# Solution to Problem 581 | Design for Flexure and Shear

**Problem 581**

A laminated beam is composed of five planks, each 6 in. by 2 in., glued together to form a section 6 in. wide by 10 in. high. The allowable shear stress in the glue is 90 psi, the allowable shear stress in the wood is 120 psi, and the allowable flexural stress in the wood is 1200 psi. Determine the maximum uniformly distributed load that can be carried by the beam on a 6-ft simple span.

**Solution 581**

$M_{max} = \frac{1}{8} w_o \, L^2$

$M_{max} = \frac{1}{8} w_o \, (6^2)$

$M_{max} = 4.5w_o \, \text{ lb·ft}$

Maximum shear for simple beam

$V_{max} = \frac{1}{2}w_o \, L$

$V_{max} = \frac{1}{2}w_o \, (6)$

$V_{max} = 3w_o \, \text{ lb}$

For bending stress of wood

$f_b = \dfrac{6M}{bd^2}$

$1200 = \dfrac{6(4.5w_o)(12)}{6(10^3)}$

$w_o = 22\,222.22 \, \text{ lb/ft}$

For shear stress of wood

$( \, f_v \, )_{wood} = \dfrac{3V}{2bd}$

$120 = \dfrac{3(3w_o)}{2(6)(10)}$

$w_o = 1600 \, \text{ lb/ft}$

For shear stress in the glued joint

$( \, f_v \, )_{glue} = \dfrac{VQ}{Ib}$

$Q = 6(4)(2.5) = 60 \, \text{ in}^3$

$I = \dfrac{bd^3}{12} = \dfrac{6(10^3)}{12} = 500 \, \text{ in}^4$

$b = 6 \, \text{ in}$

Thus,

$90 = \dfrac{3w_o (60)}{500(6)}$

$w_o = 1250 \, \text{ lb/ft}$

Use **w _{o} = 1250 lb/ft **for safe value of uniformly distributed load.

*answer*