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## Q. 13.3

A beam with the section shown in Figure 13.14(a) is subjected to bending. Calculate the maximum tensile and compressive stresses in the beam and determine the orientation of the neutral axis. ## Verified Solution

From Figure 13.14(b), we can calculate the centroidal coordinates $\left(\bar{y}_{ C }, \bar{z}_{ C }\right)$ of the section as:

$\bar{y}_{ C }=\frac{(200)(25)(12.5)+(300-25)(25)(162.5)}{(200)(25)+(275)(25)}=99.34 mm$

and          $\bar{z}_{ C }=\frac{(200)(25)(100)+(275)(25)(12.5)}{(200)(25)+(275)(25)}=49.34 mm$

Next, we calculate $\bar{I}_{y y}, \bar{I}_{z z} \text { and } \bar{I}_{y z}$ for the given section:

\begin{aligned} \bar{I}_{y y}=& {\left[\frac{1}{12}(25)^3(275)+(25)(275)(49.34-12.5)^2\right] } \\ &+\left[\frac{1}{12}(25)(200)^3+(25)(200)^3(100-49.34)^2\right] \end{aligned}

or            $\bar{I}_{y y}=39.19\left(10^6\right) mm ^4$

\begin{aligned} \bar{I}_{z z}=& {\left[\frac{1}{12}(25)(275)^3+(25)(275)(1632.5-99.34)^2\right] } \\ &+\left[\frac{1}{12}(200)(25)^3+(25)(200)(99.34-12.5)^2\right] \end{aligned}

or          $\bar{I}_{z z}=108.72\left(10^6\right) mm ^4$

and              \begin{aligned} \bar{I}_{y z}=& {[0+(25)(275)(49.34-12.5)(99.34-162.5)] } \\ &+[0+(25)(200)(99.34-12.5)(49.34-100)] \end{aligned}

$\bar{I}_{y z}=-37.99\left(10^6\right) mm ^4$

Now, bending moment on the section is

$M_z=-40 kN m =-40 \times 10^3 kN m$

and from the generalised flexure equation,

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

[refer Eq. (13.21) with $\left.M_y=0\right]$

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

Now for NA, $\sigma_{x x}=0$, so

$\frac{y}{z}=\tan \phi=\frac{\bar{I}_{y z}}{\bar{I}_{y y}}$

where $\phi$ is the angle made by neutral axis with z-axis. So,

$\tan \phi=-\frac{37.99}{39.19} \Rightarrow \phi=135.9^{\circ}$

The neutral axis is drawn in Figure 13.15:

From the above figure, we note points A, B and D are furthest from NA. Thus, noting (y, z) coordinates of A, B and D from the figure, we calculate the normal stresses developed as

$\left.\sigma_{x x}\right|_{ A }=\frac{(-24.34)(-37.99)-(200.66)(39.19)}{(39.19)(108.72)-(37.99)^2} \times \frac{1}{10^6}(-40)\left(10^6\right)$

= +98.52 MPa

$\left.\sigma_{x x}\right|_{ D }=\frac{(150.66)(-37.99)-(-74.34)(39.19)}{(39.19)(108.72)-(37.99)^2} \times(-40)$

= +39.9 MPa

and

$\left.\sigma_{x x}\right|_{ B }=\frac{(-49.34)(-37.99)-(-99.34)(39.19)}{(39.19)(108.72)-(37.99)^2} \times(-40)$

= −81.88 MPa

Thus, maximum tensile stress is $\sigma_{ A }$ = 98.52 MPa and maximum compressive stress is $\sigma_{ B }$ = 81.88 MPa. 