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ÿþbudynas_SM_ch09.qxd 12/01/2006 16:16 Page 239 FIRST PAGES Chapter 9 9-1 Eq. (9-3): F = 0.707hlÄ = 0.707(5/16)(4)(20) = 17.7 kip Ans. 9-2 Table 9-6: Äall = 21.0 kpsi f = 14.85h 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: Sut = 58 kpsi, Sy = 32 kpsi 1018 CR: Sut = 64 kpsi, Sy = 54 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: Äall = min(0.30Sut, 0.40Sy) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi for both materials. Eq. (9-3): F = 0.707hlÄall F = 0.707(5/16)(4)(12.8) = 11.3 kip Ans. 9-4 Eq. (9-3) " " 2F 2(32) Ä = = = 18.1 kpsi Ans. hl (5/16)(4)(2) 9-5 b = d = 2in F 1.414 7" (a) Primary shear Table 9-1 V F Äy = = = 1.13F kpsi A 1.414(5/16)(2) budynas_SM_ch09.qxd 12/01/2006 16:16 Page 240 FIRST PAGES 240 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design Secondary shear Table 9-1 d(3b2 + d2) 2[(3)(22) + 22] Ju = = = 5.333 in3 6 6 J = 0.707hJu = 0.707(5/16)(5.333) = 1.18 in4 Mry 7F(1) Äx = Äy = = = 5.93F kpsi J 1.18 Maximum shear 2 Ämax = Äx + (Äy + Äy )2 = F 5.932 + (1.13 + 5.93)2 = 9.22F kpsi Äall 20 F = = = 2.17 kip Ans. (1) 9.22 9.22 (b) For E7010 from Table 9-6, Äall = 21 kpsi Table A-20: HR 1020 Bar: Sut = 55 kpsi, Sy = 30 kpsi HR 1015 Support: Sut = 50 kpsi, Sy = 27.5 kpsi Table 9-5, E7010 Electrode: Sut = 70 kpsi, Sy = 57 kpsi The support controls the design. Table 9-4: Äall = min[0.30(50), 0.40(27.5)] = min[15, 11] = 11 kpsi The allowable load from Eq. (1) is Äall 11 F = = = 1.19 kip Ans. 9.22 9.22 9-6 b = d = 2in F 7" Primary shear V F Äy = = = 0.566F A 1.414(5/16)(2 + 2) Secondary shear (b + d)3 (2 + 2)3 : Table 9-1 Ju = = = 10.67 in3 6 6 J = 0.707hJu = 0.707(5/16)(10.67) = 2.36 in4 Mry (7F)(1) Äx = Äy = = = 2.97F J 2.36 budynas_SM_ch09.qxd 12/01/2006 16:16 Page 241 FIRST PAGES Chapter 9 241 Maximum shear 2 Ämax = Äx + (Äy + Äy )2 = F 2.972 + (0.556 + 2.97)2 = 4.61F kpsi Äall F = Ans. 4.61 which is twice Ämax/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 mm2 Members endurance limit: AISI 1010 steel Sut = 320 MPa, Se = 0.5(320) = 160 MPa ka = 272(320)-0.995 = 0.875 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 1 1 kf = = = 0.370 K 2.7 f s Sse = 0.875(1)(0.59)(0.37)(160) = 30.56 MPa Electrode s endurance: 6010 Sut = 62(6.89) = 427 MPa Se = 0.5(427) = 213.5MPa ka = 272(427)-0.995 = 0.657 kb = 1 (direct shear) kc = 0.59 (shear) kd = 1 kf = 1/K = 1/2.7 = 0.370 f s . Sse = 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, Fa = Äa A = 30.6(721)(10-3) = 22.1kN Ans. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 242 FIRST PAGES 242 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design 9-8 Primary shear Ä = 0 (why?) Secondary shear Table 9-1: Ju = 2Àr3 = 2À(4)3 = 402 cm3 J = 0.707hJu = 0.707(0.5)(402) = 142 cm4 M = 200F N · m (F in kN) Mr (200F)(4) Ä = = = 2.82F (2 welds) 2J 2(142) Äall 140 F = = = 49.2kN Ans. Ä 2.82 9-9 Rank Ju a3/12 a2 a2 fom = = = = 0.0833 5 lh ah 12h h a(3a2 + a2) a2 a2 fom = = = 