Torsion

156 CHAPTER 5. TORSION
The following observation gives interesting insight into the torsional behavior of thin rectangular strips.
The torque resulting from the stress distribution, eq. (5.115), is
b
+t/2
b
6M1
τ13 x2 dx2dx3 = dx3
bt3 +t/2
x 2
2 dx2 = M1
. (5.117)
2
0
−t/2
0
This means that the shearing stresses τ13 only account for half of the torque. The other half must be
associated with τ12. Though this stress component is very small, its contribution to the torque is multiplied
by a moment arm b/2, as compared to the moment are t/2 for the τ13 component.
Finally, the membrane analogy provides further insight into the behavior of bars with non circular cross
sections. Specifically, consider the distribution of shear stress in the regions on the cross section near a
convex corner, for example, a corner of a rectangular section. It should be intuitively obvious that the slope
of a membrane in this area will be relatively flat compared the the slope near the midpoints of the long edges,
for example. Since the direction of the shear stress is along a contour of the membrane and the magnitude
is proportional to the slope perpendicular to the contour, we can conclude that the corner regions do not
experience significant shear stresses. This situation is even more pronounced for a cross section in the shape
of an equilateral triangle. In fact, for this kind of shape, the shear stresses and torsional stiffness are not
significantly better than what would be obtained for a circular cross section sized to match the largest circle
that can be inscribed within the triangle.
−t/2