Page 204 - Handbook of Materials Failure Analysis

200 CHAPTER 8 Seismic risk of RC water storage elevated tanks

Table 8.6 Evaluation of Stiffness
Fundamental pulsation of vibration w 2 5.95 1/s 2
0
The stiffness K 1 134,915.35 N/m
Elastic modulus of the concrete E 32,164,195,120.34 N/m 2
Height of the supporting system 7.15 m
Moment of inertia of the supporting system cross-section I 1.17 m 4

P ¼ M total + 33 M tower g 1,488,820.03 N
0
140
Fundamental period T 0.14 s
The stiffness K 0 302,996,650.55 N/m

_

The vector of damping forces can be written as f a tðÞg ¼ C½Š XtðÞ
f
f
The vector of elastic forces can be written as f e tðÞf g ¼ K½Š XtðÞg
And {f(t)} denotes the vector of the external forces.
In our case, damping is assumed to be null and as we restrict ourselves only to the
determination the eigenmodes of vibration, we will just build two matrices [M] and
[K]. We also assume that the movement is harmonic type, described in the form:
(8.12)
ð
f XtðÞg ¼ a sin ωt + ϕÞ
fg
f
Since the system is free swinging (i.e., without external loading), therefore ftðÞg ¼
0 and is undamped, therefore f a tðÞg ¼ 0 the equation of motion can be written:
fg,
fg f
€

½Š f (8.13)
M XtðÞ + K½Š XtðÞg ¼ 0fg
Since:
d 2
2 2 (8.14)

dx 2 f XtðÞg ¼ a ω sin ωt + ϕÞ ¼ ω XtðÞg
€ XtðÞ ¼
ð
fg
f
The equation is then rewritten:
2
K ½Š ω M½Š XtðÞg ¼ 0 (8.15)
fg
f
The obvious solution is the trivial solution XtðÞg ¼ 0
fg. This solution corresponds
f
to the undeformed initial position of tank. As the structure oscillates, so deforms to a
given time t, it will have a deformation XtðÞg 6¼ 0
fg.
f
The determinant should be equal to zero, that is to say:
2
K ½Š ω M½Š ¼ 0 (8.16)

The matrix [K] and [M] can be written:

K 00 K 01 M 0 0
K ½Š ¼ and M½Š ¼
K 10 K 11 0 M 1
Where, K 00 ¼K 1 , K 11 ¼K 0 +K 1 , and K 10 ¼K 01 ¼ K 1 .
So, the calculation of the stiffness matrix elements is given in the Table 8.7.