1. Shear Force and Bending Moment Diagrams [SFD & BMD]

2. CO RELATED TO CHAPTER
C304.3 Interpret & analyze the basic principles involved in the behavior of machine parts under load by testing different materials.

3. 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
39.2 kN
3.2 kN M F
Shear force at x-x
Bending moment at x-x

4. Moment and Bending moment
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.
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’.

5. Sign Convention for shear force
+ ve shear force
- ve shear force

6. Sign convention for bending moments:
The bending moment is considered as Sagging Bending Moment if it tends to bend the beam to a curvature having convexity at the bottom as shown in the Fig. given below. Sagging Bending Moment is considered as positive bending moment.
Convexity
Fig. Sagging bending moment [Positive bending moment ]

7. Sign convention for bending moments:
Similarly the bending moment is considered as hogging bending moment if it tends to bend the beam to a curvature having convexity at the top as shown in the Fig. given below. Hogging Bending Moment is considered as Negative Bending Moment.
Convexity
Fig. Hogging bending moment [Negative bending moment ]

8. Shear Force and Bending Moment Diagrams (SFD & BMD)
Shear Force Diagram (SFD):
The diagram which shows the variation of shear force along the length of the beam is called Shear Force Diagram (SFD).
Bending Moment Diagram (BMD):
The diagram which shows the variation of bending moment along the length of the beam is called Bending Moment Diagram (BMD).

9. Point of Contra flexure [Inflection point]: It is the point on the bending moment diagram where bending moment changes the sign from positive to negative or vice versa. It is also called ‘Inflection point’. At the point of inflection point or contra flexure the bending moment is zero.

10. Relationship between load, shear force and bending moment
w kN/m x x1 dx
Fig. A simply supported beam subjected to general type loading
The above Fig. shows a simply supported beam subjected to a general type of loading. Consider a differential element of length ‘dx’ between any two sections x-x and x1-x1 as shown.

11. Neglecting the small quantity of higher orderdx v
V+dV M
M+dM
Fig. FBD of Differential element of the beam x x1
w kN/m O
Taking moments about the point ‘O’ [Bottom-Right corner of the differential element ]
M + (M+dM) – V.dx – w.dx.dx/2 = 0
V.dx = dM 
Neglecting the small quantity of higher order
It is the relation between shear force and BM

12. w kN/m M+dM M v x x1 dx Fig. FBD of Differential element of the beam
V+dV M
M+dM
Fig. FBD of Differential element of the beam x x1
w kN/m O
Considering the Equilibrium Equation ΣFy = 0
- V + (V+dV) – w dx = 0  dv = w.dx 
It is the relation Between intensity of Load and
shear force

13. Variation of Shear force and bending moments
Variation of Shear force and bending moments for various standard loads are as shown in the following Table
Table: Variation of Shear force and bending moments
Type of load
SFD/BMD
Between point loads OR for no load region
Uniformly distributed load
Uniformly varying load
Shear Force Diagram
Horizontal line
Inclined line
Two-degree curve (Parabola)
Bending Moment Diagram
Three-degree curve (Cubic-parabola)

14. Sections for Shear Force and Bending Moment Calculations:
Shear force and bending moments are to be calculated at various sections of the beam to draw shear force and bending moment diagrams.
These sections are generally considered on the beam where the magnitude of shear force and bending moments are changing abruptly.
Therefore these sections for the calculation of shear forces include sections on either side of point load, uniformly distributed load or uniformly varying load where the magnitude of shear force changes abruptly.
The sections for the calculation of bending moment include position of point loads, either side of uniformly distributed load, uniformly varying load and couple
Note: While calculating the shear force and bending moment, only the portion of the udl which is on the left hand side of the section should be converted into point load. But while calculating the reaction we convert entire udl to point load

15. Example Problem 1
Draw shear force and bending moment diagrams [SFD and BMD] for a simply supported beam subjected to three point loads as shown in the Fig. given below. E 5N
10N 8N 2m 3m 1m A C D B

16. 5N
10N 8N 2m 3m 1m A C D B E RA RB
Solution:
Using the condition: ΣMA = 0
- RB × × × × 2 =  RB = N
Using the condition: ΣFy = 0
RA =  RA = 9.75 N
[Clockwise moment is Positive]

