Shear Force and Bending Moment Diagram

Ashwin Dobariya

Vmax = +P/2

L L

V = +P

Mmax = -PL

L

P P

V = +P/2

V = -P/2

Mmax = PL/4

P = wL

Vmax = -P/2

Mmax = PL/8 = wL2/8

L

P = wL

Vmax = +P

Mmax = -PL/2 = -wL2/2

Overview of the session

• Introduction

• Learning outcomes

• Shear force

• Bending moment

• Worked examples

• Assignment questions

• Summary/conclusions

Objective of the session • At the end of this lecture, you would be able to: • Explain and calculate shear force and bending moment

in beams

• Calculate the shear force and bending moments of simply supported beams

• Draw shear force and bending moment diagrams for simply supported beams

• Solve Assignment’s questions

What is Shear Force and Bending Moment?

Shear Force: is the algebraic sum of the vertical forces at any section is known as shear force. It is briefly written as SF.

Bending Moment: The algebraic sum of the moments of all the forces acting to the right or left of the section is known as bending moment. It is written as BM.

In this session the shear force and bending moment diagrams for different types of loads acting on the beam (simply supported) will be considered.

SF and BM diagram

A SF diagram is one which shows the variation of the shear force along the length of the beam

A BM diagram is one which shows the variation of the bending moment along the length of the beam

2015/16 AD 5

Sign convention for shear force and bending moment

The force on a beam produce shearing at all sections along the length

Upward total force on the left section indicates positive shear

Downward total force on the left section indicates negative shear

Positive shear trends to make the section slide up on the left

2015/16 AD 6

Sign convention for shear force and bending moment

(a) Positive bending moment compresses the top section of the beam and stretch the lower section (sagging)

(b) Negative bending moment occurs when the loads causes the beam to stretch on its top surface (hogging)

2015/16 AD 7

Important points for SF and BM diagram

• Consider left/right section of the beam

• Add the forces (including reaction)

• The positive value of SF and BM are plotted above the base line and negative values below the base line

• The SF diagram will increase or decrease suddenly i.e. by a vertical straight line at a section where there is a vertical point load

• The SF between ay two vertical point load will be constant and hence the shear SF diagram will be horizontal

• The BM at the supports of a simply supported beam will be zero

2015/16 AD 8

Nature of SF and BM variations

load SF BM

No Load Constant Linear

UDL Linear Parabolic

Uniformly varying Parabolic Cubic

2015/16 AD 9

The bending moment is maximum/minimum wherever shear force is zero. Such points are significant points and should be indicated in BMD The point of contraflexure (where BM changes its sign) is also very important and hence should be indicated in BMD

Example

A simply supported beam of length 6 m, carries point load of 3 kN and 6 kN at a distance of 2 m and 4 m from the left end. Draw the SF and BM diagram for the beam.

2015/16 AD 10

Solution

First calculate the reaction RA and RB 𝑹𝑨 + 𝑹𝑩 = 𝟑 + 𝟔 = 𝟗 𝒌𝑵

Taking moment of the force about A we get, 𝑹𝑩 × 𝟔 = 𝟑 × 𝟐 + 𝟔 × 𝟒

𝑹𝑩× 𝟔 = 𝟑𝟎

𝑹𝑩= 𝟓 kN

Which gives 𝑹𝑨 = 𝟒 𝒌𝑵

2015/16 AD 11

6 kN

A B

3 kN

RA RB

C D

2 m 2 m

6 m

Solution Shear Force Diagram

Shear force at A = + 4 kN

Shear force between A and C is constant and equal to +4 kN

Shear force at C = 4 – 3 = +1 kN

Shear force between C and D is constant and equal to +1 kN

Shear force at D = 1 - 6 = -5 kN

Shear force between D and B is constant and equal to -5 kN

Shear Force at B = -5 kN

2015/16 AD 12

6 kN

A B

3 kN

RA RB

C D

2 m 2 m 6 m

5 kN

+

-

4 kN

5 kN

3 kN

1 kN

Solution Bending Moment Diagram

BM at A, MA = 0 kN BM at C, MC = RA × 2 = 4 × 2 = + 8 kNm BM at D, MD = RA × 4 - 3 × 2 = 4 × 4 - 3 × 2 = +10 kNm BM at B, MB = 0 kN Points to remember: The area of + ve and – ve SF diagram always remains same BM is maximum where SF is zero

2015/16 AD 13

6 kN

A B

3 kN

RA RB

C D

2 m 2 m 6 m

5 kN

+

-

4 kN

5 kN

3 kN

1 kN

8 kNm 10 kNm

SF and BM diagram for a simply supported beam carrying a UDL

Draw the shear force and bending moment diagram for a simply supported beam of length 9 m and carrying a UDL of 10 kN/m for a distance of 6 m from the left end. Also calculate the maximum BM on the section.

