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Question 46Hard
A beam carrying a UDL has two equal overhangs. For the maximum bending moment in the beam to be as small as possible, the ratio of each overhang to the total length of the beam is:
Symmetric loading: R_A = R_B = 4W/2 = 2W. The maximum shear is in the end bays: 2W.
Question 49Hard
A lever ABDC is supported on hinges at A and C (AB = BD = DC = 1 m). A vertical arm 1 m high is fixed at B, and a horizontal force of 3 kN acts at its top. The bending moment at B is:
The 3 kN force at 1 m above B applies a couple of 3 × 1 = 3 kN·m to the lever at joint B. Reactions ±1 kN; the beam moment jumps at B from 1 to 2 kN·m (total 3 kN·m).
Question 50Easy
In a cantilever beam with a UDL, the shear force varies following the:
V at distance x from the free end = wx — a straight line. (The bending moment wx²/2 follows the parabolic law.)
Question 51Medium
A simply supported beam carries a couple somewhere in the span. It would produce: 1. a rectangular SF diagram; 2. parabolic BM diagrams; 3. both +ve and −ve BM, which are maximum at the point of application of the couple. Of these statements:
Reactions ±M/L ⇒ constant SF (rectangle). No distributed load ⇒ BM linear (not parabolic), with a jump at the couple: −Ma/L and +Mb/L. Statements 1 and 3 are true.
Question 52Easy
A cantilever of span l has a load P at the free end. The bending moment at the free end is:
M at distance x from the free end = −Px. At the free end x = 0, so M = 0.
Question 53Medium
A beam ABC is supported at A and B (span L) and has an overhang BC with a couple M applied at the free end C. The shear force at the mid-span point D of AB is:
ΣM_A = 0: R_B·L = M ⇒ R_B = M/L, R_A = −M/L. Between A and B the shear is M/L (in magnitude).
Question 54Easy
A simply supported beam carries a distributed load of zero intensity over one support, increasing linearly to w over the other support. The shape of the BMD is:
M_x = (wl/6)x − wx³/(6l): an x³ term ⇒ cubic parabola.
Question 55Easy
Two identical simply supported beams of span l carry equal total loads W: one as a central point load, the other as a UDL over the entire span. The ratio of their mid-span bending moments is:
dM/dx = V: the slope of the BMD at a section equals the shear force there.
Question 59Medium
A simply supported beam of span L carries a couple M at distance a from the left support (b from the right). The shear force at the point of contraflexure is: