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

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SUMMARY

The discussion focuses on calculating the shear stress and angle of twist for a W310 x 143 steel member subjected to a torque of 4.5 kNm. The maximum shearing stress along line a-a is calculated as 37.657 MPa, while along line b-b it is 23.02 MPa, which differ from the expected values of 39.7 MPa and 24.2 MPa, respectively. The angle of twist is computed as 3.3199 degrees, but the expected answer is 4.72 degrees. The user suspects that the torque distribution between the flange and web may need to be considered for accurate results.

PREREQUISITES
  • Understanding of shear stress calculations in structural engineering
  • Familiarity with the concepts of torque and angle of twist
  • Knowledge of the properties of steel, specifically G=77 GPa
  • Ability to perform calculations involving equivalent polar moment of inertia (Jeq)
NEXT STEPS
  • Investigate the distribution of torque between the flange and web of the steel member
  • Learn about the calculation of equivalent polar moment of inertia (Jeq) for different cross-sections
  • Study the effects of varying shear stress along different lines in structural members
  • Explore methods for accurately determining the angle of twist in torsional loading scenarios
USEFUL FOR

Structural engineers, mechanical engineers, and students studying torsion in materials will benefit from this discussion, particularly those working with steel members and torque calculations.

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

[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??
 
Last edited by a moderator:
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i think i need to SOMEHOW find the torque separately on the flange and web and them somehow solve the problem,, please help someone
 

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