Solution to Problem 593 | Spacing of Rivets or Bolts in Built-Up Beams

$\Sigma M_{R2} = 0$

$12R_1 = 8P $

$R_1 = \frac{2}{3}P$
 

$\Sigma M_{R1} = 0$

$12R_2 = 4P$

$R_2 = \frac{1}{3} P$
 

 

$M_{max} = 4(\frac{2}{3}P) = \frac{8}{3}P \, \text{ lb}\cdot\text{ft}$
 

Based on allowable shearing force of beam
$f_v = \dfrac{VQ_{NA}}{Ib}$
 

Where:
V = 2/3 P
Q_NA = 6(1)(4.5) + 2 [ 5(1)(2.5) ] = 52 in³
I = 8(10³)/12 - 6(8³)/12 = 410.67 in⁴
b = 2 in
f_v = 120 psi

 

Thus,
$120 = \dfrac{\frac{2}{3}P(52)}{410.67(2)}$

$P = 2843.1 \, \text{ lb}$
 

Based on allowable shearing force of the screws
$s = \dfrac{RI}{VQ_{screw}}$
 

Where:
R = 2(300) = 600 lb
V = 2/3 P
Q_screw = 6(1)(4.5) = 27 in³
I = 410.67 in⁴
s = 5 in

 

Thus,
$5 = \dfrac{600(410.67)}{\frac{2}{3}P(27)}$

$P = 2737.8 \, \text{ lb}$
 

For safe value of P, use P = 2737.8 lb.           answer
 

Bending stress:
$f_b = \dfrac{Mc}{I} = \dfrac{\frac{8}{3}(2737.8)(12)(5)}{410.67}$

$f_b = 1066.67 \, \text{ psi}$           answer