Finite Element Analysis. Barna Szabó

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Finite Element Analysis - Barna Szabó

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u plus v comma u plus v right-parenthesis minus upper F left-parenthesis u plus v right-parenthesis 2nd Row 1st Column equals 2nd Column one half upper B left-parenthesis u comma u right-parenthesis plus upper B left-parenthesis u comma v right-parenthesis plus one half upper B left-parenthesis v comma v right-parenthesis minus upper F left-parenthesis u right-parenthesis minus upper F left-parenthesis v right-parenthesis 3rd Row 1st Column equals 2nd Column pi left-parenthesis u right-parenthesis plus ModifyingBelow upper B left-parenthesis u comma v right-parenthesis minus upper F left-parenthesis v right-parenthesis With presentation form for vertical right-brace Underscript 0 Endscripts plus one half upper B left-parenthesis v comma v right-parenthesis EndLayout"/>

      where upper B left-parenthesis v comma v right-parenthesis greater-than 0 unless double-vertical-bar v double-vertical-bar Subscript upper E left-parenthesis upper I right-parenthesis Baseline equals 0. Therefore any admissible nonzero perturbation of u will increase pi left-parenthesis u right-parenthesis.

      This important theorem, called the theorem or principle of minimum potential energy, will be used in Chapter 7 as our starting point in the formulation of mathematical models for beams, plates and shells.

      Given the potential energy and the space of admissible functions, it is possible to determine the strong form. This is illustrated by the following example.

      with ModifyingAbove upper E With tilde left-parenthesis upper I right-parenthesis equals left-brace u vertical-bar u element-of upper E left-parenthesis upper I right-parenthesis comma u left-parenthesis script l right-parenthesis equals modifying above u with caret Subscript script l Baseline right-brace.

      Since u minimizes pi left-parenthesis u right-parenthesis, any perturbation of u by v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis will increase pi left-parenthesis u right-parenthesis. Therefore pi left-parenthesis u plus epsilon v right-parenthesis is minimum at epsilon equals 0 and hence

      (1.39)period vertical-bar times times times d pi left-parenthesis right-parenthesis plus plus u times times epsilon v times times d epsilon Subscript epsilon equals 0 Baseline equals 0 period

      Therefore we have

      where the last two terms are zero because v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis. Integrating the first term by parts,

integral Subscript 0 Superscript script l Baseline kappa u prime v Superscript prime Baseline d x equals ModifyingBelow kappa u prime left-parenthesis script l right-parenthesis v left-parenthesis script l right-parenthesis minus kappa u prime left-parenthesis 0 right-parenthesis v left-parenthesis 0 right-parenthesis With presentation form for vertical right-brace Underscript 0 Endscripts minus integral Subscript 0 Superscript script l Baseline left-parenthesis kappa u prime right-parenthesis prime v d x

      (1.41)integral Subscript 0 Superscript script l Baseline left-parenthesis minus left-parenthesis kappa u Superscript prime Baseline right-parenthesis Superscript prime Baseline plus c u minus f right-parenthesis v d x equals 0 period

      Since this holds for all v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis, the bracketed expression must be zero. In other words, the solution of the differential equation

      (1.42)minus left-parenthesis kappa u Superscript prime Baseline right-parenthesis Superscript prime Baseline plus c u equals f comma left-parenthesis kappa u Superscript prime Baseline right-parenthesis Subscript x equals 0 Baseline equals k 0 left-parenthesis u left-parenthesis 0 right-parenthesis minus delta 0 right-parenthesis comma u left-parenthesis script l right-parenthesis equals modifying above u with caret Subscript script l Baseline

      Remark 1.3 Whereas the strain energy is always positive, the potential energy may be positive, negative or zero.

      The trial and test spaces defined in the preceding section are infinite‐dimensional, that is, they span infinitely many linearly independent functions. To find an approximate solution, we construct finite‐dimensional subspaces denoted, respectively, by

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