# Example 01: Spacing of Screws in Box Beam made from Rectangular Wood

**Problem**

A concentrated load *P* is carried at midspan by a simply supported 4-m span beam. The beam is made of 40-mm by 150-mm timber screwed together, as shown. The maximum flexural stress developed is 8.3 MPa and each screw can resist 890 N of shear force.

- Determine the spacing of screws at
*A*. - Determine the spacing of screws at
*B*.

**Solution**

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$I = \sum \dfrac{bd^3}{12} = \dfrac{190(190^3)}{12} - \dfrac{110(110^3)}{12}$

$I = 96\,400\,000 ~ \text{mm}^4$

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

$8.3 = \dfrac{M(190/2)(1000^2)}{96\,400\,000}$

$M = 8.422 ~ \text{kN}\cdot\text{m}$

$M = \dfrac{PL}{4}$

$8.422 = \dfrac{P(4)}{4}$

$P = 8.422 ~ \text{kN}$

$V_{max} = 0.5P = 4.211 ~ \text{kN}$

**Spacing of Screws**

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

Part 1: Spacing of screws at *A*

$R = 2 \times 890 = 1,780 ~ \text{N}$

$s = \dfrac{1,780(96\,400\,000)}{4\,211(570\,000)} = 71.49 ~ \text{mm}$

Use *s* = 70 mm *answer*

Part 2: Spacing of screws at *B*

$R = 2 \times 890 = 1,780 ~ \text{N}$

$s = \dfrac{1,780(96\,400\,000)}{4\,211(330\,000)} = 123.48 ~ \text{mm}$

Use *s* = 123 mm *answer*

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