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

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

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

Value of P based on the strength of nails

$R = 2(1300) = 2600 ~ \text{N}$

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

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

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

$P = 11.86 ~ \text{kN}$

Value of P based on shear strength of wood

$Q = 200(125)(62.5) - 150(100)(50) = 812\,500 ~ \text{mm}^3$

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

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

$P = 12.25 ~ \text{kN}$

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

Maximum Flexural Stress

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

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

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

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

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

Example 02: Maximum Concentrated Load a Box Beam Can Carry

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