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

Explanation:

• A ring beam is a closed circular beam subjected to a uniformly distributed load (UDL) along its span (such as from walls or roof loads).

• In a circular ring beam under UDL, the shear force is zero at the crown (mid-span).

• This is because the vertical component of load is balanced symmetrically at that point; no net transverse shear remains.

• In a ring beam under symmetric UDL, torsional moments are maximum at supports (e.g., quarter spans or support points).

• At the crown/mid-span, torsional effects from both sides cancel each other out, so torsion is zero there.

 Additional Information

• Shear Force in Circular Beams: In circular beams or ring beams under uniformly distributed load (UDL), shear force varies along the span. It is maximum near the supports and zero at the crown (mid-span) due to symmetry. This behavior is opposite to straight beams where maximum shear occurs at supports but not necessarily zero at mid-span. 
• Torsion in Ring Beams: Torsional moments develop in ring beams because the beam tends to twist due to the curvature and the eccentric nature of loading. Torsion is zero at the crown but maximum at support points or quarter-span locations. The torsional effect must be considered in design, especially when the beam supports walls or roofs.
• Bending Moment in Ring Beams:  Bending moment in a ring beam due to UDL is maximum at mid-span (crown). The curvature causes a moment that resists the vertical load, and this moment is symmetrically distributed. The design must accommodate this maximum moment with proper reinforcement.
• Importance of Ring Beams in Structures: Ring beams are crucial in circular structures like water tanks, silos, chimneys, or domes. They help to distribute radial loads uniformly, resist torsional and bending moments, and maintain the shape and stability of the structure. Proper detailing of shear, torsion, and bending reinforcement is essential for structural integrity.

Concept:

• SFD is 1° higher than the loading curve and BMD is 1° higher than the SF curve.

• If S.F. changes sign at a section then B.M at that section is either maximum or minimum.

• If BM changes sign at a point, then such a point is called a point of contraflexure or inflection. If BM changes sign curvature also changes.

• The rate of change of BM is equal to SF at that section or the slope of BM curve represents SF at that point.

• The negative slope of SF curve at a point represents the downward loading rate.

• If at a point concentrated load or reaction is present, then at that point ordinate of the SF curve will change suddenly and the slope of BM curve will also change.

• If at a point Concentrated moment or couple is present then ordinate of the BM curve will change suddenly.

• If over a span a constant SF is present only then such span is called shear span.

• If over a span BM is constant and SF = 0, then such a span is in pure bending and under pure bending the deflected shape will be an arc of a circle.

Additional Information

• A couple is a system of two equal and opposite forces whose lines of action do not coincide.

• The two forces create a rotational effect (moment) without causing any net translation of the body.

• The effect of a couple is to rotate the body around its center of mass or a fixed point.

Explanation:

• In pure bending, the neutral layer (or axis) is the fiber within the cross-section of a beam that experiences zero longitudinal stress.
• This layer neither stretches nor compresses and hence remains unstressed.
• Under pure bending, there are no shear forces, only bending moments.
• The resultant axial force on a transverse section is zero because the tensile and compressive stresses above and below the neutral axis cancel out.

 Additional Information

Pure Bending:

• Occurs when a beam is subjected to a constant bending moment with no shear force.
• Typical in the middle span of a simply supported beam under two-point loads.

Neutral Axis:

• A horizontal axis within the beam’s cross-section where bending stress is zero.
• Fibers above it are in compression, and those below are in tension.

Internal Equilibrium:

• The compressive and tensile forces due to bending must balance out, hence the net axial force is zero across the section. 

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.

Concept:

The relationship between shear force (V) and bending moment (M) is given by:

\( \frac{dM}{dx} = V \)

Integrating both sides between two points \( x_1 \) and \( x_2 \):

\( M_2 - M_1 = \int_{x_1}^{x_2} V \, dx \)

This integral represents the area under the shear force diagram between the two points.

