Bending Moment MCQ Quiz - Objective Question with Answer for Bending Moment - Download Free PDF

Last updated on Jun 10, 2025

Latest Bending Moment MCQ Objective Questions

Bending Moment Question 1:

Which of the following is the correct relation of shearing force (F) and bending moment (M) at a section?

  1. \(F = \frac{d^2M}{dx^2}\)
  2. \(M = \frac{d^2F}{dx^2}\)
  3. \(F = \frac{dM}{dx}\)
  4. \(M = \frac{dF}{dx}\)

Answer (Detailed Solution Below)

Option 3 : \(F = \frac{dM}{dx}\)

Bending Moment Question 1 Detailed Solution

Explanation:

Shear Force:

Shear force at any section is the summation of all transverse loading, either to the left or to the right of any section.

Bending Moment:

Bending moment at any section is a resultant algebraic sum of moments caused by transverse forces either from the left or right of the section.

Relationship between bending moment, shear force, and loading:

  • The rate of change of shear force at any section is equal to load intensity at that section.

 

\(\frac{{dF}}{{dx}} = \omega \Rightarrow dF = \omega dx \Rightarrow F = \smallint \omega dx\)

i.e. Change of shear force between two sections is equal to the area under the load intensity diagram between those two sections.

  • The rate of change of bending moment at any section is equal to shear force at that section.

\(\frac{{dM}}{{dx}} = F \Rightarrow dM = Fdx \Rightarrow M = \smallint Fdx\)

i.e. Change of bending moment between two sections is equal to the area under the shear force diagram between those two sections

The presence of a concentrated couple leads to sudden rise or fall of the Bending moment diagram depending upon the sign and direction of the couple.

Bending Moment Question 2:

A simply supported beam of length L carrying a concentrated load W at a section which is at a distance of 'x' from one end. What will be the value of the bending moment at this section?

  1. \(W(x - x^2)\)
  2. \(W(x^2 - xL)\)
  3. \(W\left(x - \frac{x^2}{L}\right)\)
  4. \(Wx\)

Answer (Detailed Solution Below)

Option 3 : \(W\left(x - \frac{x^2}{L}\right)\)

Bending Moment Question 2 Detailed Solution

Concept:

For a simply supported beam of length \( L \) with a point load \( W \) at a distance \( x \) from one end, the bending moment at that section is:

\( M = R_A \cdot x \)

Where the reaction at the left support is:

\( R_A = \frac{W(L - x)}{L} \)

Substituting:

\( M = \frac{W(L - x)}{L} \times x = W \left( x - \frac{x^2}{L} \right) \)

 

Bending Moment Question 3:

A truss is loaded and supported as shown in the figure. What will be the axial force in the member PQ, SR and TU, if a vertical load (W = 1 kN) is applied at U?

Task Id 1206 Daman (15)

  1. \(F_{T U}=0 ; F_{P Q}=\frac{W}{3}( tensile ) ; F_{R S}=\frac{2 W}{3}( tensile )\)
  2. \(F_{T U}=0 ; F_{P Q}=\frac{W}{3} (compressive) ; F_{R S}=\frac{2 W}{3}( compressive )\)
  3. \(F_{T U}=\frac{W}{3} (tensile); F_{P Q}=\frac{W}{3} (tensile); F_{P Q}=\frac{2 W}{3} (compressive)\)
  4. \(F_{T U}=0 ; F_{P Q}=\frac{W}{3} (tensile) ; F_{R S}=\frac{2 W}{3} (compressive)\)

Answer (Detailed Solution Below)

Option 2 : \(F_{T U}=0 ; F_{P Q}=\frac{W}{3} (compressive) ; F_{R S}=\frac{2 W}{3}( compressive )\)

Bending Moment Question 3 Detailed Solution

Concept:

A truss is a structure composed of members joined together at their ends to form a stable framework. The method of joints is used to determine the axial forces in the members.

Given:

Vertical load, \(W = 1~\text{kN}\) is applied at point U. The truss is symmetric, and support reactions are calculated first.

