1. UNIT –III Bending Moment and Shear Force in Beams
Subject : Mechanics of Materials
N.RAM KUMAR M.E.,
ASSISTANT PROFESSOR
DEPARTMENT OF MECHANICAL ENGINEERING
CHRIST UNIVERSITY FACULTY OF ENGINEERING
BANGALORE
N.Ram Kumar, CUFE

2. Christ University Faculty of Engineering
INTRODUCTION
A beam is a structural member used for bearing loads. It is typically used for resisting vertical loads, shear forces and bending moments.
A beam is a structural member which is acted upon by the system of external loads at right angles to the axis.
N.Ram Kumar, CUFE
N.Ram Kumar, Assistant Professor, Department of Mechanical Engineering

3. TYPES OF BEAMS
Beams can be classified into many types based on three main criteria. They are as follows:
Based on geometry:
Straight beam – Beam with straight profile
Curved beam – Beam with curved profile
Tapered beam – Beam with tapered cross section
Based on the shape of cross section:
I-beam – Beam with ‘I’ cross section
T-beam – Beam with ‘T’ cross section
C-beam – Beam with ‘C’ cross section
Based on equilibrium conditions:
Statically determinate beam – For a statically determinate beam, equilibrium conditions alone can be used to solve reactions.
Statically indeterminate beam – For a statically indeterminate beam, equilibrium conditions are not enough to solve reactions. Additional deflections are needed to solve reactions.
N.Ram Kumar, CUFE

4. Based on the type of support: Simply supported beam Cantilever beam
Fixed beam
Overhanging beam
Continuous beam
The last two beams are statically indeterminate beams.
N.Ram Kumar, CUFE

5. N.Ram Kumar, CUFE

6. TYPES OF LOADS Point Load or Concentrated Load.
Uniformly Distributed Loads.
Uniformly Varying Loads.
Externally Applied Moments.
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7. TYPES OF SUPPORTS Simple Support. Roller Support. Hinged Pin Support.
Fixed Support.
N.Ram Kumar, CUFE

8. N.Ram Kumar, CUFE

9. SHEAR FORCE
Shear Force at a section in a beam is the force that is trying to shear off the section and is obtained as the algebraic sum of all forces including the reactions acting normal to the axis of the beam either to the left or right of the section.
The procedure to find shear force at a section is to imagine a cut in a beam at a section, considering either the left or the right portion and find algebraic sum of all forces normal to the axis.
N.Ram Kumar, CUFE

10. SIGN CONVENTION FOR SHEAR FORCE
+ ve shear force
- ve shear force
N.Ram Kumar, CUFE

11. BENDING MOMENT
Bending moment at a section in a beam is the moment that is trying to bend it and is obtained as the algebraic sum of the moments about the section of all the forces(including the reaction acting on the beam either to the left or to the right of the section.
N.Ram Kumar, CUFE

12. 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)
N.Ram Kumar, CUFE

13. 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 ]
N.Ram Kumar, CUFE

14. 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.
N.Ram Kumar, CUFE

15. SHEAR FORCE DIAGRAM
A diagram in which the ordinate represents shear force and the abscissa represents the position of the section is called shear force diagram.
N.Ram Kumar, CUFE

16. BENDING MOMENT DIAGRAM
Bending moment diagram may be defined as a diagram in which ordinate represents bending moment and abscissa represents the position of the section.
N.Ram Kumar, CUFE

17. SFD AND BMD FOR THE STANDARD CASES
Cantilever Subjected to
A Point or concentrated load at free end.
Uniformly distributed load over entire span.
Uniformly varying load over entire span.
2. Simply supported beam subjected to
An external moment.
3. Overhanging beam subjected to a Point or concentrated load at free end.
N.Ram Kumar, CUFE

