This presentation summarizes shear force and bending moment diagrams. It introduces beams and the concepts of shear force and bending moment. It then provides examples of calculating the shear force and bending moment at different points for a simply supported beam and a cantilever beam under various loading conditions. Step-by-step free body diagrams and calculations are shown. Critical points where the shear force and bending moment change are identified. Finally, it discusses determining the shear force and bending moment functions for a beam with a uniform distributed load.

1. Presentation By :
Dr Muhammad Tahir
• B.E (Civil Engineering)
 Mehran University of Engineering and Technology
Jamshoro
• M.E (Structural Engineering)
 UTHM Malaysia
• PhD (Structural Engineering)
 Monash Australia
Shear Force and Bending
Moment Diagrams

2. Shear Force and Bending
Moment Diagrams
A BEAM is a long, slender, structural
member designed to support transverse
loadings.
A transverse loading is applied
perpendicular to the axis of the beam.
Beams classified by their supports.

3. Bending Moment
The algebric sum of the moments of the
forces on either side of the section of a
loaded beam is called Bending Moment.
SHEAR FORCE
The algebric sum of the vertical forces
on either side of the section of a loaded
beam is called Shearing Force

5. P
w
a
L
b
x1
x2
x3
Simply Supported Beam

6. P
w
a
L
b
x1
x2
x3
Cantilever Beam

8. 3 m
6 m
6 kN
Determine the internal forces just to the left and the right of the
external force
9 kN m
Critical Points

9. 4 ft 4 ft
50 lb/ft
A C
Critical Points

10. P
w
a
L
b
x1
x2
x3

11. P
w
a
L
b
x1
x2
x3

12. 4 ft 4 ft
50 lb/ft
A C
Critical Points

13. 3 m
6 m
6 kN
9 kN m
Shear Force and Bending
Moment Diagram

14. FBD of Beam
3 m
6 m
6 kN
9 kN m
Ay
Dy
Dx

15. x
x
y
y y
D
y
y
y
F 0
D 0
F 0
A 6 D 0
M 0
9 6(6) - A (9) 0
A 5 kN
D 1 kN



  

 






16. FBD of Beam
3 m
6 m
6 kN
9 kN m
5 kN 1 kN
x
x
9
x
3
3
x
0





17. 5 kN
x1
 
y
0 x 3
F 0
5 V 0
V 5 kN
M 0
M 5 kN x 0
M 5xkN m
 

 


 
 


V
M

18. 3 m 6 kN
5 kN
x
x
V
M
    
 
y
3 x 9
F 0
5 kN 6 kN V 0
V 1 kN
M 0
M 5 x 6 x 3 0
M 18 x kN m
 

  
 

   
  



19. V (kN)
M (kN m)
V = -1 kN
V = 5 kN
M = 5x kN m M = (18-x) kN m

20. 4 ft 4 ft
50 lb/ft
A C

21. 4 ft 4 ft
50 lb/ft
A y
Cx
Cy

22. 4 ft 4 ft
50 lb/ft
Ay
Cx
Cy
400 lb

23. x
x
y
y y
A
y
y
y
F 0
C 0
F 0
A C 50(8) 0
M 0
50(8)(4) - C (8) 0
A 200 lb
C 200 lb



  








24. x
50 lb/ft
200 lb V
M
 
y
2
0 x 8
F 0
V 200 lb 50x 0
V 50x 200 lb
M 0
x
M 200x 50(x) 0
2
M 200x 25x lb ft
 

   
 

 
  
 
 
  



25. x
50 lb/ft
50 lb/ft
200 lb
V
M
0
M
lb
200
-
V
8,
x
at
0
M
lb
200
V
0,
x
at







26. V (lb)
M (lb ft)
x
50
200
V 

2
x
25
x
200
M 


30. P
w
a
L
b
W=100 + R(0.001)
L=15
a=5
b=10
P=1000+ R(0.001)
Assignment 1 Last day 2 May 2023

31. 50 +R (0.001)kN.m
2 +R (0.001)kN/m

32. 4 +R (0.001) kip/ft

33. Thank you