Question

A fixed beam of span 8 m and flexural rigidity $EI = 51,200$ kNm$^2$ is subjected to a concentrated load of 96 kN at mid span. The maximum deflection under the load is

• A. $1.25 \times 10^{-3}$ mm
• B. $2.5 \times 10^{-3}$ mm
• C. 5 mm
• D. 2.5 mm

Hint:

The maximum deflection for a fixed beam is much smaller than for a simply supported beam under the same load.
- Fixed Beam, Center Load: $\delta_{max} = \frac{PL^3}{192EI}$
- SS Beam, Center Load: $\delta_{max} = \frac{PL^3}{48EI}$
The fixed beam is $192/48 = 4$ times stiffer.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the maximum deflection of a fixed-end beam with a concentrated load at its center.

Step 2: Key Formula or Approach:
The standard formula for the maximum deflection ($\delta_{max}$) of a fixed beam of span $L$ subjected to a point load $P$ at mid-span is:
\[ \delta_{max} = \frac{PL^3}{192EI} \]

Step 3: Detailed Explanation:
First, identify the given values. Ensure the units are consistent.

• Load ($P$) = 96 kN

• Span ($L$) = 8 m

• Flexural Rigidity ($EI$) = 51,200 kNm$^2$


All units are in kN and m. The result will be in meters.
Substitute the values into the formula:
\[ \delta_{max} = \frac{(96 \text{ kN}) \times (8 \text{ m})^3}{192 \times (51,200 \text{ kNm}^2)} \] \[ \delta_{max} = \frac{96 \times 512}{192 \times 51,200} \] We can simplify this. Note that $192 = 2 \times 96$ and $51,200 = 100 \times 512$.
\[ \delta_{max} = \frac{96 \times 512}{(2 \times 96) \times (100 \times 512)} \] Cancel the terms 96 and 512:
\[ \delta_{max} = \frac{1}{2 \times 100} = \frac{1}{200} \text{ m} \] \[ \delta_{max} = 0.005 \text{ m} \] The options are in mm. Convert the result to mm:
\[ \delta_{max} = 0.005 \text{ m} \times 1000 \frac{\text{mm}}{\text{m}} = 5 \text{ mm} \]

Step 4: Final Answer:
The maximum deflection under the load is 5 mm.