-
CHAPTER ONE
Finite Element MethodThe Big Picture
Five-bar truss example: Examples 1.4 p. 25, 1.7 p. 42, and 1.10
p. 50
The area of cross-section for elements 1 and 2 is 40 cm2, for
elements 3 and 4 is 30 cm2 and for element 5 is 20 cm2. Thefirst
four elements are made of a material with E = 200 GPa and the last
one with E = 70 GPa. The applied load P = 150 kN.
0 1 2 3 4 5Hm L
0
1
2
3
4
5
P
1
2
3 4
1
2
3
4
5
-
u1
v1
u2
v2
u3
v3
u4
v4
1
2
3
45
6
7
8
Specified nodal loads
Node dof Value
2 u2v2
0-150000
Global equations at start of the element assembly processi
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0
0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-1500000000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 1
E = 200000 A = 4000Element node Global node number x y
1 1 0 02 2 1500. 3500.
x1 = 0 y1 = 0 x2 = 1500. y2 = 3500.
L = "################################################Hx2 - x1L2
+ Hy2 - y1L2 = 3807.89Direction cosines: bs = x2 - x1L = 0.393919
ms =
y2 - y1L
= 0.919145
Substituting into the truss element equations we
getikjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.276067.2 177490. -76067.2
-177490.
-32600.2 -76067.2 32600.2 76067.2-76067.2 -177490. 76067.2
177490.
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u1
v1
u2
v2
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0.0.0.0.
y{zzzzzzzzzzzzz
The element contributes to 81, 2, 3, 4< global degrees of
freedom.
2
-
Locations for element contributions to a global vector:
ikjjjjjjjjjjjjj
1234
y{zzzzzzzzzzzzz
and to a global matrix:
ikjjjjjjjjjjjjj@1, 1D @1, 2D @1, 3D @1, 4D@2, 1D @2, 2D @2, 3D
@2, 4D@3, 1D @3, 2D @3, 3D @3, 4D@4, 1D @4, 2D @4, 3D @4, 4D
y{zzzzzzzzzzzzz
Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.2 0 0 0 076067.2 177490.
-76067.2 -177490. 0 0 0 0
-32600.2 -76067.2 32600.2 76067.2 0 0 0 0-76067.2 -177490.
76067.2 177490. 0 0 0 0
0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-150000.0000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 2
E = 200000 A = 4000Element node Global node number x y
1 2 1500. 3500.2 4 5000 5000
x1 = 1500. y1 = 3500. x2 = 5000 y2 = 5000
L = "################################################Hx2 - x1L2
+ Hy2 - y1L2 = 3807.89Direction cosines: bs = x2 - x1L = 0.919145
ms =
y2 - y1L
= 0.393919
Substituting into the truss element equations we
getikjjjjjjjjjjjjj
177490. 76067.2 -177490. -76067.276067.2 32600.2 -76067.2
-32600.2
-177490. -76067.2 177490. 76067.2-76067.2 -32600.2 76067.2
32600.2
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u2
v2
u4
v4
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0.0.0.0.
y{zzzzzzzzzzzzz
The element contributes to 83, 4, 7, 8< global degrees of
freedom.Locations for element contributions to a global vector:
ikjjjjjjjjjjjjj
3478
y{zzzzzzzzzzzzz
and to a global matrix:
ikjjjjjjjjjjjjj@3, 3D @3, 4D @3, 7D @3, 8D@4, 3D @4, 4D @4, 7D
@4, 8D@7, 3D @7, 4D @7, 7D @7, 8D@8, 3D @8, 4D @8, 7D @8, 8D
y{zzzzzzzzzzzzz
Adding element equations into appropriate locations we have
3
-
ik
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.2 0 0 0 076067.2 177490.
-76067.2 -177490. 0 0 0 0
-32600.2 -76067.2 210090. 152134. 0 0 -177490. -76067.2-76067.2
-177490. 152134. 210090. 0 0 -76067.2 -32600.2
0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 -177490. -76067.2 0 0 177490.
76067.20 0 -76067.2 -32600.2 0 0 76067.2 32600.2
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-150000.0000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 3
E = 200000 A = 3000Element node Global node number x y
1 1 0 02 3 0 5000
x1 = 0 y1 = 0 x2 = 0 y2 = 5000
L = "################################################Hx2 - x1L2
+ Hy2 - y1L2 = 5000Direction cosines: bs = x2 - x1L = 0 ms =
y2 - y1L
= 1
Substituting into the truss element equations we
getikjjjjjjjjjjjjj
0 0 0 00 120000 0 -1200000 0 0 00 -120000 0 120000
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u1
v1
u3
v3
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0000
y{zzzzzzzzzzzzz
The element contributes to 81, 2, 5, 6< global degrees of
freedom.Locations for element contributions to a global vector:
ikjjjjjjjjjjjjj
1256
y{zzzzzzzzzzzzz
and to a global matrix:
ikjjjjjjjjjjjjj@1, 1D @1, 2D @1, 5D @1, 6D@2, 1D @2, 2D @2, 5D
@2, 6D@5, 1D @5, 2D @5, 5D @5, 6D@6, 1D @6, 2D @6, 5D @6, 6D
y{zzzzzzzzzzzzz
Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.2 0 0 0 076067.2 297490.
