96
4 Multi-Degree of Freedom (MDOF) Systems
is
m =
⎡
⎣
m 0 0
0 m 0
0 0 I p
⎤
⎦ .
(4.74)
The polar moment of inertia can be calculated from
I p =
floor area
r
2
¯
m dA = ¯
m
floor area
(x
2
+ y
2 ) dA
= ¯
m
x
2 dA + ¯
m
y
2 dA = ¯
m
b d
3 /12 + b
3 d/12
= m (b
2
+ d
2 )/12,
under the assumptions that ¯
m is constant and the floor is rectangular with sides b
and d.
Equation of motion Consider the free vibration of the single-floor structure in
Fig. 4.9 under initial conditions
u 0 , v 0 , θ 0
. The equation of motion is
m¨ x + k x = 0,
(4.75)
where x = [u v θ] T is the displacement vector. We use the previous methods to find
the solution of Eq. 4.75, i.e., we calculate the modal shapes and frequencies for the
mass and stiffness matrices in Eqs. 4.73 and 4.75, represent the displacement vector
by x(t) =
3
i=1 i q i (t), and use orthogonality and Eq. 4.75 to find equations for
modal coordinates.
Example 4.7 Consider the single story in Fig. 4.11 supported by three vertical
structures with in-plane stiffnesses k A = 75 kip/ft and k B = k C = 40 kip/ft and
dimensions b = 30 ft, d = 20 ft, and e = 1.5 ft. The floor weight per unit area is
w = 100 lb/ft 2 . The stiffness matrix is
k =
⎡
⎣
k B + k C
0 (k C − k B ) (d/2)
0
k A k A e
(k C − k B ) (d/2) k A e (k C + k B ) (d/2) 2 + k A e 2
⎤
⎦ =
⎡
⎣
80 0
0
0 75
112.5
0 112.5 8168.75
⎤
⎦ .
It can be obtained from its expression in Eq. 4.73 or by direct considerations. For
example, the forces induced in the vertical supporting structures B and C by the
displacement u = 1 are k B and k C . They act in the positive direction of the x-axis.
There is no force in A. This unit displacement induces the force k B + k C in the xdirection, no force in the the y-direction, and the torque k B (d/2) − k C (d/2) under
the sign convention of Fig. 4.10. Accordingly,
(k B + k C , 0, (k B − k C ) d/2
are the
entries of the first column of the stiffness matrix k. The reader is encouraged to find
the other entries of k by using similar arguments.
4 Multi-Degree of Freedom (MDOF) Systems
is
m =
⎡
⎣
m 0 0
0 m 0
0 0 I p
⎤
⎦ .
(4.74)
The polar moment of inertia can be calculated from
I p =
floor area
r
2
¯
m dA = ¯
m
floor area
(x
2
+ y
2 ) dA
= ¯
m
x
2 dA + ¯
m
y
2 dA = ¯
m
b d
3 /12 + b
3 d/12
= m (b
2
+ d
2 )/12,
under the assumptions that ¯
m is constant and the floor is rectangular with sides b
and d.
Equation of motion Consider the free vibration of the single-floor structure in
Fig. 4.9 under initial conditions
u 0 , v 0 , θ 0
. The equation of motion is
m¨ x + k x = 0,
(4.75)
where x = [u v θ] T is the displacement vector. We use the previous methods to find
the solution of Eq. 4.75, i.e., we calculate the modal shapes and frequencies for the
mass and stiffness matrices in Eqs. 4.73 and 4.75, represent the displacement vector
by x(t) =
3
i=1 i q i (t), and use orthogonality and Eq. 4.75 to find equations for
modal coordinates.
Example 4.7 Consider the single story in Fig. 4.11 supported by three vertical
structures with in-plane stiffnesses k A = 75 kip/ft and k B = k C = 40 kip/ft and
dimensions b = 30 ft, d = 20 ft, and e = 1.5 ft. The floor weight per unit area is
w = 100 lb/ft 2 . The stiffness matrix is
k =
⎡
⎣
k B + k C
0 (k C − k B ) (d/2)
0
k A k A e
(k C − k B ) (d/2) k A e (k C + k B ) (d/2) 2 + k A e 2
⎤
⎦ =
⎡
⎣
80 0
0
0 75
112.5
0 112.5 8168.75
⎤
⎦ .
It can be obtained from its expression in Eq. 4.73 or by direct considerations. For
example, the forces induced in the vertical supporting structures B and C by the
displacement u = 1 are k B and k C . They act in the positive direction of the x-axis.
There is no force in A. This unit displacement induces the force k B + k C in the xdirection, no force in the the y-direction, and the torque k B (d/2) − k C (d/2) under
the sign convention of Fig. 4.10. Accordingly,
(k B + k C , 0, (k B − k C ) d/2
are the
entries of the first column of the stiffness matrix k. The reader is encouraged to find
the other entries of k by using similar arguments.