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 [Broken]

i know that

$$\tau$$=$$\frac{M*t}{Jeq}$$

while Jeq=$$\frac{1}{3}$$$$\sum$$hibi3

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)

$$\tau$$a-a=$$\frac{4500*0.0229}{2.7365e-6}$$

=3.7657e7
=37.657MPa

b)

$$\tau$$b-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)

$$\phi$$=$$\frac{4500*3.5}{77e9*2.7365e-6*1.29}$$