- #1
Dell
- 590
- 0
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??
Last edited by a moderator: