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643_Mechanics Homework Mechanics of Materials Solution

# 643_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 5.169 For the beam and loading showa, detemu'ne (a) the maximum value of the PROBLEM 5'169 bending moment, (b) the: maximum normal stress due to bending. SOLUTION E ‘DEM = 0 "10 E; + (‘I‘-\$K!\$)(3) -.- c: \$10x25.4 R:- f 20‘ 25 k-‘Ps +32%: 0 lo 1?, —(5.s)(:s)(3)= 0 RD = 24.75 INF: Shear: VA = 0 v5: 0 — (2K3) = - 6 Rip; v6: —6 + 20.25 =- 14. 25 kcps VD— = H.25 — (Ions) = - £5.75 ,krps VD+ = — 15.75 + 24.75 = 6! laps V3 7 q - (35(33 '-’ O era‘tS Locw‘l‘e Poi-r} E where V = 0 _e._ _ M. - 14.25 ‘ 15.73 303 ‘ ”2'5 er ms H. Io—e. = 5.25% Areas. Uncle» shear alu‘qgf‘am A +0 5 SVaht =(ir)(2\<~e\ r —e lap-H +o E Svalx *(é)(4.75)(w.25)= 33,34 kip-H +o ‘D SVAX . (§)(5.as)(15:7r)?—'-ll.3¢l lap-H +v B SVJx = mmm = 13.5 kcpﬁ Bending Mame-«+5: M = O - 6 Ma r — 5+ 33.84 Mo 3' 27.84 - ‘H. 3‘} P45 = (2 Maximum N4] = 27.34 MPH -—- : 334.: tap-in For Foiledl Si‘eef sec‘l’I'om 8 '0 V 25.4 S = Maximo-n normal? 54-12-55 5“,? mm = 33356-1 - 13.53 ks? S ...
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