ME FINITE ELEMENT ANALYSIS FORMULAS

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1 ME FINITE ELEMENT ANALYSIS FORMULAS UNIT I FINITE ELEMENT FORMULATION OF BOUNDARY VALUE PROBLEMS 01. Global Equation for Force Vector, {F} = [K] {u} {F} = Global Force Vector [K] = Global Stiffness Matrix {u} = Global Displacement Vector 02. Strain, e = du / dx = u 2 u 1 / x 2 - x 1 u 2, u 1 = Displacements & x 2 - x 1 = Actual length of the element a. Strain along x Direction, e x = u / x b. Strain along y Direction, e y = v / y c. Shear Strain along xy Direction = u/ y + v/ x 03. Stress, σ = D (e - e 0 ) = D (Bu - e 0 ) B = Stress-Strain Relationship Matix D = Strain-Displacement Relationship Matix 04. Weighted Residual Methods, General Procedures, D (δ(x x i ) R (x; a 1, a 2, a 3,., a n ) dx = 0 Where w i = Weight function, D = Domain, R = Residual a. Point Collocation Method: D (δ(x x i ) R (x; a 1, a 2, a 3,., a n ) dx = 0 & R (x; a 1, a 2, a 3,., a n ) = 0 b. Subdomain Collocation Method: D R (x; a 1, a 2, a 3,., a n ) dx = 0 c. Least Squares Method: D ([R (x; a 1, a 2, a 3,., a n )] 2 dx = 0 / min d. Galerkin s Method: D (N i (x) R (x; a 1, a 2, a 3,., a n ) dx = 0

2 UNIT II ONE DIMENSIONAL FINITE ELEMENT ANALYSIS 01. Stress Strain relationship, Stress, σ (N/mm 2 ) = Young s Modulus, E (N/mm 2 ) x Strain, e 02. Strain Displacement Relationship Strain {e} = du / dx 03. For one dimensional problem, Field variable is displacement u = N 1 u 1 Where u = Displacement 04. For two nodded one dimensional problem, Field variable is displacement n u = N i u i = N1 u 1 + N 2 u 2 Where u 1 & u 2 = Nodal Displacements i=1 05. Linear polynomial function = u = a 0 + a 1 x 06. Shape function = [N 1 N 2 ] u 1 u Shape function N 1 = l x / l N 2 = x / l 08. Strain, {e} = [B] {u} {e} = Strain Martix [B] {u} = Strain Displacement Matrix = Degrees of Freedom (Displacement) 09. Stress, {σ} = [E] {e} = [D] {e} = [D] [B] {u} [E] = [D] = Young s Modulus 10. Strain energy, U = 1/2 {e} T {σ} dv 11. General Equation for Stiffness Matrix, [K] = [B] T [D] [B] dv [B] = Strain - Displacement relationship Matrix [D] = Stress - Strain relationship Matrix 12. General Equation for Force Vector, {F} = [K] {u} {F} = Element Force Vector [K] = Stiffness Matrix {u} = Nodal Displacements

3 (i) 1D BAR ELEMENTS: 01. Stiffness Matrix for 1D Bar element [K] = AE 1-1 l Force Vector for 1D Bar element F 1 = AE 1-1 u 1 F 2 l -1 1 u Load / Force Vector, {F} E = ρ A l where, ρ = Density 04. Temperature Force {F} = E A α T -1 1 E = young s Modulus, N/mm 2 A = Area of the Element, mm 2 α = Coefficient of thermal expansion, o C T = Temperature Difference, o C 05. Thermal stress {σ} = E (du/dx) E α T Where du / dx = u 1 u 2 / l (ii) 1D TRUSS ELEMENTS: 01. General Equation for Stiffness Matrix, [k] = A e L e l 2 lm -l 2 -lm lm m 2 -lm -m Force vector for two noded Truss elements, l e -l 2 -lm l 2 lm -lm -m 2 lm m 2 F 1 A e L e l 2 lm -l 2 -lm u 1 F 2 lm m 2 -lm -m 2 u 2 F 3 l e -l 2 -lm l 2 lm u 3 F 4 -lm -m 2 lm m 2 u 4 l = Cos θ = x 2 x 1 / l e m = Sin θ = y 2 y 1 / l e l e = (x 2 x 1 ) 2 + (y 2 y 1 ) 2

