\text { Given } \quad W=75 kN \quad \sigma_{b}=165 N / mm ^{2} .

Step I Calculation of bending moment
There are two cantilever beams and the load

Select a standard rolled I-section beam from the following table:

supported by each beam is (75/2) or 37.5 kN. For a triangular load distribution, the centre of gravity of the resultant load is at a distance of (2000/3) mm from the wall. Therefore,

M_{b}=\left(37.5 \times 10^{3}\right)\left(\frac{2000}{3}\right)=25 \times 10^{6} N – mm .

\text { Step II Calculation of }\left(I_{x x} / y\right) .

From Eq. (4.12),

\sigma_{b}=\frac{M_{b} y}{I}             (4.12).

\frac{I_{x x}}{y}=\frac{M_{b}}{\sigma_{b}}=\frac{25 \times 10^{6}}{165}=151.51 \times 10^{3} mm ^{3} .

Step III Selection of beam
The cross-section of the beam is shown in Fig. 4.25
(b), (y = h/2)

Trial I  Suppose beam (ISLB 175) is suitable for the application. For this beam,

\frac{I_{x x}}{y}=\frac{1096.2 \times 10^{4}}{(175 / 2)}=125.28 \times 10^{3} mm ^{3} .

\text { Since the required }\left(I_{x x} / y\right) \text { is }\left(151.51 \times 10^{3}\right) mm ^{2} ,

beam (ISLB 175) is not suitable.
Trial II Suppose beam (ISLB 200) is suitable for the application. For this beam,

\frac{I_{x x}}{y}=\frac{1696.6 \times 10^{4}}{(200 / 2)}=169.66 \times 10^{3} mm ^{3} .

\left(I_{x x} / y\right)>\left(151.51 \times 10^{3}\right) mm ^{3} .

Therefore, the cantilever beams of standard cross-section ISLB 200 are suitable for this application.