Free-Body Diagrams, Reactions, Shear, and Moment

Free-Body Diagrams, Reactions, Shear, and Moment

For a July 2026 PE Civil: Structural problem, use the current NCEES PE Civil Reference Handbook supplied for the test date. Diagram mechanics are governed by equilibrium, but the handbook controls notation and formulas available during the CBT exam. A plausible number is not enough: its free-body diagram, units, and signs must describe the same physical system.

Start with the body and its supports

A free-body diagram isolates one body from its surroundings. Show all applied loads, dimensions, axes, and the actions exerted by removed supports or connected bodies. In a two-dimensional idealization:

Do not add a horizontal reaction to a frictionless horizontal roller or a moment to a pin merely to make the equations work. Conversely, a fixed support can develop a reaction couple even when the applied loading is vertical. If an internal hinge divides a structure, draw separate free bodies when its zero-moment condition supplies useful information. Equal-and-opposite hinge forces appear on the two pieces.

Count unknown reactions only after idealizing the supports. Then choose a convenient origin and declare positive directions. For a planar body, use ΣFx = 0, ΣFy = 0, and ΣM = 0. Taking moments about a point through multiple unknown forces often eliminates them. A negative result does not mean failure; it means the actual reaction acts opposite the assumed arrow.

Distributed loads and internal cuts

For external equilibrium, replace a distributed load w(x) by its resultant W = ∫w(x) dx acting at the centroid of the load diagram. A uniform load w over length L becomes wL at midlength. A triangular load has area wmax L/2 and acts one-third of the base length from its larger end. Keep force-per-length units distinct from force. For a trapezoid or piecewise load, split the loading into simple shapes, sum their areas, and locate the combined resultant from first moments. Verify that its line of action lies within the loaded interval.

To find internal actions, cut the member at coordinate x and isolate either side. Expose axial force N, shear V, and bending moment M. This section uses tension-positive N, positive shear consistent with dM/dx = V, and positive sagging moment. Another convention is acceptable only if every equation and diagram follows it. Forces on opposite faces of the cut must be equal and opposite.

With downward distributed load defined as positive w, the differential checks are

dV/dx = -w(x) and dM/dx = V(x).

Therefore no distributed load gives constant shear and linear moment; uniform load gives linear shear and parabolic moment; linearly varying load gives parabolic shear and cubic moment. A downward point load causes a downward jump in V equal to its magnitude. An applied concentrated couple causes a jump in M; a point force alone does not. A local maximum or minimum of a smooth moment diagram usually occurs where V = 0.

Worked beam from reactions through diagrams

A simply supported 20-ft beam has a pin at A, a roller at B, a uniform downward load of 1.2 kip/ft over the full span, and an 8-kip downward point load 6 ft from A. The uniform-load resultant is 1.2(20) = 24 kip at 10 ft. No horizontal load exists, so Ax = 0.

Taking moments about A,

By(20) - 24(10) - 8(6) = 0,

so By = 14.4 kip. Vertical equilibrium gives Ay = 24 + 8 - 14.4 = 17.6 kip. Both are upward.

For 0 < x < 6 ft,

V(x) = 17.6 - 1.2x,

so V(6-) = 10.4 kip. The point load produces an 8-kip downward jump: V(6+) = 2.4 kip. For 6 < x < 20 ft,

V(x) = 17.6 - 1.2x - 8 = 9.6 - 1.2x.

Thus V = 0 at x = 8 ft, and V(20-) = -14.4 kip. The upward reaction at B jumps shear back to zero. The net change from the distributed load, point load, and both reactions is zero, matching vertical equilibrium.

Using sagging moment as positive and the left segment,

• for 0 ≤ x ≤ 6, M(x) = 17.6x - 1.2x²/2;
• for 6 ≤ x ≤ 20, M(x) = 17.6x - 1.2x²/2 - 8(x - 6).

The moment is continuous at the point load: M(6) = 84.0 kip-ft. At the zero-shear location, M(8) = 86.4 kip-ft, the maximum. At the simple support, M(20) = 0, as required. The moment curve is parabolic throughout because the uniform load remains present, while its slope changes continuously with shear.

Fast reliability checks

Before selecting an answer, verify:

1. Reactions plus applied forces balance in each direction, and moments balance about an independent point.
2. The shear jump at each point force equals that force with sign.
3. The change in moment between two locations equals the signed area under the shear diagram.
4. Diagram slopes and curvature match w, V, and M.
5. Pin and roller end moments are zero unless an external end couple is applied; a fixed end need not be zero.
6. Units progress correctly from kip/ft to kip to kip-ft.

These checks make the sketch a second solution, not decoration.