1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
1111
2222
1111
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
00
21
0 0 0
1
0 0 0 0
0 0 0 0
000 000
11
00000000
0 0 0 0
0 0 0 0
(b) Applying boundary conditions
[K] is reduced to
[K] = k
0
2
01
1
1
x
y
f
f
0
2
01
1
1
u
v
0
10
2 0
0 1
1
1
u
v
u1 = 0
v1 =
10
k
3.20
Element 1–2
C =
2
2
; S =
3
2
; L1–2 =
L
11
AE
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
C =
2
2
; S = –
2
2
; L2–3 =
L
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
22
AE
47
Element 1–2
L1–2 =
2
3
L;
= 60°
33
11
4 4 4 4
3 3 3 3
33
11
4 4 4 4
3 3 3
1
4 4 4 4
2
L
Element 1–3
L1–3 = L;
= 90°
0 0 0 0
00
11
0 0 0 0
00
11
AE
L
Element 1–4
L1–4 =
2
3
L;
= 120°
33
11
4 4 4 4
3 3 3 3
33
11
4 4 4 4
3 3 3 3
4 4 4 4
3
2
AE
L











Applying the boundary conditions
u2 = v2 = u3 = v3 = u4 = v4 = 0
3 3 3 3
33
11
2 4 2 4
2 4 2 4
3 3 3 3 3 3 3 3
2 4 2 4 2 4 2 4
0
( ) 0 ( )
0( ) 1 ( )
AE
L
=
3
4
33
4
0
1
0
AE
L
1
1
x
y
F
F
3
41
33 1
4
0
1
0
u
AE
v
L
100
100
=
33
44
11
3 3
11
44
00
11
00
uu
AE
vv
LL
400
L
231 L
231(100)
4
(3 3 4)
AE
AE
6
30 10
3.22
22
22
22
22
C CS C CS
AE CS S CS S
LC CS C CS
CS S CS S
For element 1;
= 120°
[k(1)] =
33
11
4 4 4 4
3 3 3 3
4 4 4 4
33
11
4 4 4 4
3 3 3 3
4 4 4 4
2
AE
L
[k(3)] =
3 3 3 3
4 4 4 4
33
11
4 4 4 4
3 3 3 3
4 4 4 4
33
11
4 4 4 4
3
2
AE
L
1000
3 3 3 3 1
8 8 8 8
00
v
L
1000
1000
=
1
1
1.77 0.16
0.16 0.59
u
AE
v
L
u1 =
6
422 (100)
1 10 10
u1 = 0.00422 in.
6
1570 (100)
1 10 10
Element (1)
1
1
2
2
x
y
x
y
f
f
f
f
33
11
4 4 4 4
3 3 3 3
4 4 4 4
33
11
4 4 4 4
3 3 3 3
4 4 4 4
422
1570
0
0
L
AE
L
AE
L
f2x = 287 lb
f2y = – 497 lb
22
22xy
(1)
f
A
=
– 5741
A
psi
(1) = – 5741 psi (C)
Element (2)
1
1
3
3
x
y
x
y
f
f
f
f
422
00
11
1570
0 0 0 0
00
11 0
0 0 0 0 0
L
AE
L
AE
L
f3x = – 422 lb
f3y = 0 lb
f (2) =
22
33xy
ff
f (2) = 422 lb (T)
(2) =
(2) 422f
AA
psi
(2) = 422 psi (T)
Element (3)
1
1
4
4
x
y
x
y
f
f
f
f
3 3 3 3
4 4 4 4
33
11
4 4 4 4
3 3 3 3
4 4 4 4
33
11
4 4 4 4
422
31570
20
0
L
AE
L
AE
L
3.23
Element (1)
C =
1
2
; S =
3
2
3
1
44
33
44
3
1
44
33
44
–
–
AE
L
Element (2)
[k(2)] =
3
1
44
33
44
3
1
44
33
44
–
–
–
–
–
–
AE
L
2
0
0
1 3 1 3
u
v
E



(2) (4)
0.64 0.48 0.64 0.48 (2)
25 0.48 0.36 0.48 0.36
0.64 0.48 0.64 –0.48 (4)
0.48 0.36 –0.48 0.36
x y x y
AE 







