More Singularity Functions
Concentrated Forces
Represent a concentrated force with a singularity function that acts like a distributed
force. The singularity function that represents a unit, concentrated upward force located
at X  a .
 X  a  1
This is an improper function. Anti-differentiation of this function does not follow the
normal rules of calculus.
 Xa 
1
dX   X  a  0  C
(9.1)
In other words, this singularity function is an improper function such that its antiderivative is a Macaulay function of order 0. In the analysis of beams, th constant of integration
will be equal to zero.
Example
Consider a simply supported beam with a uniform distributed force.
q0

X
R1 
q0 L
2
L
O
R2 
q0 L
2
q ( X)  R 1  X  0   1  q 0  X  0  0
(E.1)
Anti-differentiate equation (E.1) to obtain V( X) .
V( X)   q ( X) d X  R 1  X  0  0  q 0  X  0  1  C1
(E.2)
All forces and moments acting on the beam, other than at the right end, have been
included in the distributed force, q ( X) .
C1  0
(E.3)
MAE3501 - 9 - 1
V( X)  R 1  X  0  0  q 0  X  0  1
(E.4)
Anti-differentiate equation (E.4) to obtain M( X) .
M( X )   V ( X ) d X  R 1  X  0  1 
q0  X  0  2
 C2
2
(E.5)
All forces and moments acting on the beam, other than at the right end, have been
included in the distributed force, W( X) .
C2  0
(E.6)
M( X )  R 1  X  0  1 
q0  X  0  2
2
(E.7)
Substitute equation (E.7) into the governing differential equation for beam deflection.
EI
d 2  ( X)
dX 2
q0  X  0  2
 M( X )  R 1  X  0  
2
1
(E.8)
Anti-differentiate equation (E.8).
EI
d  ( X) R 1  X  0  2 q 0  X  0  3


 C3
dX
2
6
There is no boundary condition for
(E.9)
d ( X)
. It is not yet possible to determine C 3 .
dX
Anti-differentiate equation (E.9).
E I  ( X) 
R1  X  0  3 q 0  X  0  4

 C 3 X  C4
6
24
(E.10)
There are two boundary conditions for  ( X) .
 ( X  0)  0
(E.11)
 ( X  L)  0
(E.12)
MAE3501 - 9 - 2
(E.10)  (E.11)
R1  0  0  3 q 0  0  0  4
0

 C4
6
24
(E.13)
Both Macaulay functions in equation (E.13) are equal to 0.
C4  0
E I  ( X) 
(E.14)
R1  X  0  3 q 0  X  0  4

 C3 X
6
24
(E.15)
(E.15)  (E.12)
0
R1  L  0  3 q 0  L  0  4

 C3 L
6
24
(E.16)
Evaluate the two Macaulay functions in equation (E.16).
0
R 1 L3 q 0 L4

 C3 L
6
24
(E.17)
Solve equation (E.17) for C 3 .
C3 
q 0 L3 R 1 L2 q 0 L3 q 0 L3
q L3



 0
24
6
24
12
24
(E.18)
(E.18)  (E.15)
E I  ( X) 
q  X  0  4  q 0 L3 
R1  X  0  3
 0

X
 24 
6
24


(E.19)
q 0 L  X  0  3 q 0  X  0  4  q 0 L3 


X
 24 
12
24


(E.20)
Substitute for R 1 .
E I  ( X) 
Determine  ( X) .
MAE3501 - 9 - 3
 ( X) 
q 0 L X 3 q 0 X 4 q 0 L3 X


12 E I
24 E I
24 E I
(E.21)
Example
P

R1
a
b
X
O
R2
R1 
Pb
ab
(E.1)
R2 
Pa
ab
(E.2)
q ( X)  R 1  X  0   1  P  X  a   1
(E.3)
Anti-differentiate equation (E.3) to obtain V( X) .
V( X)   q ( X) d X  R 1  X  0  0  P  X  a  0  C1
(E.4)
All forces and moments acting on the beam, other than at the right end, have been
included in the distributed force, q ( X) .
C1  0
Anti-differentiate equation (E.4) to obtain M( X) .
M( X)   V( X) d X  R 1  X  0  1  P  X  a  1  C 2
(E.5)
All forces and moments acting on the beam, other than at the right end, have been
included in the distributed force, q( X) .
C2  0
Substitute equation (E.5) into the governing differential equation for beam deflection.
MAE3501 - 9 - 4
EI
d 2  ( X)
dX
2
 M( X )  R 1  X  0  1  P  X  a  1
(E.6)
Anti-differentiate equation (E.6).
EI
d  ( X) R 1  X  0  2 P  X  a  2


 C3
dX
2
2
There is no boundary condition for
(E.7)
d
. It is not yet possible to determine C 3 .
dX
Anti-differentiate equation (E.7).
E I  ( X) 
R1  X  0  3 P  X  a  3

 C 3 X  C4
6
6
(E.8)
There are two boundary conditions for  ( X) .
 ( X  0)  0
(E.9)
 ( X  a  b)  0
(E.10)
(E.8)  (E.9)
0
R1  0  0  3 P  0  a  3

 C4
6
6
(E.11)
Both Macaulay functions in equation (E.11) are equal to 0.
C4  0
E I  ( X) 
(E.12)
R1  X  0  3 P  X  a  3

 C3 X
6
6
(E.13)
(E.13)  (E.10)
0
R1  a  b  0  3 P  a  b  a  3

 C 3 (a  b )
6
6
MAE3501 - 9 - 5
(E.14)
Evaluate the two Macaulay functions in equation (E.14).
0
R 1 (a  b ) 3 P (b ) 3

 C 3 (a  b )
6
6
(E.15)
Solve equation (E.15) for C 3 .
C3 
R (a  b ) 2
P b3
 1
6 (a  b )
6
(E.16)
(E.16)  (E.13)
E I  ( X) 
R1  X  0  3 P  X  a  3

6
6
3
 Pb
R 1 (a  b ) 2 


X
6
 6 (a  b )

(E.17)
Determine  ( X) for each region of the beam.
0 Xa
 ( X) 
R1 X3  P b 3
R (a  b ) 2 

 1
X
6EI
6EI
 6 (a  b ) E I

 P b3
P b X3
P b (a  b ) 



X
6 (a  b ) E I  6 (a  b ) E I
6 E I 
(E.18)
a X ab
 ( X) 
R 1 X 3 P ( X  a) 3  P b 3
R (a  b ) 2 


 1
X
6EI
6EI
6EI
 6 (a  b ) E I

 P b3
Pb X
P ( X  a)
P b (a  b ) 




X
6 (a  b ) E I
6EI
6 E I 
 6 (a  b ) E I
3
3
MAE3501 - 9 - 6
(E.19)
Homework
a  50 inch, b  40 inch, c  30inch, P  10 kip, Q  5 kip, E I  4.0 (10) 6 kip  inch2 .
1. Use singularity functions to determine the deflection, positive up, under each
concentrated force.
P
Q

O
a
b
c
MAE3501 - 9 - 7