Chapter 4, Energy Methods - University of Cincinnatipnagy/ClassNotes/AEEM6001 Advanced Strength...
Transcript of Chapter 4, Energy Methods - University of Cincinnatipnagy/ClassNotes/AEEM6001 Advanced Strength...
Chapter 4
Energy Methods • vectorial methods force/displacement, stress/strain equilibrium equations • energy methods deformation work, strain energy variation methods (unit load method, stationary potential energy method) Terminology • displacements: linear (translational) and rotational • forces: linear and rotational (bending and twisting moments)
{ } T, , , , ,xa ya za xa ya zaa F F F M M M⎢ ⎥= ⎣ ⎦F
{ } T, , , , ,a a a xa ya zaa u v w⎢ ⎥Δ = θ θ θ⎣ ⎦
Fyax
zy
uaθxa
va
θza
waFxaMxa
MyaFza
Mza
θya
a
Degrees of Freedom
va
vb
θa
θb
Mza
Fxa
Fya
ua
ub
Mzb
Fxb
Fyb
The overall behavior of a member (element) of a structure is defined by the displacements of the structure's joints (nodes) under the action of forces applied at the joints.
Strains and stresses within individual members can be calculated separately.
The movement of each joint can be described by three translational and three rotational displacement components.
Further reduction of the degrees of freedom (DOF) is possible by neglecting certain deformations, i.e., the DOF is not unique in approximations.
Indeterminacy
physical indeterminacy is always zero
static indeterminacy is the number of “elasticity” equations required for analysis in addition to the equations of equilibrium
kinematic indeterminacy is the number of displacements that are not predetermined
SI UR EE= −
UR R KR= −
KI KR ND= −
SI Static Indeterminacy
UR number of Unknown Reaction forces
EE number of Equilibrium Equations
R number of Reaction forces
KR number of Known Reaction forces
KI Kinematic Indeterminacy
ND Negligible Displacements
Examples of Indeterminacy (Bending Only)
loading force
unknown reaction force (UR)
kinematically indeterminate displacement (KI)
static indeterminacy: 3 - 3 = 0, kinematic indeterminacy: 3 - 1 = 2
static indeterminacy: 4 - 3 = 1, kinematic indeterminacy: 5 - 2 = 3
static indeterminacy: 6 - 3 = 3 kinematic indeterminacy: 0 – 0 = 0
static indeterminacy: 6 - 3 = 3, kinematic indeterminacy: 6 – 3 = 3
Work and Energy
Δ
F
dWF1
Δ1
W
dΔ
W*
Δ
F
F1
Δ1dΔ
F = kΔ
1
0W F d U
Δ= Δ =∫ ,
1
0* *
FW dF U= Δ =∫ , 1 1* *W W U U F+ = + = Δ
linear behavior
2 2
1 1 1 11 1 1*2 2 2
W W F k Fk
= = Δ = Δ =
vectorial form
1 1 ( )2 2 x y zW F u F v F w= ⋅ + +F Δ =
structure with n DOFs
{ } T1 2 1 2
1 1 , , , ,2 2 n nW F F F= = Δ Δ ⋅⋅ ⋅Δ ⋅ ⋅ ⋅⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦FΔ
[ ]{ }12
W U= =⎢ ⎥⎣ ⎦ KΔ Δ
free DOFs
{ } { }1 12 2f f f ff fW U⎢ ⎥ ⎢ ⎥ ⎡ ⎤= = =⎣ ⎦ ⎣ ⎦ ⎣ ⎦F KΔ Δ Δ
Method of Virtual Displacement/Work Virtual Displacement
imaginary, very small change in the configuration relative to the equilibrium condition
neither external loads nor internal forces are altered by such a virtual displacement
Total Potential
0( )loadU WΠ = − +Π or loadd dU dWΠ = −
( )internal loadd dW dWΠ = − + or d dVΠ = −
internaldU dW= −
internal loaddV dW dW= +
Π total potential of the system
U strain energy of the deformed body
Wload work done by the external loads
internalW work done by the internal forces
V virtual work done by external loads and internal forces during a virtual displacement
Stable Equilibrium
Fourier inequality 0dV ≤
Minimum potential 0dΠ ≥
Normalized Lattice Distance
Pote
ntia
l Ene
rgy
[a. u
.]