4 (iii) 1D SPRING ELEMENTS: 01. Force Vector F 1 = k 1-1 u 1 F u 2 (iv) 1D BEAM ELEMENTS: 01. Shape function for 1D Beam Element N 1 = 1 / L 3 (2 x 3 3x 2 L + L 3 ) N 2 = 1 / L 3 (x 3 L 2x 2 L 2 + x L 3 ) N 3 = 1 / L 3 (-2 x 3 + 3x 2 L) N 4 = 1 / L 3 (x 3 L x 2 L 2 ) 02. Force Vector for Two noded 1D Beam Element F 1y E e I e 12 6L -12 6L v 1 m 1 = 6L 4L 2-6L 2L 2 θ 1 F 2y L L 12-6L v 2 m 2 6L 2L 2-6L 4L 2 θ Stiffness Matrix for Two noded 1D Beam Element, [K] = E e I e 12 6L -12 6L 6L 4L 2-6L 2L 2 L L 12-6L 6L 2L 2-6L 4L 2 I = Moment Of Inertia, mm 4 L = Length of the Beam, mm

5 UNIT III TWO DIMENSIONAL FINITE ELEMENT ANALYSIS a. CST element 01. Displacement Vector, u = u (x, y) = u v 02. General Equation for Stress & Strain σ x e x Stress, σ = σ y τ xy Strain, e = e y γ xy 03. Body Force, F = F x F y 04. Polynominal function is, u = a 1 + a 2 x + a 3 y & v = a 4 + a 5 x + a 6 y 05. Shape function, N 1 + N 2 + N 3 = 1 Co-ordinates x = N 1 x 1 + N 2 x 2 + N 3 x 3 & y = N 1 y 1 + N 2 y 2 + N 3 y 3 (OR) x = N 1 (x 1 - x 3 ) + N 2 (x 2 - x 3 ) + N 3 & y = N 1 (y 1 - y 3) + N 2 (y 2 - y 3) + N Displacement Function, u 1 v 1 u (x,y) = N 1 0 N 2 0 N 3 0 u 2 0 N 1 0 N 2 0 N 3 v 2 u 3 v 3 N 1 = p 1 + q 1 x + r 1 y / 2A N 2 = p 2 + q 2 x + r 2 y / 2A N 3 = p 3 + q 3 x + r 3 y / 2A 07. Strain Displacement Matrix [B] = 1 q 1 0 q 2 0 q 3 0 2A 0 r 1 0 r 2 0 r 3 Where r 1 q 1 r 2 q 2 r 3 q 3 q 1 = y 2 - y 3 ; r 1 = x 3 x 2 q 2 = y 3 y 1 ; r 2 = x 1 x 3 q 3 = y 1 y 2 ; r 3 = x 2 - x 1

6 08. Stress Strain Relationship Matrix [D], a. FOR PLANE STRESS PROBLEM, E 1 ν 0 [D] = ν 1 0 (1 ν 2 ) ν 2 b. FOR PLANE STRAIN PROBLEM, E (1 ν) ν 0 [D] = ν (1 - ν) 0 (1 + ν) (1 2ν) ν 2 ν = Poisson s Ratio 09. Stiffness Matrix for CST Element [K] = [B] T [D] [B] A t A = Area of the triangular element = 1 1 x 1 y 1 1 x 2 y x 3 y 3 t = Thickness of the triangular element, mm 10. Temperature Force, { θ } or { f } = [B] T [D] {e 0 } t A a. For Plane Stress Problems, α T Initial Strain {e 0 } = α T 0 b. For Plane Strain Problems, α T Initial Strain {e 0 } = (1 + ν) α T Maximum Normal Stress, σ max = σ 1 = σ x + σ y + (σ x - σ y ) τ 2 xy 12. Minimum Normal Stress, σ min = σ 2 = σ x + σ y - (σ x - σ y ) τ 2 xy

