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# Lecture4-FEA.pdf

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# Lecture4-FEA.pdf

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### Lecture4-FEA.pdf

1. 1. Finite Element Methods Introduction to differential equations Dr. Muhammad Farhan
2. 2. Reading assignment: Handout + Lecture notes Summary: • Strong form of boundary value problems •Elastic bar •String in tension •Heat conduction •Flow through a porous medium •Approximate solution
3. 3. 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)
4. 4. A systematic technique of solving the differential equations Differential equations (strong) formulation (today) Variational (weak) formulation Approximate the weak form using finite elements
5. 5. 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
6. 6. 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”
7. 7. Flexible string x y x=0 x=L x S S p(x) S = tensile force in string p(x) = lateral force distribution (force per unit length) w(x) = lateral deflection of the string in the y-direction Differential equation governing the response of the bar L x p dx u d S     0 ; 0 2 2 Second order differential equations Requires 2 boundary conditions for solution
8. 8. Heat conduction in a fin Differential equation governing the response of the fin Second order differential equations Requires 2 boundary conditions for solution A(x) = cross section at x Q(x) = heat input per unit length per unit time [J/sm] k(x) = thermal conductivity [J/oC ms] T(x) = temperature of the fin at x x y x=0 x=L x Q(x) L x Q dx dT Ak dx d           0 ; 0
9. 9. Boundary conditions (examples) Dirichlet/ displacement bc L x at h dx dT k x at T      0 0 Neumann/ force bc x y x=0 x=L x Q(x)
10. 10. Fluid flow through a porous medium (e.g., flow of water through a dam) Differential equation Second order differential equations Requires 2 boundary conditions for solution A(x) = cross section at x Q(x) = fluid input per unit volume per unit time k(x) = permeability constant j(x) = fluid head x y x=0 x=L x Q(x) L x Q dx d k dx d           0 ; 0 j
11. 11. Boundary conditions (examples) Known head L x at h dx d k x at      j j 0 0 Known velocity
12. 12. 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
13. 13. 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 x x x u 6 7 6 1 ) ( 3    Analytical solution Assume that we do not know this solution.
14. 14. A generic problem in 1D A general algorithm for approximate solution: Guess where jo(x), j1(x),… are “known” functions and ao, a1, etc are constants chosen such that the approximate solution Satisfies the differential equation Satisfies the boundary conditions i.e., ... ) ( ) ( ) ( ) ( 2 2 1 1 0     x a x a x a x u o j j j 1 ... ) 1 ( ) 1 ( ) 1 ( 0 ... ) 0 ( ) 0 ( ) 0 ( 1 0 ; 0 ... ) ( ) ( ) ( 2 2 1 1 0 2 2 1 1 0 2 2 2 2 2 1 2 1 2 2 0                j j j j j j j j j a a a a a a x x dx x d a dx x d a dx x d a o o o Solve for unknowns ao, a1, etc and plug them back into ... ) ( ) ( ) ( ) ( 2 2 1 1 0     x a x a x a x u o j j j This is your approximate solution to the strong form