Question 6.6: A simply supported beam carries a concentrated load at its c...

The cross-sectional area = bh = 25 × 15 = 375 cm².
The section modulus is

Z=\frac{b h^2}{6}=\frac{15 \times 25^2}{6}=1562.5  cm ^3

By this time, you must have understood how to get maximum shear force and maximum bending moment of the beam. So, straightaway, we can write

V_{\max }=\frac{P}{2}

and        M_{\max }=\frac{P l}{4}=30 P  kg  cm

Now, equating maximum shear stress (3 / 2)  \tau_{\text {average }} to working stress in shear, we get

\frac{3}{2} \times \frac{V}{A}=11

or        \frac{3}{2} \times \frac{P}{2} \times \frac{1}{375}=11

or P = 5500 kg. Similarly, equating maximum bending (M/Z) with working stress in bending, we get

\frac{P l / 4}{b h^2 / 6}=70

or        \frac{30 P}{1562.5}=70

or P = 3647 kg. This is smaller than 5500 kg. So, the safe load is P = 3647 kg for the beam.