7 13. Principal angle, tan 2θ = 2 τ xy σ x - σ y b. Axisymmetric element 01. Displacement vector, u = u (r, z) = u w 02. Stress equation for Axisymmetric Element, σ r σ r = Radial Stress, N/mm 2 Stress {σ} = σ θ σ θ = Longitudinal Stress, N/mm 2 σ z σ z = Circumferential Stress, N/mm 2 τ rz τ rz = Shear Stress, N/mm 2 e r e r = Radial Strain Strain, {e} = e θ e θ = Longitudinal Strain e z e z = Circumferential Strain γ rz γ rz = Shear Strain 03. Stiffness matrix for Two Dimensional Scalar Variable Problems / Axisymmetric Problems, [K] = 2 π r A [B] T [D] [B] Co-ordinate r = r 1 + r 2 + r 3 / 3, mm & z = z 1 + z 2 + z 3 / 3, mm 1 1 r 1 z 1 A = Area of the Triangle = 1 r 2 z 2 (or) 1/2 *b*h 2 1 r 3 z Shape function for Axisymmetric Problems, N 1 = α 1 + β 1 r + γ 1 z / 2A N 2 = α 2 + β 2 r + γ 2 z / 2A N 3 = α 3 + β 3 r + γ 3 z / 2A

8 α 1 = r 2 z 3 r 3 z 2 ; α 2 =r 3 z 1 r 1 z 3 ; α 3 = r 1 z 2 r 2 z 1 β 1 = z 2 -z 3 ; β 2 = z 3 -z 1 ; β 3 = z 1 -z 2 γ 1 = r 3 -r 2 ; γ 2 = r 1 -r 3 ; γ 3 = r 2 -r Strain Displacement Relationship Matrix for Axisymmetric elements, [B] = β 1 0 β 2 0 β 3 0 α 1 + β 1 + γ 1 z 0 α 2 + β 2 + γ 2 0 α 3 + β 3 + γ 3 z 0 r r r r r r 0 γ 1 0 γ 2 0 γ 3 γ 1 β 1 γ 2 β 2 γ 3 β 3 α 1 = r 2 z 3 r 3 z 2 β 1 = z 2 z 3 γ 1 = r 3 r 2 α 2 = r 3 z 1 r 1 z 3 β 2 = z 3 z 1 γ 2 = r 1 r 3 α 3 = r 1 z 2 r 2 z 1 β 3 = z 1 z 2 γ 3 = r 2 r Stress Strain Relationship Matrix [D] for Axisymmetric Triangular elements, E (1 ν) ν ν 0 [D] = ν (1 - ν) ν 0 (1 + ν) (1 ν 2 ) ν ν (1 - ν) ν 2 ν = Poisson s Ratio E = Young s Modulus, N/mm Temperature Effects: For Axisymmetric Triangular elements, Temperature Force, {f} t = [B] T [D] {e} t * 2 π r A

9 F 1 u F 1 w α T { f } t = F 2 u Strain {e} = α T F 2 w 0 F 3 u α T F 3 w c. Isoparametric Element 01. Shape Function for 4 Noded Isoparametric Quadrilateral Elements (Using Natural Co-Ordinate) N 1 = ¼ (1 ε) (1 η) N 2 = ¼ (1 + ε) (1 η) N 3 = ¼ (1 + ε) (1 + η) N 4 = ¼ (1 - ε) (1 + η) 02. Displacement Vector, u = N 1 u 1 + N 2 u 2 + N 3 u 3 + N 4 u 4 ; v = N 1 v 1 + N 2 v 2 + N 3 v 3 + N 4 v 4 u 1 ` v 1 u 2 u = u = N 1 0 N 2 0 N 3 0 N 4 0 v 2 v 0 N 1 0 N 2 0 N 3 0 N 4 u To find a location/position of point P, v 3 u 4 v 4 x = N 1 x 1 + N 2 x 2 + N 3 x 3 + N 4 x 4 & y = N 1 y 1 + N 2 y 2 + N 3 y 3 + N 4 y 4 x 1 ` y 1 x 2 u = x = N 1 0 N 2 0 N 3 0 N 4 0 y 2 y 0 N 1 0 N 2 0 N 3 0 N 4 x 3 y 3 x 4 y 4

