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ASEN 5007 Introduction to Finite Element Methods Methods - Fall 2014 - Midterm Exam Exam 1 Solutions QUESTION 1. 1. 30 pts: pts: 15 for each each prob problem lem Figure Figure Q1.1 shows two flat plates of constant constant thickness thickness and uniform uniform material. They are loaded loaded as indicated. indicated. All edges are free: there are no external supports. The external boundary (B) is a circle. The crack in (A) and the square hole in (B) are centered. F
(A)
P
(B)
P
y q (uniform)
q (uniform)
thin crack
x
P
P F Figure Q1.1. Plane structures for Question 1. 1. In (B) the loads loads act at
±45◦ from x .
(a)
Try to reduce the portion to be discretized as as much as as possible by identifying symmetry and/or and/or antisymmetry lines. Sketch a diagram of the whole plate that identifi identifies such lines.
(b) (b)
Sket Sketch ch a coarse finite fi nite element mesh over the portion you picked in (a). Draw this separately from the sketch in (a). Make sure that there are several nodes over the crack in (A) and cutout in (B). Identify (for example with arrows) regions over which a fi a finer ner mesh may be desirable (but do not draw the fi the finer ner mesh). Identify the loads that are to be applied on the smaller region. region. For (A) there is no need to lump the distributed load to nodal forces.
(c)
Indicate how you would apply apply displacement displacement boundary conditions conditions on the nodes of the mesh picked picked in (b), so as to enforce symmetry and/or antisymmetry conditions identifi identified in (a).
Solution. Solution.
(A)
c
L
(B)
F
P
L
L
A
P
c
q crack
q
Symmetry and antisymmetry lines (marked in red)
L
A
F
L
c
L
c
reduce to 1/4 of original
F/2 nodes on crack must be left free to displace
reduce to 1/8 of of original
q (full load)
L
P
P
L
c
L
c
P/2 Support conditions and applied loads
c
L
A no supports on hole nodes not on symmetry or antisymmetry lines
near point load
Regions that would benefit from a finer mesh
at crack tip at entrant corner
Q1– Q1–3
near point load
QUESTION 2 35 points: 10 + 15 + 10 The plane truss problem defined in Figure Q1.2 has two elements and three nodes. Node 1 is fixed whereas 2 and 3 move over rollers as shown. The only load P has a magnitude of 40 and acts downward on node 3. Solve this problem by the Direct Stiffness Method. Start from the element stiffness equations given in (Q1.1) and (Q1.2) below. These are listed so you do not need to refer to the Notes, and already incorporate the E A / L factor in the stiffness matrices. The element stiffness equations in global coordinates are, for element (1)
−
100 0 100 0
0 0 0 0
−100
0 0 0 0
0 100 0
u x 1 (1) u y 1
(1)
f x 1 ( 1) f y 1
( 1)
(1)
f x 2 ( 1) f y 2
=
(Q1.3)
( 1)
u x 2 (1) u y 2
5 (1)
y
2
1 EA = 500
4
x
(2) 3 3
P = 40
Figure Q1.2. Structure for Questions 2 and 3.
and for element (2)
−
36 48 36 48
−
− −
48 64 48 64
−36 −48 −48 −64 36 48
48 64
u x 2 (2) u y 2
(2)
f x 2 ( 2) f y 2
( 2)
(2)
f x 3 ( 2) f y 3
=
(Q1.4)
( 2)
u x 3 (2) u y 3
(a)
Assemble the master stiffness equations.
(b)
Apply the given force and displacement BCs to get a reduced system of 2 equations and show it.
(c)
Solve the reduced stiffness system for the unknown displacements and show the complete node displacement vector.
− − − − = u = u = u = 0, f = 0, f = −40. Crossing out rows and columns 1, 2, 4
(b) BCs: u x 1 y 1 y 2 x 3 x 3 and 5 gives the reduced stiffness equation:
136 48
−
x 3
−48 64
= u x 2 u y 3
Q1–4
0 40
−
(Q1.6)
Solving:
= −3/10 = −0.30,
u x 2
u y 3
= −17/20 = −0.85
(Q1.7)
(c) Complete displacement solution: u
= [0
−0.30
0
0
−0.85 ]
0
T
(Q1.8)
QUESTION 3 35 points = 15 + 10 + 10 This question assumes the same structure solved for Question 2, with exactly the same load and roller supports . The reduced stiffness system upon applying those BCs is
∗ ∗ = ∗ u x 2 u y 3
∗ ∗
(Q1.9)
∗
where the * are the numbers you obtained in item (b) of Question 2. Suppose that an additional constraint of MFC type is given: 2u y 3 (Q1.10) u x 2
=
(a)
Starting from (Q1.11), get the modified equation by the master-slave method with u x 2 as slave, and solve for u y 3 .
(b)
Apply the MFC by the penalty function method, leaving the weight w as a free parameter. Show the modified system but do not solve.
(c)
Write down the modified system given by the Lagrange multiplier method, but do not solve.
Solution. (a) Transformation equation with u x 2 as slave:
= u x 2 u y 3
2 [ u y 3 ] 1
u
or
= T uˆ .
(Q1.11)
Modified system:
136 1] 48
[2
−48
2 1
[2
1]
0 , 40
− 64 − 5 = −0.096154. [ 4 1 6 ] [ u ] = [ −40 ] , u = − 52 The penalty element, using u − 2u = 0 as canonical form is u u 1 1 −2 0 w w = = [ 1 −2 ] u −2 −2 4 u 0 x 2
Modified system:
y 2
+w − − 2w 136 48
(Q1.12)
y 3
y 3
(b)
=
− − = x 2
x 2
y 3
y 3
48 2w 64 4w
+
u x 2 u y 3
0 . 40
−
.
(Q1.13)
(Q1.14)
(c) Lagrange multiplier system
Q1–5
128 48 1
−
−48 1 −2 64 −2 0
u x 2 u y 3 λ
0 40 0
= −
(Q1.15)
Switching signs in the last row and column is also considered correct. BONUS QUESTION. Up to 5 pts, total may not exceed 100. By inspection,
