Capter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, = 5 mm, allow = 140 MPa. F = 0.707 l allow = 0.707(5)[(50)](140)(10 ) = 49.5 kn Ans. 9- Given, b = in, d = in, = 5/16 in, allow = 5 kpsi. F = 0.707 l allow = 0.707(5/16)[()](5) =.1 kip Ans. 9- Given, b = 50 mm, d = 0 mm, = 5 mm, allow = 140 MPa. F = 0.707 l allow = 0.707(5)[(50)](140)(10 ) = 49.5 kn Ans. 9-4 Given, b = 4 in, d = in, = 5/16 in, allow = 5 kpsi. F = 0.707 l allow = 0.707(5/16)[(4)](5) = 44. kip Ans. 9-5 Prob. 9-1 wit E7010 Electrode. Table 9-6: f = 14.85 kip/in = 14.85 [5 mm/(5.4 mm/in)] =.9 kip/in =.9(4.45/5.4) = 0.51 kn/mm F = f l = 0.51[(50)] = 51. kn Ans. 9-6 Prob. 9- wit E6010 Electrode. Table 9-6: f = 14.85 kip/in = 14.85(5/16) = 4.64 kip/in Capter 9, Page 1/6
F = f l = 4.64[()] = 18.6 kip Ans. 9-7 Prob. 9- wit E7010 Electrode. Table 9-6: f = 14.85 kip/in = 14.85 [5 mm/(5.4 mm/in)] =.9 kip/in =.9(4.45/5.4) = 0.51 kn/mm F = f l = 0.51[(50)] = 51. kn Ans. 9-8 Prob. 9-4 wit E6010 Electrode. Table 9-6: f = 14.85 kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[(4)] = 7.1 kip Ans. 9-9 Table A-0: 1018 CD: S ut = 440 MPa, S = 70 MPa 1018 HR: S ut = 400 MPa, S = 0 MPa Cold-rolled properties degrade to ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(400), 0.40(0)] min(10, 88) 88 MPa for bot materials. Eq. (9-): F = 0.707l all = 0.707(5)[(50)](88)(10 ) = 1.1 kn Ans. 9-10 Table A-0: 100 CD: S ut = 68 kpsi, S = 57 kpsi 100 HR: S ut = 55 kpsi, S = 0 kpsi Cold-rolled properties degrade to ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for bot materials. Eq. (9-): F = 0.707l all = 0.707(5/16)[()](1.0) = 10.6 kip Ans. Capter 9, Page /6
9-11 Table A-0: 105 HR: S ut = 500 MPa, S = 70 MPa 105 CD: S ut = 550 MPa, S = 460 MPa Cold-rolled properties degrade to ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(500), 0.40(70)] min(150, 108) 108 MPa for bot materials. Eq. (9-): F = 0.707l all = 0.707(5)[(50)](108)(10 ) = 8. kn Ans. 9-1 Table A-0: 105 HR: S ut = 7 kpsi, S = 9.5 kpsi 100 CD: S ut = 68 kpsi, S = 57 kpsi, 100 HR: S ut = 55 kpsi, S = 0 kpsi Cold-rolled properties degrade to ot-rolled properties in te neigborood of te weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for bot materials. Eq. (9-): F = 0.707l all = 0.707(5/16)[(4)](1.0) = 1. kip Ans. 9-1 F 10010 Eq. (9-): 141 MPa Ans. l 5 50 50 9-14 F 40 Eq. (9-):.6 kpsi Ans. l 5/16 F 100 10 9-15 Eq. (9-): 177 MPa Ans. l 5 50 0 40 5/164 F 9-16 Eq. (9-): 15.1 kpsi Ans. l Capter 9, Page /6
9-17 b = d =50 mm, c = 150 mm, = 5 mm, and allow = 140 MPa. (a) Primar sear, Table 9-1, Case (Note: b and d are intercanged between problem figure and table figure. Note, also, F in kn and in MPa): V F 10.89F A 1.414550 Secondar sear, Table 9-1: J u d b d 50 50 50 8.10 mm 6 6 J = 0.707 J u = 0.707(5)(8.)(10 ) = 94.6(10 ) mm 4 Mr 175F 10 5 x 14.85F J 94.6 10 x F 14.85.89 14.85.1F (1) allow 140 F 6.06 kn Ans..1.1 (b) For E7010 from Table 9-6, allow = 1 kpsi = 1(6.89) = 145 MPa 100 HR bar: S ut = 80 MPa, S = 10 MPa 1015 HR support: S ut = 40 MPa, S = 190 MPa Table 9-, E7010 Electrode: S ut = 48 MPa, S = 9 MPa Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(40), 0.40(190) = min(10, 76) = 76 MPa Te allowable load, from Eq. (1) is allow 76 F.9 kn Ans..1.1 9-18 b = d = in, c = 6 in, = 5/16 in, and allow = 5 kpsi. Capter 9, Page 4/6
