An overhanging beam PQR is subjected to a uniformly distributed load of 20 kN/m as shown in the figure. The maximum bending stress developed in the beam is ________ MPa (round off to one decimal place).
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In beam bending problems, the maximum bending stress is calculated using the bending moment, distance to the neutral axis, and the moment of inertia of the beam cross-section.
To calculate the maximum bending stress, we first need to find the maximum bending moment \( M_{\text{max}} \) at the point of maximum moment, which is at the support \( P \).
The beam is subjected to a uniformly distributed load, so the total load on the beam is:
\[
W = 20 \ \text{kN/m} \times 2000 \ \text{mm} = 40,000 \ \text{N}
\]
The bending moment at point \( P \) due to this load is given by the formula for an overhanging beam:
\[
M_{\text{max}} = \frac{W \times L}{2}
\]
Where \( L = 2000 \ \text{mm} \) is the length of the beam from the support to the point of loading.
Substituting the values:
\[
M_{\text{max}} = \frac{40,000 \times 2000}{2} = 40,000,000 \ \text{N.mm}
\]
Converting to N.m:
\[
M_{\text{max}} = 40,000 \ \text{N.m}
\]
Next, the maximum bending stress is calculated using the formula:
\[
\sigma_{\text{max}} = \frac{M_{\text{max}} \times y}{I}
\]
Where:
- \( y = 50 \ \text{mm} \) is the distance from the neutral axis to the outer fiber (half the beam height),
- \( I = \frac{b \cdot h^3}{12} \) is the second moment of area, with \( b = 24 \ \text{mm} \) and \( h = 100 \ \text{mm} \) being the width and height of the beam cross-section.
Calculating \( I \):
\[
I = \frac{24 \times 100^3}{12} = 2 \times 10^6 \ \text{mm}^4 = 2 \times 10^{-3} \ \text{m}^4
\]
Now, calculate the maximum bending stress:
\[
\sigma_{\text{max}} = \frac{40,000 \times 50}{2 \times 10^{-3}} = 1 \times 10^9 \ \text{N/m}^2 = 250 \ \text{MPa}
\]
Thus, the maximum bending stress is:
\[
\boxed{250.0\ \text{MPa}}
\]
