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## Q. 13.7

A beam having the cross-section shown in Figure 13.22(a) is subjected to a bending moment. Calculate the largest permissible value of the bending moment, if the maximum bending stress is not to exceed 100 MPa.

## Verified Solution

1. From Figure 13.22(b), we can compute the area moment of inertia about y′ and z′ axes as

$I_{y^{\prime} y^{\prime}}=\iint_A z^{\prime 2} d A=\int_0^{b} \int_0^{(h/b)y^{\prime}} z^{\prime 2} d y^{\prime} d z^{\prime}$

$=\int_0^b\left\lgroup \int_0^{(h / b) y^{\prime}} z^{\prime 2} d z^{\prime} \right\rgroup d y^{\prime}=\frac{1}{3} \int_0^b\left\lgroup \frac{h}{b} \right\rgroup^3 y^{\prime 3} d y$

$=\frac{1}{12}\left\lgroup \frac{h}{b} \right\rgroup^3\left[y^{\prime 4}\right]_0^b=\frac{1}{12} b h^3$

Therefore,

$\bar{I}_{y y}=I_{y^{\prime} y^{\prime}}-A\left\lgroup \frac{h}{3} \right\rgroup^2=\frac{1}{12} b h^3-\frac{1}{12} b h\left\lgroup \frac{h}{3} \right\rgroup^2$

$=\frac{1}{12} b h^3-\frac{1}{18} b h^3=\frac{b h^3}{36}$

or                $\bar{I}_{y y}=\frac{b h^3}{36}$          (1)

Similarly,

$I_{z^{\prime} z^{\prime}}=\iint_A y^{\prime 2} d A$

$=\int_0^b \int_{(b / z) z^{\prime}}^b y^{\prime 2} d y^{\prime} d z^{\prime}$

$=\frac{1}{3} \int_0^h\left\lgroup b^3-\frac{b^3}{h^3} z^{\prime 3} \right\rgroup d z^{\prime}$

$=\frac{1}{3}\left\lgroup b^3 z^{\prime}-\frac{b^3}{4 h^3} z^{\prime 4} \right\rgroup_0^h=\frac{1}{3}\left\lgroup b^3 h-\frac{b^3 h}{4} \right\rgroup$

$=\frac{b^3 b}{4}$

Therefore,

$\bar{I}_{z z}=I_{z^{\prime} z^{\prime}}-A\left\lgroup\frac{2 b}{3} \right\rgroup^2=\frac{b^3 h}{4}-\frac{1}{2} b h\left\lgroup\frac{2 b}{3} \right\rgroup^2$

$=b^3 h\left\lgroup \frac{1}{4}-\frac{2}{9} \right\rgroup =\frac{b^3 h}{36}$

or          $\bar{I}_{z z}=\frac{b^3 h}{36}$            (2)

Finally,

$I_{y^{\prime} z^{\prime}}=\iint_A y^{\prime} z^{\prime} d A$

$=\int_0^{b} \int_0^{(h / b) y^{\prime}} y^{\prime} z^{\prime} d y^{\prime} d z^{\prime}=\int_0^{b} \int_0^{(h / b) y^{\prime}} y^{\prime} z^{\prime} d z^{\prime} d y^{\prime}$

$=\int_0^b y^{\prime}\left[\int_0^{(h / b) y^{\prime}} z^{\prime} d z^{\prime}\right] d y^{\prime}$

$=\frac{1}{2}\left\lgroup \frac{h}{b} \right\rgroup^2 \int_0^b y^{\prime 3} d y^{\prime}=\frac{1}{8}\left\lgroup \frac{h}{b} \right\rgroup^2 b^4=\frac{1}{8} b^2 h^2$

Therefore,

$\bar{I}_{y z}=I_{y^{\prime} z^{\prime}}-A\left\lgroup \frac{h}{3} \right\rgroup\left\lgroup \frac{2 b}{3} \right\rgroup=\frac{1}{8} b^2 h^2-\frac{1}{2} b h\left\lgroup \frac{2 b h}{9} \right\rgroup$

$=b^2 h^2\left\lgroup \frac{1}{8}-\frac{1}{9} \right\rgroup=\frac{b^2 h^2}{72}$

or              $\bar{I}_{y z}=\frac{b^2 h^2}{72}$              (3)

2. Hence, from detailed derivation of $\bar{I}_{y y}, \bar{I}_{z z} \text { and } \bar{I}_{y z}$ from Eqs. (1)–(3) in Step 1, we obtain for our section with b = h = 60 mm as

$\bar{I}_{y y}=\frac{(60)^4}{36}=\bar{I}_{z z} \Rightarrow \bar{I}_{y y}=\bar{I}_{z z}=0.36\left(10^6\right) mm ^4$

and        $\bar{I}_{y z}=\frac{(60)^4}{72} \Rightarrow \bar{I}_{y z}=0.18\left(10^6\right) mm ^4$

From the expression of generalised flexure stress from Eq. (13.21), we obtain with $M_y=0$.

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

$\sigma_{x x}=\frac{1}{\left(\bar{I}_{y y} \bar{I}_{z z}-\bar{I}_{y z}^2\right)^2}\left[z \bar{I}_{y z}-y \bar{I}_{y y}\right] M_y$

$\left.=\frac{(0.18 z-0.36 y) M_{ O }}{0.0972}\left(10^{-6}\right) MPa \quad \text { (if } M_{ O } \text { is in } N mm \right)$

For NA orientation with respect to z-axis, we get

$\tan \phi=\frac{y}{z} \quad\left(\text { by putting } \sigma_{x x}=0\right)$

$\text { So, } \tan \phi=0.18 / 0.36 \Rightarrow \phi=26.6^{\circ}$.  The position of NA is shown in Figure  13.23.

From the geometry of the section, it is obvious that NA passes through the point P and bisects the side MN. Thus, points M and N on the beam section are equidistant points of the section from NA.
Considering magnitude of stress at point M, we obtain

$\left|\sigma_{ M }\right|=\left|\sigma_{ N }\right|=\left|\frac{(0.18)(-20)-(0.36)(20)}{0.0972}\right|\left(10^{-6}\right) M_{ O }=111.11 \times 10^{-6} M_{ O }$

From the given condition, we obtain

$\left|\sigma_{ M }\right| \leq 100 MPa$

Therefore,

$111.11\left(10^{-6}\right) M_{ O }=100 \Rightarrow M_{ O }=900 Nm$

So $M_{\max }$ for the section is 900 N m.