Test: Bending Moment - Civil Engineering (CE) MCQ

The points of contraflexure (or inflection) are points of zero bending moment, i.e. where the beam changes its curvature from hogging to sagging.
In a bending beam, a point of contraflexure is a location where the bending moment is zero (changes its sign). In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero lines. 
At the point of contra flexure, the bending moment is zero.

is the equation of parabola.

L > 2a

For vertical equilibrium

Taking moment about R₁ , we have

The bending moment at the point of application of load =Area under OABC

The pillar will have the greatest tendency of overturning when the Moment of the force F induced in the rope will have a maximum moment about the base of the pillar A.
Hence, To find the greatest overturning tendency, we have to maximise M_A.
Calculation:
Given,
Length of the rope = l, 
let y = height from the ground at which rope is fixed to the pillar.
Taking moment about A we get,
M_A = F × cos θ × y
From ΔABC, y = l × sin θ 
M_A = Fl × cos θ × sin θ  = Fl/2 x sin2θ
For the maximum value of M_A, sin 2θ = 90° , θ = 45 ° 
y = l × sin θ y = l × sin 45° = l/√ 2

Here make hit and trial method
Consider first option (A) of moment diagram.

At right end B, slope must be zero as there is no shear force at B so option A is wrong.
Now, consider option (B)
Due to symmetric load intensity at left side too the shear force equal to zero so slope at left side must be zero. So option (B) wrong.

Now consider option (C)
Here at both ends, slope is zero means shear force is zero and also when we move from right to left, the rate of increase of shear force decreases due triangular shape of load intensity and at middle slope should be maximum and there after decreases so in option (C) all these criterion fulfills. Here (C) is the correct option.

As far as (D) option is concerned its middle part slope and right most part slope is strictly not agreeing with the load shown.

• When a rotating body is involved then there comes a force which is known as Centrifugal force (F_c)
  It is calculated by using formula
  F_c = mr(ω)²
• Now in the Question, it is mentioned that end Q is just lifted off the ground. So as it is just lifted off the ground there will be no reaction force from that point.

∴ FBD of mobile

M = mass of mobile
m = eccentric mass
r = eccentricity
Hence, Taking a moment about point P = O
∴ Mg × 0.06 - F_c × (0.09) = 0
∴ 90 × 10^-3 × 9.81 × 0.06 = mr(ω)² × (0.09)
90 × 10^-3 × 9.81 × 0.06 = 2 × 10^-3 × 2.19 × 10^-3 (ω)² × 0.09
∴ ω² = 134380.8964
∴ ω = 366.58
∴2πN/60 = 366.58
∴ N = 3500 rpm

Moment about (1),

Moment about (2),

From (i), (ii), (iii), we get 

Taking moment about A,