(a) Primar sear, Table 9-1(Note: b and d are intercanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F 1.1F A 1.4145 /16 Secondar sear, Table 9-1: db d Ju 5. in 6 6 J = 0.707 J u = 0.707(5/16)(5.) = 1.178 in 4 Mr 7F 1 x 5.94F J 1.178 x F 5.94 1.1 5.94 9.4F (1) allow 5 F.71 kip Ans. 9.4 9.4 (b) For E7010 from Table 9-6, allow = 1 kpsi 100 HR bar: S ut = 55 kpsi, S = 0 kpsi 1015 HR support: S ut = 50 kpsi, S = 7.5 kpsi Table 9-, E7010 Electrode: S ut = 70 kpsi, S = 57 kpsi Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(50), 0.40(7.5) = min(15, 11) = 11 kpsi Te allowable load, from Eq. (1) is allow 11 F 1.19 kip Ans. 9.4 9.4 9-19 b =50 mm, c = 150 mm, d = 0 mm, = 5 mm, and allow = 140 MPa. (a) Primar sear, Table 9-1, Case (Note: b and d are intercanged between problem figure and table figure. Note, also, F in kn and in MPa): Capter 9, Page 5/6
V F 10.89F A 1.414 5 50 Secondar sear, Table 9-1: db d 50 0 50 Ju 4.10 mm 6 6 J = 0.707 J u = 0.707(5)(4.)(10 ) = 15.(10 ) mm 4 Mr 175F 10 15 x 17.1F J 15. 10 Mr 175F 10 5 x 8.55F J 15. 10 x F 17.1.89 8.55 5.8F (1) F allow 140.91 kn Ans. 5.8 5.8 (b) For E7010 from Table 9-6, allow = 1 kpsi = 1(6.89) = 145 MPa 100 HR bar: S ut = 80 MPa, S = 10 MPa 1015 HR support: S ut = 40 MPa, S = 190 MPa Table 9-, E7010 Electrode: S ut = 48 MPa, S = 9 MPa Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(40), 0.40(190) = min(10, 76) = 76 MPa Te allowable load, from Eq. (1) is allow 76 F.1 kn Ans. 5.8 5.8 9-0 b = 4 in, c = 6 in, d = in, = 5/16 in, and allow = 5 kpsi. Capter 9, Page 6/6
(a) Primar sear, Table 9-1(Note: b and d are intercanged between problem figure and table figure. Note, also, F in kip and in kpsi): V F 0.5658F A 1.4145 /164 Secondar sear, Table 9-1: J u d b d 4 4 18.67 in 6 6 J = 0.707 J u = 0.707(5/16)(18.67) = 4.15 in 4 Mr 8F 1 x 1.99F J 4.15 Mr 8F x.879f J 4.15 x F 1.99 0.5658.879 4.85F (1) allow 5 F 5.15 kip Ans. 4.85 4.85 (b) For E7010 from Table 9-6, allow = 1 kpsi 100 HR bar: S ut = 55 kpsi, S = 0 kpsi 1015 HR support: S ut = 50 kpsi, S = 7.5 kpsi Table 9-, E7010 Electrode: S ut = 70 kpsi, S = 57 kpsi Te support controls te design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(50), 0.40(7.5) = min(15, 11) = 11 kpsi Te allowable load, from Eq. (1) is allow 11 F.7 kip Ans. 4.85 4.85 Capter 9, Page 7/6
9-1 Given, b = 50 mm, c = 150 mm, d = 50 mm, = 5 mm, allow = 140 MPa. Primar sear (F in kn, in MPa, A in mm ): V F 10 1.414F A 1.414 5 50 50 Secondar sear: Table 9-1: Maximum sear: bd 50 50 Ju 166.7 10 mm 6 6 J = 0.707 J u = 0.707(5)166.7(10 ) = 589.(10 ) mm 4 175F 10 (5) Mr x 7.45F J 589. 10 x F 7.45 1.414 7.45 11.54F allow 140 F 1.1 kn Ans. 11.54 11.54 9- Given, b = in, c = 6 in, d = in, = 5/16 in, allow = 5 kpsi. Primar sear: Secondar sear: Table 9-1: Maximum sear: V F 0.5658F A 1.414 5 /16 bd Ju 10.67 in 6 6 J = 0.707 J u = 0.707(5/16)10.67 =.57 in 4 Mr 7 F(1) x.970f J.57 x F.970 0.566.970 4.618F allow 5 F 5.41 kip Ans. 4.618 4.618 9- Given, b = 50 mm, c = 150 mm, d = 0 mm, = 5 mm, allow = 140 MPa. Capter 9, Page 8/6
Primar sear (F in kn, in MPa, A in mm ): V F 10 1.768F A 1.414 5 50 0 Secondar sear: Table 9-1: bd 50 0 Ju 85. 10 mm 6 6 J = 0.707 J u = 0.707(5)85.(10 ) = 01.6(10 ) mm 4 175F 10 (15) Mr x 8.704F J 01.6 10 Maximum sear: 175F 10 (5) Mrx 14.51F J 01.6 10 x F 8.704 1.768 14.51 18.46F allow 140 F 7.58 kn Ans. 18.46 18.46 9-4 Given, b = 4 in, c = 6 in, d = in, = 5/16 in, allow = 5 kpsi. Primar sear: Secondar sear: Table 9-1: V F 0.77F A 1.414 5 /16 4 bd 4 Ju 6 in 6 6 J = 0.707 J u = 0.707(5/16)6 = 7.954 in 4 Mr 8 F(1) x 1.006F J 7.954 Maximum sear: Mrx 8 F().01F J 7.954 x F 1.006 0.77.01.59F Capter 9, Page 9/6