[k1–3] =
(1) (3)
0.64 0.48 0.64 0.48 (1)
25 0.48 0.36 0.48 0.36
0.64 0.48 0.64 0.48 (3)
0.48 0.36 0.48 0.36
x y x y
AE
{F} = [K] {d}
1
1
2
2
3
3
4
4
?
?
0
1000
0
1000
?
?
x
y
x
y
x
y
x
y
F
F
F
F
F
F
F
F
0.0756 0.0192 0.05 0 0.0256
0.0192 0.0811 0 0 0.0192
0.05 0 0.0756 0.0192 0
0 0 0.0192 0.0811 0
0.0256 0.0192 0 0 0.0756
0.0192 0.0144 0 0.0667 0.0192
0 0 0.0256 0.0192 0.05
0 0.0667 0.0172 0.0144 0
1
1
2
2
3
3
4
4
0.0192 00 0
0
0.0144 0 0.0667
?
0 0.0256 0.0192
?
0.0667 0.0192 0.0144
?
0.0192 0.05 0
?
0.0811 0 0
0
0 0.0756 0.0192
0
0 0.0192 0.0811
u
v
u
v
u
v
u
v
0 = [0.0756 u2 – 0.0192 v2 + 0 u3 + 0 v3] AE
u2 = 0.254 v2 (1)
Adding (3) and (5)
2142.4 = [0 v2 + 0.204 v3] AE
105021
u2 = 0.254 v2 = 0.254
105021
AE
u2 =
26675
AE
105021
AE
Going back to the local stiffness matrices
2
2
x
y
f
f
=
00
11
20 0 0 0 0
AE
2
22
2
1
2
1
120
0 26675
0
x
x
uAE
fu
v
ufAE
v









AE
20
f1–2 =
22
22
( ) ( )
xy
ff
f1–2 = 1333 lb (T)
Member 1–3
3
x
f
1
1
0
0
0.64 0.48 0.64 0.48
u
v
AE
3
y
f
22
33
( ) ( )
xy
ff
f1–3 =
22
33
( ) ( )
xy
ff
f1–3 = 1667 lb (T)
Member 2–4
22
xx
ff
2
2
0.64 0.48 0.64 0.48
u
v
AE
1
1
2
2
3
3
4
4
=0
=0
=0
=0
(3) (4)
0. 0256 0.0192 0 0
0.0192 0.0144 0 0.0667
0 0 0 0
0 0.0667 0 0
0.0756 0.0192 0.05 0
0.0192 0.0811 0 0
0.05 0 0.05 0
0 0 0 0.0667
u
v
u
v
u
v
u
v
1000 = [0 u2 + 00667 v2 + 0 u3 – 00667 v3] AE (2)
Adding (2) and (4)
Substituting (3) in (5)
2000 [00192 (– 0254 v3) + 00144 v3] AE
210000
AE
AE
Substituting in (2)
 1000 =
2
210000
0.0667 0.0667vAE
AE
224993
AE
Forces on members
Member 1–2
2
x
f
AE
1 0 –1 0
2
2
0
u
v



2
0
x
f
15
AE
20
AE
1
1
x
y
f
f
1
1
x
y
f
f
=
25
AE
0.64 0.48 –0.64 –0.48
0.48 0.36 –0.48 –0.36
1
1
3
3
0
0
u
v
u
v
AE
AE
53380 210000
25
f1–3 =
22
11
( ) ( )
xy
ff
f1–3 = 3333 lb (T)
Member 2–3
2
x
f
AE
0 0 0 0
2
2
0u
v



57
3.26
Since both elements 2–4 and 1–3 are removed, the global stiffness matrix will change
1
1
2
2
3
3
4
4
?
?
0
1000
0
1000
?
?
x
y
x
y
x
y
x
y
F
F
F
F
F
F
F
F
= AE
1
1
2
2
3
3
4
(1) (2) (3) (4)
0
0.05 0 0.05 0 0 0 0 0
0
0 0.0667 0 0 0 0 0 0.0667
?
0.05 0 0.05 0 0 0 0 0
?
0 0 0 0.0667 0 0.0667 0 0
?
0 0 0 0 0.05 0 0.05 0
?
0 0 0 0.0667 0 0.0667 0 0
0
0 0 0 0 0.05 0 0.05 0
0 0.0667 0 0 0 0 0 0.0667
u
v
u
v
u
v
u
v40
0 = 005 u2 u2 = 0
Adding (1) to (2)
The matrix, therefore, is singular and we get an inconsistent equation. (The
is also
3.27
AE
AE
1–3 2–3 3–4
50 ft 50 ft 20 ft
53.13 126.87 90
cos 0.6 0.6 0
sin 0.8 0.8 1
L
AE
(1) (3)
0.36 0.48 0.36 0.48 (1)
0.48 0.64 0.48 0.64




(0.72) ( )
AE
00
00
CS
SC
00
00
CS
SC
00
00
CS
SC
[k(1)] = 280 102
(1) (2)
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
2
2
[k(2)] = 280 102
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
1 1 1 1
2 2 2 2
(1) (3)
–
–
4
4
1.76 10



9
210 10
9
210 10
1
1
0
80
x
y
F
F



0.3883 0.0866
0.0866 0.65
1
1
u
v
u1 = 3.37 10–4 m
v1 = – 15.1 10–4 m
Element stresses
9
210 10
4
4
3.37 10



7.30 10
0
0





Element stresses
(1) =
9
210 10
3
[1 0 –1 0]
4
3
16.5 10
7.30 10
0
0








(1) = 115.5 MPa (T)
9
210 10
1 3 1 3
4
3
16.5 10