2015/16 AD 14

A B

10 kN/m

RA RB

C

6 m 9 m

First calculate the reaction RA and RB

Taking moment of the forces about A, we get

𝑹𝑩 × 𝟗 = 𝟏𝟎 × 𝟔 ×𝟔

𝟐= 𝟏𝟖𝟎

𝑹𝑩 =𝟏𝟖𝟎

𝟗= 𝟐𝟎 𝒌𝑵

𝑹𝑨 + 𝑹𝑩 = 𝟏𝟎 × 𝟔 = 𝟔𝟎 𝒌𝑵 Which gives 𝑹𝑨 = 𝟒𝟎 𝒌𝑵

Solution

2015/16 AD 15

B

A

10 kN/m

RA RB

C

6 m 9 m

+

-

40

20

D C

Consider any section at a distance x form A between A and C. The shear force at the section is given by

𝐹𝑥 = 𝑅𝐴 − 10𝑥 = +40 − 10𝑥 … . . (1) Equation (1) shows that shear force varies by a straight line law between A and C At A, x =0 hence 𝐹𝐴 = 40 − 0 = +40 𝑘𝑁 At C, x = 6 m hence 𝐹𝐶 = +40 − 10 × 6 =− 20 𝑘𝑁 This means that somewhere between A and C the shear force is zero. Let SF is zero at x meter from A

0 = 40 − 10𝑥

𝑥 =40

10= 4 𝑚

Hence SF is zero at a distance 4 m from A The SF is constant between C and B At B SF is -20 kN

A

solution

2015/16 AD 16

BM Diagram The BM at any section between A and C at a distance x form A is given by

𝑀𝑥 = 𝑅𝐴𝑥 − 10. 𝑥.𝑥

2= 40𝑥 − 5𝑥2 … … .(2)

Equation (2) shows that BM varies according to parabolic law between A and C At x = 0 hence 𝑀𝐴 = 40 × 0 − 5 × 0 = 0 At x = 6 hence 𝑀𝐶 = 40 × 6 − 5 × 62 =+ 60 𝑘𝑁𝑚 At x = 4 hence 𝑀𝐷 = 40 × 4 − 5 × 42 =+ 80 𝑘𝑁𝑚 The bending moment varies between C and B varies according to linear law Maximum Bending Moment The BM is maximum at a point where shear force changes sign. Max BM = +80 kNm

Parabolic Straight line

A

B

10 kN/m

RA RB

C

6 m 9 m

+

-

40

20 D

A

C

C A D B

Base line

Extra Example

2015/16 AD 17

Simply supported beam with point load and udl

A simply supported beam of length 10 m, carries the uniformly distributed load and two point loads as shown in figure. Draw the SF and BM diagram for the beam. Also calculate the maximum bending moment.

Simply supported beam with point load and udl

Solution,

First calculate the reactions RA , RB 𝑅𝐴 + 𝑅𝐵 = 50 + (10 × 4) + 40 = 130 𝑘𝑁

Taking moments of all forces about A, we get

𝑅𝐴 × 10 = (50 × 2) + (10 × 4 × 2 +4

2) + 40(2 + 4)

𝑅𝐴 × 10 = 100 + 160 + 240 = 500

𝑅𝐵 =500

10= 50 𝑘𝑁

From 𝑅𝐴 + 𝑅𝐵 = 130 𝑘𝑁 which gives 𝑅𝐴 = 80 𝑘𝑁

SF Diagram The SF at A, RA=+80 kN

The SF will remain constant between A and C and equal to +80 kN

The SF at C = 80-50 = +30 kN

The SF just on LHS of D =80-50-10×4=-50 kN

The SF at B = -50 kN

The SF remain constant between D and B equal to -50 kN

The SF is zero at point E between C and D

Let the distance of E from point A is x,

Now SF at E = 80-50-10(x-2)

But SF at E =0

50-10x=0 which gives x = 0

BM Diagram BM diagram

BM at A is zero

BM at B is zero

BM at C, 𝑀𝑐 = RA ×2=80×2=160 kNm

BM at D, 𝑀𝐷 = 𝑅𝐴 × 6 − 50 × 4 − 10 ×

4 ×4

2= 80 × 6 − 200 − 80 = 480 −

200 − 80 = 200 𝑘𝑁𝑚

At E, x = 5 m and hence BM at E,

𝑀𝐸 = 𝐹𝐴 × 5 − 50 5 − 2 − 10 ×5 − 2

2

= 80 × 5 − 50 × 3 − 10 × 3 ×3

2= 400 − 150 − 45 = 205 𝑘𝑁𝑀

The maximum BM is at E, where SF becomes zero after changing its sign

Maximum BM = 205 kNm at E

Recap

We have covered shear force and bending moment diagram of beams

We learned about how bending moment become maximum

We calculated maximum bending moment

Next session:

Stresses in Bending and theory of bending