The tension flange of a cast iron I section beam is 240 mm wide and 50 mm deep. The compression flange is 100mm wide and 20mm deep, whereas the web is 300mm x 30mm. -

l=4m

$f_{max - compression} = 90MPa$

$f_{max - tension} = 24MPa$

Location of the centre of gravity:

$$ y = \frac{ΣA_iY_i}{ΣA_i} = \frac{2820000}{23000} = 122.609mm$$

Moment of Inertia Calculations:

$I_{N.A} = I_{Bottom \ \ Flange} + I_{Web} + I_{Top \ \ Flange}$

$I_{N.A} = [\frac{bd^3}{12} + A(y - y)^2]_{Bottom Flange} + [\frac{bd^3}{12} + A(y - y)^2]_{Web} + [\frac{bd^3}{12} + A(y - y)^2]_{Top \ \ Flange}$

$I_{N.A} = [\frac{(240)(50)^3}{12} + (12000)(97.609)^2] + [\frac{(30)(300)^3}{12} + (9000)(77.391)^2] + [\frac{(100)(20)^3}{12} + (2000)(237.391)^2]$

$I_{N.A} = 351010144.9 mm^4$

Since the bottom flange is the tension flange and the top flange is the compression flange, we assume that the bending moment is a sagging bending moment

Since the beam is simple supported, the maximum bending moment acting on the beam is given by:

$M = \frac{WI^2}{8}$

Since l = 4m, we get:

M=2W

From the flexural equation,

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

$\frac{f_{bt}}{Y_t} = \frac{f_{bc}}{Y_c} = \frac{M}{I}$

We will have to find out the value of W, by trial and error.

Considering the compression first:

The maximum stress takes place in the extreme fiber. Here, for compression,

$$y = 247.391mm$$

$\frac{f_{bcy=247.391}}{Y_c} = \frac{M}{I}$

Substituting in the formula, we have:

$\frac{90}{247.391} = \frac{2W}{351010144.9}$

$W = 63848145.33 N$

Now, considering tension:

The maximum stress takes place in the extreme fiber. For tension,

$$y = 122.609mm$$

$\frac{f_{bty = 122.609}}{Y_t} = \frac{M}{I}$

Substituting in the formula, we have:

$\frac{24}{122.609} = \frac{2W}{351010144.9}$

$$W = 34354099.12N$$

As seen, the weight that the beam can carry in tension is considerably lesser than that which can be carried under compression. Hence, the limit in tension is reached faster. Therefore, considering the smaller load W=34354099.12N as the load that can be carried by the beam as its maximum capacity.

The load that can be carried by the beam per meter run is given by :$\frac{W}{length \ \ of \ \ spam}$

$= \frac{34354099.12}{4}$

$= 8588524.78 N$

Therefore, the value of the load that the beam can carry per meter run is 8588524.78N.