Work and Energy
Consider a linear system. Linear means the following. If a slowly applied load causes a deflection at that load in the direction of that load, then a slowly applied load twice as large would cause a similar deflection twice as large. The word load refers to either a force or a moment. The word deflection refers to either a displacement or rotation.
A force with a final value of P is slowly applied to a linear system.

denotes the final displacement, at P , in the direction of P . The graph of force vs. displacement will be a straight line.
P

The area under the curve is the external work done, U
E
.
U
E

P

2
A moment, M , is slowly applied to a linear system.

denotes the final rotation, measured in radian , at M , in the direction of M .
M

U
E

2
(15.1)
(15.2)
Consider a linear, elastic system. Elastic means that, if loads are slowly applied to the system, then slowly removed, all the mechanical work put into the system is returned. None of that mechanical work shows up somewhere else, e.g., as heat. There are no elastic systems in the real world. When the tires on your car flex as you drive down the highway, they do not return all the mechanical work; some of it shows up as heat. This is why the sidewalls of your tires get hot enough to burn your hand when you travel at highway speeds on a hot day. Under-inflated tires flex more, generate more heat, and fail from overheating, which is why it is important to maintain proper tire pressure when traveling at high speed. Steel is much more elastic than rubber or artificial rubbers. That is why railroad cars have steel wheels and roll on steel rails, so that very little inelastic energy shows up as heat.
MAE3501 - 15 - 1

Slowly apply a load P
1 at some location. This load causes a final deflection

11 at the location of P
1 in the direction of P
1
. The external work done by the load.
U
E

P
1

11
2
(15.3)
Leave P
1 in place. Slowly apply a second load load causes a final deflection also causes an additional

22 at the location of deflection
P
2
at some other location. This second
P
2
in the direction of

12
at the location of P
1
P
2
. This second load
in the direction of P
1
. The additional external work done.
U
E

P
2

22
2

P
1

12
(15.4)
There is no factor of one-half in the second term because the entire load P
1 moved through the additional deflection

12
.
The total external work done in the two processes.
U
E

P
1

11
2

P
2

22
2

P
1

12
(15.5)
Define flexibility factors, denoted by f
IJ
. f
IJ
denotes the deflection, at the location of
P
I
, in the direction of P
I
, caused by a unit load at the location of P
J
, in the direction of P
J
. A flexibility factor is the reciprocal of a spring constant, k , encountered in Physics I.

11
 f
11
P
1
(15.6)

22
 f
22
P
2
(15.7)

12
 f
12
P
2
(15.8)
Substitute equations (15.6) through (15.8) into equation (15.5) .
U
E
 f
11
P
1
2
2
 f
22
P
2
2
2
 f
12
P
1
P
2
(15.9)
MAE3501 - 15 - 2

Slowly apply the loads in the reverse order.
U
E
 f
11
2
P
1
2
 f
22
2
P
2
2
 f
21
P
1
P
2
The final deflections are independent of the order in which the loads are applied.
(15.10)

1
 f
11
P
1
 f
12
P
2
(15.11)

2
 f
21
P
1
 f
22
P
2
(15.12)
The two expressions for the external work in equations (15.9) and (15.10) are equal.
Many authors argue that this equality follows directly from the fact that the deflections are the same in both cases. This argument only requires the system to be linear , and is not a valid argument. It is also necessary that the system be elasti c. For a linear system, if the loads are simultaneously and proportionally removed, then the equal final deflections in the two cases will guarantee that the same amount of work will be returned in the two cases. If the system is also elastic, then the external work done in applying the loads must have been the same in both cases. Equate equations (15.9) and (15.10) . f
12
 f
21
Equation (15.13) is referred to as Maxwell's reciprocal theorem, and is valid for
(15.13) any linear , elastic system.
Examine a few simple examples, using cantilever beams because myosotis will easily determine the flexibility factors.
MAE3501 - 15 - 3

Example
P
1
P
2 a b f
12 denotes the downward displacement at the location of P
1 caused by a unit downward force at the location of P
2
. f
12

( 1 ) a
3
3 E I

( 1
 b ) a
2
2 E I
(E.1) f
21
denotes the downward displacement at the location of P
2
caused by a unit downward force at the location of P
1
. f
21

( 1 ) a
3
3 E I

( 1 ) a
2
2 E I b (E.2)
Example
P
1
M
2 a b f
12
denotes the downward displacement at the location of P
1 caused by a unit clockwise moment at the location of M
2
. f
12

( 1 ) a
2
2 E I
(E.1) f
21
denotes the clockwise rotation at the location of M
2
caused by a unit downward force at the location of P
1
. f
21

( 1 ) a
2
(E.2)
2 E I
MAE3501 - 15 - 4

Example
P
1
M
2
L f
12
denotes the downward displacement at the location of P
1 caused by a unit clockwise moment at the location of M
2
. f
12

( 1 ) L
2
2 E I
(E.1) f
21
denotes the clockwise rotation at the location of M
2
caused by a unit downward force at the location of P
1
. f
21

( 1 ) L
2
2 E I
(E.2)
Maxwell’s Reciprocal Theorem is important in graduate-school-level derivations. It can also be used to solve some advanced engineering-mechanics problems. I have always found its primary value to be in checking my work. When working out flexibility factors or spring constants, I usually calculate all of them, even though I could, in theory, get about half of them from the other half. Then I check to make sure that all the pairs agree.
As an aside, note how nice it is to have a method of instantly calculating deflections and rotations at locations on cantilever beams. This comes in very handy when trying to understand something new.
MAE3501 - 15 - 5

Strain Energy
When loads are slowly applied to linear , elastic systems, all the work ends up as internal strain energy, U
I
.
U
I

