Shear Force and Bending Moment Concepts and Diagrams

Shear force and bending moment diagrams and calculations are clearly explained with solved examples.

1. STRENGTH OF MATERIALS
Dr. S VENKATESWARA RAO
Assistant Professor
Civil Engineering Department
10-09-2017 1
SHEAR FORCE AND
BENDING MOMENT

2. Introduction
•Structural Members are usually classified according to
the types of loads that they support (axially loaded bar,
etc.)

•Now we will begin to look at beams, which are structural
members subjected to lateral loads.

•The first beams we will investigate are called planar
structures, because they lie in a single plane.

3. •If all loads act in that same plane, and if all
deflections act in that plane, then the plane is called
the Plane of Bending.

•First we will look at shear forces and bending
moments in beams.

•Once we know these, we can find the stresses, strains
and deflections of a beam

4. Beams
•Members that are slender and support loads applied
perpendicular to their longitudinal axis.
Span, L
Distributed Load, w(x)
Concentrated Load, P
Longitudinal
Axis

5. Types of Beams
Beams are usually described by the
way they are supported.

Simply supported beam – pin
support at one end and a roller
support at the other

Cantilever beam – fixed at one end
and free at the other.

Beam with an Overhang – simply
supported beam that projects
beyond the support (similar to a
cantilever).

7. Types of Beams
Depends on the support configuration
M
F
v
F
H
Fixed
F
V
F
V
F
H
Pin
Roller
Pin Roller
F
V F
V
F
H

8. Internal Reactions in Beams
•At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
–normal force,
–shear force,
–bending moment.
L
P
a b

9. Internal Reactions in Beams
•At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
–normal force,
–shear force,
–bending moment.
Pb/L
x
Left Side of Cut
V
M
N

10. Internal Reactions in Beams
•At any cut in a beam, there are 3 possible
internal reactions required for equilibrium:
–normal force,
–shear force,
–bending moment.
Pa/L
L - x
Right Side of Cut V M
N
Positive Directions
Shown!!!

11. Finding Internal Reactions
Pick left side of the cut:
–Find the sum of all the vertical forces to the left of the cut,
including V. Solve for shear, V.

–Find the sum of all the horizontal forces to the left of the
cut, including N. Solve for axial force, N. It’s usually, but
not always, 0.

–Sum the moments of all the forces to the left of the cut
about the point of the cut. Include M. Solve for bending
moment, M

Pick the right side of the cut:
–Same as above, except to the right of the cut.

12. Review: Types of Loads
•Concentrated Loads
•Couple
•Distributed Loads
–Uniformly distributed
–Linearly varying

13. Shear Force and Bending Moment
Shear Force: is the algebraic sum of the vertical forces
acting to the left or right of a cut section along the span of
the beam

Bending Moment: is the algebraic sum of the moment of
the forces to the left or to the right of the section taken
about the section

14. Shear and Bending Moment in Beams
If we have a beam that is loaded by a system of forces all
in the y direction.




The beam is classified as a simple beam (pin support at one
end and a roller support at the other)

The beam is in equilibrium with the application of these
forces and its reactions.

15. Shear and Bending Moments
To determine the internal effects of the
applied loads we imagine a cutting
plane to isolate either the left side or
right side of the beam.

In order for the isolated section of the
beam to be in equilibrium a force (V)
and couple (M) are required at the cut.
V is termed Shear force
M is termed Bending Moment

16. Shear and Bending Moments
•The shear V and Moment M are the force and couple
applied to the left part of the beam by the right side of the
beam to maintain equilibrium.

•Equilibrium of both sides is required because the entire
beam is in equilibrium.

•The section can be made anywhere along the length of
the beam.

•Shear and Moment are a function of the distance from the
origin.

17. Shear & Moment Sign Convention
The signs associated with the shear force and
bending moment are defined in a different manner
than the signs associated with forces and moments
in static equilibrium.

The Shear Force is positive if it tends to rotate the
beam section clockwise with respect to a point
inside the beam section.

The Bending Moment is positive if it tends to bend
the beam section concave facing upward. (Or if it
tends to put the top of the beam into compression
and the bottom of the beam into tension.)

+ shear
+ moment

18. Procedure:
1. find reactions;
2. cut the beam at a certain cross section, draw F.B.D. of one
piece of the beam;
3. set up equations;
4. solve for shear force and bending moment at that cross
section;
5. draw shear and bending moment diagrams.

