Statically Indeterminate Beams
Consider two different cases, cantilever beams with a redundant support, and simply
supported beams with a redundant support.
Cantilever Beams
For a cantilever beam, there are three reactions at the wall. If there are any additional
supports, the beam is statically indeterminate. One possibility is an additional roller support.
Example
P
a
O
b
Denote the upward concentrated force at the roller by R. Determine the displacement at
the roller, and set that displacement equal to zero. Choose positive down. Because the deflection
will be set equal to zero, the product of E I can be omitted from the denominators.
P
Pb
a
R
R  0 
R  P
P a 3 ( P b) a 2 R a 3


3
2
3
3Pb
2a
(E.1)
(E.2)
The problem is now statically determinate. The region between the wall and the roller
was a single region and it was possible to determine the displacement at the reaction in one step.
MAE3501 - 14 - 1
Example
10 kip
50 inch
O 20 inch
Denote the upward concentrated force at the roller by R. Determine the displacement at
the roller, and set that displacement equal to zero. Choose positive down. Because the deflection
will be set equal to zero, the product of E I can be omitted from the denominators.
10
10 (20)  200
50
R
R  0 
10 (50) 3 ( 200) (50) 2 R (50) 3


3
2
3
(E.1)
R  0 
10 (50) ( 200) R (50)


3
2
3
(E.2)
R  16.0 kip 
(E.3)
The problem is now statically determinate. The region between the wall and the roller
was a single region and it was possible to determine the displacement at the reaction in one step.
MAE3501 - 14 - 2
Example
q0
a
O
b
Denote the upward concentrated force at the roller by R. Determine the displacement at
the roller, and set that displacement equal to zero. Choose positive down. Because the deflection
will be set equal to zero, the product of E I can be omitted from the denominators.
q0 b
q0
q0 b2
2
a
R
R
 q0 b2  2

a


4
3
q a
(q b ) a
R a3
 2 
0 0
 0


8
3
2
3
 3a
3 b 2 
R  q0 
b
 8
4 a 

(E.1)
(E.2)
The problem is now statically determinate. The region between the wall and the roller
was a single region and it was possible to determine the displacement at the reaction in one step.
MAE3501 - 14 - 3
Example
0.25 kip/ inch
40 inch
O
30 inch
Denote the upward concentrated force at the roller by R. Determine the displacement at
the roller, and set that displacement equal to zero. Choose positive down. Because the deflection
will be set equal to zero, the product of E I can be omitted from the denominators.
0.25 (30)  7.5
0.25
0.25 ( 30) 2
 112.5
2
40
R
R  0 
R
3
40 3
0.25 (40) 4 (7.5) (40) 3 112.5 (40) 2 R (40) 3



8
3
2
3
(80,000  160,000  90,000)  15.5 kip 
(E.1)
(E.2)
The problem is now statically determinate. The region between the wall and the roller
was a single region and it was possible to determine the displacement at the reaction in one step.
MAE3501 - 14 - 4
For multiple regions, the procedure is the same, although there are more calculations.
Example
q0
a
b
O
Denote the upward concentrated force at the roller by R. Determine the displacement at
the roller and set that displacement equal to zero. Omit the E I product from the denominators.
Choose deflection positive down and rotation positive clockwise.
q0
Rb
a
R
q 0 a 3 R a 2 ( R b) a


6
2
1
(E.1)
q 0 a 4 R a 3 (R b) a 2



8
3
2
(E.2)
A 
A
b
R
Deflection is positive down.
B  
R b3
3
(E.3)
MAE3501 - 14 - 5
Determine the displacement at the reaction.
R   A  b A  B
(E.4)
Substitute values and set  R = 0.
0
q a3 R a2
 R b3
q 0 a 4 R a 3 ( R b) a 2


b  0

 ( R b) a 
8
3
2
2
3
 6

(E.5)
Collect terms.
3
 a4 b a3 
 3
  R  a  b a2  b2 a  b 
0  q0 

 8
 3
6 
3 


R
 a4 b a3 

q0 

 8

6


 a3
b 3 

 b a2  b2 a 
 3
3 

The problem is now statically determinate.
MAE3501 - 14 - 6
(E.6)
(E.7)
Simply Supported Beams
Example
q0

a
O
b
O
Denote the upward concentrated force at the internal roller by R. Determine the reactions
at the end supports in terms of the external loads and R.
q0
a
b
R
R1
 M 1  0  R a  R 2 (a  b ) 
R2 
q 0 (a  b ) 2
2
q 0 (a  b ) R a

2
ab
 FY  0  R1  R 
R1 
R2
(E.1)
(E.2)
q 0 (a  b ) R a

 q 0 (a  b )
2
ab
q 0 (a  b ) R b

2
ab
(E.3)
(E.4)
Break the beam at the internal roller support. Determine the displacements at the ends of
the beam, positive up.
R1 a 3 q0 a4
1 

3EI
8EI
(E.5)
R 2 b3 q0 b4

3EI
8EI
(E.6)
2 
MAE3501 - 14 - 7
Set the displacement at the internal roller support equal to zero.
0   R  1 
 2   1  a
ab
(E.5)
Substitute numerical values and solve for R. The problem is now statically determinate.
There is a different technique that eliminates some algebra. Manipulate equation (E.5).
0  1 
0
 2   1  a
(E.6)
ab
1 b  2 a

ab ab
(E.7)
b 1   a  2
(E.8)
MAE3501 - 14 - 8
Example
0.25 kip/ inch

40 inch
O
40 inch
O
Denote the upward concentrated force at the internal roller by R. Determine the reactions
at the end supports in terms of the external loads and R.
0.25 kip/ inch
40
40
R
R1
 M1  0  40 R  80 R 2 
R2
0.25 (80) 2
2
(E.1)
R 2  10 
R
2
(E.2)
R 1  10 
R
2
(E.3)
Break the beam at the internal roller support. Determine the displacements at the ends of
the beam, positive up.
(10  0.5 R ) (40) 3 0.25 (40) 4 133,333  10,667 R
1   2 


3EI
8EI
EI
(E.4)
40  1   40  2   40  1
(E.5)
 1  0  133,333  10,667 R
(E.6)
R  12 .5 kip 
(E.7)
R1  R 2  3.75 kip 
(E.8)
MAE3501 - 14 - 9
Shortcut
Consider the following statically indeterminate problem.
P
q0
a
b
O
Denote the upward reaction at the roller with R, then solve for the deflection at the roller
and set that deflection equal to zero, and solve for R.
The point is that we are considering a multi-region cantilever beam, and only wish to
determine the deflection at its end. We need to determine the deflection and rotation at the end
of the first region, but we have no use for these values other than to calculate the deflection at the
end.
Actually, we do not need to determine the deflection and rotation at the end of the first
region.
0  R
 P a 2  R (a  b ) 3
q 0 (a  b ) 4 P a 3



 b
 2EI 
8EI
3EI
3EI


(14.1)
q0
a
0  R 
b
O
 q a 3  R (a  b ) 3
q0 a4
 b 0  
 6EI 
8EI
3EI


MAE3501 - 14 - 10
(14.2)
Homework
Determine all the reaction forces and moments. E I  5 (10) 6 kip  inch2
0.25
kip
inch
A
40 inch
50 inch
O
MAE3501 - 14 - 11