I tried the question but I didn’t get the correct answer, can anyone tell me what mistake I made?


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$$\begin{align} I &= \int y^2\text{d}A \\ &= 25^2\cdot10^{-6}\cdot200\cdot50\cdot10^{-6} \\ &= 25^2\cdot10^{-8} \\ \sigma &= \dfrac{My}{I} \\ &= \dfrac{10^3\cdot25\cdot10^{-3}}{25^2\cdot10^{-8}} \\ &= 4\cdot10^6 \end{align}$$

  • $\begingroup$ Check your y value. For the moment of inertia, y is the full depth of the section, not the distance to the neutral axis. For the stress calculation, it is convention to identify the depth to the neutral axis as c, not y. $\endgroup$ – AsymLabs Feb 5 '18 at 18:48
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    $\begingroup$ Also it is always good practice to show your units and to cancel them out. Gives you an added check on your unit conversions. $\endgroup$ – AsymLabs Feb 5 '18 at 18:52
  • $\begingroup$ Here I denotes the second moment of area not the moment of inertia so it still doesn’t work $\endgroup$ – kamalesh Feb 6 '18 at 1:39
  • $\begingroup$ Check your second moment of area here. $\endgroup$ – AsymLabs Feb 6 '18 at 4:32
  • $\begingroup$ Second Moment of Area = Area Moment of Inertia. $\endgroup$ – AsymLabs Feb 6 '18 at 5:23

While I agree with AndyT's philosophy of simply using the solved equations, it is important to learn the origin of these solutions, so here we go.

The mistake you made was that you didn't perform the integration at all. You simply got the equation for the moment of inertia, ignored the integration sign, set $y$ as half the section's height, and replaced $\text{d}A$ with the section's area. That's not integration.

This is:

$$\begin{align} I &= \int_A y^2\text{d}A \\ &= b\int_{-\frac{h}{2}}^{\frac{h}{2}} y^2\text{d}y \\ &= b \left.\dfrac{y^3}{3}\right|_{-\frac{h}{2}}^{\frac{h}{2}} \\ &= \dfrac{bh^3}{3}\left(\dfrac{1}{8}+\dfrac{1}{8}\right) \\ &= \dfrac{bh^3}{12} \end{align}$$

Once you have the correct moment of inertia, calculating the stress is straightforward.

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  • $\begingroup$ Nice, thorough explanation. $\endgroup$ – AsymLabs Feb 7 '18 at 17:36

For a rectangular section, you don't need to integrate; it's a solved solution:

$$\begin{align} I &= \dfrac{bh^3}{12} \end{align}$$

For the actual error in your integration - I don't know; I'm afraid I left that behind when I graduated 10 years ago. In the real world (as a civil/structural engineer) I've only used simple sections with solved solutions, or used software to calculate it for me.

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  • $\begingroup$ Yep. I was trying to get the OP to draw this conclusion too. His/Her Moment of Inertia was incorrectly solved; I believe it was integrated with respect to the base of the section rather than the neutral axis. Easier and safer to just refer to the closed form solution. $\endgroup$ – AsymLabs Feb 6 '18 at 9:21

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