19. Cantilever:
Case I: A cantilever beam of length l carrying a concentrated
load W at the free end. W
A
B
l
x
x
Consider a section x - x at a
distance x from free end.
Shear force at x = +W
B M at x = - Wx
S F is constant at all section of
the member between A and B
At x = 0, B M = 0
At x = l, B M = - Wl
A B
S F D
+ W
A B
B M D
-Wl

20. Case II: Cantilever of length l carrying a u d l of w /m over the
entire length.
A
B
l
w/m
V
A = wl

x
x
M
A = wl
2
/2

Consider a section x-x at a
distance x from B.
S. F at x = + wx
B.M at x = -wx.(x/2)= - (wx
2
/2)
At x = 0, S = 0 and M = 0
At x= l, S = wl and M= - (wl
2
/2)
Variation of SF is linear and
variation of BM is parabolic
law.
B
A
wl
+ ve
SFD
Linear variation
- ve
B A
BMD
- (wl
2
/2)
Parabolic variation

21. Case III: Cantilever of length l carrying a u d l of w /m over the
entire length and point load W at free end.
V
A = (wl + w)

x
x
M
A = (wl
2
/2)+wl)

Consider a section x-x at a
distance x from B.
S. F at x = W + wx
B.M at x = -{(wx
2
/2) + Wx}
At x = 0, S = W and M = 0
At x= l, S = W + wl and
M= - {(wl
2
/2) + Wl}
Variation of SF is linear and
variation of BM is parabolic law.
- ve
B A
BMD
- {(wl
2
/2) + Wl}
Parabolic variation
A
B
l
w/m
W
B
A
W+wl
+ ve
SFD
Linear variation

22. Case IV: Cantilever of length l carrying a u d l of w /m for a
distance ‘a’ from free end.
A
B
l
w/m
a
x
x
C
Consider a section x-x at a
distance x from B.
S. F at x = wx
B.M at x = -(wx
2
/2)
At x = 0, S = 0 and M = 0
At x= a, S = wa and M= - (wa
2
/2)
Consider a section y-y between C
and A .
S. F at y = wa
B.M at y = -wa(x - a/2)
V
A = wa

y
y
B A
wa
SFD
+ve
Linear variation
Straight line
At x = a, M = - (wa
2
/2)
At x = l, M= wa(l - a/2)
BMD
wa
2
/2
wa(l-a/2)
Straight line
Parabolic variation
B A
- ve

23. Case V: Cantilever of length l carrying a u d l whose intensity
varies uniformly from zero at the free end to w/m at the
fixed end.
A
B
l
w/m
V
A = (wl)/2

B
A
Parabolic variation
+ ve
SFD
B
A
- ve
BMD
Cubic curve
x
dx x

24. Case VI: Cantilever of length l carrying a u d l whose intensity
varies uniformly from zero at the fixed end to w/m at the
free end.
A
B
l
w/m
x
x
dx
x
V
A = (wl)/2

B A
SFD
+ve
Parabolic curve
B
A
- ve
BMD
Cubic curve
At x = l, S = 0 and M = 0

25. Q.1. Draw the shear force and bending moment diagrams for the
cantilever as shown in Fig.
A
B
1 kN/m
C D E
3 kN 2.5 kN
1 m 1.5 m 2 m 0.5 m
V
A = 7.5 kN

M
A = 22.5 kN-m

Let V
a is the vertical reaction at A, V
a = 3 + (1*2) + 2.5 = 7.5 kN
Reacting moment M
a =(3x1)+(2 x 1 x 3.5) +( 2.5 x 5) = 22.5 kNm
SF between A and B = 7.5 kN
SF between B and C = 7.5 – 3 = 4.5 kN
SF at D = 2.5 kN
From C to D, SF will change uniformly from 4.5 kN to 2.5 kN
a
a
b
b
c
c
d
d

26. BM at E = 0
BM at D = - 2.5 x 0.5 = - 1.25 kNm
BM at C = - (2.5 x 2.5) – (1 x 2 x 1) = - 8.25 kNm
BM at B = - (2.5 x 4) – (1 x 2 x 2.5) = - 15 kNm
BM at A = - 22.5 kNm
A
B
1 kN/m
C D E
3 kN 2.5 kN
1 m 1.5 m 2 m 0.5 m
A E
7.5 kN 7.5 kN
4.5 kN
4.5 kN
2.5 kN
2.5 kN
+ ve
SFD

27. A
B
1 kN/m
C D E
3 kN 2.5 kN
1 m 1.5 m 2 m 0.5 m
A E
- ve
22.5 kNm
15 kNm
8.25 kNm
1.25 kNm
BMD
Parabolic curve

28. Simply Supported beams
1. Simply supported beam of span l carrying a concentrated load
W at mid span.
A B
C
W
l/2 l/2
Reaction at each support
V
A = V
B = W/2

