1.9 Equilibrium of rigid bodies in 2D and support reactions
After this lesson you should be able to:
- Apply the three planar equilibrium equations to rigid bodies.
- Compute support reactions of simply supported, overhanging and cantilever beams under point, distributed and inclined loads.
- Choose moment centres that solve for one unknown at a time and check results independently.
Theory
A rigid body in a plane is in equilibrium when the resultant force and the resultant moment about any point are both zero. That gives three independent scalar equations: ΣFx = 0, ΣFy = 0 and ΣM_O = 0. They can be replaced by other independent sets, such as one force equation and two moment equations about different points, or three moment equations about three points not on one line.
The art lies in choosing equations that contain one unknown each. Taking moments about a point where two unknown reactions act removes both, so the third comes out directly. For a simply supported beam, moments about one support give the other support's reaction immediately; vertical force balance then gives the first, and a moment equation about the second support provides an independent check.
Distributed loads are replaced by their resultants for this purpose (lesson 1.6), and inclined loads are resolved into components along and across the member. A horizontal component is resisted only by supports that can take horizontal force: a pin, not a roller.
The cantilever is the standard fixed-support case. Its fixed end supplies a vertical force, a horizontal force and a moment; the fixing moment equals the moment of all loads about the support. Overhanging beams are simply supported beams with one or both ends extending past the supports; a load on the overhang can lift the far support, giving a downward (negative) reaction that must be held down by self-weight or anchorage.
Derivation
- For a rigid body, Newton's laws give ΣF = m a_G and ΣM_G = I_G α. In equilibrium a_G = 0 and α = 0.
- So ΣF = 0, giving ΣFx = 0 and ΣFy = 0 in a plane.
- If ΣF = 0 and ΣM_G = 0, the moment about any other point O is ΣM_O = ΣM_G + r_OG × ΣF = 0, so moments may be taken about any convenient point.
- Two moment equations about points A and B plus one force equation in a direction not perpendicular to AB form an independent set; three moment equations about non-collinear points also do.
ΣFx = 0, ΣFy = 0, ΣM_O = 0Simply supported beam, span L, point load P at a from A: R_B = P a / L, R_A = P (L − a) / LCantilever, length L, uniform load w and end load P: V = w L + P, M = w L² / 2 + P LUniform load on simply supported span: R_A = R_B = w L / 2
- The body is rigid and supports are ideal (pin, roller, fixed).
- Loads are static and applied slowly.
- The structure is statically determinate (three well-arranged reaction components); otherwise compatibility of deformations is also needed (structural analysis).
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 equilibrium itself. Design loads entering these calculations come from IS 875 (Parts 1 to 5) and, for bridges, IRC:6; load combinations and partial factors come from the material codes (for example IS 456:2000 Table 18). EN 1990 and ASCE 7 play the same roles elsewhere; listed for comparison, not ranked.
Worked examples
Problem. A simply supported beam AB of span 8 m (pin at A, roller at B) carries a 20 kN point load at 2 m from A, a uniform load of 5 kN/m from 4 m to 8 m, and a 10 kN load at 6 m inclined at 60° to the beam, pointing downwards and to the right. Find the reactions.
- Inclined load components: horizontal 10 cos 60° = 5 kN (to the right); vertical 10 sin 60° = 8.660 kN (downwards).
- Uniform load resultant: 5 × 4 = 20 kN at 6 m from A.
- ΣM_A = 0: R_B × 8 = 20 × 2 + 20 × 6 + 8.660 × 6 = 40 + 120 + 51.96 = 211.96, so R_B = 26.50 kN.
- ΣFy = 0: R_Ay = 20 + 20 + 8.660 − 26.50 = 22.16 kN.
- ΣFx = 0: R_Ax = 5 kN acting to the left (the roller at B cannot take horizontal force).
- Check, ΣM_B = 0: R_Ay × 8 − 20 × 6 − 20 × 2 − 8.660 × 2 = 177.3 − 120 − 40 − 17.3 ≈ 0.
Answer. R_Ax = 5 kN (leftwards), R_Ay = 22.16 kN, R_B = 26.50 kN
Problem. A 4 m cantilever balcony beam carries a uniform load of 3 kN/m (finishes and imposed load combined) and a 6 kN parapet load at its tip. Find the reactions at the fixed end.
- Total vertical load = 3 × 4 + 6 = 18 kN, so the vertical reaction V = 18 kN upwards.
- No horizontal load, so H = 0.
- Fixing moment M = 3 × 4² / 2 + 6 × 4 = 24 + 24 = 48 kN·m (hogging at the support).
Answer. V = 18 kN, H = 0, M = 48 kN·m
Step-by-step procedure
- Draw the FBD with all loads and reaction unknowns.
- Resolve inclined loads; replace distributed loads by resultants.
- Take moments about a point where two unknowns act to find the third.
- Use force equations for the rest.
- Check with an independent moment equation; resolve any negative sign into the actual direction.
Site notes
- A beam that overhangs a support (such as a cantilever slab or a scaffold beam) can lift off its back support when the overhang is loaded; check uplift and provide hold-down if needed.
- When propping a beam during construction, each prop becomes a support. Removing props early changes the reactions and moments the beam was designed for.
- For long precast elements lifted at two points, the lifting points are chosen so that the reactions (sling forces) and the resulting moments stay within the element's capacity.
- Allowing a roller to resist horizontal force.
- Taking the moment of a uniform load as w × L × L instead of w L × L/2.
- Forgetting the moment reaction at a fixed support.
- Using an inclined load's full magnitude in the vertical equation.
- Skipping the independent check equation.
For any simply supported beam with vertical loads only, R_A + R_B equals the total load, and each reaction lies between zero and the total load. A load at mid-span shares equally; a load at a support goes entirely to that support.
Summary
- Three independent equations for a planar rigid body.
- Choose moment centres to isolate one unknown at a time.
- Fixed support: V, H and M; pin: two forces; roller: one.
- Always check with an independent equation.
Practice questions
Tap a question to see the answer.
1A simply supported beam of 6 m span carries a 30 kN load at mid-span. Find the reactions.
15 kN at each support.
2A beam is supported at A (x = 0) and B (x = 4 m) and carries a 10 kN load at C (x = 6 m) on the overhang. Find the reactions.
ΣM_A: R_B × 4 = 10 × 6, so R_B = 15 kN upward; R_A = 10 − 15 = −5 kN, i.e. 5 kN downward (hold-down needed).
3A 2 m cantilever carries only a 10 kN·m couple at its free end. Find the fixed-end reactions.
V = 0, H = 0, M = 10 kN·m opposite in sense to the applied couple.
4A simply supported beam of 5 m span carries 4 kN/m over its full length. Find the reactions.
Total 20 kN; 10 kN at each support.
5Why does the roller at B carry no horizontal reaction even when an inclined load acts on the beam?
A roller allows horizontal movement, so it can supply only a force normal to its rolling surface; the pin at A takes the whole horizontal component.
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.