# Modeling Problem of Equivalent Mechanical Models of a Sloshing Fluid

Authors: Yu-Chun Li; Hong-Liang Gou
Journal: Shock and Vibration
 (2018)

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## Abstract

Because the defects in the existing modeling methods for the equivalent mechanical model of a sloshing fluid have led to incorrect or inaccurate results in the existing equivalent models, this paper discusses three different modeling methods for the equivalent models: the traditional method, Housner’s method, and the modified method. The equivalent models obtained by the three methods are, respectively, presented and compared with each other for a liquid in rectangular and upright cylindrical tanks. The results show that the traditional method cannot provide the correct location expressions of the equivalent masses because the two types of different excitations are simultaneously used in one equivalent model. An equivalent model is exclusively applicable to a certain excitation (a translational excitation in a certain direction or a rotational excitation about a certain axis). Housner’s method is based on physical intuition, instead of fluid dynamics theory, therefore the calculation precision of Housner’s solution is not satisfactory. Housner’s method is only suitable for vertical tanks with a flat bottom subjected to a horizontal excitation. Based on a conceptual mistake in the traditional method, the concept of the equivalent model is reclarified, and the modified equivalence method is therefore suggested. A supplementary solution for the equivalent model in a cylindrical tank is presented. The correct models can be acquired using the modified equivalence method, which is applicable to tanks of arbitrary shape.


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## Body

## 1. Introduction

The dynamic response analyses of a structure containing the liquid storage tank involve the fluid sloshing, the interaction between the fluid and structure, and the structural responses. In view of the structural design, engineers are not concerned with fluid sloshing because they only want to know the dynamic effects of the sloshing liquid on the structural system. The liquid dynamic effects inside a tank can be simulated by an equivalent mechanical system. The equivalent principle requires that the resultant forces and moments (acting on the tank walls) of liquid and the equivalent system are identical. If the original liquid system is replaced by the equivalent mechanical model, the dynamic response analysis of the fluid-structure system can be greatly simplified. A liquid equivalent model is needed for the analyses of the fluid-structure coupling system.The construction of the equivalent model of a sloshing fluid originated from aeronautic and astronautic fields. The existing equivalent models in the rectangular and upright cylindrical tanks are widely used in many fields such as aerospace, ship, mechanical, and civil engineering. However, certain mistakes (or defects) have been found in these equivalent models. The parameters of the equivalent model exclusively rely on the modeling method. Therefore, the modeling problem of the liquid equivalent model should be clarified.The modeling method of the equivalent model was first put forward by Graham and Rodriguez [1]. They introduced a fixed mass and a series of mass-spring oscillators to replace the real fluid in a rectangular rigid tank. Dodge [2] and Ibrahim [3] summarized this traditional modeling method for the liquid equivalent models in different liquid containers. The translational and rotational excitations of the tanks were simultaneously utilized in the traditional method to construct an equivalent model. The traditional method is applicable to arbitrary-shape tanks subjected to sinusoidal excitations. The traditional equivalence method is widely used in different engineering fields [1–3].An alternative modeling method was proposed by Housner [4]. With the aid of physical intuition, Housner derived the simpler formulae of the fixed mass and the first mode mass-spring oscillator for the vertical tanks of circular and rectangular section subjected to a horizontal translational excitation. Housner called the fixed and moving masses the “impulsive mass” and “convective mass,” respectively. Housner’s equivalent model was thought to be a good approximation to the exact solution of Graham and Rodriguez [1]. Housner’s approximate model has been widely applied in civil engineering due to its simpler form.Veletsos [5, 6] and Veletsos and Yang [7] proposed an approximate method to consider the effect of tank-wall flexibility on the seismic response. On the basis of Housner’s model, Haroun [8, 9] and Haroun and Housner [10] added an additional mass-spring oscillator for evaluating the effect of tank-wall deformations.Jaiswal et al. [11] discussed code provisions on seismic analysis of liquid storage tanks. It was noted that all the codes used equivalent mechanical models to evaluate hydrodynamic forces, particularly due to lateral base excitation. By using an added equivalent mass to replace the fluid, Livaoğlu and Doğangün [12] gave the simplified seismic analysis procedures for elevated tanks considering fluid-structure-soil interaction. Mori et al. [13] used Housner’s model to conduct a seismic assessment for two heritage-listed R/C elevated water storage tanks. Soni et al. [14] investigated the double variable frequency isolator for seismic isolation of liquid storage tanks. In their study, the liquid storage tank was modeled based on mechanical analogy proposed by Haroun and Housner [9, 10].Later, the semianalytical and seminumerical methods were introduced to build the liquid equivalent models in irregularly shaped tanks. Joshi [15] applied finite element method to evaluate equivalent mechanical model for rigid Intze tanks. Karamanos et al. [16] developed a semianalytical variational formulation to obtain the fixed mass and the first-mode mass-spring oscillator for a liquid in a horizontal cylindrical tank and a spherical tank. Damatty and Sweedan [17] applied a coupled finite-boundary element method to calculate the liquid model parameters for pure conical tanks. Drosos et al. [18] used the finite element method to obtain the decomposed equivalent masses for irregularly shaped vessels. The locations of equivalent masses were obtained by the integration of the fluid-pressure distribution. Love and Tait [19] employed the finite element method to establish an equivalent linearized mechanical model for a tuned liquid damper (TLD) with arbitrary tank geometry.Moslemi et al. [20] employed the finite element technique to investigate the seismic response of liquid-filled conical elevated tanks. They obtained the impulsive and convective response components separately. Sorace et al. [21] studied the retrofit strategies for R/C frame water towers by using passive energy dissipation method. In their study, the water was simulated with finite element method and equivalent mechanical model, respectively.At present, many different equivalent models have been proposed by different scholars (in the different fields) using different methods for the same tank. What are the differences among these equivalent models? How best to choose the equivalent model? In recent years, Li and Wang [22] have found that the position formulae of the mass-spring oscillators in the rectangular tank were not correct in the traditional equivalent model [1–3]. Because the dynamic moment of the mass-spring oscillators acting on the tank depends on the location of the oscillators, the traditional equivalent model cannot provide the identical dynamic moment with the original fluid system. The cause of this mistake was not mentioned in [22]. It might be an occasional mistake in the mathematical derivation. However, in this paper, a similar mistake is also found (see Section 4.2) in the traditional equivalent model of an upright cylindrical container. It is realized that there must have been a systemic problem in the traditional modeling method.In the present study, a systemic modeling defect in the traditional equivalent model is indicated. The concept of the equivalent model is clarified, and the modified modeling method is suggested. The equivalent models by the Housner’s theory, the traditional, and modified equivalence methods are, respectively, presented and compared with each other for a liquid in the rectangular and cylindrical tanks. The existing problems of the equivalent models in the two tanks are clarified. The applicable scope is presented for the modified equivalent models in the rectangular and upright cylindrical tanks.


