Question

If the deflection at the free end of a uniformly loaded cantilever beam is 15mm and the slope of the deflection curve at the free end is 0.02 radian, then the length of the beam is

• A. 0.8 m
• B. 1.0 m
• C. 1.2 m
• D. 1.5 m

Hint:

For beam deflection and slope problems, it's crucial to remember the standard formulas for different loading conditions (e.g., uniformly distributed load, point load) and beam types (e.g., cantilever, simply supported). When given both deflection and slope, dividing one by the other often simplifies the problem by eliminating common unknown terms like $w$, $E$, or $I$, allowing for direct calculation of the beam length. Ensure consistent units throughout the calculation.

Answer 1

The Correct Option is B

Step 1: Identify the given parameters and relevant formulas.
For a uniformly loaded cantilever beam:
Deflection at the free end ($\delta_{max}$) = 15 mm = 0.015 m Slope of the deflection curve at the free end ($\theta_{max}$) = 0.02 radian
Let $L$ be the length of the beam.
Let $w$ be the uniformly distributed load per unit length.
Let $E$ be the modulus of elasticity and $I$ be the moment of inertia of the beam's cross-section.
The standard formulas for a uniformly loaded cantilever beam are:
Deflection at the free end: $\delta_{max} = \frac{wL^4}{8EI}$
Slope at the free end: $\theta_{max} = \frac{wL^3}{6EI}$
Step 2: Use the relationship between deflection and slope to find the length.
Divide the equation for deflection by the equation for slope: $$ \frac{\delta_{max}}{\theta_{max}} = \frac{\frac{wL^4}{8EI}}{\frac{wL^3}{6EI}} $$ Simplify the expression: $$ \frac{\delta_{max}}{\theta_{max}} = \frac{wL^4}{8EI} \times \frac{6EI}{wL^3} $$ Cancel out common terms ($w$, $E$, $I$, and $L^3$): $$ \frac{\delta_{max}}{\theta_{max}} = \frac{6L}{8} $$ $$ \frac{\delta_{max}}{\theta_{max}} = \frac{3L}{4} $$
Step 3: Substitute the given values and solve for L.
Substitute $\delta_{max} = 0.015$ m and $\theta_{max} = 0.02$ radian into the simplified equation: $$ \frac{0.015}{0.02} = \frac{3L}{4} $$ $$ 0.75 = \frac{3L}{4} $$ Now, solve for $L$: $$ 3L = 0.75 \times 4 $$ $$ 3L = 3 $$ $$ L = \frac{3}{3} $$ $$ L = 1.0 \text{ m} $$
Step 4: Compare the calculated length with the given options.
The calculated length of the beam is 1.0 m, which matches option 2. The final answer is $\boxed{\text{1.0 m}}$.