Chapter 13
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.
Step-by-Step
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.