allow 5 F 9.65kip Ans..59.59 9-5 Given, b = 50 mm, d = 50 mm, = 5 mm, E6010 electrode. A = 0.707(5)(50 +50 + 50) = 50. mm Member endurance limit: From Table A-0 for AISI 1010 HR, S ut = 0 MPa. Eq. 6-19 and Table 6-, pp. 87, 88: k a = 7(0) 0.995 = 0.875 k b = 1 (uniform sear), k c = 0.59 (torsion, sear), k d = 1 Eqs. (6-8) and (6-18): S e = 0.875(1)(0.59)(1)(0.5)(0) = 8.6 MPa Electrode endurance: E6010, Table 9-, S ut = 47 MPa Eq. 6-19 and Table 6-, pp. 87, 88: k a = 7(47) 0.995 = 0.657 As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.657(1)(0.59)(1)(0.5)(47) = 8.8 MPa Te members and electrode are basicall of equal strengt. We will use S e = 8.6 MPa. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A 8.650. F 16.10 N 16. kn Ans. K.7 fs 9-6 Given, b = in, d = in, = 5/16 in, E6010 electrode. A = 0.707(5/16)( + + ) = 1.6 in Member endurance limit: From Table A-0 for AISI 1010 HR, S ut = 47 kpsi. Eq. 6-19 and Table 6-, pp. 87, 88: k a = 9.9(47) 0.995 = 0.865 k b = 1 (uniform sear), k c = 0.59 (torsion, sear), k d = 1 Eqs. (6-8) and (6-18): S e = 0.865(1)(0.59)(1)(0.5)(47) = 1.0 kpsi Electrode endurance: E6010, Table 9-, S ut = 6 kpsi Eq. 6-19 and Table 6-, pp. 87, 88: k a = 9.9(6) 0.995 = 0.657 Capter 9, Page 10/6
As before, k b = 1 (uniform sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.657(1)(0.59)(1)(0.5)(6) = 1.0 kpsi Tus te members and electrode are of equal strengt. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A 1.01.6 F 5.89 kip Ans. K.7 fs 9-7 Given, b = 50 mm, d = 0 mm, = 5 mm, E7010 electrode. A = 0.707(5)(50 +50 + 0) = 459.6 mm Member endurance limit: From Table A-0 for AISI 1010 HR, S ut = 0 MPa. Eq. 6-19 and Table 6-, pp. 87, 88: k a = 7(0) 0.995 = 0.875 k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 Eqs. (6-8) and (6-18): S e = 0.875(1)(0.59)(1)(0.5)(0) = 8.6 MPa Electrode endurance: E6010, Table 9-, S ut = 48 MPa Eq. 6-19 and Table 6-, pp. 87, 88: k a = 7(48) 0.995 = 0.58 As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.58(1)(0.59)(1)(0.5)(48) = 8.7 MPa Te members and electrode are basicall of equal strengt. We will use S e =8.6 MPa. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A 8.6459.6 F 14.110 N 14.1 kn Ans. K fs.7 9-8 Given, b = 4 in, d = in, = 5/16 in, E7010 electrode. A = 0.707(5/16)(4 +4 + ) =.09 in Member endurance limit: From Table A-0 for AISI 1010 HR, S ut = 47 kpsi. Eq. 6-19 and Table 6-, pp. 87, 88: k a = 9.9(47) 0.995 = 0.865 k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 Capter 9, Page 11/6
Eqs. (6-8) and (6-18): S e = 0.865(1)(0.59)(1)(0.5)(47) = 1.0 kpsi Electrode endurance: E7010, Table 9-, S ut = 70 kpsi Eq. 6-19 and Table 6-, pp. 87, 88: k a = 9.9(70) 0.995 = 0.58 As before, k b = 1 (direct sear), k c = 0.59 (torsion, sear), k d = 1 S e = 0.58(1)(0.59)(1)(0.5)(70) = 1.0 kpsi Tus te members and electrode are of equal strengt. For a factor of safet of 1, and wit K fs =.7 (Table 9-5) allow A 1.0.09 F 9.8 kip Ans. K.7 fs 9-9 Primar sear: = 0 (w?) Secondar sear: Table 9-1: J u = r = (1.5) = 1.1 in J = 0.707 J u = 0.707(1/4)(1.1) =.749 in 4 welds: Mr 8F 1.5 1.600F J.749 allow 1.600F 0 F 1.5 kip Ans. 9-0 l = + 4 + 4 = 10 in 1 40 4 x 10 4 4 4 0 10 1in 1.6 in M = FR = F(10 1) = 9 F r r 11 4 1.6.4 in, 1 1.6 1.077 in 1 r 1 1.6 1.887 in Capter 9, Page 1/6
1 4 JG 1 0.707 5 /16 0.147 in 1 1 JG J G 0.707 5 /16 4 1.178 in 1 J J Ar i1 i i G i 4 4 0.147 0.707 5 /16.4 1.178 0.707 5 /16 4 1.077 1.178 0.707 5 /16 4 1.887 9.0 in 1.6 tan 8.07 41 1 o r 1.6 4 1.4 in Primar sear ( in kpsi, F in kip) : V F 0.456F A 0.707 5 /16 10 Secondar sear: Mr 9F.4.19F J 9.0 o o F F F.19 sin 8.07.19 cos 8.07 0.456.74F = allow.74 F = 5 F = 6.71 kip Ans. Capter 9, Page 1/6
9-1 l = 0 + 50 + 50 = 10 mm 0 15 50 0 50 5 x 10 0 50 50 5 50 0 10 1.08 mm 1.15 mm M = FR = F(00 1.08) = 186.9 F (M in Nm, F in kn) r 1 r 15 1.08 50 1.15 8.9 mm, 1.08 5 1.15 1.6 mm r 5 1.08 1.15 4.8 mm 1 1 0.707 5 0 7.954 10 mm 4 JG 1 1 JG J G 0.707 5 50 6.8 10 mm 1 J J Ar i1 i i G i 7.954 10 0.707 5 0 8.9 6.8 10 0.707 5 50 1.6 6.8 10 0.707 5 50 4.8 07. 10 mm 4 4 1.15 tan 9.81 50 1.08 1 o r 1.15 50 1.08 4.55 mm Primar sear ( in MPa, F in kn) : V F 10.176F A 0.707 5 10 Secondar sear: Mr 186.9F 10 4.55 5.88F J 07. 10 Capter 9, Page 14/6
