6mm w := 50mmSolution: GivenAllowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, and allow equals to P /Ae .t_allowP= [1](w d) tAllowable Shear Stress: The pin is subjected to doubleshear and therefore the allowable equals to 0.5P /Apin,and the area Apin is equal to ( /4) d2.allow2Pd2= [2]Solving [1] and [2]: Initial guess: d := 20mm P := 10kNPd:= Find(P , d) P = 31.23 kN Ansd = 15.29mm Ans 105. Problem 1-104The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowabletensile stress for the bar is (t)allow = 140 MPa, and the allowable bearing stress between the pin andthe bar is (b)allow =210 MPA, determine the dimensions w and t such that the gross area of the crosssection is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the holein the bar has the same diameter as the pin.Given: t_allow := 140MPa b_allow := 210MPaA := 1250mm2 d := 25mmSolution: A = w t GivenAllowable Normal Stress: The effective cross-sectionalarea Ae for the bar must be considered here by takinginto account the reduction in cross-sectional areaintroduced by the hole. Here, effective area Ae is equalto (w - d) t, that is (A- d t) and allow equals to P /Ae .t_allowPA d t= [1]Allowable Bearing Stress: The projected area Ab is equalto (d t), and allow equals to P /Ab .b_allowPd t= [2]Solving [1] and [2]: Initial guess: t := 0.5in P := 1kipPt:= Find(P , t) P = 105.00 kN Anst = 20.00mm AnsAnd : wAt:= w = 62.50mm Ans 106. Problem 1-105The compound wooden beam is connected together by a bolt at B. Assuming that the connections atA, B, C, and D exert only vertical forces on the beam, determine the required diameter of the bolt at Band the required outer diameter of its washers if the allowable tensile stress for the bolt is (t)allow =150 MPa. and the allowable bearing stress for the wood is (b)allow = 28 MPa. Assume that the hole inthe washers has the same diameter as the bolt.Given: P1 := 3kN P2 := 1.5kN P3 := 2kNt_allow := 150MPa a := 2mb_allow := 28MPa b := 1.5mSolution:From FBD (a): GivenD=0; Cy (4b) + By (3b) + P2 (2b) + P3 (b) = 0 [1]From FBD (b):A=0; By 2 a b + ( ) Cy 2 a ( ) P1 (a) = 0 [2]Solving [1] and [2]: Initial guess: By := 1kN Cy := 2kNByCy:= Find(By , Cy)ByCy4.44.55= kNFor bolt:AraeByt_allow=4dB2 Byt_allow=dB4Byt_allow:= dB = 6.11mm AnsFor washer:AreaByb_allow=4dw2 dB2 Byb_allow=2 4dw dBByb_allow:= + dw = 15.41mm Ans 107. Problem 1-106The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 150 MPa. If a uniformdistributed load of w = 8 kN/m is placed on the bar, determine its minimum allowable position x fromB. Pins A and B each have a diameter of 8 mm. Neglect any axial force in the bar.Given: allow := 150MPa do := 8mma := 2mw 8kNm:= b := 2mSolution:A=0; By (a) w (b x) [a + x + 0.5 (b x)] = 0 [1]B=0; Ay (a) w (b x) [x + 0.5 (b x)] = 0 [2]Assume failure of pin A:AraeAyallow=4do2 Ayallow=Ay4:= (allow) Ay = 7.5398 kNdo2 Substitute value of force A into Eq [2], ( ) x1 0.5 b x1 ( ) + = 0 [2]Given Ay a ( ) w b x1Initial guess: x1 := 0.3m x1 := Find(x1) x1 = 0.480 mxcase_1 := x1Assume failure of pin B:Arae0.5Byallow=4do2 