Relation Between Shear Force and Bending Moment MCQ Quiz in বাংলা - Objective Question with Answer for Relation Between Shear Force and Bending Moment - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

The relation between shear force and bending moment is defined as,

\(V = \frac{{dM}}{{dx}}\;\;\;\;\; \ldots \left( 1 \right)\)

Explanation:

For Statement (I):

For pure bending, M = constant

\( ⇒ \frac{{dM}}{{dx}} = 0\)

By using equation (1),

⇒ Shear force, V = 0

⇒ If the bending moment along the length of a beam is constant, then the beam cross-section will not experience any shear stress.

For Statement (II):

By using equation (1),

⇒ Shear force,

\(⇒ V = \frac{{dM}}{{dx}} = 0\)

⇒ M = constant

⇒ The shear force acting on the beam will be zero everywhere along its length. 

∴ Both A and R are individually true, and R is the correct explanation of A 

Explanation:

Statement i:

• Maximum or minimum shear force occurs where the radius of curvature is zero.

  • This statement is incorrect. The radius of curvature relates to bending moments and the curvature of the beam, not shear force. Shear force is maximum at the support for a simply supported beam under a uniform load.

Statement ii:

• Maximum or minimum bending moment occurs where the shear force is zero.

  • This statement is correct. For a simply supported beam, the bending moment is maximum at the point where the shear force is zero. This typically occurs at the center of the beam under a uniform load.

Statement iii:

• Maximum or minimum bending moment occurs where the radius of curvature is zero.

  • This statement is incorrect. The maximum or minimum bending moment occurs where the shear force is zero (as mentioned above), not where the radius of curvature is zero.

Statement iv:

• Maximum bending moment and maximum shear force occur at the same section.

  • This statement is incorrect. For a simply supported beam under uniform loading, the maximum bending moment occurs at the center of the beam, while the maximum shear force occurs at the supports.

We know

Shear force, \({\rm{V = }}\frac{{{\rm{dM}}}}{{{\rm{dx}}}}\)

i.e. Shear force is nothing but a slope of the bending moment diagram. At the maxima point of the bending moment diagram, the slope changes sign. (+ve to -ve)

Concept

\(Shear\;force\;\left( V \right)\; = \frac{{dM}}{{dx}}\)

Calculation:

Given:

M = (10x² + 36x – 4) N-m

\(\therefore {\rm{Shear\;force\;}}\left( {\rm{V}} \right) = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {10{{\rm{x}}^2} + 36{\rm{x}} - 4} \right)\)

∴ Shear force (V) = 20x + 36

Now,

We have to find the value of shear force at x = 3m

∴ Shear force (V) = 20 × 3 +36

∴ Shear force (V) = 96 N

At point A there is a sudden rise in shear force by 40 kN. Therefore, a point load of 40 kN is acting on that point in an upward direction.

Between points A & B, the shear force decreases linearly. Therefore, the uniformly distributed load is acting between A & B in the downward direction.

The intensity of uniformly distributed load between A & B

\(= \frac{{40 - \left( { - 50} \right)}}{6} = 15\;kN/m\)

At point B, there is a sudden change in the value of shear force. Therefore, a point load of 80KN is acting on point B in an upward direction.

Between B & C, the shear force decreases linearly. Therefore, a uniformly distributed load is acting between B & C in the downward direction.

The intensity of uniformly distributed load between B & C is

\(= \frac{{30 - 0}}{2} = 15\;kN/m\)

Point of contraflexure: The point where a beam suffers no bending moment is also known as Point of contraflexure.

Point of Inflexion: Inflection point is a point where a function turns from concave to convex, i.e it is point shown on the curvature diagram of the structure where the curvature changes from concave to convex.

At hinge support, Bending moment is zero, Similarly at point of contraflexure bending moment is zero, therefore that point is behaving like a virtual hinge on the beam.

Concept:

The relation between bending moment (M) and shear force (V) of a beam is given as:

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

Method:

Now, the simply supported beam is shown in figure

The Shear force diagram will be represented as:

Concept:

To calculate the reaction at supports, we use equilibrium condition i.e.

ΣFx = 0, ΣFy = 0 and ΣMA = 0

Calculation:

Given:

AB = 2000 mm ⇒ 2 m, BC = 2000 mm ⇒ 2 m

ΣFy = 0

RA + RC = (3000 × 2) ⇒ 6000 N  

ΣMA = 0

RC × 4 - (3000 × 2 × 3) = 0

∴ RC = 4500 N.

∴ RA = 1500 N.

From a similar triangle:

\(\frac{{1.5}}{{\left( {2 - x} \right)}} = \frac{{4.5}}{x}\;\;\; \Rightarrow x = 6 - 3x\)

∴ x = 1.5 m from end C or 2500 mm from end A.

Concept:

Let the maximum intensity of load is, w = 360 N/m

The Bending Moment from cross-section x-x is, \(M_{x-x}=-\frac{wx^2}{2L}\frac{x}{3}=-\frac{wx^3}{6L}\)

Calculation:

Given:

w = 360 N/m, L = 6 m

\(M_{x-x}=-\frac{360~\times~x^3}{6~\times ~6}=-10x^3\)

Explanation:

i.e. Where shear force is zero, the slope of the bending moment becomes zero and the bending moment is maximum at that point and so the bending stress.

\(\therefore {\rm{S}}.{\rm{F}} = 0{\rm{\;}}\)at x = 1.5 m from A

\({{\rm{M}}_{{\rm{max}}}} = {{\rm{R}}_{\rm{C}}}\left( {1.5} \right) - 3\left( {1.5} \right)\left( {\frac{{1.5{\rm{\;}}}}{2}} \right) = 3.375\; kNm\)           

\({\sigma _b}_{max} = \frac{M}{Z}= \frac{{3.375 \times {{10}^6}}}{{\left( {30 \times \frac{{{{100}^2}}}{6}} \right)}}= 67.5\;MPa\)