Download 6. Energy Methods

ENG5312 – Mechanics of Solids II
1
6. Energy Methods
6.1 External Work
6.1.1 Work of a Force

The work done by a force is equivalent to the product of the component of the
force acting in the direction of motion and the distance travelled.
Ue 

If the force acts in the x -direction:

Ue 


 F  ds
x
 Fdx
0
If a force is applied to a prismatic beam in a gradual manner, i.e. the magnitude
of the force increases from 0 to P , and the bar stretches by  , when the material



P 
x then:
 

P
1 P2 1
xdx 
 P

2 
2
behaves in a linear-elastic manner F 

Ue 

x

0
(6-1)

6.1.2 Work of a Couple

A couple moment does work as it goes through a rotation:
Ue 

 Md
0

If a moment is applied to a body with linear-elastic material behaviour such that
the magnitude of the couple increases from 0 @   0 to M @    then:

Ue 


0

M
1
2
d  M

(6-2)

ENG5312 – Mechanics of Solids II
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6.2 Strain Energy

External work done by loads applied to a body will be converted into strain
energy. This strain energy is cause by normal and shear stresses that deform
the body.
6.2.1 Normal Stress

Consider a body deformed by a normal stress EQS:

The force on the top face is dFz   z dxdy and if it is applied gradually as the
element undergoes deformation d z  z dz the work done by the force is (using
Eq. (6-1)):



Or


1
1
dUi  dFz d z   z dxdyz dz
2
2
1
dUi   zz dV
2
So if a body is subjected to uni-axial normal stress, the strain energy is:

Ui 

V

dV
Ui 
2
 2E dV
V
Note: Ui is always positive.


2
(6-3)
For linear-elastic material behaviour, Hooke’s Law (   / E ) applies, and:




(6-4)
ENG5312 – Mechanics of Solids II
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6.2.2 Shear Stress

Consider an element subjected to shear stress,  :


The force dF  dxdy on the top face will move dz . Assuming dF is applied
gradually, and using Eq. (6-1):
1
1
dUi  dxdydz  dV
2 
2 


Or

Ui 

V


2
dV
(6-5)
For linear-elastic behaviour, Hooke’s Law (   /G) applies, and:


2
Ui  
dV
V 2G

(6-6)
ENG5312 – Mechanics of Solids II
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6.2.3 Multi-axial Stress

Consider an element subjected to a general state of stress.

Assuming linear-elastic behaviour and all loads are applied gradually, the strain
energy associated with each normal and shear stress can be added to give:
Ui 
1
  2  
x x
V


1
1
1
1
1
  yy   zz   xy  xy   yz yz   xz xz dV (6-7)

2
2
2
2
2
Using the generalized Hooke’s Law:


1
 x    y   z 
E
y 
1
 zy    x   z 
E
z 
1
 z    x   y 
E


 xy 



x 

 xy
G

;  xy 
 xy
G
;  xy 
 xy
G
The strains can be eliminated from Eq. (6-7):
Ui 
 1 2
 2E 
V
x

 1 2  2

  y2   z2 
 xy   yz   xz2   x y   y z   x z dV (6-8)


2G
E
And if only the principal stresses act on the element (i.e. 1,  2 and  3 )


Ui 
 1
 2E 
V

2
1
  22   32 

E



1 2   2 3  1 3 dV
 
(6-9)
ENG5312 – Mechanics of Solids II
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6.3 Elastic Strain Energy for Various Types of Loading
6.3.1 Axial Load

Consider a bar with a slowly changing cross-section that is loaded centroidally.

The internal load at x from one end is N , and the normal stress is N / A . using
Eq. (6-4) the strain energy is:
2
N2
U i   x dV  
dV
2
2E
2EA
V
V



The volume dV can be expressed as Adx and:




Ui 
L

0
N2
dx
2AE
(6-10)
If the cross-sectional area is constant:


Ui 
N 2L
2AE
Note:

o
L , Ui 
o
A, Ui 

o
E , Ui 

o i.e. something that is easy to distort will store more strain energy.