0.5Byallow=:= (allow) By = 15.0796 kNBy 242 doSubstitute value of force A into Eq [1], ( ) a x2 + 0.5 b x2 ( ) + = 0 [1]Given By a ( ) w b x2Initial guess: x2 := 0.3m x2 := Find(x2) x2 = 0.909 mxcase_2 := x2Choose the larger x value: x := max(xcase_1 , xcase_2)x = 0.909 m Ans 108. Problem 1-107The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 125 MPa. If x = 1 m,determine the maximum distributed load w the bar will support. Pins A and B each have a diameter of 8mm. Neglect any axial force in the bar.Given: allow := 125MPa x := 1m do := 8mma := 2m b := 2mSolution:wokNm:=GivenA=0; Bw (a) wo (b x) [a + x + 0.5 (b x)] = 0 [1]B=0; Aw (a) wo (b x) [x + 0.5 (b x)] = 0 [2]Initial guess: Bw := 1kN Aw := 1kNSolving [1] and [2]:BwAw:= Find(Bw,Aw)BwAw1.750.75= kNFor pin A: Ay w1Awwo= AraeAyallow=4do2 w1allowAwwo= w14 (allow)do2 woAw:= w1 8.378kNm=For pin B By wBwwo= Arae0.5Byallow=2do2 w2allowBwwo= w22 (allow)do2 woBw:= w2 7.181kNm=The smalleer w controls ! w := min(w1 ,w2)w 7.181kNm= Ans 109. Problem 1-108The bar is held in equilibrium by the pin supports at A and B. Note that the support at A has a singleleaf and therefore it involves single shear in the pin, and the support at B has a double leaf and thereforeit involves double shear. The allowable shear stress for both pins is allow = 125 MPa. If x = 1 m and w= 12 kN/m, determine the smallest required diameter of pins A and B. Neglect any axial force in thebar.Given: allow := 125MPa x := 1m do := 8mma := 2m b := 2mSolution:w 12kNm:=GivenA=0; By (a) w (b x) [a + x + 0.5 (b x)] = 0 [1]B=0; Ay (a) w (b x) [x + 0.5 (b x)] = 0 [2]Initial guess: By := 1kN Ay := 1kNSolving [1] and [2]:ByAy:= Find(By ,Ay)ByAy219= kNFor pin A:AraeAyallow=4dA2 Ayallow=dA4Ayallow:= dA = 9.57mm AnsFor pin BArae0.5Byallow=2dB2 Byallow=dB2Byallow:= dB = 10.34mm Ans 110. Problem 1-109The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the diameter d of the pin if the allowable shear stress is allow = 70 MPa and the load P = 40kN. Also, determine the load intensities w1 and w2 .Given: allow := 70MPa P := 40kNa := 37.5mm b := 25mmSolution:Pin: + Fy=0; P w1(a) = 0w1Pa:=w1 1066.67kNm= AnsLink: + Fy=0; P 2(0.5w2) (b) = 0w2Pb:=w2 1600.00kNm= AnsShear StressArea0.5Pallow=2d2Pallow=d2Pallow:= d = 19.073mm Ans 111. Problem 1-110The pin is subjected to double shear since it is used to connect the three links together. Due to wear,the load is distributed over the top and bottom of the pin as shown on the free-body diagram.Determine the maximum load P the connection can support if the allowable shear stress for thematerial is allow = 56 MPa and the diameter of the pin is 12.5 mm. Also, determine the load intensitiesw1 and w2 .Given: allow := 56MPa d := 12.5mma := 37.5mm b := 25mmSolution:Shear StressArea0.5Pallow=2d2Pallow=P2:= d2 (allow)P = 13.7445 kN AnsPin:+ Fy=0; P w1(a) = 0w1Pa:=w1 366.52kNm= AnsLink:+ Fy=0; P 2(0.5w2) (b) = 0w2Pb:=w2 549.78kNm= Ans 112. Problem 1-111The cotter is used to hold the two rods together. Determine the smallest thickness t of the cotter andthe smallest diameter d of the rods. All