0.3333 1 6(2a)h 3h h (2a)4 - 6a2a2 5a2 a2 fom = = = 0.2083 4 12(a + a)2ah 24h h 1 8a3 + 6a3 + a3 a4 11 a2 a2 fom = - = = 0.3056 2 3ah 12 2a + a 36 h h (2a)3 1 8a3 a2 a2 fom = = = = 0.3333 1 6h 4a 24ah 3h h 2À(a/2)3 a3 a2 a2 fom = = = = 0.25 3 Àah 4ah 4h h 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. If your area is rectangular, your goal is the same but the rankings may change. Students will be surprised that the circular weld bead does not rank first. 9-10 Iu 1 a3 1 1 a2 a2 fom = = = = 0.0833 5 lh a 12 h 12 h h Iu 1 a3 a2 fom = = = 0.0833 5 lh 2ah 6 h Iu 1 a2 1 a2 a2 fom = = = = 0.25 1 lh 2ah 2 4 h h budynas_SM_ch09.qxd 12/01/2006 16:16 Page 243 FIRST PAGES Chapter 9 243 Iu 1 a2 1 a2 a2 fom = = (3a + a) = = 0.1667 2 lh [2(2a)]h 6 6 h h b a d2 a2 a x = = , y = = = ¯ ¯ 2 2 b + 2d 3a 3 2d3 a a2 2a3 2a3 a2 a3 Iu = - 2d2 + (b + 2d) = - + 3a = 3 3 9 3 3 9 3 Iu a3/3 1 a2 a2 fom = = = = 0.1111 4 lh 3ah 9 h h Àa3 Iu = Àr3 = 8 Iu Àa3/8 a2 a2 fom = = = = 0.125 3 lh Àah 8h h The CEE-section pattern was not ranked because the deflection of the beam is out-of-plane. If you have a square area in which to place a fillet weldment pattern under bending, your objective is to place as much material as possible away from the x-axis. If your area is rec- tangular, your goal is the same, but the rankings may change. 9-11 Materials: Attachment (1018 HR) Sy = 32 kpsi, Sut = 58 kpsi Member (A36) Sy = 36 kpsi, Sut ranges from 58 to 80 kpsi, use 58. The member and attachment are weak compared to the E60XX electrode. Decision Specify E6010 electrode Controlling property: Äall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi For a static load the parallel and transverse fillets are the same. If n is the number of beads, F Ä = = Äall n(0.707)hl F 25 nh = = = 0.921 0.707lÄall 0.707(3)(12.8) Make a table. Number of beads Leg size nh 1 0.921 2 0.460 ’! 1/2" 3 0.307 ’! 5/16" 4 0.230 ’! 1/4" Decision: Specify 1/4" leg size Decision: Weld all-around budynas_SM_ch09.qxd 12/01/2006 16:16 Page 244 FIRST PAGES 244 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design Weldment Specifications: Pattern: All-around square Electrode: E6010 Type: Two parallel fillets Ans. Two transverse fillets Length of bead: 12 in Leg: 1/4 in For a figure of merit of, in terms of weldbead volume, is this design optimal? 9-12 Decision: Choose a parallel fillet weldment pattern. By so-doing, we ve chosen an optimal pattern (see Prob. 9-9) and have thus reduced a synthesis problem to an analysis problem: Table 9-1: A = 1.414hd = 1.414(h)(3) = 4.24h in3 Primary shear V 3000 707 Äy = = = A 4.24h h Secondary shear d(3b2 + d2) 3[3(32) + 32] Table 9-1: Ju = = = 18 in3 6 6 J = 0.707(h)(18) = 12.7h in4 Mry 3000(7.5)(1.5) 2657 Äx = = = = Äy J 12.7h h 1 4287 2 Ämax = Äx + (Äy + Äy )2 = 26572 + (707 + 2657)2 = h h Attachment (1018 HR): Sy = 32 kpsi, Sut = 58 kpsi Member (A36): Sy = 36 kpsi The attachment is weaker Decision: Use E60XX electrode Äall = min[0.3(58), 0.4(32)] = 