10 04. Jaccobian Matrix, [J] = x / ε y / ε x / η y / η J 11 = ¼ - (1 - η) x 1 + (1 - η) x 2 + (1 + η) x 3 (1 + η) x 4 J 12 = ¼ - (1 - η) y 1 + (1 - η) y 2 + (1 + η) y 3 (1 + η) y 4 J 21 = ¼ - (1 - ε) x 1 - (1 + ε) x 2 + (1 + ε) x 3 + (1 - ε) x 4 J 22 = ¼ - (1 - ε) y 1 - (1 + ε) y 2 + (1 + ε) y 3 + (1 - ε) y Strain Displacement Relationship Matrix for Isoparmetric elements, 06. Stiffness matrix for quadrilateral element, [K] = t [B] T [D] [B] J x y 07. Stiffness matrix for natural co-ordinates, [K] = t [B] T [D] [B] J ε η 08. Stress Strain [D] Relationship Matrix, a. FOR PLANE STRESS PROBLEM, E 1 ν 0 [D] = ν 1 0 (1 ν 2 ) ν 2

11 b. FOR PLANE STRAIN PROBLEM, E (1 ν) ν 0 [D] = ν (1 - ν) 0 (1 + ν) (1 2ν) ν Element Force vector, {F} e = [N] T F x F y N = Shape function for 4 nodded Quadrilateral elements 10. Numerical Integration (Gaussian Quadrature) Where w i = Weight function F (x i ) = values of function at pre-determined points No. of Location, x points i Corresponding weights, w i 1 x 1 = x 1 = + 1/3 = x 2 = - 1/3 = x 1 = + 3/5 = x 3 = - 3/5 = x 2 = x 1 = x 4 = x 2 = x 3 = /9 = /9 = /9 =

12 UNIT IV DYNAMIC ANALYSIS USING FINITE ELEMENT METHOD 01. 1D vibration problems, a. For Longitudinal vibration of bar element, Finite element Equation is, { [K] [M] ω 2 }{u} = {P} For free vibration, { [K] [M] ω 2 }{u} = 0 P = External Load ; [K] = stiffness matrix = A E 1-1 l -1 1 [m] = Mass Matrix Mass Matrix for consistent mass [m c ] = ρal Mass Matrix for Lumped mass [m L ] = ρal b. For Longitudinal vibration of beam element, Finite element Equation is, { [K] [M] ω 2 }{u} = {F} [K] = stiffness matrix = EI 12 6L -12 6L l 3 6L 4L 2-6L 2L L 12-6L 6L 2L 2-6L 4L 2 [m] = Mass Matrix Mass Matrix for consistent mass [m c ] = ρal 156 2l l l 4 l 2 13 l -3 l l l -13 l -3l 2-22 l 4 l 2

13 Mass Matrix for Lumped mass [m L ] = ρal For solving Eigen Value Problems, a. Determinant method (for free vibration), { [K] λ [m]} {u} = 0 If the Eigen vector is not trial, the condition is, det [K] λ [m] = 0 ie. [K] λ [m] = 0

