SlideServe
Menu
Préférences
© SlideServe · v2026

Engineering Systems

salvador-perkins avatar
24 diapositives · 388 Views · Nov 18, 2014

This document covers the essentials of lumped parameter systems in engineering mechanics. It discusses the representation of structural displacements through a finite number of state variables, using algebraic and differential equations. The focus is on deriving the stiffness matrix, exploring concepts such as equilibrium, constitutive equations, and compatibility conditions. It emphasizes transformations between local and global coordinate systems, application of boundary conditions, and the properties of the stiffness matrix. Understanding these principles is crucial for effective matrix structural analysis.

Engineering Systems - slide 1
01 / 24Engineering Systems
1 / 17

Télécharger la présentation

Engineering Systems
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation
  1. Engineering Systems Lumped Parameter (Discrete) Continuous • A finite number of state variables describe solution • Algebraic Equations • Differential Equations Govern Response

  2. Lumped Parameter Displacements of Joints fully describe solution

  3. Matrix Structural Analysis - Objectives Basic Equations Use Equations of Equilibrium Constitutive Equations Compatibility Conditions Form [A]{x}={b} Solve for Unknown Displacements/Forces {x}= [A]-1{b}

  4. Terminology Element: Discrete Structural Member Nodes: Characteristic points that define element D.O.F.: All possible directions of displacements @ a node

  5. Assumptions • Linear Strain-Displacement Relationship • Small Deformations • Equilibrium Pertains to Undeformed Configuration

  6. The Stiffness Method Consider a simple spring structural member Undeformed Configuration Deformed Configuration

  7. Derivation of Stiffness Matrix Using Basic Equations d1 d2 P1 P2

  8. Derivation of Stiffness Matrix Using Basic Equations d2 + d1 = For each case write basic equations

  9. Derivation of Stiffness Matrix Using Basic Equations P21 P11 Equilibrium X d2=0 d1 P11 Constitutive

  10. Derivation of Stiffness Matrix Using Basic Equations P12 P22 Equilibrium d1=0 d2 Constitutive

  11. Derivation of Stiffness Matrix Using Basic Equations P1 P2 Combined Action

  12. Derivation of Stiffness Matrix Using Basic Equations In Matrix Form

  13. Consider 2 Springs 1 2 3 k1 k2 2 elements 3 nodes 3 dof 1 Fix Fix 2 Fix Fix 3 Fix Fix

  14. 2-Springs Compare to 1-Spring

  15. Use Superposition d1 d1 d2 d2 0 0 d3 d3 d1 d1 X X X X d2 d2 3 1 2 1 3 2 X X X X d3 d3 DOF not connected directly yield 0 in SM

  16. Properties of Stiffness Matrix • SM is Symmetric • Betti-Maxwell Law • SM is Singular • No Boundary Conditions Applied Yet • Main Diagonal of SM Positive • Necessary for Stability

  17. Apply Boundary Conditions kii kij kik kil kim ui Pi uj Pj kji kjj kjk kjl kjm uk = Pk kki kkj kkk kkl kkm ul Pl kli klj klk kll klm um Pm kli klj klk kll klm -1 uf = Kff (Pf + Kfsus) uf Pf Kff Kfs Ksf Kss us Ps Kffuf+ Kfsus=Pf Ksfuf+ Kssus=Ps Ksfuf+ Kssus=Ps

  18. Transformations Global CS k1 k2 d3 u6 u4 u2 u4 d1 u1 u3 u5 u3 y P x x Local CS d2 d2 Objective: Transform State Variables from LCS to GCS

  19. Transformations P2x P2y Global CS 2 1 f P1x y P1x P1x = P1xcosf + P1ysinf P1y = -P1xsinf + P1ycosf P1y cosf sinf P1x x = P1x -sinf cosf P1y P1y P1 P1 = T Consider

  20. Transformations P2y P2x Global CS 1 2 f P1x or y P1x -1 -1 P1y P2 P1 P1 P2 = = T T Similarly for u x or P2 P1 u1 u2 u2 u1 P1 P2 = = = = T T T T In General

  21. Transformations P1y P2y P2x d1 1 -1 P1 k = 2 1 P2 -1 1 d2 P2 P1x f 1 0 -1 0 u1x P1x 0 0 0 0 u1y P1y P1 = k -1 0 1 0 u2x P2x 0 0 0 0 u2y P2y Element stiffness equations in Local CS Expand to 4 Local dof

  22. SM in Global Coordinate System -1 R K R u P = [T] [0] [R]= K : Element SM in global CS [0] [T] Introduce the transformed variables… Both R and T Depend on Particular Element

  23. Transformations l2 lm - l2 - lm m2 - lm - m2 K = AE/L l2 lm Symm. m2 For example for an axial element with k=AE/L l=cosf m=sinf

  24. In Summary • Derivation of element SM – Basic Equations • Structural SM by Superposition • Application of Boundary Conditions - Elimination • Solution of Stiffness Equations – Partitioning • Local & Global CS • Transformation