Simply Supported Beam with UDL A simply supported beam of span l carries a uniformly distributed load w over the whole span. What are the shear forces at the two supports (beam ends)?

Introduction:
Support reactions for standard loading cases are frequently used to sketch SFD and BMD quickly. For a symmetric UDL on a simply supported beam, reactions (and hence end shears) are equal and opposite at the two supports.

Given Data / Assumptions:

• Beam: simply supported, span l.
• Load: UDL of intensity w across entire span.
• Linear elastic behavior; negligible self-weight beyond w if not included.

Concept / Approach:
The resultant load equals w * l and acts at midspan. By symmetry, each support carries half the resultant: R_A = R_B = w * l / 2. Shear just to the right of the left support equals +R_A and just to the left of the right support equals -R_B (sign convention).

Step-by-Step Solution:
Compute resultant: W_total = w * l.Use symmetry: R_A = R_B = W_total / 2 = w * l / 2.Shear at ends: V_left = + w * l / 2, V_right = - w * l / 2.

Verification / Alternative check:
Area under SFD over the span equals the net change in bending moment, which matches the known parabolic BMD M_max = w * l^2 / 8 at midspan.

Why Other Options Are Wrong:
zero at both ends: contradicts equilibrium; supports must resist load.w l and - w l: each end carrying full load is impossible for symmetric simply supported case.w l^2 / 2 terms have wrong dimensions for shear force.

Common Pitfalls:
Confusing shear force with bending moment expressions (note power of l).Sign convention mistakes at the right support.

Final Answer:
w l / 2 at one end and - w l / 2 at the other end