Calculation:

Step 1: Support Reactions

Due to symmetry and central vertical loading, vertical reactions at P and S are:

\(R_P = R_S = \frac{W}{2} = \frac{1}{2} = 0.5~\text{kN}\)

Step 2: Joint U

At joint U, the members UR and UT are inclined. Let the force in UR and UT be \(F\).

Since the geometry shows 0.75 m vertical and 0.5 m horizontal components:

\(\sin\theta = \frac{0.75}{\sqrt{0.75^2 + 0.5^2}} = \frac{0.75}{0.902} = 0.832\)

Vertical equilibrium at U:

\(2F \cdot \sin\theta = W\)

\( \Rightarrow 2F \cdot 0.832 = 1 \)

\(\Rightarrow F = \frac{1}{2 \cdot 0.832} = 0.60~\text{kN}\)

So, \(F_{UT} = F_{UR} = 0.60~\text{kN}\) in tension.

Since TU is horizontal and not needed for vertical equilibrium:

\(F_{TU} = 0\)

Step 3: Joint Q

From method of joints, axial force in \(PQ = \frac{W}{3} = \frac{1}{3}~\text{kN}\).

Direction: Acts toward joint, hence it is compressive.

Step 4: Joint S

From equilibrium, axial force in \(SR = \frac{2W}{3} = \frac{2}{3}~\text{kN}\).

Direction: Acts toward joint, hence it is compressive.

Bending Moment Question 4:

A concentrated load P acts on a simply supported beam of span L at a distance \(\frac{\mathrm{L}}{3}\) from the left support. The bending moment at the point of application of load is given by -

  1. PL/3 
  2. 2PL/3
  3. PL/9
  4. 2PL/9

Answer (Detailed Solution Below)

Option 4 : 2PL/9

Bending Moment Question 4 Detailed Solution

Concept:

ΣFy = R1 + R2

where R1 and R2 are the reaction support.

Calculation:

Given:

558daf8c2a396511d048a7ae 16508769025051

The load P acts at the LI3 from left support.

Let the reactions at point A is "R1" and at point B is "R2".

\({R_1} = \frac{{P \times \left( {\frac{{2L}}{3}} \right)}}{L} = \frac{{2P}}{3}\) and reaction \({R_2} = \frac{{P \times \left( {\frac{{L}}{3}} \right)}}{L} = \frac{{P}}{3}\)

Take a cross section at a distance "x" from support A.

Then the Bending moment at cross section "x" is, 

Mx-x = \(R_1\times x=\frac{2P}{3}\times{x}\)

∴ M = \(\frac{2P}{3}\times \frac{L}{3}\) ⇒ M = \(\frac{2PL}{9}\)

Bending Moment Question 5:

Determine the shear force and bending moment at the fixed end, of a cantilever beam AB of span length 4 m subjected to a uniformly distributed load of 4 kN/m throughout the span of the beam. 

  1. −12 kN, −24 kN-m
  2. −18 kN, −36 kN-m
  3. −16 kN, −32 kN-m
  4. −14 kN, −32 kN-m 

Answer (Detailed Solution Below)

Option 3 : −16 kN, −32 kN-m

Bending Moment Question 5 Detailed Solution

Explanation:

Given, w = 4 kN/m, span of beam = 4 m

Shear force calculations:

SF at B = 0 kN

(SF)Total = (SF)A = 4 × 4 = 16 kN (Downward) = - 16 kN

Bending moment calculations:

BM at B = 0 kNm

BM at A = -16 × 2 = -32 kN-m

Task 956 (8)

Top Bending Moment MCQ Objective Questions

The BM diagram of the beam shown in figure is:

F9 Tabrez 12-5-2021 Swati D2

  1. A rectangle
  2. A triangle
  3. A trapezium
  4. A parabola

Answer (Detailed Solution Below)

Option 2 : A triangle

Bending Moment Question 6 Detailed Solution

Download Solution PDF

Concept:

