Section I.8 Shear Stress Distribution and Angle of Twist for Two-Cell Thin-Walled Closed Sections

This is a statically indeterminate problem. In order to find the shear flows q1 and q2, the compatibility relation between the angle of twist in cells 1 and 2 must be used. The compatibility requirement can be stated as

where

Note that each closed integral deals with the boundaries of an individual cell, which involve the exterior wall as well as the interior wall separating the two cells. Also notice that the shear flow along the interior wall is the difference between q₁ and q₂. Now by setting the right hand sides of the two angle of twist equations equal to each other, we arrive at the second equation, which together with the torque-shear flow equation result in the following closed form equations for the two shear flows.

where

and

The shear stress at a point of interest is found according to the equation

To find the angle of twist, we could use either of the two twist formulas given above. It is also possible to express the angle of twist equation similar to that for a circular section

with the polar moment of inertia expressed as

A careful scrutiny of the equations for shear flow indicates that the first term in the numerator is all that is different between the two. Therefore, if we have a two-cell thin-walled section in which a₁₀A₂ = a₂₀A₁, then the two shear flows would be equal.

• Example 1 Shear flow distribution, Maximum shear stress, and angle of twist in a two-cell section

To Index Page of Pure Torsion To Section I.7 To Section I.9