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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 -
Answer (Detailed Solution Below)
Bending Moment Question 4 Detailed Solution
Concept:
ΣFy = R1 + R2
where R1 and R2 are the reaction support.
Calculation:
Given:
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.
Answer (Detailed Solution Below)
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
Top Bending Moment MCQ Objective Questions
Answer (Detailed Solution Below)
Bending Moment Question 6 Detailed Solution
Download Solution PDFConcept:
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:
\(\sum {F_y} = 0\;\)
RA + RB = 0
\(\sum {M_A} = 0\)
M - RB × 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)
\({\left( {B.M} \right)_A} = M\)
\({\left( {B.M} \right)_B} = M - \frac{M}{L} \times L = 0\;\)
∴ 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:
Answer (Detailed Solution Below)
Bending Moment Question 7 Detailed Solution
Download Solution PDFExplanation:
Cantilever beam with uniformly distributed load:
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.
Answer (Detailed Solution Below)
Bending Moment Question 8 Detailed Solution
Download Solution PDFExplanation:
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.
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:
Points A and B are contraflexure points.
In the figure given below, the beam will be stable only if -
Answer (Detailed Solution Below)
Bending Moment Question 9 Detailed Solution
Download Solution PDFConcept-
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.
∑ 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.
Answer (Detailed Solution Below)
Bending Moment Question 10 Detailed Solution
Download Solution PDFConcept:
The bending moment and shear force diagram for a simply supported beam with UDL is shown below.
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.
Answer (Detailed Solution Below)
Bending Moment Question 11 Detailed Solution
Download Solution PDFExplanation:
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
Answer (Detailed Solution Below)
Bending Moment Question 12 Detailed Solution
Download Solution PDFConcept:
The given information in question can be represented in the given figure:
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,
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:
Answer (Detailed Solution Below)
Bending Moment Question 13 Detailed Solution
Download Solution PDFConcept:
A simply supported beam with span L and centered load P is,
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:
\({M_{\frac{L}{2}}} = {R_A}\frac{L}{2} = \frac{{PL}}{4}\)
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
Answer (Detailed Solution Below)
Bending Moment Question 14 Detailed Solution
Download Solution PDFConcept:
The maximum bending moment for a simply supported beam with a uniformly distributed load W per unit length is wL2/8.
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
Answer (Detailed Solution Below)
Bending Moment Question 15 Detailed Solution
Download Solution PDFExplanation:
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.