# Example 02: Maximum Concentrated Load a Box Beam Can Carry

**Problem**

A beam is built up by nailing together 25 mm thick planks to form a 200 mm × 250 mm box section as shown. The nails are spaced 125 mm apart and each can carry a shearing force of up to 1.3 kN. The beam is simply supported for a span of 3.6 m and to carry a concentrated load *P* at the third point of the span. The allowable shearing stress of the section is 0.827 MPa.

- Determine the largest value of
*P*that will not exceed the allowable shearing stress of the beam or the allowable shearing force of the nails. - What is the maximum flexural stress of the beam for the load
*P*computed in Part (1)?

**Solution**

$I = \dfrac{200(250^3)}{12} - \dfrac{150(200^3)}{12} = 160\,416\,666.7 ~ \text{mm}^4$

Maximum Shear

$V_{max} = \frac{2}{3}P$

Value of *P* based on the strength of nails

$Q = 150(25)(112.5) = 421\,875 ~ \text{mm}^3$

$s = \dfrac{RI}{VQ}$

$125 = \dfrac{2600(160\,416\,666.7)}{\frac{2}{3}P(421\,875)}$

$P = 11.86 ~ \text{kN}$

Value of *P* based on shear strength of wood

$F_v = \dfrac{VQ}{Ib}$

$0.827 = \dfrac{\frac{2}{3}P(812\,500)}{160\,416\,666.7(50)}$

$P = 12.25 ~ \text{kN}$

For safe value of *P*, use *P* = 11.86 kN *answer for part (1)*

Maximum Flexural Stress

$M_{max} = 1.2(\frac{2}{3}P) = 0.8(11.86)$

$M_{max} = 9.488 ~ \text{kN}\cdot\text{m}$

$f_b = \dfrac{Mc}{I}$

$f_b = \dfrac{9.488(125)(1000^2)}{160\,416\,666.7}$

$f_b = 7.39 ~ \text{MPa}$ *answer for part (2)*