# Question 7.3: Find the centroid and the second moment of area about the ho......

Find the centroid and the second moment of area about the horizontal (y) axis of the cross section shown. All dimensions are in millimeters. If a beam is constructed with the cross section shown from steel whose maximum allowable tensile stress is 400 MPa, what is the maximum bending moment that may be applied to the beam?

Given: Dimensions of beam cross section; limiting stress.
Find: Location of centroid; maximum applied moment.
Assume: Hooke’s law applies

Step-by-Step
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The symmetry of the cross section shown suggests that the horizontal coordinate of the centroid will be on the vertical centerline, as sketched. $y_c$ is then 20 cm. We need only locate the vertical location of the centroid ($z_c$ ). Several strategies are available to us. We recognize that the cross section is a large rectangle, with an inner rectangular hole. It will thus be possible for us to find the area and second moment of area of the large outer rectangle, and simply subtract off the properties of the inner rectangle. Recall that $z_c=\int z \mathrm{~d} A / \int \mathrm{d} A=\sum z \mathrm{~d} A / \sum \mathrm{d} A,$ where z and $z_c$ are measured from an arbitrarily chosen reference datum, in this case the top of the cross section. For clarity, results are tabulated as follows:

With the location of the centroid known, we can find the second moment of area with respect to that location, starting with the second moment of area of a rectangle about its own centroid from Appendix A.

If the maximum allowable normal stress is 400 MPa, we can find the maximum moment that can be applied using the relationship:

$\sigma_{\max }=\frac{M c}{I}$.

We have found the second moment of area I, and c is the maximum distance from the centroid attainable on the cross section, in this case 31.7 mm. Solving for M:

$M=\frac{\sigma_{\max } I}{c}=\frac{\left(400 \mathrm{~N} / \mathrm{mm}^2\right)\left(655,000 \mathrm{~mm}^4\right)}{31.7 \mathrm{~mm}}=8.26 \mathrm{kNm}$

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