Training Civil Engineering Bending Stress & the Flexure Formula
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Bending Stress & the Flexure Formula

24 min Civil Engineering

Bending Stress & the Flexure Formula

When a beam bends under load, internal stresses develop. The flexure formula relates bending moment to stress.

Flexure Formula

$$\sigma = \frac{M \cdot c}{I}$$

where $\sigma$ = bending stress (Pa), $M$ = bending moment (N·m), $c$ = distance from the neutral axis to the outermost fiber (m), $I$ = moment of inertia of the cross section (m⁴).

Moment of Inertia — Common Shapes
Shape$I$
Rectangle ($b \times h$)$\dfrac{bh^3}{12}$
Circle (diameter $d$)$\dfrac{\pi d^4}{64}$
Hollow circle ($d_o$, $d_i$)$\dfrac{\pi(d_o^4 - d_i^4)}{64}$
Example 1

A rectangular beam is 200 mm wide and 400 mm deep. The maximum bending moment is 50 kN·m. Find the maximum bending stress.

$I = \frac{0.2 \times 0.4^3}{12} = \frac{0.2 \times 0.064}{12} = 1.067 \times 10^{-3}$ m⁴

$c = 0.2$ m

$$\sigma = \frac{50{,}000 \times 0.2}{1.067 \times 10^{-3}} = 9.37 \text{ MPa}$$

Example 2

A steel I-beam has $I = 8.5 \times 10^{-5}$ m⁴ and depth of 310 mm. If yield stress is 250 MPa, what is the maximum moment it can carry?

$c = 0.155$ m

$$M = \frac{\sigma \cdot I}{c} = \frac{250 \times 10^6 \times 8.5 \times 10^{-5}}{0.155} = 137 \text{ kN·m}$$

Example 3

A circular shaft of diameter 50 mm is subjected to a bending moment of 800 N·m. Find the maximum bending stress.

$I = \frac{\pi (0.05)^4}{64} = 3.068 \times 10^{-7}$ m⁴, $c = 0.025$ m

$$\sigma = \frac{800 \times 0.025}{3.068 \times 10^{-7}} = 65.2 \text{ MPa}$$

Practice Problems

1. Rectangular beam 150 mm × 300 mm, $M = 30$ kN·m. Find $\sigma_{\max}$.
2. Circular beam, $d = 80$ mm, $M = 2$ kN·m. Find $\sigma_{\max}$.
3. Rectangular beam 250 mm × 500 mm, $\sigma_{\text{allow}} = 150$ MPa. Find $M_{\max}$.
4. Find $I$ for a hollow circular section with $d_o = 100$ mm, $d_i = 60$ mm.
5. A W310×21 steel beam has $I = 37 \times 10^{-6}$ m⁴ and depth 303 mm. Find $\sigma$ when $M = 25$ kN·m.
6. If the allowable stress is 165 MPa for the beam in #5, find the maximum moment.
7. A timber beam 100 mm × 200 mm spans 3 m with a central load. The allowable stress is 10 MPa. Find the maximum load $P$.
8. A cantilever of length 2 m carries a point load at the free end. Rectangular section 120 mm × 240 mm, $\sigma_{\text{allow}} = 12$ MPa. Find $P_{\max}$.
9. Compare $I$ for a 100 × 200 mm rectangle oriented tall vs. flat.
10. A steel pipe, $d_o = 200$ mm, $d_i = 180$ mm, $M = 15$ kN·m. Find $\sigma_{\max}$.
11. Why is a beam oriented tall (large $h$) more efficient in bending?
12. A beam must resist $M = 80$ kN·m with $\sigma_{\text{allow}} = 200$ MPa. Find the required section modulus $S = I/c$.
Show Answer Key

1. $I = 3.375 \times 10^{-4}$ m⁴, $c = 0.15$ m; $\sigma = 13.3$ MPa

2. $I = 2.011 \times 10^{-6}$ m⁴, $c = 0.04$ m; $\sigma = 39.8$ MPa

3. $I = 2.604 \times 10^{-3}$ m⁴, $c = 0.25$ m; $M = 1{,}563$ kN·m

4. $I = \frac{\pi(0.1^4 - 0.06^4)}{64} = 4.27 \times 10^{-6}$ m⁴

5. $c = 0.1515$ m; $\sigma = 102.4$ MPa

6. $M = 165 \times 10^6 \times 37 \times 10^{-6}/0.1515 = 40.3$ kN·m

7. $M_{\max} = PL/4$; $I = 6.667 \times 10^{-5}$ m⁴, $c=0.1$ m; $M_{\text{allow}} = 6.667$ kN·m; $P = 4M/L = 8.89$ kN

8. $M = PL$; $I = 1.382 \times 10^{-4}$ m⁴, $c=0.12$ m; $M_{\text{allow}} = 13.82$ kN·m; $P = 6.91$ kN

9. Tall: $I = \frac{100 \times 200^3}{12} = 6.667 \times 10^7$ mm⁴. Flat: $I = \frac{200 \times 100^3}{12} = 1.667 \times 10^7$ mm⁴. Tall is 4× stiffer.

10. $I = \frac{\pi(0.2^4 - 0.18^4)}{64} = 2.7 \times 10^{-5}$ m⁴; $\sigma = 27.8$ MPa

11. $I \propto h^3$; doubling height increases $I$ by 8× while $c$ only doubles — net gain of 4× in section modulus.

12. $S = M / \sigma = 80{,}000 / (200 \times 10^6) = 4 \times 10^{-4}$ m³ = 400 cm³

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