12.8 kpsi 4287 Ämax = Äall = = 12 800 psi h 4287 h = = 0.335 in 12 800 Decision: Specify 3/8" leg size Weldment Specifications: Pattern: Parallel fillet welds Electrode: E6010 Type: Fillet Ans. Length of bead: 6 in Leg size: 3/8 in budynas_SM_ch09.qxd 12/01/2006 16:16 Page 245 FIRST PAGES Chapter 9 245 9-13 An optimal square space (3" × 3") weldment pattern is or or . In Prob. 9-12, there was roundup of leg size to 3/8 in. Consider the member material to be structural A36 steel. Decision: Use a parallel horizontal weld bead pattern for welding optimization and convenience. Materials: Attachment (1018 HR): Sy = 32 kpsi, Sut = 58 kpsi Member (A36): Sy = 36 kpsi, Sut 58 80 kpsi; use 58 kpsi From Table 9-4 AISC welding code, Äall = min[0.3(58), 0.4(32)] = min(16.6, 12.8) = 12.8 kpsi Select a stronger electrode material from Table 9-3. Decision: Specify E6010 Throat area and other properties: A = 1.414hd = 1.414(h)(3) = 4.24h in2 x = b/2 = 3/2 = 1.5in ¯ y = d/2 = 3/2 = 1.5in ¯ d(3b2 + d2) 3[3(32) + 32] Ju = = = 18 in3 6 6 J = 0.707hJu = 0.707(h)(18) = 12.73h in4 Primary shear: V 3000 707.5 Äx = = = A 4.24h h x y rx r x y ry x Secondary shear: Mr Ä = J Mr Mrx Äx = Ä cos 45æ% = cos 45æ% = J J 3000(6 + 1.5)(1.5) 2651 Äx = = 12.73h h 2651 Äy = Äx = h budynas_SM_ch09.qxd 12/01/2006 16:16 Page 246 FIRST PAGES 246 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design 2 Ämax = (Äx + Äx)2 + Äy 1 = (2651 + 707.5)2 + 26512 h 4279 = psi h Relate stress and strength: Ämax = Äall 4279 = 12 800 h 4279 h = = 0.334 in ’! 3/8in 12 800 Weldment Specifications: Pattern: Horizontal parallel weld tracks Electrode: E6010 Type of weld: Two parallel fillet welds Length of bead: 6 in Leg size: 3/8 in Additional thoughts: Since the round-up in leg size was substantial, why not investigate a backward C weld pattern. One might then expect shorter horizontal weld beads which will have the advan- tage of allowing a shorter member (assuming the member has not yet been designed). This will show the inter-relationship between attachment design and supporting members. 9-14 Materials: Member (A36): Sy = 36 kpsi, Sut = 58 to 80 kpsi; use Sut = 58 kpsi Attachment (1018 HR): Sy = 32 kpsi, Sut = 58 kpsi Äall = min[0.3(58), 0.4(32)] = 12.8 kpsi Decision: Use E6010 electrode. From Table 9-3: Sy = 50 kpsi, Sut = 62 kpsi, Äall = min[0.3(62), 0.4(50)] = 20 kpsi Decision: Since A36 and 1018 HR are weld metals to an unknown extent, use Äall = 12.8 kpsi Decision: Use the most efficient weld pattern square, weld-all-around. Choose 6" × 6" size. Attachment length: l1 = 6 + a = 6 + 6.25 = 12.25 in Throat area and other properties: A = 1.414h(b + d) = 1.414(h)(6 + 6) = 17.0h b 6 d 6 x = = = 3in, y = = = 3in ¯ ¯ 2 2 2 2 budynas_SM_ch09.qxd 12/01/2006 16:16 Page 247 FIRST PAGES Chapter 9 247 Primary shear V F 20 000 1176 Äy = = = = psi A A 17h h Secondary shear (b + d)3 (6 + 6)3 Ju = = = 288 in3 6 6 J = 0.707h(288) = 203.6h in4 Mry 20 000(6.25 + 3)(3) 2726 Äx = Äy = = = psi