Any section between A & C
SF = W/2

Any section between C & D
SF = - W/2
At any section between A &
C at a distance x from A,
M
x = (W/2) x (sagging)
At x = 0, M = 0 and x = l/2,
M = Wl/4
V
A = W/2
V
B = W/2

A B
+ ve
- ve
SFD
BMD
+ ve
A B
Wl/4

29. 2. Simply supported beam of span l carrying a concentrated load
W placed eccentrically on the span.
A B
C
W
a b
l
V
A = Wb/l

V
B = Wa/l

A
B
+ ve
- ve
SFD
BMD
+ ve
A B
Wab/l

30. 3. Simply supported beam carrying no. of concentrated loads
A B
C
4 kN
1.5m
D E
10 kN 7 kN
2.5m 2m 2m
Let V
A and V
B are the vertical reactions at the supports A and B.

Taking moments about A
V
B x 8 = (4 x 1.5) + (10 x 4) + (7 x 6)
V
B = 11 kN and V
A = 21 – 11 = 10 kN

SF between A & C = 10 kN
SF between C &D = 10 – 4 = 6 kN
SF between D & E = 10 – 4 – 10 = - 4 kN
SF between E & B = 10 – 4 – 10 – 7 = - 11 kN
V
A = 10 kN
V
B = 11 kN

31. BM at A = 0, BM at C = 10 x 1.5 = 15 kN m (sagging)
BM at D = (10 x 4) – (4 x 2.5) = 30 kN m (sagging)
BM at E = (11 x 2) = 22 kN m (sagging)
A B
C
4 kN
1.5m
D E
10 kN 7 kN
2.5m 2m 2m
A
B
10 kN
11 kN
6 kN
4 kN
+ ve
- ve
SFD
15 kN m
30 kN m
22 kN m
BMD
A B

32. 4. Simply supported beam of span l carrying u d l of w per unit
run over the entire span.
A B
l
w/m x
x
B
A
SFD
B A
BMD

33. 5. Simply supported beam of span l carrying u d l of w per unit
run for a certain distance from one end.
A B
4 m 5 m
C
18 kN/m
Let V
A and V
B are the vertical reactions at A and B
Taking moments about A, V
B x 9 = 18 x 4 x 2
V
B = 16 kN and V
A = (18 x 4) – 16 = 56 kN
SF between C & D = - 16 kN
Consider a section between A & C at a distance x from A
S
x = 56 – 18x
At x = 0, SF = 56 kN At x = 4 m, SF = 56 – (18 x 4) = - 16 kN

Let SF is zero at x m from A. Equating SF to zero
56 – 18x = 0 then x = 3.11 m from A
V
A = 56 kN
V
B = 16 kN

x
x

34. M
x = 56x – 18.x
2
/2 = 56x – 9x
2
At x = 0, M = 0 At x = 4 m, M = 80 kN m
At x = 3.11 m, M = 56 x 3.11 – ( 9 x 3.11
2
) = 87.1 kN m
Maximum BM occurs at D where SF is zero or SF changes sign
B A
16 kN
56 kN
+ ve
- ve
3.11 m
SFD 87.1 kN m
80 kN m
BMD
A B
4 m 5 m
C
18 kN/m
D

35. 6. Simply supported beam of span l carrying u d l of w per unit
run on an intermediate part of the span.
A
B
4 m 3 m
C
1.8 kN/m
D
2 m
Let V
A and V
B are the vertical reactions at A and B
Taking moments about A, V
B x 9 = 1.8 x 4 x 4
V
B = 3.2 kN and V
A = (1.8 x 4) – 3.2 = 4.0 kN
At a section between A & C, SF = 4 kN
At a section between B & D, SF = - 3.2 kN
Consider a section between C & D at a distance x from A
S
x = 4 – 1.8(x -2)
At x = 2m, SF = 4 kN and at x = 6 m,SF = 4 – 1.8(6 -2) = -3.2 kN
Let the shear forece is zero at x from A, then 4 – 1.8(x -2) = 0
So x = 4.22 m
V
A = 4 kN

V
B = 3.2 kN

a
b
b
b
c
c

36. A
B
4 m 3 m
C
1.8 kN/m
D
2 m
B A
9.6 kN m 8 kN m
12.44 kN m
B A
E
3.2 kN
4 kN
- ve
+ ve
4.22 m

37. Q. A Simply supported beam AB of span 10 m. It carries a point
load of 5 kN at 3 m from A and a point load of 5 kN at 7 m
from A and u d l of 1 kN/m between the point loads. Draw
SFD and BMD for the beam.
A
B
4 m 3 m
C
1kN/m
D
3 m
5kN 5kN
V
A = V
B = (5 + 5 + 4)/2 = 7 kN

