MECH-321 Mechanics of deformable solidsPROBLEM 2.101 Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. A force P is applied to the rod and then removed to give it a permanent set δ p = 2 mm. Determine the maximum value of the force P and the maximum amount δ m by which the rod should be stretched to give it the desired permanent set. SOLUTION AAB = ABC = π 4 (30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2 (40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2 π 4 Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N Pmax = 176.7 kN δ′ = P′LAB P ′LBC (176.715 × 103 )(0.8) (176.715 × 103 )(1.2) + = + EAAB EABC (200 × 109 )(706.86 × 10−6 ) (200 × 109 )(1.25664 × 10−3 ) = 1.84375 × 10−3 m = 1.84375 mm δ p = δ m − δ ′ or δ m = δ p + δ ′ = 2 + 1.84375 δ m = 3.84 mm 715 × 103 )(0.84375 × 10−3 m = 1.16 mm .68 × 10−6 ) (200 × 109 )(1.8) (176.715 × 103 )(1. A force P is applied to the rod until its end A has moved down by an amount δ m = 5 mm. it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. SOLUTION AAB = ABC = Pmax π 4 (30)2 = 706.MECH-321 Mechanics of deformable solids PROBLEM 2.7 kN π δ′ = P′LAB P ′LBC (176.86 × 10−6 m 2 (40) 2 = 1.84375 = 3.715 × 103 N Pmax = 176.25644 × 10−3 m 2 4 = Aminσ Y = (706. Determine the maximum value of the force P and the permanent set of the rod after the force has been removed.86 × 10−6 )(250 × 106 ) = 176.86 mm 2 = 706.25664 × 10−3 ) = 1.16 mm δ p = 3.102 Rod ABC consists of two cylindrical portions AB and BC.2) + = + EAAB EABC (200 × 109 )(706.25664 × 103 mm 2 = 1.84375 mm δ p = δ m − δ ′ = 5 − 1. 5 kN δ perm = 0.705 mm δ m = 310.0 mm − 3.795 mm AE E δ P = δm − δ ′ (a) δ m = 4.103 The 30-mm square bar AB has a length L = 2.5 mm > δY Pm = 310.5 × 103 N Unloading: δ ′ = Pm L σ Y L = = δY = 3.2 m.795 mm δ m = 310. Determine the maximum value of the force P and the permanent set of the bar after the force has been removed.5 × 103 N δ perm = 8. SOLUTION A = (30)(30) = 900 mm2 = 900 × 10−6 m 2 δY = Lε Y = Lσ Y (2. knowing that (a) δ m = 4.795 mm .5 mm.2)(345 × 106 ) = = 3.5 mm − 3.5 kN δ perm = 4. it is made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa.795 × 10−3 = 3.MECH-321 Mechanics of deformable solids PROBLEM 2.5 × 103 N δ perm = 4. A force P is applied to the bar until end A has moved down by an amount δ m . (b) δ m = 8 mm.795 mm E 200 × 109 If δ m ≥ δ Y . Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.205 mm (b) δ m = 8 mm > δY Pm = 310. Determine the maximum value of the force P and the maximum amount δ m by which the bar should be stretched if the desired value of δ p is (a) 3.5 m. the maximum force is Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.81 mm = 10.104 The 30-mm square bar AB has a length L = 2.5)(345 × 106 ) = = 4.5 kN δ p = δ m − δ ′ = δ m − δY ∴ δ m = δ p + δY (a) (b) δ p = 3. thus producing a permanent stretch of δ p .3125 mm δ p = 6.3125 mm 9 E 200 × 10 When δ m exceeds δ Y .3125 mm = 7.5 × 103 N = 310.5 mm. SOLUTION A = (30)(30) = 900 mm 2 = 900 × 10−6 m 2 δY = Lε Y = Lσ Y (2.MECH-321 Mechanics of deformable solids PROBLEM 2.3125 × 103 m = 4.5 mm.5 mm + 4.81 mm . it is made of mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa.5 mm δ m = 6.5 mm δ m = 3.5 mm + 4. (b) 6. A force P is applied to the bar and then removed to give it a permanent set δ p . After the rod has been attached 3 to the rigid lever CD. . SOLUTION Since the rod AB is to be stretched permanently. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed. E 29 × 106 θ= δB 22 = δC 33 ∴ δC = 33 δ B = 0. A vertical force Q is then applied at C until this point has moved to position C ′. −22 δ C = 0.65 kips During unloading. the spring back at B is δ B = LABε Y = From the deformation diagram.65 kips Q= P= 33 33 Q = 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi.0745 in. it is found that end C is 8 in. where PY = Aσ Y = π 3 2 (36) = 3.976 kips 4 8 Referring to the free body diagram of lever CD.1117 in. ΣM D = 0: 33 Q − 22 P = 0 22 (22)(3.MECH-321 Mechanics of deformable solids PROBLEM 2.1117 in. too high. the peak force in the rod is P = PY .976) = 2. Slope: LABσ Y (60)(36 × 103 ) = = 0. too high. . After the rod 3 has been attached to the rigid lever CD.1034 in. where PY = Aσ Y = π 3 2 (50) = 5.522 kips 4 8 Referring to the free body diagram of lever CD. it is found that end C is 8 in. the peak force in the rod is P = PY .522) P= = 3.105. ΣM D = 0: 33Q − 22 P = 0 Q= 22 (22)(5. A vertical force Q is then applied at C until this point has moved to position C′.1552 in. SOLUTION Since the rod AB is to be stretched permanently. Slope: θ= δB 22 = δC 33 ∴ δC = 33 δB 22 δ C = 0. Determine the required magnitude of Q and the deflection δ1 if the lever is to snap back to a horizontal position after Q is removed.106 Solve Prob. assuming that the yield point of the mild steel is 50 ksi.MECH-321 Mechanics of deformable solids PROBLEM 2.68 kips 33 33 LABσ Y (60)(50 × 103 ) = = 0. the spring back at B is δ B = LAB ε Y = From the deformation diagram. E 29 × 106 Q = 3. PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi.68 kips During unloading. 2. 107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. θ= δ BD LAB = δ CE LAC δ BD = Equilibrium of bar ABC. Knowing that the cables were initially taut. . cable BD is elastic when Q = 50 kN. (c) the final displacement of point C. Let θ = rotation angle of rigid bar ABC. LAB 1 δ CE = δ CE LAC 2 (1) M A = 0 : LAB FBD + LAC FCE − LAC Q = 0 Q = FCE + LAB 1 FBD = FCE + FBD 2 LAC (2) Assume cable CE is yielded. (Hint: In Part c.5 × 103 N. determine (a) the maximum stress that occurs in cable BD. From (2).0 × 103 N Since FBD < Aσ Y = 34.) SOLUTION Elongation constraints for taut cables. cable CE is not taut.MECH-321 Mechanics of deformable solids PROBLEM 2. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 ) = 31. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. (b) the maximum deflection of point C.5 × 103 N FBD = 2(Q − FCE ) = (2)(50 × 103 − 34. 