(6-11)
ENG5312 – Mechanics of Solids II
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6.3.2 Bending Moment

Application of a bending moment to a straight prismatic member results in a
normal stress.

Consider the element of area dA, y from the neutral axis, then (  My /I) , and
using Eq. (6-4):
 Ui

 2E
dV 
V

V
2
1 My 
  dV
2E  I  
The volume dV can be written as dV  dAdx , so:



2

Ui 
L

0
M2
2EI 2
 y dAdx
2
A
Will give the strain energy in the member, and since
2
A

Ui 
L

0

 y dA  I :
M2
dx
2EI

(6-12)
Note: The bending moment needs to be expressed as a function of x , then Eq.
(6-12) can be integrated.


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6.3.3 Transverse Shear

Consider a prismatic beam with an axis of symmetry y .


The internal shear force at x is V , and the shear stress on the element of area
dA is  



VQ
. Using Eq. (6-6) the strain energy is:
It
2
 
2
1 VQ
Ui  
dV  
  dAdx
 It 
2G
2G
V
V
Or

Ui 
L

0

fs 

A
(6-13)
A
I2

A
Q2
dA
t2
(6-14)
The strain energy can be written as;

Ui 
L

0


Q2
dAdx
t2
Defining the form factor, f s , which is a function of geometry:


V2
2GI 2
f sV 2
dx
2GA
(6-15)
An example of the form factor calculation is given in the text. For a rectangular
cross-section f s  6/5 .


Note: Ui due to shear is usually much less than Ui for bending (se e.g.14.4,
Hibbeler, 6e) and the shear strain energy stored in beams is usually neglected.



ENG5312 – Mechanics of Solids II
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6.3.4 Torsional Moment

Consider a shaft with a gradually changing cross-section:

If the shaft is subjected to an internal torque T at x from one end, the shear
stress on the element dA at  from the centroid is   T /J , and using Eq. (6-6)
the strain energy is;


 Ui 

V
 
2
1 T 
dV      dAdx
 J 
2G
V 2G

Or

L

Ui 
0

T 2  2 
  dAdx
2GJ 2 A

(6-16)
But the polar moment of inertia, J, is defined as:

J
  dA
2
(6-17)
A

Using Eq. (6-17) the strain energy can be written:

Ui 
L

0

T2
dx
2GJ
(6-18)
If the shaft (or tube) has constant cross-sectional area:


Ui 
T 2L
2GJ
(6-19)
ENG5312 – Mechanics of Solids II
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6.4 Conservation of Energy

The principal of conservation of energy states: Energy is a conserved property.
It can neither be created nor destroyed; only its form can be altered from one
form of energy to another.

Only mechanical energy will be considered, but kinetic energy will be neglected
since all loadings will be gradual.

Conservation of energy would require that the external work done by applied
loads (i.e. applied loads that cause deflections) must be equivalent to the strain
energy developed in a body as it deforms.
Ue  Ui

(6-20)
If the loads are removed the stored strain energy will restore the body to its
undeformed state (if the elastic limit has not been exceeded).

6.4.1 Trusses

Consider a truss subjected to the load P .


If the point of application of the load P deflects  in the direction of P , and the
load is increased gradually from 0 to P , then from Eq. (6-1):
1
 Ue  P

2




(6-21)
This external work done on the body is stored as strain energy. If, due to P , the
axial force N develops in a member, the strain energy stored in that member is

N 2L
from Eq. (6-11). To determine the total strain energy stored in the truss:
2AE


N 2L
Ui  
(6-22)
2AE

ENG5312 – Mechanics of Solids II
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
Where the summation is over all the members in the truss.

Conservation of energy requires Ue  Ui , therefore:


1
N 2L
P  
2
2AE
(6-23)
The deflection  caused by P can be evaluated after the axial forces in each
member of the truss has been determined using statics.



6.4.2 Vertically Loaded Beams

Consider a beam loaded with the vertical force P.

