To determine the deflection at the free end of a cantilever beam under a load applied at the free end, we use the standard formula:

$\delta = \frac{P \cdot L^3}{3 \cdot E \cdot I}$

where:

• $\delta$ is the deflection at the free end,
• $P$ is the applied load,
• $L$ is the length of the beam,
• $E$ is the modulus of elasticity,
• $I$ is the moment of inertia.

Given values:

• $P = 300 \, \text{N}$,
• $L = 2 \, \text{m}$,
• $E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{Pa}$,
• $I = 50 \times 10^{-9} \, \text{m}^4$.

Substituting these into the formula:

$\delta = \frac{300 \times (2)^3}{3 \times 200 \times 10^9 \times 50 \times 10^{-9}}$

Calculating:

$\delta = \frac{300 \times 8}{3 \times 200 \times 50}$

$\delta = \frac{2400}{30000} \, \text{m}$

$\delta = 0.08 \, \text{m}$

So, the deflection at the free end of the beam is 0.080 m.

The correct answer is 0.080 m.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How would the deflection change if the length of the beam were doubled?
2. What happens to the deflection if the applied load is halved?
3. How does the moment of inertia influence the deflection of the beam?
4. Can the superposition formula be used for beams with varying cross-sections?
5. How would you calculate the deflection if the load was applied at the midpoint instead of the free end?

Tip: The modulus of elasticity ($E$) is a material property that indicates the material's stiffness. The higher the $E$, the less a material deforms under a given load.