Question

A distributed load of w N/m is applied over an entire length of a simply supported beam 8m long. The beam section is made up of four 20mmx100mm rectangles arranged as shown. A. Determine the maximum value of w without exceeding the permissible flexure stress of 10MPa. B. Determine the maximum value of w without exceeding the permissible shear stress of 1.2MPa.C. Determine the maximum value of w without exceeding the permissible screw strength of 800N for a pitch of 75mm. * 

Answer 1

$\sum N_A\varOmega=0$

$w\times8\times4-R_B\times8=0$

$R_B=4W$

and ,$R_A=R_B=4W$

Moment at any distance $\eta,$

$N\eta=4W\eta-w\eta,\frac{\eta}{2}$

$\frac{dN}{dm}=4w-w\eta=0$

$\eta=4m$

$M_{max}=4W.4-w.4.\frac{4}{2}$

$=16w-8w=8w$

$M_{max}=8w NM$

$\tau=\frac{VA\bar{y}}{Ib}=\frac{4w\times(A\bar{y_1}+A_2\bar{y}_2)}{Ib}$

$1.2=\frac{4w\times(100\times20\times60+40\times50\times25}{I\times 40}$

$I=\frac{100\times140^3}{12}-\frac{2\times30\times100^3}{12}=1.786\times10^7mm$

$1.2=\frac{4\times w\times(100\times20\times60+40\times50\times25}{1.786\times10^2\times40}$

$w=1260.71 N/m$

$\frac{VA\bar{y}}{I}.P$

$\frac{4w\times100\times20\times60\times75}{1.786\times10^7}=2\times800$

$w=793.78 N/m$