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A 3.5 m long steel member with a W310 x 143 cross- section is subjected to 4.5KNm torque. Knowing that G=77GPa, determine (a) the maximum shearing stress along the line a-a, (b) the maximum shearing stress along the line b-b, (c) the angle of twist
http://lh4.ggpht.com/_H4Iz7SmBrbk/Sysfg6WJmLI/AAAAAAAACBI/SsE4jsJMYAE/22.jpg
i know that
\tau=\frac{M*t}{Jeq}
while Jeq=\frac{1}{3}\sumhibi3
Jeq=1/3*((0.309*0.02293) +(0.309*0.02293) + (0.2872*0.0143))
Jeq=2.7365e-6
\tau=\frac{4500*t}{2.7365e-6}
a)
\taua-a=\frac{4500*0.0229}{2.7365e-6}
=3.7657e7
=37.657MPa
b)
\taub-b=\frac{4500*0.014}{2.7365e-6}
= 2.3022e7
=23.02MPa
(according to my book the correct answers are meant to be 39.7MPa and 24.2MPa)
for the angle
\phi=M*L/(G*Jeq*\eta)
\eta=1.29)
\phi=\frac{4500*3.5}{77e9*2.7365e-6*1.29}
=0.0579rad
= 3.3199 degrees,
but again the correct answer is meant to be 4.72 degrees,
what am i doing wrong??
http://lh4.ggpht.com/_H4Iz7SmBrbk/Sysfg6WJmLI/AAAAAAAACBI/SsE4jsJMYAE/22.jpg
i know that
\tau=\frac{M*t}{Jeq}
while Jeq=\frac{1}{3}\sumhibi3
Jeq=1/3*((0.309*0.02293) +(0.309*0.02293) + (0.2872*0.0143))
Jeq=2.7365e-6
\tau=\frac{4500*t}{2.7365e-6}
a)
\taua-a=\frac{4500*0.0229}{2.7365e-6}
=3.7657e7
=37.657MPa
b)
\taub-b=\frac{4500*0.014}{2.7365e-6}
= 2.3022e7
=23.02MPa
(according to my book the correct answers are meant to be 39.7MPa and 24.2MPa)
for the angle
\phi=M*L/(G*Jeq*\eta)
\eta=1.29)
\phi=\frac{4500*3.5}{77e9*2.7365e-6*1.29}
=0.0579rad
= 3.3199 degrees,
but again the correct answer is meant to be 4.72 degrees,
what am i doing wrong??
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