Question

A simply supported beam of span L and flexural rigidity EI is subjected to a concentrated load of W at mid span. The ratio of maximum deflection to maximum slope anywhere in the beam is

• A. $L/2$
• B. $L/3$
• C. $L/4$
• D. $L/5$

Hint:

Memorizing the standard deflection and slope formulas for simply supported and cantilever beams is essential for solving these types of problems quickly.
- SS Beam, Center Load P: $\delta_{max} = PL^3/(48EI)$, $\theta_{max} = PL^2/(16EI)$.
- SS Beam, UDL w: $\delta_{max} = 5wL^4/(384EI)$, $\theta_{max} = wL^3/(24EI)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the ratio of the maximum deflection to the maximum slope for a simply supported beam with a point load at the center.

Step 2: Key Formula or Approach:
We need the standard formulas for maximum deflection ($\delta_{max}$) and maximum slope ($\theta_{max}$) for this loading case.
1.

Maximum Deflection: Occurs at the center, under the load.
\[ \delta_{max} = \frac{WL^3}{48EI} \] 2.

Maximum Slope: Occurs at the supports (A and B).
\[ \theta_{max} = \frac{WL^2}{16EI} \]

Step 3: Detailed Explanation:
Now we need to find the ratio $\frac{\delta_{max}}{\theta_{max}}$.
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{WL^3 / (48EI)}{WL^2 / (16EI)} \] We can cancel the common terms $W$, $L^2$, and $EI$ from the numerator and denominator:
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{L/48}{1/16} \] \[ \frac{\delta_{max}}{\theta_{max}} = \frac{L}{48} \times 16 \] \[ \frac{\delta_{max}}{\theta_{max}} = \frac{16L}{48} \] Simplifying the fraction:
\[ \frac{\delta_{max}}{\theta_{max}} = \frac{L}{3} \]

Step 4: Final Answer:
The ratio of maximum deflection to maximum slope is $L/3$.