parts are made of steel for which the failure tensile stress is fail= 500 MPa and the failure shear stress is fail = 375 MPa. Use a factor of safety of (F.S.)t = 2.50 intension and (F.S.)s = 1.75 in shear.Given: fail := 500MPa fail := 375MPa P := 30kNd2 := 40mm h := 10mmFSt := 2.50 FSs := 1.75Solution:Allowable Normal Stress : Design of rod sizeallowfailFSt:= allow = 200MPaAreaPallow=4d2Pallow=d4Pallow:= d = 13.82mm AnsAllowable Shear Stress : Design of cotter sizeallowfailFSs:= allow = 214.29MPaArea0.5Pallow= h t0.5Pallow=t1h0.5Pallow:= t = 7mm Ans 113. Problem 1-112The long bolt passes through the 30-mm-thick plate. If the force in the bolt shank is 8 kN, determinethe average normal stress in the shank, the average shear stress along the cylindrical area of the platedefined by the section lines a-a, and the average shear stress in the bolt head along the cylindrical areadefined by the section lines b-b.Given: P := 8kN dshank := 7mmda_a := 18mm db_b := 7mmhhead := 8mm hplate := 30mmSolution:Average Normal Stress:Ashank4dshank2 :=tPAshank:= t = 207.9MPa AnsAverage Shear Stresses:Aa_a := (da_a)hplatea_avgPAa_a:= a_avg = 4.72MPa AnsAb_b := (db_b)hheadb_avgPAb_b:= b_avg = 45.47MPa Ans 114. Problem 1-113The bearing pad consists of a 150 mm by 150 mm block of aluminum that supports a compressiveload of 6 kN. Determine the average normal and shear stress acting on the plane through section aa.Show the results on a ifferential volume element located on the plane.Given: P := 6kNd := 150mm := 30degSolution: := 90deg Equations of Equilibrium:+ Fx=0; Va_a Pcos() = 0Va_a := Pcos() Va_a = 3 kN+ Fy=0; Na_a P sin() = 0Na_a := P sin() Na_a = 5.196 kNAt inclined plane:Ad2sin() :=a_aNa_aA:= a_a = 0.2MPa Ansa_aVa_aA:= a_a = 0.115MPa Ans 115. Problem 1-114Determine the resultant internal loadings acting on the cross sections located through points D and E ofthe frame.Given: w 2.5kNm:= a := 1.2m b := 0.9mc := 1.5m e := 0.45mSolution: d := a2 + b2 d = 1.5mvad:= hbd:=Support Reactions: L := b + c+ A=0; By (b) (wL)(0.5L) = 0By (wL) 0.5Lb:= By = 8 kN+ Fy=0; Ay By + wL = 0Ay := By wLAy = 2 kNBy = FBC (v) FBCByv:=FBC = 10 kN+ Fx=0; FBC (h) Ax = 0Ax := FBC (h)Ax = 6 kNSegment AD:+ Fx=0; ND Ax = 0 ND := Ax ND = 6 kN Ans+ Fy=0; Ay + w (e) + VD = 0 VD := Ay w (e) VD = 3.13 kN Ans+ D=0; MD + [w (e)] (0.5e) + Ay (e) = 0MD := [w (e)] (0.5e) Ay (e) MD = 1.153 kNm AnsSegment CE:+ Fx=0; NE + FBC = 0 ND := FBC ND = 10 kN Ans+ Fy=0; VD := 0 VD = 0 kN Ans+ E=0; MD := 0 MD = 0 kNm Ans 116. Problem 1-115The circular punch B exerts a force of 2 kN on the top of the plate A. Determine the average shearstress in the plate due to this loading.Given: P := 2kNdpunch := 4mm hplate := 2mmSolution:Average Shear Stresses:Aa_a := (dpunch)hplatea_avgPAa_a:=a_avg = 79.58MPa Ans 117. Problem 1-116The cable has a specific weight (weight/volume) and cross-sectional area A. If the sag s is small, sothat its length is approximately L and its weight can be distributed uniformly along the horizontal axis,determine the average normal stress in the cable at its lowest