1
Numerical methods – Topics
Table of Contents
NUMERICAL METHODS – TOPICS
1
1. PIN JOINTED FRAME IN 3 DIMENSIONS
2. RIGID JOINTED FRAME IN 2 DIMENSIONS
3. FINITE ELEMENT ANALYSIS OF 2D HEAT FLOW
4. FINITE ELEMENT ANALYSIS OF PLANE STRESS
5. FINITE DIFFERENCE HEAT FLOW
6. FINITE DIFFERENCE PLATE BENDING
7. DYNAMIC RELAXATION OF 3D PIN JOINTED TRUSSES AND CABLE NETWORKS
8. DYNAMIC RELAXATION FORMFINDING OF MINIMAL SURFACE
9. DISCRETE FOURIER TRANSFORM
10. ‘SOCIAL FORCE’ SIMULATION OF PEOPLE MOVEMENT
11. B-SPLINE CURVES
12. PARAMETRIC DESIGN OF TOWER STRUCTURE USING ROBOT FOR ANALYSIS
1
3
6
7
9
10
10
11
11
13
13
14
1. Pin jointed frame in 3 dimensions
x1, y1 , z1  to x2 , y2 , z2 
2
2
2
x2  x1  y2  y1  z2  z1  .
Consider a member from
L
with cross-sectional area A and Young’s modulus E . Its unstressed length is
2
x1  x1 , y1  y1 , z1  z1 and x2  x2 , y2  y2 , z2  z2 , where the displacements are small, the length becomes
x  x  x  x   y  y  y  y   z  z  z  z 
If the ends move to
L  L 

2
2
x
 L
1
2
1
2
1
2
2
1
2
1
2
1
 x1  y2  y1   z2  z1  2x2  x1x2  x1  2y2  y1 y2  y1 2z2  z1z2  z1  in which the products of small
2
2
2
x
2
2
 x1
L
x
2
 x1
2
y
2
 y1 
L
y
2
 y1 
z
2
 z1
L
z
2
 z1
quantities have been neglected in the binomial series.
Hence the tension in the member is

 EA
y2  y1
z2  z1 


EA x2  x1 

T  EA 
x2  x1 L y2  y1 L z2  z1 L2  X
L
L 
L


L
Y
Z
X
Y
x1
 
y1
z 
Z  1 
x2
y2
 
z2 
where X  x2  x1,Y  y2  y1 ,Z  z2  z1 .


If p x1, p y1 , pz1 is the force that node 1 exerts on the member and p x2 , p y2 , pz2 is the force that node 2 exerts on the member, then
px1 
 X 
 
 
p y1 
Y 
p z1  T Z 
   .
px2  L  X 
p y2 
 Y 
 
 
 Z 
p z2 
3
Thus
k11 k12

k21 k22
k31 k32

k 41 k42
k51 k52

k61 k62
k13
k14
k15
k23
k33
k24 k25
k34 k35
k 43
k53
k44
k54
k 45
k55
k63
k64
k65
px1 
 X 
 
 
p y1
Y 
p z1  EA Z 
  3   X Y Z X
px2  L  X 
p y2
 Y 
 
 
 Z 
p z2 
 XX
k16 
XY
XZ  XX


k26 
YY
YZ
YX
 YX
k36  EA  ZX
ZY
ZZ ZX
 3 
k46 L  XX  XY  XZ XX
YX YY YZ YX
k56 


k66 
ZX ZY ZZ ZX
x1  k11
  
y1  k21
z  k
Y Z  1   31
x2  k41
y2  k51
  
z2  k61
 XY  XZ

YY YZ 
ZY ZZ 
.
XY
XZ 
YY
YZ 

ZY
ZZ 
k12
k13
k14
k15
k22
k23
k24
k25
k32
k33
k34
k35
k 42
k52
k43
k53
k 44
k54
k45
k55
k62
k63
k64
k65
k16 x1
 
k26 y1
k36 z1 
 
k 46x2
k56 y2
 
k66 z2 
where
2. Rigid jointed frame in 2 dimensions

EA
e where e is the elongation. The end moments and shear forces as applied to the beam are
L
EI 
6 
EI 
6 
M1  M 2 EI 
12 
M1 
 2 61  6 2 
41  2 2 
, M 2 
21  4 2 
 and F 
 in which anticlockwise moments and rotations
L 
L 
L 
L 
L
L 
L 

are positive
 and  is the relative displacement of the two ends in the direction corresponding to the shear force.
Tension in beam, T 
In matrix form:




