The vertical force acting o the beam or any structure is known as shear force and the turning effect on the structure is known as the bending moment.

The diagram which represents the different effective force on the beam is known as Shear and Moment diagrams. I have given in the example below.

Let us see some numerical example of beam for better understanding.

Also Read, Shear force diagram and Bending Moment diagram for Different Load

Table of Contents

## Draw the Shear and Moment diagrams for the beam

**1)** **Calculate the shear force and bending moment for the beam subjected to concentrated load as shown in the figure. Also, draw the shear force diagram (SFD) and the** **bending moment diagram (BMD).**

Solution;

Free body diagram of the given figure,

Taking moment about point B,

R_{Ay} x 4 – 20 x 2 = 0

4 R_{Ay} = 40

R_{Ay} = 40/4 = 10 KN

Sum of vertical force is equal to zero i.e. Ʃ V = 0

R_{Ay} + R_{By} – 20 = 0

10 + R_{By} -20 = 0

R_{By} = 10 KN

Sum of horizontal force is equal to zero i.e. ƩH = 0

R_{Ay} + R_{By} – 20 = 0

10 + R_{By} -20 = 0

R_{By} = 10 KN

Sum of horizontal force is equal to zero i.e. ƩH = 0

R_{Ax} = 0

**Shear force calculation**

Shear force at A left (F_{AL}) = 0

Shear force at A right ( F_{AR}) = 10 KN

Shear force at C left (F_{CL}) = 10 KN

Shear force at C right (F_{CR}) = 10 – 20 = -10 KN

Shear force at B left (F_{BL}) = -10 KN

Shear force at right (F_{BR}) = 10 – 10 = 0

**Bending moment calculation**

Moment at A (M_{A}) = 0

Moment at C (M_{c}) = 10 x 2 = 40 KNm

Moment at B (M_{B}) = 10 x 4 – 20 x 2 = 0

**2) Calculate the shear force and bending moment diagram of the beam as shown in the figure. Also draw shear force diagram (SFD) and bending moment diagram (BMD).**

Solution;

Free body diagram of the given figure is given below;

Taking moment about point B , we get

ƩM_{B} = 0

R_{Ay} x 6 – 10 x 4 – 10 x 2 = 0

6 R_{Ay} = 40 + 20

R_{Ay} = 60/6 = 10 KN

Sum of vertical force is equal to zero i.e. Ʃ V = 0

R_{Ay} + R_{By} – 10 – 10 =0

10 + R_{By} -20 = 0

R_{by } = 10 KN

Sum of horizontal force is equal to zero i.e. ƩH = 0

R_{Ax} = 0

**Shear force calculation**

Shear force at A left ( F_{AL }) = 0

Shear force at A right (F_{AR}) = 10 KN

Shear force at C left (F_{CL}) = 10 KN

Shear force at C right (F_{CR}) = 10 – 10 = 0

Shear force at D left ( F_{DL}) = 0

Shear force at D right (F_{DR}) = 0 – 10 = -10 KN

Shear force at B left ( F_{BL}) = -10 KN

Shear force at B right (F_{BR}) = – 10 + 10 = 0

**Bending moment calculation**

Moment at A (M_{A}) = 0

Moment at C(M_{C }) = 10 x 2 = 20 KNm

Moment at D (M_{D}) = 10 x 4 – 10 x 2 = 20 KNm

Moment at B (M_{B}) = 10 x 6 – 10 x 4 – 10 x 2 = 0

**3) Find the shear force and bending moment of the given figure. Also, calculate shear force diagram and bending moment diagram.**

Solution;

Free body diagram of the given figure is given below;

Taking moment about point B, we get

R_{Ay}x 5 – 10 x 4 – 5x 2 x 2/2 = 0

5 R_{Ay} -50 = 0

R_{Ay} = 10 KN

Sum of vertical force is equal to zero i.e. Ʃ V = 0

R_{Ay} + R_{By}– 10 – 5 x 2 = 0

10 + R_{By} – 20 = 0

R_{By} = 10 KN

Sum of horizontal force is equal to zero i.e. ƩH = 0

R_{Ax} = 0

**Shear force calculation**

Shear force at A left (F_{AL}) = 0

Shear force at A right (F_{AR}) = 10 KN

Shear force at C left (F_{CL}) = 10 KN

Shear force at C right (F_{CR}) = 10 -10 = 0

Shear force at D left (F_{DL}) = 0

Shear force at D right (F_{DR) = 0}

Let, x be the distance from point D.

Shear force at (x = 1) = 0 – 5 x 1 = -5 KN

Shear force at (x = 2 ) left = -5 x 2 = -10 KN

Shear force at B right (F_{BR}) =-10 + 10 = 0

**Bending moment calculation**

Moment at A = 0

Moment at C = 10 x 1 = 10 KNm

Moment at D = 10 x 3 – 10 x 2 = 10 KNm

Moment at B = 10 x 5 – 10 x 4 – 5 x 2 x2/2 = 0

**4) Calculate the bending moment and shear force of the given beam. Also draw shear force diagram and bending moment diagram.**

Solution,

Free body diagram of the given figure is given below;

Taking moment about point B, we get

ƩM_{B} = 0

R_{Ay} x 4 – 10 x 4 x 4/2 = 0

4 R_{Ay} – 80 = 0

R_{Ay} = 80/4 = 20 KN

Sum of vertical force is equal to zero i.e. Ʃ V = 0

R_{Ay} + R_{By} -10 x 4 = 0

20 + R_{By} – 40 = 0

R_{By} = 20 KN

Sum of horizontal force is equal to zero i.e. ƩH = 0

R_{Ax} = 0

**Shear force calculation**

Let, x be the distance from point A.

Shear force at A left ( F_{AL}) = 0

Shear force at A right ( F_{AR}) = 20 KN

Shear force at (x =1 ) = 20 – 10 x 1 = 10 KN

Shear force at (x = 2 m) = 20 – 10 x 2 = 0

Shear force at ( x = 3 ) = 20 – 10 x 3 = – 10 KN

Shear force at B left ( F_{BL}) = 20 – 10 x 4 = -20 KN

Shear force at B right (F_{BR}) = -20 +20 = 0

**Bending moment calculation**

Moment at A ( M_{A}) = 0

Moment at (x = 1) = 20 x 1 – 10 x 1 x ½ = 15 KNm

Moment at (x = 2 ) = 20 x 2 – 10 x 2 x 2/2 = 20 KNm

Moment at (x = 3) = 20 x 3 – 10 x 3 x 3/2 = 15 KNm

Moment at B (M_{B}) = 20 x 4 – 10 x 4 x 4/2 = 0

In this way we can Draw the Shear and Moment diagrams for the beam.

Read Also,

Point of contraflexure – Zero bending moment at a section of Beam

Moment of inertia formula | Definition for moment of inertia

Shear Force Diagram and Bending Moment Diagram

Kinematic indeterminacy and Static indeterminacy – Beam, Frame etc