beam - 2

[logistic_guy]

If the beam is subjected to a bending moment of \(\displaystyle M = 50 \ \text{kN} \cdot \text{m}\), determine the maximum bending stress in the beam.

 

[logistic_guy]

The maximum bending stress is given by:

\(\displaystyle \sigma_{\text{max}} = \frac{Mc}{I}\)

where \(\displaystyle I\) is the moment of inertia about the neutral axis. So far we don't have it, and we need to derive it!

 

[khansaheb]

and we need to derive it!

This is a very "primitive" problem and can be found as a "solved example" problem in any reputable "solid mechanics" text book. Study it ....

 

[logistic_guy]

This is a very "primitive" problem and can be found as a "solved example" problem in any reputable "solid mechanics" text book. Study it ....

There is no moment of inertia for the shape \(\displaystyle H\).



But I think that we can derive it with some tricks!

 

[logistic_guy]

I will try to derive the moment of inertia of a rectangular cross section that has height \(\displaystyle h\) and width \(\displaystyle w\) about its horizontal centroidal axis \(\displaystyle \rightarrow x\)-axis.

\(\displaystyle I_x = \int y^2 \ dA = \int_{-w/2}^{w/2}\int_{-h/2}^{h/2} y^2 \ dy \ dx = 4\int_{0}^{w/2}\int_{0}^{h/2} y^2 \ dy \ dx = 4\int_{0}^{w/2} \frac{y^3}{3}\bigg|_{0}^{h/2}\ dx\)


\(\displaystyle = 4\frac{h^3}{24}\int_{0}^{w/2} \ dx = 4\frac{h^3}{24}\frac{w}{2} = \textcolor{red}{\frac{wh^3}{12}}\)

 

[logistic_guy]

Now all I can see is that I can imagine that the cross-section of the beam is all rectangular rather than an \(\displaystyle H\) shape. In this way I can calculate the moment of inertia easily, then subtract whatever is missing from the big rectangular cross-section to form the \(\displaystyle H\) shape again. I can also divide the \(\displaystyle H\) cross-section into three parts, and calculate each part separately, but the top and bottom parts will need the parallel axis theorem.

Here is my first approach,

determine the maximum bending stress in the beam.

\(\displaystyle \sigma_{\text{max}} = \frac{Mc}{I} = \frac{50000 \times 0.15}{\frac{1}{12}0.2(0.3)^3 - 2\bigg[\frac{1}{12}0.09(0.26)^3\bigg]} = 4.024 × 10^7 \ \text{Pa} = \textcolor{blue}{40.24 \ \text{MPa}}\)