A simply supported beam having 4 m span is subjected to UDL of 20 kN/m.
Cross section of beam is 250 mm X 400 mm. Calculate maximum bending
stress produced in the beam and stress produced at the layer 80 mm far from
neutral axis. Also draw bending stress diagram.

To solve the problem of finding the maximum bending stress and the stress at a specific layer in a simply supported beam subjected to a uniformly distributed load (UDL), we will proceed step by step.

### Given Data:
- Span of the beam, \( L = 4 \) m
- Uniformly Distributed Load (UDL), \( w = 20 \) kN/m
- Cross-sectional dimensions of the beam: 
  - Width, \( b = 250 \) mm
  - Height, \( h = 400 \) mm
- Distance from the neutral axis where stress is to be calculated, \( y = 80 \) mm

### Step 1: Calculate the maximum bending moment

For a simply supported beam subjected to a uniformly distributed load, the maximum bending moment occurs at the center of the span and is given by:

\[
M_{\text{max}} = \frac{w \cdot L^2}{8}
\]

Substituting the given values:

\[
M_{\text{max}} = \frac{20 \times 4^2}{8} = \frac{20 \times 16}{8} = 40 \text{ kNm}
\]

Convert the bending moment into Nmm:

\[
M_{\text{max}} = 40 \times 10^6 \text{ Nmm}
\]

### Step 2: Calculate the moment of inertia (I)

The moment of inertia for a rectangular section is given by:

\[
I = \frac{b \cdot h^3}{12}
\]

Substituting the given values:

\[
I = \frac{250 \times 400^3}{12} \text{ mm}^4 = \frac{250 \times 64 \times 10^6}{12} \text{ mm}^4 = 1333.33 \times 10^6 \text{ mm}^4
\]

### Step 3: Calculate the maximum bending stress

The maximum bending stress occurs at the outermost fiber (i.e., at a distance of \( y = h/2 \) from the neutral axis). The formula is:

\[
\sigma_{\text{max}} = \frac{M_{\text{max}} \times y}{I}
\]

Here, \( y = \frac{h}{2} = \frac{400}{2} = 200 \) mm.

Substituting the values:

\[
\sigma_{\text{max}} = \frac{40 \times 10^6 \times 200}{1333.33 \times 10^6} \text{ N/mm}^2 = 6 \text{ N/mm}^2
\]

### Step 4: Calculate the stress at a layer 80 mm from the neutral axis

The stress at a distance \( y_1 \) from the neutral axis is given by:

\[
\sigma_{y_1} = \frac{M_{\text{max}} \times y_1}{I}
\]

Substituting \( y_1 = 80 \) mm:

\[
\sigma_{y_1} = \frac{40 \times 10^6 \times 80}{1333.33 \times 10^6} \text{ N/mm}^2 = 2.4 \text{ N/mm}^2
\]

### Step 5: Draw the Bending Stress Diagram

To draw the bending stress diagram, we note that:

- The stress distribution is linear across the depth of the beam.
- The maximum stress (6 N/mm²) occurs at the outermost fibers.
- The stress varies from zero at the neutral axis to maximum at the extreme fibers.

The diagram will show a linear increase in stress from 0 at the neutral axis to the maximum stress at the outer edges.

### Summary:

- **Maximum bending stress, \( \sigma_{\text{max}} = 6 \) N/mm².**
- **Stress at 80 mm from the neutral axis, \( \sigma_{80} = 2.4 \) N/mm².**

Would you like the bending stress diagram visualized? Additionally, do you need any more details or explanations?

---

**Relative Questions:**

1. How does the bending stress distribution change with varying beam height?
2. What effect would an increased UDL have on the maximum bending stress?
3. How does the span length influence the maximum bending moment?
4. How would changing the material of the beam affect the stress distribution?
5. What is the significance of the neutral axis in beam bending analysis?

**Tip:** Always ensure that the cross-sectional properties like the moment of inertia are accurately calculated, as they significantly influence the stress distribution.

Learn how to calculate maximum bending stress and stress at a specific layer in a simply supported beam subjected to a UDL. Includes step-by-step solutions and bending stress diagram. Suitable for engineering students and professionals.

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