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Lecture Note 4Lecture Note 4
Virtual Work & Energy Method
Second Semester, Academic Year 2012,Department of Mechanical Engineering
Chulalongkorn University
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Objectives
Use the energy method to analyze structures Describe the characteristics and properties as well as Describe the characteristics and properties as well as
determine strain energy and complementary energy and potential energyD ib th i i l f i t l k d th i i l Describe the principle of virtual work and use the principle to determine equilibrium, stability and analyze simple elasticity problems with emphasis on bending problems
A simple statically indeterminate problems with emphasis on bending
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Topics
Virtual Work Strain energy complementary and potential energy Strain energy, complementary and potential energy Deflections Statically indeterminate problems
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4
Work By a Force
cosFW F dr
F dr cosF dr
workFW
force that done the workdisplacement
F
Fdr
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5
Work By a Couple
( ) ( ) ( )2 2Mr rW F F Fr
W M magnitude of couple that do the workM
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MW M small angle of rotation
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Virtual Work Virtual Movements
Imaginary or virtual movements is assumed and does not actually exist. Virtual displacement Virtual rotation Virtual rotation Virtual deformation
Virtual movements are infinitesimally small and does not violate physical constraints.
Principle of virtual work is an alternative form ofPrinciple of virtual work is an alternative form of Newton’s laws that can analyze the system in equilibrium under work and energy concepts.
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Virtual Work Principle of Virtual Work
Consider an object in equilibriumTh i t l k d b ll f t th bj t ith The virtual work done by all forces to move the object with a virtual displacement
1
cosr
F k v kk
W F
In equilibrium, 0FW
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Virtual Work Principle of Virtual Work for Rigid Bodies
t e iW W W
total virtual work doneexternal work done
tWW
external work doneinternal virtual work
e
i
WW
e iW W
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9
Exercise Virtual Work for Rigid Bodies #1
Determine the support reactions
a
, ,
00
v B v C
t
LWR W
, ,
, ,
0
0
C v C v B
C v C v C
R WaR WL
C
LaR WL
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10
Exercise Virtual Work for Rigid Bodies #2
0( ) ( ) 0
( ) ( ) 0
t
A v v v C v v
WR W a R LR R W R L W
10
( ) ( ) 00 and 0
A C v C v
A C C
R R W R L WaR R W R L Wa
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Virtual Work Virtual Work for Deformable Bodies
t e iW W W
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Virtual Work Internal Virtual Work from Axial Load
NN A AA
, ( )i N vA
ANw dA xA
,
, v
i N v
i N N x
w N dx
w
,
L
i N vL
w N dx
v vv
NE EA
A vi N
N Nw dx
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,i NL
w dxEA
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Virtual Work Internal Virtual Work from Torsion
,A v
i TL
T Tw dxGJ
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Virtual Work Internal Virtual Work from Bending
,A v
i ML
M Mw dxEI
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Virtual Work Internal Virtual Work from Shear Force
S A
, ( )i S vA
S A
w dA x
, ( ) ( )vA
i SS dA xA
w
,
, v
i S
i S
vL
S x
w S dx
w
,A v
i SS Sw dxGA
L
v vv G
SGA
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L GA
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Virtual Work Virtual Work from External Loads
w W P
, ,
( )
e v y v x
Ve vw
w W P
M T
w x xw d ,( )L
e v yw x xw d
( ( ) )e v y v x V v v yW W P M T w x dx
, , ,( ( ) )
( )
e v y v x V v v yL
A v A v A v A vi A v
N N S S M M T TW dx dx dx dx MEA GA EI GJ
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L L L LEA GA EI GJ
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Exercise Virtual Work for Deformable Bodies #1
Determine the bending moment at B
a b
,v B a babb
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Exercise Virtual Work for Deformable Bodies #2
Determine the bending moment at B
(1 )a L
(1 )
0
B b b
W
,
00
t
v B B B
WW M
L
B
B
LWa Mb
WabM
18
BM L
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Exercise Virtual Work for Truss #1
Determine the force in AB
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Exercise Virtual Work for Truss #2
, 3
4v B
,44
30
v BC
tW
,30 0
40 kN
t
C BA v B
BA
FF
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Exercise Virtual Work for Cantilever Beam #1
Determine the end deflection
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Exercise Virtual Work for Cantilever Beam #2
2( )2
1( )
A
v
wM L x
M L x
,
3
( )1 (1)
( )
v
i M B
A v
W vM M wW dx L x dx
,
4, 0
( )2
( )8
i ML L
L
i M
W dx L x dxEI EIwW L xEI
, 0
4
,
8
From (1), 18
i M
B i M
EIwLv WEI
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Strain Energy DefinitionEnergy
Strain energy U: energy stored in member Complementary energy C: no physical meaning but obeys the Complementary energy C: no physical meaning but obeys the
law of energy conservation
y
U Pdy P
C ydP
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0U Pdy 0C ydP
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Strain Energy RelationshipsEnergy
dU dC
,
Assuming function n
dU dCP ydy dP
P by
1/
0 0
1 ( )y P n
P y n
PU Pdy dPn b
0 0
D t i d f
y nC ydP n by dy
U C
Determine and for linear elastic material
U C
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Complementary Energy PrincipleEnergy
For an elastic body in equilibrium under the action ofFor an elastic body in equilibrium under the action of applied forces, the true internal forces (or stresses) and reactions are those for which the total complementary energy has a stationary valueenergy has a stationary value.
