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ME 221 Statics
Lecture #27
Section 7.4
ME221
Lecture 27
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Homework #9
Chapter 5 problems:
– 54, 56, 62, 64, 69, 71 & 73
• Due Today
ME221
Lecture 27
2
Homework #10
Chapter 7 problems:
– 5, 8, 19, 21, 24, 26 & 35
• Due Wednesday, November 13
ME221
Lecture 27
3
Last Lecture:
Internal Forces in Structures
• Reviewed internal/external forces
• Found internal forces
• Started shear & moment diagrams
ME221
Lecture 27
4
Shear and Moment Diagrams using Sectioning Method
Generate a shear / bending diagram as follows:
1. Find reaction forces
2. Take a section on each side of an applied force or
moment and inside a distributed load
(take a new section whenever there is a change in the load
or shape of the beam)
- draw a FBD and sum forces / moments
3. Repeat 2 along the length of the beam
w(x) distributed load
V(x) shear force
M(x) moment
ME221
Lecture 27
5
Sign Convention
M
M
V
V
Positive Shear and Positive Moment
ME221
Lecture 27
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Effect of External Forces
Positive Shear
M
M
Positive Moment
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Lecture 27
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Relations Between w, V, and M
In balancing forces, we can come up with differential
equations relating w, V, and M. These are as follows:
F
 0  V  (V  dV )  w( x)dx  0
M
w(x)
M+dM
MV
 0 M  Vdx  ( M  dM )  0
Thus,
V  x 
dM  x 
dx
, w x 
dV  x 
dx
V+dV
dx
This means you can integrate the shear diagram to obtain
the moment diagram.
ME221
Lecture 27
8
Shear and Moment Diagrams Using
Integration Method
1 - Distributed Load:
w(x)
M+dM
MV
xj
x j+1
dx
dV ( x)
w( x) 
dx

V  V j 1  V j 
ME221
V+dV
x j 1
V j 1
 w( x ) dx   dV
xj
Vj
Area under the load intensity diagram
between xj and x j+1
Lecture 27
9
Similarly:
dM ( x) 
V ( x) 
dx
x j 1
M j 1
 V ( x ) dx   dM
xj
Mj
M  M j 1  M j  Area under the shear diagram
between xj and x j+1
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Lecture 27
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2 - Concentrated Load:
P
P
M+dM
MV
C
V+dV
V   P, M  0
3 - Concentrated Moment:
MV
V  0, M  C
M+dM
C
V+dV
ME221
Lecture 27
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Notes
•Concentrated force will cause a jump in the shear diagram by an
amount equal to the applied load
•Concentrated moment will cause a jump in the moment diagram by an
amount equal to the negative of the applied moment
Connecting points in the shear and moment diagrams
xj
Negative
Positive
xj+1
xj
increasing slope
decreasing slope
xj
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xj+1
Positive
xj
Negative
decreasing slope
xj+1
increasing slope
Lecture 27
xj+1
12
Developing Shear and Bending Diagrams
1. Show FBD and statics for each section
2. Determine equation for V(x) and M(x)
3. Draw shear and bending diagrams indicating linear or
parabolic
4. Label end points of diagram as well as every region
endpoint
ME221
Lecture 27
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125 lb
20 lb/in
9 in.
12 in.
125 lb
12 in.
12in.
V
x
M
x
ME221
Lecture 27
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