point C.Solution:Equations of Equilibrium:MA=0; T sAL2L4 = 0AL8 s=Average Normal Stresses:TA=L8 s= Ans 118. Problem 1-117The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to holdup the frame. If AB weighs 2.0 kN/m and the column FC has a weight of 3.0 kN/m, determine theresultant internal loadings acting on cross sections located at points D and E. Neglect the thickness ofboth the beam and column in the calculation.Given: wb 2.0kNm:= L := 3.6m d := 1.8mwc 3.0kNm:= H := 4.8m e := 1.2ma := 3.6m b := 3.6m c := 1.2mSolution:Beam AB: Lc := L2 + c2 vbcLc:= hbLLc:=+ A=0; By (L) wAB (L)(0.5L) = 0By wb (L) 0.5LL:= By 7936.64ms2= lb+ F Ay := By + wb (L) Ay = 3.6 kN y=0; Ay By + wb (L) = 0By = FBC (vb) FBCByvb:= FBC = 11.38 kN+ Fx=0; FBC (h) + Ax = 0 Ax := FBC (hb) Ax = 10.8 kNSegment AD:+ Fx=0; ND + Ax = 0 ND := Ax ND = 10.8 kN Ans+ Fy=0; Ay + wb (d) + VD = 0 VD := Ay wb (d) VD = 0 kN Ans+ D=0; MD wb d ( ) + (0.5d) Ay (d) = 0:= (0.5d) + Ay (d) MD = 3.24 kNm AnsMD wb d ( ) Member CG: Hb := H2 + b2 vcHHb:= hcbHb:=Column FC:+ C=0; Fx (H) Axc = 0 Fx AxcH:= Fx = 2.7 kN+ Fx=0; FBC (hb) Ax + Fx FCG (hc) = 0+ Fy=0; FCGFBC (hb) Ax + Fx:= FCG = 4.5 kNhcFy + By + wc (H) + FBC (vb) + FCG (vc) = 0Fy := By + wc (H) + FBC (vb) + FCG (vc)Fy = 25.2 kNSegment FE:+ Fx=0; VE Fx = 0 VE := Fx VE = 2.7 kN Ans+ Fy=0; NE + wc (e) Fy = 0 NE := wc (e) + Fy NE = 21.6 kN Ans+ E=0; ME + Fy (e) = 0 ME := Fy (e) ME = 30.24 kNm Ans 119. Problem 1-118The 3-Mg concrete pipe is suspended by the three wires. If BD and CD have a diameter of 10 mm andAD has a diameter of 7 mm, determine the average normal stress in each wire.Given: M := 3000kg g 9.81ms2:=h := 2m r := 1m := 120degdBD := 10mm dCD := 10mm dAD := 7mmSolution: W := Mg W = 29.43 kN := 0.5 = 60 degL := r2 + h2 vhL:=Equations of Equilibrium: x=0; 2F (rcos()) FAD (r) = 0FBD = FCD = F=0; FBD (r sin()) FCD (r sin()) = 0 yFAD = FFz=0; 3 [F (v)] W = 0FW3v:=F = 10.97 kNAllowable Normal Stress :BD = CD = 1 1FArea= 1F:= 1 = 139.65MPa Ans4dBD2 AD = 2 2FArea= 2F:= 2 = 285MPa Ans4dAD2 120. Problem 1-119The yoke-and-rod connection is subjected to a tensile force of 5 kN. Determine the average normalstress in each rod and the average shear stress in the pin A between the members.Given: P := 5kN dpin := 25mmd30 := 30mm d40 := 40mmSolution:Average Normal Stress:A404d402 :=40PA40:= 40 = 3.979MPa AnsA304d302 :=30PA30:= 30 = 7.074MPa AnsAverage Shear Stresses:Apin4dpin2 :=avg0.5PApin:= avg = 5.093MPa Ans 121. Problem 2-1An air-filled rubber ball has a diameter of 150 mm. If the air pressure within it is increased until theball's diameter becomes 175 mm, determine the average normal strain in the rubber.Given: d0 := 150mm d := 175mmSolution:d d0d0:= 0.1667mmmm= Ans 122. Problem 2-2A thin strip of rubber has an unstretched length of 375 mm. If it is stretched around a pipe having anouter diameter of 125 mm, determine the average normal strain in the strip.Given: L0 := 375mmSolution:L := (125) mmL L0L0:= 0.0472mmmm= Ans 123. Problem 2-3The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes theend