hwk4_5 - F = f l = 4.64[2(2)] = 18.6 kip Ans.

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Unformatted text preview: F = f l = 4.64[2(2)] = 18.6 kip Ans. ______________________________________________________________________________ 9-7 Prob. 9-3 with E7010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85 [5 mm/(25.4 mm/in)] = 2.923 kip/in = 2.923(4.45/25.4) = 0.512 kN/mm F = f l = 0.512[2(50)] = 51.2 kN Ans. ______________________________________________________________________________ 9-8 Prob. 9-4 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in F = f l = 4.64[2(4)] = 37.1 kip Ans. ______________________________________________________________________________ 9-9 Table A-20: 1018 CD: S ut = 440 MPa, S y = 370 MPa 1018 HR: S ut = 400 MPa, S y = 220 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: W all min(0.30Sut , 0.40S y ) min[0.30(400), 0.40(220)] min(120, 88) 88 MPa for both materials. Eq. (9-3): F = 0.707hlW all = 0.707(5)[2(50)](88)(103) = 31.1 kN Ans. ______________________________________________________________________________ 9-10 Table A-20: 1020 CD: S ut = 68 kpsi, S y = 57 kpsi 1020 HR: S ut = 55 kpsi, S y = 30 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: W all min(0.30Sut , 0.40S y ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi for both materials. Eq. (9-3): F = 0.707hlW all = 0.707(5/16)[2(2)](12.0) = 10.6 kip Ans. ______________________________________________________________________________ Chapter 9, Page 2/36 9-17 b = d =50 mm, c = 150 mm, h = 5 mm, and W allow = 140 MPa. (a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kN and W in MPa): F 103 V Wc y A 1.414 5 50 Secondary shear, Table 9-1: Ju d 3b 2  d 2 6 2.829 F 50 ª3 502  502 º ¬ ¼ 6 83.33 103 mm3 J = 0.707 h J u = 0.707(5)(83.33)(103) = 294.6(103) mm4 cc y W x W cc Mry J 175 F 103 25 294.6 103 2 14.85 F W max F cc W x 2  W c  W cc y y F 14.852  2.829  14.85 2 23.1F (1) W allow 23.1 140 6.06 kN Ans. 23.1 (b) For E7010 from Table 9-6, W allow = 21 kpsi = 21(6.89) = 145 MPa 1020 HR bar: S ut = 380 MPa, S y = 210 MPa 1015 HR support: S ut = 340 MPa, S y = 190 MPa Table 9-3, E7010 Electrode: S ut = 482 MPa, S y = 393 MPa The support controls the design. Table 9-4: W allow = min(0.30S ut , 0.40S y ) =min[0.30(340), 0.40(190) = min(102, 76) = 76 MPa The allowable load, from Eq. (1) is 76 Ans. 3.29 kN 23.1 23.1 ______________________________________________________________________________ F W allow 9-18 b = d =2 in, c = 6 in, h = 5/16 in, and W allow = 25 kpsi. Chapter 9, Page 4/36 9-21 Given, b = 50 mm, c = 150 mm, d = 50 mm, h = 5 mm, W allow = 140 MPa. Primary shear (F in kN, W in MPa, A in mm2): F 103 V Wc 1.414 F y A 1.414 5 50  50 Secondary shear: Table 9-1: 6 6 J = 0.707 h J u = 0.707(5)166.7(103) = 589.2(103) mm4 cc y W x W cc Maximum shear: Mry J 175 F 103 (25) 589.2 103 7.425 F Ju b  d 3 50  50 3 166.7 103 mm3 W max F cc W x 2  W c  W cc y y 2 F 7.4252  1.414  7.425 