Week 3: Shear and Moment
I.
Introduction
Definition of shear – it is the net unbalanced vertical force acting at the exploratory section of the
left or right section of the beam.
Definition of bending moment – it is the net unbalanced moment acting at the centroid of the
exploratory section of the left or right section of the beam.
+
+
Note: subscript L indicates left section and subscript R indicates right section
II.
Objectives
After completion of the course the student should be able to:
1. Define the shear and bending moment on structures.
2. Formulate the shear and moment equations used to compute the shear and bending
moment values of the structures.
3. Draw the shear and moment diagram of the beam and frames using area method.
4. Compute the shear and bending moment values of the structures by the use of the short
method.
III. Shear and Moment Equations
Shear and Moment Equations are developed by passing exploratory sections at distance x from the left
support at points of loading changes. Usually the left support is taken as the origin (0, 0).
Shear equations are formed by using the definition of shear expressed as a function of x.
Moment equations are formed by using the definition of moment expressed as a function of x.
Signs of bending moment - upward external forces tend to produce positive bending moment on the right
or left section of the beam, downward external forces tend to produce negative moment on either
section. Positive moment tends to bend the beam concave upward; negative moment tends to bend the
same beam concave downward.
IV. Shear and Moment Diagrams
Shear and Moment diagrams are simply visualization of the shear and moment equations plotted on the
V-x axes and M-x axes.
The discontinuities in the shear diagram are joined by vertical lines drawn up or down to represent abrupt
change in shear. The point of zero shears corresponds to the point of maximum moment.
Shear and moment at points of loading change can be computed by substituting appropriate values of x in
the shear and moment equation or direct computation by using the definition of shear and moment. The
point where the value of the moment (except supports) is zero is called point of inflection.
This is where the shape of the bending of the beam changes from concave up to concave down. The shape
of the shear and moment diagrams conforms to the kind of equation the shear and moment equations are
made of. Example if the moment equation is the equation of a straight line/parabola/3rd degree curve the
shape of the moment diagram is a straight line/parabola/3rd degree curve.
Example 3.1
Determine the shear and moment in the beam shown in the figure as a function of x.
   Fy  0
1 x
30  ( )( x )  V  0
2 15
V  30  0.0333x 2
M
s
0
1 x
 x
600  30 x   ( )( x )( )  M  0
 2 15
 3
M  600  30 x  0.0111x 3
Picture source: Structural Analysis Eigth Edition by R.C. Hibbeler
Example 3.2
Determine the shear and moment in the beam shown in the figure as a function of x.
0  x1  12 ft
   Fy  0
108  4 x1  V  0
V  108  4 x1
M
s
0
1588  108 x1  4 x1 (
x1
)M 0
2
M  1588  108 x1  2 x1
2
12 ft  x 2  20 ft
   Fy  0
108  48  V  0
V  60k
M
s
0
1588  108 x 2  48( x 2  6)  M  0
M  60 x 2  1300
Picture source: Structural Analysis Eigth Edition by R.C. Hibbeler
V.
Shear and Moment Diagram of Beams Using Area Method
Relation among Load, Shear and Moment Diagrams:
The relation among the load, shear and moment diagrams provide a method of constructing
shear and moment diagrams without writing shear and moment equations. These relations are
not independent of the basic definition of shear and moment; they supplement them and are
used in conjunction with them.
Figure shows a differential segment of the beam of length dx. The load on the left side is shear V
and moment M; on the right side V + dV and M+ dM( due to the change in shear and moment
between the left and right side of the segment). The uniform load W remains constant since the
length of the segment is a very small quantity.
V2 - V1 = ΔV= Area(load)
M2 - M1 = ΔM= Area(Shear)
Change in moment b/w any two points = Area of shear diagram b/w these two points
Intensity of loading = slope of shear diagram
Shear at any point = the slope of the moment diagram about the same point
General Procedure
Support Reactions:
1. Determine the reactions of the supports and resolve the forces acting on the beam into its
parallel and perpendicular components.
Shear Diagram:
1. Establish the V and x axes and plot the shear values at the ends of the beam. Since the slope of
the shear diagram at any point is equal to the strength of the load distributed at the point. (Note
that when w acts upward, therefore, it is positive.
2. You can find the value of the shear using method of sections that states that the change in shear
force is equal to the area under the load diagram.
3. Since w(x) is integrated to obtain V, if w(x) is a curve of degree n, then V(x) will be a curve of
degree For example, if w(x) is uniform, V(x) will be linear.
