Description
The objective of this chapter is to explain the principle of minimum potential energy and its application in the elastic analysis of structures. Two fundamental notions of the finite element method viz. discretization and numerical approximation of the exact solution are also explained.
2.1 The principle of Minimum Potential Energy (MPE)
Deformation and stress analysis of structural systems can be accomplished using the principle of Minimum Potential Energy (MPE), which states that
For conservative structural systems, of all the kinematically admissible deformations, those corresponding to the equilibrium state extremize (i.e., minimize or maximize) the total potential energy. If the extremum is a minimum, the equilibrium state is stable.
Let us first understand what each term in the above statement means and then explain how this principle is useful to us.
A constrained structural system, i.e., a structure that is fixed at some portions, will deform when forces are applied on it. Deformation of a structural system refers to the incremental change to the new deformed state from the original undeformed state. The deformation is the principal unknown in structural analysis as the strains depend upon the deformation, and the stresses are in turn dependent on the strains. Therefore, our sole objective is to determine the deformation. The deformed state a structure attains upon the application of forces is the equilibrium state of a structural system. The Potential energy (PE) of a structural system is defined as the sum of the strain energy (SE) and the work potential (WP).

PE = SE +WP
(1)
The strain energy is the elastic energy stored in deformed structure. It is computed by integrating the strain energy density (i.e., strain energy per unit volume) over the entire volume of the structure.

SE = ∫ (strain energy density) dV
(2)
V
The strain energy density is given by

Strain energy density =
1
(stress)(strain)
(2a)
2
Ananthasuresh, IISc
The work potential WP, is the negative of the work done by the external forces acting on the structure. Work done by the external forces is simply the forces multiplied by the displacements at the points of application of forces. Thus, given a deformation of a structure, if we can write down the strains and stresses, we can obtain SE, WP, and finally PE. For a structure, many deformations are possible. For instance, consider the pinnedpinned beam shown in Figure 1a. It can attain many deformed states as shown in Figure 1b. But, for a given force it will only attain a unique deformation to achieve equilibrium as shown in Figure 1c. What the principle of MPE implies is that this unique deformation corresponds to the extremum value of the MPE. In other words, in order to determine the equilibrium deformation, we have to extremize the PE. The extremum can be either a minimum or a maximum. When it is a minimum, the equilibrium state is said to be stable. The other two cases are shown in Figure 2 with the help of the classic example of a rolling ball on a surface.
(a) (b) (c)
Figure 1 The notion of equilibrium deformed state of a pinnedpinned beam
Stable Unstable Neutrally stable
Figure 2 Three equilibrium states of a rolling ball
There are two more new terms in the statement of the principle of MPE that we have not touched upon. They are conservative system and the kinematically admissible deformations. Conservative systems are those in which WP is independent of the path taken from the original state to the deformed state. Kinematically admissible deformations are those deformations that satisfy the geometric (kinematic) boundary conditions on the structure. In the beam example above (see Figure 1), the boundary conditions include zero displacement at either end of the beam. Now that we have defined all the terms in the statement, it is a good time to read it again to make more sense out of it before we apply it.
2.2 Application of MPE principle to lumpedparameter uniaxial structural systems
Consider the simplest model of an elastic structure viz. a mass suspended by a linear spring shown in Figure 3. We would like to find the static equilibrium position of the mass when a force F is applied. We will first use the familiar forcebalance method, which gives
F = spring force = kx 
at equilibrium ( k is the spring constant) 

∴ 
^{x}equilibrium 
= δ = 
F 
(3) 

k 

x
F
δ
Figure 3 Simplest model of an elastic structural system
We can arrive at the same result by using the MPE principle instead of the forcebalance method. Let us first write the PE for this system.

