An overhanging beam 𝑷𝑸𝑹 is subjected to uniformly distributed load 20 kN/m as shown in the figure.

The maximum bending stress developed in the beam is ___________ MPa (round off to one decimal place).

The correct answer is 250.

Here is the detailed step-by-step derivation to find the maximum bending stress developed in the beam:

1. Beam Configuration and Loading Analysis
Based on the provided schematic, we have an overhanging beam $PQR$ where:
- The segment $PQ$ has a span length of $2000\text{mm}=2\text{m}$ and is supported by a simple hinge support at $P$ and a roller support at $Q$.
- The segment $QR$ is an overhang of length $1000\text{mm}=1\text{m}$.
- A uniformly distributed load (UDL) of intensity $w=20\text{kN/m}$ acts along the entire length of the beam $PR$ (total length $L=3\text{m}$).

2. Determination of Support Reactions
Let $R_P$ and $R_Q$ be the vertical support reactions at $P$ and $Q$, respectively.
Taking the moment about point $P$:
$\sum M_P=0$

$\Rightarrow R_Q\times 2-(20\times 3)\times 1.5=0$

$\Rightarrow 2R_Q-90=0$

$\Rightarrow R_Q=45\text{kN}$

Using vertical force equilibrium ($\sum F_y=0$):
$R_P+R_Q=20\times 3=60\text{kN}$

$\Rightarrow R_P+45=60$

$\Rightarrow R_P=15\text{kN}$

3. Finding the Maximum Bending Moment
We need to determine the bending moment distribution along the beam to find its maximum absolute value.
Let $x$ be the distance from the left support $P$ ($0\le x\le 2\text{m}$):
$M\left(x\right)=R_P·x-\frac{w{x}^2}{2}=15x-10x^2$

The maximum positive (sagging) bending moment occurs where the shear force is zero:
$V\left(x\right)=15-20x=0\Rightarrow x=0.75\text{m}$

At $x=0.75\text{m}$, the local maximum bending moment is:
$M_{\text{sagging}}=15\left(0.75\right)-10{\left(0.75\right)}^2=11.25-5.625=5.625\text{kNm}$

At support $Q$ ($x=2\text{m}$), the hogging bending moment is:
$M_Q=-w\times 1\times 0.5=-20\times 0.5=-10\text{kNm}$

Comparing the absolute magnitudes:
$\left|M_{\text{max}}\right|=10\text{kNm}=10\times {10}^6\text{Nmm}$

4. Cross-Sectional Properties
The rectangular cross-section has the following dimensions visible in the beam cross-section diagram:
- Width, $b=24\text{mm}$
- Depth, $h=100\text{mm}$

The area moment of inertia $I$ about the horizontal neutral axis $x$ is calculated as:
$I=\frac{b{h}^3}{12}=\frac{24\times {100}^3}{12}=2\times {10}^6{\text{mm}}^4$

The maximum distance from the neutral axis to the outermost fiber is:
$y_{\text{max}}=\frac{h}{2}=\frac{100}{2}=50\text{mm}$

5. Maximum Bending Stress Calculation
Applying the flexure formula:
${\sigma }_{\text{max}}=\frac{\left|M_{\text{max}}\right|}{I}·y_{\text{max}}$

$\Rightarrow {\sigma }_{\text{max}}=\frac{10\times {10}^6}{2\times {10}^6}\times 50=5\times 50=250\text{MPa}$