Cross section and torque- please help

  • Thread starter 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 [Broken]

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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Answers and Replies

  • #2
590
0
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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