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

**Problem 593**

A box beam, built up as shown in Fig. P-593, is secured by screws spaced 5 in. apart. The beam supports a concentrated load *P* at the third point of a simply supported span 12 ft long. Determine the maximum value of *P* that will not exceed f_{v} = 120 psi in the beam or a shearing force of 300 lb in the screws. What is the maximum flexural stress in the beam?

**Solution 593**

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$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

$f_v = \dfrac{VQ_{NA}}{Ib}$

*V*= 2/3

*P*

*Q*

_{NA}= 6(1)(4.5) + 2 [ 5(1)(2.5) ] = 52 in

^{3}

*I*= 8(10

^{3})/12 - 6(8

^{3})/12 = 410.67 in

^{4}

*b*= 2 in

*f*= 120 psi

_{v}

Thus,

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

$P = 2843.1 \, \text{ lb}$

Based on allowable shearing force of the screws

$s = \dfrac{RI}{VQ_{screw}}$

*R*= 2(300) = 600 lb

*V*= 2/3

*P*

*Q*

_{screw}= 6(1)(4.5) = 27 in

^{3}

*I*= 410.67 in

^{4}

*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*

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