4
EA
0

L
 T  
   0 12EI
 F  
L3
M1  
6EI
   0
L2
M 2  
6EI
 0

L2
0
6EI
L2
4EI
L
2EI
L

0 
 
6EI e 
L2  
2EI 1 

L 
 2

4EI

L 
We have to resolve the end displacements and forces parallel and perpendicular to the beam:
 f1x  c s 0 0
  
 
 f1 y  s c 0 0 T 
M1  0 0 1 0 F 
  
  and
f
c
s
0
0
 2x  
M1 
f 2 y  s c 0 0M 2 
  

M 2  0 0 0 1
x2  x1
y  y1
and s  2
.
L
L
EA
 f1x 


0
1x

 
 
L

 f1 y 
1 y 
 0 12EI


M1  T 1 
L3
Thus finally,   Q PQ in which P  

6EI
f 2x 
2x 
 0
L2
f 2 y 
2 y 

6EI
 
 
 0
M 2 
1 

L2
1x 
e  c s 0 c s 01 y 
 
  

 

s
c
0
s
c
0
  
 1
1  0 0 1 0 0 02x 
 
  


0
0
0
0
0
1
2 y 
 2 
 
1 
in which c 

0
6EI
L2
4EI
L
2EI
L




0 

c s 0 c s 0
6EI 


L2 , Q  s c 0 s c 0 and QT is the transpose of Q .
0 0 1 0 0 0
2EI 



L
0 0 0 0 0 1

4EI



L 
5
The matrix multiplication giving the element stiffness matrix K  QTPQ can be done using the computer: Kij 
using 0 to 3 for the summations). Alternatively:
c 2
s 2 cs
cs 0 c 2 cs 0


  2
s 2 0 cs s 2 0
cs
cs c


EA 0
0 0 0
0 0 12EI  0
0
K
 2
 3  2
2
L c cs 0 c
cs 0 L s
cs
2
cs s 2 0 cs


s
0
cs c 2



0 0 0
0 0
0
 0
 0
0 0 s 0 0 s 
0 0 0 0



0 0 c 0 0 c
0 0 0 0
6EI s c 0 s c 0  2EI 0 0 2 0
 2 


L 0 0 s 0 0 s L 0 0 0 0
0 0 c 0 0 c 
0 0 0 0



s c 0 s c 0 
0 0 1 0
2
2
Using the fact that c  s 1, this can be written


0 s 2 cs
0 cs c 2
0 0
0
2
0 s
cs
0 cs c 2
0 0
0
0 0

0 0
0 1

0 0
0 0

0 2
0

0
0

0
0

0

 4

Qmi PmnQnj  (maybe

m1
n1
4
6
c 2

cs
EA 12EI  0
K    3  2
L c
 L
cs

 0
0 0 s

0 0 c
6EI s c 0
 2 
L 0 0 s
0 0 c

s c 0
1
0 c 2 cs 0


0 cs s 2 0
0
0 0
0 0 12EI 0
 3 
0 c2
cs 0 L 1
0
0 cs
s 2 0


0 0
0 0
0
0 0 0 0
0 0 s 


0 0 c
0 0 0 0
s c 0  2EI 0 0 2 0


0 0 s L 0 0 0 0
0 0 0 0
0 0 c 


s c 0 
0 0 1 0
0 1 0 0

0 0 1 0
0 0 0 0

0 1 0 0
0 0 1 0

0 0 0 0
0

0
1

0
0

2
0
1
0
0
1
0
cs
s2
0
cs
s 2
0
0
0
0
0
0
0
3. Finite Element Analysis of 2D Heat Flow

3
L2
3
L1
1
1
2
L3
2





Consider triangular element with nodes at vertices x1, y1 , x2 , y2 and x3 , y3 . Temperatures at nodes are T1 , T2 and T3 . Temperature at any
point within element is



 

7
 x  x y  y  y  y x  x    x  x y  y  y  y x  x    x  x y  y  y  y x  x  
2
3
2
3
1
3
1
2
1
2
T1   3
T2   1
T3
T  

x2  x1 y3  y1  y2  y1x3  x1
 
x3  x2 y1  y2  y3  y2 x1  x2 
 
x1  x3 y2  y3  y1  y3 x2  x3 


y
2
 y3 x  x2  x3 y  x2 y3  y2 x3
2
T  y
3
 y1 x  x3  x1 y  x3 y1  y3 x1
2
1
T  y  y x  x  x y  x y
1
2
1
2
1 2
 y1 x2
2
2
T
3
where   12 L1L2 sin 3  12 L2 L3 sin 1  12 L3 L1 sin 2 is the area of the triangle.

