Solution to Problem 581 | Design for Flexure and Shear

Maximum moment for simple beam
$M_{max} = \frac{1}{8} w_o \, L^2$

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

$M_{max} = 4.5w_o \, \text{ lb·ft}$
 

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

Where:
$Q = 6(4)(3.0) = 72 \, \text{ in}^3$

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

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

 

Thus,
$90 = \dfrac{3w_o (72)}{500(6)}$

$w_o = 1250 \, \text{ lb/ft}$
 

Use w_o = 1250 lb/ft for safe value of uniformly distributed load.           answer