In one line
Draw the FBD, replace distributed loads by equivalent point loads, then use \( \Sigma M = 0 \) about one support and \( \Sigma F = 0 \) to get the reactions.
Explained simply
Two people carrying a long pole with a bag hung on it: the one closer to the bag carries more. Finding support reactions is working out exactly how much each support 'carries'.
Technically, a simply supported or overhanging beam has three unknowns (\( H_A, V_A \) at the hinge, \( R_B \) at the roller), found from \( \Sigma F_x = 0, \Sigma F_y = 0, \Sigma M = 0 \).
How it works
- Draw the FBD: hinge A with \( H_A, V_A \); roller B with \( R_B \).
- Convert UDL/UVL into point loads; resolve inclined loads.
- \( \Sigma M_A = 0 \) gives \( R_B \) directly (only one unknown).
- \( \Sigma F_y = 0 \) gives \( V_A \); \( \Sigma F_x = 0 \) gives \( H_A \).
- Check with \( \Sigma M_B = 0 \).
- For an overhanging beam, loads on the overhang give moments of opposite sense; a reaction may come out negative (acting downward).
Key formula
\( \Sigma M_A = 0 \Rightarrow R_B = \dfrac{\Sigma (P_i a_i) + \Sigma C}{L} \), with \( a_i \) the distance of each load from A, L the span, C the applied couples with sign.
Quick example
Beam AB, 6 m span, 12 kN at 2 m from A, UDL 2 kN/m over the last 3 m. UDL = 6 kN at 4.5 m from A. \( \Sigma M_A = 0 \): \( 6R_B = 12(2) + 6(4.5) = 51 \), \( R_B = 8.5 \) kN. \( \Sigma F_y = 0 \): \( V_A = 18 - 8.5 = 9.5 \) kN. \( H_A = 0 \). Check about B: \( 6V_A = 12(4) + 6(1.5) = 57 \), \( V_A = 9.5 \) kN.
Common mistakes
- Using the span length instead of the actual distance of each load from the moment centre.
- Ignoring the horizontal reaction when an inclined load acts.
- Not checking with a second moment equation.
How MU asks it
A guaranteed 6 to 8-mark numerical: 'Find the support reactions of the beam loaded as shown'. Full marks need: FBD, equivalent loads, three equilibrium equations with sign convention, results with units and directions, and the check. Internal-hinge beams are excluded.