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solution-hw7 - 8-15 z A ={0.1054(00'5)1 = 0.442 in2 Am =...

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Unformatted text preview: 8-15 (.) z A, = {0.1054(00'5)1 = 0.442 in2 Am = 0.1054040:1 - 0.753) = 0.552 in2 _ AJE 0.442(30x106) _ h— W = 13 -1.02[10‘)10flm Am. k... =- AmE = °——'552(3°)“°6} =1.27(10‘)1Wm m. 13 13 . 1.02 C — m = 0.445 A)“. 1 I 1 . (b) p—ugb—n-I 5=—~u=—=0.020331l1 - 1" 3 ‘8 A _ fl“ (Ia—0.02033) _ _._, , mm EB Iabl — AE _—D.442{30)(10‘)lpl —9.T9(10 )[le h - _|_{E= {Plfll’o _ _, . -‘ ”ml” AE 0.552(30){106}_7‘35(1° “P '1" lab] + Ifiml=5=0.02083 9.7% 10-013,; + ?.85(ID'T)|P| = 0.020 33 0.020 33 (0) At opening [006% 939(10-‘590 = 0.020 33 0.02033 Pu——-9'79(m_1)—212801bf Ans. Asacheckusefi={l—C)Pg Pn=—f‘—= “810 =212801bf l—C 1—0.445 8-30 (It) 150 M 20 x 2.5 grade 3.3 coarse pitch bolts. lubricated Tablc 8-2 A, = 245 m3 Table 3-11 5,, = 600 MPa A: = :r(20)2/4 = 314.2 mm2 FF = 245mm) = 14? m F.- = 0.905}, = 0.9m147): 132.3 [IN T = 0.18{132.3}(20} = 476 N - m Ans. 01)]. 2: Lo + H = 43 +13 = 66 m.Therefore. seal. = 30mm perTahleA-l?. Lr =2LI+6= 2(2o)+6=45mm Id =L~Lr =30—46=34nnn k, = fl = 314.2(245)(on) = 1251.9 MNIm Adi, + Add 314.2(14) + 245(34) Us: “film at at. Eq. (3-23) A = 0.18715. B = 0.1523 73 ' k... 3:! 20)] —— = — = . 7 0.62873 — = 1.0229 Ed Acxp(LG) 073 15mp[ (43 k... = l.0229[207)(20} = 4235 MN}!!! 1251.9 C = 1251.9 + 4235 Bolts carry 0.228 of the cxlernal loud; members carry 0.772 of the cxtamal load. Am. 111115.th actual loads an: F5. = C? + F.- = 0.223(20) + 132.3 = 136.9 m ,. ={1— C)? — F; = (1 —D.228)20— 132.3 = —116.9kN = 0.228 ...
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