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10.2 Find the lumped mass matrices (element and global) of the frame in Problem 10.1
with and without rotary inertia.
Answer:
Global matrix, based on physical lumping of mass at the ends of two
elements, without rotary inertia
M
10
−
3
diag
≈
1
.
259
×
(
220110
)
where
V
=
c
]
T
10
−
3
diag
[
U
Xb
U
Zb
b
U
Xc
U
Zc
with rotary inertia
M
≈
1
.
259
×
(
2217118
.
5
)
10.3 Use Guyan reduction to condense out the DOF related to the rotation for the frame
in Problem 10.1. Find the stiffness, consistent, and lumped mass matrices.
Answer:
[
U
Xb
U
Zb
U
Xc
U
Zc
]
1
.
2065
−
0
.
0026
−
0
.
0017
0
−
0
.
0026
1
.
2117
0
.
0026
−
1
.
2048
10
5
K
=
8
.
3
×
−
0
.
0017
0
.
0026
0
.
0017
0
0
−
1
.
2048
0
1
.
2048
315
.
3469
−
35
.
8776
64
.
6531
0
=
ρ
420
−
35
.
8776
267
.
1020
−
9
.
1224
70
.
0000
M
64
.
6531
−
9
.
1224
115
.
3469
0
0
70
.
0000
0
140
.
0000
Lumped mass matrix without rotary inertia:
2
.
0572
0
0
0
0
2
.
0572
0
0
10
−
3
M
=
1
.
259
×
0
0
1
.
0286
0
0
0
0
1
.
0286
Lumped mass matrix with rotary inertia:
2
.
2321
0
.
0159
−
0
.
1749
0
0
.
0159
2
.
2003
−
0
.
0159
0
10
−
3
M
=
1
.
259
×
−
0
.
1749
−
0
.
0159
1
.
2035
0
0
0
0
1
.
0286
10.4 Use three point Gauss quadrature to find the mass matrix for beams using the shape
functions of Eq. (10.8)
Answer:
444
−
62
156
38
2
2
ρ
1200
−
62
11
−
38
−
9
m
i
=
156
−
38
444
62
2
2
38
−
9
62
11
10.5 Approximate mass matrices can be obtained in many ways. Show that the higher
order matrix of Eq. (10.15) can be produced by e
mpl
oying a special numerical inte-
gration scheme with the integration points at
±
√
2
3 and the weights equal to one in
the integral for the element mass matrix of Eq. (10.5).
/
Hint:
Use the shape functions of Chapter 4, Example 4.1.
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