12 Variational Formulation of Plane Beam Element IFEM Ch 12 - PDF document

Introduction to FEM 12 Variational Formulation of Plane Beam Element IFEM Ch 12 – Slide 1

Introduction to FEM A Beam is a Structural Member Designed to Resist Primarily Transverse Loads IFEM Ch 12 – Slide 2

Introduction to FEM Transverse Loads are Transported to Supports by Flexural Action Neutral surface Compressive stress Tensile stress IFEM Ch 12 – Slide 3

Introduction to FEM Beam Configuration Spatial (General Beams) Plane (This Chapter) Beam Models Bernoulli-Euler Timoshenko (more advanced topic: described in Chapter 13 but not covered in course) IFEM Ch 12 – Slide 4

Introduction to FEM Plane Beam Terminology y, v q(x) y, v Beam cross section x, u z Neutral axis Neutral L Symmetry surface plane IFEM Ch 12 – Slide 5

Introduction to FEM Common Support Conditions ������ Simply Supported ������ ������ Cantilever IFEM Ch 12 – Slide 6

Introduction to FEM Basic Relations for Bernoulli-Euler Model of Plane Beam � − v ′ � � − θ � � − ∂v ( x ) y y y u ( x , ) � � � y ∂ x = = = y v ( x , ) v ( x ) v ( x ) v ( x ) ∂ x = − ∂ 2 v ∂ x 2 = − d 2 v e = ∂ u y y dx 2 = − κ y σ = Ee = − E d 2 v y dx 2 = − E κ y M = E I κ Plus equilibrium equation M'' = q (not used specifically in FEM) IFEM Ch 12 – Slide 7

Introduction to FEM Kinematics of Bernoulli-Euler Beam Cross section P' ( x+u,y+v ) θ =dv/dx = v' y, v v ( x,y )= v ( x, 0) x, u y x P ( x,y ) IFEM Ch 12 – Slide 8

Introduction to FEM Tonti Diagram for Bernoulli-Euler Model of Plane Beam (Strong Form) Displacement Prescribed Distributed Transverse BCs end transverse load displacements v(x) displacements q(x) κ = v'' Kinematic M''=q Equilibrium M = EI κ Bending Force BCs Prescribed Curvature moment end loads κ (x) Constitutive M(x) IFEM Ch 12 – Slide 9

Introduction to FEM Total Potential Energy of Beam Member � = U − W Internal External � L � L � 2 � ∂ 2 v � U = 1 σ xx e xx dV = 1 M κ dx = 1 E I dx 2 2 2 ∂ x 2 V 0 0 � L E I κ 2 dx = 1 Internal energy due to bending 2 0 � L W = q v dx . External energy due to transverse load q 0 IFEM Ch 12 – Slide 10

Introduction to FEM Degrees of Freedom of Plane Beam Element θ 2 v 2 θ 1 v 1 2 1 v 1   θ 1 u e =   v 2   θ 2 IFEM Ch 12 – Slide 11

Introduction to FEM Bernoulli-Euler Kinematics of Plane Beam Element θ 2 P' ( x+u,y+v ) y, v v 2 θ 1 v E, I 1 x, u 1 2 x ℓ P ( x,y ) IFEM Ch 12 – Slide 12

Introduction to FEM Shape Functions in Terms of Natural Coordinate ξ v 1   θ 1 v e = [ N e  = N u e v 1 N e θ 1 N e v 2 N e θ 2 ]   v 2  θ 2 ξ = 2 x Introduce the natural ℓ − 1 (isoparametric) coordinate N (ξ) = (1 − ξ) (2 + ξ) N (ξ) = (1 + ξ) (2 − ξ) e 2 2 e 1 1 v 1 v 2 4 4 N (ξ) = − (1 + ξ) (1 − ξ) N (ξ) = (1 − ξ) (1 + ξ) 2 e 2 e 1 1 ℓ ℓ θ 1 θ2 8 8 Plots on next slide IFEM Ch 12 – Slide 13

Introduction to FEM Element Shape Function Plots v = 1 N (ξ) e 1 N (ξ) = (1 − ξ) (2 + ξ) 2 v 1 e 1 v 1 4 θ = 1 1 N (ξ) e N (ξ) = (1 − ξ) (1 + ξ) e 2 θ 1 1 ℓ θ 1 8 v = 1 N (ξ) e 2 N (ξ) = (1 + ξ) (2 − ξ) e 2 v 2 1 v 2 4 θ = 1 2 N (ξ) = − (1 + ξ) (1 − ξ) 2 e 1 ℓ θ2 N (ξ) e 8 θ2 ξ = 1 ξ = −1 IFEM Ch 12 – Slide 14