o o F F F 5.88 sin 9.81 5.88 cos 9.81.176 7.79F = allow 7.79 F = 140 F = 5.04 kn Ans. 9- Weld Pattern Figure of merit Rank 1. Ju a /1 a a fom 0.08 l a 1 5.. 4. 5. 6. a a a a a fom 0. 1 6a 6 5 1 4 4 a a a a a fom 0.08 a a a 4 4 1 8a 6a a a a fom 0.056 a 1 a a a 1 8a a fom 0. 6 4a 4a 1 a a a / fom 0.5 a 4a Capter 9, Page 15/6
9- Weld Pattern Figure of merit Rank I a /1 u a 1. fom 0.08 6 l a.. 4.* 5. & 7. 6. & 8. 9. a /6 a fom 0.08 a 6 aa / a fom 0.5 a 1 a /1 6a a 7a a fom 0.1944 a 6 x a a, a a a a a a a Iu a aa I a / 1 u a a fom 0.1111 l a 9 a /6 a a 1 a a fom 0.1667 4a 6 a / a a fom 0.15 a 8 4 5 *Note. Because tis section is not smmetric wit te vertical axis, out-of-plane deflection ma occur unless special precautions are taken. See te topic of sear center in books wit more advanced treatments of mecanics of materials. 9-4 Attacment and member (1018 HR), S = 0 MPa and S ut = 400 MPa. Te member and attacment are weak compared to te properties of te lowest electrode. Decision Specif te E6010 electrode Controlling propert, Table 9-4: all = min[0.(400), 0.4(0)] = min(10, 88) = 88 MPa For a static load, te parallel and transverse fillets are te same. Let te lengt of a bead be l = 75 mm, and n be te number of beads. Capter 9, Page 16/6
F n 0.707 l all all F 100 10 n 1.4 0.707l 0.707 75 88 were is in millimeters. Make a table Number of beads, n Leg size, (mm) 1 1.4 10.71 7.14 4 5.6 6 mm Decision Specif = 6 mm on all four sides. Weldment specification: Pattern: All-around square, four beads eac side, 75 mm long Electrode: E6010 Leg size: = 6 mm 9-5 Decision: Coose a parallel fillet weldment pattern. B so-doing, we ve cosen an optimal pattern (see Prob. 9-) and ave tus reduced a sntesis problem to an analsis problem: Table 9-1, case, rotated 90: A = 1.414d = 1.414()(75) = 106.05 mm Primar sear V 1 10 11. A 106.05 Secondar sear: d( b d ) Ju 6 75[(75 ) 75 ] 81. 10 mm 6 J 0.707( )(81.) 10 198.8 10 mm Wit = 45, 4 Capter 9, Page 17/6
o cos 45 Mr 1 10 (187.5)(7.5) 44.4 Mr x J J 198.8 10 1 684.9 x 44.4 (11. 44.4) Attacment and member (1018 HR): S = 0 MPa, S ut = 400 MPa Decision: Use E60XX electrode wic is stronger all min[0.(400), 0.4(0)] 684.9 all 88 MPa 684.9 7.78 mm 88 88 MPa Decision: Specif 8 mm leg size Weldment Specifications: Pattern: Parallel orizontal fillet welds Electrode: E6010 Tpe: Fillet Lengt of eac bead: 75 mm Leg size: 8 mm 9-6 Problem 9-5 solves te problem using parallel orizontal fillet welds, eac 75 mm long obtaining a leg size rounded up to 8 mm. For tis problem, since te widt of te plate is fixed and te lengt as not been determined, we will explore reducing te leg size b using two vertical beads 75 mm long and two orizontal beads suc tat te beads ave a leg size of 6 mm. Decision: Use a rectangular weld bead pattern wit a leg size of 6 mm (case 5 of Table 9-1 wit b unknown and d = 75 mm). Materials: Attacment and member (1018 HR): S = 0 MPa, S ut = 400 MPa From Table 9-4, AISC welding code, all = min[0.(400), 0.4(0)] = min(10, 88) = 88 MPa Select a stronger electrode material from Table 9-. Decision: Specif E6010 Solving for b: In Prob. 9-5, ever term was linear in te unknown. Tis made solving for relativel eas. In tis problem, te terms will not be linear in b, and so we will use an iterative solution wit a spreadseet. Troat area and oter properties from Table 9-1: A = 1.414(6)(b + 75) = 8.484(b + 75) (1) Capter 9, Page 18/6