SF between A & C = 7 kN
SF just on RHS of C = 7 – 5 = 2 kN
SF just on RHS of D = -7 + 5 = - 2 kN

Consider a section a-a at a distance x from C ( between C & D)
S
x = 7 – 5 – 1x then at x = 2, SF = 0 kN
V
A = 7 kN
V
B = 7 kN

a
a

38. BM at A = BM at B = 0
BM at C = BM at D = 7 x 3 = 21 kN m
BM at E = (7 x 5) – (5 x 2) – (1 x 2
2
/2) = 23 kNm
A
B
4 m 3 m
C
1kN/m
D
3 m
5kN 5kN
B A
21 kN m 21 kN m
23 kN m
B A
E
7 kN
7 kN
- ve
+ ve
5 m
2 kN
2 kN

39. 7. Simply supported beam of span l carrying a load whose
intensity varies uniformly from zero at each end to w/m at the
mid span.
A B
C
l
w/m
V
B = wl/4

B A
V
A = wl/4

a
a

40. A B
C
l
w/m
B A
BMD
wl
2
12

A

41. 8. Simply supported beam of span l carrying a load whose
intensity varies uniformly from zero at one end to w/m at the
other end.
A B
l
w/m x
x
B
A C
+ve
-ve

42. A B
l
w/m
A
B
+ ve
BMD
x
x

43. 9. The intensity of loading on a simply supported beam of 5 m
span increases uniformly from 8 kN/m at one end to 16 kN/m
at the other end. Find the position and magnitude of the
maximum bending moment. Also draw shear force and
mending moment diagram.
A B
5 m
16kN/m 8kN/m
x
x
Let V
A and V
B are the
vertical reactions at A and B
Taking moments about A,

V
B x 5 = (8 x 5 x 2.5) + (0.5 x 8 x 5 x (2/3*5))
V
B = 33.33 kN and V
A = 26.67 kN
Consider a section x-x at a distance x from A
Loading intensity at x-x = 8 + (8x/5) = (8 + 1.6x)
SF at x-x Sx = 26.67 – 8x – (½. 1.6x) = 26.67 – 8x – 0.8x
2
At x = 0, SF = 26.67kN
At x =5m, SF = - 33.33 kN
V
A = 26.67
V
B = 33.33

44. A B
5 m
16kN/m 8kN/m
B
A C
+ve
-ve SFD
A
B
+ ve
BMD

45. 10. A beam of length (l + 2a) has supported l apart with an
overhang on each side. The beam carries a concentrated load
W at each end . Draw SFD and BMD.
A B C D
W W
a a l
V
A = V
B = W
SF between C & A = - W
SF between B & D = + W
SF between A & B = 0
BM at C = 0
BM at A = - Wa
Consider a section x-x at a
distance x from C.
M
x = -Wx + W(x-a) = -Wa
BM at B = -Wa
BM at D = 0
V
A = W
V
B = W

- ve
+ ve
SFD
A B C D
-W
+W
- ve
BMD
A B
C
D
-Wa

46. 11. A simply supported beam with overhang at one end carrying
a u d l over the entire length.
A
B
C
2 m 6 m
15 kN/m
Let V
A and V
B are the
vertical reactions.
Taking moments about A
V
B x 6 = 15 x 8 x 4
V
B = 80 kN and V
A = 40 kN
SF at the left end = 40 kN
SF at just on LHS of B
= (40 – 15x6) = - 50 kN
SF at just on RHS of B
= 15x2 = 30 kN
SF at C = 0
SF is zero at x m from A
S
x = 40 – 15x = 0
Then x = 2.67 m from A
V
A = 40 kN
V
B = 80 kN

A
B
C
D
+ ve + ve
- ve
40 kN
50 kN
30 kN
2.67 m
S F D
x
x

47. BM at a section x from A, M
x = 40x – 15x
2
/2 = 40x – 7.5x
2

BM at A & C = 0
At x = 6 m, BM = (40 x 6) – (7.5 x 6
2
) = - 30 kN m
At x = 2.67 m, BM = (40 x 2.67) – (7.5 x 2.67
2
) = 53.3 kN m
There must be a section where BM is zero. M
x = 0
M
x = 40x – 7.5x
2
= 0 then x = 0 and 5.33 m
A
B
C
2 m 6 m
15 kN/m
A
B
C
D E
53.3 kN m
30 kN m
+ ve
- ve
5.33 m
B M D

48. (H.W)Draw SFD and BMD for the beam shown in Fig.
A B
C
1m 3 m
10 kN/m
D
10 m
V
A = 84 kN
V
B = 56 kN

