Question

A 3 m × 1 m signboard is subjected to a wind pressure of 7.5 kPa. It is supported by a hollow square pole of outer dimension 250 mm and inner dimension $d$ (unknown). The yield strength is 240 MPa. Find $d$ (round off to nearest integer).

Hint:

For wind-load bending, maximum moment occurs at the ground; use section modulus of hollow square tube.

Answer 1

Correct Answer: 232

Wind force:
\[ F = pA = 7.5\ \text{kPa} \times (3\times1)\ \text{m}^2 = 22500\ \text{N} \] Bending moment at pole base:
Total height = 1/2 board height + 5.5 = 0.5 + 5.5 = 6.0 m
\[ M = F \times 6 = 22500 \times 6 = 135000\ \text{N·m} \] Outer dimension: \(b = 250\ \text{mm} = 0.25\ \text{m}\). Inner dimension: \(d\). Section modulus: \[ Z = \frac{b^4 - d^4}{6b} \] Yield stress relation: \[ \sigma = \frac{M}{Z} = 240\ \text{MPa} \] Convert moment to N·mm: \[ M = 135000\ \text{N·m} = 1.35\times10^8\ \text{N·mm} \] Section modulus required: \[ Z = \frac{M}{\sigma} = \frac{1.35\times10^8}{240} = 5.625\times10^5\ \text{mm}^3 \] Set equal to hollow square expression: \[ \frac{250^4 - d^4}{6(250)} = 5.625\times10^5 \] \[ 250^4 - d^4 = 8.4375\times10^7 \] \[ d^4 = 3.90625\times10^8 - 8.4375\times10^7 \] \[ d^4 = 3.0625\times10^8 \] \[ d = (3.0625\times10^8)^{1/4} \approx 235.8\ \text{mm} \] Rounded to nearest integer: \[ \boxed{236\ \text{mm}} \]