18. CANTILEVER BEAM
A cantilever is a beam whose one end is fixed and the other end is free.
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19. EXAMPLES OF CANTILEVER BEAM PAMBAN BRIDGE
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20. CAR PARKING ROOF
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21. BUS STOP SHELTER
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22. TRAFFIC SIGNAL POST
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23. TRAFFIC SIGNAL POST
The above traffic signal post is a best example of cantilever beam. Its a cantilever beam with three point loads.
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24. Crane
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25. CANTILEVER BEAM SUBJECTED TO POINT LOAD
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26. CANTILEVER BEAM WITH UNIFORMLY DISTRIBUTED LOAD
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27. SIMPLY SUPPORTED BEAM
A simply supported beam is one whose ends freely rest on walls or columns.
In all such cases, the reactions are always upwards.
N.Ram Kumar, CUFE

28. EXAMPLE
The above shaft could be a best example of a simply supported Beam.
In that shaft both ends are supported by bearings and two loads are given by the pulley which could be taken as point load acting downwards.
N.Ram Kumar, CUFE

29. SIMPLY SUPPORTED BEAM WITH A POINT LOAD
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30. SIMPLY SUPPORTED BEAM WITH UNIFORMLY DISTRIBUTED LOAD
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31. SIMPLY SUPPORTED BEAM WITH UNIFORMLY VARYING LOAD
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32. OVERHANGING BEAM
A overhanging beam is one which has the loads beyond the supports.
Eg: Car.
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33. EXAMPLE
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34. Common Relationships Load Constant Linear Shear Parabolic Moment Cubic
Constant
Linear
Shear
Parabolic
Moment
Cubic
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35. Common Relationships Load Constant Shear Linear Moment Parabolic M
Constant
Shear
Linear
Moment
Parabolic M
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36. Example: Draw Shear & Moment diagrams for the following beam
12 kN
8 kN A C D B
1 m
3 m
1 m
RA = 7 kN  RC = 13 kN 
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37. 12 kN 8 kN 1 m 3 m 1 m 2.4 m 8 7 8 7 V -15 -5 7 M -8 A C D B (kN)
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38. Relationship between load, shear force and
bending moment L
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.
N.Ram Kumar, CUFE

39. 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
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
N.Ram Kumar, CUFE

40. 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.
N.Ram Kumar, CUFE

41. 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)
N.Ram Kumar, CUFE

42. 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
N.Ram Kumar, CUFE

43. Using the condition: ΣMA = 0 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]
N.Ram Kumar, CUFE

44. 5N 10N 8N 2m 3m 1m Shear Force Calculation:1 8 9 2 3 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)
N.Ram Kumar, CUFE

45. 10N 5N 8N B A C E D 2m 2m 1m 3m 9.75N 9.75N 4.75N 4.75N 5.25N SFD
N.Ram Kumar, CUFE
13.25N
13.25N

46. 10N 5N 8N B A C E D 2m 2m 1m 3m 9.75N 9.75N 4.75N 4.75N 5.25N SFD
N.Ram Kumar, CUFE
13.25N
13.25N

47. 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]
N.Ram Kumar, CUFE

48. 10N 5N 8N A B C D E 2m 2m 1m 3m 29Nm 19.5Nm 13.25Nm BMD
N.Ram Kumar, CUFE

49. 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
N.Ram Kumar, CUFE

50. E 5N 10N 8N 2m 3m 1m A C D B 9.75N 4.75N 5.25N SFD 13.25N 29Nm BMD
N.Ram Kumar, CUFE

51. 2. Draw SFD and BMD for the double side overhanging
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
N.Ram Kumar, CUFE

52. Calculation of Reactions: 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
N.Ram Kumar, CUFE

53. 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
N.Ram Kumar, CUFE

54. 2m 3m
5kN
10kN
2kN/m A B C D E +
5kN
11kN
SFD
5kN
N.Ram Kumar, CUFE

55. Bending Moment Calculation:
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
N.Ram Kumar, CUFE

56. 2m 3m
5kN
10kN
2kN/m A B C D E
14kNm
BMD
10kNm
10kNm
N.Ram Kumar, CUFE

57. 2m 3m 5kN 10kN 2kN/m A B C D E + 5kN 11kN SFD 10kNm 14kNm BMD
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58. 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
N.Ram Kumar, CUFE

