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## Q. 4.11

A cantilever beam of rectangular cross-section is used to support a pulley as shown in Fig. 4.38 (a). The tension in the wire rope is 5 kN. The beam is made of cast iron FG 200 and the factor of safety is 2.5. The ratio of depth to width of the cross-section is 2. Determine the dimensions of the cross-section of the beam.

## Verified Solution

Given P = 5 kN        $S_{u t}=200 N / mm ^{2}$.

(fs) = 2.5 d /w = 2
Step I Calculation of permissible bending stress

$\sigma_{b}=\frac{S_{u t}}{(f s)}=\frac{200}{2.5}=80 N / mm ^{2}$

Step II Calculation of bending moments
The forces acting on the beam are shown in Fig. 4.38(b). Referring to the figure,

$\left(M_{b}\right)_{ at B }=5000 \times 500=2500 \times 10^{3} N – mm$.

$\left(M_{b}\right)_{\text {at A }}=5000 \times 500+5000 \times 1500$.

$=10000 \times 10^{3} N – mm$.

Step III Calculation of dimensions of cross-section
The bending moment diagram is shown in Fig. 4.38(c).
The cross-section at A is subjected to maximum bending stress. For this cross-section,

$y=\frac{d}{2}=w \quad I=\frac{1}{12}\left[(w)(2 w)^{3}\right]=\frac{2}{3} w^{4} mm ^{4}$.

$\sigma_{b}=\frac{M_{b} y}{I} \quad \text { or } \quad 80=\frac{\left(10000 \times 10^{3}\right)(w)}{\left(\frac{2}{3} w^{4}\right)}$.

Therefore,
w = 57.24 mm or 60 mm d = 2w = 120 mm.