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Equilibrium of Rigid Body Example

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A simply supported beam of length 6 m carries a single vertical point load P at 3 m from the left support. Find the reactions at the two supports.

For a rigid body in equilibrium in 2D: the sum of forces in the x and y directions is zero, and the sum of moments about any point is zero. Use \sum M_{A} = 0 to find one reaction, then \sum F_{y} = 0 to find the other (for a beam with no horizontal loads, \sum F_{x} = 0 is satisfied if supports are vertical). \sum F_{x} = 0, \sum F_{y} = 0, \sum M_{O} = 0 Notation: R_{A}, R_{B} = reactions, P = applied load, M_{A} = moment about point A.

A beam is simply supported at A (left) and B (right), 6 m apart. A vertical point load P = 12 kN acts downward at the midpoint (3 m from A). Find: Reactions at A and B (assume vertical reactions only).

Span 6 m; load P = 12 kN at midspan (3 m from A) Supports at A and B (pin and roller, or both vertical reactions)

Taking moments about A eliminates R_A. Clockwise positive: \sum M_{A} = 0 : - P (3) + R_{B} (6) = 0 \Rightarrow R_{B} = \frac{P}{2}

\sum F_{y} = 0 : R_{A} + R_{B} - P = 0 \Rightarrow R_{A} = P - \frac{P}{2} = \frac{P}{2} P = 12 kN \Rightarrow R_{A} = R_{B} = 6 kN ↑

R_{A} = R_{B} = 6 kN upward (each support carries half the load by symmetry).

\sum F = 0, \sum M = 0 (2D rigid body)