Equilibrium equations for rigid bodies
A rigid body is in equilibrium only if the forces balance and the moments balance. Choosing where to take moments can make the algebra almost trivial.
The three equations
For a rigid body in a plane, equilibrium requires \(\sum F_x = 0\), \(\sum F_y = 0\), and \(\sum M_O = 0\) about any point O. Unlike particles, rigid bodies can rotate, so the moment equation is essential.
Simply supported beam
A 6 m beam is pinned at A and on a roller at B, with a 12 kN load 2 m from A. Taking moments about A: By x 6 = 12 x 2, so By = 4 kN. Then the vertical force balance gives Ay = 12 - 4 = 8 kN, and Ax = 0 since there are no horizontal loads. The support nearer the load carries more.
Choosing the moment point
Take moments about a point where two unknown lines of action intersect, such as a pin, so that the moment equation contains only one unknown. Each equation then gives one answer directly, avoiding simultaneous equations.
Cantilevers
A 2 m cantilever beam fixed at a wall carries 5 kN at its free end. The wall's reactions are 5 kN upward and a couple of 5 x 2 = 10 kN·m resisting rotation. The fixed end must be designed for this moment, which is why balconies and diving boards need strong anchorage.
Distributed loads on beams
Replace distributed loads with their resultants before writing equilibrium equations. A 6 m simply supported beam with a uniform 3 kN/m load has an 18 kN resultant at its centre, so each support carries 9 kN.
Statical indeterminacy
If a body has more unknown reactions than independent equations, such as a beam with three supports, it is statically indeterminate, and the reactions depend on how the beam deforms. Mechanics of materials courses solve such cases. Redundant supports make structures safer, since one can fail without collapse.
Improper constraints
Even with enough reactions, supports can be arranged so the body can still move, such as three parallel reactions or three reactions all passing through one point. Such a body is improperly constrained and will not stay in equilibrium under general loads.
Worked example
A 4 m diving board, pinned at A and on a roller 1.2 m from A, carries a 700 N diver at the tip and its own 300 N weight at its midpoint, 2 m from A. Moments about A: By x 1.2 = 700 x 4 + 300 x 2 = 3,400, so By is about 2,833 N upward. Vertical balance gives Ay = 1,000 - 2,833 = -1,833 N, so the pin pulls the board down with 1,833 N.
Key takeaways
A planar rigid body is in equilibrium when sum Fx, sum Fy, and the sum of moments about any point all equal zero. Take moments about a point where unknowns intersect to isolate one unknown per equation. A simply supported beam shares a point load in inverse proportion to distances; a cantilever's fixed end carries both the load and a moment. Replace distributed loads with resultants first. More unknowns than equations means statically indeterminate; poorly arranged supports mean improper constraint.
Flashcard
Sum Fx = 0, sum Fy = 0, sum M = 0
Check your understanding
A 6 m beam is pinned at A and on a roller at B, with 12 kN at 2 m from A. What is the reaction at B?