The deflection at the point of application of P can be determined from the
conservation of energy, Eq. (6-20), using Eqs. (6-1) and (6-12), for U e and Ui ,
respectively:
1
P 
2
L

0
M2
dx
2EI


(6-24)

The bending moment would be written as a function of x .

 due to bending moment and shear, however, the strain
Note: the beam deflects
energy due to shear is usually neglected, thus the deflection can be written as a
function of bending moment only.

ENG5312 – Mechanics of Solids II
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6.4.3 Beams Loaded with a Couple

Consider a cantilever beam subjected to an applied moment Mo .


The couple moment will cause the rotation  at the point of application, and it
does work due to this rotation: Ue  Mo /2 from Eq. (6-2).

The strain energy would be caused by the bending moment M , and
Ui 
L

0


M2
dx from Eq.
(6-12).

2EI

Conservation of energy, Eq. (6-20), would require:

1
M 0 
2

L

0
M2
dx
2EI
(6-25)
Where M is a function of x .
Note: Application ofthe conservation of energy is limited to situations where only
one applied load exists. For multiple applied loads, each load would have an
 associated external
 work and deflection, but there is only one conservation
equation, so only one unknown deflection can be solved.

ENG5312 – Mechanics of Solids II
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6.5 Impact Loading

Remember Mechanics II? Remember work-energy and conservation of energy
methods?

E.g. A weight is dropped from rest from a height h on to a linear spring, with
spring constant k . What is the maximum deflection of the spring?
o
Conservation of energy:


T1  V1  T2  V2
1
W h   max   k2max
2


o
Or Work-Energy: Ue  Ui  strain energy in the spring.
W h   max  

o
1
kxdx  k2max
2
The result can be rearranged to give:
2max 
2W
2W
 max 
h 0
k
k
The quadratic equation can be solved to give the maximum root:

o

0

o
 max
 max
W 2 W 
W
     2 h
 k 
 k 
k
(6-26)
If the weight is applied statically (i.e. gradually) W  k st or  st  W /k , and
Eq. (6-26) can be written as:

 max   st  2st  2 st h


o
Or



h 
 max   st 1 1 2 
 st 

(6-27)
ENG5312 – Mechanics of Solids II

13
o
Where the term in the square root is the extra displacement due to dynamic
loading.
o
Note: if h  0 , i.e. the weight W is released while it just touches the spring,
 max  2 st .
E.g. A weight W travelling with velocity v on a frictionless horizontal surface

impacts
a linear spring, with spring constant k . What is the maximum deflection
of the spring?

o
Conservation of Energy; 
T1  V1  T2  V2

1 W  2 1 2
 v  k max
2  g 
2


o
Or
2max 
o
Wv 2
gk
(6-28)
A statically loaded spring would deflect  st  W /k , so Eq. (6-28) can be
written as:

  max 
 stv 2
g
(6-29)

How to convert this information into deflections of dynamically loaded members?
i.e. How is impact loading simulated?

Assume:

i.
The moving body is rigid.
ii. The stationary body deforms in a linear-elastic manner (i.e. it behaves as
a linear spring).
iii. No energy is lost during the collision.
iv. The bodies remain in contact during the collision.
ENG5312 – Mechanics of Solids II
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
These are conservative assumptions, which lead to overestimates of forces (i.e.
good for design purposes).

With these assumptions, the deformable body behaves like a linear spring.

i.e. an effective spring constant can be defined and Eqs. (6-27) or (6-29) can be
used to determine  max .

An equivalent spring constant is not required. All that is needed is the static
deflection,  st , for use in Eq. (6-27).  st can be obtained from the equation of the
 Hooke’s Law, Appendix C, or conservation of energy and strain
elastic curve,
energy.

An
impact factor, n , can bedefined from Eq. (6-27):
 h 
n  1 1 2 
 st 




So:
 

 max  n st  n W k

Pmax  nW  k max
And
And the maximum stress is then:  max  n st .


(6-30)