CIVL 3121
Shear Force and Bending Moment Diagrams
1/8
Shear and Moment Diagrams
Shear and Moment Diagrams
 If the variation of V and M are written as functions
of position, x, and plotted, the resulting graphs
are called the shear diagram and the moment
diagram.
 We will develop a simpler method for constructing
shear and moment diagrams.
 We will derive the relationship between loading,
shear force, and bending moment.
 Developing the shear and moment functions for
complex beams can be quite tedious.
Shear and Moment Diagrams
Shear and Moment Diagrams
 Consider the beam shown below subjected to an
arbitrary loading.
 We will assume that distributed loadings will be
positive (+) if they act upward.
 Let’s draw a free body diagram of the small
segment of length x and apply the equations of
equilibrium.
x
w = w(x)
w = w(x)
x
x
x
x
Shear and Moment Diagrams
Shear and Moment Diagrams
F
 Since the segment is chosen at a point x where
there is no concentrated forces or moments, the
result of this analysis will not apply to points of
concentrated loading
y
w(x)
V  w ( x )x
x
2
M
x
x
 0  V  w ( x )x  (V  V )
w(x)x
M+M
V
w = w(x)
x
x
V+V
x
M
0
 0  M  (M  M )  V x
 x 
w ( x )x 

 2 
M Vx  w(x)
(x)2
2
CIVL 3121
Shear Force and Bending Moment Diagrams
2/8
Shear and Moment Diagrams
Shear and Moment Diagrams
 Dividing both sides of the V and M expressions
by x and taking the limit as x tends to 0 gives:
 The slope of the shear diagram at a point is equal
to the intensity of the distributed loading w(x) at
that point
dV
 w(x)
dx
Slope of
shear curve
=
Intensity of the
loading
Slope of
moment curve
=
Intensity of the
shear
Shear and Moment Diagrams
 The slope of the moment diagram at a point is
equal to the intensity of the shear at that point.
dV
 w(x)
dx
Slope of
shear curve
=
Intensity of the
loading
dM
V
dx
Slope of
moment curve
=
Intensity of the
shear
Shear and Moment Diagrams
 The change in shear between any two points is
equal to the area under the loading curve
between the points.
V   w ( x )dx
Change in
shear
=
Area under the
loading
dV
 w(x)
dx
dM
V
dx
M   V ( x )dx
Change in
moment
=
Area under the
shear diagram
Slope of
shear curve
=
Intensity of the
loading
dM
V
dx
Slope of
moment curve
=
Intensity of the
shear
Shear and Moment Diagrams
 If we multiply both sides of each of the above
expressions by dx and integrate:
V   w ( x )dx
Change in
shear
=
Area under the
loading
M   V ( x )dx
Change in
moment
=
Area under the
shear diagram
Shear and Moment Diagrams
 The change in moment between any two points is
equal to the area under the shear diagram
between the points.
V   w ( x )dx
Change in
shear
=
Area under the
loading
M   V ( x )dx
Change in
moment
=
Area under the
shear diagram
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams
Shear and Moment Diagrams
 Let’s consider the case where a concentrated
force and/or a couple are applied to the segment.
 Let’s consider the case where a concentrated
force and/or a couple are applied to the segment.
F
P
y
M
3/8
M + M
V
V  P
M’
x
M
P
 0  V  P  (V  V )
M
Shear and Moment Diagrams
 Therefore, when a force P acts downward on a
beam, V is negative so the “jump” in the shear
diagrams is downward. Likewise, if P acts
upward, the “jump” is upward.
 When a couple M ’ acts clockwise, the resulting
moment M is positive, so the “jump” in the
moment diagrams is up, and when the couple
acts counterclockwise, the “jump” is downward.
 0  M  (M  M )  V x  M '
M + M
V
M  M '
M’
V + V
0
x
V + V
Shear and Moment Diagrams
 Procedure for analysis - the following is a
procedure for constructing the shear and moment
diagrams for a beam .
1. Determine the support reactions for the
structure.
Shear and Moment Diagrams
Shear and Moment Diagrams
 Procedure for analysis - the following is a
procedure for constructing the shear and moment
diagrams for a beam .
 Procedure for analysis - the following is a
procedure for constructing the shear and moment
diagrams for a beam .
2. To construct the shear diagram, first, establish
the V and x axes and plot the value of the
shear at each end of the beam.
Since the dV/dx = w, the slope of the shear
diagram at any point is equal to the intensity of
the applied distributed loading.
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams
4/8
Shear and Moment Diagrams
 Procedure for analysis - the following is a
procedure for constructing the shear and moment
diagrams for a beam .
 Draw the shear and moment diagrams for the
following beam
P
P
The change in the shear force is equal to the
area under the distributed loading.
B
A
L
If the distributed loading is a curve of degree n,
the shear will be a curve of degree n+1.
Shear and Moment Diagrams
P
A
 Establish the V and x axes and plot the value of
the shear at each end.
P
P
In this case, the values are: at x = 0, V = P; and at x = 3L, V = -P.
B
Ax
L
L
L
L
L
A
By
F
x
By  P
y
y
V
L
Ay
 M  0  P (L  2L)  B (3L)
 F  0  A  B  2P
y
L
Shear and Moment Diagrams
 Find the support reactions
P
L
P
(k)
Ay  P
y
x
-P
L
L
L
 0  Ax
Shear and Moment Diagrams
Shear and Moment Diagrams
 The slope of the shear diagram over the interval 0
< x < L is the equal to the loading. In this case
w(x) = 0.
 At a point x = L, a concentrated load P is applied.
The shear diagram is discontinuous and “jumps”
downward (recall V = -P).
V
V
P
P
(k)
x
(k)
x
-P
L
L
L
-P
L
L
L
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams
 The slope of the shear diagram over the interval
L < x < 2L is zero since, w(x) = 0.
Shear and Moment Diagrams
 At 2L, P is applied and the shear diagram
“jumps” downward (recall V = -P).
V
V
P
P
(k)
x
5/8
(k)
x
-P
L
L
-P
L
L
L
L
Shear and Moment Diagrams
Shear and Moment Diagrams
 The slope of the shear diagram over the interval
2L < x < 3L is zero since, w(x) = 0.
 The slope of the shear diagram over the interval
2L < x < 3L is zero since, w(x) = 0.
V
V
P
P
(k)
x
(k)
x
-P
L
L
-P
L
L
L
L
The resulting shear diagram matches the shear at the
right end determined from the equilibrium equations.
The resulting shear diagram matches the shear at the
right end determined from the equilibrium equations.
Shear and Moment Diagrams
Shear and Moment Diagrams
 Establish the M and x axes and plot the value of
the moment at each end.
In this case, the values are: at x = 0, M = 0; and at x = 3L,
M = 0.
M
 The slope of the moment diagram over the interval 0 < x < L
is the equal to value of the shear; in this case V = P. This
indicates a positive slope of constant value.
M
PL
1
P
(k ft.)
x
L
L
L
(k ft.)
x
L
L
L
The change in moment is equal to the area under the
shear diagram, in this case, M = PL.
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams
 The slope of the moment diagram over the interval L < x <
2L is the equal to value of the shear; in this case V = 0.
M
6/8
Shear and Moment Diagrams
 The slope of the moment diagram over the interval
2L < x < 3L is the equal to value of the shear, V = -P.
M
PL
PL
1
1
P
1
P
(k ft.)
x
L
L
-P
(k ft.)
x
L
L
L
L
The change in moment is equal to the area under the
shear diagram, in this case, M = -PL.
Shear and Moment Diagrams
 The slope of the moment diagram over the interval 2L < x
< 3L is the equal to value of the shear, V = -P.
M
Shear and Moment Diagrams

