ME 422
Do all 5 problems
Name
Problem
Problem
Problem
Problem
Problem
TOTAL
1
2
3
4
5
(15)
(8)
(26)
(26)
(25)
(100)
1
Problem 1 (15 points): Consider the weak form given by
Z
|V
∇T · k∇T dV +
{z
I
}
Z
|
T hT dS3 −
S3
{z
II
}
Z
|
T QdS2 −
S2
{z
III
}
Z
|
T hTa dS3 −
S3
{z
IV
a. Term I will ultimately result in contributions to (circle one):
The stiffness matrix
The forcing vector
Both the stiffness matrix and forcing vector
b.Term II will ultimately result in contributions to (circle one):
The stiffness matrix
The forcing vector
Both the stiffness matrix and forcing vector
c.Term III will ultimately result in contributions to (circle one):
The stiffness matrix
The forcing vector
Both the stiffness matrix and forcing vector
d.Term IV will ultimately result in contributions to (circle one):
The stiffness matrix
The forcing vector
Both the stiffness matrix and forcing vector
e.Term V will ultimately result in contributions to (circle one):
The stiffness matrix
The forcing vector
Both the stiffness matrix and forcing vector
2
}
Z
T gdV = 0
| V {z
V
}
Problem 2 (8 points): Consider the two 2-D finite elements shown below:
y
x
a. The isoparametric shape functions for the two elements will be (circle one)
The same
Different
b. The element stiffness matrix for the two elements will be (circle one)
The same
Different
Problem 3 (26 points): Suppose we have a two-node linear finite element of length L
with a constant internal heat generation g. The element’s area varies linearly from 1 at its
left edge to 2 at its right edge. Using isoparametric shape functions set up the integral for
the forcing function for the element. Set up all integrals completely but do not integrate.
3
Problem 4 (26 points): Consider the 4-node finite element shown below. Find the Jacobian matrix J.
y
(2,2)
1
2
(0,1)
4
3
(0,0)
(2,0)
4
x
Problem 5 (25 points): Consider the two-dimensional block shown below, which is insulated on its base and experiences convection on the three exposed sides.
convection
(0,1)
(1,1)
2
1
convection
convection
4
3
(0,0)
(1,0)
insulated base
The convection coefficient (film coefficient) h=3, the ambient temperature Ta =20. We discretize the block with a single 4-node conduction element and appropriate convection elements. The global stiffness matrix will be 4×4 and the forcing function will be a column
vector with 4 entries. Recall that
Kconv
hL
=
6
and
Fconv
"
hTa
=
2
2 1
1 2
"
1
1
#
#
Add the contributions due to convection to the empty matrix shown below:


































5

T1 











T2 




=



T3 








T4  
