1 × 10−3 m EA (200 × 109 )(100 × 10−6 ) 6.107 (Continued) (a) Maximum stresses. FBD = 2(Q − FCE ) = 2(0 − 0) = 0 FBD LBD = 0. FBD LBD (31.2 × 10−3 m σ Y LCE E Permanent elongation of cable CE: (δ CE ) p = (δ CE ) − (δ CE ) P = (δ CE ) max − FCE LCE σ L = (δ CE ) max − Y CE EA E 6 (345 × 10 )(2) = 6.MECH-321 Mechanics of deformable solids PROBLEM 2. δ BD = From (1). δ BD = .20 × 10−3 − = 2.75 × 10−3 m 9 200 × 10 (c) Unloading. σ CE = σ Y = 345 MPa σ BD = FBD 31.0 × 103 )(2) = = 3. Cable CE is slack ( FCE = 0) at Q = 0.0 × 103 = = 310 × 106 Pa −6 A 100 × 10 σ BD = 310 MPa (b) Maximum of deflection of point C. From (2).20 mm ↓ δ C = δ CE = 2δ BD = 6. EA Since cable BD remained elastic. 5 × 103 N From (2).) SOLUTION Elongation constraints. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34. Let θ = rotation angle of rigid bar ABC. A force Q is applied at C to the rigid bar ABC and is gradually increased from 0 to 50 kN and then reduced to zero. PROBLEM 2. (c) the final displacement of point C.MECH-321 Mechanics of deformable solids PROBLEM 2.107. (Hint: In Part c. assuming that the cables are replaced by rods of the same cross-sectional area and material. (b) the maximum deflection of point C. Knowing that the cables were initially taut. FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31. 2.0 × 103 N Since FBD < Aσ Y = 34.108 Solve Prob. δ BC LAB = δ CE LAC LAB 1 δ CE = δ CE LAC 2 (1) M A = 0: LAB FBD + LAC FCE − LAC Q = 0 Q = FCE + LAB 1 FBD = FCE + FBD LAC 2 (2) Assume cable CE is yielded. determine (a) the maximum stress that occurs in cable BD.5 × 103 N. cable BD is elastic when Q = 50 kN. θ= δ BD = Equilibrium of bar ABC. . Further assume that the rods are braced so that they can carry compressive forces.107 Each cable has a cross-sectional area of 100 mm2 and is made of an elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. cable CE is not taut. 5 × 106 δ C = 50 × 103 ′ δ C = 4 × 10−3 m From (2). FBD LBD (31. σ CE = σ Y = 345 MPa σ BD = FBD 31. ′′ Elastic FBD = ′ FCE = ′ ′ δ BD = 1 δ C 2 9 −6 ′ ′ EAδ BD (200 × 10 )(100 × 10 )( 1 δ C ) 2 ′ = = 5 × 106 δ C LBD 2 ′ ′ EAδ CE (200 × 109 )(100 × 10−6 )(δ C ) ′ = = 10 × 106 δ C LCE 2 ′ ′ ′ Q′ = FCE + 1 FBD = 12.0 × 103 )(2) = = 3.20 mm ↓ .20 mm ↓ δ C = δ CE = 2δ BD = 6. δ CE = δ C From (1). Equating expressions for Q′.2 × 10−3 m ′ ′ Unloading.MECH-321 Mechanics of deformable solids PROBLEM 2.1 × 10−3 m EA (200 × 109 )(100 × 10−6 ) 6. δ BD = From (1).5 × 106 δ C 2 ′ 12.2 × 10−3 − 4 × 10−3 = 2. Q′ = 50 × 103 N.2 × 10−3 m 2. (c) Final displacement. ′ δ C = (δ C ) m − δ C = 6.0 × 103 = = 310 × 106 Pa −6 A 100 × 10 σ BD = 310 MPa (b) Maximum of deflection of point C.108 (Continued) (a) Maximum stresses. 5 × 103 N 0.5 × 103 )(0. AC = 250 MPa FBC 537.190)(250 × 106 ) δ C .109 Rod AB consists of two cylindrical portions AC and BC.292 mm 250 MPa −307 MPa (b) Maximum stresses: σ AC = σ Y . AC = = 0. portion AC yields. