A cast iron beam with dimension as given here. (Top Flange: 80x20 mm, Web: 20 mm x 200 mm and Bottom flange : 160 mm x 40 mm). The beam is simply supported over a span of 5 m. If the tensile stress is not to exceed 20 N/mm2, Find the safe udl which the beam can carry. Find also the maximum compressive stress.

1. Calculation of $\bar y$ and $I_N-A$:

$y_b=\bar y_{bottom}=\frac{160*40*\frac{40}{2}+200*20*(40+\frac{200}{2})+20*80*$ $(40+200+\frac{20}{2})}{160*40+200*20+20*80}$

$y_b=\bar y_{bottom}=90.666=90.67$

$\sum A=160*40+200*20+20*80$

$\sum A=12000mm^2$

$y_t=\bar y_{top}=20+200+40-y_b=260-90.67=169.33mm$

$I_{base}=\frac{160*40^3}{3}+\frac{20*200^3}{12}+20*200*140^2+\frac{80*20^3}{12}+80*20*250^2$

$I_{base}=195.12*10^6 mm^4$

1. Calculation of maximum compressive stress and safe UDL (w) which the beam can carry:

Bending equation:

$\frac{M}{I}=\frac{y}{y}$

Now, $\frac{M}{I}=\frac{f_t}{y_b}=\frac{f_c}{y_t} i.e \frac{f_t}{y_b}=\frac{f_c}{y_t}$

Let $f_c$ be maximum compressive stress

$f_c=\frac{y_t}{y_b}*{y_t}=\frac{20}{90.67}*169.33=37.35 N/mm^2$

Let w be the UDL the beam carried,

maximum bending moment=$M=\frac{wL^2}{8}=\frac{w*5^2}{8}$

$M=3.125 w (KNm) sagging$

In sagging, top fibre is in compression, bottom fibre is in tension.

$\frac{M}{I}=\frac{f_t}{y_b}$

i.e $3.125w=\frac{y_t*I}{y_b}$

i.e $w=\frac{20*96.55*10^6}{90.67*10^6*3.125}$

$w=6.815 KN/m$