17. Shear Force Calculation:2m 3m 1m 1 9 2 3 8 4 5 6 7 8 9 1 1 2 3 4 5 6 7
RA = 9.75 N
RB=13.25N
Shear Force at the section 1-1 is denoted as V1-1
Shear Force at the section 2-2 is denoted as V2-2 and so on...
V0-0 = 0; V1-1 = N V6-6 = N
V2-2 = N V7-7 = 5.25 – 8 = N
V3-3 = – 5 = 4.75 N V8-8 =
V4-4 = N V9-9 = = 0
V5-5 = – 10 = N (Check)

18. 10N 5N 8N B A C E D 2m 2m 1m 3m
9.75N
9.75N
4.75N
4.75N
5.25N
SFD
5.25N
13.25N
13.25N

19. 10N 5N 8N B A C E D 2m 2m 1m 3m
9.75N
9.75N
4.75N
4.75N
5.25N
SFD
5.25N
13.25N
13.25N

20. Bending Moment Calculation
Bending moment at A is denoted as MA
Bending moment at B is denoted as MB
and so on…
  MA = 0 [ since it is simply supported]
MC = 9.75 × 2= 19.5 Nm
MD = 9.75 × 4 – 5 × 2 = 29 Nm
ME = 9.75 × 7 – 5 × 5 – 10 × 3 = Nm
MB = 9.75 × 8 – 5 × 6 – 10 × 4 – 8 × 1 = 0
or MB = 0 [ since it is simply supported]

21. 10N 5N 8N A B C D E 2m 2m 1m 3m
29Nm
19.5Nm
13.25Nm
BMD

22. E 5N 10N 8N 2m 3m 1m A C D B VM-34 9.75N Example Problem 1 4.75N 5.25N
BMD
19.5Nm
29Nm
13.25Nm
9.75N
4.75N
5.25N
13.25N
SFD
Example Problem 1

23. E 5N 10N 8N 2m 3m 1m A C D B 9.75N 4.75N 5.25N SFD 13.25N 29Nm BMD

24. Example Problem 2
2. Draw SFD and BMD for the double side overhanging
beam subjected to loading as shown below. Locate points
of contraflexure if any. 2m 3m
5kN
10kN
2kN/m A B C D E
5kN

25. 2m 3m
5kN
10kN
2kN/m A B C D E RA RB
Solution:
Calculation of Reactions:
Due to symmetry of the beam, loading and boundary conditions, reactions at both supports are equal.
.`. RA = RB = ½( × 6) = 16 kN

26. Shear Force Calculation: V0-0 = 0 V1-1 = - 5kN V6-6 = - 5 – 6 = - 11kN
2kN/m 4 5 2 3 6 7 1 8 9 8 9 2 3 4 5 7 1 6 2m 3m 3m 2m
RA=16kN
RB = 16kN
Shear Force Calculation: V0-0 = 0
V1-1 = - 5kN V6-6 = - 5 – 6 = - 11kN
V2-2 = - 5kN V7-7 = = 5kN
V3-3 = = 11 kN V8-8 = 5 kN
V4-4 = 11 – 2 × 3 = +5 kN V9-9 = 5 – 5 = 0 (Check)
V5-5 = 5 – 10 = - 5kN

27. 2m 3m
5kN
10kN
2kN/m A B C D E +
5kN
11kN
SFD
5kN

28. 2m 3m
5kN
10kN
2kN/m A B C D E
RA=16kN
RB = 16kN
Bending Moment Calculation:
MC = ME = 0 [Because Bending moment at free end is zero]
MA = MB = - 5 × 2 = - 10 kNm
MD = - 5 × × 3 – 2 × 3 × 1.5 = +14 kNm

29. 2m 3m
5kN
10kN
2kN/m A B C D E
14kNm
BMD
10kNm
10kNm

30. 2m 3m
5kN
10kN
2kN/m A B C D E +
5kN
11kN
SFD
10kNm
14kNm
BMD

31. 2m 3m 5kN 10kN 2kN/m A B C D E 10kNm 10kNm x = 1 or 10
Points of contra flexure
Let x be the distance of point of contra flexure from support A
Taking moments at the section x-x (Considering left portion)
x = 1 or 10
.`. x = 1 m