59. 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
N.Ram Kumar, CUFE

60. Solution : Calculation of Reactions: ΣMA = 0
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
N.Ram Kumar, CUFE

61. 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
N.Ram Kumar, CUFE

62. 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
N.Ram Kumar, CUFE

63. 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
N.Ram Kumar, CUFE

64. 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
N.Ram Kumar, CUFE

65. 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
N.Ram Kumar, CUFE

66. 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
N.Ram Kumar, CUFE

67. 4m 1m 2m 2 kN 5kN 10kN/m A B C D X x RA=22.4kN RB=24.6kN
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
N.Ram Kumar, CUFE

68. 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
N.Ram Kumar, CUFE

69. 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
N.Ram Kumar, CUFE A

70. Draw SFD and BMD for the single side overhanging beam
Example Problem 4
Draw SFD and BMD for the single side overhanging beam
subjected to loading as shown below. Mark salient points on
SFD and BMD.
60kN/m
20kN
20kN/m A C B D 3m 2m 2m
N.Ram Kumar, CUFE

71. 60kN/m 20kN 20kN/m A B 3m 2m 2m C D Solution:RA 3m 2m RB 2m
Solution:
Calculation of reactions:
ΣMA = 0
RB × 5 + ½ × 3 × 60 × (2/3) × × 4 × × 7 = 0  RB =144kN
ΣFy = 0
RA – ½ × 3 × 60 – 20 × =  RA = 46kN.
N.Ram Kumar, CUFE

72. 60kN/m 20kN 20kN/m 3m 2m 2m Shear Force Calculations:1 4 2 3 5 6 3 4 1 2 5 6
RB = 144kN RA
RA = 46kN 3m 2m 2m
Shear Force Calculations:
V0-0 =0 ; V1-1 = kN V4-4 = = + 60kN
V2-2 = +46 – ½ × 3 × 60 = - 44 kN V5-5 = +60 – 20 × 2 = + 20 kN
V3-3 = - 44 – 20 × 2 = - 84 kN V6-6= 20 – 20 = 0 (Check)
N.Ram Kumar, CUFE

73. Example Problem 4 60kN/m 20kN 20kN/m 3m 2m 2m 46kN 44kN 84kN 60kN 20kN1 4 2 3 5 6 3 4 1 2 5 6
RB = 144kN RA
RA = 46kN 3m 2m 2m
46kN
44kN
84kN
60kN
20kN
SFD
Parabola
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74. 60kN/m 3m
20kN/m
20kN 2m A B
RA =46kN C D
RB=144kN 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 expression at that section should be equated to zero. i.e.,
Vx-x = 46 – ½ .x. (60/3)x = 0  x = m
N.Ram Kumar, CUFE

75. 60kN/m 20kN 20kN/m A C B D 3m 2m Calculation of bending moments:
RA =46kN C D
RB=144kN
Calculation of bending moments:
MA = MD = 0
MC = 46 × 3 – ½ × 3 × 60 × (1/3 × 3) = 48 kNm[Considering LHS of section]
MB = × 2 – 20 × 2 × 1 = - 80 kNm [Considering RHS of section]
Absolute Maximum Bending Moment, Mmax = 46 × – ½ × ×(2.145 × 60/3) × (1/3 × 2.145) = kNm
N.Ram Kumar, CUFE

76. 60kN/m 20kN 20kN/m A C B D 3m 2m BMD RB=144kN RA =46kN 48kNm 65.74kNm
Parabola
Cubic
parabola
Point of
Contra flexure
BMD
Parabola
N.Ram Kumar, CUFE
80kNm

77. 60kN 46kN 20kN 44kN SFD 84kN BMD Parabola 65.74kNm Parabola Cubic
Point of
Contra flexure
BMD
Parabola
N.Ram Kumar, CUFE
80kNm

78. 60kN/m 3m
20kN/m
20kN 2m A B
RA =46kN C D
RB=144kN X
x=2.145m
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 expression at that section should be equated to zero. i.e.,
Vx-x = 46 – ½ .x. (60/3)x = 0  x = m
BM at X- X ,
Mmax = 46 × – ½ × ×(2.145 × 60/3) × (1/3 × 2.145)=65.74 kNm.
N.Ram Kumar, CUFE

79. 60kN/m 20kN 20kN/m A C B D 3m 2m BMD RB=144kN RA =46kN 48kNm 65.74kNm
Parabola
Cubic
parabola a
Point of
Contra flexure
BMD
Parabola
N.Ram Kumar, CUFE
80kNm

80. Point of contra flexure:
BMD shows that point of contra flexure is existing in the portion CB. Let ‘a’ be the distance in the portion CB from the support B at which the bending moment is zero. And that ‘a’ can be calculated as given below.
ΣMx-x = 0
a = m
N.Ram Kumar, CUFE

81. 5. Draw SFD and BMD for the single side overhanging beam
Example Problem 5
5. Draw SFD and BMD for the single side overhanging beam
subjected to loading as shown below. Mark salient points on
SFD and BMD.
40kN
0.5m
30kN/m
20kN/m
0.7m A B C D E 2m 1m 1m 2m
N.Ram Kumar, CUFE

82. 20kN/m 30kN/m 40kN 2m A D 1m B C E 20kN/m 30kN/m 40kN 2m A D 1m B C E
40x0.5=20kNm
20kN/m
30kN/m
40kN 2m A D 1m B C E
N.Ram Kumar, CUFE

83. 40kN 30kN/m 20kN/m 20kNm A B C D E 2m 1m 1m 2mRA 2m 1m 1m RD 2m
Solution: Calculation of reactions:
ΣMA = 0
RD × × 2 × × ½ × 2 × 30 × (4+2/3) = 0  RD =80kN
ΣFy = 0
RA + 80 – 20 × ½ × 2 × 30 =  RA = 30 kN
N.Ram Kumar, CUFE

84. 20kNm
40kN
30kN/m
20kN/m 1 2 3 4 6 5 7 7 1 2 5 3 4 6
RD =80kN
RA =30kN 2m 1m 1m 2m
Calculation of Shear Forces: V0-0 = 0
V1-1 = 30 kN V5-5 = - 50 kN
V2-2 = 30 – 20 × 2 = - 10kN V6-6 = = + 30kN
V3-3 = - 10kN V7-7 = +30 – ½ × 2 × 30 = 0(check)
V4-4 = -10 – 40 = - 50 kN
N.Ram Kumar, CUFE

85. 20kNm 40kN 30kN/m 20kN/m 2m 1m SFD 1 2 7 5 4 6 3 RD =80kN RA =30kN
Parabola
SFD
x = 1.5 m
N.Ram Kumar, CUFE

86. 40kN 30kN/m 20kN/m 20kNm A B C D E 2m 1m 1m 2m X x = 1.5 m X RA RD
Calculation of bending moments:
MA = ME = 0
MX = 30 × 1.5 – 20 × 1.5 × 1.5/2 = 22.5 kNm
MB= 30 × 2 – 20 × 2 × 1 = 20 kNm
MC = 30 × 3 – 20 × 2 × 2 = 10 kNm (section before the couple)
MC = = 30 kNm (section after the couple)
MD = - ½ × 30 × 2 × (1/3 × 2) = - 20 kNm( Considering RHS of the section)
N.Ram Kumar, CUFE

87. 40kN 30kN/m 20kN/m 20kNm A B C D E 2m 1m 1m 2m BMD X x = 1.5 m X RA RD
Parabola
20kNm
10kNm
Point of contra flexure
BMD
Cubic parabola
20kNm
N.Ram Kumar, CUFE

88. SFD BMD 30kN 10kN 50kN Parabola x = 1.5 m Parabola 20kNm 10kNm
Point of contra flexure
BMD
Cubic parabola
20kNm
N.Ram Kumar, CUFE

89. 6. Draw SFD and BMD for the cantilever beam subjected
to loading as shown below.
40kN
0.5m
300
20kN/m
0.7m A 3m 1m 1m
N.Ram Kumar, CUFE

90. 40kN 20kN/m A 3m 1m 1m 20kN/m A 3m 1m 1m 0.5m 300 0.7m 40Sin30 = 20kN
40Cos30 =34.64kN
0.7m A 3m 1m 1m
N.Ram Kumar, CUFE

91. 20kN/m 3m 1m 20kN/m 3m 1m 40Sin30 = 20kN 0.5m 40Cos30 =34.64kN 0.7m
20x0.5 – 34.64x0.7=-14.25kNm
20kN/m 3m 1m
20kN
34.64kN
N.Ram Kumar, CUFE

92. Calculation of Reactions (Here it is optional):
20kN
20kN/m
14.25kNm HD
34.64kN A B C 3m 1m D 1m MD VD
Calculation of Reactions (Here it is optional):
ΣFx = 0  HD = kN
ΣFy = 0  VD = 20 × = 80 kN
ΣMD = 0  MD × 3 × 3.5 – 20 × 1 – = kNm
N.Ram Kumar, CUFE

93. 20kN/m HD 3m 1m 1m MD Shear Force Calculation: V1-1 =06 2 3 4 5 HD
34.64kN 3m 2 1m 3 4 1m 1 5 MD 6
VD=80kN
Shear Force Calculation:
V1-1 =0
V2-2 = -20 × 3 = - 60kN
V3-3 = - 60 kN
V4-4 = - 60 – 20 = - 80 kN
V5-5 = - 80 kN
V6-6 = = 0 (Check)
N.Ram Kumar, CUFE

94. 20kN/m HD 3m 1m 1m SFD 20kN 1 14.25kNm 6 2 3 4 5 34.64kN 2 3 4 1 5 MD
VD=80kN
60kN
60kN
SFD
80kN
80kN
N.Ram Kumar, CUFE

95. 20kN/m 3m 1m A B C D Bending Moment Calculations: MA = 0MD
Bending Moment Calculations:
MA = 0
MB = - 20 × 3 × 1.5 = - 90 kNm
MC = - 20 × 3 × 2.5 = kNm (section before the couple)
MC = × 3 × 2.5 – = kNm (section after the couple)
MD = - 20 × 3 × – 20 × 1 = kNm (section before MD) moment)
MD = = 0 (section after MD)
N.Ram Kumar, CUFE

96. 20kN/m A B C D 3m 1m 1m BMD 20kN 14.25kNm 34.64kN 90kNm 150kNm
N.Ram Kumar, CUFE

97. L/2 W
wkN/m L W
wkN/m
N.Ram Kumar, CUFE

98. Exercise Problems 20kN/m 5kNm 15kN/m 10kN 3m 1m 2m
Draw SFD and BMD for a single side overhanging beam
subjected to loading as shown below. Mark absolute
maximum bending moment on bending moment diagram and
locate point of contra flexure.
20kN/m
5kNm
15kN/m
10kN 3m 1m 2m
[Ans: Absolute maximum BM = kNm ]
N.Ram Kumar, CUFE

99. Exercise Problems
2. Draw shear force and bending moment diagrams [SFD and BMD] for a simply supported beam subjected to loading as shown in the Fig. given below. Also locate and determine absolute maximum bending moment.
10kN
16kN
4kN/m
600 B A 1m 2m 1m 1m 1m
[Ans: Absolute maximum bending moment = kNm
Its position is 3.15m from Left hand support ]
N.Ram Kumar, CUFE

