Chapter Three
Chapter three discusses load and stress analysis. This is obviously a very important
subject and we will spend a number of lectures discussing this topic.
Section 3 – 2, Shear Force and Bending Moment in Beams
Consider shear and moment diagrams. Consider a beam loaded transversely. Loads
include concentrated forces, couples (concentrated moments), uniform distributed forces, and
distributed forces that vary.
P
q1
q2
MO

O
P is expressed in units of kip or kN.
M O is expressed in units of kip inch or kN  m .
q 1 and q 2 are expressed in units of
kip
kN
or
.
inch
m
Shigley uses q to represent a distributed force, positive upwards.
This is a simply-supported beam, pinned at one end, with a roller at the other end.
Simply supported beams are statically determinate.
The best way to teach shear and moment diagrams is to work examples.
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Example
Consider a simply supported beam with a non-centered, concentrated force.
P
A

a
b
B
O
X
Determine the external reactions. This can be accomplished using the equations of static
equilibrium in two dimensions. Or by using the following relationships.
AY 
Pb
ab
(E.1)
BY 
Pa
ab
(E.2)
Make a mathematical cut, a section, through the beam and consider a FBD of everything
to one side of the section. Make a section to the left of the concentrated force, and consider a
FBD of the portion of the beam to the left of the section, because that portion of the beam is
subjected to fewer loads.
X
M( X)
A
Pb
ab
V( X)
There are required sign conventions for V( X) and M( X) . V( X) and M( X) are shown
positive in the above FBD.
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 FY  0 
(E.3)
V( X) 
Pb
 V( X)
ab
(E.3)
Pb
ab
(E.4)
 Pb 
X
ab
(E.5)
 M X  0  M( X )  
(E.5)
Pb X
ab
(E.6)
M( X ) 
V( X) and M( X) have been determined for X < a. In general, a new section is required on
each side of a concentrated force, a concentrated moment, and each end of a distributed force.
Make a section to the right of the concentrated force, and consider a FBD of the simpler
portion of the beam, this time the portion to the right of the section. Define a new position
variable.
X
M( X)
B
Pa
ab
V( X  )
V( X) and M( X) are shown positive in the above FBD.
 FY  0 
(E.7)
V( X )  
Pa
 V( X )
ab
Pa
ab
(E.8)
 Pa 
 X
ab
(E.9)
(E.7)
 M X  0   M( X)  
(E.9)
P a X
ab
(E.10)
M( X  ) 
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We could replace X with a  b  X, and determine analytic relationships for V( X) and
M( X) everywhere on the beam. Instead, graph the results.
V( F )
Pb
ab
X(L)
a
Pa

ab
M(F  L)
Pab
ab
a
X(L)
The following statements are always true.
V( X) is discontinuous only at a concentrated force. The magnitude of the discontinuity
is equal to the magnitude of the concentrated force. V( X) at the left end of a beam is equal to
the upward concentrated force at that location. In this example, that force is the vertical reaction
at the left end of the beam, equal to A Y . V( X) at the right end of a beam is equal to the negative
of the upward concentrated force at that location. In this example that is the negative of the
vertical reaction at the right end of the beam, equal to  B Y .
M( X) is discontinuous only at a concentrated moment (couple). The magnitude of the
discontinuity is equal to the magnitude of the concentrated moment.
The slope of M( X) is discontinuous only at a concentrated load.
The following relationship is always true wherever both terms are defined.
V( X) 
dM ( X)
dX
(2.1)
In this example, neither term is defined at X  a .
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Example
The following example problem demonstrates how to create properly labeled, to-scale
shear and moment diagrams.
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Example
Consider a simply supported beam with a uniform distributed force over its entire length.
q0
A

O B
X
L
Determine the reactions. Use symmetry.
AY  BY 
q0 L
2
(E.1)
Make a section at an arbitrary value of X, and consider a FBD of everything to the left of
the section.
q0
M( X)
X
q0 L
2
V( X)
 FY  0 
(E.2)
V( X) 
q0 L
 q 0 X  V( X)
2
q0 L
 q0 X
2
(E.3)
 X   q0 L 
X
2 
 M X  0  M( X)  (q 0 X)  2   
(E.4)
M( X ) 
(E.2)
q0 L X q0 X2

2
2
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(E.4)
(E.5)
Graph the results.
V( F )
q0 L
2
X(L)