J 203.6h h 1 4760 2 Ämax = Äx + (Äy + Äy)2 = 27262 + (2726 + 1176)2 = psi h h Relate stress to strength Ämax = Äall 4760 = 12 800 h 4760 h = = 0.372 in 12 800 Decision: Specify 3/8 in leg size Specifications: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3/8 in Attachment length: 12.25 in 9-15 This is a good analysis task to test the students understanding (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d. (3) Find h and round and output all parameters on a single screen. Allow return to Step 1 or Step 2. (4) When the iteration is complete, the final display can be the bulk of your adequacy assessment. Such a program can teach too. 9-16 The objective of this design task is to have the students teach themselves that the weld patterns of Table 9-3 can be added or subtracted to obtain the properties of a comtem- plated weld pattern. The instructor can control the level of complication. I have left the budynas_SM_ch09.qxd 12/01/2006 16:16 Page 248 FIRST PAGES 248 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design presentation of the drawing to you. Here is one possibility. Study the problem s opportuni- ties, then present this (or your sketch) with the problem assignment. Section AA A b1 1" 2 8" d 1018 HR 8" A A36 b a Body welds Attachment weld 10000 lbf not shown pattern considered Use b1 as the design variable. Express properties as a function of b1. From Table 9-3, category 3: A = 1.414h(b - b1) x = b/2, y = d/2 ¯ ¯ bd2 b1d2 (b - b1)d2 Iu = - = 2 2 2 I = 0.707hIu V F Ä = = A 1.414h(b - b1) Mc Fa(d/2) Ä = = I 0.707hIu Ämax = Ä 2 + Ä 2 Parametric study Let a = 10 in, b = 8 in, d = 8 in, b1 = 2in, Äall = 12.8 kpsi, l = 2(8 - 2) = 12 in A = 1.414h(8 - 2) = 8.48h in2 Iu = (8 - 2)(82/2) = 192 in3 I = 0.707(h)(192) = 135.7h in4 10 000 1179 Ä = = psi 8.48h h 10 000(10)(8/2) 2948 Ä = = psi 135.7h h 1 3175 Ämax = 11792 + 29482 = = 12 800 h h from which h = 0.248 in. Do not round off the leg size  something to learn. Iu 192 fom = = = 64.5 hl 0.248(12) A = 8.48(0.248) = 2.10 in2 I = 135.7(0.248) = 33.65 in4 budynas_SM_ch09.qxd 12/01/2006 16:16 Page 249 FIRST PAGES Chapter 9 249 h2 0.2482 vol = l = 12 = 0.369 in3 2 2 I 33.65 = = 91.2 = eff vol 0.369 1179 Ä = = 4754 psi 0.248 2948 Ä = = 11 887 psi 0.248 4127 . Ämax = = 12 800 psi 0.248 Now consider the case of uninterrupted welds, b1 = 0 A = 1.414(h)(8 - 0) = 11.31h Iu = (8 - 0)(82/2) = 256 in3 I = 0.707(256)h = 181h in4 10 000 884 Ä = = 11.31h h 10 000(10)(8/2) 2210 Ä = = 181h h 1 2380 Ämax = 8842 + 22102 = = Äall h h Ämax 2380 h = = = 0.186 in Äall 12 800 Do not round off h. A = 11.31(0.186) = 2.10 in2 I = 181(0.186) = 33.67 884 0.1862 Ä = = 4753 psi, vol = 16 = 0.277 in3 0.186 2 2210 Ä = = 11 882 psi 0.186 Iu 256 fom = = = 86.0 hl 0.186(16) I 33.67 eff = = = 121.7 (h2/2)l (0.1862/2)16 Conclusions: To meet allowable stress limitations, I and A do not change, nor do Ä and à . To meet the shortened bead length, h is increased proportionately. However, volume of bead laid down increases as h2. The uninterrupted bead is superior. In this example, we did not round h and as a result we learned something. Our measures of merit are also sensitive to rounding. When the design decision is made, rounding to the next larger standard weld fillet size will decrease the merit. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 250 FIRST PAGES 250 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design Had the weld bead gone around the corners, the situation would change. Here is a fol- lowup task analyzing an alternative weld pattern. b1 d1 d b 9-17 From Table 9-2 For the box A = 1.414h(b + d) Subtracting b1 from b and d1 from d A = 1.414 h(b - b1 + d - d1) 3 d2 d1 b1d2 Iu = (3b + d) - - 6 6 2 1 1 3 = (b - b1)d2 + d3 - d1 2 6 length of bead l = 2(b - b1 + d - d1) fom = Iu/hl 9-18 Computer programs will vary. 9-19 Äall = 12 800 psi. Use Fig. 9-17(a) for general geometry, but employ beads and then beads. Horizontal parallel weld bead pattern 6" b = 6in 8" d = 8in From Table 9-2, category 3 A = 1.414 hb = 1.414(h)(6) = 8.48 h in2 x = b/2 = 6/2 = 3in, y = d/2 = 8/2 = 4in ¯ ¯ bd2 6(8)2 Iu = = = 192 in3 2 2 I = 0.707hIu = 0.707(h)(192) = 135.7h in4 10 000 1179 Ä = = psi 8.48h h budynas_SM_ch09.qxd 12/01/2006 16:16 Page 251 FIRST PAGES Chapter 9 251 Mc 10 000(10)(8/2) 2948 Ä = = = psi I 135.7h h 1 3175 Ämax = Ä 2 + Ä 2 = (11792 + 29482)1/2 = psi h h Equate the maximum and allowable shear stresses. 3175 Ämax = Äall = = 12 800 h from which h = 0.248 in. It follows that I = 135.7(0.248) = 33.65 in4 The volume of the weld metal is h2l 0.2482(6 + 6) vol = = = 0.369 in3 2 2 The effectiveness, (eff)H , is I 33.65 (eff)H = = = 91.2in vol 0.369 Iu 192 (fom )H = = = 64.5in hl 0.248(6 + 6) Vertical parallel weld beads 6" b = 6in 8" d = 8in From Table 9-2, category 2 A = 1.414hd = 1.414(h)(8) = 11.31h in2 x = b/2 = 6/2 = 3in, y = d/2 = 8/2 = 4in ¯ ¯ d3 83 Iu = = = 85.33 in3 6 6 I = 0.707hIu = 0.707(h)(85.33) = 60.3h 10 000 884 Ä = = psi 11.31h h Mc 10 000(10)(8/2) 6633 Ä = = = psi I 60.3 h h 1 Ämax = Ä 2 + Ä 2 = (8842 + 66332)1/2 h 6692 = psi h budynas_SM_ch09.qxd 12/01/2006 16:16 Page 252 FIRST PAGES 252 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design Equating Ämax to Äall gives h = 0.523 in. It follows that I = 60.3(0.523) = 31.5in4 h2l 0.5232 vol = = (8 + 8) = 2.19 in3 2 2 I 31.6 (eff)V = = = 14.4in vol 2.19 Iu 85.33 (fom )V = = = 10.2in hl 0.523(8 + 8) The ratio of (eff)V/(eff)H is 14.4/91.2 = 0.158. The ratio (fom )V/(fom )H is 10.2/64.5 = 0.158. This is not surprising since I I 0.707 hIu Iu eff = = = = 1.414 = 1.414 fom vol (h2/2)l (h2/2)l hl The ratios (eff)V/(eff)H and (fom )V/(fom )H give the same information. 9-20 Because the loading is pure torsion, there is no primary shear. From Table 9-1, category 6: Ju = 2Àr3 = 2À(1)3 = 6.28 in3 J = 0.707 hJu = 0.707(0.25)(6.28) = 1.11 in4 Tr 20(1) Ä = = = 18.0 kpsi Ans. J 1.11 9-21 h = 0.375 in, d = 8in, b = 1in From Table 9-2, category 2: A = 1.414(0.375)(8) = 4.24 in2 d3 83 Iu = = = 85.3in3 6 6 I = 0.707hIu = 0.707(0.375)(85.3) = 22.6in4 F 5 Ä = = = 1.18 kpsi A 4.24 M = 5(6) = 30 kip · in c = (1 + 8 + 1 - 2)/2 = 4in Mc 30(4) Ä = = = 5.31 kpsi I 22.6 Ämax = Ä 2 + Ä 2 = 1.182 + 5.312 = 5.44 kpsi Ans. