AL
AL
AL
AL
AL
AL
Chapter 16
16.1
1 2 2 3
AL
21
AL
21
6
AL
00
21
0
1 4 1
01 4 1
0012
16.3
[K] =
AE
L
0
11
1 2 1
011






6
AL
0
21
([K] –

=
0
0
2 =
E
22
60 60 36
7 7 7
4 (1)
2
=
8.571 7.273
2
1 = 0.649
2 = 7.922
1
2
1
2
2
16.4 (a) Two equal length elements
From Problem 16.3 results
1
1
[M] =
00
21
0
1 4 1
601 4 1
0012
AL
2 [M]) {X} = 0
1 1 2
01
L
or
0
1 4 1
0
2
1 1 1 4 1
260
1 1 2
01
= 0
2
36
2
20
22
22
3 3 3
2
2
6 6 3 6 6 3
= 0
2
2 2 2
84
11
2
13
rad
s
16.5
t0 = 0 d0 = d0dot = 0
1
2
2
ft
s
d–1 = 0 – 0 +
2
0.03
2
25 = 0.01125 ft
2
= 0.01125 ft
d2 =
1
2
{(0.03)2 (66.67) + (2 2 – 0.032 2000 0.01125}
1
2
ft
1
2
1
2
ft
2
2
2
2
t3 = 0.09 s
1
2
ft
1
1
0.06 0.039319 1.084121 2.347196 83.33333 602.8986
0.12 0.081832 –0.13384 –44.3317 75 1254.753
0.18 0.011632 –1.62917 13.3684 50 178.3512
16.6
(a) Using central difference
d0 = 0, d0dot = 0
t = 0.02 s
1
2
[K] = k = 1200
lb
ft
d0dotdot =
1
2
[20 – 1200 (0)] = 10
2
ft
s
{d–1} = 0 – (0.02) (0) +
2
0.02
2
(10) = 0.002 ft
1
2
= 0.002 ft
1
2
= 0.00672 ft
1
2
2
ft
s
0.00672 0
ft
1
2
d3 = 0.01223 ft
d2dotdot =
1
2
(12 – 1200 (0.00672)) = 1.968
2
ft
s
d2dot =
0.01223 – 0.002
2 0.02
= 0.2558
ft
s
Step 3 t = 0.06 s F3 = 8 lb
1
2
= 0.0164 ft
1
2
2
ft
s
0.0164 – 0.00672
2 0.02
ft
s
Step 4 t = 0.08 s F4 = 4 lb
1
2
= 0.01743 ft
1
2
2
ft
s
d4dot =
0.01743 – 0.01223
2 0.02
= 0.13
ft
s
Step 5 t = 0.10 s F5 = 0
1
2
= 0.01428 ft
1
2
2
ft
s
0.01428 0.01640
2(0.02)
ft
s
Summary
t, s
d, ft
ddot,
ft
s
ddotdot,
2
ft
s
0
0
0
10
0.02
0.002
0.168
6.8
0.04
0.00672
0.2558
1.968
0.06
0.01223
0.242
–3.338
0.08
0.01640
0.130
–7.89
0.10
0.01743
–0.053
–10.46
(b) Newmark’s time integration method (Mathcad solution)
F0 = 20 lb K = 1200
lb
ft
M = 2 slug M = 2
2
lb s
ft
t = 0.02s
1
1
d2 =
2prime
prime
F
K
d2 = 6.252 10–3 ft
Acceleration at t = 0.04
2
11
2
11
3
t
4
5
F1prime = F1 +
2
M
t
[6d0 + 6(t)d0dot + 2(t)2d0dotdot]
F1prime = 56 lb
d1 =
1prime
prime
F
K
d1 = 1.795 10–3 ft
Acceleration at t = 0.02
d1dotdot =
2
2
6
t
(d1 – d0) –
6
t
d0dot – 2 d0dotdot d1dotdot = 6.923
2
ft
s
Velocity at t = 0.02
d1dot =
3
t
(d1 – d0) – 2 d0dot –
2
t
d0dotdot d1dot = 0.169
ft
s
d2 =
2prime
prime
F
K
d2 = 6.252 10–3 ft
Acceleration at t = 0.04
d2dotdot =
2
2
6
t
(d2 – d1) –
6
t
d1dot – 2 d1dotdot d2dotdot = 2.249
2
ft
s
ft
6
6
ft
F4 =
1
5
F0 F4 = 4 lb
F4prime = F4 +
2
M
t
[6 d3 + 6(t)d3dot + 2(t)2d3dotdot]
3
t
ft
= 0.167 K 31200
= 0.5
K31200
16.10
[K] =
1 1 0
1 2 1
0 1 1
AE
L
= 104
1 1 0
1 2 1
0 1 1
AL
1 0 0
1 0 0
2
400 2 200
= 25.7
rad
s
t
max
32
4
=
32
4 25.7
= 0.058 s
1
1
16.11
Global stiffness matrix
v1
v2
2 v3
3
EI
22
12 6 12 6 0 0
6 4 6 2 0 0
12 6 24 0 12 6
LL
L L L L
LL
[m] =
2
AL
1 0 0 0 0 0
0 0 0 0 0 0
002000
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 0