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa.5 × 103 =− = −307.Y = LAC ε Y . Assuming both steels to be elastoplastic.26464 × 103 m −6 9 (200 × 10 )(1750 × 10 ) 0.26464 × 10−3 = 0.5 × 103 − 975 × 103 N = −537. SOLUTION Displacement at C to cause yielding of AC.2375 × 10−3 ) =− = −437.190) = = 0.5 × 103 N EAδ C (200 × 109 )(1750 × 10−6 )(0.5 × 10−3 N δ′ = (487.190 LCB PY = FAC − FCB = 875 × 103 N For equilibrium of element at C.5 × 103 N (a) δC = − FCB LCD (537.02715 × 10−3 m . each with a cross-sectional area of 1750 mm2.29179 × 10−3 m EA (200 × 109 )(1750 × 10−6 ) 0. FCB = − FAC = Aσ Y . AC = AC Y .0272 mm δ p = δ m − δ ′ = 0. and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa. P′ L P′ L L ′ ′ ′ δ ′ = AC AC = − CB CB ∴ PCB = − PAC AC = − PAC EA EA LAB σ BC = (c) ′ ′ ′ ′ P′ = 975 × 103 = PAC − PCB = 2 PAC PAC = 487. AC = (1750 × 10−6 )(250 × 106 ) = 437. A load P is applied at C as shown.5 × 103 )(0. FCB = FAC − P = 437. FAC − ( FCB + PY ) = 0 Since applied load P = 975 × 103 N > 875 × 103 N. L σ (0.190) = 0.MECH-321 Mechanics of deformable solids PROBLEM 2. (c) the permanent deflection of C.2375 × 10−3 m 9 E 200 × 10 Corresponding force.29179 × 10−3 − 0. (b) the maximum stress in each portion of the rod. determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero.14 × 106 Pa = −307 MPa −6 A 1750 × 10 Deflection and forces for unloading. determine (a) the maximum value of P. (b) the maximum stress in each portion of the rod. if P is gradually increased from zero until the deflection of point C reaches a maximum value of δ m = 0. Portion AC is made of a mild steel with E = 200 GPa and σ Y = 250 MPa.725 × 10−3 (elastic) 200 × 109 = Forces: FAC = Aσ Y = (1750 × 10−6 )(250 × 106 ) = 437.5789 × 10−3 190 mm =− 0. Assuming both steels to be elastoplastic.25 × 10−3 (yielding) 200 × 109 345 × 106 =− = −1.6 × 103 =− = −316 × 106 Pa −6 A 1750 × 10 . A load P is applied at C as shown.5789 × 10−3 ) = −552.30 mm = −1. and portion CB is made of a high-strength steel with E = 200 GPa and σ Y = 345 MPa.5789 × 10−3 190 mm ε CB = − Strains at initial yielding: δm LCB ε Y. BC E 250 × 106 = 1.5 × 103 + 552.3 mm and then decreased back to zero. SOLUTION Displacement at C is δ m = 0. 2.MECH-321 Mechanics of deformable solids PROBLEM 2. AC CB : σ CB = FCB 552. (c) the permanent deflection of C. CB = (a) σ Y. AC E σ Y.6 × 10−3 N For equilibrium of element at C. PROBLEM 2. each with a cross-sectional area of 1750 mm2.30 mm = 1. The corresponding strains are ε AC = δm LAC = 0.5 × 10−3 N FCB = EAε CB = (200 × 109 )(1750 × 10−6 )(−1.6 × 103 = 990.30 mm. (c) the permanent deflection of C after the load is removed.109 Rod AB consists of two cylindrical portions AC and BC.110 For the composite rod of Prob. FAC − FCB − P = 0 P = FAC − FCD = 437. (b) the maximum stress in each portion of the rod. determine (a) the maximum deflection of C if P is gradually increased from zero to 975 kN and then reduced back to zero. AC = ε Y.1 × 103 N = 990 kN = 250 MPa −316 MPa (b) Stresses: AC : σ AC = σ Y.109. 