1
2
+ (− Fx) =
1
2
PE = (SE) + (WP) =
kx
kx
− Fx
(4)
2
2
As per the MPE principle, we have to find the value of x that extremizes PE. The condition for extremizing PE is that the first derivative of PE with respect to x is zero.

d (PE)
= 0 ⇒ kx − F = 0 ⇒ x_{equilibrium} = δ =
F
(5)
dx
k
We got the same result as in Equation (3). Further, verify that the second derivative of PE with respect to x is positive in this case. This means that the extremum is a minimum and therefore the equilibrium is stable.
Figure 4 pictorially illustrates the MPE principle: of all possible deformations (i.e., the values of x here), the stable equilibrium state corresponds to that x which minimizes PE. For the assumed values of k

5, and F = 10, equilibrium deflection is 2 which is consistent with Figure 4. As illustrated in Figure 3, the MPE principle is an alternative way to write the equilibrium equations for elastic systems. It is, as we
will see, more efficient than the forcebalance method. Let us now consider a second example of a springmass system with three degrees of freedom viz. q_{1}, q_{2}, and q_{3}. The number of degrees of freedom of a system refers to the minimum number of independent scalar quantities required to completely specify the system. It is easy to see that the system shown in Figure 5 has three degrees of freedom because we can independently move the three masses to describe this completely.
PE of a springmass system





10
5
k = 5 and F = 10
PE
0
5
10
0
2 _{x}
4




Figure 4 PE of a springmass system
k_{1 }^{ q}^{1}






1
F_{1}
2
q_{2}
q_{3}





k_{2}
3
k_{3 }^{ k}^{4}
Figure 5 A springmass system with three degrees of freedom
We will use the MPE principle to solve for the equilibrium values of q_{1}, q_{2}, and q_{3} when forces F_{1} and F_{3} are applied (Note that one can also apply F_{2}, but in this problem we assume that there is no force on mass 2). In order to write the SE for the springs, we need to write the deflection (elongation or contraction) of the springs in terms of the degrees of freedom q_{1}, q_{2}, and q_{3}.

^{u}1
= q_{1} − q_{2}
u_{2}
= q_{2}
(6)
^{u}3
= q_{3} − q_{2}
u_{4}
= −q_{3}
The PE for this system can now be written as
(bold letters indicate that they are either vectors or matrices.)
The matrix K is referred to as the stiffness matrix of a structural system. Any linear elastic structural system can be represented as Equation (9b). We will see later that the finite element method enables us to construct the matrix K, and vectors q and F systematically for any complex structure.
Exercise 2.1
Use MPE principle and the forcebalance method to obtain the equilibrium equations shown in the matrix representation in Figure 7.



1
k_{4}F_{3}
3
F_{1}
2
F_{3}
k3
F2
k_{1}


k5
k2

A, E, l
F
k = (AE/l)
x
F
δ
Figure 8 Axially loaded bar as a springmass system
Now, we can also analyze a stepped bar (a bar with two different crosssection areas) under two concentrated forces F_{1} and F_{2} as shown in Figure 9.
segments, the number of degrees of freedom increases (i.e., more q’s) and the size of the stiffness matrix increases. But, the procedures for doing two bar segments in Figure 9 or many bar segments in Figure 10 are exactly the same, except that it is repetitive and tedious as the number of segments increases. However, it is ideal for implementation on a computer. Notice that k for each segment is of the identical form.






Tapering bar loaded
A lumped model
with its own weight
.
.
.





Figure 10 A tapering bar loaded with its own weight, and its lumped springmass model
This example illustrates two important concepts.

Continuous systems can be approximated as lumped segments. This is called discretization an important concept in FEM. The segments are called “finite elements”.

