hw6-2008sol

hw6-2008sol - CHAPTER 7 SOLUTION (7.1) Table 7.2: x,...

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< 6S0,!raclure controls. Thus = 265.5(350x 15) = 1,394 kN net = ~(350 X 15) = 2.844 MN able 7.1: with alw = O.07~ A = 1.12 Then It follows t = a(wt) = 81.93(125 x 25) = 256 kN By Eq.(7.3 ,with n = 1: - 4- - 23JiOOO - 81 93 1 L D a - J.... J1f1l - 1.12J1f(20) - lYlrl -1.- 256(10') 975 M"n = (w-tl)t = (0.125-0.02)0.025 = · rQ = 97.5 MPa < 8, 74 Thus, e have = u(2wt) = 208.4(0.5 x 0.025)10 6 = 2,605 kN The DOmin 1stress at fracture - 2.605(10') - 231 6 ) In - 0.025(0.5-0.05) - lYJra This is wel below the yield strength of 1503 MPa. SOLUTION(7.3) FromTabl 7.2: s, = 23.JI000 MPa.Jmm and s, =444 MPa Case B of able 7.1: aJw = 0.16, it = CHAPTER 7 SOLUTION (7.1) Table 7.2: x, = 59.JI000MPa.Jmm and s, = 1503MPa CaseAof able7.1:with a/w=O.l, .,1,=1.01 From Eq.( .3), with n = 1: - - S9.J"iOOO - 208 4 MPa - A...Jlftl - (I.Ol)JK(25) - SOLUTION (7.2)
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:. Fails :. Fails er s = ~(143.3) = 124.9 MPa ± 164.2> 160 '( = .isz. = 16(0.4F) = 31 831F 1C D J 11'(0.04)' , U x = aa + (ib = 143,239.5F 77 82.1 > 80 M=O.8F = 6O(103X78.6)(10~) == 53.3MPa t = ~ = 64.2MPa 13.4(10- 6)(0.0066) 'c Ib Alternativel ) using Eq.(7.11), we have Hence Apply Eq.(7.16): s: =[ ; +3i2]1; 2S01~~O') =F[(143,239.5)2 +3(31,831)2]1 or .~ F=L 63 leN and We have QB = 02(103)(74.85) = 78.6 x 10 3 mm' Q c = B + 6.6(69.7)(34.85) = 94.6 x 10 3 Thus or O"IB = 44.6 MPa U 2B = -19.65 MPa ( t"max) = 82.1 MPa We have Stresses at fixed d: a = 3 M = 32(O.8F) = 127 324F b D 3 11'(0.04)3 , eTa = (O.~)2 4 = 15,915.5F SOLUTION (7.11) SOLUTION (7.10)
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CHAPTERS SOLUTION(8.1)
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This note was uploaded on 05/19/2008 for the course MAE 314 taught by Professor Rabiei,heeter during the Spring '08 term at N.C. State.

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hw6-2008sol - CHAPTER 7 SOLUTION (7.1) Table 7.2: x,...

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