A simply supported beam of span l, carrying point load W at centre of span, deflection at centre will be

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Option B: $\frac{Wl^3}{48EI}$

Explanation & Concepts:
For a simply supported beam of span $l$ subjected to a central point load $W$, the maximum deflection occurs at the center. This deflection is determined using the double integration method or Macaulay's method, resulting in the standard formula $\delta = \frac{Wl^3}{48EI}$.

Working Principle: The beam undergoes bending due to the applied moment induced by the point load. The curvature $\kappa$ is related to the bending moment $M$ by the elastic curve equation $\frac{d^2y}{dx^2} = \frac{M}{EI}$. Integrating this equation twice with appropriate boundary conditions at the supports (where deflection $y=0$ at $x=0$ and $x=l$) yields the deflection at any point $x$, which peaks at $x = l/2$.

Key Formulas: $\delta_{max} = \frac{Wl^3}{48EI}$ — Maximum deflection at center; $	heta = \frac{Wl^2}{16EI}$ — Slope at supports

Key Points: Maximum deflection occurs at the location of the point load.; $E$ is the Young's Modulus and $I$ is the Moment of Inertia.; The deflection is inversely proportional to the flexural rigidity $EI$.

Advantages: Standardized design parameter for structural stiffness.; Simple derivation using Euler-Bernoulli beam theory.

Disadvantages: Applicable only for linear elastic behavior.; Assumes small deflections (small angle approximation).

Applications: Civil structural design of floor beams.; Mechanical design of drive shafts and load-bearing members.

Additional Info: Option A: $\frac{5wl^4}{384EI}$ represents the deflection of a simply supported beam under a Uniformly Distributed Load (UDL).. Option C: $\frac{wl^3}{24EI}$ is a dimensional variant, not standard for this case.. Option D: $\frac{Wl^2}{16EI}$ relates to the slope at the support for a central point load, not deflection.

Answer: B is correct — The deflection at the center of a simply supported beam with a central point load $W$ is given by $\frac{Wl^3}{48EI}$.

Exam Tip: Remember that UDL produces a higher power of length ($l^4$) in the numerator, while point loads produce $l^3$.

Examoogle · Answer

Option A: $\frac{5wl^4}{384EI}$

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Option C: $\frac{wl^3}{24EI}$

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Option D: $\frac{Wl^2}{16EI}$

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