23
4 4 4 8
24 12 28 288
2
L L L L
3
48
12 120
LL
23
6 12 48 648
4 4 4 6 12
22
L L L L L
48
2
LL
22 3 4
4 4 4 8 12
12 56 12 864 2448
22
L L L L L
= 0
12
1 =
1
22
1.875
2
EI
AL
=
1
2
2
0.879 EI
A
L
(a)
2=
1
22
4.694
2
EI
AL
=
1
2
2
5.5 EI
A
L
2
22
2
2
24 0 12 6
8 6 2
24 0
Symmetry 8
l
l l l
l
Rewrite 4 4 determinant as
2
2
2
22
24 0 12 6
24 0 6 30
6 0 24 24
6 2 0 8
l
l l l
l
l l l
= 0
2
2
24 12 6
l
EI
22
24 0 6
l
16.13
16.14 Note that even though damping data was not given in the problem. This solution includes
damping.
Let the time step or increment = 0.002 s and the number of time steps = 200.
16.16
620
16.17
16.18
Using Equation (16.8.16)
1[ ] [ ]MK
1(1 )MK
[K =
1000 500 0 0 0
1000 500 0 0
1000 500 0
1000 500
Symmetry 500
8000 0 0 0 0
8000 0 0 0
16.19
Two element solutions
0.4 cm diameter
W
1 1 2 1
AK h L
0.50893 0.49951 0
Symmetry 0.50893
[M =
0.1398 0.06991 0
0.2796 0.06991
Symmetry 0.1398
0.2356
0.2356
Using Equation (16.8.16) with
25°C
25°C
Suddenly end (left) temperature becomes 85°C (
=
2
3
)
For t = 0.1 s
1[ ] [ ]
tMK
{T1} =
1[ ] (1 )[ ]
tMK
{T0} + {F0}
As {Fi} = {Fi} + 1 for all time t
1.7374 0.3660 0
3.4747 0.3660
Symmetry 1.7374
2
3
85 C
t
t
=
1.228 0.865 0 25 0.2356
2.457 0.865 25 0.4712
1.228 25 0.2356
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
=
85
134.0
52.687
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
=
85
207.7
95.28
1
2
3
1 0 0
3.475 0.366
1.737
t
t
t
627
2.6
85
73.96766
69.78261
2.7
85
74.47531
70.50053
2.8
85
74.94336
71.16246
2.9
85
75.3749
71.77274
3
85
75.77277
72.33542