2 11.54 F 140 12.1 kN Ans. 11.54 11.54 ______________________________________________________________________________ W allow 9-22 Given, b = 2 in, c = 6 in, d = 2 in, h = 5/16 in, W allow = 25 kpsi. Primary shear: V A F 1.414 5 /16 2  2 3 Wc y Secondary shear: Table 9-1: Ju 0.5658 F b  d 2  2 3 6 6 J = 0.707 h J u = 0.707(5/16)10.67 = 2.357 in4 cc y W x W cc Mry J 7 F (1) 2.357 2.970 F 10.67 in 3 Maximum shear: W max F cc W x 2  W c  W cc y y 2 F 2.9702  0.566  2.970 2 4.618 F 25 5.41 kip Ans. 4.618 4.618 ______________________________________________________________________________ W allow 9-23 Given, b = 50 mm, c = 150 mm, d = 30 mm, h = 5 mm, W allow = 140 MPa. Chapter 9, Page 8/36 9-43 F = 0, T = 15 kip˜in. Table 9-1: J u = 2S r 3 = 2S (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4 Tr 15 1 13.5 kpsi Ans. J 1.111 ______________________________________________________________________________ W max 9-44 F = 2 kip, T = 0. Table 9-2: A = 1.414 S h r = 1.414 S (1/4)(1) = 1.111 in2 I u = S r 3 = S (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 Wc V A 2 1.80 kpsi 1.111 W cc Mr I 2 6 1 0.5553 21.6 kpsi Ans. ______________________________________________________________________________ 9-45 F = 2 kip, T = 15 kip˜in. Bending: Table 9-2: A = 1.414 S h r = 1.414 S (1/4)(1) = 1.111 in2 I u = S r 3 = S (1)3 = 3.142 in3, I = 0.707(1/4) 3.142 = 0.5553 in4 W max = (Wc 2 + Wcc 2)1/2 = (1.802 + 21.62)1/2 = 21.7 kpsi Wc V A 2 1.80 kpsi 1.111 W cc M Torsion: Table 9-1: Mr I 2 6 1 0.5553 21.6 kpsi J u = 2S r 3 = 2S (1)3 = 6.283 in3, J = 0.707(1/4) 6.283 = 1.111 in4 W cc T Tr J 15 1 13.5 kpsi 1.111 Chapter 9, Page 26/36 W max 9-46 W c2  W cc M  W cc T 2 2 1.802  21.62  13.52 25.5 kpsi Ans. ______________________________________________________________________________ F = 2 kip, T = 15 kip˜in. Bending: Table 9-2: A = 1.414 S h r = 1.414 S h (1) = 4.442h in2 I u = S r 3 = S (1)3 = 3.142 in3, I = 0.707 h (3.142) = 2.221h in4 Wc V A 2 4.442h 0.4502 kpsi h W cc M Torsion: Table 9-1: Mr I 2 6 1 2.221h 5.403 kpsi h J u = 2S r 3 = 2S (1)3 = 6.283 in3, J = 0.707 h (6.283) = 4.442 in4 W cc T W max 2 2 Tr J 15 1 4.442h 2 3.377 kpsi h 2 2 2 W c  W cc M  W cc T Ÿ § 0.4502 · § 5.403 · § 3.377 · ¨ ¸ ¨ ¸ ¨ ¸ ©h¹©h¹©h¹ 20 Ÿ h 0.319 in Ans. 6.387 kpsi h W max W all 6.387 h 3 Should specify a 8 in weld. Ans. ______________________________________________________________________________ 9-47 h 9 mm, d 200 mm, b 25 mm From Table 9-2, case 2: A = 1.414(9)(200) = 2.545(103) mm2 Iu d3 6 2003 6 1.333 106 mm3 I = 0.707h I u = 0.707(9)(1.333)(106) = 8.484(106) mm4 Chapter 9, Page 27/36 Wc V A 2 ª 0.707 6  9 150 º ¬ ¼ F 103 0.3143F W cc W max Materials: Mc I 200 F 103 225 53.69 106 0.8381F W c2  W cc2 F 0.31432  0.83812 0.8951F 1015 HR (Table A-20): S y = 190 MPa, E6010 Electrode(Table 9-3): S y = 345 MPa Eq. (5-21), p. 225 F W all = 0.577(190) = 109.6 MPa W all / n 109.6 / 2 61.2 kN Ans. 0.8951 0.8951 ______________________________________________________________________________ 9-52 In the textbook, Fig. Problem 9-52b 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) F y = 1200 sin 30q = 600 lbf Ans. (c) F x = 1200 cos 30q = 1039 lbf Ans. (d) From Table 9-2, case 6: A 1.414(0.25)(0.25  2.5) 0.972 in 2 d2 2.52 Iu (3b  d ) [3(0.25)  2.5] 3.39 in 3 6 6 The second area moment about an axis through G and parallel to z is I 0.707hI u 0.707(0.25)(3.39) 0.599 in 4 Ans. (e) Refer to Fig. Problem 9-52b. The shear stress due to F y is W1 Fy A 600 0.972 617 psi The shear stress along the throat due to F x is W2 Fx A 1039 0.972 1069 psi The resultant of W 1 and W 2 is in the throat plane Chapter 9, Page 32/36 Wc 2 W 12  W 2 617 2  10692 1234 psi The bending of the throat gives W cc Mc I 439(1.25) 0.599 916 psi The maximum shear stress is W max W c2  W cc2 1234 2  916 2 1537 psi Ans. (f) Materials: 1018 HR Member: E6010 Electrode: n S sy S y = 32 kpsi, S ut = 58 kpsi (Table A-20) S y = 50 kpsi (Table 9-3) 0.577S y 0.577(32) 1.537 12.0 Ans. W max W max (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 in 2 Fx 1039 1662 psi W xy A1 0.625 I bd 2 0.25(2.5) 2 0.260 in 3 c 6 6 At location A, Fy M Vy  A1 I /c 600 439 2648 psi Vy  0.625 0.260 The von Mises stress V c is Vc 2 2 V y  3W xy 26482  3(1662) 2 32 3.912 3912 psi Thus, the factor of safety is, Sy n Vc 8.18 Ans. The clip on the mooring line bears against the side of the 1/2-in hole. If the clip fills the hole Chapter 9, Page 33/36 V n F 1200 td 0.25(0.50) Sy 32(103 )   9600 Vc 9600 psi 3.33 Ans. 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 HR with S ut = 58 kpsi, Eq. (6-8), p. 282, gives Se c Eq. (6-19), p. 287: 0.504Sut 0.504(58) 29.2 kpsi k a = 14.4(58)-0.718 = 0.780 For the size factor estimate, we first employ Eq. (6-25), p. 289, for the equivalent diameter d e 0.808 0.707hb 0.808 0.707(2.5)(0.25) 0.537 in Eq. (6-20), p. 288, is used next to find k b kb § de · ¨ © 0.30 ¸ ¹ -0.107 § 0.537 · ¨ © 0.30 ¸ ¹ -0.107 0.940 Eq.(6-26), p. 290: k c = 0.59 From Eq. (6-18), p. 287, the endurance strength in shear is S se = 0.780(0.940)(0.59)(29.2) = 12.6 kpsi From Table 9-5, the shear stress-concentration factor is K f s = 2.7. The loading is repeatedly-applied W 1.537 2.07 kpsi W a W m K f s max 2.7 2 2 Table 6-7, p. 307: Gerber factor of safety n f , adjusted for shear, with S su = 0.67S ut nf 2 1 § S su · W a ª « 1  ¨ ¸ 2 © W m ¹ S se « ¬ 2 2 ­ ª 2(2.07)(12.6) º ½ 1 ª 0.67(58) º § 2.07 · ° ° « 2.07 » ¨ 12.6 ¸ ® 1  1  « 0.67(58)(2.07) » ¾ 5.55 Ans. 2¬ ¼© ¹° ¬ ¼° ¯ ¿ Attachment metal should be checked for bending fatigue. ______________________________________________________________________________ 9-53 (a) Use b = d = 4 in. 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This note was uploaded on 10/26/2010 for the course ME 433 taught by Professor Cupschalk,s during the Spring '08 term at Old Dominion.

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