Find the Reaction force at A and B and draw the bending moment diagram

\({\rm{\Sigma }}{{\rm{F}}_{\rm{x}}} = 0,\;\;{\rm{\Sigma }}{{\rm{F}}_y} = 0,\;\;{\rm{\Sigma M}}_{\left( {{\rm{about\;a\;point}}} \right)} = 0\)

Calculation:

Given:

F9 Tabrez 12-5-2021 Swati D2

\(\sum {F_y} = 0\;\)

RA + RB = 0

\(\sum {M_A} = 0\)

M - R× L = 0

\({R_B} = \frac{M}{L}\) and \({R_A} = - \frac{M}{L}\)

Shear force:

\({\left( {S.F} \right)_B} ={\left( {S.F} \right)_A} = - \frac{M}{L} \)

Bending movement:

\({\left( {B.M} \right)_{X - X}} = M - \frac{M}{L}x\)   (x taken from the left side)

Clockwise bending moment  -ve, Anticlockwise bending moment  +ve 

\(% MathType!Translator!2!1!LaTeX.tdl!LaTeX 2.09 and later! % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaqadaWdaeaaieWapeGaa8Nqaiaac6cacaWFnbaacaGLOaGaayzk % aaWdamaaBaaaleaapeGaa8hwaiabgkHiTiaa-Hfaa8aabeaak8qacq % GHDisTcaWFybaaaa!3F7A! {\left( {B.M} \right)_{X - X}} \propto X% MathType!End!2!1! \)  (Bending moment varies linearly)

F9 Tabrez 12-5-2021 Swati D3   

\({\left( {B.M} \right)_A} = M\)

\({\left( {B.M} \right)_B} = M - \frac{M}{L} \times L = 0\;\)

∴ bending moment diagram will be a TRINGLE.

Important Points

  • If SFD is constant throughout the span of the beam then BMD will be linear.
  • If at a point a couple is acting then there will be a sudden jump in the BMD.

A cantilever 9 m long has uniformly distributed load over the entire length. The maximum bending moment is 8100 N-m, the rate of loading is:

  1. 200 N/m
  2. 100 N/m
  3. 400 N/m
  4. 900 N/m

Answer (Detailed Solution Below)

Option 1 : 200 N/m

Bending Moment Question 7 Detailed Solution

Download Solution PDF

Explanation:

Cantilever beam with uniformly distributed load:

RRB JE ME SOM 3 123

So, the cantilever beam has a maximum bending moment at the fixed end. and it is given as, \(M=\frac{wL^2}{2}\)

where w = rate of loading

Calculation:

Given:

M = 8100 N-m, L = 9 m

\(8100=\frac{w~\times~ 9^2}{2}\)

w = 200 N/m

A bending moment causing concavity upward will be taken as _____ and called as ______ bending moment.

  1. positive, sagging
  2. positive, hogging
  3. negative, sagging
  4. negative, hogging

Answer (Detailed Solution Below)

Option 1 : positive, sagging

Bending Moment Question 8 Detailed Solution

Download Solution PDF

Explanation:

Let us assume a simply supported beam.

The uniformly distributed load (UDL) of w/length is acted on the beam.

Due to downward load, the beam is sagging.

We also know that when a simply supported beam is subjected to  UDL the bending moment will be positive.

F1 S.S Madhu 27.11.19 D14

Sagging Or Positive Bending Moment

We take bending moment at a section as positive if

  • Force tends to bend the beam at that considered point.
  • This bending forms to curvature having concavity at the top
  • Concavity at the top indicates compression in the top fibers of the beam. 
  • Hence bottom fibers of the beam would have tension.
  • Here due to UDL simply supported beam will give us the positive value of bending moment that indicates sagging.
  • Due to this, the upper layer fibers are getting compressive stresses and the bottom layer fibers are getting tensile stresses.

Hogginng or Negative Bending Moment

We take bending moment at a section as positive if 

  • Force tends to bend the beam at that considered point.
  • This bending forms to curvature having convexity​ at the top
  • Convexity at the top indicates tension in the top fibers of the beam. 
  • Hence top fibers of the beam would have compression.

