- 1. 8/29/2023 1 Principles of minimum potential energy and Rayleigh-Ritz So far, structural mechanics using Direct Stiffness approach Finite element method is used to solve physical problems Solid Mechanics Fluid Mechanics Heat Transfer Electrostatics Electromagnetism …. Physical problems are governed by differential equations which satisfy Boundary conditions Initial conditions One variable: Ordinary differential equation (ODE) Multiple independent variables: Partial differential equation (PDE) A systematic technique of solving the differential equations Differential equations (strong) formulation (today) Variational (weak) formulation Approximate the weak form using finite elements Axially loaded elastic bar x y x=0 x=L A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus u(x) = displacement of the bar at x x Differential equation governing the response of the bar L x b dx du AE dx d 0 ; 0 Second order differential equations Requires 2 boundary conditions for solution
- 2. 8/29/2023 2 x y x=0 x=L x Boundary conditions (examples) L x at u x at u 1 0 0 Dirichlet/ displacement bc L x at F dx du EA x at u 0 0 Neumann/ force bc Differential equation + Boundary conditions = Strong form of the “boundary value problem” Observe: 1. All the cases we considered lead to very similar differential equations and boundary conditions. 2. In 1D it is easy to analytically solve these equations 3. Not so in 2 and 3D especially when the geometry of the domain is complex: need to solve approximately 4. We’ll learn how to solve these equations in 1D. The approximation techniques easily translate to 2 and 3D, no matter how complex the geometry A generic problem in 1D 1 1 0 0 1 0 ; 0 2 2 x at u x at u x x dx u d Approximate solution strategy: Guess Where jo(x), j1(x),… are “known” functions and ao, a1, etc are constants chosen such that the approximate solution 1. Satisfies the boundary conditions 2. Satisfies the differential equation Too difficult to satisfy for general problems!! ... ) ( ) ( ) ( ) ( 2 2 1 1 0 x a x a x a x u o j j j
- 3. 8/29/2023 3 Potential energy W loading of energy potential (U) energy Strain The potential energy of an elastic body is defined as F u F x k k u k 1 Hooke’s Law F = ku Strain energy of a linear spring F = Force in the spring u = deflection of the spring k = “stiffness” of the spring Strain energy of a linear spring F u u+du Differential strain energy of the spring for a small change in displacement (du) of the spring Fdu dU For a linear spring kudu dU The total strain energy of the spring 2 u 0 u k 2 1 du u k U dU Strain energy of a nonlinear spring F u u+du Fdu dU The total strain energy of the spring curve nt dispalceme force the under Area du F U u 0 dU
- 4. 8/29/2023 4 Potential energy of the loading (for a single spring as in the figure) Fu W Potential energy of a linear spring W loading of energy potential (U) energy Strain Fu ku 2 1 Π 2 F x k k u Example of how to obtain the equlibr Principle of minimum potential energy for a system of springs For this system of spring, first write down the total potential energy of the system as: 3x 2 2 3 2 2 2 1 Fd ) d d ( k 2 1 ) d ( k 2 1 x x x system Obtain the equilibrium equations by minimizing the potential energy 1 k F x 2 k 1x d 2x d 3x d ) 2 ( 0 ) d d ( k d ) 1 ( 0 ) d d ( k d k d 2 3 2 3 2 3 2 2 1 2 Equation F Equation x x x system x x x x system Principle of minimum potential energy for a system of springs In matrix form, equations 1 and 2 look like F x x 0 d d k k k k k 3 2 2 2 2 2 1 Does this equation look familiar? Also look at example problem worked out in class Axially loaded elastic bar x y x=0 x=L A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus u(x) = displacement of the bar at x x F dx du ε Axial strain Axial stress dx du E Eε Strain energy per unit volume of the bar 2 dx du E 2 1 σε 2 1 dU Strain energy of the bar Adx σε 2 1 dV σε 2 1 dU U L 0 x since dV=Adx
- 5. 8/29/2023 5 Axially loaded elastic bar Strain energy of the bar L L 0 2 0 dx dx du EA 2 1 dx σεA 2 1 U Potential energy of the loading L) Fu(x dx bu W 0 L Potential energy of the axially loaded bar L) Fu(x dx bu dx dx du EA 2 1 0 0 2 L L 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 Admissible displacements: these are any reasonable displacement that you can think of that satisfy the displacement boundary conditions of the original problem (and of course certain minimum continuity requirements). Example: Exact solution for the displacement field uexact(x) Any other “admissible” displacement field w(x) L 0 x Lets see what this means for an axially loaded elastic bar A(x) = cross section at x b(x) = body force distribution (force per unit length) E(x) = Young’s modulus x y x=0 x=L x F Potential energy of the axially loaded bar corresponding to the exact solution uexact(x) L) (x Fu dx bu dx dx du EA 2 1 ) (u exact 0 exact 0 2 exact exact L L Potential energy of the axially loaded bar corresponding to the “admissible” displacement w(x) L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2 L L Exact solution for the displacement field uexact(x) Any other “admissible” displacement field w(x) L 0 x