100. Exercise Problems
3. Draw shear force and bending moment diagrams [SFD and BMD] for a single side overhanging beam subjected to loading as shown in the Fig. given below. Locate points of contra flexure if any.
25kN/m
50kN
10kN/m
10kNm A B 1m 3m 1m 2m
[Ans : Position of point of contra flexure from RHS = 0.375m]
N.Ram Kumar, CUFE

101. Exercise Problems
4. Draw SFD and BMD for a double side overhanging beam subjected to loading as shown in the Fig. given below. Locate the point in the AB portion where the bending moment is zero.
16kN
8kN
8kN
4kN/m A B 2m 2m 2m 2m
[Ans : Bending moment is zero at mid span]
N.Ram Kumar, CUFE

102. Exercise Problems
5. A single side overhanging beam is subjected to uniformly distributed load of 4 kN/m over AB portion of the beam in addition to its self weight 2 kN/m acting as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate the inflection points if any. Also locate and determine maximum negative and positive bending moments.
4kN/m
2kN/m A B 6m 2m
[Ans :Max. positive bending moment is located at 2.89 m from LHS.
and whose value is kNm ]
N.Ram Kumar, CUFE

103. Exercise Problems
6. Three point loads and one uniformly distributed load are acting on a cantilever beam as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate and determine maximum shear force and bending moments.
10kN
5kN
20kN
2kN/m A 1m 1m 1m B
[Ans : Both Shear force and Bending moments are maximum
at supports.]
N.Ram Kumar, CUFE

104. Exercise Problems
7. One side overhanging beam is subjected loading as shown below. Draw shear force and bending moment diagrams [SFD and BMD] for beam. Also determine maximum hogging bending moment.
200N
100N
30N/m A B 4m 4m 3m
[Ans: Max. Hogging bending moment = 735 kNm]
N.Ram Kumar, CUFE

105. Exercise Problems
8. A cantilever beam of span 6m is subjected to three point loads at 1/3rd points as shown in the Fig. given below. Draw SFD and BMD for the beam. Locate and determine maximum shear force and hogging bending moment.
10kN
5kN
8kN
5kN
0.5m
300 A 2m 2m 2m B
[Ans : Max. Shear force = 20.5kN, Max BM= 71kNm
Both max. shear force and bending moments will occur
at supports.]
N.Ram Kumar, CUFE

106. Exercise Problems
9. A trapezoidal load is acting in the middle portion AB of the double side overhanging beam as shown in the Fig. given below. A couple of magnitude 10 kNm and a concentrated load of 14 kN acting on the tips of overhanging sides of the beam as shown. Draw SFD and BMD. Mark salient features like maximum positive, negative bending moments and shear forces, inflection points if any.
14kN
40kN/m B 2m
10kNm 1m A 4m
20kN/m
600
[Ans : Maximum positive bending moment = kNm
N.Ram Kumar, CUFE

107. Exercise Problems
10. Draw SFD and BMD for the single side overhanging beam subjected loading as shown below.. Mark salient features like maximum positive, negative bending moments and shear forces, inflection points if any.
24kN
4kN/m
0.5m
6kN/m 1m 1m 3m 2m 3m
Ans: Maximum positive bending moment = 41.0 kNm
N.Ram Kumar, CUFE

108. ACKNOWLEDGEMENT
My Acknowledgement to innumerable websites in the internet, and to all those who have uploaded their knowledge, imaginations, ideas, graphic skills etc., on these websites.
Also, to all those(including my parents, teachers, friends, and relatives), from pre-historic days to to-day, who have registered their knowledge, imaginations, thoughts etc., through different means and mediums.
N.Ram Kumar, CUFE

109. THANK YOU
N.Ram Kumar, CUFE