Question

A simply supported beam of span 2L and flexural rigidity EI is subjected to a uniformly distributed load of w/m throughout the span. The deflection at mid span is

• A. $\frac{5 wL^4}{384 EI}$
• B. $\frac{5 wL^4}{48 EI}$
• C. $\frac{5 wL^4}{24 EI}$
• D. $\frac{1 wL^4}{48 EI}$

Hint:

Be very careful when a problem uses non-standard notation for span length (like 2L instead of L).
Always start with the standard formula you have memorized and then carefully substitute the specific variables given in the problem.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the formula for the maximum deflection (at mid-span) of a simply supported beam of a given length and under a UDL.

Step 2: Key Formula or Approach:
The standard formula for the maximum deflection ($\delta_{max}$) of a simply supported beam of span 'S' under a uniformly distributed load 'w' over its entire length is:
\[ \delta_{max} = \frac{5 w S^4}{384 EI} \]

Step 3: Detailed Explanation:
In this problem, the variables are given with slightly different notation. We must be careful to substitute correctly.
- The given span is $S = 2L$.
- The given load intensity is $w$.
Now, we substitute $S=2L$ into the standard formula:
\[ \delta_{max} = \frac{5 w (2L)^4}{384 EI} \] \[ \delta_{max} = \frac{5 w (16L^4)}{384 EI} \] \[ \delta_{max} = \frac{80 wL^4}{384 EI} \] Now, simplify the fraction 80/384. We can divide both by 16:
$80 / 16 = 5$
$384 / 16 = 24$
So the expression becomes:
\[ \delta_{max} = \frac{5 wL^4}{24 EI} \]

Step 4: Final Answer:
The deflection at mid span is $\frac{5 wL^4}{24 EI}$.