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

Explanation:

The cantilever beam with a span length of L meters carrying a uniform moment of intensity 'M' N-m/m throughout its length is an interesting case to analyze in structural mechanics. To understand the correct statement regarding the shear force distribution along the length of the cantilever beam, we need to delve into the fundamental concepts of beam theory and moment distribution.

Understanding Shear Force and Moment Distribution:

In beam theory, the shear force and bending moment are two critical parameters that describe the internal forces within a beam subjected to external loads. The shear force at any section of the beam is the sum of all vertical forces acting on the beam to the left or right of the section. The bending moment at any section is the sum of moments about that section caused by external loads.

For a cantilever beam subjected to a uniform moment, the moment distribution is constant along the length of the beam. A uniform moment means that at every point along the beam's length, there is a moment of intensity 'M' N-m/m acting.

Shear Force Calculation:

To determine the shear force distribution in the beam, we need to consider the equilibrium conditions of the beam. In a cantilever beam subjected to a uniform moment, the moment is constant along the entire length. Since the moment is uniformly distributed, there are no point loads or distributed loads causing a change in the shear force along the beam.

Key Point: In the absence of vertical loads, the shear force in the beam remains zero throughout its length.

This can be further clarified by considering the differential relationships between shear force and bending moment in beam theory. The shear force (V) and bending moment (M) are related by the following differential equation:

dM/dx = V

Where:

• dM/dx is the rate of change of bending moment with respect to the length of the beam.
• V is the shear force at a particular section.

Since the bending moment 'M' is constant along the length of the beam (uniform moment), its rate of change dM/dx is zero. Therefore, the shear force V must be zero throughout the length of the beam.

Conclusion:

Based on the above analysis, we can conclude that the shear force throughout the length of the cantilever beam carrying a uniform moment of intensity 'M' N-m/m is zero.

The correct answer is option 3.

Concept:

For a simply supported beam of length \( L \) subjected to a uniformly distributed load \( w \), the support reactions are:

\( R_A = R_B = \frac{wL}{2} \)

Shear Force at \( x = L/4 \):

\( V_{x=L/4} = R_A - \text{UDL Load up to } x \)

\( V_{x=L/4} = \frac{wL}{2} - \frac{wL}{4} \)

\( V_{x=L/4} = \frac{wL}{4} \)

Explanation:

Cantilever beam carries a uniformly distributed load:

Shear force:

SFxx = wx

Shear force diagram is a triangle with zero at the free end and WL at the fixed end, ∴ it is linear.

Bending Moment:

\(B{M_{xx}} = w\frac{{{x^2}}}{2}\)

Shear Force at a distance x for a cantilever beam subjected to uniformly distributed load w is given by

SFxx = w × x

S.F at mid span (i.e.x = 1.5m) is 6kN

6 kN = w × 1.5 ⇒ w = 4 kNIm

Explanation:

Concentrated Moment (Clockwise) M acts on the middle of the supported beam as given in the question.

Sign Convention Assumed:

Shear Force: +ve (Upwards from left) and vice-versa

Bending Moment: +ve (Clockwise from left) and vice-versa

Calculations:

Using equations of static equilibrium, we have

\(\Sigma V_y =0 \)

V_A + V_B = 0 

and \(\Sigma M_z= 0\) at B

\(V_A L +M = 0\)

∴ V_A = -M/L

and, VB = M/L

Now, 

Shear Force at L/4 from the left support (Support A) = V_A +0 =  -M/L

Concept:

Shear force at any section X-X is given by:

Cantilever beam:

F_x-x = wx

At x = 4 m (mid-point)

F_x-x = 12 kN

w × 4 = 12 kN

Explanation:

Concentrated Moment (Clockwise) M acts on the middle of the supported beam as given in the question.

Sign Convention Assumed:

Shear Force: +ve (Upwards from left) and vice-versa

Bending Moment: +ve (Clockwise from left) and vice-versa

Calculations:

Using equations of static equilibrium, we have

\(\Sigma V_y =0 \)

V_A + V_B = 0 

and \(\Sigma M_z= 0\) at B

\(V_A L +M = 0\)

∴ V_A = -M/L

and, VB = M/L

Now, 

Shear Force at L/4 from the left support (Support A) = V_A +0 =  -M/L

Concept:

The figure shows the simply supported beam with the point loads.

Total vertical force:

∑F_y = R_A + R_B = 20

Calculation:

From figure,

R_A = R_B = 10 N and C is the midpoint of the beam AB.

From SFD, at point "C", the magnitude of Shear force is zero.

Explanation:

∑F_y = 0

R_A + R_B = 0

consider equilibrium, ∑M_B = 0

so, R_A × L + M = 0

R_A = -M/L

R_B = M/L

So, shear force is uniform throughout section.

Explanation:

∑M = 0

⇒ RB × L = M

⇒ \({R_B} = \frac{M}{L}\)

∑Fy = 0 

⇒RA = RB \( = \frac{M}{L}\)

The SED & BMD will work like this.

Explanation:

Relationship between Shear force, bending moment and Loading rate

• The slope of Shear force diagram at any section will be equal to the load intensity at that section

\(\dfrac{{{\rm{dS}}}}{{{\rm{\;dx}}}} = {\rm{W}}\)

Hence statement 2 is true.

• The slope of Bending moment diagram at any section of a loaded beam will be equal to Intensity of shear force at that section

\(\dfrac{{{\rm{dM}}}}{{{\rm{\;dx}}}} = {\rm{S}}\)

Hence statement 1 is true.

Curvature is given by

\(\dfrac{{\rm{M}}}{{{\rm{EI}}}} = \dfrac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \dfrac{1}{{\rm{R}}}\)

Slope of curvature is given by

\(\dfrac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}} = \dfrac{d}{{dx}}\dfrac{{\rm{M}}}{{{\rm{EI}}}} = \dfrac{V}{{EI}}\)

Hence statement 3 is false.

y is the deflection. The second derivative is given by

\(\dfrac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = \dfrac{1}{{\rm{R}}}\)

Explanation:

The shear force diagram for a cantilever beam with a point concentrated load at the free end is shown below in the figure

As we can see from the diagram irrespective of the length the shear force will remain the same i.e. W.

Explanation:

Shear force is the internal transverse force that is developed to maintain free body equilibrium in either the left portion or the right portion of the section.

Sign convention for shear force

Shear force having an upward direction to the left-hand side of the section or clockwise shear taken as positive.

Shear force having a downward direction to the right-hand side of the section or anticlockwise shear will be taken as negative.

Explanation:

The bending moment diagram shows the variation of bending moment along the length of the beam. 

The point of contra flexure is defined as a point where the bending moment changes its sign from positive to negative and vice-versa.

There is a sudden change in the bending moment diagram i.e. from positive to negative when there is a concentrated moment acting on a point.

Similarly, if the uniformly distributed load is acting on the beam, the BMD becomes parabolic (2°)

If equal and opposite moments are acting on supports the BMD becomes a rectangle.

Explanation:

∑Fx = 0

RA + RB = 50 kN

By taking moment A equal to zero,

RB × 5 = 50 × 3

∴ RB = 30 kN & RA = 20 kN

Note: Shortcut method for simple beams as given in the question

Reaction = load × opposite distance / Total distance

RA = 50 × 2/5 = 20 kN

RB = 50 × 3/5 = 30 kN