Compatibility
1
0n
t e i r rVr
W W W ydP P
1
01
( ) ( ) 0
rnP
i e r rVr
C C ydP P
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Example Deflection #1Energy
Determine the deflection, 2cross sectional area A = 1800 mm2,
E = 200 GPa.
k FL FC
212 2
0i i i
i i i
FL FCP AE P
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Example Deflection #2Energy
Real loadImaginary load
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Example Deflection #3Energy
6
, 2 5 21
6
1 1268 10 N mm 3.52 mm(1800 mm )(2 10 N/mm )
1 880 10 N
ki
B v i ii
k
FFLAE P
F
6
, 2 5 21
1 880 10 N mm 2.44 mm(1800 mm )(2 10 N/mm )
ki
D h i ii
FFLAE P
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Example SI Problem #1Energy
Redundant
01
ik F
i ii
C dF P
member
1
10
ik
ii
i
dFCR R
1
1 10 (4.83 2.707 ) 0
0 56
ki
i ii
FFL RL PLAE R AE
R P 0.56R P
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Unit Load Description Energy
,0 ,1k
i i iF F L
0
With applied dummy load i
fk nF
i i r r
P
C dF P
1
Ci i iAE
Real load
01 1
10
i i r ri r
ki
i Cif f
FCP P
Imaginary load
M M
1
1
if fk
iC i
i f
FP
0 1
0 1
M
T
M Mdz
EIT T
dzAssume unit load instead of fP
T dzGJ
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Example Unit Load #1Energy
Determine displacement at D
0 1 0 1x
M M T Tds ds
EI GJ x EI GJ
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Example Unit Load #1Energy
2 4l wl x wldx
04
211 1( )
24 2
x
y
dxEI EI
wlEI GJ
4
24 21 1( )
6 2z
EI GJ
wlEI GJ
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Flexibility Method Description Energy
Remove a redundant member to formulate a SD problem Solve for displacements of the SD problem Determine redundant load that negate the same
displacementsp
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Exercise Flexibility Method #1Energy
0, 1,j j jn n F F LF F L
, ,
, 1,
1 1
2
j j jn na j j j
BDj j
F F LAEAE
F L
1,
2
1
jn
jBD
j
F La
AE
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0BD BD BDX a
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Exercise Flexibility Method #2Energy
2 71 4 82PL L
2.71 4.82,
From 0
BD BD
BD BD BD
PL LaAE AE
X a 0.56 AnsBDX P
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Potential Energy Total Potential EnergyEnergy
Total potential energy TPE is the sum of its strain (internal) energy U and the potential energy V of the applied external loadsgy p gy pp
Zero potential energy at the unloaded state
1 1
( )r
n n
r rr r
V V P
0
TPE
TPE
( )
y
n
U V P
U PU V
dy Py
1
TPE ( )rr
rU PU V
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Potential Energy StabilityEnergy
( ) 0U V
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Exercise TPE #1Energy
Assume sinB
zv vL
0 at 0 and and / 0 at / 2B
Lv z z Lv v dv dx z L
2 2
2
2 4 2 4
2M d vU dz EIEI dz
EIEI
38
2 4 2 42
4 3sin2 4
B Bv v EIEI zU dzLL L
39
Exercise TPE #2Energy
2 4
3
44B
Bv EITPE U V WvL
4
3
3 3
( ) 04
2
BB
B
v EIU V vv L
WL WL
39
4
2 as compared to exact solution 48B
WL WLvEIEI
40
Principle of Superposition DescriptionEnergy
If th b d i li l l ti thIf the body is linearly elastic, the effect of a number of forces is the sum of the effects of the forces applied separately.
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Reciprocal Theorem DescriptionEnergy
1Total deflection at point 1 in the direction of from all loadsP
influence or flexibility coefficientija
a P a P a P
1 11 1 12 2 1
2 21 1 22 2 2
......
n n
n n
a P a P a Pa P a P a P
1 1 2 2 ...n n n nn na P a P a P
1 11 12 1 1
2 21 22 2 2
n
n
a a a Pa a a P
41
1 2n n n nn na a a P
ij jia a
42
Exercise Reciprocal Theorem #1Energy
The 800 mm-long beam is propped at 500 mm, giving
, g g0 mm at 0 mm
0.3 mm at 100 mmv xv x
1.4 mm at 200 mm2.5 mm at 300 mm1 9 mm at 400 mm
v xv xv x
1.9 mm at 400 mm0 mm at 500 mm2.3 mm at 600 mm
v xv xv x 4.8 mm at v x
700 mm10.6 mm at 800 mmv x
42BDetermine when the applied loads change.
43
Exercise Reciprocal Theorem #2Energy
due to 40 N at 1.4 mm due to 40 N at 1.4 mm
D
C
v Cv D
due to 30 due to 30 N at 1.4 (3 / 4)
1.05 mmN at C
C
v Dv D
due to 10 N at 2.4 (1/due to 10 N at
4)0 6 mm
Cv Ev E
,
due to 10 N at 0.6 mm1.05 0.6 1.65 C tot
C
al
v Ev
1
mm1.65tan
43
1tan300B
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