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TABLE 10.5
Procedure for Computer Implementation of the Newmark Method
Initial Step
1.
Select
t
,
γ
, and
β
2.
Calculate the constants
c
0
=
1
/(β(
t
)
2
)
c
1
=
(γ /β) (
1
/
t
)
c
2
=
(
1
/γ )
c
1
1
2
−
β)/β
1
2
−
β)(
γ
t
c
3
=
(
c
4
=
γ/β
−
1
c
5
=
(
β
)
+
(γ
−
1
)
t
Initialize
V
0
,
V
0
, and
V
0
3.
Form the effective stiffness matrix
K
4.
=
K
+
c
0
M
+
c
1
C
Decompose (triangularize)
K
LDL
T
5.
=
At Each Step
c
2
V
n
+
c
3
V
n
)
+
c
4
V
n
+
c
5
V
n
)
+
1.
Calculate
P
n
+
1
=
M
(
c
0
V
n
+
C
(
c
1
V
n
+
P
n
+
1
KV
n
+
1
=
2.
Solve
P
n
+
1
V
n
+
1
and
V
n
+
1
from Eq. (10.115)
Calculate
Once
V
n
+
1
is obtained from Eq. (10.116), the corresponding velocity and acceleration
vectors can be computed using
V
n
+
1
−
1
2
−
β
V
n
(
2
(10.118a)
V
n
+
γ
β
1
V
n
+
1
=
V
n
+
(
)
V
n
−
1
−
γ )(
t
)
V
n
−
(
t
t
)
t
V
n
+
1
−
1
2
−
β
2
V
n
1
β(
V
n
+
1
=
)
V
n
+
V
n
+
(
t
(
t
)
(10.118b)
t
)
2
The procedure for computer implementation is outlined in Table 10.5.
The stability and accuracy of the Newmark method, which is controlled by two pa-
rameters
, can be evaluated by examining the eigenvalues associated with the
amplification matrix. This process, however, becomes tedious and lengthy because of the
parameters
β
and
γ
. Newmark (1959) established the stability and accuracy of this inte-
gration scheme by examining a single-DOF system without damping, and demonstrated
that the method is unconditionally stable for
β
and
γ
1
2
1
4
.
γ
=
and
β
≥
Wilson
θ
Method
10.5.4
In the Wilson
θ
method [Bathe and Wilson, 1973], the acceleration is assumed to be linear
from time
n
t
to time
(
n
+
θ)
t
, with
θ
≥
1
.
0 as shown in Fig. 10.11. With this assumption,
τ
≤
τ
≤
θ
the acceleration at any time
t
, where 0
, can be obtained by linear interpolation,
i.e.,
V
n
+
τ
V
n
+
τ
=
V
n
+
θ
−
V
n
)
θ
(
Integrate with respect to
τ
t
to obtain
+
(τ
t
)
2
V
n
+
τ
=
V
n
+
V
n
τ
V
n
+
θ
−
V
n
)
t
(
(10.119)
2
θ
t
where the first term on the right-hand side is the constant of integration evaluated at
τ
=
0. Also,
+
(τ
t
)
2
V
n
+
(τ
t
)
3
V
n
+
V
n
τ
V
n
+
θ
−
V
n
)
V
n
+
τ
=
t
(
(10.120)
2
6
θ
t
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