Free-body diagrams and equilibrium of rigid bodies

Why it matters

Every calculation in statics, strength of materials and machine design starts with one picture: the free-body diagram (FBD). If the FBD is right, the equations of equilibrium almost solve themselves; if a force is missing or points the wrong way, every later stress, deflection and safety factor is wrong. Support reactions on beams, forces in truss members, friction on ladders and loads on machine parts are all found this way.

Key ideas

Rigid body. A body whose particles keep fixed distances from each other. Deformations are assumed small enough not to change the geometry used in the equilibrium equations. This is the assumption of statics; strength of materials then studies the small deformations separately.

Free-body diagram. Isolate the body from everything it touches. Replace every contact and support by the force (and moment) it can exert, and add every applied load and the self-weight. Steps:

1. Choose the body (whole structure, one member, one joint, or one part of a beam cut at a section).
2. Sketch it alone, with its dimensions.
3. Draw known loads: point loads, distributed loads, couples, weight at the centre of gravity.
4. Draw unknown reactions according to the support type, with an assumed direction. A negative answer simply means the actual direction is opposite.
5. Choose axes and a positive sense for moments.

Support reactions in 2D.

• Roller or smooth surface: one force, normal to the surface.
• Pin (hinge): two force components (Rx, Ry), no moment.
• Fixed (built-in) support: Rx, Ry and a moment M.
• Cable or rope: one tension, pulling away from the body along the cable.
• Two-force member (pinned at both ends, no load in between): the force acts along the line joining the pins.

Equilibrium. A rigid body is in static equilibrium when the resultant force and the resultant moment about any point are both zero. A body moving with constant velocity (no rotation) also satisfies these equations. In a plane there are exactly three independent equations, so at most three unknowns can be found from one FBD.

Alternative sets of equations. Instead of ΣFx, ΣFy and one moment equation, you may use one force equation plus two moment equations (the two moment centres must not lie on a line perpendicular to the force direction used), or three moment equations about three non-collinear points. Taking moments about a point where two unknowns meet removes them from the equation, which is the main trick for fast solutions.

Special cases.

• Two-force body: the two forces are equal, opposite and collinear.
• Three-force body: three non-parallel forces must be concurrent (meet at one point); if parallel, they must balance as a parallel system. This is the basis of Lami's theorem.

Static determinacy. If the number of unknown reactions equals the number of independent equilibrium equations (3 in a plane) and the supports prevent all motion, the body is statically determinate. More unknowns make it indeterminate (compatibility of deformation is then needed, see the strength-of-materials topics). Fewer, or badly arranged supports (for example three parallel reactions or three concurrent reactions), leave it a mechanism, which is improperly constrained.

Distributed loads. Replace a distributed load by its resultant for computing reactions only: a uniform load w over length a gives w·a acting at the middle of that length; a triangular load of peak w over length a gives w·a/2 acting at a/3 from the larger end. Do not use the resultant to find internal shear and bending inside the loaded region.

Formulas

ΣFx = 0, ΣFy = 0, ΣM_O = 0

• Forces in N, moments in N·m; O is any point. Valid for a rigid body in static equilibrium in a plane.

M_O = F · d

• M_O = moment of force F about O (N·m), d = perpendicular distance from O to the line of action of F (m).

M_O = x·Fy − y·Fx

• Moment of a force with components (Fx, Fy) applied at (x, y) relative to O; positive anticlockwise.

F1 / sin α = F2 / sin β = F3 / sin γ (Lami's theorem)

• For three concurrent coplanar forces in equilibrium; α, β, γ are the angles between the other two forces (the angle opposite each force).

W = w·a acting at a/2 (uniform load); W = w·a/2 acting at a/3 from the peak end (triangular load)

• w in N/m, a in m.

Worked examples

Example 1 (standard). A simply supported beam AB of span 6 m (pin at A, roller at B) carries a point load of 30 kN at 2 m from A and a uniform load of 4 kN/m over the whole span. Find the reactions.

Given: L = 6 m, P = 30 kN at x = 2 m, w = 4 kN/m.

1. Resultant of the uniform load: W = w·L = 4 × 6 = 24 kN, acting at 3 m from A.
2. No horizontal loads, so ΣFx = 0 gives the horizontal reaction at A as zero.
3. Moments about A (anticlockwise positive): R_B × 6 − 30 × 2 − 24 × 3 = 0, so R_B = (60 + 72)/6 = 22 kN.
4. Vertical forces: R_A + R_B = 30 + 24 = 54 kN, so R_A = 54 − 22 = 32 kN.
5. Check with moments about B: R_A × 6 = 30 × 4 + 24 × 3 = 120 + 72 = 192, so R_A = 32 kN. ✓

Answer: R_A = 32 kN (upward), R_B = 22 kN (upward).

Example 2 (GATE level). A uniform ladder 5 m long, weight 200 N, rests against a smooth vertical wall with its foot on a rough floor, making 60° with the floor. A person of weight 600 N stands 4 m up the ladder (measured along it). Find the wall reaction, the friction force at the floor, and the least coefficient of friction that prevents slipping.

Given: L = 5 m, W_L = 200 N at 2.5 m, W_P = 600 N at 4 m, θ = 60°.

1. FBD: at the wall, a horizontal normal force N_w only (smooth). At the floor, a normal force N_f upward and friction F toward the wall. Weights act downward.
2. Vertical forces: N_f = 200 + 600 = 800 N.
3. Moments about the foot. Horizontal lever arm of a weight at distance s along the ladder is s·cos θ; vertical height of the top is L·sin θ: N_w × 5 sin 60° = 200 × 2.5 cos 60° + 600 × 4 cos 60° N_w × 4.330 = (500 + 2400) × 0.5 = 1450 N·m, so N_w = 334.9 N.
4. Horizontal forces: F = N_w = 334.9 N.
5. No slip requires F ≤ μ·N_f, so μ_min = 334.9 / 800 = 0.419.

Answer: N_w ≈ 334.9 N, F ≈ 334.9 N, μ_min ≈ 0.42. (Without the person, μ_min = 57.7/200 = 0.289; the person raises it because the load moves up the ladder.)

Common mistakes

• Leaving out a force: self-weight, the moment at a fixed support, or friction at a rough contact.
• Drawing internal forces between parts of the same body on the FBD; only external forces belong there.
• Using the slant length instead of the perpendicular distance as the lever arm.
• Showing friction at a smooth surface, or a roller reaction that is not normal to its surface.
• Replacing a distributed load by its resultant and then using that resultant to find bending moment inside the loaded span.
• Treating a negative answer as an error. It only means the assumed direction was wrong; keep the sign consistent in later steps.
• Mixing kN and N, or kN·m and N·mm, in the same equation.

For GATE PI

Expect short numerical questions on support reactions of simply supported, overhanging and cantilever beams with point, uniform and triangular loads; ladders and blocks with friction; Lami's theorem with cables and pulleys; and conceptual questions on two-force and three-force members and determinacy. Practise taking moments about the point that eliminates the most unknowns, and always check the answer with a second, independent moment equation.

Quick check

1. How many unknowns does a fixed support introduce in a planar problem?
2. A uniform 4 m beam of weight 400 N is simply supported at its ends. What is each reaction?
3. Three non-parallel forces keep a body in equilibrium. What must be true of their lines of action?
4. A triangular load has a peak of 6 kN/m over 3 m. What is its resultant and where does it act?

Answers: 1. Three (Rx, Ry, M). 2. 200 N each. 3. They must meet at one point (be concurrent). 4. 9 kN, acting 1 m from the peak end (one third of the length).