C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.Given: a := 3m LCE := 4mb := 4m LBD := 4mLCE := 10mmSolution:LBDaa + b:= LCE LBD = 4.2857mmCELCELCE:= CE 0.00250mmmm= AnsBDLBDLBD:= BD 0.00107mmmm= Ans 124. Problem 2-4The center portion of the rubber balloon has a diameter of d = 100 mm. If the air pressure within itcauses the balloon's diameter to become d = 125 mm, determine the average normal strain in therubber.Given: d0 := 100mm d := 125mmSolution:d d0d0:= 0.2500mmmm= Ans 125. Problem 2-5The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam is displace10 mm downward, determine the normal strain developed in wires CE and BD.Given: a := 3m b := 2m c := 2mLCE := 4m LBD := 3mtip := 10mmSolution:LBDaLCEa + b=LCEa + b tipa + b + c=LCEa + ba + b + c:= tip LCE = 7.1429mmLBDaa + b + c:= tip LBD = 4.2857mmAverage Normal Strain:CELCELCE:= CE 0.00179mmmm= AnsBDLBDLBD:= BD 0.00143mmmm= Ans 126. Problem 2-6The rigid beam is supported by a pin at A and wires BD and CE. If the maximum allowable normalstrain in each wire is max = 0.002 mm/mm, determine the maximum vertical displacement of the loadP.Given: a := 3m b := 2m c := 2mLCE := 4m LBD := 3mallow 0.002mmmm:=Solution:LBDaLCEa + b=LCEa + b tipa + b + c=Average Elongation/Vertical Displacement:LBD := LBD allow LBD = 6.00mm tipa + b + ca:= LBDtip = 14.00mmLCE := LCE allow LCE = 8.00mm tipa + b + ca + b:= LCEtip = 11.20mm (Controls !) Ans 127. Problem 2-7The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2mm, determine the normal strain developed in each wire.Given:a := 300mm := 30degA := 2mmSolution:Consider the triangle CAA':A := 180deg A = 150 deg+ 2 2a := ( A)cos(A)LCA' a2 ALCA' = 301.734mmCALCA' aa:=CA 0.00578mmmm= Ans 128. Problem 2-8Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If aforce is applied to the end D of the member and causes it to rotate by = 0.3, determine the normalstrain in the cable. Originally the cable is unstretched.Given:a := 400mm b := 300mm c := 300mm := 0.3degSolution:LAB := a2 + b2 LAB = 500mmConsider the triangle ACB':C := 90deg + C = 90.3 degLAB' := a2 + b2 2abcos(C)LAB' = 501.255mmABLAB' LABLAB:=AB 0.00251mmmm= Ans 129. Problem 2-9Part of a control linkage for an airplane consists of a rigid member CBD and a flexible cable AB. If aforce is applied to the end D of the member and causes a normal strain in the cable of 0.0035 mm/mmdetermine the displacement of point D. Originally the cable is unstretched.Given: a := 400mm b := 300mm c := 300mmAB 0.0035mmmm:=Solution:LAB := a2 + b2 LAB = 500mmLAB' := LAB (1 + AB) LAB' = 501.750mmConsider the triangle ACB':C = 90deg + LAB' = a2 + b2 2abcos(C)C acos(a2 + b2) LAB' 22ab:=C = 90.419 deg := C 90deg = 0.41852 deg = 0.00730 radD := (b + c) D = 4.383mm Ans 130. Problem 2-10The wire AB is unstretched when = 45. If a vertical load is applied to bar AC, which causes =47, determine the normal strain in the wire.Given: := 45deg := 2degSolution:LAB = L2 + L2 LAB = 2LLCB = (2L)2 + L2 LCB = 5LFrom the triangle CAB:A := 180deg A = 135.00 degsin(B)Lsin(A)LCB=B asinL sin(A)5L:= B = 18.435 degFrom the triangle