MEM427_Lecture_4_One-D_Elements

# MEM427_Lecture_4_One-D_Elements - (PMTPE AQuickReview px P...

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MEM427 Introduction to Finite Element Method ct res 4 Lectures 4 One Dimensional Elements (Trusses, Beams, Frames) Chapter 4 One Dimensional Elements 1 MEM427 Introduction to Finite Element Method Principle of Minimum Total Potential Energy (PMTPE) Qik Ri A Quick Review Example: A S.S. beam subjected to a load P  P x p inematicBC onsider mmetry): ethod 1: Solve the D.E. (MEM330): 2 L 2 L Kinematic B.C. (Consider symmetry): 0 2 , 0 0 L v v 1 v i v n v ex v x M dx v d EI x p dx v d EI 2 2 4 4 or , Method 1: Solve the D.E. (MEM330): ethod 2: MTPE ex v •There are many possible solutions, e.g., v 1 ( x ) ,…, v i ( x ) ,…, v n ( x ) , and v ex ( x ) . All of them satisfy the kinematic B.C. v v v v v v n i , , , , 1 i i i P v U v : Energy Potential Total OTE: ach of ) ) ) an be Method 2: PMTPE y NOTE: Each of v 1 (x) ,…, v i (x) ,…, v n (x) can be considered as an approximate solution having a total potential energy greater than that of the exact solution, v ex (x) . The smaller the value of the •The one solution, i.e., v ex ( x ) , that yields the minimum total potential energy as compared to all the th ’ i th th t ti fi Chapter 4 One Dimensional Elements 2 total potential energy is, the closer the approximate solution is to the exact solution. other’s is the one that satisfies the D.E., hence is the exact solution. MEM427 Introduction to Finite Element Method Approximate Solution Methods An approximate solution can be either displacement (kinematical) based, force (stress) based , or a combination of both. In a displacement based approach we assume a displacement solution that satisfies the displacement (kinematics) boundary conditions whereas in a force (stress) based approach we assume a force (stress) solution that satisfies the force boundary conditions . Example: A simply supported beam subject to a uniform load Exact solutions: Possible approximate solutions: “Force” boundary conditions: M (0) = M ( L ) = 0 x A M pp sin 2 qx qLx M Bending moment: Kinematic boundary conditions: v (0) = v ( L ) = 0 L app 2 2 Deflection: Chapter 4 One Dimensional Elements 3 L x B v app sin ( ) ( ) 3 2 3 2 24 x Lx L EI qx v MEM427 Introduction to Finite Element Method Approximate Solution Methods Example: A fixed end beam Kinematical boundary conditions: ? app v v ( 0 ) = v’ ( 0 ) = v ( L ) = v’ ( L ) = 0 here is no rce boundary conditions Example: A cantilever beam There is no force boundary conditions Kinematical boundary conditions: v ( L ) = v’ ( L ) = 0 ? app v “Force” boundary conditions: = 0, = Chapter 4 One Dimensional Elements 4 M (0) 0, V ( L ) P ? app M

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MEM427 Introduction to Finite Element Method Approximate Solution Methods L x A v sin 1 1 Example: A simply supported beam subject to a concentrated force   P x p Assume approx. solution: 4 1 0 2 2 1 2 1 2 I P dx dx v d EI V U L OTE 1 2 1 3 4 PA A L EI v ex v 1 Objective: Find the value of A that minimizes 1 1 2 : NOTE A L x v EI PL EI PL A
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MEM427_Lecture_4_One-D_Elements - (PMTPE AQuickReview px P...

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