Question 7.15: A cast iron beam is of I-section as shown in Fig. 7.20. The ...

Given :
Length of beam,                 L = 5 m
Maximum tensile stress,  σ_t = 20  N/mm^2
First calculate the C.G. of the section. Let \bar{y} is the distance of the C.G. from the bottom face. As the section is symmetrical about y-axis, hence \bar{y} is only to be calculated.

Now \bar{y}=\frac{A_1y_1+A_2y_2+A_3y_3}{(A_1+A_2+A_3)} \\ \space \\ \quad \quad \quad \quad =\frac{(160\times 40) \cdot \frac{40}{2}+(200\times 20)\Big(40+\frac{200}{2}\Big)+(80\times 20)\cdot \Big(40+200+\frac{20}{2}\Big)}{160\times 40+200\times 20 + 80 \times 20} \\ \space \\ \quad \quad \quad \quad =\frac{128000+560000+400000}{6400+4000+1600}=\frac{1088000}{12000}=90.66  mm

N.A. lies at a distance of 90.66 mm from the bottom face or 260 – 90.66 = 169.34 mm from the top face.
Now moment of inertia of the section about N-axis is given by,

I=I_1+I_2+I_3

where I_1 = M.O.I. of bottom flange about N.A.

= \text{ M.O.I. of bottom flange about its C.G. } + A_1 × \text{ (Distance of its C.G. from N.A.)}^2 \\ = \frac{160 \times 40^3}{12} +160 \times 40 \times (90.66-20)^2 \\ = 853333.33 + 31954147.84 = 32807481.17  mm^4 \\ I_2 = \text{ M.O.I. of web about N.A.} \\  = \text{ M.O.I. of web about its C.G. }+ A_2 × \text{ (Distance of its C.G. from N.A.)}^2 \\ = \frac{20 \times 200^3}{12} + 200 \times 20 \times (140-90.66)^2 \\ = 13333333.33 + 9737742.4 = 23071075.73  mm^4 \\ I_3 = \text{ M.O.I. of top flange about N.A.} \\ = \text{M.O.I. of top flange about its C.G. + }A_3 \times \text{ (Distance of its C.G. from N.A.)}^2\\ =\frac{80\times 20^3}{12}+80\times 20 \times (250-90.66)^2 \\ = 53333.33 + 40622776.96 = 40676110.29  mm^4

∴            I = 32807481.17 + 23071075.75 + 40676110.29 = 96554667.21  mm^4.

For a simply supported beam, the tensile stress will be at the extreme bottom fibre and compressive stress will be at the extreme top fibre.
Here maximum tensile stress = 20 N/mm²
Hence for the maximum tensile stress,

y = 90.66 mm

[i.e., y is the distance of the extreme bottom fibre (where the tensile stress is maximum) from the N.A.]
Using the relation, \frac{M}{I}=\frac{\sigma}{y}

∴                                 M=\frac{\sigma}{y}\times I \\ =\frac{20}{90.66} \times 96554667.21  \quad (∵  σ = σ_t = 20  N/mm^2) \\ = 21300389.85  Nmm            …(i)

Let w = Uniformly distributed load in N/m on the simply supported beam.
The maximum B.M. is at the centre and equal to \frac{wL^2}{8}

∴               M=\frac{w\times 5^2}{8}  Nm=\frac{w \times 25 \times 1000}{8}  Nmm=3125w  Nmm        …(ii)

Equating the two values of M, given by equations (i) and (ii), we get

3125w = 21300389.85

∴                w=\frac{21300389.85}{3125}=\pmb{6816.125  N/m.}

Maximum Compressive Stress
Distance of extreme top fibre from N.A.,

y_c = 169.34 mm
M = 21300389.85
I = 96554667.21

Let  σ_c = Max. compressive stress

Using the relation, \frac{M}{I}=\frac{\sigma}{y}

∴              \sigma=\frac{M}{I}\times y

or             \sigma_c=\frac{M}{I}\times y_c =  \frac{21300389.85}{96554667.21} \times 169.34 = \pmb{37.357  N/mm^2.}