Chapter 2
2.1
(a)
[k(1)] =
11
11
0 – 0
0000
– 0 0
0000
kk
kk
22
22
0 0 0 0
0 0 0 0
0 0 –
0 0 –
kk
kk
[k 3(3)] =
33
33
0 0 0 0
0 0 –
0 0 0 0
0 – 0
kk
kk
1 2 1 3 2 3
k k k k k k
K] –1 {F} = {d}
[K –1] =
[]
det
T
C
K
2 3 2
2 1 2
2
1 2 2 3 2
( )( ) –
k k k
k k k
k k k k k
2 3 2
2 1 2
1 2 1 3 2 3
k k k
k k k
k k k k k k
3
4
u
u
=
2 3 2
2 1 2
1 2 1 3 2 3
0
k k k
k k k P
k k k k k k
u3 =
2
1 2 1 3 2 3
kP
k k k k k k
12
()k k P
lb
in.
lb
in.
lb
in.
lb
in.
lb
in.
(1) (2) (2) (3)
(1)
kk
(2)
kk
(2)
2
(2)
3
x
x
f
f
=
22
33
0.5
11
uu
k k
uu
k k
(2)
2
(2)
3
– 500 lb
500 lb
x
x
f
f
2.3
(a) [k(1)] = [k(2)] = [k(3)] = [k(4)] =
kk
kk
By the method of superposition we construct the global [K] and knowing {F} = [K] {d}
we have
1
2
?
0
x
x
F
F
000
2 0 0
kk
kkk
1
2
0
u
u
P
P
8
4
4
F5x = – k u4 + k
= – k
3
4
+ k
F5x =
4
k
2.5
u1 u2 u2 u4
[k (1)] =
11
11
; [k (2)] =
22
22
0 0 5 5 in.
0 9 5 14
2.6 Now apply + 3 kip at node 2 in spring assemblage of P 2.5.
9
0 0 5 5
0 9 5 14
3
4
0
u
u
3
0
F
where u1 = 0, u3 = 0 as nodes 1 and 3 are fixed.
Using Equations (1) and (3) of (A)
2
4
10 9
9 14
u
u
=
3
0
2.7
f = – k
= – k(u2 – u1)
–k k
tensile element
2.8
11
11
11
11
So
110
5000 5000
5000 5000
(3) (4)
5000 5000
5000 5000
(1) (2) (3) (4)
5000 5000 0 0
5000 10000 5000 0
0 5000 10000 5000
0 0 5000 5000
1
2
3
4
?
1000
0
4000
x
x
x
x
F
F
F
F
=
5000 5000 0 0
5000 10000 5000 0
0 5000 10000 5000
0 0 5000 5000
1
2
3
4
0u
u
u
u
[k(1)] =
1000 1000
1000 1000
500 500
500 500
1
2
3
4
?
– 8000
?
?
x
x
x
x
F
F
F
F
=
1
2
3
4
0
1000 1000 0 0
?
1000 2000 500 500
0
0 500 500 0
0 500 0 500 0
u
u
u
u
8000
1
2
3
4
x
x
x
x
F
F
F
F
=
0
1000 1000 0 0
1000 2000 500 500 4
0 500 500 0 0
0 500 0 500 0
1
2
3
4
x
x
x
x
F
F
F
F
=
4000
8000
2000
2000
lb
(1)
2
x
f
1000 1000 –4
(1)
2
x
f
4000
Element (2)
(2)
2
(2)
3
x
x
f
f
=
500 500 4
500 500 0
(2)
2
(2)
3
x
x
f
f
=
– 2000
2000
lb
Element (3)
(3)
2
(3)
4
x
x
f
f
500 500 4
500 500 0
(3)
2
(3)
4
x
x
f
f
2000
2000
2.11
[k(1)] =
1000 1000
1000 1000
; [k(2)] =
3000 3000
3000 3000