Singularity Functions
It is possible to represent concentrated forces, concentrated moments (couples), and
distributed forces acting on a beam with special functions that act like distributed forces. These
special functions are referred to as singularity functions. Some of these singularity functions are
improper functions, because they do not follow the normal rules of calculus for antidifferentiation.
It is possible to combine these singularity functions and represent all the loads and
reactions on a beam as though they were distributed forces. Successive anti-differentiations then
yield shear force, bending moment, slope, and deflection, as functions of position on the beam.
Macaulay Functions
Macaulay functions are proper functions, and are used to represent actual distributed
forces. Macaulay functions are defined by the following relationships. n defines the order of the
Macaulay function. n  0, 1, 2,
 X  a n  0
Xa
(8.1)
 X  a  n  ( X  a) n
Xa
(8.2)
Express the above two equations verbally. If the argument of a Macaulay function is
negative, the value of the function is equal to zero. If the argument of a Macaulay function is
non-negative, the value of the function is equal to the argument raised to the order of the
Macaulay function..
Graphs of the first three Macaulay functions, for n  0, 1, 2 .
n1
n2
n0
1
a
a1
X
We will only consider distributed forces that can be represented by Macaulay functions
of order 0 and 1. However, higher-order Macaulay functions will occur in the calculations.
MAE3501 - 8 - 1
Anti-Differentiation
Macaulay functions are proper functions. The anti-derivatives of these functions follow
the ordinary rules of calculus.
Example
 Xa 
0
dX   X  a 1  C
(E.1)
The constant of integration is evaluated using an appropriate boundary condition. In this
example, assume C  0 .
 X  a 0
1
X
a
L
 X  a 1
La
X
a
MAE3501 - 8 - 2
L
Example
1
  X  a  dX 
 X  a 2
C
2
(E.2)
The constant of integration is evaluated using an appropriate boundary condition. In this
example, assume C  0 .
 X  a 1
La
a
L
 X  a 2
2
(L  a) 2
2
a
MAE3501 - 8 - 3
L
Distributed Forces
Develop Macaulay functions that represent all possible distributed forces that can be
represented by Macaulay functions of order 0 and 1. The purpose of this exercise is to furnish
you with a “handbook” of Macaulay functions with which to represent distributed forces on
beams. Number the examples for easy reference later.
1.
Consider a uniform, distributed force, magnitude q 0 , with its left edge at X  a , that
extends to the right end of a beam.
q0

O
a
X
q ( X)   q 0  X  a  0
(8.3)
2.
Consider a uniform, distributed force, magnitude q 0 , with its left edge at X  a , and
right edge at X  b .
q0

a
b
O
X
q ( X)   q 0 (  X  a  0   X  b  0 )
(8.4)
q0

a
b
X
MAE3501 - 8 - 4
O
3.
Consider a triangular, distributed force, increasing to the right, with maximum magnitude
q 0 , with its left edge at X  a , that extends to the right end of a beam at X  L .
q0

O
L
a
X
 q 
q ( X)    0   X  a  1
La
(8.5)
4.
Consider a triangular, distributed force, increasing to the right, with maximum magnitude
q 0 , with its left edge at X  a , and right edge at X  b .
q0

a
b
O
X
 q 
q ( X)    0  (  X  a  1   X  b  1 )  q 0  X  b  0
ba
q0

a
b
X
MAE3501 - 8 - 5
O
(8.6)
5.
Consider a triangular, distributed force, decreasing to the right, with maximum magnitude
q 0 , with its left edge at X  a , that extends to the right end of a beam at X  L .
q0

a
OL
X
 q 
q( X)   q 0  X  a  0   0   X  a  1
La
(8.7)
q0

a
O L
X
6.
Consider a triangular, distributed force, decreasing to the right, with maximum magnitude
q 0 , with its left edge at X  a ,and right edge at X  b .
q0

a
b
O
X
 q 
q ( X)   q 0  X  a  0   0  (  X  a  1   X  b  1 )
ba
q0

a
b
X
MAE3501 - 8 - 6
O
(8.8)
7.
Consider a trapezoidal, distributed force, increasing to the right, with edge magnitudes
q 1 and q 2 , with its left edge at X  a , that extends to the right end of a beam at X  L .
q1

