Chapter_9_Solutions

# Chapter_9_Solutions - Chapter 9 Figure for Probs 9-1 to 9-4...

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Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hl allow = 0.707(5)[2(50)](140)(10 3 ) = 49.5 kN Ans . ______________________________________________________________________________ 9-2 Given, b = 2 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hl allow = 0.707(5/16)[2(2)](25) = 22.1 kip Ans . ______________________________________________________________________________ 9-3 Given, b = 50 mm, d = 30 mm, h = 5 mm, allow = 140 MPa. F = 0.707 hl allow = 0.707(5)[2(50)](140)(10 3 ) = 49.5 kN Ans . ______________________________________________________________________________ 9-4 Given, b = 4 in, d = 2 in, h = 5/16 in, allow = 25 kpsi. F = 0.707 hl allow = 0.707(5/16)[2(4)](25) = 44.2 kip Ans . ______________________________________________________________________________ 9-5 Prob. 9-1 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 . ______________________________________________________________________________ 9-6 Prob. 9-2 with E6010 Electrode. Table 9-6: f = 14.85 h kip/in = 14.85(5/16) = 4.64 kip/in Chapter 9, Page 1/36

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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: all min(0.30 , 0.40 ) min[0.30(400), 0.40(220)] min(120, 88) 88 MPa ut y SS  for both materials. Eq. (9-3): F = 0 . 707 hl all = 0 . 707(5)[2(50)](88)(10 3 ) = 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: all min(0.30 , 0.40 ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi ut y for both materials. Eq. (9-3): F = 0 . 707 hl all = 0 . 707(5 / 16)[2(2)](12 . 0) = 10.6 kip Ans. ______________________________________________________________________________ Chapter 9, Page 2/36
9-11 Table A-20: 1035 HR: S ut = 500 MPa, S y = 270 MPa 1035 CD: S ut = 550 MPa, S y = 460 MPa Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.30 , 0.40 ) min[0.30(500), 0.40(270)] min(150, 108) 108 MPa ut y SS  for both materials. Eq. (9-3): F = 0 . 707 hl all = 0 . 707(5)[2(50)](108)(10 3 ) = 38.2 kN Ans. ______________________________________________________________________________ 9-12 Table A-20: 1035 HR: S ut = 72 kpsi, S y = 39.5 kpsi 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: all min(0.30 ) min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi ut y for both materials. Eq. (9-3): F = 0 . 707 hl all = 0 . 707(5 / 16)[2(4)](12 . 0) = 21.2 kip Ans. ______________________________________________________________________________ 9-13 Eq. (9-3):     3 2 100 10 2 141 MPa . 5 2 50 50 F Ans hl   ______________________________________________________________________________ 9-14 Eq. (9-3):   24 0 2 22.6 kpsi . 5/16 2 2 2 F Ans hl ______________________________________________________________________________ 9-15 Eq. (9-3):    3 2 100 10 2 177 MPa . 5 2 50 30 F Ans hl ______________________________________________________________________________ 9-16 Eq. (9-3):   0 2 15.1 kpsi . 5/16 2 4 2 F Ans hl ______________________________________________________________________________ Chapter 9, Page 3/36

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9-17 b = d = 50 mm, c = 150 mm, h = 5 mm, and 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 in MPa):      3 10 2.829 1.414 5 50 y F V F A  Secondary shear , Table 9-1:   22 33 50 3 50 50 3 83.33 10 mm 66 u db d J   J = 0.707 h J u = 0.707(5)(83.33)(10 3 ) = 294.6(10 3 ) mm 4    3 3 175 10 25 14.85 294.6 10 y xy F Mr F J   2 2 max 14.85 2.829 14.85 23.1 y F F   (1) allow 140 6.06 kN .
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## This note was uploaded on 11/20/2011 for the course MCE 325 taught by Professor Ahm during the Spring '11 term at American Dubai.

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Chapter_9_Solutions - Chapter 9 Figure for Probs 9-1 to 9-4...

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