3 2 3 2
22
3 2 3 2
22
66
12 12
66
42
66
12 12
66
24
0 0 0 0
00
00
0 0 0 0
00
00
AE AE
LL
EI EI
EI EI
L L L L
EI EI
EI EI
LL
LL
AE AE
LL
EI EI
EI EI
L L L L
EI EI
EI EI
LL
LL
Functional equation for transformation matrix between local and global coordinates.
0 0 0 0
0 0 0 0
0 0 1 0 0 0
CS
SC
1
1
1
2
2
2
u
v
u
v
1
1
1
2
2
2
x
y
x
y
f
f
m
f
f
m
Forces are in Newtons, moments in N m.
f1y comes back out of the equations as expected.
i = 2
klocal2 = klocal(Ai, Ci, Si, E, Ii, Li) Local k matrix for a element 2
99
66
67
99
66
66
2.8 10 0 0 2.8 10
0 1.4 10 4.2 10 0
0 4.2 10 1.68 10 0
2.8 10 0 0 2.8 10
0 1.4 10 4.2 10 0
0 4.2 10 8.4 10 0
66
66
66
67
00
1.4 10 4.2 10
4.2 10 8.4 10
00
1.4 10 4.2 10
4.2 10 1.68 10
T2 = T (Ci, Si) Transformation matrix for element 2.
0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1
Calculate local forces/moments in element 2.
22
22
22
33
33
33
uu
vv
uu
vv
Forces are in Newtons, moments in N m.
f2y and f3y again are of same magnitude as applied load at node 1 as expected.
m2 and m3 also have correct magnitude. m3 must be equal and opposite of applied moment.
f2 =
2
2
2
3
3
3
10000
0
30000
10000
0
30000
x
y
x
y
f
f
m
f
f
m



















246
5.13
Boundary conditions u1 = v1 =

1 = 0, u4 = v4 =
4 = 0
Element (1) by Equation (5.1.11)
C = 0; S = 1
E
L
2
2 2 2
12 6
6
0
00
04
II
L
L
I
L
uv
A
I
Element (2)
[k(2)] =
E
L
22
2
2 2 2 3 3 3
12
66
12
6
6
12
0 0 0 0
0
4 0 2
00
Symmetry 4
I
II
I
LL
LL
I
L
I
I
L
L
u v u v
AA
II
A
I
Element (3)
C = 0, S = – 1
E
E
2
3 3 3
6
12 0
I
I
L
L
uv
248
PLANE FRAME PROBLEM 5.13
NUMBER OF ELEMENTS = 4
NUMBER OF NODES = 4
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1,K)
1 1 1 1 0.000000 0.000000 0.000000 0.000000
ELEMENTS
K NODE(I, K) E(K) G(K) A(K) XI(K)
1 1 2 3.0000000E+07 0.0000000E+00 1.0000000E+01 2.0000000E+02
NODE DISPLACEMENTS Z-ROTATION
X Y THETA
1 0.00000E+00 0.00000E+00 0.00000E+00
ELEMENTS
K NODE(I, K) X-FORCE Y-FORCE Z-MOMENT X-FORCE Y-FORCE
1 1 2 – 0.4771E+04 0.1976E+03 0.2746E+05 0.4771E+04 – 0.1976E+03
2 2 3 0.1972E+03 – 0.1663E+03 – 0.1997E+05 – 0.1972E+03 0.1663E+03
0.2739E+05
0.3950E+02
5.14
Element (1)
C =
2
2
, S =
2
2
24
Symmetry 800
Element (2)
C = 1, S = 0
E
22
2
2
2
6
22
12 12
22
6
12
I
II
L
L
L
I
I
AC S A CS S
Equivalent nodal forces
f2x = 0 f2y =
2
2
wL
= – 22,500 lb
2
2
–
wL
675000


2
2
27,475,000
Solving
u2 = 0.0261 in.
Element forces
(1)
(1) (1) (1)
Element one (1)
(1)
1
(1)
1
(1)
(1)
2
(1)
2
(1)
2
(1)
x
y
x
y
f
f
f
f
m
5
5
4 2 4 2
1
55
1
44
48 2 48 2
5
5
42
5
1
4
48 2
[ ]{ }
0 0 0 0 0
00
0
100 2 0 50 2
0.03250
00
0.06945
0.002475
Symmetry
100 2
Td














(1)
1x
f
= 28740 lb =
(1)
2
–x
f
(1)
2y
f
= – 2077.7 lb =
(1)
2
–y
f
(1)
1
m
= – 8.8578 × 104 lb in.
m
(1)
(2)
{}f
=
(2) (2) (2) (2)
0
[ ] [ ] { } – { }k T d f
[T ] (2) {d}(2) =
0.0261
– 0.0721
– 0.0025
0
0
0









(2)
2
(2)
2
(2)
2
xe
ye
e
f
f
m
55
66
10 10
11
81 9 81 9
400 10 200
3 9 3
0 0 0 0
0
0
0.0261
0.0721
0.0025




5.15
Element (1)
2
12 I
L
=
4
2
12 (2 10 )
(4)
= 1.5 10–4 m2
6I
L
= 3.0 10–4 m3
E
L
=
6
70 10
4
= 1.75 107
kN
m
[k(1)] = 1.75 107
2 2 2
44
2
4
1.5 10 0 3.0 10
4 10 0
Symmetry 8 10
uv
[k(2)] = 1.75 107
2 2 2 3 3
22
4 4 4
44
2
4
4 10 0 0 4 10 0
1.5 10 3 10 0 3 10
8 10 0 4 10
4 10 0
Symmetry 8 10
u v u
1.75 107
2 4 2
2
2 4 4
2
34
2
23
43
4.015 10 0 3 10 4 10 0
4.015 10 3 10 0 3 10
1.6 10 0 4 10
4 10 0
Symmetry 8 10
u
v
u




















