Problem 4 · MMD May 2019
16 m beam, max BM and contraflexure
A uniform beam AB 16 m long weighs 160 kg. It is supported at points two metres from each end. A load of 15 kg is at a point 6 m from end A. Draw the SF and BM diagram for the beam. What is the maximum value of the BM and the position on the beam where it occurs? Calculate the point on the beam between the load and end B where the BM is nil.
- Kilogrammes treated as kgf, as written in the paper.
Loading diagram
R1
90 kgf
2 m from A
R2
85 kgf
14 m from A
Max sagging BM
191.244 kg·m
at 7.533 m
Max hogging BM
-20 kg·m
at 2 m
Asked in the paper
SF and BM diagrams
Drag across the plots to read values. Sagging BM is drawn positive.
Step 1
Reactions
w = 10 kg/m. ΣV = 175 kg. Moments about C (2 m): 160×6 + 15×4 = 1020 = RD×12, so RD = 85 kg and RC = 90 kg.
Step 2
Maximum BM
Sagging BM is maximum where SF crosses zero between the 15 kg load and D. SF = 75 − 10x = 0 ⇒ x = 7.5 m from A. Mmax = 191.25 kg·m sagging.
Step 3
Point of contraflexure
Between the load and B, set 75x − 90 − 5x² = 0 ⇒ x² − 15x + 18 = 0. The root in range is x = (15 + √153)/2 ≈ 13.685 m from A (2.315 m from B).