Question

Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable shearing force in each nail is 740 N, determine the largest permissible spacing s between the nails.

Answer 1

Answer:

100.49 mm

Explanation:

Given that :

The vertical shear force = 1200 N

Allowable shearing force in each nail = 740 N

From the diagram attached below, We first determine the horizontal force per unit length on the lower surface of the upper flange

$q = \frac{VQ_{1-1}}{I_{NA}} \ \ \ N/mm$

where;

$I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( \bar {y}^2]$

$I_{NA} = \frac{50*100^3}{12}+2[\frac{150*50^3}{12}+ (150*50)*( 75)^2]$

$I_{NA} = 91.667*10^6 \ mm^4$

Also;

$Q_{1-1} = A \bar{y}$

where A = area above (1-1)

$Q_{1-1} = 50*150*75$

$Q_{1-1} =56.25*10^4 \ \ \ mm^3$

Therefore ;

$q = \frac{1220*56.25*10^4}{91.667*10^6}$

$q = 7.3636 \ N/mm$

Now; the largest permissible spacing s between the nails.  $S_{max} = \frac{740}{7.3636} = 100.49 \ \ \ mm$