14 UNIT V APPLICATIONS IN HEAT TRANSFER & FLUID MECHANICS I. HEAT TRANSFER PROBLEMS One Dimensional Heat Transfer Problems, 01. General equation for Force Vector {F} = [K C ] {T} 02. Stiffness Matrix for 1D Heat conduction Element, [K C ] = A k 1-1 k = Thermal conductivity, w/mk l D Heat conduction, the FE equation is F 1 = Ak 1-1 T 1 F 2 l -1 1 T D Heat conduction with free end convection, the FE equation is A k h A 0 0 T 1 = h T A 0 l T 2 1 k = Thermal conductivity of element, W/mK A = Area of the element, m 2 l = Length of the element, m h = Heat transfer Coefficient, W/m 2 K T = fluid Temperature, K T 1, T 2 = Temperatures, K 05. 1D Heat conduction, convection with internal heat generation, the FE equation is A k hpl 2 1 T 1 = QA l + P h T l 0 l T 2 2 1

15 P = Perimeter, m ; Q = heat Generation, W II. HEAT TRANSFER Two Dimensional Heat Transfer Problems, 01. Shape Functions: N 1 = a 1 + b 1 x + c 1 y / 2A ; N 2 = a 2 + b 2 x + c 2 y / 2A ; N 1 = a 3 + b 3 x + c 3 y / 2A a 1 = x 2 y 3 - x 3 y 2 b 1 = y 2 - y 3 c 1 =x 3 - x 2 a 2 = x 3 y 1 - x 1 y 3 b 2 = y 3 - y 1 c 2 =x 1 - x 3 a 3 = x 1 y 2 - x 2 y 1 b 2 = y 1 - y 2 c 2 =x 2 - x Temperature at any point T (x, y) = N 1 T 1 + N 2 T 2 + N 3 T Thermal Stiffness matrix, [K] = [K c] + [K h] = [K c] + [K h ] f + [K h ] e [K] = Thermal Stiffness matrix due to heat conduction by body particles. [K h] = Thermal Stiffness matrix due to heat convection by body particles. [K h ] f = Thermal Stiffness matrix due to heat convection through face areas. [K h ] e = Thermal Stiffness matrix due to heat convection through edge surface. a. [K c] = k (b 1 2 +c 1 2 ) (b 1 b 2 +c 1 c 2 ) (b 1 b 3 +c 1 c 3 ) 4A (b 1 b 2 +c 1 c 2 ) (b 1 2 +c 1 2 ) (b 2 b 3 +c 2 c 3 ) (b 1 b 3 +c 1 c 3 ) (b 2 b 3 +c 2 c 3 ) (b 3 2 +c 3 2 ) b. Thermal Stiffness matrix due to heat convection through edge areas 1-2, [K h ] e1-2 = h 1-2 S

16 c. Thermal Stiffness matrix due to heat convection through edge areas 2-3, [K h ] e2-3 = h 2-3 S d. Thermal Stiffness matrix due to heat convection through edge areas 3-1, [K h ] e3-1 = h 3-1 S e. overall convect part of stiffness matrix by edge convection, [K h ] e = [K h ] e1-2 + [K h ] e2-3 + [K h ] e3-1 f. Thermal Stiffness matrix due to heat convection through face areas. [K h ] f = ha Thermal Stiffness matrix, {F} = {F h} + {F Q} = {F h} e + {F h } f + {F Q} {F h} e = Thermal force vector due to heat convection through edge surface. {F h } f = Thermal force vector due to heat convection through face areas. {F Q} = Thermal force vector due to heat convection due to internal heat generation. a. Thermal force vector due to heat convection through edge areas 1-2, {F h} e 1-2 = h 1-2 T S Edge Length, S 1-2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2

