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624_Mechanics Homework Mechanics of Materials Solution

# 624_Mechanics Homework Mechanics of Materials Solution - 5...

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Unformatted text preview: 5. 149 A cantilever beam AB consisting ofa steel plate of unifonn depth h and variable width b is to support a coneentrated load P at pointA. (0:) Knowing that the beam is to be of mnstant strength, express b in terms of x, L, and be. (b) Determine the smallest allowable value of h if . L =12 in., be =15 in.,P = 3.2 kips, and a." = 24 ksi. PROBLEM 5.149 P SOLUTION {32M}. 2 0 —M- PCL-x\= o M:-P(L-x3 lMl 3-.- PCL-x3 MI if x V :- ._.._. = PCL-X ) M I: *P 8 GA 6242 GLLPX 9' A For a. realm. "30.94% cross Seal-(0M S =- El—b l,“ . ..L. 1. POL-H _ Q? L- 3 E704+Lni G bk " “—63!— b“ .aih—i A+ x=O lo= 5.: 25%“ b=loa(l--§\ .. .n r : gm. , /CG)(3.2}(:2) = _ m .. Solv. % 'Fo In ‘0 6‘" ho q (‘QLHCLg'D O 300 5.150 A cantilever beamAB consisting of a steel plate of Imifonn depth 11 and variable PROBLEM 5'15" width b is to support a distributed load w along its center line AB. (a) Knowing that the beam is to be of constant sh‘engﬂl, express (1 in terms of x, L, and bu, (b) Determine the maximum allowable value ofw ifL =15 in, bo= 18 in, h = 0,75 in, and 0;” = 24 ksi SOLUTION Dip/[3:0 ~M— wCL—xﬁL—"I : o 2 M: -—(—1W L'“ a [M]: “K L-x)‘ 2 2 m. wa-vr S:6;u- 26'.” [email protected] For a. Melange/Paw cross 56°41'01" S = '6'." bk; .r z " A a 1 ﬁ wCL-x} = awn—x) L-|__x o] C. U" ' 262:: b 62;: lo‘ A+x=o 5:59:35: mono-3})“ serving 42°. xw W: M‘ t W“: 0.3a: mm 1. 3 ‘- 31. . ( )05) = 86° Ila/in ...
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