30 kN
A
B
C D
+ ve + ve
- ve
54 kN
46 kN
10 kN
5.4 m
S F D
- ve
E
4.6 m
45 kN m
A B
C D
E
100.8 kN m
5 kN m
+ ve
- ve
0.91 m B M D
- ve
0.11 m
F

49. Q. Draw the SFD and BMD for the overhanging beam carrying
loads as shown in Fig. Mark the values of the principal
ordinates and locate the point of contra flexure.
A
B
C
2 m 5 m
20 kN/m
10 kN
Let V
A and V
B are the
vertical reactions.
Taking moments about A
V
Bx 5 =(20 x 7 x 3.5)+10x7
V
B =112 kN and V
A = 38 kN
SF at just RHS of A = 38kN
SF at just LHS of B
= 38 – (20 x 5) = - 62 kN
SF at just RHS of B
= 10 + ( 20 x 2) = 50 kN
SF at just on LHS of C
= 10 kN
V
A = 38 kN
V
B = 112 kN

A
B
C
D
+ ve + ve
- ve
38 kN
62 kN
50 kN
1.9 m
S F D
10 kN

50. Consider a section x-x at a distance from A
S
x = 38 – 20x and M
x = 38x – 20x
2
/2
Shear force is zero at S
x = 38 – 20x = 0 then x = 1.9 m
At x = 0, BM = 0 and At x = 5, BM at B = - 60 kN m
At x = 1.9 m, BM = 36.1 kN m
A
B
C
2 m 5 m
20 kN/m
10 kN
A
B
C
D E
36 kN m
60 kN m
+ ve
- ve
3.8 m
B M D
1.9 m
x
x
Point of contra flexure:
BM is zero at
M
x = 38x – 20x
2
/2 = 0

then x = 3.8 m from A

51. HW. Calculate the reactions at the supports A & B of the beam
as shown in Fig. Draw SFD and BMD. Also determine the
point of contra flexure.
A B C D
1k N 1.5 kN
2m 3 m 7 m
2.4 kN
4m
HW. A beam AB, 20 m long supported on two intermediate
props 12 m apart carries a u d l of 6 kN/m together with
concentrated loads of 30 kN at the left end A and 50 kN at
the right end B. The props are so located that the reactions
is same of each support. Determine the position of the
props and draw SFD and BMD. Mark the values of
Maximum SF and BM.

52. Q. Find the reactions at the fixed end of the cantilever loaded as
shown in Fig. Also draw the SFD and BMD.
M
A = 29 kNm

Vertical reaction V
A = 3 + 2 = 5 kN
Taking moments about A, (2 x 8) + 3 + (3 x 4) – 2 = 29 kNm
At fixed end, the support will provide a balancing moment of
29 kN m

SF between A & C = 5 kN
SF between C & E = 5 – 3 = 2 kN
BM at E = 0
BM at just on RHS of D = - 2 x 2 = - 4 kN m

V
A = 5 kN

A
E
2 kN 3 kN
2 kN m
3 kN m

B C D
2 m 2 m 2 m 2 m

53. BM at just on LHS of D = - 4 – 3 = - 7 kN m
BM at C = -(2 x 4) – 3 = - 11 kN m
BM at just on the RHS of B = - (2 x 6) – 3 – (3 x 2) = - 21 kNm
BM at just on RHS of B = - 21 + 2 = - 19 kN m
BM at A = - (2 x 8) – 3 – (3 x 4) + 2 = - 29 kN m
A
E
2 kN 3 kN
2 kN m
3 kN m

B C D
2 m 2 m 2 m 2 m
- ve
B M D
2 kN
5 kN + ve
+ ve
A
E
S F D

54. HW. Find the reactions at the fixed end of the cantilever loaded
as shown in Fig. Also draw the SFD and BMD.
A
E
20 kN 10 kN
120 kN m

B C D
2 m 2 m 2 m 2 m
20 kN
20 kN 10 kN + ve
+ ve
A
E
S F D
20 kN
40 kN
80 kN
40 kN
B M D

55. Q. Draw SFD and BMD for the beam as shown in fig.
B A
2.5 kN
52.5 kN
+ ve
+ ve
SFD
150 kN m

A B
5 m 5 m
C
10 kN/m
2.5 m 2.5 m
D
137.5 kNm
143.75 kNm
6.25 kNm B M D
+ ve
- ve

56. Q. Draw SFD and BMD for the beam as shown in fig.
A
B
2 m
C
3 kN/m
2 m 2 m
D E
10 kN
1 m
1 m
1 m
1 m
S F D
B M D

57. THANK YOU
10-09-2017 57