Answer to Question #85470 in Mechanics | Relativity for ABC

Question

To find the optimum shape for a cross-section of a beam against bending, beam sections of a square, circle and equilateral triangle with the same cross-sectional area (A) are considered. If the beams have the same length (L), Young’s Modulus (E) and support condition, determine which shape of the beam is the best against bending

Answer 1

Beams must withstand two types of loads: elastic bending (defined by the deflection $\delta$) and failure to bending (defined by the stress on the top and bottom surface $\sigma$). So calculate these values for a square, circle, and triangle with the same are A. Let's begin with deflection (smaller $\delta$ is better):

$$\delta=\frac{FL^2}{kEI}=\frac{G}{I},$$

where G is a coefficient equal for all cross-sections and I is moment of inertia of cross-section:

$$\delta_{square}=\frac{G}{A^2/12},$$ $$\delta_{circle}=\frac{G}{A^2/4\pi},$$

$$\delta_{triangle}=\frac{G}{A^2/6\sqrt{3}}.$$

We see that triangles withstand elastic bending better than squares, and squares - better than circles.

Now compute the stress inside the beams that defines failure in bending, and smaller $\sigma$ is better:

$$\sigma=\frac{Mc}{I}=\frac{M}{Z},$$

where c - the smallest distance from the axis of a cross-section to its surface, I - moment of inertia that we already acquainted with. M - moment.

$$\sigma_{square}=\frac{M}{A^{3/2}/6},$$

$$\sigma_{circle}=\frac{M}{A^{3/2}/4\sqrt{\pi}},$$

$$\sigma_{triangle}=\frac{M}{A^{3/2}/12\sqrt[4]{3}},$$

that is why square resists failure in bending better than circle, circle is better than triangle.

Overall result: squares are the best, but actually one needs to choose the cross-section by what load is being considered.