A 4-meter cantilever beam supports a uniform load of 25 kN/m throughout its length. The beam deflects 16 mm at the free end.

Given the following formulas for deflection at free-end of a cantilever beam:

δ = (Pa2/6EI)(3L-a)

a. What force at the free end is required to eliminate the deflection at that end?

b. What force at the free end is required to reduce the deflection at that end by 10 mm?

c. What force at the midspan is required to eliminate the deflection at that point?

Let's solve the problem step by step. The problem involves a cantilever beam subjected to a uniform load, and we need to find the forces required to achieve specific deflection criteria.

### Given Data:

- Length of the beam, \( L = 4 \) meters
- Uniform load intensity, \( w = 25 \, \text{kN/m} \)
- Deflection at the free end due to the uniform load, \( \delta_0 = 16 \, \text{mm} \)
- Elastic modulus, \( E \) (unknown)
- Moment of inertia, \( I \) (unknown)

### Formulas:

The deflection at the free end \( \delta \) due to a point load \( P \) applied at distance \( a \) from the fixed end is given by:

\[
\delta = \frac{P \cdot a^2}{6EI} (3L - a)
\]

For a uniform load \( w \) over the length \( L \), the deflection at the free end is:

\[
\delta_0 = \frac{wL^4}{8EI}
\]

### a. Force at the Free End to Eliminate Deflection:

To eliminate the deflection at the free end, the point load \( P \) at the free end must counteract the deflection caused by the uniform load.

Set the total deflection to zero:

\[
\delta_0 - \delta_P = 0
\]

Where \( \delta_P \) is the deflection caused by the point load \( P \) at the free end:

\[
\delta_P = \frac{P \cdot L^3}{3EI}
\]

So:

\[
\frac{wL^4}{8EI} = \frac{P \cdot L^3}{3EI}
\]

Solving for \( P \):

\[
P = \frac{3wL}{8}
\]

Substituting the values:

\[
P = \frac{3 \times 25 \times 4}{8} = 37.5 \, \text{kN}
\]

### b. Force at the Free End to Reduce Deflection by 10 mm:

The new deflection should be \( \delta_0 - 10 \, \text{mm} \):

\[
\delta' = \delta_0 - 10 \, \text{mm}
\]

The new deflection due to the point load \( P \) at the free end:

\[
\delta_P' = \frac{P \cdot L^3}{3EI}
\]

Set up the equation:

\[
\frac{wL^4}{8EI} - \frac{P \cdot L^3}{3EI} = 10 \, \text{mm}
\]

Converting 10 mm to meters:

\[
10 \, \text{mm} = 0.01 \, \text{m}
\]

Now, solve for \( P \):

\[
\frac{wL^4}{8EI} - \frac{P \cdot L^3}{3EI} = 0.01
\]

Substituting the known values:

\[
\frac{25 \times 4^4}{8EI} - \frac{P \times 4^3}{3EI} = 0.01
\]

Let \( C \) be the constant \( \frac{L^4}{EI} \):

\[
\frac{25C}{8} - \frac{64P}{3EI} = 0.01
\]

Solve for \( P \):

\[
P = \frac{25 \times 4}{8} - \frac{0.01 \times 3EI}{64}
\]

\[
P = 37.5 \, \text{kN} - \frac{0.01 \times 3EI}{64}
\]

We need more data or assumptions to solve this directly, but this setup is correct.

### c. Force at Midspan to Eliminate Deflection:

The deflection at midspan can be found using superposition:

\[
\delta_{mid} = \delta_0 - \delta_P
\]

To eliminate the deflection at midspan:

\[
\delta_{mid} = 0
\]

Setting up the equation, the force required can be found in a similar manner. However, it will require the integral of the moment equation or the exact boundary conditions for the moment or displacement.

---

Would you like further details or clarification on any part of the above calculations?

### Related Questions:
1. What is the value of \( E \) and \( I \) for this specific beam?
2. How would the solution change if the beam were not cantilevered but simply supported?
3. What is the significance of the moment of inertia \( I \) in beam deflection problems?
4. Can you derive the deflection formula for a uniformly loaded beam?
5. How does the deflection change if the load is not uniform but concentrated?

### Tip:
Always check the units when working with deflection formulas to ensure consistency and correctness.

Learn how to solve cantilever beam deflection problems caused by uniform loads. Explore step-by-step solutions to calculate forces required to eliminate or reduce deflection at the free end and midspan of a 4-meter beam under a 25 kN/m load. Suitable for engineering students and professionals.

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