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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
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