Shear Force & Bending Moment - 3 - Free MCQ Practice Test with solutions,

C) Uniformly varying:
Because the shear force is determined by the location of the load and varies from one end to the point of load application.

From compatibilty equation, deflection due to R + deflection to w = 0

Next, from the overall equilibrium (balancing the total vertical forces on the beam):

V_A + R + V_C = wL

2V + 5/8 wL = wL [from symmetry V_A = V_C = V ]

Thus,
V = (3wL) / (8×2) = (3wL) / 16.

Now, let at a section x distance apart from support B, the bending moment at the section is:
Mx = (3wL / 16) × x - (wx²) / 2 = 0.

Thus,
x / 2 = 3 / 16 L
Therefore,
x = 3L / 8 (from both ends).

The bending moment due to 50 kNm moment would be hogging. Similarly the bending moment due to 30 kNm would be sagging. Therefore the bending moment diagram is,

∑F_V = 0
⇒ R_A + R_C = 20
M_B = 0 [from right]
∴ R_C x 4 - 20 x 6 = 0
⇒ R_c x 4 = 20 x 6 

M_B = 0 [from left]
R_A x 4 - M_A = 0
⇒ M_A = R_A x 4 = -10 x 4
= - 40 kNm
The fixed end moment at A will be in clockwise direction, opposite to the direction shown in the figure.
∴ M_A = -40 kNm

Let the overhang percentage in 100p.
So overhang = pl
The shear force diagram will be,

For B.M. at mid point to be zero, area of shear force diagram under (1) should be equal to that of (2),
1/2w(pl)² = 1/2w[(0.5 - P)l]²
or p² = (0.5 - p)² = 0.25 - p + p²
∴ p = 0.25 or 25%

Due to hinge at C the part CB of beam be have s as simply supported beam. Thus reaction at B will be in upward direction and equals half of the load in span BC.
Span CA be haves as a cantilever beam. There fore B.M.D. will be sagging in BC and hogging in AC and it will be parabolic.
For a cantiiever B.M.D. will be as shown below.

Combining all the arguments answer will be (d) which is given with BM plotted on tension side.

In the case of (-section, nearly 80-90% shear force, is resisted by web and only 10-20% is resisted by flanges, whereas in bending. 80-90% moment is resisted by flanges and only 10-20% by the web.