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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

Tau=M*t/J

while J=1/3*sigma(h*b^3)

i get J=1/3*(287.2*14^3 + 2*309*22.9)*10^-12

=2.736e-6 m^4

therefore in a-a the shear stress will be

4500*0.0229/(2.736e-6)=37.66MPa

and for b-b

4500*0.014/(2.736e-6)=23.03MPa

now for the angle

phi=M*L/(G*eta*J)

=4500*3.5/(77e9*1.29*2.736e-6)

and i get 0.058rad=3.32degrees

but the correct answers are meant to be

a-a 39.7Mpa

b-b 24.2Mpa

phi 4.72degrees

where have i gone wrong??

someone told me i need to consider the web and the flanges separately and obtain a proportion between the torque exerted on the web and a flange, respectively, by assuming that the resulting angles of twist are equal

how can i do this??

http://lh4.ggpht.com/_H4Iz7SmBrbk/Sysfg6WJmLI/AAAAAAAACBI/SsE4jsJMYAE/22.jpg

i know that

Tau=M*t/J

while J=1/3*sigma(h*b^3)

i get J=1/3*(287.2*14^3 + 2*309*22.9)*10^-12

=2.736e-6 m^4

therefore in a-a the shear stress will be

4500*0.0229/(2.736e-6)=37.66MPa

and for b-b

4500*0.014/(2.736e-6)=23.03MPa

now for the angle

phi=M*L/(G*eta*J)

=4500*3.5/(77e9*1.29*2.736e-6)

and i get 0.058rad=3.32degrees

but the correct answers are meant to be

a-a 39.7Mpa

b-b 24.2Mpa

phi 4.72degrees

where have i gone wrong??

someone told me i need to consider the web and the flanges separately and obtain a proportion between the torque exerted on the web and a flange, respectively, by assuming that the resulting angles of twist are equal

how can i do this??

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