Budynas ch09

# Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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Chapter 9 9-1 Eq. (9-3): F = 0 . 707 hl τ = 0 . 707(5 / 16)(4)(20) = 17 . 7 kip Ans. 9-2 Table 9-6: τ all = 21 . 0 kpsi f = 14 . 85 h kip/in = 14 . 85(5 / 16) = 4 . 64 kip/in F = f l = 4 . 64(4) = 18 . 56 kip Ans. 9-3 Table A-20: 1018 HR: S ut = 58 kpsi, S y = 32 kpsi 1018 CR: S ut = 64 kpsi, S y = 54 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: τ all = min(0 . 30 S ut , 0 . 40 S y ) = min[0 . 30(58), 0 . 40(32)] = min(17 . 4, 12 . 8) = 12 . 8 kpsi for both materials. Eq. (9-3): F = 0 . 707 hl τ all F = 0 . 707(5 / 16)(4)(12 . 8) = 11 . 3 kip Ans. 9-4 Eq. (9-3) τ = 2 F hl = 2(32) (5 / 16)(4)(2) = 18 . 1 kpsi Ans. 9-5 b = d = 2 in (a) Primary shear Table 9-1 τ y = V A = F 1 . 414(5 / 16)(2) = 1 . 13 F kpsi F 7" 1.414

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240 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design Secondary shear Table 9-1 J u = d ( 3 b 2 + d 2 ) 6 = 2[ ( 3 )( 2 2 ) + 2 2 ] 6 = 5 . 333 in 3 J = 0 . 707 hJ u = 0 . 707 ( 5 / 16 )( 5 . 333 ) = 1 . 18 in 4 τ x = τ y = Mr y J = 7 F ( 1 ) 1 . 18 = 5 . 93 F kpsi Maximum shear τ max = τ 2 x + ( τ y + τ y ) 2 = F 5 . 93 2 + (1 . 13 + 5 . 93) 2 = 9 . 22 F kpsi F = τ all 9 . 22 = 20 9 . 22 = 2 . 17 kip Ans. (1) (b) For E7010 from Table 9-6, τ all = 21 kpsi Table A-20: HR 1020 Bar: S ut = 55 kpsi, S y = 30 kpsi HR 1015 Support: S ut = 50 kpsi, S y = 27 . 5 kpsi Table 9-5, E7010 Electrode: S ut = 70 kpsi, S y = 57 kpsi The support controls the design. Table 9-4: τ all = min[0 . 30(50), 0 . 40(27 . 5)] = min[15, 11] = 11 kpsi The allowable load from Eq. (1) is F = τ all 9 . 22 = 11 9 . 22 = 1 . 19 kip Ans. 9-6 b = d = 2 in Primary shear τ y = V A = F 1 . 414(5 / 16)(2 + 2) = 0 . 566 F Secondary shear Table 9-1: J u = ( b + d ) 3 6 = (2 + 2) 3 6 = 10 . 67 in 3 J = 0 . 707 hJ u = 0 . 707(5 / 16)(10 . 67) = 2 . 36 in 4 τ x = τ y = Mr y J = (7 F )(1) 2 . 36 = 2 . 97 F F 7"
Chapter 9 241 Maximum shear τ max = τ 2 x + ( τ y + τ y ) 2 = F 2 . 97 2 + (0 . 556 + 2 . 97) 2 = 4 . 61 F kpsi F = τ all 4 . 61 Ans. which is twice τ max / 9 . 22 of Prob. 9-5. 9-7 Weldment, subjected to alternating fatigue, has throat area of A = 0 . 707(6)(60 + 50 + 60) = 721 mm 2 Members’endurance limit: AISI 1010 steel S ut = 320 MPa, S e = 0 . 5(320) = 160 MPa k a = 272(320) 0 . 995 = 0 . 875 k b = 1 (direct shear) k c = 0 . 59 (shear) k d = 1 k f = 1 K f s = 1 2 . 7 = 0 . 370 S se = 0 . 875(1)(0 . 59)(0 . 37)(160) = 30 . 56 MPa Electrode’s endurance: 6010 S ut = 62(6 . 89) = 427 MPa S e = 0 . 5(427) = 213 . 5 MPa k a = 272(427) 0 . 995 = 0 . 657 k b = 1 (direct shear) k c = 0 . 59 (shear) k d = 1 k f = 1 / K f s = 1 / 2 . 7 = 0 . 370 S se = 0 . 657(1)(0 . 59)(0 . 37)(213 . 5) = 30 . 62 MPa . = 30 . 56 Thus, the members and the electrode are of equal strength. For a factor of safety of 1, F a = τ a A = 30 . 6(721)(10 3 ) = 22 . 1 kN Ans.

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242 Solutions Manual Instructor’s Solution Manual to Accompany Mechanical Engineering Design 9-8 Primary shear τ = 0 (why?) Secondary shear Table 9-1: J u = 2 π r 3 = 2 π (4) 3 = 402 cm 3 J = 0 . 707 hJ u = 0 . 707(0 . 5)(402) = 142 cm 4 M = 200 F N · m ( F in kN) τ = Mr 2 J = (200 F )(4) 2(142) = 2 . 82 F (2 welds) F = τ all τ = 140 2 . 82 = 49 . 2 kN Ans. 9-9 Rank fom = J u lh = a 3 / 12 ah = a 2 12 h = 0 . 0833 a 2 h 5 fom = a (3 a 2 + a 2 ) 6(2 a ) h = a 2 3 h = 0 . 3333 a 2 h 1 fom = (2 a ) 4 6 a 2 a 2 12( a + a )2 ah = 5 a 2 24 h = 0 . 2083 a 2 h 4 fom = 1 3 ah 8 a 3 + 6 a 3 + a 3 12 a 4 2 a + a = 11 36 a 2 h = 0 . 3056 a 2 h 2 fom = (2 a ) 3 6 h 1 4 a = 8 a 3 24 ah = a 2 3 h = 0 . 3333 a 2 h 1 fom = 2 π ( a / 2) 3 π ah = a 3 4 ah = a 2 4 h = 0 . 25 a 2 h 3 These rankings apply to fillet weld patterns in torsion that have a square area a × a in which to place weld metal. The object is to place as much metal as possible to the border.
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