Question

A cantilever beam of span 2 m carries a concentrated load of 48 kN at the free end. If the flexural rigidity of beam EI = 20,000 kNm$^2$, the slope at the free end is

• A. $3.2 \times 10^{-3}$ radians
• B. $4.8 \times 10^{-3}$ radians
• C. $48 \times 10^{-3}$ radians
• D. $64 \times 10^{-3}$ radians

Hint:

Memorize the standard formulas for cantilever beam slope and deflection:
- Load at Free End (P):
- Max Slope = $PL^2 / (2EI)$
- Max Deflection = $PL^3 / (3EI)$
- UDL over entire span (w):
- Max Slope = $wL^3 / (6EI)$
- Max Deflection = $wL^4 / (8EI)$

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to calculate the slope at the free end of a cantilever beam subjected to a point load at its tip.

Step 2: Key Formula or Approach:
For a cantilever beam of length $L$ with a point load $P$ at the free end, the standard formula for the slope ($\theta$) at the free end is:
\[ \theta_{free-end} = \frac{PL^2}{2EI} \]

Step 3: Detailed Explanation:
First, identify the given values. It's important to ensure the units are consistent.

• Load ($P$) = 48 kN

• Span ($L$) = 2 m

• Flexural Rigidity ($EI$) = 20,000 kNm$^2$


All units (kN and m) are consistent, so no conversion is needed. The resulting slope will be in radians.
Now, substitute the values into the formula:
\[ \theta = \frac{(48 \text{ kN}) \times (2 \text{ m})^2}{2 \times (20,000 \text{ kNm}^2)} \] \[ \theta = \frac{48 \times 4}{40,000} \] \[ \theta = \frac{192}{40,000} \] \[ \theta = 0.0048 \text{ radians} \] To express this in the format of the options, we write it in scientific notation:
\[ \theta = 4.8 \times 10^{-3} \text{ radians} \]

Step 4: Final Answer:
The slope at the free end is $4.8 \times 10^{-3}$ radians.