Guide
How these diagrams are drawn
The same sign convention and equilibrium steps used in MMD written papers. Worked into every problem in this studio.
Sign convention
- Shear force at a cut is the vertical resultant of all forces to the left. Upward on the left is positive.
- Bending moment is positive when the beam sags (hogging is negative). This is the marine / Capt. Subramaniam convention.
- Point loads and UDLs are drawn downward. Knife-edge reactions are drawn upward unless the support is holding the beam down.
Finding reactions
- Replace self-weight with a UDL w = W / L, or a single load W at the centroid if you only need reactions.
- ΣV = 0 gives RA + RB = total downward load.
- ΣM about one support gives the other reaction. Check by summing moments about the second support.
- A beam on a rigid flat bed has no concentrated reactions. The bed pushes up at total load / length, uniformly.
- A cantilever wall supplies both a vertical force and a fixing moment.
Walking the beam
SF jumps by a concentrated force (up for a reaction, down for a load). Under a UDL, SF changes linearly. BM is the running integral of SF, so a linear SF makes a parabolic BM. Maximum sagging or hogging BM occurs where SF crosses zero, or at a concentrated load if the crossing is a jump.
| Load | SF shape | BM shape |
|---|---|---|
| Point load | Vertical jump | Kink, still continuous |
| UDL | Straight slope | Parabola |
| No load (between points) | Horizontal | Straight slope |
Standard results
- Simply supported, UDL w over ℓ: RA = RB = wℓ/2, Mmax = wℓ²/8 at mid-span.
- Simply supported, point load W at mid-span: RA = RB = W/2, Mmax = Wℓ/4.
- Cantilever, tip load W: SF at wall = W, BM at wall = −Wℓ.
- Cantilever, UDL w: SF at wall = wℓ, BM at wall = −wℓ²/2.
Units in these papers
MMD questions mix N, kN, kg and tonnes. Masses such as 61.162 kg are chosen so that × 9.81 gives a round newton load. Where the paper writes kg or t as a load, this studio keeps those as kgf / tonne-force so the printed numbers match. Convert with g = 9.81 when the examiner wants SI.