MEM427_Lecture_3_Energy_Principles

MEM427_Lecture_3_Energy_Principles -...

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MEM427 Introduction to Finite Element Method Lecture 3 nergy rinciples Energy Principles Chapter 3 Energy Principles 1 MEM427 Introduction to Finite Element Method Outline of Lecture •Strain Energy and Strain Energy Density •Castigliano’s Theorems inciple of Minimum Total Potential Energy •Principle of Minimum Total Potential Energy Chapter 3 Energy Principles 2 MEM427 Introduction to Finite Element Method Strain Energy For Axially Loaded Members W : Work done by the load 0 1 1 d P W U : Strain energy stored in the member no energy is dissipated, then Chapter 3 Energy Principles 3 0 1 1 d P W U If no energy is dissipated, then MEM427 Introduction to Finite Element Method Strain Energy For Axially Loaded Members For Linearly Elastic Materials P P 2 P W P Inelastic L EA P EA PL or , Recall for axially loaded members Elastic strain energy  EA L P P W U 2 2 2 strain energy Chapter 3 Energy Principles 4 L EA P W U 2 2 2 Elastic deformation Permanent deformation
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MEM427 Introduction to Finite Element Method Example: Determine the displacement at joint B cos H L L EA EA L P U 2 2 2 2 B P W L Work Done by P: B B AB AB AB AB AB A E L S 2 Strain Energy in i th Member: Free body BC S BA S i i i i i A E L S U 2 2 AB EA L S U U i i 2 2 2 2 1 diagram for joint B P B n n Total Strain Energy: 0 cos cos P S S F BC BA y H 0 sin sin BC BA x S S F 3 2 cos 4 EA H P Chapter 3 Energy Principles 5 U W 3 cos 2 EA PH B cos 2 P S S S BC BA MEM427 Introduction to Finite Element Method Example HW #1a Revisit BC D ft .5 1 ft .0 1 j P i P P A 45 sin 45 cos : at Force j v i u d A A A A : at nt Displaceme ft 2 2 6 psi 10 9 . 1 E  A i i d P P W 2 1 2 1 : Done Work           45 sin 45 cos 1 v P u P A x, u y, v 45 in 0 . 8 A 3 2 : Energy Strain i i L S U lb in 1606 . 0 2 A A in 83 . 26 in, 0 . 24 in, 0 . 30 AD AC AB L L L Member S (lb) L (in) U (in lb) AB 364.30 30.00 0.1310 1 2 i i i E A P = 400 lb 0 894 in 10 9459 . 0 3 3 A u 20 lb 3 . 364 C AB S AC 120.00 24.00 0.0114 AD 143.70 26.83 0.0182 Recall the solution for HW #1a: Chapter 3 Energy Principles 6 in 10 1894 . 0 A v lb 7 . 143 lb 0 . 120 AD AC S S W U lb in 0.1606 i U MEM427 Introduction to Finite Element Method Homework Assignment #3(a) Use the solution of Lab #1 to verify that the strain energy ( U ) stored in the truss is equal to the work done ( W ) by the loads. y 5 y P 4 y P 5 y 4 lb 500 5 4 y y P P component and 5 4 y y y member in Force : th i i S nt displaceme of 6 1 2 6 1 2 : energy Strain i i i i i i i A E L S U U 1 1 y one ork y y y y P P W 5 5 4 4 2 2 : P by Done Work Chapter 3 Energy Principles 7
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This note was uploaded on 11/14/2010 for the course MEM 427 taught by Professor Tein-mintan during the Fall '10 term at Drexel.

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MEM427_Lecture_3_Energy_Principles -...

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