Question

a. Using shear flow ideas, explain why the HSS section is torsionally stiffer than the W section

b. Compute the torsional constant for a W14X34 section and compare with the value from the AISC table

c. What is the role of warping restraint on torsional stiffness? Explain with a figure

Answer 1

Ans a) The equation for torsional constant(J) for HSS section can be given by,

J = t b³

where, t = wall thickness

b = width between wall centerline

From this equation we can observe that accounting for the corner radii increases J. The rectilinear corner produces stress concentration and the curved corner produce a more uniform torsional stress flow and hence a a torsionally stiffer section .This is why hollow square section is torsionally stiffer then W section

Ans b) W 14 x 34 section has following dimensions:

Depth (d) = 13.98 in

Width (b) = 6.745 in

Web thickness(t_w) = 0.285 in

Flange thickness( t_f) = 0.455 in

We know, J = (2bt_f³ + d' t_w³) /3

d' = d - t_f

d' = 13.98 - 0.455  

d '= 13.525

Putting values,

J = [2(6.745)(0.455)³ + 13.525(0.285)³ ] /3

J = 0.53 in⁴

Value of J from AISC table for W14x34 section = 0.57

Difference = 0.57 - 0.53

= 0.04

Calculated value differ from value given in table by 7%

Ans c) Warping occurs when twisting of member results in cross section disortation out of plane along the direction of member longitudinal axis .If this out of plane disortation is restrained at any particular cross section , longitudinal shear stress and strain are developed in a member .These induced stress helps to resist torque .Hence, if warping restrain is applied to member , the torsional stiffness may be considerably greater then it would be if the section is allowed to warp freely.