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To solve this beam design problem, we need to determine the dimensions of the beam using the given constraints and stresses. Here's a step-by-step breakdown:

### Given Data:

- Length of the beam \(L = 4.20 \, \text{m}\)
- Uniform distributed load \(w = 1.5 \, \text{kN/m}\)
- Concentrated load at midspan \(P = 4 \, \text{kN}\)
- Maximum allowable bending stress \(\sigma_{\text{max}} = 11.60 \, \text{MPa}\)
- Allowable shearing stress \(\tau_{\text{max}} = 1.30 \, \text{MPa}\)
- Maximum deflection \(\delta_{\text{max}} = \frac{L}{240}\)
- Modulus of elasticity \(E = 14300 \, \text{MPa}\)
- Beam depth-to-breadth ratio \(d = 3b\)

We need to design the section of the beam by determining \(b\) (breadth) and \(d\) (depth).

### Step 1: Calculate the Maximum Bending Moment

For a simply supported beam with both a uniform load and a concentrated load, the maximum bending moment occurs at the midspan.

- **Moment due to uniform distributed load:**

\[
M_w = \frac{wL^2}{8}
\]

Substituting values:

\[
M_w = \frac{1.5 \times (4.20)^2}{8} = 3.3075 \, \text{kNm}
\]

- **Moment due to concentrated load at midspan:**

\[
M_P = \frac{P \times L}{4}
\]

Substituting values:

\[
M_P = \frac{4 \times 4.20}{4} = 4.2 \, \text{kNm}
\]

- **Total maximum bending moment:**

\[
M_{\text{max}} = M_w + M_P = 3.3075 + 4.2 = 7.5075 \, \text{kNm}
\]

Convert this to Nmm:

\[
M_{\text{max}} = 7.5075 \times 10^6 \, \text{Nmm}
\]

### Step 2: Use Bending Stress Formula

The maximum bending stress is given by:

\[
\sigma = \frac{M_{\text{max}}}{Z}
\]

Where \(Z\) is the section modulus, and for a rectangular section:

\[
Z = \frac{b d^2}{6}
\]

Given that \(\sigma_{\text{max}} = 11.60 \, \text{MPa}\), and substituting \(d = 3b\) into the formula for \(Z\):

\[
Z = \frac{b (3b)^2}{6} = \frac{9b^3}{6} = 1.5 b^3
\]

Now, using the bending stress formula:

\[
11.60 = \frac{7.5075 \times 10^6}{1.5 b^3}
\]

Solve for \(b\):

\[
b^3 = \frac{7.5075 \times 10^6}{1.5 \times 11.60}
\]

\[
b^3 = \frac{7.5075 \times 10^6}{17.4}
\]

\[
b^3 = 431469.83 \, \text{mm}^3
\]

\[
b = \sqrt[3]{431469.83} = 75.66 \, \text{mm}
\]

### Step 3: Calculate Depth \(d\)

Since \(d = 3b\):

\[
d = 3 \times 75.66 = 226.98 \, \text{mm}
\]

### Step 4: Check for Shear Stress

The maximum shear force occurs at the supports. The total load on the beam is:

\[
V = \frac{wL}{2} + \frac{P}{2} = \frac{1.5 \times 4.20}{2} + \frac{4}{2} = 3.15 + 2 = 5.15 \, \text{kN}
\]

The shear stress is given by:

\[
\tau = \frac{1.5 V}{b d}
\]

Substituting values:

\[
\tau = \frac{1.5 \times 5.15 \times 10^3}{75.66 \times 226.98}
\]

\[
\tau = \frac{7.725 \times 10^3}{17170.08} = 0.45 \, \text{MPa}
\]

Since \(0.45 \, \text{MPa}\) is less than the allowable shear stress of \(1.30 \, \text{MPa}\), the design is safe in shear.

### Step 5: Check for Deflection

The deflection for a beam with both a uniform load and concentrated load is given by:

\[
\delta = \frac{5 w L^4}{384 E I} + \frac{P L^3}{48 E I}
\]

The moment of inertia \(I\) for a rectangular section is:

\[
I = \frac{b d^3}{12} = \frac{75.66 \times (226.98)^3}{12} = 7342871.12 \, \text{mm}^4
\]

Now calculate the deflection:

\[
\delta_w = \frac{5 \times 1.5 \times (4.2)^4}{384 \times 14300 \times 7342871.12}
\]

\[
\delta_w \approx 0.000191 \, \text{m}
\]

\[
\delta_P = \frac{4 \times (4.2)^3}{48 \times 14300 \times 7342871.12}
\]

\[
\delta_P \approx 0.000102 \, \text{m}
\]

Total deflection:

\[
\delta = 0.000191 + 0.000102 = 0.000293 \, \text{m} = 0.293 \, \text{mm}
\]

The allowable deflection is:

\[
\delta_{\text{allow}} = \frac{L}{240} = \frac{4200}{240} = 17.5 \, \text{mm}
\]

Since \(0.293 \, \text{mm}\) is much less than the allowable deflection of \(17.5 \, \text{mm}\), the design is safe in deflection.

### Final Dimensions:
- Breadth \(b = 75.66 \, \text{mm}\)
- Depth \(d = 226.98 \, \text{mm}\)

The beam design satisfies all the given criteria.

---

Would you like further clarification or detailed steps for any part of this calculation?

Here are 5 related questions:
1. How does the bending moment vary along the length of the beam?
2. What is the importance of the section modulus in beam design?
3. How would the design change if the concentrated load was placed off-center?
4. How do different boundary conditions affect the deflection of a beam?
5. Why is the deflection limit often set as \( \frac{L}{240} \)?

**Tip:** Always check for the worst-case scenario when designing for combined loads to ensure safety.

A 4.20 M simply supported beam carries a uniform distributed load of 1.5KN/M over its entire span and concentrated load of 4KN at the Midspan. If the maximum allowable bending stress is 11.60 MPa, the allowable shearing stress is 1.30MPa, the maximum deflection is L/240, and E= 14300 MPa. Design the section of beam by using the ratio of d=3b.

This problem involves the design of a 4.20 m simply supported beam subjected to a uniform distributed load of 1.5 kN/m and a concentrated load of 4 kN at midspan. It requires calculating the bending stress, shear stress, and deflection using modulus of elasticity (E= 14300 MPa) and the section ratio of d = 3b. Explore key concepts in beam theory and structural analysis.

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