q2
a
OL
X
 q  q1 
q ( X)   q 1  X  a  0   2
  X  a 1
L

a


(8.9)
q 2  q1
q1

a
O L
X
8.
Consider a trapezoidal, distributed force, increasing to the right, with edge magnitudes
q 1 and q 2 , with its left edge at X  a , and right edge at X  b .
q1

q2
a
b
O
X
q ( X)   q 1  X  a  0  q 2  X  b  0
 q  q1 
 2
 (  X  a 1   X  b 1 )
 ba 
(8.10)
q 2  q1
q1

a
X
b
MAE3501 - 8 - 7
O
9.
Consider a trapezoidal, distributed force, decreasing to the right, with edge magnitudes
q 1 and q 2 , with its left edge at X  a , that extends to the right end of a beam at X  L .
q1

q2
a
OL
X
 q  q2 
q ( X)   q 1  X  a  0   1
  X  a 1
L

a


q1  q 2
q1

(8.11)
a
O L
X
10.
Consider a trapezoidal, distributed force, decreasing to the right, with edge magnitudes
q 1 and q 2 , with its left edge at X  a , and right edge at X  b .
q1

q2
a
b
O
X
q ( X)   q 1  X  a  0  q 2  X  b  0
 q  q2 
 1
 (  X  a 1   X  b 1 )
b

a


(8.12)
q1  q 2
q1
q2

a
b
X
MAE3501 - 8 - 8
O
Example
Create q( X) that represents the following distributed force on the beam.
q0
L
L
L
Use the areas shown in the following drawing.
q0
L
L
L
q 
The largest triangle is example 3 with a = 0. q ( X)    0   X  0  1
 L
q 
The medium triangle is example 3 with a = L. q ( X)    0   X  L  1
 L
q 
The smallest triangle is example 3 with a = 2L. q ( X)    0   X  2L  1
 L
Desired area equals largest triangle minus medium triangle minus smallest triangle.

q 
q( X)    0   X  0  1   X  L  1   X  2 L  1
 L
MAE3501 - 8 - 9

Example
Repeat the previous example.
Use the areas shown in the following drawing.
q0
L
L
L
The triangle on the left is example 4 with a = 0 and b = L.
q 
q ( X)    0  (  X  0  1   X  L  1 )  q 0  X  L  0
 L
The central rectangle is example 2 with a = L and b = 2L.
q ( X)   q 0 (  X  L  0   X  2 L  0 )
The triangle on the right is example 6 with a = 2L and b = 3L.
q 
q ( X)   q 0  X  2 L  0   0   X  2 L  1
 L
Desired area equals the sum of the three areas.
q 
q ( X)    0  (  X  0  1   X  L  1 )  q 0  X  L  0
 L
q 
 q0 (  X  L  0   X  2 L  0 )  q0  X  2 L  0   0   X  2 L 1
 L

q 
q( X)    0   X  0  1   X  L  1   X  2 L  1
 L
MAE3501 - 8 - 10

Example
Repeat the previous example.
Use the areas shown in the following drawing.
q0
L
L
L
The large rectangle is example 1 with a = 0.
q ( X)   q 0  X  0  0
The triangle on the left is example 6 with a = 0 and b = L.
q 
q ( X)   q 0  X  0  0   0  (  X  0  1   X  L  1 )
 L
The triangle on the right is example 3 with a = 2L.
q 
q ( X)    0   X  2 L  1
 L
Desired area equals large rectangle minus the two triangles.
q 
q ( X)   q 0  X  0  0  q 0  X  0  0   0  (  X  0  1   X  L  1 )
 L
q 
  0   X  2 L 1
 L

q 
q( X)    0   X  0  1   X  L  1   X  2 L  1
 L
MAE3501 - 8 - 11

Homework
Create q( X) that represents the distributed force on the beam.
2q0
2q0
q0
L
L
L
MAE3501 - 8 - 12