U
E

U (15.14)
Bending Moments
Examine a differential length of beam, d S , to which a bending moment is slowly applied.
M denotes the final value of the bending moment. d

denotes the final relative rotation of the two cross sections of the differential length of beam. Shown are before and after sketches. d

The final internal strain energy of the differential length of beam. dU

M d


2
M
2


M d S
E I



M
2 d S
2 E I
(15.15)
The final internal strain energy of the entire beam.
U
  M
2 d S
2 E I
(15.16)
In this analysis, S is a generalized coordinate. For a straight beam, S is measured along the beam and is usually denoted by X . For a curved beam, S is measured along the beam.
MAE3501 - 15 - 6

Method of Virtual Work
Develop the fundamental relationships using a beam in bending.
Consider a beam with external loads.
P Q

O
We wish to determine the deflection, at some location, in some direction. This could be a rotation, at some location, about some axis. Suppose we wish to determine the downward deflection at the center of the beam. Slowly apply a virtual (fictitious) unit load, 1 kip , 1 kN , 1 of whatever unit is being used, at that location, in that direction.
1
 
O
As a result of this virtual load, the beam deflects a distance

, at the location of the virtual load, in the direction of the virtual load. The bending moment in the beam, resulting from the virtual load, as a function of X , is denoted by m ( X ) .
Leave that virtual unit load in place. Slowly apply the real loads.
P
1
Q
 

O
The real loads cause an additional deflection,

, at the location of the virtual load, in the direction of the virtual load. The real loads also cause an additional bending moment in the beam, as a function of X , denoted by M ( X ) .
MAE3501 - 15 - 7

Determine the final internal strain energy in the beam.
U
 
 m ( X )

M ( X )

2 d X
2 E I
(15.17)
U
  m
2
( X ) d X
2 E I
  M
2
( X ) d X
2 E I
  m ( X ) M ( X ) d X
E I
(15.18)
There is no factor of 2 in the denominator of the third term.
The first term represents the internal energy in the beam caused by the application of the virtual load. This term will not be used.
U
V

( 1 )

2
(15.19)
The second term represents the internal energy caused by the application of the real loads. This term will also not be used.

Q f
PQ
)
U
R

P ( P f
PP
2

Q ( P f
Q P
2

Q f
Q Q
)
(15.20)
The third term will be used. That term represents the internal energy caused by the virtual moments, m ( X ) , as they are rotated through the additional angles caused by the real moments, M ( X ) . However, this term is also equal to the work done by the virtual unit load, as it moves through the real displacement,

, caused by the real loads. This equality leads directly to the virtual-work relationship to determine

.
( 1 )
     m ( X ) M ( X ) d X
E I
(15.21)
When determining the moment distributions, m ( X ) and M ( X ) , it is necessary to use the same sign convention for both moments. This presents no problem because there is a required sign convention for bending moments in beams.
MAE3501 - 15 - 8

Example
P
 a a O
X
Determine

, the deflection, positive down, at the center of the beam.
The problem is symmetric, so it is possible to perform the integration from 0 to a , then multiply by 2 . Determine the left reaction, and M ( X ) for 0

X
 a .
R
1

P
2
(E.1)
M ( X )

R
1
X

P X
(E.2)
2 and m (
Apply a downward unit virtual load at the center of the beam, determine the left reaction,
X ) for 0

X
 a .
1 a O

X a
R
1

1
2 m ( X )

R
1
X

X
2
Apply the theorem of virtual work,
 
2
E I
 a
0


P X
2




X
2

 d X
P a
3
  
6 E I
(E.3)
(E.4)
(E.5)
(E.6)
MAE3501 - 15 - 9

Example q
0
 a a O
X
Determine

, the deflection, positive down, at the center of the beam.
The problem is symmetric, so it is possible to perform the integration from 0 to a , then multiply by 2 . Determine the left reaction, and M ( X ) for 0

X
 a .
R
1
 q
0 a
2
Make a section at X and draw a FBD of everything to the left of the section. q
0 a
X
(E.1)
X M ( X ) q
0 a
2
V ( X )

M
X

0

M ( X )


 q
0 a
X


X
2
X
3


 q
0
2 a


( X ) (E.2)
M ( X )
 q
0 a X
2
 q
0
X
3
6 a
(E.3)
MAE3501 - 15 - 10

and m (
Apply a downward unit virtual load at the center of the beam, determine the left reaction,
X ) for 0

X
 a .
1
 a a O
X
R
1

1
2 m ( X )

R
1
X

X
2
Apply the theorem of virtual work,
 
2
E I
 a
0 q
0 a X
2
 q
0
X
3
6 a
 
2
E I
 a
0 q
0 a X
2
4
 q
0
X
4
12 a
X
2 d X d X
(E.4)
(E.5)
(E.6)
(E.7)
 
2
E I q
0 a
4
12
 q
0 a
4
60
(E.8)
 
2 q
0 a
4
15 E I

(E.9)
MAE3501 - 15 - 11

Example
Re-work the previous example using the extended myositis formulas from lesson 13. q
0

END

END
L q
0
L
3
24 E I q
0
L
4
30 E I
(13.4)
L q
0
L
3
8 E I
11 q
0
L
4
120 E I
(13.5) center.
Examine one of the two identical cantilever beams resulting from breaking the beam at its q
0 a q
0 a
2
  
2


 q
0
2 a


( a )
3
3 E I
 q
0 a
4
30 E I

2 q
0 a
4
15 E I
(E.1)
MAE3501 - 15 - 12

Homework
Use virtual work to determine the downward deflection at the center of the beam. q
0
 a a O
MAE3501 - 15 - 13