Beams, Supports & Reactions

Go through the supports one at a time and ask a single question of each: which motions does it forbid? That count is the number of unknowns the support contributes, and the forbidden directions fix where those unknowns point. A roller, rocker or smooth bearing blocks motion into its contact surface only, so it supplies one force perpendicular to that surface and lets the beam slide and turn. A pin blocks translation in both directions but lets the beam rotate, so it supplies Aₓ and A_y and no moment. A fixed end, such as a beam built into a wall or welded to a column, also blocks rotation and adds a couple M_A in N m. A cable or slender link supplies one force along its own line, and a cable can only pull. Bridges put a roller at one end for a reason: a 40 m steel span warming by 30 K grows by 12 × 10⁻⁶ K⁻¹ × 40 m × 30 K = 14 mm, which a roller allows and a second pin would fight.

One rigid body in a plane has three independent equations, ΣFₓ = 0, ΣF_y = 0 and ΣM = 0, so count the reaction components r first. A pin plus a roller, or a single fixed end, gives r = 3: statically determinate, with every reaction fixed by statics for any load. Pin the beam at both ends and r = 4. The vertical reactions still follow, but the two horizontal ones appear only as Aₓ + Bₓ = 0, so any equal and opposite axial push satisfies every equation; the thermal squeeze of a pinned bridge is exactly that undetermined pair. A propped cantilever, a fixed end plus a roller, also has r = 4. Too few is worse: two rollers give r = 2, and the first sideways nudge moves the beam. And r = 3 is not enough on its own. Three parallel reactions cannot resist a sideways load, and three whose lines meet at one point cannot resist a moment about it.

A moment equation ignores every force whose line of action passes through the chosen point, so choose the point that removes the most unknowns. Take a 6.0 m beam with a pin at A, a roller at B on the far end, 12 kN at 2.0 m and 8.0 kN at 5.0 m. Moments about A remove both pin components at once: 6.0 B_y = 12 × 2.0 + 8.0 × 5.0 = 64 kN m, so B_y = 10.7 kN. Vertical balance leaves A_y = 20 − 10.7 = 9.3 kN. Then check with moments about B, a route that shares no arithmetic with the first: 6.0 A_y = 12 × 4.0 + 8.0 × 1.0 = 56 kN m, so A_y = 9.3 kN again. The idea scales up. When three unknown lines of action cross in pairs at three separate points, a moment equation about each crossing contains exactly one unknown, and no simultaneous solving is needed.

A cable's line of action is known from the geometry, so it adds only one unknown, its tension. Take a boom 4.0 m long, pinned to a wall at A, with a cable from its tip E to an anchor D a height h above A and 6.0 kN hanging 3.0 m out. Moments about A see only the cable's vertical component: T sinθ × 4.0 m = 6.0 kN × 3.0 m. Moments about D, where the lines of T and A_y meet, give Aₓ h = 18 kN m. Moments about E give 4.0 A_y = 6.0 × 1.0, so A_y = 1.5 kN whatever the anchor height. With h = 3.0 m, sinθ = 0.60, so T = 7.5 kN and Aₓ = 6.0 kN. Lower the anchor to h = 1.0 m and sinθ = 0.243, so T = 18.6 kN and Aₓ = 18 kN. A shallow cable buys its lift with a large horizontal pull, which the pin and the wall must resist. A negative tension would mean the cable has to push, so it has gone slack.

Assume a direction for every unknown and let the sign report back. A negative A_y at a pin means the pin pulls down, which a bolt can do. A support that works only by contact cannot. A 4.0 m plank weighing 300 N rests on trestles 0.5 m and 2.5 m from its left end, and a 700 N painter walks out along the right-hand overhang. Moments about the right trestle give the left reaction: R_left × 2.0 m = 300 N × 0.5 m − 700 N × s, where s is the painter's distance beyond that trestle. R_left reaches zero at s = 150/700 = 0.21 m. One step further, the algebra returns a trestle that would have to hold the plank down. The real plank lifts off the left trestle and rotates about the right one. The equations did not fail; the contact model did, and the sign said so.

An indeterminate structure is not unsolvable, only unsolvable by statics. Rest a uniform beam on three equally spaced supports. Equilibrium says the reactions add up to the weight, symmetry says the two ends match, and nothing says how much the middle support takes. Elastic beam theory, which lets the beam bend, puts 62.5% of the weight on the middle support and 18.75% on each end. Lower the middle support by a few millimetres and its share falls, because the split is set by stiffness and fit, the very things a rigid body leaves out. The ideal supports are models too. A bolted base plate resists some moment, so it sits between a pin and a fixed end, and a corroded roller carries some sideways force. State the support model you chose, and treat the reactions as belonging to that model.