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

Concept:

The BMD is linear between A and B. It implies that there is no load acting on these segment. there is sudden change of moment and similarily there is sudden change of moment at point. So, couple will act at these points. The magnitude of couple is calculated by calculating the change in BM at these points.

Calculation:

Couple acting at C = Change in BM at C

= 0 - (-M) = M

and, couple acting at D = change in BM at D

= M - (-M) = 2M

Hence, there are couples of M at C and 2 M at D.

Explanation:

Consider a cantilever of length l carries a concentrated load ‘w' at its free end:

∵ Reaction at the fixed end will be equal to the net load on the beam, i.e., W.

So, the shear force at the fixed end is W (Upward) and at the free end, it is W (downward).

Taking moments about the section:

M = \(W~\times~x\)

The maximum bending moment occurs at the fixed end i.e. M = \(W~\times~L\)

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.

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:

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

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.

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:

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

R_A + R_B = 0

\(\sum {M_A} = 0\)

M - R_B × 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.

Explanation:

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

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.

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.

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.

∑ M_B = 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

Concept:

The bending moment due to point load P at a distance of x is

= P × x

Calculation:

Given:

Load at point C = 15 kN

Distance of application of load from left support A = 1 m

Let’s assume reaction at the supports A and B as R_A and RB respectively.

From the equilibrium of forces in y-direction

ΣFy = 0

RA + R­B = 15 kN        …(1)

Taking moment about point A and using the equilibrium condition

ΣMA = 0

-(15 × 1) + (RB × 3) = 0

R­B = 5 kN         …(2)

From eq. (1), we get

RA = 10 kN         …(3)

The bending moment diagram is shown as follow

The bending moment is maximum at point C with a value of 10 kNm

• Be careful while taking the sign conventions for shear force and bending moment diagrams.

Concept:

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 R_A from L/4 distance - Moment due to load w from L/4 distance

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

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

Concept:

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

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

Concept:

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, \({M} = \frac{PL}{4}\)

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

Concept:

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