-76067.2 -177490. 0 -120000 0 0
-32600.2 -76067.2 210090. 152134. 0 0 -177490. -76067.2-76067.2
-177490. 152134. 210090. 0 0 -76067.2 -32600.2
0 0 0 0 0 0 0 00 -120000 0 0 0 120000 0 00 0 -177490. -76067.2 0
0 177490. 76067.20 0 -76067.2 -32600.2 0 0 76067.2 32600.2
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-150000.0000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 4
E = 200000 A = 3000
4
-
Element node Global node number x y1 3 0 50002 4 5000 5000
x1 = 0 y1 = 5000 x2 = 5000 y2 = 5000
L = "################################################Hx2 - x1L2
+ Hy2 - y1L2 = 5000Direction cosines: bs = x2 - x1L = 1 ms =
y2 - y1L
= 0
Substituting into the truss element equations we
getikjjjjjjjjjjjjj
120000 0 -120000 00 0 0 0
-120000 0 120000 00 0 0 0
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u3
v3
u4
v4
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0000
y{zzzzzzzzzzzzz
The element contributes to 85, 6, 7, 8< global degrees of
freedom.Locations for element contributions to a global vector:
ikjjjjjjjjjjjjj
5678
y{zzzzzzzzzzzzz
and to a global matrix:
ikjjjjjjjjjjjjj@5, 5D @5, 6D @5, 7D @5, 8D@6, 5D @6, 6D @6, 7D
@6, 8D@7, 5D @7, 6D @7, 7D @7, 8D@8, 5D @8, 6D @8, 7D @8, 8D
y{zzzzzzzzzzzzz
Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.2 0 0 0 076067.2 297490.
-76067.2 -177490. 0 -120000 0 0
-32600.2 -76067.2 210090. 152134. 0 0 -177490. -76067.2-76067.2
-177490. 152134. 210090. 0 0 -76067.2 -32600.2
0 0 0 0 120000 0 -120000 00 -120000 0 0 0 120000 0 00 0 -177490.
-76067.2 -120000 0 297490. 76067.20 0 -76067.2 -32600.2 0 0 76067.2
32600.2
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-150000.0000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEquations for element 5
E = 70000 A = 2000Element node Global node number x y
1 2 1500. 3500.2 3 0 5000
x1 = 1500. y1 = 3500. x2 = 0 y2 = 5000
L = "################################################Hx2 - x1L2
+ Hy2 - y1L2 = 2121.32Direction cosines: bs = x2 - x1L = -0.707107
ms =
y2 - y1L
= 0.707107
5
-
Substituting into the truss element equations we
getikjjjjjjjjjjjjj
32998.3 -32998.3 -32998.3 32998.3-32998.3 32998.3 32998.3
-32998.3-32998.3 32998.3 32998.3 -32998.3
32998.3 -32998.3 -32998.3 32998.3
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u2
v2
u3
v3
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0.0.0.0.
y{zzzzzzzzzzzzz
The element contributes to 83, 4, 5, 6< global degrees of
freedom.Locations for element contributions to a global vector:
ikjjjjjjjjjjjjj
3456
y{zzzzzzzzzzzzz
and to a global matrix:
ikjjjjjjjjjjjjj@3, 3D @3, 4D @3, 5D @3, 6D@4, 3D @4, 4D @4, 5D
@4, 6D@5, 3D @5, 4D @5, 5D @5, 6D@6, 3D @6, 4D @6, 5D @6, 6D
y{zzzzzzzzzzzzz
Adding element equations into appropriate locations we havei
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
32600.2 76067.2 -32600.2 -76067.2 0 0 0 076067.2 297490.
-76067.2 -177490. 0 -120000 0 0
-32600.2 -76067.2 243089. 119136. -32998.3 32998.3 -177490.
-76067.2-76067.2 -177490. 119136. 243089. 32998.3 -32998.3 -76067.2
-32600.2
0 0 -32998.3 32998.3 152998. -32998.3 -120000 00 -120000 32998.3
-32998.3 -32998.3 152998. 0 00 0 -177490. -76067.2 -120000 0
297490. 76067.20 0 -76067.2 -32600.2 0 0 76067.2 32600.2
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzi
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
u1
v1
u2
v2
u3
v3
u4
v4
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz=
i
k
jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
000
-150000.0000
y
{
zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzEssential boundary
conditions
Node dof Value
1 u1v1
00
4 u4v4
00
Remove 81, 2, 7, 8< rows and columns.After adjusting for
essential boundary conditions we haveikjjjjjjjjjjjjj
243089. 119136. -32998.3 32998.3119136. 243089. 32998.3
-32998.3-32998.3 32998.3 152998. -32998.3
32998.3 -32998.3 -32998.3 152998.
y{zzzzzzzzzzzzzikjjjjjjjjjjjjj
u2
v2
u3
v3
y{zzzzzzzzzzzzz =
ikjjjjjjjjjjjjj
0-150000.
00
y{zzzzzzzzzzzzz
6
-
Solving the final system of global equations we get8u2 =
0.538954, v2 = -0.953061, u3 = 0.264704, v3 = -0.264704
