Answer to Question #85420 in Mechanics | Relativity for ABC

Question

Q1.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

The deflection of beams in the vertical direction with the length $L$, the moment of inertia of cross-section $I$ under load $F/k$ is defined by

$$\delta=\frac{FL^3}{kEI}.$$

Thus, since all other properties are equal, the higher $I$, the more stress a beam can withstand. So we will just compare $I$, higher - better.

For a square cross-section with side $a$ it is medium:

$$I_{\Box}=\frac{a^4}{12}=\frac{A^2}{12}.$$

For a circle with radius $r$ it is the smallest:

$$I_{\bigcirc}=\frac{\pi r^4}{4}=\frac{A^2}{4\pi}=\frac{A^2}{12.57}.$$

For a triangle with side $b$ it is the highest:

$$I_{\bigtriangleup}=\frac{b^4}{32\sqrt{3}}=\frac{A^2}{6\sqrt{3}}=\frac{A^2}{10.39}.$$

Triangles rule.