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265_Mechanics Homework Mechanics of Materials Solution

# 265_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 3.136 A 3-m-long steel member has an W 250 x 58 cross section. Knowing that G PROBLEM 3-1-36 = 77 GPa and that the allowable shearing stress is 35 MPa, determine (a) the largest torque T which may be applied, (11) the corresponding angle of twist. Refer to Appendix C for the dimensions of the cross section and neglect the effect of stress concentrations. (See hint of Prob. 3.135.) SOLUTION Pygmy: 0.: 203 mm) b? l':'>.\$'trrwmJ Caz-C2: éCi‘O.é30£)'-'O.Squ _ TL . _ 3&2 _ GE CpF‘EEfFé ..TF-czab L - KFL KP = (O.3l‘l£ll(0.203)(0.0135‘)3 = 159.53x/o-q m" : 252 -(2)(l3.53= 223 MM J 15‘- 3 mm I For ma'io‘ninaj {wt/{9} anqpes Tom mm: T = 27;+T., =(2KF++(.,)§§ .E'ﬂ-z T rJ—KT .“ '- 21¢+I¢2 > 77‘ mew» :' “W " .. Tw QVF‘lKvU M T - AWouaJJe Vapue ‘For T leased on aﬂﬂenmiie Venn)? -F)I\ T;- 1}. = (2.01917 = (o.SHQXO.2033(0.0I36‘l‘(35‘xio‘l= 4233 MW, T = (L4H ii??? 37'5“ (was) = Wi-S NW Alﬁe-#11:}: Valid? gar T Ease On @89de Vex/Pile -For Tu! T“, = cab“: = (o.3m)(o.qzs)(o.oo2)‘(3mo‘) = 164.25 N-m T =erms) = 1560 N-m 37.51} Smaller wide 9'25' N-M z-L—hﬂq'sﬁxao") :- I0l.owo'gw = 5.7‘i' (356.6 xto“ )(17 no“) ...
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