Free body diagram checklist
Almost every statics error starts with a wrong or missing force on the diagram. Work through this list every time, in this order.
The checklist
- Isolate the body. Draw the beam, truss or object on its own, with no supports, walls or ground.
- Replace every support with its reactions.
- Pin: a horizontal and a vertical reaction, \(H\) and \(V\).
- Roller: one reaction, perpendicular to the surface it rolls on.
- Fixed support: \(H\), \(V\) and a moment \(M\).
- Add every applied load at the point where it acts. Replace a uniformly distributed load \(w\) (in kN/m) over length \(L\) with its resultant \(wL\), acting at the middle of that length.
- Include self-weight if the question gives a mass or weight for the body.
- Mark dimensions between every force and the supports. You need them for moments.
- Choose a sign convention and write it down: for example, up and right positive, anticlockwise moments positive.
- Count the unknowns. A two-dimensional body has three equilibrium equations, so three unknowns at most if the structure is statically determinate.
Then apply equilibrium:
\[ \sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M = 0. \]Take moments about a support with an unknown reaction. That reaction drops out of the equation.
Worked examples
Example 1: point load
A 6 m beam is pinned at \(A\) and on a roller at \(B\). A 12 kN load acts 2 m from \(A\). Find the reactions.
No horizontal loads act, so \(H_A = 0\).
Moments about \(A\) (anticlockwise positive):
\[ R_B \times 6 - 12 \times 2 = 0 \implies R_B = 4 \text{ kN} \]Vertical forces:
\[ R_A + R_B - 12 = 0 \implies R_A = 8 \text{ kN} \]\(R_A = 8\) kN up, \(R_B = 4\) kN up
Check: the reaction nearer the load is larger. It should be.
Example 2: distributed and point loads
An 8 m beam is pinned at \(A\) and on a roller at \(B\). It carries a uniformly distributed load of 3 kN/m along its full length and a 10 kN point load 6 m from \(A\). Find the reactions.
Replace the distributed load with its resultant: \(3 \times 8 = 24\) kN, acting 4 m from \(A\).
Moments about \(A\):
\[ R_B \times 8 - 24 \times 4 - 10 \times 6 = 0 \] \[ 8R_B = 156 \implies R_B = 19.5 \text{ kN} \]Vertical forces:
\[ R_A + 19.5 - 24 - 10 = 0 \implies R_A = 14.5 \text{ kN} \]Check with moments about \(B\): \(14.5 \times 8 - 24 \times 4 - 10 \times 2 = 116 - 96 - 20 = 0\).
\(R_A = 14.5\) kN up, \(R_B = 19.5\) kN up