The shape of the shear and moment diagrams for selected loadings
P
Loading
w(x)
Shear Diagram
dV
w
dx
Moment Diagram
dM
V
dx
PL
V
V = -P
(k ft.)
x
x
M
L
L
L
The change in moment is equal to the area under the
shear diagram, in this case, M = -PL.
x
Shear and Moment Diagrams

The shape of the shear and moment diagrams for selected loadings
Shear and Moment Diagrams

The shape of the shear and moment diagrams for selected loadings
w
M0
Loading
w(x)
V
Loading
w(x)
Shear Diagram
dV
w
dx
Moment Diagram
dM
V
dx
V
1
-w
x
Shear Diagram
dV
w
dx
M
x
M
M = -M0
Moment Diagram
x
dM
V
dx
Smaller
(+) slope
Large
(+) slope
x
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams

Shear and Moment Diagrams
The shape of the shear and moment diagrams for selected loadings

The shape of the shear and moment diagrams for selected loadings
w(x)
Loading
V
Larger
(-) slope
x
Larger
(-) slope
dV
w
dx
Shear Diagram
w(x)
Shear Diagram
dV
w
dx
Large
(+) slope
Moment Diagram
dM
V
dx
Small
(-) slope
x
Smaller
(+) slope
M
Smaller
(+) slope
M
Moment Diagram
x
dM
V
dx
Large
(+) slope
x
Shear and Moment Diagrams
Shear and Moment Diagrams

Draw the shear and moment diagrams for the following beam
Draw the shear and moment diagrams for the following beam
P
w0
M0
L
L
w0
L
V
V
L
V
V
w 0L
w 0L
P
(k)
x
(k)
(k ft.)
x
(k)
x
M
M0
x

-PL
Shear and Moment Diagrams
(k ft.)
x
w 0 L2
x
w 0L
2
2
3
Shear and Moment Diagrams
 Draw the shear and moment diagrams for the
following beam
Draw the shear and moment diagrams for the following beam
P
x
M
(k ft.)
(k ft.)
L
2
(k)
x
M
M

Loading
V
Small
(-) slope

7/8
w0
2
4 k/ft.
L
L
V
V
P
(k)
M
(k ft.)
B
A
w0L
2
2
L
2
x
(k)
x
L/2
P 2
PL
M
4
(k ft.)
x
w 0 L2
8

w 0L
2
x
18 ft.
CIVL 3121
Shear Force and Bending Moment Diagrams
Shear and Moment Diagrams

Draw the shear and moment diagrams for the following beam
8/8
Shear and Moment Diagrams

Draw the shear and moment diagrams for the following beam
600 lb.
60 k
4 k/ft.
4,000 lb. ft.
A
B
A
10 ft.
12 ft.
8 ft.
100 k ft.
End of Internal Loads – Part 3
Any questions?
5 ft.
5 ft.