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1336_Mechanics Homework Mechanics of Materials Solution

# 1336_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROBLEM ll 38 11.38 RodAC is made of aluminmn (G = 73 GPa) and is subjected to a torque ' T applied at end C. Knowing that portion BC ofthe rod is hollow and has an inside diameter of 16 mm, determine the strain energy of the rod for a minimum shearing stress of 120 MP3. SOLUTION Co=%‘ IQMM: C5=é7ri=8mm TTCJT E02)“: 32.572vioswmq= 32.572’16qh.‘ 3" Jae * 7f'§r(c:.,“-c:.-‘§= ELOT— 8*)= 26.!381‘10‘wm' : 2a.:33xlo‘” m” 1w=§i T:.~Img.£~£:£’%'3lg———%~——W: 261.33 Now -2 7 1 .s ”as: 2‘6 ET: = (Z)£§:Ixi:))g::.o;7l20wl?b") = 5'7“ J U“: 2,13: 7(naggiﬁggﬁgigw) = 3-95" J I Tow U= me + U“ r (4.70 I - 11.39- In the assembly shown torques TA and T, are exerted on disks A and B respectively. Knowing that both shafts are solid and made of aluminum (G = 73 GPa), determine the total energy acquired by the assembly. . PROBLEM 11.39 SOLUTION Ouer Par‘h’om AB The 7 T. "' 300 NW“ JAB? g C" = g’ (129),. 7 79.52‘!03 M»: 5 77-5/1“”; ”1' LAB: 0.9 m U r T .J 1-49.: M '5 ZGJ»: (21(73ﬁloqx7‘f.52x10'43 _ = 6.q77 J Over P°"+"°" BC= Ta = T. +Ts .= 300+qu = 700 u-m ,, Lac =o.7\$m J}.= E015)“ 437.57,:103 mm” : 439.57xlo“m” U _ T‘L _ (700)2(035) u- _£S-—J-ﬁ- _ : 5.72.6 J '2 9 Joe. (2 )(73 “aﬂoat. 57:40“) — U” + [53¢ = e.€m+ 572s - 12.70 J" 4 ...
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