![2.10
% Brass tube
E2 = 100E9; % Pa
A2 = (pi/4)*( (150E-3)^2 - (100E-3)^2 ) ; % m^2
L2 = 1.25; % m
% Steel pipe
E3 = 210E9; % Pa
A3 = (pi/4)*( (200E-3)^2 - (125E-3)^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/m3
), and has a circular cross-section
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](https://image.slidesharecdn.com/minimumpotentialenergy-180423140827/85/Minimum-potential-energy-10-320.jpg)
![2.11
clear all
clc
%clg
%hold off
axis normal
% Tapering aluminum bar under its own weight
E = 73E9; % Pa
A0 = (pi/4)*(100E-3)^2; % m^2
At = (pi/4)*(20E-3)^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 -(A0-At)/N*(i-1);
Abot = A0 - (A0-At)/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:N-1,
K(i,i-1) = -k(i);
K(i,i) = k(i) + k(i+1);
K(i,i+1) = -k(i+1);
end
K(N,N-1) = -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](https://image.slidesharecdn.com/minimumpotentialenergy-180423140827/85/Minimum-potential-energy-11-320.jpg)
Minimum potential energy
- 1. Ananthasuresh, IISc Chapter 2 The Principle of Minimum Potential Energy 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). WPSEPE += (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. ∫= V dVdensityenergystrainSE )( (2) The strain energy density is given by ))(( 2 1 strainstressdensityenergyStrain = (2a)
- 2. 2.2 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 pinned-pinned 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 pinned-pinned 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 lumped-parameter uniaxial structural systems
- 3. 2.3 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 force-balance method, which gives kxforcespringF == at equilibrium ( k is the spring constant) ∴ k F x mequilibriu == δ (3) x δ F Figure 3 Simplest model of an elastic structural system We can arrive at the same result by using the MPE principle instead of the force-balance method. Let us first write the PE for this system. ( ) FxkxFxkxWPSEPE −=−+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =+= 22 2 1 2 1 )()( (4) 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. k F xFkx dx PEd mequilibriu ==⇒=−⇒= δ00 )( (5) 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 force-balance method. Let us now consider a second example of a spring- mass system with three degrees of freedom viz. q1, q2, and q3. 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.
- 4. 2.4 0 2 4 -10 -5 0 5 10 PE of a spring-mass system x PE k = 5 and F = 10 Figure 4 PE of a spring-mass system q1 F1 q3 q2 k1 k2 k3 k4 1 2 3 Figure 5 A spring-mass system with three degrees of freedom We will use the MPE principle to solve for the equilibrium values of q1, q2, and q3 when forces F1 and F3 are applied (Note that one can also apply F2, 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 q1, q2, and q3. 34 233 22 211 qu qqu qu qqu −= −= = −= (6) The PE for this system can now be written as ( )3311 2 44 2 33 2 22 2 11 2 1 2 1 2 1 2 1 qFqFukukukukPE −−+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +++=
- 5. 2.5 ( ) ( ) ( )3311 2 34 2 233 2 22 2 211 2 1 2 1 2 1 2 1 qFqFqkqqkqkqqkPE −−+⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−++−= (7) For equilibrium, PE should be an extremum with respect to all three q’s. That is, 0 )( = ∂ ∂ iq PE for i = 1, 2, and 3. (8a) i.e., 0)( )( 1211 1 =−−= ∂ ∂ Fqqk q PE (8b) 0)()( )( 23322211 2 =−−+−−= ∂ ∂ qqkqkqqk q PE (8c) 0)( )( 334233 3 =−+−= ∂ ∂ Fqkqqk q PE (8d) Noting the relationship between q’s and u’s from Equation (6), we can readily see that the equilibrium equations obtained in Equations (8) can be directly obtained from force-balance on the three masses as shown in Figure 6. F1 k1 u1 k2 u2 1 2 3 k1 u1 k3 u3 k3 u3 k4 u4 Figure 6 Force-balance free-body-diagrams for the system in Figure 5 It is important to note that Equations (8) were obtained routinely from the MPE principle where as force-balance method requires careful thinking about the various forces (including the internal spring reaction forces and their directions. Thus, for large and complex systems, the MPE method is clearly advantageous, especially for implementation on the computer. The linear Equations (8) can be written in the form of matrix system as follows: ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ +− −++− − 3 1 3 2 1 433 33211 11 0 0 0 F F q q q kkk kkkkk kk (9a) or FKq = (9b)
