Potential Energy
Potential Energy
Potential Energy
Summary:
• Potential energy of a system
• Elastic bar
• String in tension
• Principle of Minimum Potential Energy
• Rayleigh-Ritz Principle
A generic problem in 1D
d 2u
2
x 0; 0 x 1
dx
u 0 at x 0
u 1 at x 1
F x
k k
1 F
u k
u
F = Force in the spring
u = deflection of the spring
k = “stiffness” of the spring
Hooke’s Law
F = ku
Strain energy of a linear spring
dU dU Fdu
F The total strain energy of the spring
u u+du
u
U F du Area under the force dispalceme nt curve
0
Potential energy of the loading (for a single spring as in the figure)
W Fu
x
k
F
k
u
For this system of spring, first write down the total potential
energy of the system as:
1 1
system k1 (d 2 x ) k 2 (d 3 x d 2 x ) 2 Fd 3x
2
2 2
Obtain the equilibrium equations by minimizing the potential
energy
system
k1d 2 x k 2 (d 3 x d 2 x ) 0 Equation (1)
d 2 x
system
k 2 (d 3 x d 2 x ) F 0 Equation (2)
d 3 x
Principle of minimum potential energy for a system of springs
k1 k 2 k 2 d 2 x 0
k
2 k 2 d 3 x F
du
Axial strain ε
dx
du
Axial stress Eε E
dx 2
1 1 du
Strain energy per unit volume of the bar dU σε E
2 2 dx
Strain energy of the bar
1 L 1
U dU σε dV σε Adx since dV=Adx
2 x 0 2
Axially loaded elastic bar
2
1 L du exact L
(u exact ) EA dx bu exact dx Fu exact (x L)
2 0 dx 0
Potential energy of the axially loaded bar corresponding to the
“admissible” displacement w(x)
2
1 dw
L L
(w) EA dx bw dx Fw(x L)
2 0 dx 0
d 2u
AE 2 b 0; 0 xL
dx
u 0 at x 0
du
EA F at x L
dx
Notice
7
since 1
6
(u exact ) (w)
Principle of Minimum Potential Energy
Among all admissible displacements that a body can have, the one
that minimizes the total potential energy of the body satisfies the
strong formulation
The exact solution (uexact) that satisfies the strong form, renders the
potential energy of the system a minimum.
Task is to find the function ‘w’ that minimizes the potential energy
of the system
2
1 dw
L L
(w) EA dx 0 bw dx Fw(x L)
2 0
dx
From the Principle of Minimum Potential Energy, that function
‘w’ is the exact solution.
Rayleigh-Ritz Principle
w( x) a 0 o ( x) a11 ( x) a 2 2 ( x) ...
2
1 dw
L L
(w) EA dx 0 bw dx Fw(x L)
2 0
dx
2
1
L d 0 d1
(a 0 , a 1 ,...) EA a 0 a1 ... dx
2 0
dx dx
b a 0 0 a11 ... dx
L
F a 0 0 ( x L) a11 ( x L) ...
Rayleigh-Ritz Principle
(w)
0, i 0,1,2,...
ai
The approximate solution is
u ( x) a 0 o ( x) a11 ( x) a 2 2 ( x) ...
E=A=1
F x F=2
x=1
x=0 x=2
u a0 a1 x a2 x 2
a2 2 x x 2
Now we apply Rayleigh Ritz principle, which says that if I plug
this approximation into the expression for the potential energy P, I
can obtain the unknown (in this case a2) by minimizing P
2
1 du
2
(u) dx Fu(x 1)
2 0 dx
2
1 2 d
2 0 dx
a2 2 x x 2 dx F a2 2 x x 2
evaluated
at x 1
4 2
a2 2a2
3
0
a2
8
a2 2 0
3
3
a2
4
Hence the approximate solution to this problem, using the
Rayleigh-Ritz principle is
u a0 a1 x a2 x 2
a2 2 x x 2
3
2x x2
4
Notice that the exact answer to this problem (can you prove this?) is
x for 0 x 1
u exact
2 x for 1 x 2
The displacement solution :
Exact solution
1
0.8 Approximate
solution
0.6
0.4
0.2
0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
x
1.5
Exact Stress
0.5
Approximate
stress
Stress
-0.5
-1
-1.5