Heat flow vector is f  

  k x
k y2  y3 T1  y3  y1T2  y1  y2 T3 i
2

 x3 T1  x3  x1T2  x1  x2 T3 j
2
 The change of sign is because the flow is in the direction of decreasing temperature.
2
where k is the thermal conductivity.
Rate of heat flow out of element across side 1 – 2 is equal to


k y2  y3 T1  y3  y1T2  y1  y2 T3 y1  y2  k x2  x3 T1  x3  x1 T2  x1  x2 T3 x1  x2 

x1  x2 i  y1  y2 j  f  k 
2
2


 



k L T  L3 L1 cos2T1  L3 L2 cos1T2
2
3 3
2
 



 

4
Q1 , Q2 and Q3 are the rates of heat flow into the nodes,
2
3
2
1
2
2
2
2
2
1
2
2
2
2
2
3
2
1
1
2
1
2
2
2
3
2
2
3
2
3
3

Triangular constant strain elements are applied to the linear elastic analysis of a homogeneous isotropic material.
The displacements u and v are given by


.
L  L  L  L  L  L T 
be taken as the absolute value of the area of the triangle.
 2L
L  L  L T . Note must


 L  L  L  2L T 
4. Finite element analysis of plane stress



k 2L23T3  L22  L23  L12 T1  L12  L22  L23 T2
This is equal to twice the heat flow into node 3. Thus if

2L12
Q1 

  k
2
2
2
Q2  8  L3  L1  L2
 2 2 2

Q3 


 L2  L3  L1

8
u
 
v 

y
2

 y3 x  x2  x3 y  x2 y3  y2 x3 u1
 
2
v1 
y
3

 y1 x  x3  x1 y  x3 y1  y3 x1 u2
 
2
v2 
y  y x  x  x y  x y
1
2
1
2
2
1 2

 y1 x2 u3
 .
v3 
 u  u1 

 v 
x   x   1 
 
v  u2 
G
The strains are  y  
 y  v2 
 

 xy 
 v u   
u3
  

x y  
v 
3


y  y 

 x  x 
0
0
0
y3  y1
y1  y2 
2
3

   
1 
G
x2  x3 
0
x3  x1 
0
x1  x2 .
where
The
stresses
are
where
 0
 y  S y 

2 


 xy
 
 xy 

x2  x3  y2  y3  x3  x1  y3  y1  x1  x2  y1  y2 


 x 
1 
0 
 
E 
t


S

1
0 , E is Young’s modulus and  is Poison’s ratio. The strain energy of the element isU  x  y  xy  y  where
2

 1  
2

0 0 1  
 xy




2 
u1


 
v1 
x 
u 
  t
t
t is the thickness of the plate. Thus U  x  y  xy S y  u1 v1 u2 v2 u3 v3 GTSG 2.
2
2
v2 

 xy 

u3
 
v3 

9
U 
 
u1 
U 
f x1   
u1 
  v1 
 
f y1  U 
v1 
f x2  u 
u 
2
 tGTSG 2 . Hence the element stiffness matrix is k  tGTSG. Again  must be taken as the absolute value
The nodal forces,  
 
f y2  U 
v2 

v
f x3   2 
u3 
  U 
 
v3 

f y3 

u3 
U 
 
v3 
of the area of the triangle.
5. Finite difference heat flow

 2  2
dy 
d y  x  .....
Consider a series of points in order along a curve, x i, yi . We have y  y  y i  
 
x   2 
dx i
dx i 2
 2  x  x 2
dy 
d y  i1 i
yi1  y i  
dx 
 xi1  xi   2 
2
 i
dx 
y i1  yi yi  y i1

 2 
d y  x i1  x i xi  x i1
i


so that
and therefore 
.
2 
x i1  x i1
 2  x  x 2
dx


dy 
i
i1 
d y   i
yi  yi1  
2
dx 
 xi  x i1   2 
2
 i
dx i
In two dimensions, with lines of constant i having constant x and lines of constant j having constant y ,
10
i1, j   i, j  i, j   i1, j  i, j1  i, j i, j   i, j1


 2 
 2 
x

x
x

x
y

y
yi, j  y i, j1




i1,
j
i,
j
i,
j
i1
i,
j1
i,
j
     
.