q0 L
2
M(F  L)
q 0 L2
8
X(L)
As a result of equation (2.1) on page 4, M( X) will have a local maximum or minimum
where V( X)  0 . In this example, that occurs at X  0.5 L . The value of M( X) at X  0.5 L is
determined by substituting X  0.5 L into (E.5), the relationship for M( X) determined above.
The shape of the curve is a downward-opening parabola.
q ( X) denotes the value of a distributed force as a function of X. q ( X) is positive
upward. The following relationship is always true wherever both terms are defined.
d V( X)
 q ( X)
dX
(2.2)
The bending moment at each end of this beam is zero. This will always be true at an end
of a beam at which no concentrated moment acts. It will generally not be true for a cantilever
beam at the built-in support.
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We have determined V( X) and M( X) for a simply-supported beam with a single, noncentered, concentrated force and for a simply-supported beam with a uniform distributed force
over its entire length. These functions were graphed.
It can be necessary to develop analytic expressions, mathematical relationships, for V( X)
and M( X) . It is therefore necessary to be comfortable with both the formulas and the diagrams.
Example
Consider a simply-supported beam with a non-uniform distributed force over its entire
length.
q0
A 
O B
X
L
Determine the reactions. In this step it is generally necessary to replace a distributed
force with a statically equivalent concentrated force. The magnitude of the concentrated force,
denoted by P, is equal to the area under the distributed-force diagram. The location of the
concentrated force, denoted by X P , is the X - coordinate of the centroid of that area.
P
q0 L
2
(E.1)
2L
3
(E.2)
XP 
Determine the vertical reactions.
1 q L
AY  P    0
6
 3
(E.3)
 2 q L
BY  P    0
3
 3
(E.4)
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Make a section and consider a FBD of everything to one side of the section. In this step
it is not permissible to replace a distributed force with a statically-equivalent concentrated force.
All distributed forces must be present. Failure to understand this concept will cause major
problems.
q0 X
L
M( X)
X
q0 L
6
V( X)
The magnitude of the distributed force at the section is not equal to q 0 . That magnitude
must be determined from the geometry of the distributed-force diagram, in this case through
similar triangles. This is a very important step, and can also cause major problems.
Determine V( X) and M( X) . This is a problem involving static equilibrium, and it is now
permissible to replace the distributed force with a statically equivalent concentrated force. Use
the same notation.
2
q X  X q X
P 0   0
2L
 L  2
XP 
(E.5)
2X
3
(E.6)
Consider a FBD of everything to the left of the section.
q0 X2
2L
2X
3
q0 L
6
X
3
V( X)
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M( X)
q0 L q0 X2

 V( X)
6
2L
 FY  0 
V( X) 
(E.7)
q0 L q0 X2

6
2L
(E.8)
 q0 X2   X   q0 L 


M

0

M
(
X
)


X
 X
 2 L   3   6 


(E.9)
q0 L X q0 X3
M( X ) 

6
6L
(E.10)
Graph V( X) and M( X).
V( F )
q0 L
6
X (L)

q0 L
3
V( X  0) is equal to the left reaction force. V( X  L) is equal to the negative of the
right reaction force. The shape is a parabola, opening downward. The slope at X = 0 is equal to
q ( X  0) , zero in this example. It is necessary to know where V( X)  0 .
0  V( X) 
X2 
q0 L q0 X2

6
2L
L2
3
(E.11)
(E.12)
X  0.577 L
(E.13)
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M(F  L)
M MAX
0.577 L
M MAX
q 0 L (0.577 L) q 0 ( 0.577 L) 3
 M( X  0.577 L) 

6
6L
M MAX  0.128 q 0 L2
X(L)
(E.14)
(E.15)
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Example
Re-work the previous example, using the differential and integral relationships
among q ( X) , V( X) , and M( X) . It is necessary to determine the reactions at A and B. Use the
values determined in the previous example.
AY 
q0 L
6
(E.1)
BY 
q0 L
3
(E.2)
Determine the analytic expression for q ( X) .
q ( X)  
q0 X
L
(E.3)
In order to use direct integration, it is necessary to determine V( X  0) .
V( X  0 )  A Y 
q0 L
6
(E.4)
Use equation (2.2).
 q X 
V( X)  V( X  0 )     0
 d X
L


0
X
(E.5)
X is a limit of the integration, and X' is a dummy variable of integration. Many people
get sloppy and use X for both. It sort of works, on simple problems like this one, but it will
cause a great deal of difficulty in more complex problems. Learn to use correct notation.
q0 L q0 X2
V( X) 

6
2L
(E.6)
The result obtained in the previous example.
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Determine M( X) .
M( X  0 )  0
(E.7)
Use equation (2.1).
X
q L q X 2 
M( X )  M( X  0 )    0  0
d X
 6

2
L

0 
M( X ) 
q0 L X q0 X3

6
6L
(E.8)
(E.9)
The result obtained in the previous example.
In this particular example, it was easier to use equations rather than sections, FBDs, and
the equations of static equilibrium. However, there was only one region, and it was possible to
determine one analytic expression for q ( X) that was valid over the entire length of the beam. In
general, there will be multiple regions. Using the equations will be more complicated, because it
will be necessary to match boundary conditions at every location that connects two regions.
In addition, there is a fundamental drawback to using the equations. When the FBDs and
equations of static equilibrium were used, V( X) and M( X) were determined independently. If
there is an error in one of the terms, equation (2.1) will not be satisfied and the error can be
discovered rather easily. However, if V( X) is determined incorrectly, and M( X) is then
determined from the incorrect V( X) using equation (2.1), the error will not be easily discovered.
Homework
Read sections 3 – 4, 3 – 5, and 3 – 6.
Problems 3 – 4, 3 – 5, and 3 – 6. The beam in problem 3 – 6 is a cantilever beam.
.
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