1. MESF593 Finite Element Methods HW #2 Solutions

2. Prob. #1 (25%) The element equations of a general tapered beam with a rectangular cross- section are given above. Derive the explicit expression of each term of K ij in terms of the dimensions (a, b, c, L) and Young’s modulus (E) of the beam. x = 0 q(x) x = L Q 1, v 1 Beam Width: b, Beam Thickness: a+(c-a)x/L v x Q 2, v 2 Q 4, v 4 Q 3, v 3

3. Prob. #1 Solution x = 0 q(x) x = L Q 1, v 1 Beam Width: b, Beam Thickness: a+(c-a)x/L v x Q 2, v 2 Q 4, v 4 Q 3, v 3

4. Prob. #1 Solution

5. Prob. #2 (15%) For the truss structure shown above at the left hand side, is it possible to use the finite elements shown at the right hand side for modeling? Why? 3 Linear Bar Elements: 3 Quadratic Bar Elements: 6 Nodes: F R 1, u 1 S 1, v 1 R 3, u 3 S 2, v 2 R 2, u 2 S 3, v 3

6. Prob. #2 Solution Connection to the mid node of a higher order bar element is not allowed. This is because the degree of freedom (i.e., displacement) of the mid node must be along the axial direction of the bar element as shown in Fig. 1. However, if the mid node is tied to the end node of another linear element, that means this common node can move in any direction. As a result, the higher order bar element may not be a “straight” element. u1u1 u3u3 U2?U2? P3P3 P1P1 P2?P2?  u1u1 u3u3 u2u2 P3P3 P1P1 P2P2  Fig. 1 Fig. 2

7. Prob. #3 (30%) Derive the explicit form of the stiffness matrix K ij of a combined bar-beam element arbitrarily oriented on a 2-D plane. E, A, I, L Y 1, v 1 Y 2, v 2 M 1,  1 X 1, u 1 M 2,  2 X 2, u 2 xx - Note: The plane coordinate transformation matrix for a rotation about the z-axis is

8. Prob. #3 Solution E, A, I, L Y’ 1, v’ 1 Y’ 2, v’ 2 M 1,  1 X’ 1, u’ 1 M 2,  2 X’ 2, u’ 2 From the lecture notes

9. Prob. #3 Solution

11. Prob. #4 (30%) Use the finite element method to solve for all reaction forces and moments at the boundaries of the anti-symmetric Z-frame structure given above. (All three segments have the same E, A, I, L) F F

12. Prob. #4 Solution For Element I (from the lecture notes) F F 1 2 3 4 I II III X 4, u 4 Y 4, v 4 X 3, u 3 Y 3, v 3 X 1, u 1 Y 1, v 1 X 2, u 2 Y 2, v 2 M 4,  4 M 3,  3 M 2,  2 M 1,  1

13. Prob. #4 Solution For Element III For Element II (use the result of Prob. #4, let  = -  /2) (from the lecture notes)

14. Prob. #4 Solution

15. (due to anti-symmetry, u 2 = -u 3, v 2 = v 3,  2 = -  3 )

16. Prob. #4 Solution