Question

Compare the percentage increase in the bending moment that can be carried to produce
the same maximum bending stress when a rectangular beam section is
1) Doubled in breadth, and 2) Doubled in depth.

Answer 1

Question #78923

Compare the percentage increase in the bending moment that can be carried to produce the same maximum bending stress when a rectangular beam section is

1) Doubled in breadth, and 2) Doubled in depth.

Answer:

The maximum bending moment that can be carried by a beam is given by:

$$M = \sigma W,$$

where $\sigma$ is a bending stress,

$W = \frac{1}{6} bh^2$ – the elastic section module of a rectangular beam section,

$b$ and $h$ are, respectively, the breadth and the depth of the cross section of a beam.

From (1) we can determine the increase in the maximum bending moment with the increase of the cross section sizing as follow:

$$\frac{M_i}{M} = \frac{\sigma W_i}{\sigma W},$$ $$\frac{M_i}{M} = \frac{b_i}{b} \left(\frac{h_i}{h}\right)^2.$$

The percentage increase could be determined by:

$$\varepsilon_M = \left(\frac{b_i}{b} \left(\frac{h_i}{h}\right)^2 - 1\right) \cdot 100\%.$$

Thus, in case 1) with doubled breadth ($b_i = 2b, h_i = h$) we have increase in the maximum bending moment by:

$$\varepsilon_M = \left(\frac{2b}{b} \left(\frac{h}{h}\right)^2 - 1\right) \cdot 100\% = (2 - 1) \cdot 100\% = 100\%.$$

In case 2) with doubled depth ($b_i = b, h_i = 2h$) we have increase in the maximum bending moment by:

$$\varepsilon_M = \left(\frac{b}{b} \left(\frac{2h}{h}\right)^2 - 1\right) \cdot 100\% = (4 - 1) \cdot 100\% = 300\%.$$