=
2
2
2
3
3
– 30
0
30
0
0
x
y
x
F
F
M
F
M









(1)
{}f
=
(1) (1) (1)
[ ] [ ] { }k T d
0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1
1
1
1
2
2
2
3
2
0
0
0
2.43 10
0
6.4 10
u
v
u
v











[T ](1) {d }(1) =
2
3
0
0
0
0
2.43 10
6.4 10











(1)
{}f
=
(1) (1) (1)
[ ] [ ] { }k T d
=
55
3 3 3 3
4 3 3
5
33
4
7 10 0 0 7 10 0 0
2.625 10 5.25 10 0 2.625 10 5.25 10
1.4 10 0 5.25 10 7 10
7 10 0 0
2.625 10 5.25 10
Symmetry 1.4 10
2
0
2.43 10






0
0
0





Element (1)
C = 0, S = 1
92
200 10 (1 10 )
EA
(2) (1)
0 0 0 0
AE
L
(2) (4)
1 0 1 0
1 0 1 0
0 0 0 0
Element (4)
C = 1, S = 0
AE
L
(1) (3)
1 0 1 0
1 0 1 0
0 0 0 0
Element (5)
2
2
Similarly assembling [k(6)] through [k(10)] we obtain
3 3 4 4 5 5 6 6
1 0 0 0 1 0 0 0
[] 0 1 0 1 0 0 0 0
0 0 1.3535 0.3535 0.3535 0.3535 1 0
0 1 0.3535 1.3535 0.3535 0.3535 0 0
[] 1 0 0.3535 0.3535 1.3535 0.3535 0 0
[] 0 0 0.3535 0.3535 0.3535 1.3535 0 1
0 0 1 0 0 0 1 0
0 0 0 0 0 1 0 1
ee
ie
u v u v u v u v
K
K
K
9
[]
2 10
[]
ei
ii
K
K
Now [Kii] – [Kie]
1
[]
ee
K
[Kei] {ui} = {fi} – [Kie]
1
[]
ee
K
{fe}
1.3535 0.3535 0 0 1 0 0.3535 0.3535
0.3535 1.3535 0 1 0 0 0.3535 0.3535
4
4
0 0 1.3535 0.3535 0.3535 0.3535 1 0
0 1 0.3535 1.3535 – 0.3535 0.3535 0 0
u
v


=
0
0
0
0
–
1 0 0.3535 0.3535
0 0 0.3535 0.3535
0 0 1 0
0 0 0 0
10
1 0 0 0
0 1 0 1 20000
0 0 1.3535 0.3535 0
0 1 0.3535 1.3535 0
2 109
3
3
4
4
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
u
v
u
v
=
20000
20000
20000
0
(b)
Adding two sections (a) and (b)
1.3535 0.3535 0 0
0.3535 2.3535 0 2
0 0 1 0
0 2 0 2
3
3
4
4
u
v
u
v
20000
20000
20000
0
Solving
u3 = 2.832 10–11 m, v3 = – 2.828 10–5 m
u4 = 1.0 10–5 m, v4 = – 2.828 10–5 m
5.17
[k]’s for each element are
33
29 10 10
(1) (2)
12 720 12 720
720 57600 720 28800
16.78
24 0 12 720
115200 720 28800
12 720
Symmetry 57600
2
2
3
3
v
v
=
10
0
10
0
16.78
12 720 12 720
720 57600 720 28800
12 720 24 0
720 28800 0 115200
3
3
2
2
v
v
=
10
0
10
0
1
ee
1
ee
1
12 720 –12 720 24 0 12 720
16.78 720 57600 720 28800 0 115200 720 28800
3
v
10
1
12 720 24 0 10
u1 = v1 = 0
C = 0, S = 1
[k(1)] {d}
0.0008 0.0012 0 0.0004
0.0024 0 0.0012
0.01 0
Symmetry 0.0008
1
2
2
2
u
v
d
3
0
0
0
80 10






Rearranging equations (rows and columns) we place interface displacements first

2
2
0.0024 0 0.0012 0.0012
0.01 0 0
u
v
0
0



261
2
2
2
2.0 0 0
Symmetry 0.12
u
0
0
Adding (1) and (2)
1 109
2
2
2
2.12 0 0.12
2.12 0.12
Symmetry 0.24
u
v
=
3
0
0
80 10





Solving
5.19
NUMBER OF ELEMENTS = 3
NUMBER OF NODES = 4
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1, K)
4 1 1 1 96.000000 0.000000 0.000000 0.000000
FORCE(2, K) FORCE (3, K)
0.000000 180000.000000
0.000000 0.000000
ELEMENTS
K NODE(I, K) E(K) G(K) A(K) XI(K)
1 3 1 3.0000000E+07 1.0000000E+00 1. 0000000E+01 2.0000000E+02
262
DISPLACEMENT Z-ROTATION
X Y THETA
ELEMENTS
K NODE(I, K) X-FORCE Y-FORCE Z-MOMENT X–FORCE Y-FORCE Z-MOMENT
3 2 4 0.1993E+05 0.6903E+04 0.6471E+06 – 0.1993E+05 – 0.6903E+04 0.6783E+06
5.20
NUMBER OF NODES = 6
NODE POINTS
K IFIX XC(K) YC(K) ZC(K) FORCE(1, K)
1 1 1 0 0.000000 0.000000 0.000000 0.000000
4 0 0 0 300.000000 360.000000 0.000000 0.000000
FORCE(2, K) FORCE(3, K)
0.000000 0.000000
0.000000 0.000000
0.000000 0.000000
ELEMENTS