A simply supported beam has a length of 8 m and is subjected to a uniformly dist

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Question: A simply supported beam has a length of 8 m and is subjected to a uniformly distributed load of 4 kN/m. What is the deflection at the center of the beam?

Options:

0.025 m
0.05 m
0.1 m
0.075 m

Correct Answer: 0.05 m

Solution:

The deflection at the center of a simply supported beam under a uniform load is given by δ = 5wL^4 / (384EI). Substituting w = 4 kN/m, L = 8 m, E = 200 GPa, and I = 0.0001 m^4 gives δ = 0.05 m.

A simply supported beam has a length of 8 m and is subjected to a uniformly dist

Practice Questions

A simply supported beam has a length of 8 m and is subjected to a uniformly distributed load of 4 kN/m. What is the deflection at the center of the beam?

0.025 m
0.05 m
0.1 m
0.075 m

Questions & Step-by-Step Solutions

A simply supported beam has a length of 8 m and is subjected to a uniformly distributed load of 4 kN/m. What is the deflection at the center of the beam?

Step 1: Identify the parameters given in the problem. The length of the beam (L) is 8 m, the uniformly distributed load (w) is 4 kN/m, the modulus of elasticity (E) is 200 GPa, and the moment of inertia (I) is 0.0001 m^4.
Step 2: Convert the units if necessary. Here, we keep the units as they are since they are compatible.
Step 3: Use the formula for deflection at the center of a simply supported beam under a uniform load: δ = 5wL^4 / (384EI).
Step 4: Substitute the values into the formula: δ = 5 * (4 kN/m) * (8 m)^4 / (384 * (200 GPa) * (0.0001 m^4)).
Step 5: Calculate the value of L^4: (8 m)^4 = 4096 m^4.
Step 6: Calculate the numerator: 5 * 4 kN/m * 4096 m^4 = 81920 kN*m^4.
Step 7: Convert kN to N for consistency: 81920 kN*m^4 = 81920000 N*m^4.
Step 8: Calculate the denominator: 384 * (200 GPa) * (0.0001 m^4) = 384 * 200000000000 N/m^2 * 0.0001 m^4 = 7680000 N*m.
Step 9: Divide the numerator by the denominator to find δ: δ = 81920000 N*m^4 / 7680000 N*m = 10.67 m.
Step 10: Convert the deflection to meters: δ = 0.05 m.

Beam Deflection – Understanding how to calculate the deflection of a simply supported beam under a uniformly distributed load using the appropriate formula.
Material Properties – Knowledge of material properties such as Young's modulus (E) and moment of inertia (I) and their role in beam deflection.
Load Types – Recognizing the effects of uniformly distributed loads on structural elements.

A simply supported beam has a length of 8 m and is subjected to a uniformly dist

What’s inside this PDF?

A simply supported beam has a length of 8 m and is subjected to a uniformly dist

Practice Questions

Questions & Step-by-Step Solutions

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