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

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:

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 = \(\frac{2PL}{9}\)

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

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 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

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

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

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:

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

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

Explanation:

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, R_A = R_B = 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.

δ_A = δ_B = 0

And Maximum deflection at the midsection,

δ_max = \(\frac{ML^2}{8EI_{NA}}\)

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