J u b 75, J = 0.707 (6) J u = 0.707(b +75) () 6 Primar sear ( in MPa, in mm): V A 1 10 A () Secondar sear (See Prob. 9-5 solution for te definition of ) : Mr J Mr Mr 110 150 b / (7.5) x cos cos (4) J J 0.707b 75 Mr Mr 110 150 b / ( b / ) x sin sin (5) J J 0.707 b 75 x Enter Eqs. (1) to (6) into a spreadseet and iterate for various values of b. A portion of te spreadseet is sown below. (6) b (mm) A (mm ) J (mm 4 ) ' (Mpa) " (Mpa) " x (Mpa) (Mpa) 41 984.144 11055.5 1.194 69.554 8.007 90.149 4 99.68 1140.4 1.0891 67.9566 8.05569 88.6156 4 1001.11 11616.6 11.98667 66.4718 8.09065 87.18485 < 88 Mpa 44 1009.596 1191407.4 11.88594 64.96518 8.1191 85.788 We see tat b 4 mm meets te strengt goal. Weldment Specifications: Pattern: Horizontal parallel weld tracks 4 mm long, vertical parallel weld tracks 75 mm long Electrode: E6010 Leg size: 6 mm 9-7 Materials: Member and attacment (1018 HR): S kpsi, Sut 58 kpsi Table 9-4: all min[0.(58), 0.4()] 1.8 kpsi Capter 9, Page 19/6
Decision: Use E6010 electrode. From Table 9-: S 50 kpsi, S 6 kpsi, all min[0.(6), 0.4(50)] 0 kpsi Decision: Since 1018 HR is weaker tan te E6010 electrode, use all 1.8 kpsi Decision: Use an all-around square weld bead track. Primar sear l 1 = 6 + a = 6 + 6.5 = 1.5 in Troat area and oter properties from Table 9-1: A 1.414 b ( d) 1.414( )(6 6) 16.97 V F 0 10 1179 psi A A 16.97 Secondar sear ( bd) (66) Ju 88 in 6 6 4 J 0.707 (88) 0.6 in Mr 0 10 (6.5 )() 76 psi x J 0.6 1 476 x ( ) 76 (1179 76) psi Relate stress to strengt 476 all 1.810 476 0.7 in 1.8 10 Decision: Specif/8in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Tpe of weld: Fillet Weld bead lengt: 4 in Leg size: /8in Attacment lengt: 1.5 in ut Capter 9, Page 0/6
9-8 Tis is a good analsis task to test a student s understanding. (1) Solicit information related to a priori decisions. () Solicit design variables b and d. () Find and round and output all parameters on a single screen. Allow return to Step 1 or Step. (4) Wen te iteration is complete, te final displa can be te bulk of our adequac assessment. Suc a program can teac too. 9-9 Te objective of tis design task is to ave te students teac temselves tat te weld patterns of Table 9- can be added or subtracted to obtain te properties of a contemplated weld pattern. Te instructor can control te level of complication. We ave left te presentation of te drawing to ou. Here is one possibilit. Stud te problem s opportunities, and ten present tis (or our sketc) wit te problem assignment. Use b as te design variable. Express properties as a function of b. 1 1 From Table 9-, case : A1.414 ( bb) bd bd ( b b) d Iu I 0.707I u V F A 1.414 b ( b1 ) Mc Fa( d /) I 0.707I u 1 1 1 Parametric stud a10 in, b8 in, d 8 in, b in, 1.8 kpsi, l (8 ) 1 in Let 1 all Capter 9, Page 1/6
A1.414 (8 ) 8.48in I (8 )(8 / ) 19 in u 4 I 0.707( )(19) 15.7 in 10 000 1179 psi 8.48 10 000(10)(8 / ) 948 psi 15.7 1 175 1179 948 1 800 psi from wic 0.48 in. Do not round off te leg size someting to learn. I 19 fom ' u 64.5 in l 0.48(1) A 8.48(0.48).10 in 4 I 15.7(0.48).65 in 0.48 vol l 1 0.69 in I.65 eff 91. in vol 0.69 1179 4754 psi 0.48 948 11 887 psi 0.48 175 1 800 psi 0.48 Now consider te case of uninterrupted welds, b1 0 A 1.414( )(8 0) 11.1 Iu (8 0)(8 / ) 56 in 4 I 0.707(56) 181 in 10 000 884 11.1 10 000(10)(8 / ) 10 181 1 80 884 10 all 80 0.186 in all 1 800 Do not round off. Capter 9, Page /6
