2.094 F E A
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2.094 FINITE ELEMENT ANALYSIS OF SOLIDS AND FLUIDS SPRING 2008 Homework 1 - Solution Instructor: Assigned: Due: Prof. K. J. Bathe 02/07/2008 02/14/2008 Problem 1 (10 points): In this problem, the principle of virtual work reduces to π π ππ½ x2 πππ ππππ π ππ½ = ππΊ ππ (a) πππ π ππΊπ π x2 x1 πΊπ π x2 x1 (b) x1 (c) configuration at time “t” virtual displacement Figure 1. Three simple independent virtual displacement patterns. (a) a simple tension in the x 1 direction; (b) a simple tension in the x2 direction; (c) a simple shear Page 1 of 5 π (a) A simple extension in ππ direction; ππ = ππ − π and ππ = π ππππ π πΊ ππ π =π =π Therefore, π πππ ππππ π ππ½ = π ππ½ π ππΊ ππ πΊπ π πππ π ππΊπ = π π Check! π (b) A simple extension in ππ direction; ππ = π and ππ = π ππ πΊπ ππππ ππ + π =π π = π (on the top surface) and ππ πΊπ = π (on the bottom surface) Therefore, π ππ½ π ππΊ ππ πππ ππππ π ππ½ = ππ × π π ππ½ = ππ × ππ½ πΊπ π πππ π ππΊπ = ππΊ π π × ππ π ππΊπ = ππ × π π½ = ππ π πΊπ = ππ π Check! (c) A simple shear; ππ = π and ππ = π − πππ π ππ ππππ πΊπ =π = π − πππ Therefore, π ππ½ π ππΊ ππ π πΊπ π πππ π ππΊπ = ππΊ πππ ππππ π ππ½ = π ππ × π − πππ π ππΊπ + ππ π π = ππ π π − πππ π πππ − ππ (on the top surface) Check! Page 2 of 5 π π ππΊ −ππ × π − πππ π ππΊπ ππ π − πππ π πππ = π (on the bottom surface) π Problem 2 (20 points): A B C D Figure 2. Plate with a hole in tension In this solution, the horizontal symmetry line corresponds to the line CD and the vertical symmetry line corresponds to the line AB. The distance of the horizontal line is measured from the point C and the distance of the vertical line is measured from the point A. (See Figure 2.) The stresses are shown through Figures 3 and 6. We can see that the solutions obtained with 9-node elements are better than those obtained with 4-node elements. One possible way, in general, to check calculated solutions is to see whether the stress boundary conditions are satisfied. Here, we know that we should have πππ π¨ = ππ due to the applied traction and πππ π© = πππ = πππ = π because no tractions are applied at the πͺ π« points B,C, and D. The solutions obtained with 9-node elements are closer to these exact values than those obtained with 4-node elements. In this problem we also have the analytical solutions for the πππ at point C and D, see reference 1. The solutions are compared in Figure 7. [1] Timoshenko, S.P. and Goodier, J.N., Theory of Elasticity, Third Edition, McGraw-Hill, 1970, pp. 94-95. C D Figure 3. Stresses on the horizontal symmetry line solved using 4-node elements Page 3 of 5 C D Figure 4. Stresses on the horizontal symmetry line solved using 9-node elements A B Figure 5. Stresses on the vertical symmetry line solved using 4-node elements A B Figure 6. Stresses on the vertical symmetry line solved using 9-node elements Page 4 of 5 110 STRESS-ZZ along the horizontal symmetry line 4.3 P 4-node elements 9-node elements Analytical solution 100 90 80 ο΄ 70 zz 60 50 40 30 20 10 -1 0.75 P 0 1 2 3 DISTANCE 4 Figure 7. ππ³π³ on the horizontal symmetry line Page 5 of 5 5 6 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.