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 253 FIRST PAGES Chapter 9 253 9-22 h = 0.6cm, b = 6cm, d = 12 cm. Table 9-3, category 5: 6 A = 0.707h(b + 2d) B = 0.707(0.6)[6 + 2(12)] = 12.7cm2 4.8 d2 122 G y = = = 4.8cm ¯ b + 2d 6 + 2(12) 7.2 2d3 Iu = - 2d2y + (b + 2d)y2 ¯ ¯ A 3 2(12)3 = - 2(122)(4.8) + [6 + 2(12)]4.82 3 = 461 cm3 I = 0.707hIu = 0.707(0.6)(461) = 196 cm4 F 7.5(103) Ä = = = 5.91 MPa A 12.7(102) M = 7.5(120) = 900 N · m cA = 7.2cm, cB = 4.8cm The critical location is at A. McA 900(7.2) ÄA = = = 33.1MPa I 196 Ämax = Ä 2 + Ä 2 = (5.912 + 33.12)1/2 = 33.6MPa Äall 120 n = = = 3.57 Ans. Ämax 33.6 9-23 The largest possible weld size is 1/16 in. This is a small weld and thus difficult to accom- plish. The bracket s load-carrying capability is not known. There are geometry problems associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern  Table 9-2, category 6: A = 1.414 h(b + d) = 1.414(1/16)(1 + 7.5) = 0.751 in2 7.5" x = b/2 = 0.5in ¯ d 7.5 y = = = 3.75 in ¯ 2 2 1" budynas_SM_ch09.qxd 12/01/2006 16:16 Page 254 FIRST PAGES 254 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design d2 7.52 Iu = (3b + d) = [3(1) + 7.5] = 98.4in3 6 6 I = 0.707hIu = 0.707(1/16)(98.4) = 4.35 in4 M = (3.75 + 0.5)W = 4.25W V W Ä = = = 1.332W A 0.751 Mc 4.25W (7.5/2) Ä = = = 3.664W I 4.35 Ämax = Ä 2 + Ä 2 = W 1.3322 + 3.6642 = 3.90W Material properties: The allowable stress given is low. Let s demonstrate that. For the A36 structural steel member, Sy = 36 kpsi and Sut = 58 kpsi. For the 1020 CD attachment, use HR properties of Sy = 30 kpsi and Sut = 55. The E6010 electrode has strengths of Sy = 50 and Sut = 62 kpsi. Allowable stresses: A36: Äall = min[0.3(58), 0.4(36)] = min(17.4, 14.4) = 14.4 kpsi 1020: Äall = min[0.3(55), 0.4(30)] Äall = min(16.5, 12) = 12 kpsi E6010: Äall = min[0.3(62), 0.4(50)] = min(18.6, 20) = 18.6 kpsi Since Table 9-6 gives 18.0 kpsi as the allowable shear stress, use this lower value. Therefore, the allowable shear stress is Äall = min(14.4, 12, 18.0) = 12 kpsi However, the allowable stress in the problem statement is 0.9 kpsi which is low from the weldment perspective. The load associated with this strength is Ämax = Äall = 3.90W = 900 900 W = = 231 lbf 3.90 If the welding can be accomplished (1/16 leg size is a small weld), the weld strength is 12 000 psi and the load W = 3047 lbf. Can the bracket carry such a load? There are geometry problems associated with sheet metal folding. Load placement is important and the center of twist has not been identified. Also, the load-carrying capability of the top bend is unknown. These uncertainties may require the use of a different weld pattern. Our solution pro- vides the best weldment and thus insight for comparing a welded joint to one which em- ploys screw fasteners. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 255 FIRST PAGES Chapter 9 255 9-24 y x F FB Rx A 60 A Ry B A F = 100 lbf, Äall = 3 kpsi FB = 100(16/3) = 533.3 lbf x FB =-533.3 cos 60æ% =-266.7 lbf y FB =-533.3 cos 30æ% =-462 lbf y It follows that RA = 562 lbf and Rx = 266.7 lbf, RA = 622 lbf A M = 100(16) = 1600 lbf · in 100 462 266.7 16 3 266.7 562 The OD of the tubes is 1 in. From Table 9-1, category 6: A = 1.414(Àhr)(2) = 2(1.414)(Àh)(1/2) = 4.44h in2 Ju = 2Àr3 = 2À(1/2)3 = 0.785 in3 J = 2(0.707)hJu = 1.414(0.785)h = 1.11h in4 V 622 140 Ä = = = A 4.44h h Tc Mc 1600(0.5) 720.7 Ä = = = = J J 1.11h h The shear stresses, Ä and Ä , are additive algebraically 1 861 Ämax = (140 + 720.7) = psi h h 861 Ämax = Äall = = 3000 h 861 h = = 0.287 ’! 5/16" 3000 Decision: Use 5/16 in fillet welds Ans. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 256 FIRST PAGES 256 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design 9-25 y g g 1" 3" G 4 8 x 1" 3" 4 8 B g g 9" 7" For the pattern in bending shown, find the centroid G of the weld group. 6(0.707)(1/4)(3) + 6(0.707)(3/8)(13) x = ¯ 6(0.707)(1/4) + 6(0.707)(3/8) = 9in I1/4 = 2 IG + A2 x ¯ 0.707(1/4)(63) = 2 + 0.707(1/4)(6)(62) 12 = 82.7in4 0.707(3/8)(63) I3/8 = 2 + 0.707(3/8)(6)(42) 12 = 60.4in4 I = I1/4 + I3/8 = 82.7 + 60.4 = 143.1in4 The critical location is at B. From Eq. (9-3), F Ä = = 0.189F 2[6(0.707)(3/8 + 1/4)] Mc (8F)(9) Ä = = = 0.503F I 143.1 Ämax = Ä 2 + Ä 2 = F 0.1892 + 0.5032 = 0.537F Materials: A36 Member: Sy = 36 kpsi 1015 HR Attachment: Sy = 27.5 kpsi E6010 Electrode: Sy = 50 kpsi Äall = 0.577 min(36, 27.5, 50) = 15.9 kpsi Äall/n 15.9/2 F = = = 14.8 kip Ans. 0.537 0.537 9-26 Figure P9-26b is a free-body diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 1200(0.366) = 439 lbf · in Ans. (b) Fy = 1200 sin 30æ% = 600 lbf Ans. (c) Fx = 1200 cos 30æ% = 1039 lbf Ans. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 257 FIRST PAGES Chapter 9 257 (d) From Table 9-2, category 6: A = 1.414(0.25)(0.25 + 2.5) = 0.972 in2 d2 2.52 Iu = (3b + d) = [3(0.25) + 2.5] = 3.39 in3 6 6 The second area moment about an axis through G and parallel to z is I = 0.707hIu = 0.707(0.25)(3.39) = 0.599 in4 Ans. (e) Refer to Fig. P.9-26b. The shear stress due to Fy is Fy 600 Ä1 = = = 617 psi A 0.972 The shear stress along the throat due to Fx is Fx 1039 Ä2 = = = 1069 psi A 0.972 The resultant of Ä1 and Ä2 is in the throat plane 1/2 2 2 Ä = Ä1 + Ä2 = (6172 + 10692)1/2 = 1234 psi The bending of the throat gives Mc 439(1.25) Ä = = = 916 psi I 0.599 The maximum shear stress is Ämax = (Ä 2 + Ä 2)1/2 = (12342 + 9162)1/2 = 1537 psi Ans. (f) Materials: 1018 HR Member: Sy = 32 kpsi, Sut = 58 kpsi (Table A-20) E6010 Electrode: Sy = 50 kpsi (Table 9-3) Ssy 0.577Sy 0.577(32) n = = = = 12.0 Ans. Ämax Ämax 1.537 (g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. . A1 = bh = 0.25(2.5) = 0.625 in2 Fx 1039 Äxy = = = 