In one line
At a fixed end, the vertical and horizontal reactions balance the loads, and a fixing moment balances the total moment of all loads about the support.
Explained simply
When you hold a heavy bag at arm's length, your shoulder pushes up to hold the weight and also has to stop your arm from rotating down. A fixed support does both jobs: a force and a moment.
Technically, a fixed support supplies \( H_A \), \( V_A \) and \( M_A \), so a cantilever is statically determinate and can be solved from the free end without needing another support.
How it works
- Draw the FBD with \( H_A \), \( V_A \) and \( M_A \) at the fixed end.
- Convert distributed loads into point loads.
- \( \Sigma F_x = 0 \Rightarrow H_A \); \( \Sigma F_y = 0 \Rightarrow V_A = \Sigma \) vertical loads.
- \( \Sigma M_A = 0 \Rightarrow M_A = \Sigma (P_i a_i) \pm \) applied couples.
- State the sense of \( M_A \): for downward loads on a beam fixed at its left end, \( M_A \) is anticlockwise.
Key formula
\( V_A = \Sigma P_i \), \( M_A = \Sigma P_i a_i \) (kN·m), where \( a_i \) is the distance of load \( P_i \) from the fixed end.
Quick example
A 3 m cantilever fixed at A with 5 kN at the free end B and a UDL of 2 kN/m over its full length. UDL = 6 kN at 1.5 m. \( V_A = 5 + 6 = 11 \) kN upward; \( H_A = 0 \); \( M_A = 5(3) + 6(1.5) = 24 \) kN·m anticlockwise.
Common mistakes
- Forgetting the fixing moment entirely.
- Taking moments about the free end and then calling that value \( M_A \).
- Giving \( M_A \) without its sense.
How MU asks it
Asked as part of 5 to 8-mark beam numericals, often in a combined figure (cantilever with UVL and a couple). Show the FBD with all three unknowns and state each answer with direction.