PPT: Shear Force & Bending Moment Diagrams (SFD & BMD) | Strength of Materials (SOM) - Mechanical Engineering PDF Download

 Page 1


Shear Force and Bending Moments 
Consider a section x-x at a distance 6m from left hand support A 
5kN
10kN
8kN
4m 5m 5m 1m
A
C
D
B
R
A 
= 8.2 kN R
B
=14.8kN
E
x
x
6 m
Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
Page 2


Shear Force and Bending Moments 
Consider a section x-x at a distance 6m from left hand support A 
5kN
10kN
8kN
4m 5m 5m 1m
A
C
D
B
R
A 
= 8.2 kN R
B
=14.8kN
E
x
x
6 m
Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
To find the forces experienced by the section, consider any one portion of the
beam. Taking left hand portion
Transverse force experienced = 8.2 – 5 = 3.2 kN (upward)
Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise)
If we consider the right hand portion, we get
Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward)
Moment experienced  = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m       
(anticlockwise)
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
4 m
6 m
9 m
1 m
5 m
Page 3


Shear Force and Bending Moments 
Consider a section x-x at a distance 6m from left hand support A 
5kN
10kN
8kN
4m 5m 5m 1m
A
C
D
B
R
A 
= 8.2 kN R
B
=14.8kN
E
x
x
6 m
Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
To find the forces experienced by the section, consider any one portion of the
beam. Taking left hand portion
Transverse force experienced = 8.2 – 5 = 3.2 kN (upward)
Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise)
If we consider the right hand portion, we get
Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward)
Moment experienced  = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m       
(anticlockwise)
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
4 m
6 m
9 m
1 m
5 m
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
3.2 kN
3.2 kN
39.2 kN-m
39.2 kN-m
Thus the section x-x considered is subjected to forces 3.2 kN and
moment 39.2 kN-m as shown in figure. The force is trying to shear off
the section and hence is called shear force. The moment bends the
section and hence, called bending moment.
Page 4


Shear Force and Bending Moments 
Consider a section x-x at a distance 6m from left hand support A 
5kN
10kN
8kN
4m 5m 5m 1m
A
C
D
B
R
A 
= 8.2 kN R
B
=14.8kN
E
x
x
6 m
Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
To find the forces experienced by the section, consider any one portion of the
beam. Taking left hand portion
Transverse force experienced = 8.2 – 5 = 3.2 kN (upward)
Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise)
If we consider the right hand portion, we get
Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward)
Moment experienced  = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m       
(anticlockwise)
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
4 m
6 m
9 m
1 m
5 m
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
3.2 kN
3.2 kN
39.2 kN-m
39.2 kN-m
Thus the section x-x considered is subjected to forces 3.2 kN and
moment 39.2 kN-m as shown in figure. The force is trying to shear off
the section and hence is called shear force. The moment bends the
section and hence, called bending moment.
Shear force at a section: The algebraic sum of the vertical forces 
acting on the beam either to the left or right of the section is 
known as the shear force at a section.
Bending moment (BM) at section: The algebraic sum of the moments
of all forces acting on the beam either to the left or right of the
section is known as the bending moment at a section
3.2 kN
3.2 kN
F
F
Shear force at x-x
M
Bending moment  at x-x
39.2 kN
Page 5


Shear Force and Bending Moments 
Consider a section x-x at a distance 6m from left hand support A 
5kN
10kN
8kN
4m 5m 5m 1m
A
C
D
B
R
A 
= 8.2 kN R
B
=14.8kN
E
x
x
6 m
Imagine the beam is cut into two pieces at section x-x and is separated, as
shown in figure
To find the forces experienced by the section, consider any one portion of the
beam. Taking left hand portion
Transverse force experienced = 8.2 – 5 = 3.2 kN (upward)
Moment experienced = 8.2 × 6 – 5 × 2 = 39.2 kN-m (clockwise)
If we consider the right hand portion, we get
Transverse force experienced = 14.8 – 10 – 8 =-3.2 kN = 3.2 kN (downward)
Moment experienced  = - 14.8 × 9 +8 × 8 + 10 × 3 = -39.2 kN-m = 39.2 kN-m       
(anticlockwise)
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
4 m
6 m
9 m
1 m
5 m
5kN
A
8.2 kN
10kN
8kN
B
14.8 kN
3.2 kN
3.2 kN
39.2 kN-m
39.2 kN-m
Thus the section x-x considered is subjected to forces 3.2 kN and
moment 39.2 kN-m as shown in figure. The force is trying to shear off
the section and hence is called shear force. The moment bends the
section and hence, called bending moment.
Shear force at a section: The algebraic sum of the vertical forces 
acting on the beam either to the left or right of the section is 
known as the shear force at a section.
Bending moment (BM) at section: The algebraic sum of the moments
of all forces acting on the beam either to the left or right of the
section is known as the bending moment at a section
3.2 kN
3.2 kN
F
F
Shear force at x-x
M
Bending moment  at x-x
39.2 kN
Moment and Bending moment
Bending Moment (BM): The moment which causes the
bending effect on the beam is called Bending Moment. It is
generally denoted by ‘M’ or ‘BM’.
Moment: It is the product of force and perpendicular
distance between line of action of the force and the point
about which moment is required to be calculated.