1662 psi A1 0.625 I bd2 0.25(2.5)2 = = = 0.260 in3 c 6 6 At location A Fy M Ãy = + A1 I/c 600 439 Ãy = + = 2648 psi 0.625 0.260 budynas_SM_ch09.qxd 12/01/2006 16:16 Page 258 FIRST PAGES 258 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design The von Mises stress à is 1/2 2 2 à = Ãy + 3Äxy = [26482 + 3(1662)2]1/2 = 3912 psi Thus, the factor of safety is, Sy 32 n = = = 8.18 Ans. à 3.912 The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole F -1200 à = = =-9600 psi td 0.25(0.50) Sy 32(103) n =- =- = 3.33 Ans. à -9600 Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 with Sut = 58 kpsi Se = 0.5Sut = 0.5(58) = 29 kpsi Table 7-4: ka = 14.4(58)-0.718 = 0.780 For the size factor estimate, we first employ Eq. (7-24) for the equivalent diameter. " de = 0.808 0.707hb = 0.808 0.707(2.5)(0.25) = 0.537 in Eq. (7-19) is used next to find kb -0.107 -0.107 de 0.537 kb = = = 0.940 0.30 0.30 The load factor for shear kc , is kc = 0.59 The endurance strength in shear is Sse = 0.780(0.940)(0.59)(29) = 12.5 kpsi From Table 9-5, the shear stress-concentration factor is K = 2.7. The loading is f s repeatedly-applied. Ämax 1.537 Äa = Äm = K = 2.7 = 2.07 kpsi f s 2 2 Table 7-10: Gerber factor of safety n , adjusted for shear, with Ssu = 0.67Sut f ñø üø 2 òø 2 ýø 1 0.67(58) 2.07 2(2.07)(12.5) nf = -1 + 1 + = 5.52 Ans. óø þø 2 2.07 12.5 0.67(58)(2.07) Attachment metal should be checked for bending fatigue. budynas_SM_ch09.qxd 12/01/2006 16:16 Page 259 FIRST PAGES Chapter 9 259 9-27 Use b = d = 4 in. Since h = 5/8 in, the primary shear is F Ä = = 0.283F 1.414(5/8)(4) The secondary shear calculations, for a moment arm of 14 in give 4[3(42) + 42] Ju = = 42.67 in3 6 J = 0.707hJu = 0.707(5/8)42.67 = 18.9in4 Mry 14F(2) Äx = Äy = = = 1.48F J 18.9 Thus, the maximum shear and allowable load are: Ämax = F 1.482 + (0.283 + 1.48)2 = 2.30F Äall 20 F = = = 8.70 kip Ans. 2.30 2.30 From Prob. 9-5b, Äall = 11 kpsi Äall 11 Fall = = = 4.78 kip 2.30 2.30 The allowable load has thus increased by a factor of 1.8 Ans. 9-28 Purchase the hook having the design shown in Fig. P9-28b. Referring to text Fig. 9-32a, this design reduces peel stresses. 9-29 (a) l/2 1 PÉ cosh(Éx) Ä = dx ¯ l 4b sinh(Él/2) -l/2 l/2 = A1 cosh(Éx) dx -l/2 l/2 A1 = sinh(Éx) É -l/2 A1 = [sinh(Él/2) - sinh(-Él/2)] É A1 = [sinh(Él/2) - (-sinh(Él/2))] É 2A1 sinh(Él/2) = É PÉ = [2 sinh(Él/2)] 4bl sinh(Él/2) P Ä = Ans. ¯ 2bl budynas_SM_ch09.qxd 12/01/2006 16:16 Page 260 FIRST PAGES 260 Solutions Manual " Instructor s Solution Manual to Accompany Mechanical Engineering Design PÉ cosh(Él/2) PÉ (b) Ä(l/2) = = Ans. 4b sinh(Él/2) 4b tanh(Él/2) (c) Ä(l/2) PÉ 2bl K = = Ä 4b sinh(Él/2) P ¯ Él/2 K = Ans. tanh(Él/2) For computer programming, it can be useful to express the hyperbolic tangent in terms of exponentials: Él exp(Él/2) - exp(-Él/2) K = Ans. 2 exp(Él/2) + exp(-Él/2) 9-30 This is a computer programming exercise. All programs will vary.

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