A 11.1(0.186).10 in I 181(0.186).67 in 884 0.186 475 psi, vol 16 0.77 in 0.186 10 1188 psi 0.186 I 56 fom ' u 86.0 in l 0.186(16).67 eff I 11.7 in ( / ) l (0.186 / )16 4 Conclusions: To meet allowable stress limitations, I and A do not cange, nor do τ and σ. To meet te sortened bead lengt, is increased proportionatel. However, volume of bead laid down increases as. Te uninterrupted bead is superior. In tis example, we did not round and as a result we learned someting. Our measures of merit are also sensitive to rounding. Wen te design decision is made, rounding to te next larger standard weld fillet size will decrease te merit. Had te weld bead gone around te corners, te situation would cange. Here is a follow up task analzing an alternative weld pattern. 9-40 From Table 9- For te box A 1.414 b ( d) Subtracting b1 from b and d1 from d A 1.414bbd d 1 1 d d1 bd 1 1 1 ( ) 1 1 Iu bd bb d d d 6 6 6 Lengt of bead l ( bb1d d1) fom Iu / l Capter 9, Page /6
9-41 Computer programs will var. 9-4 Note to te Instructor. In te first printing of te nint edition, te loading was stated incorrectl. In te fourt line, bending moment of 100 kip in in sould read, 10 kip bending load 10 in from. Tis will be corrected in te printings tat follow. We apologize if tis as caused an inconvenience. all = 1 kpsi. Use Fig. 9-17(a) for general geometr, but emplo beads. beads and ten Horizontal parallel weld bead pattern b = in, d = 6 in Table 9-: A1.414b1.414( )() 4.4 in bd (6) Iu 54 in I 0.707I 0.707( )(54) 8. in u 10.58 kpsi 4.4 Mc 10(10)(6 / ) 7.85 kpsi I 8. 1 8.199.58 7.85 kpsi Equate te imum and allowable sear stresses. 4 8.199 all 1 from wic 0.68 in. It follows tat I 8.(0.68) 6.1 in Te volume of te weld metal is l (0.68) ( ) vol 1.40 in Te effectiveness, (eff) H, is 4 Capter 9, Page 4/6
I 6.1 (eff) H 18.6 in vol 1.4 54 (fom ') I u H 1. in l 0.68( ) Vertical parallel weld beads b in d 6 in From Table 9-, case A1.414d 1.414( )(6) 8.48 in d 6 Iu 7 in 6 6 I 0.707Iu 0.707( )(7) 50.9 10 1.179 psi 8.48 Mc 10(10)(6 / ) 5.894 psi I 50.9 1 6.011 1.179 5.894 kpsi Equating to all gives 0.501 in. It follows tat 4 I 50.9(0.501) 5.5 in l 0.501 vol (6 6) 1.51 in I 5.5 (eff ) V 16.7 in vol 1.51 7 (fom') I u V 1.0 in l 0.501(6 6) Te ratio of (eff ) V / (eff ) His 16.7 /18.6 0.898. Te ratio (fom') V / (fom') H is 1.0 /1. 0.909. Tis is not surprising since I I 0.707Iu Iu eff 1.414 1.414fom' vol ( /) l ( /) l l Te ratios (e ff ) V / (eff ) H and (fom') V / (fom') H give te same information. Capter 9, Page 5/6
9-4 F = 0, T = 15 kipin. Table 9-1: J u = r = (1) = 6.8 in, J = 0.707(1/4) 6.8 = 1.111 in 4 Tr 15 1 1.5 kpsi Ans. J 1.111 9-44 F = kip, T = 0. Table 9-: A = 1.414 r = 1.414 (1/4)(1) = 1.111 in I u = r = (1) =.14 in, I = 0.707(1/4).14 = 0.555 in 4 V 1.80 kpsi A 1.111 Mr 6 1 1.6 kpsi I 0.555 = ( + ) 1/ = (1.80 + 1.6 ) 1/ = 1.7 kpsi Ans. 9-45 F = kip, T = 15 kipin. Bending: Table 9-: A = 1.414 r = 1.414 (1/4)(1) = 1.111 in Torsion: I u = r = (1) =.14 in, I = 0.707(1/4).14 = 0.555 in 4 V 1.80 kpsi A 1.111 6 1 Mr 1.6 kpsi M I 0.555 Table 9-1: J u = r = (1) = 6.8 in, J = 0.707(1/4) 6.8 = 1.111 in 4 15 1 Tr 1.5 kpsi T J 1.111 Capter 9, Page 6/6
1.80 1.6 1.5 5.5 kpsi Ans. M T 9-46 F = kip, T = 15 kipin. Bending: Table 9-: A = 1.414 r = 1.414 (1) = 4.44 in I u = r = (1) =.14 in, I = 0.707 (.14) =.1 in 4 V 0.450 kpsi A 4.44 Mr 6 1 5.40 M kpsi I.1 Torsion: Table 9-1: J u = r = (1) = 6.8 in, J = 0.707 (6.8) = 4.44 in 4 Tr 15 1.77 T kpsi J 4.44 0.450 5.40.77 6.87 M T kpsi 6.87 all 0 0.19 in Ans. Sould specif a in weld. Ans. 8 9-47 9 mm, d 00 mm, b5mm From Table 9-, case : A = 1.414(9)(00) =.545(10 ) mm d 00 6 Iu 1.10 mm 6 6 I = 0.707 I u = 0.707(9)(1.)(10 6 ) = 8.484(10 6 ) mm 4 Capter 9, Page 7/6
