EML 5526 Finite Element Analysis and Design Exam-III

Last Name:Gandlapalleramana
First Name: Dimpu Dharshan Gowda

EML 5526 Finite Element Analysis and Design
Exam-III
50 minutes

Problem 1:
The cylindrical test specimen shown in the figure is loaded axially. Assume that the forces are applied
uniformly along the top and bottom surfaces. Describe the simplest finite element model that makes use of
symmetry and can determine the stress distribution though the length of the specimen. Show a sketch of
the model showing the mesh, forces, BCs.

Sketch showing mesh, loads and BCs: (10 pts)




Boundary conditions:
𝑈𝑥 = 0

𝑈𝑦 ≠ 0

The loads are applied axially along the x-axis in lateral direction on top surface is P/2.




Type of Element: (Analysis type and the interpolation used) (5pts)


Triangular or Quadrilateral plane stress element.


Material properties needed: (5 pts)

Youngs modulus, yield strength to predict failure and poisons ratio.


Problem 2:
The figure shows a 4-node quadrilateral element in parametric space where
two of the nodes are collapsed or merged to form a triangle in the real
space. Assuming global node 1 is local node 1, the nodal displacements are
found to be: {u1, v1, u2, v2, u3, v3, u4, v4} = {1, 0, 0, 2, 1, 0, 1, 0}x10-3in.

(a) Compute the displacement field u(s,t) and v(s,t) (15pts)

For the 4 node element:

1
𝑁1 = (1 − 𝑠)(1 − 𝑡);
4



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, Last Name:Gandlapalleramana
First Name: Dimpu Dharshan Gowda
1
𝑁2 = (1 + 𝑠)(1 − 𝑡);
4

1
𝑁3 = (1 + 𝑠)(1 + 𝑡);
4

1
𝑁4 = (1 − 𝑠)(1 + 𝑡)
4

U(s,t) = 𝑈1 𝑁1 + 𝑈2 𝑁2 + 𝑈3 𝑁3 + 𝑈4 𝑁4

V(s,t) = 𝑉1 𝑁1 + 𝑉2 𝑁2 + 𝑉3 𝑁3 + 𝑉4 𝑁4

𝑈1 = 1, 𝑈2 = 0. 𝑈3 = 1, 𝑈4 = 1

𝑉1 = 0, 𝑉2 = 2, 𝑉3 = 0, 𝑉4 = 0


U(s,t) =( 𝑁1 +𝑁2 + 𝑁3 )*10−3
1 1 1
= [ 4 (1 − 𝑠)(1 − 𝑡) + 4 (1 + 𝑠)(1 + 𝑡) + 4 (1 − 𝑠)(1 + 𝑡)] *10−3

1
= (3+t-s+st) *10−3
4

3 𝑡 𝑠 𝑠𝑡
=(4+4−4+ ) *10−3
4

V(s,t) = 2𝑁2 *10−3

1
= 2* 4 (1 + 𝑠)(1 − 𝑡)*10−3

= ½ (1+ s – t – st ) *10−3

(b) Compute the (x, y) coordinates of a point P whose parametric coordinates (s, t) = (0,0) (10 pts)


We know that from the figure,
𝑥1 , 𝑦1 = (1,1)
𝑥2 , 𝑦2 = (−1,1)
𝑥3 , 𝑦3 = (0,0)
𝑥4 , 𝑦4 = (0,0)
Then,

X = 𝑁1 𝑥1 + 𝑁2 𝑥2 + 𝑁3 𝑥3 + 𝑁4 𝑥4

= 𝑁1 ∗ 1 + 𝑁2 ∗ −1 + 0 + 0
1 1
= (1 − 𝑠)(1 − 𝑡) − (1 + 𝑠)(1 − 𝑡)
4 4


= ¼ ( 1- t-s+st-1+t-s+st)

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