A simply supported beam of span l, carrying udl of w on entire span, slope at both end will be

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Examoogle · Accepted Answer

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

Explanation & Concepts:
For a simply supported beam of span $l$ subjected to a uniformly distributed load (udl) of $w$ per unit length over its entire span, the maximum slope occurs at the supports. The slope ($ heta$) at both ends is given by the expression $	heta = \frac{wl^3}{24EI}$, where $E$ is Young's Modulus and $I$ is the Moment of Inertia.

Working Principle: According to the double integration method or Macaulay's method, the governing differential equation for beam deflection is $EI\frac{d^2y}{dx^2} = M_x$. By integrating this equation twice and applying boundary conditions (deflection $y=0$ at $x=0$ and $x=l$), we derive the expression for the slope $\frac{dy}{dx}$. At the supports ($x=0$ and $x=l$), this slope reaches its maximum magnitude, resulting in the standard formula.

Key Formulas: $	heta = \frac{wl^3}{24EI}$ — Slope at the ends of a simply supported beam with UDL; $\delta_{max} = \frac{5wl^4}{384EI}$ — Maximum deflection at the center of a simply supported beam with UDL

Key Points: The slope is maximum at the supports and zero at the mid-span for a symmetric UDL.; The units of slope are in radians (dimensionless).; Flexural rigidity ($EI$) inversely affects both slope and deflection.

Advantages: Predictable deformation behavior; Standardized coefficients for structural analysis

Applications: Design of floor beams; Structural analysis of simply supported girders

Additional Info: Option A ($\frac{Wl^2}{16EI}$) is incorrect; it relates to different boundary conditions.. Option B ($\frac{5wl^4}{384EI}$) represents the maximum central deflection, not the slope.. Option D ($\frac{Wl^3}{48EI}$) is the maximum deflection for a simply supported beam under a central point load $W$.

Answer: C is correct — The slope at the ends of a simply supported beam under UDL is determined by the formula $\frac{wl^3}{24EI}$.

Exam Tip: Remember that 'slope' terms typically contain $l^3$ while 'deflection' terms contain $l^4$ for distributed loads; use this dimensional analysis to verify your formulas during the exam.

Examoogle · Answer

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

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

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

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