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Dell
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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
[tex]\tau[/tex]=[tex]\frac{M*t}{Jeq}[/tex]
while Jeq=[tex]\frac{1}{3}[/tex][tex]\sum[/tex]hibi3
Jeq=1/3*((0.309*0.02293) +(0.309*0.02293) + (0.2872*0.0143))
Jeq=2.7365e-6
[tex]\tau[/tex]=[tex]\frac{4500*t}{2.7365e-6}[/tex]
a)
[tex]\tau[/tex]a-a=[tex]\frac{4500*0.0229}{2.7365e-6}[/tex]
=3.7657e7
=37.657MPa
b)
[tex]\tau[/tex]b-b=[tex]\frac{4500*0.014}{2.7365e-6}[/tex]
= 2.3022e7
=23.02MPa
(according to my book the correct answers are meant to be 39.7MPa and 24.2MPa)
for the angle
[tex]\phi[/tex]=M*L/(G*Jeq*[tex]\eta[/tex])
[tex]\eta[/tex]=1.29)
[tex]\phi[/tex]=[tex]\frac{4500*3.5}{77e9*2.7365e-6*1.29}[/tex]
=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
[tex]\tau[/tex]=[tex]\frac{M*t}{Jeq}[/tex]
while Jeq=[tex]\frac{1}{3}[/tex][tex]\sum[/tex]hibi3
Jeq=1/3*((0.309*0.02293) +(0.309*0.02293) + (0.2872*0.0143))
Jeq=2.7365e-6
[tex]\tau[/tex]=[tex]\frac{4500*t}{2.7365e-6}[/tex]
a)
[tex]\tau[/tex]a-a=[tex]\frac{4500*0.0229}{2.7365e-6}[/tex]
=3.7657e7
=37.657MPa
b)
[tex]\tau[/tex]b-b=[tex]\frac{4500*0.014}{2.7365e-6}[/tex]
= 2.3022e7
=23.02MPa
(according to my book the correct answers are meant to be 39.7MPa and 24.2MPa)
for the angle
[tex]\phi[/tex]=M*L/(G*Jeq*[tex]\eta[/tex])
[tex]\eta[/tex]=1.29)
[tex]\phi[/tex]=[tex]\frac{4500*3.5}{77e9*2.7365e-6*1.29}[/tex]
=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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