F 5 10 9.8 MPa A.545(10 ) M = 5(150) = 750 Nm Mc I 8.484(10 ) 750(100) 6 10 44.0 MPa 9.8 44.0 45. MPa Ans. 9-48 Note to te Instructor. In te first printing of te nint edition, te vertical dimension of 5 in sould be to te top of te top plate. Tis will be corrected in te printings tat follow. We apologize if tis as caused an inconvenience. = 0.5 in, b =.5 in, d = 5 in. Table 9-, case 5: A = 0.707 (b +d) = 0.707(0.5)[.5 + (5)] =.09 in d 5 in bd.5 5 d Iu d bd 5 5.5 5. in I = 0.707 I u = 0.707(1/4)(.) = 5.891 in 4 Primar sear: F 0.905 kpsi A.09 Secondar sear (te critical location is at te bottom of te bracket): = 5 = in M 5 5.09 kpsi I 5.891 0.905 5.09 5.17 kpsi all 18 n.48 Ans. 5.17 Capter 9, Page 8/6
9-49 Te largest possible weld size is 1/16 in. Tis is a small weld and tus difficult to accomplis. Te bracket s load-carring capabilit is not known. Tere are geometr problems associated wit seet metal folding, load-placement and location of te center of twist. Tis is not available to us. We will identif te strongest possible weldment. Use a rectangular, weld-all-around pattern Table 9-, case 6: A 1.414 ( b d) 1.414(1 / 16)(1 7.5) 0.751 in x b / 0.5 in d / 7.5 /.75 in d 7.5 Iu ( b d) [(1) 7.5] 98.44 in 6 6 I 0.707Iu 0.707(1 / 16)(98.44) 4.50 in M (.75 0.5) W 4.5W V W 1.1W A 0.751 Mc 4.5 W (7.5 / ).664W I 4.50 W 1.1.664.90W 4 Material properties: Te allowable stress given is low. Let s demonstrate tat. For te 100 CD bracket, use HR properties of S = 0 kpsi and S ut = 55. Te 100 HR support, S = 7.5 kpsi and S ut = 68. Te E6010 electrode as strengts of S = 50 and S ut = 6 kpsi. Allowable stresses: 100 HR: all = min[0.(55), 0.4(0)] = min(16.5, 1) = 1 kpsi 100 HR: all = min[0.(68), 0.4(7.5)] = min(0.4, 15) = 15 kpsi E6010: all = min[0.(6), 0.4(50)] = min(18.6, 0) = 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as te allowable sear stress, use tis lower value. Terefore, te allowable sear stress is all = min(14.4, 1, 18.0) = 1 kpsi However, te allowable stress in te problem statement is 1.5 kpsi wic is low from te weldment perspective. Te load associated wit tis strengt is all.90w 1500 1500 W 85 lbf.90 Capter 9, Page 9/6
If te welding can be accomplised (1/16 leg size is a small weld), te weld strengt is 1 000 psi and te load associated wit tis strengt is W = 1 000/.90 = 077 lbf. Can te bracket carr suc a load? Tere are geometr problems associated wit seet metal folding. Load placement is important and te center of twist as not been identified. Also, te load-carring capabilit of te top bend is unknown. Tese uncertainties ma require te use of a different weld pattern. Our solution provides te best weldment and tus insigt for comparing a welded joint to one wic emplos screw fasteners. 9-50 F F F F B x B B 100 lbf, all kpsi 100(16 / ) 5. lbf 5.cos 60 66.7 lbf 5.cos0 46 lbf x It follows tat RA 56 lbf and RA 66.7 lbf, RA = 6 lbf M = 100(16) = 1600 lbf in Te OD of te tubes is 1 in. From Table 9-1, case 6: A 1.414( r) (1.414)( )(1 / ) 4.44 in Ju r (1 / ) 0.7854 in J (0.707) J 1.414(0.7854) 1.111 in u 4 Capter 9, Page 0/6
V 6 140.0 A 4.44 Tc Mc 1600(0.5) 70.1 J J 1.111 Te sear stresses, and, are additive algebraicall 1 860 (140.0 70.1) psi 860 all 000 860 0.87 5 / 16 in 000 Decision: Use 5/16 in fillet welds Ans. 9-51 For te pattern in bending sown, find te centroid G of te weld group. 6mm 6150 9150 75 6 150 5 9 150 x I I Ax G 6mm 5 mm 0.707 6 150 0.70761505 75 1.010 mm 1 6 4 I 9mm 0.707 9 150 0.7079150175 75.6710 mm 1 6 4 I = I 6 mm + I 9 mm = (1.0 +.67)(10 6 ) = 5.69(10 6 ) mm 4 Te critical location is at B. Wit in MPa, and F in kn Capter 9, Page 1/6
