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.

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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.