681_Mechanics Homework Mechanics of Materials Solution

681_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: '_'”1 L I 1—. L l __ .. i L Aw; Ll ,4 } mm PROBLEM 6.20 V 11,- ul- H H x .16: -341- 4 4 (a) (in) Solving ‘Fov h SoDw'na Etna) ‘Fof‘ b maximum values of when}: audL are, respectively, thedepthhanduddthboflhebeam,knowingthatL= Sm,w= 8 kNlm, 1,, = 1.08 MPa, and a_=12 MP3. RA’RE : From sleev- oli'aflmm [VLF “+3- (11 Fm" Y‘ea‘iunswpaw SQW'HOVI A '-' b"! (2) , _g._ V... _ 3wL 73'- ‘ 1 7K ' 8L»?— Ls) Frown Bending momen'i’ anagram h =. Lfm = (530.03MO‘) ‘9 ZZS’IID‘S “'1 26'... (Zulauo‘i '= 9.25 MM -‘ l: IMLnr 3wL 620 A timber beam AB of length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio 1'. la, of the the shearing and normal stresses' in the beam is equal to 2h/L, wL "E" 1 "—"—'-‘- m Din'rJina 33.843 [:7 €7.(C) “—1!“ 8h'I’... the depth and the length of the beam. (1:) Determine 32 3) 8x105 )(5) (3}(2’1910’3 )(Los uto‘) cm «16% = 6mm. ...
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