The simply supported beam carries a uniformly distributed load w over its whole span. Find the maximum bending moment Mmax in the beam.
Shear force & bending moment worksheet with answers
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
- 1A (x = 0 m)
- 2B (x = 2 m)
- 3C (x = 4 m)
- 4D (x = 6 m)
- 5E (x = 8 m)
- 6F (x = 10 m)
- 7G (x = 12 m)
The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. It carries a point load P1 between the supports and a point load P2 at C. Find the bending moment Mx under the load P1. Use sagging moments positive and upward shear on the left face positive.
Find the bending moment Mx at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x0 from A to the section where the bending moment is largest.
The beam rests on a pin at A and a roller at B and overhangs B by 2 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the bending moment MB at support B. Use sagging moments positive and upward shear on the left face positive.
Answer key
- 1. Mmax = 9.375 kN·m
- 2. C (x = 4 m)
- 3. Mx = 67.5 kN·m
- 4. Mx = 94.5 kN·m
- 5. x0 = 4.55 m
- 6. MB = −6 kN·m
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