- 6. 8/29/2023 6 Example: L x b dx u d AE 0 ; 0 2 2 L x at F dx du EA x at u 0 0 Assume EA=1; b=1; L=1; F=1 Analytical solution is 2 2 2 x x uexact 6 7 1) (x u dx u dx dx du 2 1 ) (u exact 1 0 exact 1 0 2 exact exact Potential energy corresponding to this analytical solution Now assume an admissible displacement x w Why is this an “admissible” displacement? This displacement is quite arbitrary. But, it satisfies the given displacement boundary condition w(x=0)=0. Also, its first derivate does not blow up. 1 1) w(x dx w dx dx dw 2 1 (w) 1 0 1 0 2 Potential energy corresponding to this admissible displacement Notice exact 7 1 6 (u ) (w) since 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 Mathematical statement: If ‘uexact’ is the exact solution (which satisfies the differential equation together with the boundary conditions), and ‘w’ is an admissible displacement (that is quite arbitrary except for the fact that it satisfies the displacement boundary conditions and its first derivative does not blow up), then (w) ) (uexact unless w=uexact (i.e. the exact solution minimizes the potential energy) The Principle of Minimum Potential Energy and the strong formulation are exactly equivalent statements of the same problem. The exact solution (uexact) that satisfies the strong form, renders the potential energy of the system a minimum. So, why use the Principle of Minimum Potential Energy? The short answer is that it is much less demanding than the strong formulation. The long answer is, it 1. requires only the first derivative to be finite 2. incorporates the force boundary condition automatically. The admissible displacement (which is the function that you need to choose) needs to satisfy only the displacement boundary condition
- 7. 8/29/2023 7 Finite element formulation, takes as its starting point, not the strong formulation, but the Principle of Minimum Potential Energy. Task is to find the function ‘w’ that minimizes the potential energy of the system From the Principle of Minimum Potential Energy, that function ‘w’ is the exact solution. L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2 L L Step 1. Assume a solution ... ) ( ) ( ) ( ) ( 2 2 1 1 0 x a x a x a x w o j j j Where jo(x), j1(x),… are “admissible” functions and ao, a1, etc are constants to be determined from the solution. Rayleigh-Ritz Principle The minimization of the potential energy is difficult to perform exactly. The Rayleigh-Ritz principle is an approximate way of doing this. Step 2. Plug the approximate solution into the potential energy L) Fw(x dx bw dx dx dw EA 2 1 (w) 0 0 2 L L Rayleigh-Ritz Principle ... ) ( ) ( F dx ... b dx ... dx d dx d EA 2 1 ,...) a , (a 1 1 0 0 0 1 1 0 0 0 2 1 1 0 0 1 0 L x a L x a a a a a L L j j j j j j Step 3. Obtain the coefficients ao, a1, etc by setting ,... 2 , 1 , 0 , 0 (w) i ai Rayleigh-Ritz Principle The approximate solution is ... ) ( ) ( ) ( ) ( 2 2 1 1 0 x a x a x a x u o j j j Where the coefficients have been obtained from step 3
- 8. 8/29/2023 8 Example of application of Rayleigh Ritz Principle x x=0 x=2 x=1 F E=A=1 F=2 1 2 0 2 1) Fu(x dx dx du 2 1 (u) x at applied F load of Energy Potential Energy Strain The potential energy of this bar (of length 2) is Let us assume a polynomial “admissible” displacement field 2 2 1 0 u x a x a a Note that this is NOT the analytical solution for this problem. Example of application of Rayleigh Ritz Principle For this “admissible” displacement to satisfy the displacement boundary conditions the following conditions must be satisfied: 0 4 2 2) u(x 0 0) u(x 2 1 0 0 a a a a Hence, we obtain 2 1 0 2 0 a a a Hence, the “admissible” displacement simplifies to 2 2 2 2 1 0 2 u x x a x a x a a Now we apply Rayleigh Ritz principle, which says that if I plug this approximation into the expression for the potential energy , I can obtain the unknown (in this case a2) by minimizing 4 3 0 2 3 8 0 2 3 4 2 F dx 2 dx d 2 1 1) Fu(x dx dx du 2 1 (u) 2 2 2 2 2 2 1 2 2 2 0 2 2 2 2 0 2 a a a a a x x a x x a x at evaluated 2 2 2 2 2 1 0 2 4 3 2 u x x x x a x a x a a Hence the approximate solution to this problem, using the Rayleigh-Ritz principle is Notice that the exact answer to this problem (can you prove this?) is 2 1 2 1 0 uexact x for x x for x
- 9. 8/29/2023 9 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 0.6 0.8 1 x Exact solution Approximate solution The displacement solution : How can you improve the approximation? 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 -1.5 -1 -0.5 0 0.5 1 1.5 x Stress Approximate stress Exact Stress The stress within the bar: