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394_Mechanics Homework Mechanics of Materials Solution

394_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 4.131 Knowing that the allowabIc stress is 150 MPa in section a-a of the hanger shown, dctennine (a) the largest vertical force P that can be applied at point A, (b) the corresponding location ofthc neutral axis of section a—a. PROBLEM 4.132 4.132 Solve Prob. 4.13], assuming that the vertical forcc P is applied at point B. SOLUTION Loans}: cani'won'ai , _l20wl03 2400 ‘: 50 MM The CGM‘i‘V‘oiei it" .50 MM 4*: +4.: N‘sl'} 0'? H1: 92'“ ecije o'P ‘rhe sec-Hun. Pe Section a-a Bananas couloje M ‘-‘ e '-" JD - :20 e -0.07D m L: 11- I: gnome? +(laoo\(203‘ r mm: n...“ q fi(éo)(2o)3+(l2°o)(2o)‘ = 520x In“ M I, 4 I; = 1.360 no" t...“ = Latovto" m“ a A =2¥oov/d‘m" (03 Baseci Oh when ail pe'H' eels: d'F-SOO‘iI'oW: 3:430 MM 2 —0.06‘o,m %- car- Km _ _ e _ _§-o.a701§-O.°:a) _ _ . -1. KL— 3L 3! - “moon” LSGano" - :2. 1569x103 w. 6' ~15'ouo‘ '- E- : 4.1561qu 84.584 an sires: wi N3 H ”be a? seal-{am -‘ KR P = 69.61103 N d '-‘ 30mm = 0.030m .. hex _ I _[—9,o7o)(o-0301 2 v03 -1 K? _ i. I - 240mm" [.360Xlo" 1.47903 ( m 6- _ i-S'DKIO‘ 3 P R "R" " LQGO‘BII 76.5 KID iv CLooSe Ha smflw value P: emewos N =' 67.; an (5) Locati't‘om a? Hed‘i‘v‘a) «M‘s: 6'10 P _ e 6‘ = 7A3 — EQI — o .3! 1i" .. I __ L360 we“ 31 _. AQ. - (2400x[0" )(-.0.070J Nev-in) and: fixes 5'0- 8-[0 = d ll = - 8.10 .xlo'sm = - 8.10 Wm 43.4 um “From fleP-l edge, g} ...
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