potential energy, - indian institute of technology...
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POTENTIAL ENERGY, ∏.
The total potential energy of an elastic body , is defined as the sum of total strain energy (U) and the work potential (WP) .
∏ = U + WP
For linear elastic materials , the strain energy per unit volume in the body is “ ”
For elastic body total strain energy (U) is
12
Tσ ε
12
TU dvσ ε= ∫
The work potential is given by
The total potential energy for the general elastic body is
. .T T Ti iV S
iWP u fdV u Tds u P= − − −∑∫ ∫
. .12
T T T Ti iV S
i
dv u fdV u Tds u Pπ σ ε= − − −∑∫ ∫ ∫
Principal of minimum potential energy
For conservative systems, of all the kinematically admissible displacement
fields, those corresponding to equilibrium extremize the total potential energy . If the
extremum condition is a minimum , the equilibrium state is stable
2
1
3 4
2
1
3
K2`K1`
K3K4
q1
q3
q2
Example
Figure-1
Figure 1 ,shows a system of spring .
The total potential energy is given by
where δ1 , δ2 , δ3 , and δ4 are extensions of four spring . since
δ1 = q 1 - q 2
δ2 = q 2
δ3 = q 3 - q 2
δ4 = - q 3
2 2 2 21 1 2 2 3 3 4 4 1 1 3 3
1 1 1 12 2 2 2
k k k k F q F qπ δ δ δ δ= + + + − −
we have
where q 1 , q 2 , and q 3 are the displacements of nodes 1 , 2 and 3 respectively
( ) ( )2 22 21 1 2 2 2 3 3 2 4 3 1 1 3 3
1 1 1 12 2 2 2
k q q k q k q q k q F q F qπ = − + + − + − −
• For equilibrium of this 3- DOF system , we need to minimize to ∏ with respect to q 1 , q 2 , and q 3 the three equations are given by
i = 1 ,2 ,3
which are
= k1 (q 1 - q 2) - F1 = 0
0iq
π∂ =∂
1qπ∂∂
= -k1 (q 1 - q 2) + k2 q2 - k3 (q 3 - q 2) = 0
= k3 (q 3 - q 2) + k4 q3 – F3 = 0
Equilibrium equation can be put in the form of
K q = F as follows
K1
-K3
-K1
-K1 K1+ K2+ K3 -K3
0
0 K3+ K4
=
q1
q2
q3
F1
0
F3
.. ….. 1 1
2qπ∂∂
3qπ∂∂
If on the other hand , we proceed to write the equilibrium of the system by considering the equilibrium of each separate node as shown in figure 2We can write
K1δ1 = F1
K2δ2 - K1δ1 - K3δ3 = 0
K3δ3 - K4δ4 = F3
Which is precisely the set of equations represented in Eq-1
We see clearly that the set of equation “1”is obtained in a routine manner using the potential energy approach, without any reference to the free body diagrams .
This make the potential energy approach attractive for large and complex problems .
RAYLEIGH-RITZ METHOD
Rayleigh-Ritz method involves the construction of an assumed displacement field, say
u = ∑ai Φi ( x, y, z) i = 1 to L
v = ∑ajΦj( x, y, z) j = L + 1 to M
w = ∑ak Φk( x, y, z) k = M + 1 to N
N > M > L
Eq-1
The functions Φi are usually taken as polynomials. Displacements u, v, w must satisfy boundary conditions.
Introducing stress-strain and strain-displacement relation Substituting equation – 1 in to ∏ (PE)
∏= ∏ ( a1, a2, a3, ….…….. ar )
where r = no of independent unknowns the extremum with respect to ai,( i = 1 to r) yields the set of r equation
i = 1, 2, 3……… ,r0ia
π∂ =∂
Example
The potential energy of for the linear 1-Drod with body force is neglected , is
where u1= u (x = 1)
( )2
10
1 22
ld uE A d x ud xπ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫
1 1
E = 1, A= 1
X
Y
21
let as consider a polynomial function u = a1 + a2x + a3x3 this must satisfy
u = 0, at x = 0u = 0 at x = 2
thus 0 = a1
0 = a1 +2 a2+ 4a3
Figure- 2
Hence a2 = -2a3u = a3 (-2x + x2)u1 = -a3
then , and
( )2
10
1 22
lduEA dx udxπ ⎡ ⎤= −⎢ ⎥⎣ ⎦∫
23 3
22 23
a a⎛ ⎞= +⎜ ⎟⎝ ⎠
( )32 1du a xdx
= − +
we set
Resulting in a3 = -0.75u1 = - a3 = 0.75
the stress in bar given by
exact solution is obtained if piecewise polynomial interpolation is used in the construction of u .
30a
π∂ =∂
( )1.5 1duE xdx
σ = = −
GALERKIN’S METHOD• Galerkin’s method uses the set of governing
equations in the development of an integral form.
• It is usually presented as one of the weighted residual methods.
• Let us consider a general representation of a governing equation on a region “V”
Lu = PWhere , “L” as operator operating on “u”
For the one-dimensional rid considered in previous example
Governing equation
( ) 0d duEAdx dx =
( )dd EAdx dx⎛ ⎞⎜ ⎟⎝ ⎠
We may consider L as operator, operating on “u”
• The exact solution needs to satisfy “Lu=P”at every point x .• If we seek an approximate solution , if
introduces an error , called the residual
• The approximate methods revolve around setting the residual relative to a weighting function Wi ,i = 0 to n
( )xεu
( )x L u Pε = −
( ) 0iW Lu P dV− =∫
The weighting function Wi are chosen from the basis functions used for constructing
• Here ,we choose the weighting function to be linear combination of the basis function Gi . Specifically ,consider an arbitrary function φ
1
n
i ii
u Q G=
= ∑
Where the coefficient are arbitrary , except for requiring that satisfy boundary conditions were is prescribed.
iφ
uφ
1
n
i ii
Gφ φ=
= ∑
Given by
For elastic materials
∫ =+Φ+∂∂
+∂
∂+
∂∂
Vxx
xzxyx dVfzyx
0......])[( ττσ
∫ ∫ ∫+∂∂
−=∂∂
V V Sx dSndV
xdV
xαθθαθα
∫ ∫ ∫ ∑−−−V V S i
TTTT PTdSfdVdV φφφφεσ )(
Example let us consider the problem of the previous example and solve it by Galerkin’s approach.The equilibrium equation is
• u=0 at x=0• u=0 at x=0Multiplying this differential equation by Integrating by pars, we get
0d duEAdx dx =
Figure- 2
Where is zero at x = 0 and x = 2.
is the tension in the rod ,which takes a jump of magnitude 2 at x = 1 , thus
( ) ( )2 1 2
0 10
0ddu du duEA EA EAdx dx dx dxφ φ φ− + + =∫
φ
duEA dx
2
10
2 0dduEA dx dxφ φ− + =∫
Now we use the same polynomial (basis ) for u and if u1 and are the value at x = 1 ,thus
• Substituting these and E = 1, A = 1 in the previous integral yields
φφ
( )212u x x u= −
( )212x xφ φ= −
( )2
21
0
2 2 2 0u x d xφ⎡ ⎤− − + =⎢ ⎥
⎣ ⎦∫
This is to be satisfied for every .We get
( )1 18 2 03 uφ − + =
1 0.75u =
1φ