More Simply-Supported Beams
During the last lesson we considered simply supported beams that could be broken into
two single-region cantilever beams. Today we will discuss simply supported beams that require
at least one multiple-region cantilever beam.
Example
Determine the downward deflections under each concentrated force, denoted by
 P and  Q .
P

Q
a
b
c
R1
O
R2
Determine the reaction forces.
R1 
P (b  c )
Qc

abc abc
(E.1)
R2 
Pa
Q (a  b )

abc abc
(E.2)
The beam can be broken at P or at Q. Choose P, and sketch the deflection curves,
labeling the end deflections as shown. Draw the true un-deflected line of the beam.
1
2
Q
a
b
c
R1
R2
Determine deflection at the free end of each cantilever beam. The left end is
straightforward. Deflection is positive upward..
1 
R1 a 3
3EI
(E.3)
MAE3501 - 13 - 1
Determine  2 by analyzing a two-region cantilever beam. Deflection is positive upward
and rotation is positive counter-clockwise.
b
R2 c
R2  Q
A
( R 2  Q ) b 2 ( R 2 c) b


2EI
EI
A 
( R 2  Q ) b 3 ( R 2 c) b 2

3EI
2EI
(E.4)
(E.5)
c
R2
B 
R 2 c3
3EI
(E.6)
Determine  2 .
2   A  c A  B
(E.7)
Determine  P by linear interpolation between the deflections at the ends.
 P  1 
( 2   1 ) a
abc
(E.8)
MAE3501 - 13 - 2
There are two different methods to determine  Q . The first method is to rework the
problem, breaking the simply-supported beam at Q. This demonstrates nothing new. The
second method is to examine the deflection curve and determine two new displacements. The
force Q has been removed from the sketch for clarity. Both new displacements, denoted
by  3 and  4 , are measured at the location of the force Q.
2
3
1
4
a
b
c
Determine the new displacements.  4 is the upward deflection of the right cantilever
beam at the concentrated load Q. This deflection is equal to  A and has already been
determined.
4   A
(E.9)
Determine  3 by linear interpolation between the deflections at the ends.
 3  1 
(  2   1 ) (a  b )
abc
(E.10)
Determine  Q by subtraction.
Q   3  4
(E.11)
MAE3501 - 13 - 3
Example
Determine the downward deflections under each concentrated force, denoted by
 L and  R . E I  1.0 (10) 7 kip  inch2
10 kip

45
10 kip
50
40
R1
O
R2
Determine the reaction forces.
R1 
10 (90) 10 (40)

 9.63 kip
135
135
(E.1)
R 2  20  9.63  10.37 kip
(E.2)
The beam can be broken at either concentrated force. Choose the left concentrated force,
and sketch the deflection curves, labeling the end deflections as shown. Draw the true undeflected line of the beam.
1
2
10
45
9.63
50
40
10.37
Determine deflection at the free end of each cantilever beam. The left end is
straightforward. Deflection is positive upward..
1 
9.63 (45) 3
3 (10) 7
 0.0293 inch
MAE3501 - 13 - 4
(E.3)
Determine  2 by analyzing a two-region cantilever beam. Deflection is positive upward
and rotation is positive counter-clockwise.
10.37 (40)  415
50
10.37  10  0.37
A 
A 
0.37 (50) 2
2 (10)

7
0.37 (50) 3

3 (10) 7
415 (50)
(10) 7
 0.00212
415 (50) 2
2 (10) 7
 0.0534 inch
(E.4)
(E.5)
40
10.37
B 
10.37 (40) 3
 0.0221 inch
(E.6)
 2  0.0534  40 (0.00212)  0.0221  0.160 inch
(E.7)
3 (10) 7
Determine  2 .
Determine  L by linear interpolation between the deflections at the ends.
 L  0.0293 
(0.160  0.0293) 45
 0.0729 inch 
135
MAE3501 - 13 - 5
(E.8)
There are two different methods to determine  R . The first method is to rework the
problem, breaking the simply-supported beam at the right concentrated force. This demonstrates
nothing new. The second method is to examine the deflection curve and determine two new
displacements. The right concentrated force has been removed from the sketch for clarity. Both
new displacements, denoted by  3 and  4 , are measured at the location of the right concentrated
force.
2
3
1
4
45
50
40
Determine the new displacements.  4 is the upward deflection of the right cantilever
beam at the right concentrated force. This deflection is equal to  A and has already been
determined.
 4   A  0.0534 inch
(E.9)
Determine  3 by linear interpolation between the deflections at the ends.
 3  0.0293 
(0.160  0.0293) 95
 0.121 inch
135
(E.10)
Determine  R by subtraction.
 R   3   4  0.0676 inch 
MAE3501 - 13 - 6
(E.11)
Example
Determine  , the downward displacement under the right edge of the distributed force,
and  P , the downward deflection under the concentrated force.
P
q0
a
b
c

O
R1
R2
Determine the reactions.
R1 
R1 
q 0 a  a  b  c 
2
abc

Pc
abc
(E.1)
q 0 a  a 
 2   P (a  b )
abc
abc
(E.2)
Break the beam at the right end of the distributed force and sketch the deflection curves,
labeling the end deflections as shown. Draw the true un-deflected line of the beam.
1
q0
2
P
a
b
c
R1
R2
Determine deflection at the free end of each cantilever beam. The left end is
straightforward. Deflection is positive upward.
R1 a 3 q0 a4
1 

3EI
8EI
(E.3)
MAE3501 - 13 - 7
Determine  2 by analyzing a two-region cantilever beam. Deflection is positive upward
and rotation is positive counter-clockwise.
b
R2 c
R2  P
A
( R 2  P ) b 2 ( R 2 c) b


2EI
EI
A 
( R 2  P ) b 3 ( R 2 c) b 2

3EI
2EI
(E.4)
(E.5)
c
R2
B 
R 2 c3
3EI
(E.6)
Determine  2 .
2   A  c A  B
(E.7)
Determine  by linear interpolation between the deflections at the ends.
  1 
( 2   1 ) a
abc
(E.8)
MAE3501 - 13 - 8
Determine  P in the same manner that was used to determine the second deflection in the
previous example. The concentrated force P has been removed from the sketch for clarity. Both
new displacements, denoted by  3 and  4 , are measured at the location of the force P.
2
3
1
4
a
b
c
Determine the new displacements.  4 is the upward deflection of the right cantilever
beam at the concentrated load Q. This deflection is equal to  A and has already been
determined.
4   A
(E.9)
Determine  3 by linear interpolation between the deflections at the ends.
 3  1 
(  2   1 ) (a  b )
abc
(E.10)
Determine  Q by subtraction.
Q   3  4
(E.11)
MAE3501 - 13 - 9
Expanding Myosotis
I once consulted for a firm that manufactures yurts, tent-like structures that utilize
medieval Mongolian designs and modern materials, e.g., multi-strand, aircraft-strength
aluminum cables rather than braided yak hair. In particular, I was analyzing snow loads on
rafters that radiate outward like the spokes of a wheel from a central hub to which they are
attached. Because the distance between adjacent rafters increases linearly with distance from the
center of the structure, the distributed force per unit length also increases linearly with distance
from the center of the structure. This led to the following problem.
O
I was intensely frustrated that I would not be able to use myosotis to analyze the problem,
when I realized that it is possible to “extend” the classic myosotis method. I am not aware of
anyone else having done this.
 END
q0
L
 END
q 0 L3
24 E I
q 0 L4
30 E I
(13.1)
q 0 L3
8EI
11 q 0 L4
120 E I
(13.2)
q0
L
The first pair of formulas I found in some Mechanics of Materials text. I worked out the
second pair using double-integration.
Trapezoidal loads can be considered as a superposition of a uniform and a triangular load.
MAE3501 - 13 - 10
Homework
Determine the displacements, positive down, at the left end of the distributed load and
under the concentrated force. E I  5 (10) 6 kip  inch2
10 kip
0.25
 40 inch
50 inch
30 inch O
MAE3501 - 13 - 11
kip
inch