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208_Mechanics Homework Mechanics of Materials Solution

# 208_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 3.53 The composite shaﬁ shown is to be twisted by applying a torque T at end A. PROBLEM 3-53 Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine the largest angle through which end A may be rotated if the following allowable stresses are not to be exceeded 1;”. = 60 MP3 and rm 45 MP3. SOLUTION 72 mm Tm: GK“ : GCMd-xﬂ 54mm L 39 Steel core I 2‘5 m <9 = .324. .- ‘f G Chi-1t. \$0, 3-4011 “def “ﬂ Aluminumjncket2£17 Sleeﬂ com: 214 = sow" Pa.) Cm=ﬁdr 0.027% G = 77m” P4 6 30:! =_._ﬂ___ :- 23,350x10'3 raj/m L (77xtoq)(o.027j Mommy... Jaded-z tawsxw‘t’q, cwz-galro.ossm) G=27xlo" Pq \$1: ”5*‘06 = 46.274 «10'5 rad/m L- (27wto‘iio.osc) smﬂﬂw uaﬁue jouems 1:3! = 23.3eox/o“ rad/M Apﬂowdpe anjpe o'P +u€s+ C?“ = L («£3 -: (2.5)(28360ﬂ04) = 72.15xzo‘srwl -- 4-:3° 4- ...
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