# Solution to Problem 595 | Spacing of Rivets or Bolts in Built-Up Beams

**Problem 595**

A concentrated load P is carried at midspan of a simply supported 12-ft span. The beam is made of 2-in. by 6-in. pieces screwed together, as shown in Fig. P-595. If the maximum flexural stress developed is 1400 psi, find the maximum shearing stress and the pitch of the screws if each screw can resist 200 lb.

**Solution 595**

$V_{max} = \frac{1}{2}P$

$M_{max} = \frac{1}{4}PL = \frac{1}{4}P(12) = 3P$

From the cross section shown:

$I =\dfrac{8(8^3)}{12} - \dfrac{4(4^3)}{12} = 320 \, \text{ in}^4$

$Q_{NA} = 4(2)(2) + 6(2)(3) + 2(2)(1) = 56 \, \text{ in}^3$

From bending stress

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

$1400 = \dfrac{3P(12)(4)}{320}$

$P = 3111.11 \, \text{ lb}$

Maximum shear stress

$f_v = \dfrac{VQ_{NA}}{Ib} = \dfrac{\frac{1}{2}(3111.11)(56)}{320(4)}$

$f_v = 68.06 \, \text{ psi}$ *answer*

Pitch or Spacing of Screws, e

$R = \dfrac{VQ_\text{screws}}{I}e$

$e = \dfrac{RI}{VQ_\text{screws}}$

For Horizontal Bolts

$Q_h = 4(2)(3) = 24 ~ \text{in.}^3$

$e_h = \dfrac{2(200)(320)}{(3111.11/2)(24)} = 3.42 ~ \text{in.}$ *answer*

For Vertical Bolts

$Q_v = 8(2)(3) = 48 ~ \text{in.}^3$

$e_v = \dfrac{2(200)(320)}{(3111.11/2)(48)} = 1.734 ~ \text{in.}$ *answer*

## Comments

## Old Solution for Spacing of

Old Solution for Spacing of Screws

From strength of screws

$R = \dfrac{VQ_{screws}}{I}e$

$2(200) = \dfrac{\frac{1}{2}(3111.11)(36)}{320}s$

$s = 2.28 \, \text{ in}$

answer