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:

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. 

Concept:

For calculating load at support for any transverse loaded beam:

ΣF_x = 0, ΣF_y = 0 and ΣM_{any point} = 0

Calculation:

Given:

No load is acted externally on horizontal direction. ∴ hinge supports R_AX = 0.

ΣFy = 0

∴ R_AY + R_BY = (10 × 4) + 12

∴ RAY + RBY = 52 kN ..... eq(1)

ΣM_A = 0

∴ (R_BY × 4)  - (10 × 4 × 2) - (12 × 5) + 3 = 0

∴ 4R_BY - 80 - 60 + 3 = 0

∴ 4R_BY = 137

∴ R_BY = 34.25 kN

From eq(1)

RAY + RBY = 52 kN

∴ R_AY = 17.75 kN.