Chapter 9: Derivative Boundary Conditions and Point SourcesSinks

1
Chapter 9 Derivative Boundary Conditions and
Point Sources/Sinks

• Chapter 8 evaluation of shear stresses in a
  noncircular section ? ø 0 on the boundary of
  the section.
• Most physical problems have a mixture of boundary
  conditions, however. You may know the value of ø
  or how ø is changing with position (dF/dx and
  dF/dy).
• Some problems have a point source (ex heat
  added) or sink (ex water removed).

2
We will discuss two types of boundary conditions

• Where F is known on the boundary.
• F is changing on the boundary, given by
• Dx(dF/dx)cos? Dy(dF/dy)sin? - MFb S

(9.1)
M S 0 for insulated (heat problems) or
impermeable boundaries (water flow) and axis of
symmetry. If Dx Dy then Dx (dF/dn) - MFb S
where dF/dn is the partial derivative normal to
the boundary.
a.
b.
3
Recall R(e) I(e) k(e) ?(e) f (e)

• Where I(e) is the inter-element requirement. It
  was mentioned in Chapter 3, often viewed as an
  error term and generally deleted, but not if
  there is a b.c.
• I(e) - ?G NT(Dx(dF/dx)cos? Dy(dF/dy)sin?)
  dG

Where G is the boundary and I (e) is evaluated
around the boundary, in a counterclockwise
direction.
4
The derivative boundary condition is included in
the finite element model using the inter-element
vector I(e). Ibc(e) kM(e) F(e)
fS(e) (9.12) Where kM(e) ?Gbc MNTNdG
fS(e) ?Gbc SNTdG
For any 2-D element

• Notes
• The Ibc(e) integrals are line integrals, not
  area integrals, because they are only defined on
  the element boundary.
• The text gives the derivation of equation (9.12).

5
Note that Dx (dF/dx) cos? Dy(dF/dy) sin? -
MFb S (9.12) has different signs on different
sides of the rectangular region. ? is the angle
from the x-axis to the outward normal so the sign
of M and S will change
6
Evaluation of Element Integrals a. bilinear
rectangular elements
k
m

• fS(e)

j
i
side i-j side j-k side k-m side i-m 1
0 0 1 1
1 0 0 0
1 1 0 0
0 1 1
7
kM(e)

• side i-j side j-k

0
0
1
2
0
0
0
0
MLij 6
MLjk 6
0
1
2
0
0
0
2
1
0
2
1
0
0
0
0
0
0
0
0
0
0
0
0
0
side k-m side i-m
0
0
0
0
MLim 6
0
0
0
0
MLkm 6
1
2
0
0
2
1
0
0
8
Evaluation of Element Integrals b. triangular
elements
k
j

• fS(e)

i
side i-j side j-k side k-i 1
0 1 1 1
0 0 1 1



9
kM(e)
side i-j side j-k
MLij 6
MLjk 6
side k-i
MLki 6
10
Point Sources Sinks

• Q occurs over a very small area.
• It is best (easiest) to put point sources/sinks
  at a node and add the contribution to the row of
  f corresponding to the node number.
• Sources are added and sinks are subtracted.
• If Q is located at point (x0,y0) (not at a node)
  you must calculate fQ(e) by distributing the
  contribution over all nodes of the element it
  occurs in

Ni (x0, y0) Nj (x0, y0) Nk (x0, y0)
f Q
11
Assignment