CA'B:'B := B + 'B = 20.435 degsin('B)Lsin(180deg A')LCB=A' 180deg asin5L sin('B)L:= A' = 128.674 deg'C := 180deg 'B A''C = 30.891 degsin('B)Lsin('C)LA'B= LA'Bsin('C)sin('B) := L LAB := 2LABLA'B LABLAB:= AB = 0.03977 Ans 131. Problem 2-11If a load applied to bar AC causes point A to be displaced to the left by an amount L, determine thenormal strain in wire AB. Originally, = 45.Given: := 45degSolution: ACLL=LAB = L2 + L2 LAB = 2LLCB = (2L)2 + L2 LCB = 5LFrom the triangle A'AB:A := 180deg A = 135.00 degLA'B L2 LAB+ 2 2 (L) L= ( AB)cos(A)LA'B L2 = + 2L2 + 2 (L)LABLA'B LABLAB=ABL2 + 2L2 + 2 (L)L 2L2L=AB12LL 2+ 1LL= + 1Neglecting the higher-order terms,AB 1LL+ 0.5= 1+ + ..... AB 112LL= 1 (Binomial expansion)AB12LL= AnsAlternatively,ABLA'B LABLAB= ABL sin()2L=AB12LL= Ans 132. Problem 2-12The piece of plastic is originally rectangular. Determine the shear strain xy at corners A and B if theplastic distorts as shown by the dashed lines.Given:a := 400mm b := 300mmAx := 3mm Ay := 2mmCx := 2mm Cy := 2mmBx := 5mm By := 4mmSolution:Geometry : For small angles,Cxb + Cy:= = 0.00662252 radBy Cya + (Bx Cx):= = 0.00496278 radBx Axb + (By Ay):= = 0.00662252 radAya + Ax:= = 0.00496278 radShear Strain :xy_B := + xy_B = 11.585 10 3 rad Ansxy_A := ( + ) xy_A = 11.585 10 3 rad Ans 133. Problem 2-13The piece of plastic is originally rectangular. Determine the shear strain xy at corners D and C if theplastic distorts as shown by the dashed lines.Given:a := 400mm b := 300mmAx := 3mm Ay := 2mmCx := 2mm Cy := 2mmBx := 5mm By := 4mmSolution:Geometry : For small angles,Cxb + Cy:= = 0.00662252 radBy Cya + (Bx Cx):= = 0.00496278 radBx Axb + (By Ay):= = 0.00662252 radAya + Ax:= = 0.00496278 radShear Strain :xy_D := + xy_D = 11.585 10 3 rad Ansxy_C := ( + ) xy_C = 11.585 10 3 rad Ans 134. Problem 2-14The piece of plastic is originally rectangular. Determine the average normal strain that occurs along thdiagonals AC and DB.Given:a := 400mm b := 300mmAx := 3mm Ay := 2mmCx := 2mm Cy := 2mmBx := 5mm By := 4mmSolution:Geometry :LAC := a2 + b2 LAC = 500mmLDB := a2 + b2 LDB = 500mmLA'C' (a + Ax Cx)2 := + (b + Cy Ay)2LA'C' = 500.8mmLDB' (a + Bx)2 := + (b + By)2LDB' = 506.4mmAverage Normal Strain :ACLA'C' LAC:= AC 1.601 10 3LACmmmm= AnsBDLDB' LDBLDB:= BD 12.800 10 3mmmm= Ans 135. Problem 2-15The guy wire AB of a building frame is originally unstretched. Due to an earthquake, the two columnsof the frame tilt = 2. Determine the approximate normal strain in the wire when the frame is in thiposition. Assume the columns are rigid and rotate about their lower supports.Given:a := 4m b := 3m c := 1m := 2degSolution:180= = 0.03490659 radGeometry : The vertical dosplacement is negligible.Ax := c Ax = 34.907mmBx := (b + c) Bx = 139.626mmLAB := a2 + b2 LAB = 5000mmLA'B' := (a + Bx Ax)2 + b2LA'B' = 5084.16mmAverage Normal Strain :LA'B' LABABLAB:=AB 16.833 10 3mmmm= Ans 136. Problem 2-16The corners of the square plate are given the displacements indicated. Determine the shear strain alongthe edges of the plate at A and B.Given: ax := 250mm ay := 250mmv := 5mm h := 7.5mmSolution:At A :tan'A2ax hay + v='A 2 atanax hay + v:= 'A = 1.52056 radnt_A2:= 'A nt_A = 0.05024 rad AnsAt B :tan'B2ay + vax h='B 2 atanay + vax h:= 'B = 1.62104 radnt_B 'B2:= nt_B = 0.05024 rad Ans 137. Problem 2-17The corners of the square plate are given the displacements indicated. Determine the average normalstrains along side AB and diagonals AC and DB.Given: ax := 250mm ay := 250mmv := 5mm h := 7.5mmSolution:For AB :LAB ax:= + 22 ayLA'B' (ax h)2 := + (ay + v)2ABLA'B' LABLAB:=LAB = 353.55339mmLA'B' = 351.89665mmAB 4.686 10 3mmmm= AnsFor AC :LAC := 2(ay) LAC = 500mmLA'C' := 2 (ay + v) LA'C' = 510mmACLA'C' LAC:= AC 20.000 10 3LACmmmm= AnsFor DB :LDB := 2(ax) LDB = 500mmLD'B' := 2 (ax h) LD'B' = 485mmDBLD'B' LDB:= DB 30.000 10 3LDBmmmm= Ans 138. Problem 2-18The square deforms into the position shown by the dashed lines. Determine the average normal strainalong each diagonal, AB and CD. Side D'B' remains horizontal.Given:a := 50mm b := 50mmBx := 3mmCx := 8mm Cy := 0mm'A := 91.5deg LAD' := 53mmSolution:For AB :By := LAD'cos('A 90deg) b By = 2.9818mmLAB := a2 + b2 LAB = 70.7107mmLAB' (a + Bx)2 := + (b + By)2 LAB' = 70.8243mmABLAB' LABLAB:=AB 1.606 10 3mmmm= AnsFor CD :Dy := By Dy = 2.9818mmLCD := a2 + b2 LCD = 70.7107mmLC'D' (a + Cx)2 := + (b + Dy)2 2 (a + Cx) (b + Dy)cos('A)LC'D' = 79.5736mmCDLC'D' LCDLCD:=CD 125.340 10 3mmmm= Ans 139. Problem 2-19The square deforms into the position shown by the dashed lines. Determine the shear strain at each ofits corners, A, B, C, and D. Side D'B' remains horizontal.Given: a := 50mm b := 50mmBx := 3mmCx := 8mm Cy := 0mm'A := 91.5deg LAD' := 53mmSolution: 'A = 1.597 radGeometry :By := LAD'cos('A 90deg) b By = 2.9818mmDy := By Dy = 2.9818mmDx := LAD' sin('A 90deg) Dx = 1.3874mmIn triangle C'B'D' :LC'B' := (Cx Bx)2 + (b + By)2LC'B' = 54.1117mmLD'B' := a + Bx Dx LD'B' = 48.3874mmLC'D' := (a + Cx Dx)2 + (b + Dy)2LC'D' = 79.586mm2 cos(B') LC'B'+ LD'B'2 LC'D'22 (LC'B') (LD'B')=B' acosLC'B'2 LD'B'+ 2 LC'D' 22 (LC'B') (LD'B'):= B' = 101.729 deg B' = 1.7755 radD' := 180deg 'A D' = 88.500 deg D' = 1.5446 radC' := 180deg B' C' = 78.271 deg C' = 1.3661 radShear Strain :xy_A := 0.5 ('A) xy_A = 26.180 10 3 rad Ansxy_B := 0.5 (B') xy_B = 204.710 10 3 rad Ansxy_C := 0.5 (C') xy_C = 204.710 10 3 rad Ansxy_D := 0.5 (D') xy_D = 26.180 10 3 rad Ans 140. Problem 2-20The block is deformed into the position shown by the dashed lines. Determine the average normalstrain along line AB.Given: xBA := (70 30)mm yBA := 100mmxB'A := (55 30)mmyB'A := ( 1102 152)mmSolution:For AB :LAB xBA:= 2 + yBA2 LAB = 107.7033mm:= + 2 LAB' = 111.8034mmLAB' xB'A2 yB'AABLAB' LABLAB:=AB 38.068 10 3mmmm= Ans 141. Problem 2-21A thin wire, lying along the x axis, is strained such that each point on the wire is displaced x = k x2along the x axis. If k is constant, what is the normal strain at any point P along the wire?Given: x kx 2= Solution:d xdx= = 2kx Ans 142. Problem 2-22The rectangular plate is subjected to the deformation shown by the dashed line. Determine the averagshear strain xy of the plate.Given: a := 150mm b := 