Question

A simply supported beam of length 'L' having the flexural rigidity of 'EI', subjected to a uniformly distributed load of 'w'. Then the maximum bending moment and maximum deflection are

• A. \( \frac{\text{wL}^2}{2} \, \text{and} \, \frac{5\text{wL}^4}{384 \, \text{EI}} \)
• B. \( \frac{\text{wL}^2}{8} \, \text{and} \, \frac{\text{wL}^4}{192 \, \text{EI}} \)
• C. \( \frac{\text{wL}^2}{8} \, \text{and} \, \frac{5\text{wL}^4}{384 \, \text{EI}} \)
• D. \( \frac{\text{wL}^2}{2} \, \text{and} \, \frac{\text{wL}^4}{192 \, \text{EI}} \)

Hint:

For common beam configurations and loading types, the formulas for maximum bending moment and maximum deflection are standard results. It's crucial to memorize or be able to quickly derive these for frequently encountered scenarios, such as simply supported beams with uniformly distributed loads or concentrated loads. Remember that \(M_{max} = \frac{\text{wL}^2}{8}\) and \(\delta_{max} = \frac{5\text{wL}^4}{384 \, \text{EI}}\) for a simply supported beam with UDL over its entire span.

Answer 1

The Correct Option is C

Step 1: Identify the type of beam and loading condition.
The problem specifies a "simply supported beam of length 'L'".
It is "subjected to a uniformly distributed load of 'w'".
"EI" represents the flexural rigidity of the beam.
A simply supported beam is a structural element that is supported at its two ends in such a way that it can freely rotate at the supports but cannot undergo vertical displacement. A uniformly distributed load (UDL), denoted by 'w', implies that the load is spread evenly across the entire length of the beam.
Step 2: Determine the maximum bending moment for a simply supported beam with UDL.
For a simply supported beam of length 'L' subjected to a uniformly distributed load 'w' over its entire span, the maximum bending moment (\(M_{max}\)) occurs at the mid-span of the beam.
This is a standard formula in the study of mechanics of materials. The formula for the maximum bending moment in this specific case is: \[ M_{max} = \frac{\text{wL}^2}{8} \]
Step 3: Determine the maximum deflection for a simply supported beam with UDL. For the same simply supported beam of length 'L' with a uniformly distributed load 'w' over its entire span, the maximum deflection (\(\delta_{max}\)) also occurs at the mid-span of the beam. This is also a standard formula in the study of mechanics of materials. The formula for the maximum deflection in this specific case is: \[ \delta_{max} = \frac{5\text{wL}^4}{384 \, \text{EI}} \] Here, EI is the flexural rigidity, which is a measure of the beam's resistance to bending.
Step 4: Compare the derived formulas with the given options. Now, let's compare the established formulas for \(M_{max}\) and \(\delta_{max}\) with the provided options: Calculated Maximum Bending Moment: \( \frac{\text{wL}^2}{8} \)
Calculated Maximum Deflection: \( \frac{5\text{wL}^4}{384 \, \text{EI}} \) Let's check each option: % Option (1) \( \frac{\text{wL}^2}{2} \, \text{and} \, \frac{5\text{wL}^4}{384 \, \text{EI}} \) - The maximum bending moment is incorrect (should be \( \frac{\text{wL}^2}{8} \)). % Option (2) \( \frac{\text{wL}^2}{8} \, \text{and} \, \frac{\text{wL}^4}{192 \, \text{EI}} \) - The maximum deflection is incorrect (should be \( \frac{5\text{wL}^4}{384 \, \text{EI}} \)). % Option (3) \( \frac{\text{wL}^2}{8} \, \text{and} \, \frac{5\text{wL}^4}{384 \, \text{EI}} \) - This option correctly matches both the maximum bending moment and the maximum deflection formulas. % Option (4) \( \frac{\text{wL}^2}{2} \, \text{and} \, \frac{\text{wL}^4}{192 \, \text{EI}} \) - Both the maximum bending moment and maximum deflection are incorrect. Therefore, Option (3) provides the correct formulas for both the maximum bending moment and the maximum deflection for a simply supported beam subjected to a uniformly distributed load. The final answer is \( \boxed{\text{3}} \).