What Are the Shear Stress and Angle of Twist for a Steel Member Under Torque?

AI Thread Summary
The discussion focuses on calculating the shear stress and angle of twist for a steel member subjected to a torque of 4.5 kNm. The user applies the formula for shear stress and calculates Jeq, resulting in values of 37.657 MPa for line a-a and 23.02 MPa for line b-b, which differ from the expected answers of 39.7 MPa and 24.2 MPa. The angle of twist is calculated as 3.32 degrees, while the expected answer is 4.72 degrees. The user suspects that the discrepancy may stem from not accounting for the torque distribution between the flange and web. The conversation seeks clarification on the correct approach to resolve these calculations.
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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??
 
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i think i need to SOMEHOW find the torque seperately on the flange and web and them somehow solve the problem,, please help someone
 

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