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

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

$120 = \dfrac{6(4.5w_o)(12)}{6(10^3)}$

$w_o = 2\,222.22 \, \text{ lb/ft}$

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:

$Q = 6(4)(2.5) = 60 \, \text{ in}^3$
$I = \dfrac{bd^3}{12} = \dfrac{6(10^3)}{12} = 500 \, \text{ in}^4$

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

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*

that should be 3 not 2.5

.: Q= (6)(4)(3)= 72 in.^3

Thanks... This page is flagged for revision.