Bending Moment MCQ Quiz in বাংলা - Objective Question with Answer for Bending Moment - বিনামূল্যে ডাউনলোড করুন [PDF]

Explanation:

Case-I:

The bending moment diagram for a fixed beam subjected to UDL throughout the span is as shown below:

The maximum bending moment is given as, \(M=\frac{wL^2}{24}\).

Case-II:

Net weight of the UDL =  wL = W

RA + RB = wL

Due to symmetry,

\({R_A} = \frac{{wL}}{2};{R_B} = \frac{{wL}}{2}\)

Taking moment about B

\(M = {R_A}x - \frac{{w{x^2}}}{{2}} = \frac{{wL}}{2}x - \frac{{w{x^2}}}{2}\)

For maximum B.M,

\(\frac{{dM}}{{dx}} = 0\;i.e.\;\frac{{wL}}{2} - \frac{{w.2x}}{2} = 0\)

\(x = \frac{L}{2}\)

\({M_{max}} = \frac{{wL}}{2} \times \frac{L}{2} - \frac{{w{L^2}}}{8} = \frac{{w{L^2}}}{8}\)

The ratio between case-I and case-II will be:

\(\frac{\frac{wL^2}{8}}{\frac{wL^2}{24}}=\frac{wL^2}{8}\times\frac{24}{wL^2}=3\)

Concept:

Equilibrium of a beam is satisfied by following there equations

∑f_x = 0

∑f_y = 0

∑M = 0

And for a distributed load 

Total load = wL, where w is the distributed load/unit length, and L is the length of that section of the beam where distributed load is acting

And acts at the centroid of the section in which distributed load is acting.

Calculation:

Total load adding on the beam \(=4~\left( \frac{kN}{m} \right)\times 1\left( m \right)=~4\text{ }\!\!~\!\!\text{ kN}\)

From Vertical Equilibrium

R_A + R_C = 4 kN

Now from moment equilibrium

∑M_A = 0

R_C × 2 – 4 × 0.5 = 0

R_C = 1 kN

R_A = 3 kN

Explanation:

Explanation:

M_A 

Using equilibrium equations:

∑F_y = 0 ⇒ V_A = 20 + 30 + 10 + 30 = 90 KN

∑MA = 0 ⇒ 20 × 0.5 + 30 × 1.5 + 10 × 2.2 + 30 × 3 = M_A

_MA = 167 kN.m

So, the bending moment at A is 167 kN.m in the anticlockwise direction.

Concept:

R_P+R_Q=10

ΣM_P = 0, R_Q × 1 - 10 × 0.5 = 0

R_Q = 5 kN and R_P 5 kN (upward force)

Since this is a case of a propped cantilever, deflection δ at point P will be zero, which will give an opposite 5 kN force acting in the downward direction at point P as shown in the last diagram.

Thus, M_O = R_P × OP = 5 × 2 = 10 kN-m

Concept

Moment:

• It is the measure of the tendency of a force to cause a body to rotate about a specific point or axis.
• It occurs every time when the force is applied such that it does not passes through the centroid of the body.
• Hence, It is defined as the product of the magnitude of force (F) and the perpendicular distance (d) and is given by:

Moment = Force × ⊥r distance

⇒ M = F × d

Here,

F – Magnitude of force

d – lever arm or moment arm

Sign convention:

• For the left end, the clockwise moment is taken as the positive and anticlockwise moment is taken as negative.
• For the right end, the anticlockwise moment is taken as the positive and clockwise moment is taken as negative

Fixed Moment → Are always considered as negative

Twisting Moment → Can be positive or negative.

Normal Moment → No such moment exists

Zero Moment → Can be considered as positive.

Concept:

Moment at hinged point is always zero. Therefore M_A = M_B = 0

Total M at C = R_A × 0.5 + M_C = R_B × 0.5 + MC  

Force Balance ⇒ RA + R_B = 100 N......(1)

Calculation:

Moment equation about  A ⇒ 100 × 0.5 + 10 = R_B × 1

⇒ 50 + 10 = R_B

⇒ R_B = 60 N

R_A = 100 – 60 = 40 N

Now, MA = MB = 0

M_C = 100 N × 100 mm = 100 × 0.1 m = 10 Nm

Therefore, Total M at point C = RA × 0.5 + M_C = 40 × 0.5 + 10 = 30 N

And Total Moment at point C = R_B × 0.5 + MC = 60 × 0.5 - 10 = 20 N

So, from above the maximum moment at point C is 30 N

Concept:

The maximum bending moment at a cantilever beam subjected to UDL is given by

\(M=\frac{wL^2}{2}\)

where M = Maximum bending moment of cantilever beam, L = Length of the beam, w = load per unit length of the beam

Calculation:

Given:

w = 1 kN/m, L = 2m

∴ \(M=\frac{1\times (2)^2}{2}\)

M = 2 kN-m

Hence the maximum bending moment of beam will be 2 kN-m.

Explanation:

Let AB be the fixed beam and W is the unit load acting at the mid of the beam and let M_fAB and M_fBA be the fixed end moments at A and B respectively. So, the deflected shape of the beam will be as follow

Clockwise fixed end moment  = Positive and Anticlockwise fixed end moment = Negative

First, suppose only unit load acts on the beam and make its BMD and deflected shape then apply fixed end moment and then superimpose both of them and find the value of fixed end moment

 Fig 1

 Fig 2

But, in reality, the beam is neither sagging (fig 1) or hogging (fig 2), Which implies that,

Area of BMD in fig 1 = Area of BMD of fig 2

0.5 × \(\frac{WL}{4}\) × L = M_fAB × L

M_fAB = \(\frac{WL}{8}\)

Moment at mid point = M

M = Moment at midpoint in fig 1 - Moment at midpoint in fig2

M = \(\frac{WL}{4}\) -  M_fAB

_{M = \(\frac{WL}{4}\) - \(\frac{WL}{8}\)}

_{M = \(\frac{WL}{8}\)}

Note: Remember the formula for fixed end moment for various loading from exam point of view, it will save your time.

Explanation:

By symmetry:

Reaction at supports is RM = W, RN = W

For SFD: (From left side)

Shear force at C, SFM = RM = W

Shear force at A (just left of A), SFA = W

Shear force at A (just right of A), SFA = W - W = 0

Shear force at B (just left of B), SFB = 0

Shear force at B (just right of B), SFB = -W

Shear force at C, SFN = RN = -W

For BMD: (From left side)

BM at A, BMA = Wa

BM at B, BMB = Wa

From SFD and BMD it is clear that there will be no shear force between A and B, but there will be a uniform bending moment between A and B.