200mma := 0mm b := 3mmSolution: atanba:= = 1.146 deg = 19.9973 10 3 radShear Strain :xy := xy = 19.997 10 3 rad Ans 143. Problem 2-23The rectangular plate is subjected to the deformation shown by the dashed lines. Determine the averagshear strain xy of the plate.Given: a := 200mm a := 3mmb := 150mm b := 0mmSolution: atanab:= = 1.146 deg = 19.9973 10 3 radShear Strain :xy := xy = 19.997 10 3 rad Ans 144. Problem 2-24The rectangular plate is subjected to the deformation shown by the dashed lines. Determine theaverage normal strains along the diagonal AC and side AB.Given: a := 200mm b := 150mmAx := 3mm Ay := 0mmBx := 0mm By := 0mmCx := 0mm Cy := 0mmDx := 3mm Dy := 0mmSolution:Geometry :LAC := a2 + b2 LAC = 250mmLAB := b LAB = 150mmLA'C := (a + Cx Ax)2 + (b + Cy Ay)2LA'C = 252.41mmLA'B := Ax2 + b2LA'B = 150.03mmAverage Normal Strain :ACLA'C LACLAC:= AC 9.626 10 3mmmm= AnsABLA'B LABLAB:= AB 199.980 10 6mmmm= Ans 145. Problem 2-25The piece of rubber is originally rectangular. Determine the average shear strain xy if the corners Band D are subjected to the displacements that cause the rubber to distort as shown by the dashed linesGiven: a := 300mm b := 400mmAx := 0mm Ay := 0mmBx := 0mm By := 2mmDx := 3mm Dy := 0mmSolution:AB atanBya:=AB = 0.38197 degAB = 6.6666 10 3 radAD atanDxb:=AD = 0.42971 degAD = 7.4999 10 3 radShear Strain :xy_A := AB + ADxy_A = 14.166 10 3 rad Ans 146. Problem 2-26The piece of rubber is originally rectangular and subjected to the deformation shown by the dashedlines. Determine the average normal strain along the diagonal DB and side AD.Given: a := 300mm b := 400mmAx := 0mm Ay := 0mmBx := 0mm By := 2mmDx := 3mm Dy := 0mmSolution:Geometry :LDB := a2 + b2 LDB = 500mmLAD := b LAD = 400mmLD'B' := (a + Bx Dx)2 + (b + Dy By)2 LD'B' = 496.6mmLAD' := Dx2 + (b + Dy)2 LAD' = 400.01mmAverage Normal Strain :LD'B' LDBBD:= BD 6.797 10 3LDBmmmm= AnsADLAD' LAD:= AD 28.125 10 6LADmmmm= Ans 147. Problem 2-27The material distorts into the dashed position shown. Determine (a) the average normal strains xand y , the shear strain xy at A, and (b) the average normal strain along line BE.Given: a := 80mm Bx := 0mm Ex := 80mmb := 125mm By := 100mm Ey := 50mmAx := 0mm Cx := 10mm Dx := 15mmAy := 0mm Cy := 0mm Dy := 0mmSolution:xAC := Cx Ax xAC = 10.00mmyAC := Cy Ay yAC = 0.00mmAC atanCxb:= AC = 79.8300 10 3 radSince there is no deformation occuring along the y- and x-axis,x_A := yAC x_A = 0 Ansy_A2 + b2 bbxAC:=y_A = 0.00319 Ansxy_A := ACxy_A = 79.830 10 3 rad AnsGeometry :BxByBy= BxCxbb:= Cx Bx = 8mmBy := 0mmExDxEyb= ExEyb:= Dx Ex = 6mmEy := 0mmLBE := (Ex Bx)2 + (Ey By)2 LBE = 94.34mmLB'E' (a + Ex Bx)2 := + (Ey + Ey By)2 LB'E' = 92.65mmBELB'E' LBE:= BE 17.913 10 3LBEmmmm= AnsNote: Negative sign indicates shortening of BE. 148. Problem 2-28The material distorts into the dashed position shown. Determine the average normal strain that occursalong the diagonals AD and CF.Given: a := 80mm Bx := 0mm Ex := 80mmb := 125mm By := 100mm Ey := 50mmAx := 0mm Cx := 10mm Dx := 15mmAy := 0mm Cy := 0mm Dy := 0mmSolution:xAC := Cx Ax xAC = 1