Shear Force MCQ Quiz in मराठी - Objective Question with Answer for Shear Force - मोफत PDF डाउनलोड करा

Concept:

Draw the SFD for finding the shear force distribution along the beam

Calculation:

From the Free body diagram (FBD)

\({R_A} = \frac{{2P}}{3}\)

\({R_B} = \;\frac{P}{3}\)

Now,

Drawing the SFD

Shear force is maximum in the section AB

\({\tau _{avg}} = \frac{{2P}}{{3A}}\)     -----(1)

We know that for rectangular cross - section

\({\tau _{max}} = \;\frac{3}{2} \times {\tau _{avg}}\)      -----(2)

Substituting (1) in (2) gives

\(\therefore {\tau _{max}} = \frac{P}{A}\)

Misconception: Generally, one may assume that (\({\tau _{avg}} = \frac{P}{A}\)) without drawing SFD

Explanation:

Let w be load intensity, V be the shear force and M be the bending moment.

w = dv/dx

V = dM/dx

For finding the maximum value of Bending Moment:

dM/dx = 0 ⇒ V = 0

Bending moment is maximum where shear force is zero or its changes sign (positive to negative or vice-versa).

Mistake Points

While solving the problems of Bending Moment, only consider the magnitude of the Shear force and Bending Moment. In the above, when Shear force is zero the Magnitude of Bending moment is Maximum only if it bends in the upper or lower direction.

Concept:

\(S{F_{xx}} = - Wx\)

\(S{F_{x = 0}} = 0;S{F_{x = L}} = wL\)

Shear force at free end is zero and at fixed end it will be wL.

R₁ + R₂ = P

R₂ × L = P × L/4

R₂ = P/4

R₁ = 3P/4

Bending moment at the point of applied load:

\({M_P} = {R_1} \times \frac{L}{4} = \frac{{3P}}{4} \times \frac{L}{4} = \frac{{3PL}}{{16}}\)

Explanation:

Using equilibrium equations

∑F_H = 0

H_A = 0

∑F_V = 0

V_A = 0

∑M_A = 0

-M + M_A = 0

M_A = M

So, the maximum shear force is equal to V_A which is zero.

Shear force diagram for the given beam van be drawn as follows

This diagram is drawn only for a cantilever beam subjected to the concentrated load at the free end.

Calculation:

ΣF_y = 0,

R_A + R_B = 10 × 6 = 60 kN

Taking moment about point A, 

ΣM = 0

R_B × 10 - 10 × 6 × 3 = 0

R_B = 18 kN 

Concept:

Condition for equilibrium on the beam:

ΣFy = 0  ( Beam must be in equilibrium in Y-direction )

ΣMA = 0  ( Point is hinged so net moment about A must be zero )

Calculation:

Given;

L = 8 m, w = 4kN/m,  Uniformly distributed load only over 4 m of length from the right end of a beam

RA + RB = 16 kN

16 × 6 - R_B × 8 = 0  ( Consider clockwise as positive and anticlockwise as negative moment )

R_B = 12 kN

R_A = 4kN

Additional Information

• Either the load is UDL ( Uniformly distributed load ) or VDL ( Varying distributed load the total load is found by calculating its geometric area. 
• For calculation of support reaction we assume UDL or VDL as point load acting on its geometric centre.

Concept:

The slope of the bending moment diagram at a point gives shear force at that point.

Calculation:

\({\left( {Shear\;Force} \right)_{1 - 1}} = {\left( {\frac{{dM}}{{dx}}} \right)_{1 - 1}}\)

(Shear Force)_1-1 = \({\left( {\frac{{dM}}{{dx}}} \right)_{1 - 1}} = 0\;N\)

\({\left( {Shear\;Force} \right)_{2 - 2}} = {\left( {\frac{{dM}}{{dx}}} \right)_{2 - 2}}\)

(Shear Force)_2-2 = \({\left( {\frac{{dM}}{{dx}}} \right)_{2 - 2}} = \frac{{200 - 100}}{2}\)

(Shear Force)2-2 = \( {\left( {\frac{{dM}}{{dx}}} \right)_{2 - 2}} = 50\;N\)

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.