V F 10 0.14F A 0.7076 9150 Mc 00F 10 5 0.881F 6 I 5.69 10 F 0.14 0.881 0.8951 F Materials: 1015 HR (Table A-0): S = 190 MPa, E6010 Electrode(Table 9-): S = 45 MPa Eq. (5-1), p. 5 all = 0.577(190) = 109.6 MPa / 109.6 / all n F 61. kn Ans. 0.8951 0.8951 9-5 In te textbook, Fig. Problem 9-5b is a free-bod diagram of te bracket. Forces and moments tat act on te welds are equal, but of opposite sense. (a) M = 100(0.66) = 49 lbf in Ans. (b) F = 100 sin 0 = 600 lbf Ans. (c) F x = 100 cos 0 = 109 lbf Ans. (d) From Table 9-, case 6: A 1.414(0.5)(0.5.5) 0.97 in d.5 Iu ( b d) [(0.5).5].9 in 6 6 Te second area moment about an axis troug G and parallel to z is I 0.707I 4 0.707(0.5)(.9) 0.599 in Ans. u (e) Refer to Fig. Problem 9-5b. Te sear stress due to F is F 600 1 A 0.97 617 psi Te sear stress along te troat due to F x is F x 109 1069 psi A 0.97 Te resultant of 1 and is in te troat plane Capter 9, Page /6
1 617 1069 14 psi Te bending of te troat gives Mc 49(1.5) I 0.599 916 psi Te imum sear stress is 14 916 157 psi Ans. (f) Materials: 1018 HR Member: S = kpsi, S ut = 58 kpsi (Table A-0) E6010 Electrode: S = 50 kpsi (Table 9-) n Ss 0.577S 0.577() 1.0 A. 1.57 ns (g) Bending in te attacment near te base. Te cross-sectional area is approximatel equal to b. A1 b 0.5(.5) 0.65 in Fx 109 x 166 psi A1 0.65 I bd 0.5(.5) 0.60 in c 6 6 At location A, F M A1 I / c 600 49 648 psi 0.65 0.60 Te von Mises stress is x 648 (166) 91 psi Tus, te factor of safet is, S n 8.18 Ans..91 Te clip on te mooring line bears against te side of te 1/-in ole. If te clip fills te ole Capter 9, Page /6
F 100 9600 psi td 0.5(0.50) S (10 ) n. Ans. 9600 Furter investigation of tis situation requires more detail tan is included in te task statement. () In sear fatigue, te weakest constituent of te weld melt is 1018 HR wit S ut = 58 kpsi, Eq. (6-8), p. 8, gives S e 0.504S 0.504(58) 9. kpsi ut Eq. (6-19), p. 87: k a = 14.4(58) -0.718 = 0.780 For te size factor estimate, we first emplo Eq. (6-5), p. 89, for te equivalent diameter d 0.808 0.707b 0.808 0.707(.5)(0.5) 0.57 in e Eq. (6-0), p. 88, is used next to find k b k b -0.107-0.107 de 0.57 0.0 0.0 0.940 Eq.(6-6), p. 90: k c = 0.59 From Eq. (6-18), p. 87, te endurance strengt in sear is S se = 0.780(0.940)(0.59)(9.) = 1.6 kpsi From Table 9-5, te sear stress-concentration factor is K f s =.7. Te loading is repeatedl-applied 1.57 a m K f s.7.07 kpsi Table 6-7, p. 07: Gerber factor of safet n f, adjusted for sear, wit S su = 0.67S ut 1S su a m S se n f 1 1 m Sse Ssua 1 0.67(58).07 (.07)(1.6) 1 1 5.55 Ans..07 1.6 0.67(58)(.07) Attacment metal sould be cecked for bending fatigue. 9-5 (a) Use b = d = 4 in. Since = 5/8 in, te primar sear is Capter 9, Page 4/6
F 0.89F 1.414(5 / 8)(4) Te secondar sear calculations, for a moment arm of 14 in give J u J x 4[(4) 4] 4.67 in 6 0.707Ju 0.707(5 / 8)4.67 18.85 in Mr 14 F() 1.485F J 18.85 4 Tus, te imum sear and allowable load are: F 1.485 (0.89 1.485).09F all 5 F 10.8 kip Ans..09.09 Te load for part (a) as increased b a factor of 10.8/.71 =.99 Ans. (b) From Prob. 9-18b, all = 11 kpsi F all 11.09.09 all 4.76 kip Te allowable load in part (b) as increased b a factor of 4.76/1.19 = 4 Ans. 9-54 Purcase te ook aving te design sown in Fig. Problem 9-54b. Referring to text Fig. 9-9a, tis design reduces peel stresses. 9-55 (a) l / 1 l/ Pcos( x) l/ A1 dx A / 1 cos( x) dx sin( x) l l 4bsin( l / ) l/ l / A1 A1 [sin( l / ) sin( l / )] [sin( l / ) ( sin( l / ))] A1 sin( l / ) P P [sin( l / )] Ans. 4bl sin( l / ) bl P cos( l/ ) P (b) (/) l Ans. 4bsin( l/ ) 4btan( l/ ) Capter 9, Page 5/6
(c) K ( l / ) P bl l / 4btan( l / ) P tan( l / ) Ans. For computer programming, it can be useful to express te perbolic tangent in terms of exponentials: l exp( l / ) exp( l / ) K Ans. exp( l / ) exp( l / ) 9-56 Tis is a computer programming exercise. All programs will var. Capter 9, Page 6/6