1.7 Free-body diagrams
After this lesson you should be able to:
- Isolate a body and draw a complete, correct free-body diagram (FBD).
- Show the reactions provided by common supports and connections, with the right number of unknowns.
- Use the FBD to count unknowns and decide whether equilibrium equations can solve the problem.
Theory
A free-body diagram is a sketch of a body cut free from everything it touches, showing every external force and couple acting on it. It is the single most important step in statics: almost every error in an equilibrium problem can be traced to a missing or wrong force on the FBD. Drawing it well forces you to decide exactly what the body is and what acts on it.
External forces fall into two groups. Applied forces are known loads: weights, imposed loads, wind, water pressure, prestress. Reactions are the forces that supports and connections exert to stop motion; they are unknown until equilibrium is applied. Each support prevents certain movements, and it provides one reaction component for every movement it prevents.
In two dimensions, a roller or a smooth surface prevents movement in one direction only, so it gives one force normal to the surface. A cable or a link gives one force along its own line; a cable can only pull. A pin or hinge prevents movement in both directions but allows rotation, so it gives two force components. A fixed (built-in) support prevents both movements and rotation, so it gives two force components and a moment. Friction at a rough surface adds a tangential force opposing impending motion.
Once the FBD is drawn, count the unknowns. A planar rigid body has three independent equilibrium equations. If there are three unknown reactions, properly arranged, the body is statically determinate. More unknowns mean static indeterminacy; fewer, or a poor arrangement such as three parallel or three concurrent reactions, mean the body is a mechanism and cannot be in equilibrium under general loads.
Derivation
- Each support restrains some of the three rigid-body movements in a plane (translation in x, translation in y, rotation).
- By Newton's third law, a restrained movement is matched by a reaction of the support on the body: a force for a translation, a moment for a rotation.
- Hence reactions per support: roller 1, link or cable 1, pin 2, fixed 3.
- Equilibrium of a planar rigid body supplies three equations (ΣFx = 0, ΣFy = 0, ΣM = 0). With r reaction unknowns: r = 3 and a proper arrangement gives a determinate body; r > 3 is indeterminate to degree r − 3; r < 3 is a mechanism.
Roller or smooth contact: 1 force normal to the surfaceCable or link: 1 force along its axis (cable in tension only)Pin or hinge: 2 force componentsFixed support: 2 force components + 1 momentDeterminacy of a single rigid body in 2D: r = 3 (determinate), r > 3 (indeterminate, degree r − 3)
- Supports are idealised: real bearings, bolted joints and footings fall between the ideal pin, roller and fixed cases.
- Self-weight acts at the centre of gravity.
- Unknown reactions may be drawn in any assumed sense; a negative answer means the actual sense is opposite.
Codes and standards
Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.
- No design code governs drawing FBDs. Support idealisation matters in design: for example, IS 456:2000 cl. 22.2 sets the effective span of beams from support conditions, and IS 800:2007 and EN 1993-1-8 classify steel joints as simple, semi-rigid or rigid. Listed for comparison, not ranked.
Worked examples
Problem. A 50 kg block is held at rest on a smooth plane inclined at 30° by a rope parallel to the plane. Draw the FBD and find the rope tension and the normal reaction.
- FBD: weight W = 50 × 9.81 = 490.5 N vertically down; normal reaction N perpendicular to the plane; tension T up the plane. No friction (smooth).
- Resolve along the plane: T − W sin 30° = 0, so T = 490.5 × 0.5 = 245.25 N.
- Resolve perpendicular to the plane: N − W cos 30° = 0, so N = 490.5 × 0.8660 = 424.79 N.
Answer. T = 245.3 N; N = 424.8 N
Problem. A 6 m beam has a pin at A (left end) and a roller at B (right end). It carries a 12 kN point load at 2 m from A and its self-weight of 2 kN/m. Draw the FBD, count the unknowns and find the reactions.
- FBD: A has Ax and Ay; B has By (vertical); applied loads are 12 kN at 2 m and the self-weight resultant 2 × 6 = 12 kN at 3 m.
- Unknowns: 3 (Ax, Ay, By) with 3 equations: statically determinate.
- ΣFx = 0 gives Ax = 0 (no horizontal load).
- ΣM_A = 0: By × 6 − 12 × 2 − 12 × 3 = 0, so By = 60 / 6 = 10 kN.
- ΣFy = 0: Ay = 12 + 12 − 10 = 14 kN.
Answer. Ax = 0, Ay = 14 kN, By = 10 kN
Step-by-step procedure
- Decide which body (or part of a structure) to isolate and draw its outline alone.
- Mark all applied loads with values, directions and positions.
- At each cut support or connection, draw the reaction components that support provides.
- Add dimensions and the coordinate axes; label unknowns.
- Count unknowns against available equations before solving.
Site notes
- Temporary works failures often start with a wrong support assumption, such as treating a prop on soft ground as a fixed support. Draw the FBD with the support you can actually guarantee.
- A bearing that is meant to slide (roller) but is jammed by debris becomes a pin and attracts horizontal forces it was not designed for; inspect bearings for freedom of movement.
- Cables, slings and ties can only pull. If equilibrium gives a negative tension, the arrangement will not work as drawn.
- Leaving out the self-weight of the body.
- Including internal forces between parts of the same isolated body.
- Drawing a moment at a pin or a horizontal reaction at a roller.
- Showing a cable pushing on the body.
- Forgetting the reaction of the ground or wall that the body touches.
Count before you calculate: in 2D a single rigid body needs exactly three well-arranged reaction components to be statically determinate. Three parallel reactions, or three meeting at one point, cannot hold a general load.
Summary
- An FBD shows the isolated body with every external force and couple.
- Supports supply one reaction per prevented movement: roller 1, pin 2, fixed 3.
- Unknown senses may be assumed; signs sort them out.
- Compare unknowns with the three planar equilibrium equations.
Practice questions
Tap a question to see the answer.
1How many reaction components does a fixed support provide in a plane problem?
Three: two force components and a moment.
2How many unknowns do a pin and a roller provide together, and is a beam on them determinate?
2 + 1 = 3 unknowns; with three equations it is statically determinate (if the roller is not parallel to the line through the pin).
3A 20 kg lamp hangs from a single vertical cord. Find the tension.
T = 20 × 9.81 = 196.2 N.
4A 3 m cantilever carries a 5 kN load at its free end. Find the reactions at the fixed end.
Vertical force 5 kN upward and a moment of 5 × 3 = 15 kN·m (anticlockwise if the cantilever extends to the right).
5What reaction does a smooth surface provide?
A single force normal to the surface at the contact point.
6A beam has two pins and one roller. How many reaction unknowns does it have and what is its degree of indeterminacy?
2 + 2 + 1 = 5 unknowns; degree of indeterminacy 5 − 3 = 2.
References
Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.
- F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
- J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
- S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.