(2)
2
(2)
3
x
x
f
f
=
6000
7500
{F} =
11500
3000 6000
7500
x
F
11500
9000
7500
x
F
=
6
(2)(30 10 )
30
1
2
3
0
1 1 0
1 2 1
0 1 1
u
u
u
E
1
u
v
6
30 10
2
0.00825
0
u
v
AE
AE
AE
1
AE
109
1
2
lb
in.
F1x = 18000 lb
P(x) = 18000 –
1
2
(10 x) x
P(x) = 18000 – 5 x2
u(x) =
2
0
(18000 5 )
xx
AE
dx
3
15
AE
u(0) = 0 = C
3
5
3
6
18000
60 10
x
Analytical comparison with FEM
Element stress
Exact
(x)
()Px
A
3.55
Analytical solution
x = displacement
()fx
18000 300
x
2
E
Applying the boundary conditions
(x = 0) = 0 = C
Finite element solutions
(i) One element
A
2
(2) Two elements
f1x =
300 30
2
300 30 (1 1)
2
2
300 30
E
u
112
Computer solutions
One element
NUMBER OF ELEMENTS (NELE) = 1
NUMBER OF NODES (KNODE) = 2
NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD) = 5
DISPLACEMENTS X Y Z
NODE NUMBER 1 0.0000E+00 0.0000E+00 0.0000E+00
NODE NUMBER 2 0.9000E–02 0.0000E+00 0.0000E+00
STRESSES IN ELEMENTS (IN CURRENT UNITS)
FORCE (1, K) FORCE (2, K) FORCE (3, K)
0.000000E+00 0.000000E+00 0.000000E+00
ELEMENTS
K MODE (I, K) K(K) A(K)
NUMBER OF NONZERO UPPER CO-DIAGONALS (MUD) = 5
DISPLACEMENTS X Y Z
STRESSES IN ELEMENT (IN CURRENT UNITS)
ELEMENT NUMBER STRESS
1 = 0.67500E+04
Four elements
NUMBER OF ELEMENTS (NELE) = 4
NUMBER OF MODES (KNODE) = 5
NODE POINTS
K IFIX XC(K) YC(K) ZC(K)
1 1 1 1 0.000000E+00 0.000000E+00 0.000000E+00
FORCE (1, K) FORCE (2, K) FORCE (3, K)
0.000000E+00 0.000000E+00 0.000000E+00
4.500000E+03 0.000000E+00 0.000000E+00
ELEMENTS
114
NODE NUMBER 3 0.6750E–02 0.0000E+00 0.0000E+00
STRESSES IN ELEMENTS (IN CURRENT UNITS)
ELEMENT NUMBER STRESS
Analytical comparison with FEM
Analytical comparison with FEM
3.56
1
3
1500 lb
1500
x
x
F
F
[k(1)] =
30
AE
11
11
[k(2)] =
30
AE
11
11
Global equations
1
6
2
3
0
1 1 0
(2)(30 10 ) 1 1 1 1
30 0 1 1 0
u
u
1
3
1500
1500
x
x
F
F
Solving Equation (2)
2 106 (2 u2) = 3000
Element stresses
(1) = [C {d} =
E
L
[– C – S C S]
1
1
2
2
u
v
u
v
6
30 10
0
0
2
3.57 Bar hanging under own weight
Two element solution
AE
1 1 0
1
u
4
AL
AL
Wx =
V(x) =
Ax = P(x)
()
xPx
xAx
118
2
A L A E
2
3
L
3.58
1
2 1 1
L
f1x u1 + f2x u2 =
10
0
(100 + 5 x)
12
1
10
uux u dx
10 2
21
( – )
uu
(b)
x
1
1
2 1 1
( ) ( ) x
u x u u x
L
f1x u1 + f2x u2 =
4221
1
0(5 ) 4
uu
x x u
5
3
1
5 (4)
u
f1x = – (16) 5 +
3
5(4)
3
= 26.67 kN
3
= 106.67 kN total force
3.59
Exact solution
P(x) = 18000 – 300 x
1
x
1