 2 
 2 
xi1, j  x i1, j
y i, j1  yi, j 1
x i, j y i, j
2
2
i1, j
xi1, j  xi, j
 i1, j

i, j 1

 i, j 1
x i, j  x i1 yi, j 1  y i, j yi, j  y i, j 1

xi1, j  xi1, j
y i, j1  yi, j 1
2
2
 
2
2
Thus if  2  2  2  0 , then i, j 
. If the spacing is constant and equal to a in
1
1
1
1
x y


xi1, j  xi, j x i, j  x i1 yi, j 1  y i, j yi, j  y i, j 1

xi1, j  xi1, j
y i, j1  yi, j 1

2
2

i1, j  i1, j  i, j1  i, j1
both the x and y directions, i, j 
.
4
6. Finite difference plate bending


2
 4w
 4 w  4 w   2  2 
Et 3
 p
4
  w  4  2 2 2  4   2  2  w where D 
As above but solving
.
D
x
x y y x y 
12 1  2


7. Dynamic relaxation of 3D pin jointed trusses and cable networks
i.
ii.
Apply
loads to nodes.

Calculate member lengths using Pythagoras’s theorem, L  x 2  y 2  z2 . Hence member tensions. Add forces from members to
nodes.

11
iii.
Move nodes using the Verlet algorithm:
Ft Resultant force

m
Mass
Velocity  v t  v t  a tt
Acceleration  a t 
t
2
t
2
Position  rt t  rt  v
t
t
t
2
The carry over,  1.0 to represent damping.
iv.
Go back to i.

8. Dynamic relaxation formfinding of minimal surface

That is a soap film surface, good shape for fabric structures. Use triangular element. If T is the surface tension and the corners of an element
are at ri ,i  0,1,2 , the force on node i is fi 



T
r
k
 ri  rk  ri 
2 qi  qi

qi where
qi  rk  rj  rk  ri  rj  ri  rj  rk  rj  ri  rk  ri 


qi  rk  ri  0

j  i  1 If j  2, j  i  2
k  i  2 If k  2, k  i 1
9. Discrete Fourier transform

Also known as fast Fourier transform

12
Write
pt  
N
P e
i
2 nt
T
n
n N


N
 A  iB e
n
N
 An  iBn e
i
2 nt
T
n
N 
2nt
2nt 
 A0  2An cos
 Bn sin

T
T 
n1 
A0   p
An  An
Bn  Bn

Given p t find An and Bn .


2 nt
T
n N
n N

i

13
10. ‘Social force’ simulation of people movement
11. B-spline curves
Uniform cubic B-spline:
1 3 3 1pi1 

 
1 3 6 3 0 pi 
3
2
ru  u u u 1
6 3 0 3 0pi1 

 
1 4 1 0pi2 
pi1 
1 3 3 1pi1 
 

 
p
1 3 6 3 0 pi  1
r0 0 0 0 1
 1 4 1 0 i 
pi1 
6 3 0 3 0pi1  6
 

 
1 4 1 0pi2 
pi2 
pi1 
1 3 3 1pi1 


 


p
1 3 6 3 0 pi  1
r
1  1 1 1 1
 0 1 4 1 i 



6 3 0 3 0 pi1 6
pi1 
 

 
1 4 1 0pi2 
pi2 



14
1 3 3 1pi1 

 
dr
1 3 6 3 0 pi 
2
 ru u  3u 2u 1 0
ru
6 3 0 3 0pi1 

 
1 4 1 0pi2 
pi1 
1 3 3 1pi1 
 

 
p
1 3 6 3 0 pi  1
ru 0 0 0 1 0
 3 0 3 0 i 
pi1 
6 3 0 3 0pi1  6


 


1 4 1 0pi2 
pi2 
pi1 
1 3 3 1pi1 
 

 
p
1 3 6 3 0 pi  1
ru 
1  3 2 1 0
 0 3 0 3 i 
pi1 
6 3 0 3 0pi1  6
 

 
1 4 1 0pi2 
pi2 


12. Parametric design of tower structure using Robot for analysis
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