32. Example Problem
Example Problem 3
3. Draw SFD and BMD for the single side overhanging beam subjected to loading as shown below. Determine the absolute maximum bending moment and shear forces and mark them on SFD and BMD. Also locate points of contra flexure if any.
5kN
10kN/m
2 kN A D C B 4m 1m 2m

33. 5kN
10kN/m
2 kN A B RA 4m RB 1m 2m
Solution : Calculation of Reactions:
ΣMA = 0
- RB × × 4 × × × 7 = 0  RB = 24.6 kN
ΣFy = 0
RA – 10 x 4 – =  RA = 22.4 kN

34. Shear Force Calculations:
5kN
10kN/m
2 kN 5 2 4 1 3 6 7 4 5 1 2 3 6 7
RA=22.4kN 4m 1m 2m
RB=24.6kN
Shear Force Calculations:
V0-0 =0; V1-1 = 22.4 kN V5-5 = = 5 kN
V2-2 = 22.4 – 10 × 4 = -17.6kN V6-6 = 5 kN
V3-3 = – 2 = kN V7-7 = 5 – 5 = 0 (Check)
V4-4 = kN

35. 5kN 10kN/m 2 kN 4m 1m 2m SFD A C B D RA=22.4kN RB=24.6kN 22.4kN 5 kN
x = 2.24m
17.6kN
19.6kN
19.6kN
SFD

36. 4m 1m 2m
2 kN
5kN
10kN/m A B C D
RA=22.4kN
RB=24.6kN X x
Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance x
from support A as shown above.
The shear force at that section can be calculated as
Vx-x = x = 0  x = 2.24 m

37. MB = -5 × 2 = -10 kNm (Considering Right portion of the section)A B C D
RA=22.4kN
RB=24.6kN
Calculations of Bending Moments:
MA = MD = 0
MC = 22.4 × 4 – 10 × 4 × 2 = 9.6 kNm
MB = 22.4 × 5 – 10 × 4 × 3 – 2 × 1 = - 10kNm (Considering Left portion
of the section)
Alternatively
MB = -5 × 2 = -10 kNm (Considering Right portion of the section)
Absolute Maximum Bending Moment is at X- X ,
Mmax = 22.4 × 2.24 – 10 × (2.24)2 / 2 = 25.1 kNm

38. 5kN 10kN/m 2 kN A D C B 4m 1m 2m BMD X x = 2.24m X RA=22.4kN RB=24.6kN
Mmax = 25.1 kNm
9.6kNm
Point of
contra flexure
BMD
10kNm

39. 4m 1m 2m 2 kN 5kN 10kN/m A B C D SFD BMD X x = 2.24m RA=22.4kN
RB=24.6kN X
x = 2.24m
22.4kN
19.6kN
17.6kN
5 kN
SFD
x = 2.24m
9.6kNm
10kNm
BMD
Point of
contra flexure

40. 4m 1m 2m
2 kN
5kN
10kN/m A B C D
RA=22.4kN
RB=24.6kN X x
Calculations of Absolute Maximum Bending Moment:
Max. bending moment will occur at the section where the shear force is
zero. The SFD shows that the section having zero shear force is available
in the portion AC. Let that section be X-X, considered at a distance x
from support A as shown above.
The shear force at that section can be calculated as
Vx-x = x = 0  x = 2.24 m
Max. BM at X- X ,
Mmax = 22.4 × 2.24 – 10 × (2.24)2 / 2 = 25.1 kNm

41. 5kN 10kN/m 2 kN A D C B 4m 1m 2m BMD X x = 2.24m X RA=22.4kN RB=24.6kN
Mmax = 25.1 kNm
9.6kNm
Point of
contra flexure
BMD
10kNm

42. BMD Let a be the distance of point of contra flexure from support B
Taking moments at the section A-A (Considering left portion) A
a = 0.51 m
Mmax = 25.1 kNm
9.6kNm
Point of
contra flexure
BMD
10kNm a A