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AAE321/ME345 – Midterm exam
Wednesday, March 17, 2004 – 8:30am to 9:50pm
Closed notes/closed books/no calculator
- Shape functions of triangular element Using the triangular coordinates
Li ( x , y ) ( i = 1,2,3), write the shape functions for nodes 3, 6 and 10 for the 10-node (global) triangular element shown in Figure 1.
- Finite element formulation for anti-plane shear problems Consider the structural problem shown in Figure 2.
z
y
x
L
Figure 2
•^10
Figure 1
The “slab” of linearly elastic material is subjected to shear loads applied on the external surface and parallel to the z - axis. The cross-section is assumed to be independent of z and the length L of the slab is assumed to be very big (i.e., the slab is almost infinite in the z - direction). For this type of problem, it is possible to show that the 3D theory of elasticity reduces to a much simpler scalar problem. There is only one nonzero displacement component: the displacement u 3 ( x, y ) parallel to the z - axis. Since it depends only the in-plane coordinates x and y , we just have a 2-D problem to solve on the cross-section S of the slab (Figure 3).
x
y
S
!
Figure 3
The displacement u 3 ( x, y )(which is perpendicular to the plane of the cross-section) can be shown to satisfy the following scalar equation
μ!
(^2) u 3 ! x^2
+^!^
(^2) u 3 ! y^2
' (^) + f 3 ( x,y ) = 0 on S ,
where μ denotes the shear modulus and f 3 is the (distributed) body force parallel to the z - axis. We have two types of boundary conditions u 3 = u 3 * along! u (imposed displacement), μ! u^3 ! n
= T 3 * along! T (imposed traction),
where u 3 * and T 3 * denote the applied (and therefore known) out-of-plane displacement and traction, respectively, and n is the outward normal to the boundary.
For this class of problems (called anti-plane shear problems), there are only two non- zero stress and strain components
! 13 =^1 2
" u 3 " x
,! 23 =^1
" u 3 " y
! 13 = 2 μ" 13 ,! 23 = 2 μ " 23..
The potential energy Π for anti-plane shear problems is given by
