Calculate the safe load W.

Given:

For a rectangular S.S wooden beam -

b = 150mm, d = 250mm, L = 1.3m ,

Central point load = 'W' N

$\sigma_{b} = 7 N/mm^{2} and q_{max} = 1 N/mm^{2}$

Solution:

M = Max. B.M. = $\frac{WL}{4} = \frac{w \times 1.3}{4} = 0.325 W - N.m$

S = Max S.F =. Reaction = $\frac{W}{2}N = 0.5 WN$

For rectangular section,

A = b x d = 150 x 250 = 37500 $mm^{2}$

I = $\frac{bd^{3}}{12} = \frac{150 \times 250^{3}}{12} = 195.31 \times 10^{6} mm^{4}$

$y_{max} = d/2 = \frac{250}{2} = 125 mm$

Value of 'W' for bending stress criteria

$\frac{M}{I} = \frac{\sigma}{y} \therefore M = \frac{\sigma}{y} \times y$

$\therefore 325 W = \frac{7 \times 195.31 \times 10^{6}}{125}$

$\therefore W = 33653.41 N = 33.65kN$

Value of 'W' for shear stress criteria

$q_{max} = \frac{1.55}{A}$

$\therefore 1 = \frac{1.5 \times 0 w}{37500} = 33.65kN$

$\therefore W = 50000N - 50kN$

Safe Value of W = min, of A & B = 33.65 kN