All elements have the identical form. So, a general method can be developed to handle large and complex structures. That is, by discretizing the structure into identical elements, the whole structure can be analyzed in a repetitive manner systematically.
What we have done in this Chapter is not FEM yet. It suffices to note at this point that FEM provides a systematic way of discretizing a complex structure to get an approximate solution. In addition to the intuitive notion presented in this chapter, there is a firm theoretical basis for FEM. We will examine
that in the later Chapters. Before we embark upon FEM formulation, it is worthwhile to discuss another important concept method called the RayleighRitz method. That is the topic of the next Chapter.
2.4 Implementation of massspring systems in Matlab
The finite element method is a numerical method. It is important to understand the practical implementation of it in addition to gaining a theoretical understanding of it. This notes emphasizes this aspect and includes finite element programs written in Matlab. You can do this in Maple, Mathematica, MathCad or anything else you are comfortable with. In order to be prepared to handle the finite element programs later, let us get started here with a simple Matlab script to solve the problem shown in Figure 11.
Exercise 2.3
Write down the matrix equation system for the system shown in Figure 11 and study its implementation in the attached Matlab script. Run the script to get experience with Matlab.
% Brass tube 

E2 
= 100E9; 
% Pa 

A2 
= (pi/4)*( (150E3)^2 – (100E3)^2 ) ; 
% m^2 

L2 
= 1.25; 
% m 

% Steel pipe 

E3 
= 210E9; 
% Pa 

A3 
= (pi/4)*( (200E3)^2 – (125E3)^2 ); 
% m^2 

L3 
= 0.75; 
% m 

% Forces 

F = [650 850 1500]*1e3; 
% N 

Compute the spring constants k1 = A1*E1/L1;
k2 = A2*E2/L2;
k3 = A3*E3/L3;

Construct the stiffness matrix of the system K(1,1) = k1;
K(1,2) = k1;
K(1,3) = 0;
K(2,1) = k1;
K(2,2) = k1+k2;
K(2,3) = k2;
K(3,1) = 0;
K(3,2) = k2;
K(3,3) = k2+k3;

Solve for displacements u = inv(K)*F’
____________________________________
Exercise 2.4
Solve the linearly tapering bar problem by using a Matlab script. The advantage of writing in Matlab (or other similar software) is that we can vary the number of elements (i.e., the “fineness” of discretization) and observe what happens. Assume the following data.
The bar is made of aluminum (E = 73 GPa, mass density = 2380 Kg/m^{3}), and has a circular crosssection with beginning diameter of 100 mm and tip diameter of 20 mm. The length of the bar is 1 m.
____________________________________
Matlab script 2 for Exercise 2.4
clear all 

clc 

%clg 

%hold off 

axis 
normal 

% Tapering aluminum bar under its own weight 

E = 
73E9; 
% Pa 
A0 = 
(pi/4)*(100E3)^2; 
% m^2 
At = 
(pi/4)*(20E3)^2; 
% m^2 
L = 
1.0; 
% m 
rho 
= 2380*9.81; 
% N/m^3 
echo 
on 

N = 
2; 
% Number of elements 

Change the number of elements and see the how the accuracy

of the solution improves. You need to run the script many

times by changing the number of element N, above.

Note that the hold on graphics is on.
echo off

Compute element length, area, k and force Le = L/N;
for i = 1:N,
Atop = A0 (A0At)/N*(i1); Abot = A0 – (A0At)/N*i; A(i) = (Atop+Abot)/2; x(i) = L/N*i;
k(i) = A(i)*E/Le; F(i) = A(i)*Le*rho;
end

Assembly of the stiffness matrix using k’s. K = zeros(N,N);
K(1,1) = k(1) + k(2);
K(1,2) = k(2);
for i = 2:N1,
K(i,i1) = k(i);
K(i,i) = k(i) + k(i+1);
K(i,i+1) = k(i+1);
end
K(N,N1) = k(N);
K(N,N) = k(N);

Solve for displacements {q}. It is a column vector. q = inv(K)*F’;
plot([0 x],[0; q],’w’,x,q,’c.’);
hold on
title(‘Effect of Discretization’); xlabel(‘X — the length of the bar (m)’); ylabel(‘Axial deformation (m)’);
____________________________________
Figure 12 Effect of discretization in the lumpedmodel
It can be seen in the figure that as the number of elements increases, the solution begins to converge to the exact solution. More about this in the next chapter.