Calculation:

Given: Area under shear force diagram = P

\( \Rightarrow M_2 - M_1 = P \)

Hence, the difference in the bending moments at those points is: P

Explanation:

The relation between shear force (V) and loading rate (w) is:

\(\frac{dV}{dx}=w\)

it means a positive slope of the shear force diagram represents an upward loading rate.

The relation between shear force (V) and bending moment (M) is:

\(\frac{dM}{dx}= V\)

it means the slope of a bending moment diagram will represent the magnitude of shear force at that section.

The relation between loading rate and shear force can be written as:

\(\frac{d^2M}{dx^2}= w\)

Additional Information

If y is the deflection then relation with moment M, shear force V and load intensity w.

\(EI\frac{d^2y}{dx^2}= M\)

\(EI\frac{d^3y}{dx^3}= V\)

\(EI\frac{d^4y}{dx^4}= w\)

Concept:

The area under the shear-force diagram gives a bending moment between those two points and the area under the load diagram gives shear-force between those two points. i.e.

\({{M}_{B}}-{{M}_{A}}=\mathop{\int }_{A}^{B}{{F}_{x-x}}dx\)

\({{F}_{B}}-{{F}_{A}}=-\mathop{\int }_{A}^{B}{{w}_{x-x}}dx\)

Calculation:

Given:

\({{M}_{C}}-{{M}_{A}}=\frac{1}{2}\times \left( 14+2 \right)\times 2\)

M_C = 16 kN-m   (M_A = 0)

Similarly;

Point of contraflexure in a beam is a point at which bending moment changes its sign from positive to negative and vice versa.

Bending moment diagram for a fixed beam subjected to udl throughout the span is as shown below:

The point of contraflexure lies at a distance of L/(2√3) from centre of the beam.

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.

Concept:

Composite Beam is the one in which the beam is made up of two or more material and rigidly connected together in such a way that they behave as one piece. In the composite beam, there is a common neutral axis through the centroid of the equivalent homogeneous section.

While the compound beam is a beam of different element assembled together to form a single unit.

Calculation:

Bending moment at hinge = 0

Considering right hand side of hinge

Bending moment at hinge = 0 kN

R_Q × (L/4) = 0

R_Q  = 0 kN

∴ The vertical reaction at support Q is 0.0 KN

Concept:

The maximum bending stress in a beam is given by

\({\sigma _b} = \frac{{Mc}}{I}\)

Where

c – distance from the neutral axis to extreme end (maximum),

I – Area moment of inertia w.r.t centroidal axis perpendicular to the plane of the bending moment M.

∴For given stress (σ_b), it takes a bigger Moment (M) to bend a beam with a bigger Area moment of inertia (I).

Additional Information

• Polar moment of inertia is a measure of the beam’s ability to resist torsion
• Mass moment of inertia is a measure of an object’s resistance to change in the rotation direction.

Explanation:

Given data:

Calculating shear force at mid-span (or at point C):

Shear force at mid-span (or at point C) = \(w\times {({L\over 3})\over 2}\)

Shear force at mid-span (or at point C) = \(w\times {L\over 6}\) = \({wL\over 6}\)

Calculating shear force at the fixed end (or at point A):

Shear force at the fixed end (or at point A) = \(w\times ({L\over 3})\)

Shear force at the fixed end (or at point A) =\({wL\over 3}\)

The shear force at its mid-span is \({wL\over 6}\) and the fixed end is \({wL\over 3}\).

Concept-

Shear force:

• It is defined as the algebraic sum of all the vertical forces, either to the left or to the right-hand side of the section. 

Bending moment:

• It is defined as the algebraic sum of the moments of all the forces either to the left or to the right of a section.

Calculation:

As there is no vertical and horizontal load acting on the beam, the Vertical and horizontal reaction at fixed support is zero.

The FBD of the beam is as shown in the figure.

V_A = 0 = H_A

The SFD and BMD of the beam are shown in the figure.

So the bending moment at the center is M kN-m and the shear force at the center is zero. 

Explanation:

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