2.094 F E A
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2.094 FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS SPRING 2008 Homework 2 - Solution Instructor: Assigned: Due: Prof. K. J. Bathe 02/14/2008 02/21/2008 Problem 1 (20 points): y z U1 x, u U2 1 U3 2 x=20 x=100 x=160 In this problem, the global coordinate system is used. Here, the non-zero strain components are [πΊππ , πΊππ ]. Hence the stress-strain law represented by the matrix πͺ is πͺ= π¬ π π π−π π π π And πΊπ» = πΊππ πΊππ = πΌπ» = πΌπ πΌπ ππ π ππ π πΌπ The applied force is given by π ππ© π π = πππ Page 1 of 3 The thickness of each element is given by π(π) = π (πππ − π) ππ π(π) = π. π For each element, the interpolation functions are π―(π) = π π (πππ − π) − (ππ − π) π ππ¨π« ππ ≤ π± ≤ πππ ππ ππ π π (πππ − π) − (πππ − π) ππ¨π« πππ ≤ π± ≤ πππ ππ ππ π―(π) = π Then, π©π» = π©(π) π©(π) ππ― π― ππ π π π π ππ ππ = (πππ − π) (ππ − π) − π πππ πππ − π π ππ ππ = (πππ − π) (πππ − π) π − πππ πππ π − The stiffness matrices are πππ π²= π© ππ πππ = π π» πͺ π© π π π½ π πππ + πππ π© π π» πͺ π© π π π» π© πͺ π© πππ π(π) ππ ππ π + π π½ π πͺ π© π ππππ π π½ π π© ππ π π» π π(π) ππ ππ π πππ The external force vector is πππ πΉ= π― ππ πππ = π π» ππππ π π½ π πππ + π― πππ π― π π» πππ ππππ π(π) ππ ππ π + ππ π π» π― πππ Therefore we have π²πΌ = πΉ Page 2 of 3 π π» ππππ π(π) ππ ππ π After integrations using Matlab, π²= ππ. ππππ − ππ. πππππ π¬ −π. ππππ + π. πππππ π − ππ π −π. ππππ + π. πππππ ππ. ππππ + π. πππππ −ππ. πππ π −ππ. πππ ππ. ππππ + π. πππππ ππππππ πΉ = πππ πππππππ πππππππ The same results are obtained using a local coordinate system for each element (like the example 4.5 in the textbook). For your reference, a sample solution using a local coordinate system is attached. Page 3 of 3 MIT OpenCourseWare http://ocw.mit.edu 2.094 Finite Element Analysis of Solids and Fluids II Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