Moment Diagram:
1. Establish the M and x axes and plot the values of the moment at the ends of the beam. Since the
slope of the moment diagram at any point is equal to the strength of the shear at the point.
2. The point of maximum or minimum moment is where the zero shears is located.
3. You can find the value of the moment using the method of sections that states that the change in
moment is equal to the area under the shear diagram.
4. Since V(x) is integrated to obtain M, if V(x) is a curve of degree n, then M(x) will be a curve of
degree For example, if V(x) is linear, M(x) will be parabolic.
Example 3.3
Without writing shear and moment equations, draw the shear and moment diagrams of the
beams and frames specified as follows:
M
A
0
RE (5)  722   48(3)  16( 2)4  0
RE  83.2kN
F
y
0
RB  83.2  72  48  16( 2)  20( 4)  0
RB  148.8kN
VA  0
VB  20( 2)  40
VB  40  148.8  108.8
VC  108.8  20(2)  68.8
VC  68.8  72  3.2
VD  3.2  48  51.2
VD  51.2  16( 2)  83.2
M B  M A  A1
1
M B  0   (40)2
2
M B  40
M C  M B  A2
M C  ( 40) 
M C  137.6
(108.8  68.8)
( 2)
2
M D  M C  A3
M D  137.6  3.2(1)
M D  134.4
M E  M D  A4
(51.2  83.2)
M E  134.4  
( 2)
2
ME  0
Example 3.4
Determine the shear and moment diagram of the beam.
VA  0
1
VB  48(3)  72
2
72  24 x  0
x3
VD  24(3)  72
1
VE  72  48(3)
2
VE  0
M B  M A  A1
1
M B  0  (72)(3)
3
M B  72
M C  M B  A2
M C  (72) 
M C  180
(1)
(72)(3)
2
M D  M C  A3
1
M D  (180)   72(3)
2
M D  72
M E  M D  A4
1
M E  72  (72)(3)
3
ME  0
VI. Shear and Moment Diagram of Frames Using Area Method
Convention of Signs for Shear
Convention of Signs for Moment
Example 3.5
Draw the shear and moments diagrams of the frame shown.
M
A
0
6 RD  12(6)3  24( 2)  0
RD  44kN 
F
x
0
RAx  24  0
RAx  24kN 
F
y
0
RAy  RB  12(6)  0
RAy  44  12(6)  0
RAy  28kN 
M
B
0
24( 4)  24( 2)  M B  0
M B  48kN .m
Member BC
M B ' M B  A1
1
M B '48  ( 28)( 2.33)
2
M B '  80.62
M c  M B '  A2
M c  80.62 
Mc  0
 1( 44)
(3.67)
2
Example 3.6
Draw the shear and moment diagrams of the frame shown in figure.
M
A
0
RD (10)  15(15)(7.5)  12(9)( 4.5)  0
RD  217.35kN 
F
x
0
Ax  12(9)  0
Ax  108kN 
F
y
0
Ay  RD  15(15)  0
Ay  217.35  225  0
Ay  7.65kN 
15x=7.65
x =0.51
-9.49(15)=-142.35
217.35-142.35=75
15(5)=75
M
B
0
108(9)  12(9)( 4.5)  M B  0
M B  486kN .m
M B ' M B  A1
1
M B '486  (0.5)(7.65)(0.51)
2
M B '  487.95
M c  M B '  A2
1
M c  487.95   (142.35)(9.49)
2
M c  187.5
M E  M C  A3
M E  187.5 
ME  0
1
(75)(5)
2
Example 3.7
Draw the shear and moment diagrams of the frame shown in figure.
M
0
A
RD 8  9( 4)( 4)  12(8)( 4)  24( 4)  0
RD  42kN 
F
x
0
Ax  24  36  0
Ax  12kN 
F
y
0
Ay  RD  12(8)  0
Ay  42  12(8)  0
Ay  54kN 
M
B
0
12(6)  24( 2)  M B  0
M B  120kN .m
F
x
0
Bx  12  24  0
Bx  36kN 
M A ' M A  A1
M A '0  ( 12)( 4)
M A '  48
M B  M A '  A2
M B  ( 48)  36( 2)
M B  120
M
B
0
12(8)( 4)  M B  42(8)  M C  0
12(8)( 4)  120  42(8)  M C  0
M C  72kN .m
VC  3.5(12)  42
M B ' M B  A1
M B ' 120 
1
(54)( 4.5)
2
M B '  1.5
M C  M B '  A2
1
M C  1.5   42(3.5)
2
M C  72