Points of zero bending moment

  • The points of contra flexure (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 contra flexure 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. 

for example:

F1 Ram S 1-12-2020 Swati D3

Points A and B are contraflexure points.

In the figure given below, the beam will be stable only if -

F18 Abhishek M 26-4-2021 Swati D1

  1. x = √2y
  2. x = 2y
  3. √2x = y
  4. 2x = y

Answer (Detailed Solution Below)

Option 3 : √2x = y

Bending Moment Question 9 Detailed Solution

Download Solution PDF

Concept-

For a system to be in stable condition, it should be in stable equilibrium.

So to be in stable equilibrium the moment about the support point should be zero.

F21 Chandramouli 29-4-2021 Swati D4

∑ MB = 0

\(⇒ 2W \times x \times \dfrac{x}{2} = W \times y \times \dfrac{y}{2}\)

\(⇒ x^2 = \dfrac{y^2}{2}\)

⇒ y2 = 2x2

⇒ y = √2 x

Find the bending moment at a distance L/4 from end A of the simply supported beam as shown below.

F1 Savita ENG 15-12-23 D4

  1. Zero
  2. \(\rm \frac{3.wL^2}{32}\)
  3. \(\rm \frac{wL^2}{8}\)
  4. \(\rm \frac{wL^2}{32}\)

Answer (Detailed Solution Below)

Option 2 : \(\rm \frac{3.wL^2}{32}\)

Bending Moment Question 10 Detailed Solution

Download Solution PDF

Concept:

The bending moment and shear force diagram for a  simply supported beam with UDL is shown below.

04.11.2017.021

Calculation:

Calculating the support reactions, due to loading symmetry, will be equal at A and B.

So, \(R_A=R_B=\frac{wL}{2}\)

The bending moment at a distance L/4 from end A of the simply supported beam is given by

M = Moment due to RA from L/4 distance - Moment due to load w from L/4 distance

\(M=\frac{wL}{2}\times\frac{L}{4}-w\times\frac{L}{4}\times(\frac{1}{2}\times\frac{L}{4})\)

\(M=\frac{3wL^2}{32}\)

An overhanging beam CADEBF is shown in the figure below. Calculate the sum of the bending moment values at A and B. Ignore the sign conventions. 

F2 Savita ENG 27-12-23 D1 V2

  1. 13.50 kN-m  
  2. 10.00 kN-m 
  3. 12.50 kN-m 
  4. 16.78 kN-m 

Answer (Detailed Solution Below)

Option 1 : 13.50 kN-m  

Bending Moment Question 11 Detailed Solution

Download Solution PDF

Explanation:

Concept:

  • It is the algebraic sum of moments acting on either side of a section along the length of the beam. 

Calculation:

The bending moment at A ( from left) = 9×1.5×\( \frac{1.5}{2}\) = 10.125 kN-m

The bending moment at B (from right) = 3×1.5×\( \frac{1.5}{2}\) = 3.375 kN-m

the Thus sum of moments = 10.125 +3.375 = 13.50 kN-m

Two people weighing 'W' each are sitting on a plank of length 'L' floating on water at L/4 from either end, neglecting the weight of the plank the bending moment at center of plank is

  1. Zero
  2. \(\frac{ WL }{16}\)
  3. \(\frac{WL}{8}\)
  4. \(\frac{WL}{12}\)

Answer (Detailed Solution Below)

Option 1 : Zero

Bending Moment Question 12 Detailed Solution

Download Solution PDF

Concept:

The given information in question can be represented in the given figure:

F1 A.M Madhu 09.05.20 D6

The water will exert uniform pressure at the bottom plank in the upward direction.

Calculation:

Given:

Since it is given that Plank is floating on water. We get,

Upward force (buoyant force) exerted by water = Downward weight of pressure standing on Plank

(W')(L) = 2W

\(W' = \frac{{2W }}{L}\)

Consider midsection (left) of plank,

F1 A.M Madhu 09.05.20 D7

Consider Moment at the midpoint to be zero.

