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## Q. 13.4

A steel-rolled section shown in Figure 13.16 is used as a cantilever beam of length 2.0 m. For the loading, calculate the maximum flexural stress developed in the section.

## Verified Solution

The section is antisymmetric with C being its centroid. Therefore,

$\bar{I}_{y y}=2\left[\frac{1}{12}(20)(100)^3+(100)(20)(40)^2\right]+\frac{1}{12}(160)(20)^3 mm ^4$

$=9.84 \times 10^6 mm ^4$

Again,

$\bar{I}_{z z}=2\left[\frac{1}{12}(100)(20)^3+(100)(20)(90)^2\right]+\frac{1}{12}(20)(160)^3 mm ^4$

$=39.36 \times 10^6 mm ^4$

and                  $\bar{I}_{y z}=2[0+(100)(20)(-40)(90)] mm ^4$

$=-14.4 \times 10^6 mm ^4$

Obviously, the fixed end of the beam is subjected to following bending moments:

$M_z=-14\left(10^3\right) \sin 71.6^{\circ} \times 2.0 Nm$

$=-26.57 \times 10^6 Nmm$

and        $M_y=-(14)\left(10^3\right) \cos 71.6^{\circ} \times 2.0 Nm$

$=-8.84 \times 10^6 Nmm$

From generalised flexure equation (refer to Eq. 13.21), we get

$\sigma_{x x}=\left\lgroup \frac{z \bar{I}_{z z}-y \bar{I}_{y z}}{\bar{I}_{y y} \bar{I}_{z z}-\bar{I}_{y z}^2} \right\rgroup M_y+\left\lgroup \frac{z \bar{I}_{y z}-y \bar{I}_{y y}}{\bar{I}_{y y} \bar{I}_{z z}-\bar{I}_{y z}^2} \right\rgroup M_z$              (13.21)

For the neutral axis orientation with respect to z-axis, we set $\sigma_{x x}=0 \text {, so }$

$\frac{y}{z}=\frac{\bar{I}_{z z} M_y+M_z \bar{I}_{y z}}{\bar{I}_{y z} M_y+M_z \bar{I}_{y y}}=\tan \phi$          (Let)

or            $\tan \phi=\frac{(-8.84)(39.36)+(-26.57)(-14.4)}{(-14.4)(-8.84)+(-26.57)(9.84)}=-0.258$

or        $\phi=165.5^{\circ}$

Let us draw the neutral axis in the section as shown in Figure 13.17:

By simple geometry, we note points A and B are furthest from neutral axis and note their ( y, z) coordinates. Thus,

$\left.\sigma_{x x}\right|_{ A }=\frac{1}{(9.84)(39.36)-(-14.4)^2}\left\{\begin{array}{l} {[(16)(39.36)-(100)(-14.4)](-8.89)} \\ +[(10)(-14.4)-(100)(9.84)](-26.57) \end{array}\right\}=76.5 MPa$

We see that identical stress but compressive in nature is developed at point B.