Simply Supported Rectangular Beam — Deflection under Central Point Load A rectangular beam of span l, simply supported at both ends, carries a central point load W. Where along the beam does the maximum deflection occur?

Introduction:
Deflection shapes provide insight into stiffness and serviceability. For statically determinate beams with symmetric loading, the location of maximum deflection can often be inferred from symmetry and slope conditions without solving the full differential equation.

Given Data / Assumptions:

• Simply supported beam, span l.
• Central point load W.
• Prismatic beam; linear elastic bending; small deflection theory.

Concept / Approach:
The loading and boundary conditions are symmetric about midspan, so the deflected shape is symmetric and slope is zero at the center. The point of zero slope is typically an extremum of deflection, giving the location of maximum deflection at the center.

Step-by-Step Solution:
Write standard result for midspan deflection: y_max = (W * l^3) / (48 * E * I).Symmetry dictates dy/dx = 0 at x = l / 2.Boundary conditions y(0) = y(l) = 0 at the supports; thus the interior extremum occurs at midspan.

Verification / Alternative check:
Area-moment method or conjugate-beam method both yield the same midspan location and value.

Why Other Options Are Wrong:
at the ends: deflection is constrained to zero at simple supports.at l / 3: not an extremum for this loading; applies to other cases (e.g., certain distributed loads) only under special conditions.none of these: incorrect since midspan is correct.

Common Pitfalls:
Confusing the location of maximum moment (also at midspan here) with other load cases where it shifts.

Final Answer:
at the centre