17 b. Thermal force vector due to heat convection through edge areas 2-3, {F h} e 2-3 = h 2-3 T S Edge Length, S 2-3 = (x 3 x 2 ) 2 + (y 3 y 2 ) 2 c. Thermal force vector due to heat convection through edge areas 3-1, {F h} e 3-1 = h 3-1 T S Edge Length, S 1-2 = (x 3 - x 1 ) 2 + (y 3 - y 1 ) 2 d. Thermal force vector due to heat convection through edge surface, {F h} e = {F h} e {F h} e {F h} e 3-1 e. Thermal force vector due to heat convection through face areas, {F h } f = h T A f. Thermal force vector due to heat convection due to internal heat generation, {F Q } = QA Thermal Force Vector due to Heat Transfer {F} = [K] {T} ie. {F h} e + {F h } f + {F Q } = [K c] + [K h ] e + [K h ] f {T}

18 III. FLUID MECHANICS One Dimensional Fluid Mechanics Problems 01. For One Dimensional Fluid Mechanics Problems, Pressure (P) = N 1 P 1 + N 2 P 2 N 1 = x 2 x 1 / l ; N 2 = x / l P(x) = [N 1 N 2 ] P 1 = [N] {P} P Nodal Fluid Flow Rate, Q = [K] {P} {P} = Nodal Potential Vector [K] = Stiffness matrix for Fluid Flow = K` 1-1 = ka l -1 1 K` = Flow Stiffness = ka (m 2 /s) L k = Permeability of pipe element, (m/s) Q = Discharge (Flow Rate) = Q 1 Q 2 (m 3 /s) {P} = Potential Vector = P 1 P 2 (m) IV. FLUID MECHANICS Two Dimensional Fluid Mechanics Problems 01. P(x,y) = [N 1 N 2 N3] P 1 = [N] {P} = N 1 P 1 + N 2 P 2 + N 3 P 3 P 2 P Shape Functions: N 1 = a 1 + b 1 x + c 1 y / 2A ; N 2 = a 2 + b 2 x + c 2 y / 2A ; N 1 = a 3 + b 3 x + c 3 y / 2A a 1 = x 2 y 3 - x 3 y 2 b 1 = y 2 - y 3 c 1 =x 3 - x 2 a 2 = x 3 y 1 - x 1 y 3 b 2 = y 3 - y 1 c 2 =x 1 - x 3 a 3 = x 1 y 2 - x 2 y 1 b 2 = y 1 - y 2 c 2 =x 2 - x 1

19 03. Stiffness matrix, [K], = k x t (b 2 1 ) (b 1 b 2 ) (b 1 b 3 ) k y t (c 2 1 ) (c 1 c 2 ) (c 1 c 3 ) 4A (b 1 b 2 ) (b 2 1 ) (b 2 b 3 ) + 4A (c 1 c 3 ) (c 2 c 3 ) (c 2 3 ) (m 2 /s) (b 1 b 2 ) (b 2 1 ) (b 2 b 3 ) (c 1 c 3 ) (c 2 c 3 ) (c 2 3 ) 04. Gradient Matrix, {g} = [B] {P} {P} = Pressure Gradient [B] = 1 b 1 b 2 b 3 2A c 1 c 2 c 3 {g} = g x g y 05. Velocity Gradient {v} = v x = [D] {g] = [D] [B]{P} v y Strain displacement Matrix [B] = 1 b 1 b 2 b 3 2A c 1 c 2 c 3 Stress-Strain Matrix [D] = k x 0 k x, k y = Permeability on x, y direction, m/s 0 k y 06. Force vector Matrix, a. Force due to constant volumetric flow rate per unit volume over the whole element, {F Q } = QA b. Force on edges, Along edge surface 1-2, {F q } 1-2 = q L 1-2 t Along edge surface 2-3, {F q } 2-3 = q L 2-3 t Along edge surface 3-1, {F q } 3-1 = q L 3-1 t

JEPPIAAR ENGINEERING COLLEGE

JEPPIAAR ENGINEERING COLLEGE Jeppiaar Nagar, Rajiv Gandhi Salai 600 119 DEPARTMENT OFMECHANICAL ENGINEERING QUESTION BANK VI SEMESTER ME6603 FINITE ELEMENT ANALYSIS Regulation 013 SUBJECT YEAR /SEM: III