0 1 2
typical
parabolic
potential well
Reciprocity
flexibility and stiffness matrices are symmetric
local
[ ] [ ]T=d d or ij jid d=
[ ] [ ]T=k k or ij jik k=
global
[ ] [ ]T=D D or ij jiD D=
[ ] [ ]T=K K or ij jiK K=
1
2
F1
F2
Δ1
Δ2
1 11 12 1
2 21 22 2
d d Fd d F
Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ⎩ ⎭ ⎩ ⎭⎣ ⎦
1
2
F1
Δ11
Δ21
Δ12
Δ22
1
2
F2
11 11 12 1
21 21 22 0d d Fd d
Δ⎧ ⎫ ⎡ ⎤ ⎧ ⎫=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ ⎩ ⎭⎩ ⎭ ⎣ ⎦
12 11 12
22 21 22 2
0d dd d F
Δ⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥Δ⎩ ⎭ ⎩ ⎭⎣ ⎦
1,2 11 1 12 1 22 21 12 2
W F F F= Δ + Δ + Δ
2,1 22 2 21 2 11 11 12 2
W F F F= Δ + Δ + Δ
1,2 2,1W W=
12 1 21 2F FΔ = Δ
12 21d d=
Maxwell's Reciprocal Theorem
andij ji ij jid d k k= =
Strain Energy of Beams
oV
U U dV= ∫
2 2 2
2 2 2
1 [ 2 ( )]21 ( )
2
o x y z x y y z x z
xy yz xz
UE
G
= σ + σ + σ − ν σ σ + σ σ + σ σ
+ τ + τ + τ
• straight (approximately) prismatic bar • bisymmetrical cross section
symmetrical bisymmetrical
Cx
y
S
x
y
C, S
cross sectional area A dx dy= ∫∫
first moments xQ y dx dy= ∫∫ and yQ x dx dy= ∫∫
second moments 2xI y dx dy= ∫∫ , 2
yI x dx dy= ∫∫ , and xyI x y dx dy= ∫∫
centroid 0x yQ Q= =
symmetrical 0xyI = ( x xMθ ∝ and y yMθ ∝ )
shear center ( ) 0z zy zxM x y dx dy= τ − τ =∫∫ in shear
extension
2 2 2
20 0 02 22
L L L
A A
P PU dAdx dAdx dxE EAEA
σ= = =∫ ∫∫ ∫ ∫∫ ∫
torsion
2 2 22
20 0 02 22
L L L
A A
T TU dAdx r dAdx dxG GJGJ
τ= = =∫ ∫∫ ∫ ∫∫ ∫
2
0 2
L TU dxGK
= ∫
bending
2 2 22
20 0 02 22
L L L
A A
M MU dAdx y dAdx dxE EIEI
σ= = =∫ ∫∫ ∫ ∫∫ ∫
shear-bending
2 2 2 2 2
2 2 2 20 0 02 2 2
L L L
A A A
V Q V A QU dAdx dAdx dAdxG G GAI t I t
τ= = =∫ ∫∫ ∫ ∫∫ ∫ ∫∫
2
0 2
L VU k dxGA
= ∫
2
2 2A Qk dy dzI t
= ∫∫
combined loading
2 22 22 2
0 2 2 2 2 2 2
L y yz zy z
y z
M VM VP TU k k dxEA GK EI EI GA GA
⎧ ⎫⎪ ⎪= + + + + +∫ ⎨ ⎬⎪ ⎪⎩ ⎭
Shear Stress Distribution in Bending
τ
τ = 0
τ = 0
VV
x
z
y
z z + dz
yτyz C
σz
A'
t
σzτzy
equilibrium in the z-direction
' '( , ') ' ( , ') ' ( ) ( ) 0z z zy
A Az dz y dA z y dA y t y dzσ + − σ − τ =∫∫ ∫∫
( )( , ') 'x
zx
M zz y yI
σ =
'
( ) ( ) ' ' ( ) ( ) 0x xzy
x A
M z dz M z y dA y t y dzI
+ −− τ =∫∫
'( ) ' '
AQ y y dA= ∫∫
( ) ( ) ( )( )
( )x x
zyM z dz M z Q yy
dz I t y+ −
τ =
xdMV
dz=
( )( )( )xy
V Q yyI t y
τ =
Transverse Shear Factor
2
2 2A Qk dy dzI t
= ∫∫
rectangular cross section
h/2
z
y
y( /2 + )/2h y
b
ht
Qh/2h/2
z
y
y( /2 + )/2h y( /2 + )/2h y
b
ht
Q
A hb= 3
12h bI =
22( ) ( ) ( ) ( )
2 2 2 2 4 2h h b h h bQ y b y y y y= − = − + = −
t b=
2 2 2/ 22 2 2 2
2 2 5 5/ 2
36 36( ) ( )4 4
h
h
A Q h hk dy dz y dy dz y dyI t h b h −
= = − = −∫∫ ∫∫ ∫
/ 24 2 2 4 2 3 5/ 2 4
5 50 0
72 72( )16 2 16 6 5
hh h h y h y h y yk y dyh h
⎡ ⎤= − + = − +⎢ ⎥∫
⎢ ⎥⎣ ⎦
65
k =
Unit Load Method
A
ΔA
loadingforces
PA
, , 0A AV
W W U U dV= = = ∫
2 2 2
2 2 2
1 [ 2 ( )]21 ( )
2
o x y z x y y z x z
xy yz xz
UE
G
= σ + σ + σ − ν σ σ + σ σ + σ σ
+ τ + τ + τ
A
x x xσ = σ + σ , Ay y yσ = σ + σ , ...