110 (Continued) (c) Deflection and forces for unloading.26874 mm 0.05 × 103 )(0.031mm .30 mm − 0.26874 mm (200 × 109 )(1750 × 10−6 ) ′ ′ ′ ′ P′ = PAC − PCB = 2 PAC = 990.190) = 0.1 × 103 N ∴ PAC = 495.05 × 103 N δ′ = δ p = δ m − δ ′ = 0.MECH-321 Mechanics of deformable solids PROBLEM 2. δ′ = ′ PAC LAC P′ L L ′ ′ = − CB CB ∴ PCB = − PAC AC = − PAC EA EA LAB (495.26874 × 10−3 m = 0. and then decreased back to zero. A2 = 2 (2) = 0.00906 in. Tempered steel is elastic. thick. This composite bar is subjected as shown to a centric axial load of magnitude P.75)(0. for the tempered and mild steel.14 × 103 )(14) = = 0. The load P is gradually increased from zero until the deformation of the bar reaches a maximum value δ m = 0. EA (29 × 106 )(1.04 in.86 ksi A2 0. The mild steel yields.75 in 2 16 δY1 = δY 2 = Lσ Y 1 (14)(50 × 103 ) = = 0. . are bonded to a 1 -in.00)(50 × 103 ) = 50 × 103 lb 1 P2 = EA2δ m (29 × 103 )(0.00906in.86 × 103 psi = 82. (c) the permanent set after the load is removed.03094 in.04) = = 62. A1 = (2) = 1.75) 0.04 − 0.00 in 2 2 3 For the tempered steel.75 82. Both steels are elastoplastic with E = 29 × 106 and with yield strengths equal to 100 ksi and 50 ksi.75 in 2 δ Y 1 < δ m < δY 2 . SOLUTION 1 For the mild steel.03094 = 0.024138 in.111 3 Two tempered-steel bars.1 kips P = P + P2 = 112.MECH-321 Mechanics of deformable solids PROBLEM 2. E 29 × 106 Lσ Y 2 (14)(100 × 103 ) = = 0.86 ksi σ2 = Unloading: (c) δ′ = PL (112. Determine (a) the maximum value of P. (a) Forces: P = A1σ Y 1 = (1.048276 in. mild-steel 2 bar. (b) the maximum stress in the tempered-steel bars.1kips 1 (b) Stresses: σ1 = P 1 = σ Y 1 = 50 × 103 psi = 50 ksi A1 P2 62.14 × 103 lb = 112. Permanent set: δ p = δ m − δ ′ = 0. respectively. each 16 -in.14 × 103 = = 82. E 29 × 103 Total area: A = A1 + A2 = 1.14 × 103 lb L 14 P = 112. each 16 -in. P2 = P − P = 98 × 103 − 50 × 103 = 48 × 103 lb 1 1 (a) (b) δm = σ2 = P2 L (48 × 103 )(14) = EA2 (29 × 106 )(0.112 For the composite bar of Prob. Let P = force carried by mild steel. = 64 ksi Unloading: δ ′ = (c) δ P = δ m − δ ′ = 0. 1 P2 = force carried by tempered steel. (c) the permanent set after the load is removed.MECH-321 Mechanics of deformable solids PROBLEM 2. (b) the maximum stress in the tempered-steel bars.111.00 in 2 2 3 A2 = 2 (2) = 0. 3 PROBLEM 2. mild steel yields. respectively.75) = 0. EA (29 × 106 )(1.03090in. therefore.50 × 103 lb A P > PY . are bonded to a 1 -in. for the tempered and mild steel.00387 in. determine (a) the maximum deformation of the bar.75 in 2 16 A = A1+ A2 = 1.111 Two tempered-steel bars. thick. SOLUTION Areas: Mild steel: Tempered steel: Total: Total force to yield the mild steel: 1 A1= (2) = 1.02703 = 0.75 in 2 σY1 = PY ∴ PY = Aσ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb 1 P + P2 = P. This composite bar is subjected as shown to a centric 2 axial load of magnitude P.75 PL (98 × 103 )(14) = = 0. 2. Both steels are elastoplastic with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi.75)(50 × 103 ) = 87.02703 in.03090 − 0.75) P2 48 × 103 = = 64 × 103 psi A2 0. . P = A1σ1 = (1. mild-steel bar. if P is gradually increased from zero to 98 kips and then decreased back to zero. 89 × 10−6 ) = 255.640 m.88 × 106 ] θ = 317. of uniform 37.80 × 106 + (4.89 × 10−6 QY = (317.640θ σ AD PBE = (26.640 σ BE From Statics.06 × 106 )(804. (b) the maximum deflection of point B.80 × 106 θ P = BE = 128 × 109 θ A 2.125PAD 0.64θ . Knowing that a = 0.64 PAD = 0 δ A = 2. