A cantilever beam of cross section 150 mm x 200 mm is 2 m long.
If deflection at free end is 2 mm, calculate u.d.l. that the beam can carry
over whole span. Also calculate maximum slope of the beam. Take E = 210 
GPa.

To solve this problem, we will use the principles of beam deflection and slope for a cantilever beam subjected to a uniformly distributed load (UDL). Let's break down the problem into the required steps.

### Given Data:
- Cross-sectional dimensions of the beam:
  - Width, \( b = 150 \) mm
  - Height, \( h = 200 \) mm
- Length of the beam, \( L = 2 \) m
- Deflection at the free end, \( \delta = 2 \) mm
- Young's Modulus, \( E = 210 \) GPa = \( 210 \times 10^3 \) N/mm²

### Step 1: Calculate the moment of inertia (I)

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

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

Substituting the given values:

\[
I = \frac{150 \times 200^3}{12} \text{ mm}^4 = \frac{150 \times 8 \times 10^6}{12} \text{ mm}^4 = 100 \times 10^6 \text{ mm}^4
\]

### Step 2: Relate deflection to UDL

For a cantilever beam subjected to a UDL \( w \), the deflection at the free end is given by:

\[
\delta = \frac{w \cdot L^4}{8 \cdot E \cdot I}
\]

Rearranging to solve for \( w \):

\[
w = \frac{8 \cdot E \cdot I \cdot \delta}{L^4}
\]

Substituting the known values:

\[
w = \frac{8 \cdot 210 \times 10^3 \cdot 100 \times 10^6 \cdot 2}{(2 \times 10^3)^4} \text{ N/mm}
\]

Simplifying:

\[
w = \frac{8 \cdot 210 \times 10^3 \cdot 100 \times 10^6 \cdot 2}{16 \times 10^{12}} \text{ N/mm}
\]

\[
w = \frac{336 \times 10^{9}}{16 \times 10^{12}} \text{ N/mm} = 21 \times 10^{-3} \text{ N/mm} = 21 \text{ kN/m}
\]

Thus, the UDL that the beam can carry is \( w = 21 \) kN/m.

### Step 3: Calculate the maximum slope of the beam

The maximum slope in a cantilever beam subjected to a UDL occurs at the fixed end and is given by:

\[
\theta_{\text{max}} = \frac{w \cdot L^3}{6 \cdot E \cdot I}
\]

Substituting the known values:

\[
\theta_{\text{max}} = \frac{21 \times 10^3 \times (2 \times 10^3)^3}{6 \cdot 210 \times 10^3 \cdot 100 \times 10^6} \text{ radians}
\]

Simplifying:

\[
\theta_{\text{max}} = \frac{21 \times 8 \times 10^9}{126 \times 10^9} \text{ radians} = \frac{168 \times 10^9}{126 \times 10^9} \text{ radians}
\]

\[
\theta_{\text{max}} = 1.333 \times 10^{-3} \text{ radians} = 1.333 \text{ mrad}
\]

### Summary:

- **The UDL that the beam can carry is \( w = 21 \) kN/m.**
- **The maximum slope of the beam is \( \theta_{\text{max}} = 1.333 \) mrad.**

Would you like more details or explanations on any of these steps?

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**Relative Questions:**

1. How would the deflection change if the length of the beam were increased?
2. What would be the effect of changing the material's Young's modulus on the deflection and slope?
3. How does the moment of inertia influence the deflection of a beam?
4. What happens to the maximum slope if the UDL is doubled?
5. How would the deflection and slope change if the beam were simply supported instead of cantilevered?

**Tip:** When dealing with beam deflection problems, always double-check the units of the inputs to ensure consistent and accurate calculations.

Learn how to calculate the uniformly distributed load (UDL) that a cantilever beam can carry and its maximum slope. This problem involves beam deflection principles, moment of inertia calculations, and Young's modulus. Suitable for undergraduate engineering students.

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