Calculating Beam Deflection Under Point Load

This problem requires us to calculate the deflection of a simply supported beam under a specific point load using the given material and geometric properties.

We are given the following information about the beam and the load:

• Span of the beam (L) = 4.5 metres
• Point load (W) = 30 KN = $\text{30} \times \text{10}^\text{3}$ N
• Position of the load from the left support (a) = 3 metres
• Position of the load from the right support (b) = L - a = 4.5 - 3 = 1.5 metres
• Moment of Inertia (I) = 54.97 $\times$ $\text{10}^\text{-6}$ $\text{m}^\text{4}$
• Young's Modulus (E) = 200 GN/$\text{m}^\text{2}$ = $\text{200} \times \text{10}^\text{9}$ N/$\text{m}^\text{2}$

Formula for Deflection

For a simply supported beam carrying a point load W at a distance 'a' from the left support and 'b' from the right support, the deflection under the load is given by the formula:

$\delta = \frac{Wa^2 b^2}{3EIL}$

Where:

• $\delta$ is the deflection under the load
• W is the point load
• a is the distance from the left support to the load
• b is the distance from the right support to the load
• E is the Young's Modulus of the beam material
• I is the Moment of Inertia of the beam section
• L is the total span of the beam

Step-by-Step Calculation of Deflection

Now, let's substitute the given values into the formula to find the deflection under the load:

$\delta = \frac{(30 \times 10^3 \text{ N}) \times (3 \text{ m})^2 \times (1.5 \text{ m})^2}{3 \times (200 \times 10^9 \text{ N/m}^2) \times (54.97 \times 10^{-6} \text{ m}^4) \times (4.5 \text{ m})}$

Calculate the terms in the numerator:

Numerator = $(30 \times 10^3) \times 9 \times 2.25 = 607.5 \times 10^3 \text{ N.m}^4$

Calculate the terms in the denominator:

Denominator = $3 \times (200 \times 10^9) \times (54.97 \times 10^{-6}) \times 4.5$

Denominator = $3 \times 200 \times 54.97 \times 4.5 \times 10^9 \times 10^{-6}$

Denominator = $600 \times 54.97 \times 4.5 \times 10^3$

Denominator = $148419 \times 10^3 \text{ N.m}^3$

Now, divide the numerator by the denominator to find the deflection in meters:

$\delta = \frac{607.5 \times 10^3 \text{ N.m}^4}{148419 \times 10^3 \text{ N.m}^3}$

$\delta = \frac{607.5}{148419} \text{ m}$

$\delta \approx 0.004093 \text{ m}$

Convert the deflection from meters to millimetres (1 m = 1000 mm):

$\delta = 0.004093 \times 1000 \text{ mm}$

$\delta \approx 4.093 \text{ mm}$

Conclusion

The calculated deflection under the load is approximately 4.09 mm. This value matches one of the given options.