More Singularity Functions
Concentrated Moments
Represent a concentrated moment with a singularity function that acts like a distributed
force. The singularity function that represents a unit, concentrated clockwise moment located
at X  a .
 X  a  2
This is an improper function. Anti-differentiation of this function does not follow the
normal rules of calculus.
 Xa 
2
dX   X  a   1  C
(10.1)
In other words, this singularity function is a function such that its anti-derivative is a
singularity function with exponent 1. In the analysis of beams, the constant of integration will
be equal to zero.
MAE3501 - 10 - 1
Example
M0

a
O
X
R1
R1 
b
R2
M0
ab
R2  
(E.1)
M0
ab
(E.2)
q ( X)  R 1  X  0   1  M 0  X  a   2
(E.3)
V( X)   q ( X) d X  R 1  X  0  0  M 0  X  a   1  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
M( X)   V( X) d X  R 1  X  0  1  M 0  X  a  0  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, W( X) .
C2  0
EI
d 2  ( X)
dX
2
 M( X )  R 1  X  0  1  M 0  X  a  0
d  ( X) R 1  X  0  2
EI

 M 0  X  a 1  C3
dX
2
There is no boundary condition for
d
. It is not yet possible to determine C 3 .
dX
MAE3501 - 10 - 2
(E.6)
(E.7)
E I  ( X) 
R1  X  0  3 M 0  X  a  2

 C3 X  C4
6
2
(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 M 0  0  a  2

 C4
6
2
(E.11)
Both Macaulay functions in equation (E.11) are equal to 0.
C4  0
E I  ( X) 
(E.12)
R1  X  0  3 M 0  X  a  2

 C3 X
6
2
(E.13)
(E.13)  (E.10)
R1  a  b  0  3 M 0  a  b  a  2
0

 C 3 (a  b )
6
2
(E.14)
Evaluate the two Macaulay functions in equation (E.14).
0
R 1 (a  b ) 3 M 0 (b ) 2

 C 3 (a  b )
6
2
(E.15)
Solve equation (E.15) for C 3 .
C3 
M0 b2
R (a  b ) 2
 1
2 (a  b )
6
MAE3501 - 10 - 3
(E.16)
(E.16)  (E.13)
E I  ( X) 
R1  X  0  3 M 0  X  a  2

6
2
 M b2
R (a  b ) 2 
 0
 1
X
6
 2 (a  b )

(E.17)
Determine  ( X) for each region of the beam.
0 Xa
R1 X3  M 0 b 2
R 1 (a  b ) 2 
 ( X) 


X
6EI
6EI
 2 (a  b ) E I

 M0 b2
M0 X3
M (a  b ) 


 0
X
6 (a  b ) E I  2 (a  b ) E I
6 E I 
(E.18)
a X ab
R 1 X 3 M 0 ( X  a) 2  M 0 b 2
R (a  b ) 2 


 1
X
6EI
2EI
6EI
 2 (a  b ) E I

M0 X3
M 0 ( X  a) 2  M 0 b 2
M (a  b ) 



 0
X
6 (a  b ) E I
2EI
6 E I 
 2 (a  b ) E I
 ( X) 
MAE3501 - 10 - 4
(E.19)
Example
Consider the following statically indeterminate beam. Create a FBD showing all the
reaction forces, including the vertical reaction at the roller.
F
q0
L
M
O
X
R
It is not possible to determine the three unknown reactions from the equations of static
equilibrium. However, it is possible to determine two of the reactions in terms of the third
reaction. Determine F and M in terms of R.
 FY  0  F  R  q 0 L
(E.1)
F  q0 L  R
(E.2)
 M ( X  0)  0  M  R L 
M
q 0 L2
2
(E.3)
q 0 L2
 RL
2
(E.4)
Create the singularity function that represents the distributed force in this problem.
q ( X)   M  X  0  2  F  X  0   1  q 0  X  0  0
(E.5)
Anti-differentiate twice to obtain the bending moment. Both constants of integration are
equal to zero.
V ( X)   M  X  0  1  F  X  0  0  q 0  X  0  1
(E.6)
 X  0 2
2
(E.7)
M ( X)   M  X  0  0  F  X  0  1  q 0
MAE3501 - 10 - 5
Substitute into the governing differential equation.
EI
d 2  ( X)
dX 2
  M  X  0  0  F  X  0 1  q0
 X  0 2
2
(E.8)
Anti-differentiate.
EI
d  ( X)
 X  0 2
 X  0 3
  M  X  0 1  F
 q0
 C3
dX
2
6
(E.9)
The constant of integration is equal to zero. Anti-differentiate again
E I  ( X)   M
 X  0 2
 X  0 3
 X  0 4
F
 q0
 C4
2
6
24
(E.9)
The constant of integration is equal to zero. The deflection at the roller must be equal to
zero.
 q L2
 L2 
 3
 4
  (q 0 L  R )  L   q 0  L 
0 0
 R L 
 6 
 24 
 2
 2 



 


(E.9)
Solve equation (E.9) for R, then substitute the result into equations (E.2) and (E.4) to
determine F and M.
There is nothing fundamentally difficult about a problem such as this, but it is certainly
not trivial.
MAE3501 - 10 - 6
Homework
a  50 inch, b  40 inch, 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
a
Q
b
MAE3501 - 10 - 7