- 6. 2.6 (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 force-balance method to obtain the equilibrium equations shown in the matrix representation in Figure 7. F3 1 2 3 k1 k2 k3 k4 k5 F3 F2 F1 ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ +−− −++−− −−−+++ 3 2 1 3 2 1 5454 553232 4324321 F F F q q q kkkk kkkkkk kkkkkkk Figure 7 A spring-mass system and its equilibrium matrix equation 2.3 Modeling axially loaded bars using the spring-mass models The spring-mass model is useful in arriving at the equilibrium equations for an axially loaded bar as shown in Figure 8. For a bar of uniform cross-section A, homogeneous material with Young’s modulus E, and total length l, the spring constant k is given by l AE k = (10) In order to see how we wrote Equation (10), consider the following equations. ( )δδ δ δ k l AE F A Fl strain stress E l strain A F stress =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =⇒==== ;;; (11)
- 7. 2.7 x δ F F A, E, l k = (AE/l) Figure 8 Axially loaded bar as a spring-mass system Now, we can also analyze a stepped bar (a bar with two different cross-section areas) under two concentrated forces F1 and F2 as shown in Figure 9. q1 A1, E1, l1 k1 = (A1E1/l1)F2 F1 A2, E2, l2 k2 = (A2E2/l2) q2 F1 F2 Figure 9 Axially loaded stepped bar and its lumped spring-mass model Exercise 2.2a Solve for q1 and q2 for the system shown in Figure 9 using the MPE principle. Answer: ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − −+ = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ − 2 1 1 22 221 2 1 F F kk kkk q q (12) Exercise 2.2b Repeat Exercise 2.2a when there are three segments. That is, determine the displacements q1, q2, and q3. Answer: ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − −+− −+ = ⎪ ⎭ ⎪ ⎬ ⎫ ⎪ ⎩ ⎪ ⎨ ⎧ − 3 2 1 1 33 3322 221 3 2 1 0 0 F F F kk kkkk kkk q q q (13) Do you see any pattern emerging after working out Exercises 2.2a and b? If you work through more number of segments, we will see the tridiagonal pattern in the stiffness matrix. Let us now proceed to use this for a more realistic problem. Consider the linearly tapering bar loaded with its own weight. This can be easily modeled as a spring mass system. This type of lumped modeling gives only an approximate solution, and as you can imagine, the accuracy improves with increased number of segments. As we increase the number of
- 8. 2.8 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 with its own weight A lumped model Figure 10 A tapering bar loaded with its own weight, and its lumped spring-mass 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
- 9. 2.9 that in the later Chapters. Before we embark upon FEM formulation, it is worthwhile to discuss another important concept method called the Rayleigh-Ritz method. That is the topic of the next Chapter. 2.4 Implementation of mass-spring 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. 850kN 650 kN 1500 kN Aluminum bar Brass tube Steel pipe The steel pipe ID and OD are 125 mm and 200 mm. The brass tube ID and OD are 100 mm and 150 mm. The aluminum bar cross-section diameter is 100 mm. E-steel = 210 GPa E-brass = 100 GPa E-aluminum = 73 GPa Find the displacements at the points where forces are applied. Figure 11 A composite axially loaded system ____________________________________ Matlab script 1 for Exercise 2.3 clear all clc clg hold off axis normal % Aluminum bar E1 = 73E9; % Pa A1 = (pi/4)*(100E-3)^2; % m^2 L1 = 1.0; % m
- 10. 2.10 % Brass tube E2 = 100E9; % Pa A2 = (pi/4)*( (150E-3)^2 - (100E-3)^2 ) ; % m^2 L2 = 1.25; % m % Steel pipe E3 = 210E9; % Pa A3 = (pi/4)*( (200E-3)^2 - (125E-3)^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/m3 ), and has a circular cross-section 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
- 11. 2.11 clear all clc %clg %hold off axis normal % Tapering aluminum bar under its own weight E = 73E9; % Pa A0 = (pi/4)*(100E-3)^2; % m^2 At = (pi/4)*(20E-3)^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 -(A0-At)/N*(i-1); Abot = A0 - (A0-At)/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:N-1, K(i,i-1) = -k(i); K(i,i) = k(i) + k(i+1); K(i,i+1) = -k(i+1); end K(N,N-1) = -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
- 12. 2.12 title('Effect of Discretization'); xlabel('X -- the length of the bar (m)'); ylabel('Axial deformation (m)'); ____________________________________ 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 x 10-7 Effect of Discretization X -- the length of the bar (m) Axialdeformation(m) 2 3 5 10 20 50 100 Figure 12 Effect of discretization in the lumped-model 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.