∑ Mmid = 0

\({M_{mid}} + W \times \frac{L}{4} = \frac{{{W'}\times L}}{2} \times \frac{L}{4}\)

\({M_{mid}} + \frac{{W \times L}}{4} = \left( {\frac{{2\times W }}{L}}\right)\times \left ( {\frac{L}{2}} \right)\times \left( {\frac{L}{4}} \right)\)

∴ Mmid = 0

A simply supported beam of span length 4m, carries a concentrated load of 8 kN at mid span, the value of maximum bending moment is:

  1. 8 kN.m
  2. 16 kN.m
  3. 32 kN.m
  4. 128 kN.m

Answer (Detailed Solution Below)

Option 1 : 8 kN.m

Bending Moment Question 13 Detailed Solution

Download Solution PDF

Concept:

A simply supported beam with span L and centered load P is,

SSC JE ME 25 Jan 18 Evening Uploading.docx   Shraddha 10

RA + RB = P      ---(1)

∑MB = 0

\({R_A} \times L - P \times \frac{L}{2} = 0\)

\({R_A} = \frac{P}{2}\)

\({R_B} = \frac{P}{2}\)

The BM will be maximum on the point at which shear force changes its sign.

So the value of bending moment at a distance x = L/2 is:

\({M_{\frac{L}{2}}} = {R_A}\frac{L}{2} = \frac{{PL}}{4}\)

SSC JE ME Strength of Material Part test (1-35) images Q2

Calculation:

Given:

P = 8 kN, L = 4 m

Maximum bending moment, \({M} = \frac{PL}{4}\)

\({M} = \frac{8\times 4}{4}=8~kN.m\)

In a simply supported beam of 10 m span having udl of 8 kN/m, the maximum Bending Moment shall be

  1. 80 kNm
  2. 100 kNm
  3. 0.8 kNm
  4. 10 kNm

Answer (Detailed Solution Below)

Option 2 : 100 kNm

Bending Moment Question 14 Detailed Solution

Download Solution PDF

Concept:

The maximum bending moment for a simply supported beam with a uniformly distributed load W per unit length is wL2/8.

F1 N.M D.K 13.08.2019 D 4

Calculation:

Given:

w = 8 kN/m

L = 10 m

Maximum bending moment = \(\frac{{w{L^2}}}{8} = \frac{{8 \times \left( {{{10}^2}} \right)}}{8} = 100kNm\)

Consider the following statements:

If a simply supported beam of uniform cross-section is subjected to a clockwise moment at the left support and an equal anticlockwise moment at the right support, then the:

1. B.M.D. will be in the shape of a rectangle 

2. S.F.D. will be a straight line coinciding with the base 

3. Deflection curve will be in the shape of a circular arc 

Of these statements 

  1. 1, 2 and 3 are correct 
  2. 1 and 2 are correct 
  3. 1 and 3 are correct 
  4. 2 and 3 are correct 

Answer (Detailed Solution Below)

Option 1 : 1, 2 and 3 are correct 

Bending Moment Question 15 Detailed Solution

Download Solution PDF

Explanation:

F5 Savita Engineering 2-5-22 D6

As given in the question that the simply supported beam is subjected to a clockwise moment on the left side and an anti-clockwise moment of the same magnitude on the right side.

∴ Net moments on the beam = +M - M = 0

Reaction of the beam = \(\frac{Net~moments~on~the~beam}{Span~of~the~beam}\)

So, RA = RB = 0.

∴ Shear force = 0

⇒ The sheer force diagram will be a straight line coinciding with the base.

Concentrated moment at A = M (Clockwise)

The concentrated moment at B = M (Anti-Clockwise)

And these concentrated moments are shown as a vertical straight line in the bending moment diagram.

 B.M.D. will be in the shape of a rectangle.

δA = δB = 0

And Maximum deflection at the midsection,

δmax = \(\frac{ML^2}{8EI_{NA}}\)

So, the deflection curve will be in the shape of a circular arc.

Get Free Access Now
Hot Links: teen patti gold download apk teen patti pro teen patti rummy teen patti cash happy teen patti