Question 15.14: Determine the direct stress distribution in the thin-walled ...

The section is antisymmetrical, with its centroid at the mid-point of the vertical web. Therefore, the direct stress distribution is given by either of Eq. (15.17) or (15.18), in which M_{y}=0. From Eq. (15.18),

\sigma_{z}=\left(\frac{M_{y} I_{x x}-M_{x} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}}\right) x+\left(\frac{M_{x} I_{y y}-M_{y} I_{x y}}{I_{x x} I_{y y}-I_{x y}^{2}}\right) y  (15.17)

\sigma_{z}=\frac{M_{x}\left(I_{y y} y-I_{x y} x\right)}{I_{x x} I_{y y}-I_{x y}^{2}}+\frac{M_{y}\left(I_{x x} x-I_{x y} y\right)}{I_{x x} I_{y y}-I_{x y}^{2}}  (15.18)

\sigma_{z}=\frac{M_{x}\left(I_{y y} y-I_{x y} x\right)}{I_{x x} I_{y y}-I_{x y}^{2}}  (i)

The section properties are calculated as follows:

Substituting these values in Eq. (i),

\sigma_{z}=\frac{M_{x}}{h^{3} t}(6.86 y-10.30 x)  (ii)

On the top flange, y=h / 2,0 \leq x \leq h / 2 and the distribution of direct stress is given by

which is linear. Hence,

In the web, -h / 2 \leq y \leq h / 2 \text { and } x=0 Again, the distribution is of linear form and is given by the equation

from which

and

The distribution in the lower flange may be deduced from antisymmetry; the complete distribution is as shown in Fig. 15.35.