A A
o o o oU U U U= + +
1 [ 2 ( )]
1 ( )
A A A A A A Ao x x y y z z x y y z z x
A A Axy xy yz yz zx zx
UE
G
= σ σ + σ σ + σ σ − ν σ σ + σ σ + σ σ
+ τ τ + τ τ + τ τ
A A
o A AV
U U dV P= = Δ∫
For a unit load (PA = 1):
A AA o A
VD U U dV= = = Δ∫
General case of combined tension torsion, bending, and shear:
0
L y y y y yz z z z z
y z
M m k V vN n T t M m k V vD dxE A G K E I E I G A G A
⎛ ⎞= + + + + +⎜ ⎟∫ ⎜ ⎟
⎝ ⎠
Example 1:
x
q0
L
y, v
P = 1A
x
q0
L
y, v
P = 1AP = 1A
202
q xM = , m x=
2 2 2
0 0 0 0
( )2 2 2
L L L LM m M m M mU dx dx dx dxEI EI EI EI+
= = + +∫ ∫ ∫ ∫
4
30 0
0 02 8
L LAA
q q LM mv U dx x dxE I E I E I
= = = =∫ ∫
Example 2:
1P =B
x
q0
L
y, v
1P =B 1P =BP =B
x
q0
L
y, v
20
0 2q xM q L x= − , m x= −
42 3 4
0 0 0
0 0 0
5( )2 3 8 24
LL LB
q q q LM m x L x xv dx L x x dxE I E I E I E I
⎡ ⎤= = − − = − − = −⎢ ⎥∫ ∫
⎢ ⎥⎣ ⎦
Example 3:
45°B1P =B
x
a
y, v
aP
A C
D
k
Nn
from A to B:
2P xM = ,
2xm = − ,
2PN = , 1
2n = −
for the whole system:
32
0 02
2 2 2 6
a aB
N n M m P P P P av dx x dxk E I k E I k E I
= + = − − = − −∫ ∫
Curved Members
circular ring in the x-y plane
zx
y
R
Fx
Cx
Fz
CzFy
Cy
VR
MR
VzMz
T
N
φz
x
y
R
Fx
Cx
Fz
CzFy
Cy
VR
MR
VzMz
T
N
φ
Equilibrium Equations:
sin cosx yN F F= φ − φ
cos sinR x yV F F= − φ − φ
z zV F= −
sin cos sinR z x yM F R C C= φ − φ − φ
sin (1 cos )z x y zM F R F R C= − φ − − φ −
(1 cos ) sin cosz x yT F R C C= − − φ + φ − φ
Slender Uniform Half-Ring
P, 1
A
B
1•R
1•R
φ
P
R
A
B
C1
P, 1
A
B
1•R
1•R
φ
P
R
A
B
C1
0 0
L LR R
CM m T tD d d
E I G K= +∫ ∫
/ 2 / 2
0 0
R RC
M m T tD R d R dE I G K
π π= φ + φ∫ ∫
Load effect:
sinRM P R= − φ
(1 cos )T P R= − φ
Unit load effect:
sin ( )cos ( )sin cosRm R R R R= − φ − φ + φ = − φ1 1i i
(1 cos ) ( )sin ( ) cos (1 sin )t R R R R= − φ + φ + φ = + φ1 1i i
Displacement at C:
3 3/ 2 / 2
0 0sin cos (1 cos ) (1 sin )C
P R P RD d dE I G K
π π= φ φ φ + − φ + φ φ∫ ∫
3 312 2CP R P RD
E I G Kπ −
= +
Castigliano's Theorems
Castigliano's first theorem
0loadd dU dWΠ = − =
kk
U P∂=
∂Δ, k
k
U M∂=
∂θ
Castigliano's second theorem
*k
k
UP
∂= Δ
∂,
*k
k
UM
∂= θ
∂
Example:
x
q0
L
y, v
PA
x
q0
L
y, v
PAPA
2202
0 0
( )( ) 2*2 2
L L Aq x P xM mU dx dx
EI EI
++= =∫ ∫
2
0
0
0
( )* 2
A
L AA
A
P
q x P x xUv dxP EI
=
+∂= = ∫
∂
4
30 0
02 8
LA
q q Lv x dxE I E I
= =∫
Finite Element Method
va
vb
θa
θb
Mza
Fxa
Fya
ua
ub
Mzb
Fxb
Fyb
The overall behavior of a member (element) of a structure is defined by the displacements of the structure's joints (nodes) under the action of forces applied at the joints.
Strains and stresses within individual members can be calculated separately.
Structural analysis:
1. Basic mechanics; fundamental constitutive, compatibility, and equilibrium equations
2. Finite element mechanics; exact or approximate solutions of the
fundamental equations for an element 3. Equation formulation; establishment of the governing algebraic
equations for the structure 4. Equation solution; computational methods and algorithms 5. Solution interpretation; presentation of the results for design or
analysis
Example: Beam in Bending
va
vbθa
L
θb
Vb
Va
MaMb
1. Basic mechanics
x,u
y,v
V(x+dx)V(x)
M(x) M(x+dx)
compatibility relationship: y ′′ε = − v
constitutive relationship:
Eσ = ε
fundamental bending equation:
''M E I v=
equilibrium equations:
' '''M E I v V= = −
' '''' 0V E I v= − =
general solution
2 30 1 2 3v a a x a x a x= + + +
boundary conditions
0(0)av v a= =
1'(0)a v aθ = =
2 3
0 1 2 3( )bv v L a a L a L a L= = + + +
21 2 3'( ) 2 3b v L a a L a Lθ = = + +
specific solution
0 aa v=
1 aa = θ
2 2 23 3 2 1
a b a ba v vL LL L
= − + − θ − θ
3 3 3 2 22 2 1 1
a b a ba v vL L L L
= − + θ + θ
2 3
0 1 2 3v a a x a x a x= + + +
internal displacement
2 3 2 3 2 3 2 3
2 3 2 2 3 23 2 2 3 2(1 ) ( ) ( ) ( )a a b b
x x x x x x x xv v x vL LL L L L L L
= − + + − + θ + − + − + θ
2 3( ) (2 6 )M x E I a a x= +
3( ) 6V x E I a x= −
local stiffness matrix
2 2
3
2 2
12 6 12 6
6 4 6 212 6 12 6
6 2 6 4
a a
a a
b b
b b
L LV vM L L L LE IV vL LLM L L L L
−⎡ ⎤⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪θ−⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥− − −⎪ ⎪ ⎪ ⎪⎢ ⎥
⎪ ⎪ ⎪ ⎪θ⎢ ⎥⎩ ⎭ ⎩ ⎭−⎣ ⎦
bending strain energy
2
0 2
L MU dxEI
= ∫
2
2 22 3 ( ) 3b a b aa a b b a b
E I v v v vUL L L
⎡ ⎤− −⎛ ⎞= θ + θ θ + θ − θ + θ +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦
Example
P
a b
1 2 3
U WΠ = −
1 1 3 3 0v v= θ = = θ =
2 22 22 2 2 22 2 2 2 22 2
2 23 3 3 3E I v v E I v v P va a b ba b
⎡ ⎤ ⎡ ⎤Π = θ − θ + + θ + θ + −⎢ ⎥ ⎢ ⎥
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
20
v∂Π
=∂
and 2
0∂Π=
∂θ
2 23 3 2 21 1 1 112 6E I v E I Pb a b a
⎛ ⎞ ⎛ ⎞+ + − θ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2 22 21 1 1 16 4 0E I v E I
b ab a⎛ ⎞ ⎛ ⎞− + + θ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
3 3
2 33 ( )P a bv
E I a b= −
+
2 2
2 3( )
2 ( )P a b a b
E I a b−
θ =+