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN.113 The rigid bar ABC is supported by two links. Q = PBE + = [28. AD and BE.125)(69.2 × 103 N .640θ ) = 28.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa. SOLUTION Statics: Deformation: Elastic analysis: A = (37.640(Q − PBE ) − 2.47 × 106 )(2.6 × 109 θ θY = 804.88 × 106 θ P = AD = 310.7 ΣM C = 0 : 0.06 × 106 θ θY at yielding of link AD: σ AD = σ Y = 250 × 106 = 310.47 × 106 δ A LAD 1. determine (a) the value of the normal stress in each link.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD = EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.MECH-321 Mechanics of deformable solids PROBLEM 2.64 PAD = PBE + 4.64θ ) = 69. δ B = aθ = 0.0 LBE = (45 × 106 )(0.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1. 53 × 10−6 m EA (200 × 109 )(225 × 10−6 ) . link AD yields.MECH-321 Mechanics of deformable solids PROBLEM 2. PBE = Q − 4.97 × 103 N PBE 27. PAD = Aσ Y = (225 × 10−6 )(250 × 10−6 ) = 56.113 (Continued) (a) Since Q = 260 × 103 > QY .3 MPa δ B = 0.97 × 103 = = 124.3 × 106 Pa −6 A 225 × 10 σ BE = (b) σ BE = 124.125PAD = 260 × 103 − (4.97 × 103 )(1.622 mm ↓ δB = PBE LBE (27.25 × 103 ) PBE = 27.0) = = 621.25 × 103 N σ AD = 250 MPa From Statics.125)(56. 62 × 106 θ θY at yielding of link BE: σ BE = σ Y = 250 × 106 = 352 × 109 θY θY = 710.76(Q − PBE ) − 2.0 LBE = (45 × 106 )(1. Q = PBE + = [73.8 × 106 + (1.64θ .76 m and that the magnitude of the force Q applied at B is gradually increased from zero to 135 kN.88 × 106 θ P = AD = 310. knowing that a = 1.64 PAD = 0 δ A = 2.47 × 106 )(2.113 The rigid bar ABC is supported by two links. Knowing that a = 0. link BE yields.76 σ BE From Statics.76θ σ AD PBE = (26.62 × 106 )(710.113.88 × 106 ] θ = 178.64 PAD = PBE + 1.23 × 10−6 ) = 126.47 × 106 δ A LAD 1.500 PAD 1.86 × 103 N Since Q = 135 × 103 N > QY .114 Solve Prob. (b) the maximum deflection of point B.5 × 6-mm rectangular cross section and made of a mild steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.500)(69.5)(6) = 225 mm 2 = 225 × 10−6 m 2 PAD = EA (200 × 109 )(225 × 10−6 ) δA = δ A = 26.25 × 103 N σ BE = σ Y = 250 MPa . of uniform 37. AD and BE.MECH-321 Mechanics of deformable solids PROBLEM 2.640 m.23 × 10−6 QY = (178. 2. PROBLEM 2. determine (a) the value of the normal stress in each link. PBE = Aσ Y = (225 × 10−6 )(250 × 106 ) = 56.76θ ) = 79. δ B = 1. The magnitude of the force Q applied at B is gradually increased from zero to 260 kN.7 ΣM C = 0 : 1. SOLUTION Statics: Deformation: Elastic Analysis: A = (37.64θ ) = 69.2 × 106 θ P = BE = 352 × 109 θ A 2.6 × 109 θ A (200 × 109 )(225 × 10−6 ) EA = δB = δ B = 45 × 106 δ B 1. 5 × 103 N 1.322 × 10−3 m δ B = 1.MECH-321 Mechanics of deformable solids PROBLEM 2.5 × 103 = = 233.114 (Continued) From Statics.3 × 106 −6 A 225 × 10 PAD = 751.29 × 10−3 rad 69.76θ = 1.500 σ AD = From elastic analysis of AD. PAD 52. PAD = (a) 1 (Q − PBE ) = 52.322 mm ↓ .88 × 106 σ AD = 233 MPa θ= (b) δ B = 1.