Introduction to FEM Getting Curvatures from Displacement Interpolation 0 2 2 2 2 e d v ( x ) d N d u d N κ = = u + N = u e e dx 2 dx 2 dx 2 dx 2 v 1 e e e e 2 2 2 2 d N d N d N d N θ 1 θ2 = = B u θ 1 def v 1 v 2 � � e dx 2 v 2 dx 2 dx 2 dx 2 θ 2 Applying the chain rule 1 x 4 curvature- d f ( x ) d f ( ξ ) d ξ 2 d f ( ξ ) displacement = matrix = dx d ξ dx ℓ d ξ 0 2 ր d f ( x ) d (2/ ℓ ) d f ( ξ ) 2 d d f ( ξ ) 4 d f ( ξ ) 2 2 � � = + = dx dx d ξ ℓ dx d ξ ℓ d ξ 2 to differentiate the shape functions we get 6 ξ − 6 ξ B = 1 3 ξ − 1 3 ξ + 1 � � ℓ ℓ ℓ IFEM Ch 12 – Slide 15

Introduction to FEM Element Stiffness and Consistent Node Forces Varying the element TPE 2 u e T K e u e − u e T f e � e = 1 we get � ℓ � 1 E I B T B dx = E I B T B 1 K e 2 ℓ d ξ = 0 − 1 � ℓ � 1 N T q dx = N T q 1 f e = 2 ℓ d ξ 0 − 1 IFEM Ch 12 – Slide 16

Introduction to FEM Analytical Computation of Prismatic Beam Element Stiffness ("prismatic" means constant EI ) 36 ξ 2 6 ξ (3 ξ − 1)` − 36 ξ 2 6 ξ (3 ξ + 1) ℓ   � 1 (3 ξ − 1) 2 ℓ 2 − 6 ξ (3 ξ − 1) ℓ (9 ξ 2 − 1) ℓ 2 = E I   K e  d ξ   36 ξ 2 − 6 ξ (3 ξ + 1) ℓ 2 ℓ 3   − 1  (3 ξ + 1) 2 ℓ 2 symm 12 6 ℓ − 12 6 ℓ   4 ℓ 2 2 ℓ 2 = E I − 6 ℓ   ℓ 3 12 − 6 ℓ   4 ℓ 2 symm IFEM Ch 12 – Slide 17

Introduction to FEM Mathematica Script for Symbolic Computation of Prismatic Plane Beam Element Stiffness ClearAll[EI,l, Ξ ]; B={{6* Ξ ,(3* Ξ -1)*l,-6* Ξ ,(3* Ξ +1)*l}}/l^2; Ke=(EI*l/2)*Integrate[Transpose[B].B,{ Ξ ,-1,1}]; Ke=Simplify[Ke]; Print["Ke for prismatic beam:"]; Print[Ke//MatrixForm]; Ke for prismatic beam: �  EI  EI �  EI  EI � � � � � l  l  �������� ������� �������� l  ���� �������� ������� �������� l  ���� � � � � � � � � � �  EI �  EI �  EI  EI � � � � � l  l  �������� ���� �������� ����� �������� ���� �������� ���� � � l l � � � � � � � � �  EI �  EI  EI �  EI � � � � � l  l  l  l  � �������� ������� �������� ���� �������� ������� �������� ���� � � � � � � � �  EI  EI  EI �  EI � � l  l  �������� ���� �������� ���� �������� ���� �������� ����� l l Corroborates the result from hand integration. IFEM Ch 12 – Slide 18

Introduction to FEM Analytical Computation of Consistent Node Force Vector for Uniform Load q 4 ( 1 − ξ 2 2 + ξ 1 ) ) (   � 1 � 1 8 ℓ ( 1 − ξ ) 2 1 + ξ 1 ( ) d ξ   f e = 1 d ξ = 1 2 q ℓ N 2 q ℓ   1 + ξ ) 2 2 − ξ 1   ( ) 4 ( − 1 − 1   2 1 − ξ 8 ℓ ( 1 + ξ ) − 1 ) ( 1   2 1 12 ℓ   = q ℓ   1 "fixed end moments"    2  − 1 12 ℓ IFEM Ch 12 – Slide 19

Introduction to FEM Mathematica Script for Computation of Consistent Node Force Vector for Uniform q ClearAll[q,l, Ξ ] Ne={{2*(1- Ξ )^2*(2+ Ξ ), (1- Ξ )^2*(1+ Ξ )*l, 2*(1+ Ξ )^2*(2- Ξ ),-(1+ Ξ )^2*(1- Ξ )*l}}/8; fe=(q*l/2)*Integrate[Ne,{ Ξ ,-1,1}]; fe=Simplify[fe]; Print["fe^T for uniform load q:"]; Print[fe//MatrixForm]; fe^T for uniform load q: l  q l  q � l q l q  � ������� ����������� ������� � �����������    Force vector printed as row vector to save space. IFEM Ch 12 – Slide 20