## 2. Governing Equations of Liquid Sloshing under Translational and Rotational Excitations

Figure1 shows a moving rigid tank that is partially filled with fluid. The reference frame oxyz is set on the tank. The xoy plane coincides with the static liquid-free surface. The origin of the coordinates is at the geometrical center of the static liquid-free surface. The symbols Ω, ∂Sf, and ∂Sw represent the fluid domain, the liquid-free surface, and wetted boundaries, respectively. The o′x′y′z′ system is the Earth-fixed coordinate system. It is assumed that the oxyz coordinate system coincides with the o′x′y′z′ system at the initial time. According to potential flow theory (the fluid is assumed to be inviscid, incompressible, and irrotational), the linearized equations of liquid sloshing can be written as follows [23]:(1)∇2Φx,y,z,t=0x,y,z∈Ω,(2)∂Φ∂n∂Sw=v0+ω×r⋅n,(3)∂Φ∂n∂Sf=v0+ω×r⋅n+∂h/∂t∂Sf,(4)∂Φ∂t−v0+ω×r⋅∇Φ∂Sf+gh=0,where t is the time, Φ=Φx,y,z,t is a function of the total velocity potential, hx,y,t is the vertical-displacement function on the liquid-free surface, g is the acceleration of gravity, n is the normal vector on the wetted boundary, ∂/∂n is the normal-direction differentiation, r=xe1+ye2+ze3 is the radius vector of a fluid particle P in which ei (i = 1, 2, 3) are the unit vectors of the oxyz frame, respectively, v0=dr0′/dt is the translational (excitation) velocity of the tank, and ω=ωxe1+ωye2+ωze3 is the rotational (excitation) angular velocity of the tank in which ωii=x,y,z are the components of ω about the x, y, and z-axes. By analytically (or numerically) solving Equations (1)–(4), one can obtain the velocity potential Φ. The hydrodynamic pressure can be determined by Bernoulli’s equation:(5)p=px,y,z,t=−ρ∂Φ∂t+ρv0+ω×r⋅∇Φ,where ρ is the liquid mass density. The resultant force and moment that act on the tank are further computed by(6)Ft=∫∂SwpndS,(7)Mt=∫∂Swpr×ndS.Figure 1
Moving rigid tank partially filled with fluid.


## 3. Equivalence Methods for the Equivalent Model

Figure2(a) shows an original fluid system in which H is the depth of liquid and 2a is the inner width (or diameter) of the tank. The tank is subjected to an excitation of translational velocity G˙t or a rotational angular velocity θ˙t regarding the y-axis. When the fluid is assumed to be of small displacement on the free surface, substituting v0=dr0′/dt=G˙te1 (or ω=θ˙te2) in Equations (1)–(5), one can obtain the resultant force and moment that exert on the tank through the integrals of Equations (6) and (7).Figure 2
Fluid system and equivalent model. (a) Fluid system; (b) equivalent model.The corresponding equivalent model, as shown in Figure2(b), includes a fixed mass M0 and a series of mass-spring oscillators Mi,Kii=1,2,3,…,n. The h0 and h1,h2,h3,…,hn represent the heights of the fixed mass and mass-spring oscillators above the tank bottom, respectively. hc denotes the position of the static fluid mass center C. The equivalent model is subjected to the same excitation as the original fluid system.The traditional method for constructing the equivalent model is based on the following conditions [2, 3]:(1)
The sum of all the equivalent masses must be equal to the mass of liquidM (mass conservation), that is,(8)M0+∑n=1∞Mn=M.(2)
For the small-amplitude lateral sloshing, because the free surface wave takes the form of antisymmetric profile, the vertical position of mass center of the actual liquid remains unchanged. The mass center of the equivalent model must be equal to the mass center of the actual liquid (mass-center conservation), that is,(9)M0h0+∑n=1∞Mnhn=Mhc.(3)
Thenth modal frequency ωn (n = 1, 2, 3, …) of the sloshing fluid must equal to the corresponding frequency of the spring-mass oscillator, that is,(10)ωn=KnMn.(4)
Under a certain excitation (translational or rotational excitation), the fluid system and the equivalent model must have the identical resultant force and moment that exert on the tank, that is,(11a)Foriginal=Fequivalentunder  translational  excitation,(11b)Moriginal=Mequivalentunder  translational  excitation,(12a)Foriginal′=Fequivalent′under  rotational  excitation,(12b)Moriginal′=Mequivalent′under  rotational  excitation,where Foriginal (Foriginal′) and Moriginal (Moriginal′) are the resultant force and the moment of the contained fluid under the translational (rotational) excitation, respectively, Fequivalent (Fequivalent′) and Mequivalent (Mequivalent′) are the resultant force and the moment produced by the equivalent model under the translational (rotational) excitation, respectively. By means of the analytical or numerical method, the sloshing frequencies, resultant force, and moment of fluid can be in advance obtained by solving Equations (1)–(7).The traditional equivalence steps [2, 3] are as follows:(i)
Use force Equation (11a) (under the translational excitation) to obtain the sloshing masses Mn (n = 1, 2, 3, …) and fixed mass M0(ii)
Use Equation (10) to compute the spring constants Kn (n = 1, 2, 3, …)(iii)
Apply force Equation (12a) (under the rotational excitation) to acquire the location expressions of the sloshing masses hn (n = 1, 2, 3, …)(iv)
Obtain the positionh0 of the fixed mass by solving Equation (9)The computed results showed that the position expressionshn (n = 0, 1, 2, 3, …) obtained by the traditional equivalence method appear apparently irrational (see Section 4). The main reason is that the translational and rotational excitations (two different types of excitations) were simultaneously used in an equivalent model. The position formulae hn (n = 0, 1, 2, 3, …), which are determined by force Equation (12a) under rotational excitation, generally do not satisfy moment Equation (11b) under translational excitation. In other words, the equivalent system cannot provide the same dynamic moment as the original system under translational excitation because of the incorrect results of positions hn (n = 0, 1, 2, 3, …).In consideration of the confusion of the traditional equivalence step, the concept of the equivalent model must be clarified as follows:(i)
Under a certain excitation, the equivalent model must have the same (or approximately same) modal frequencies, force, and moment (which exert on the tank) as the actual fluid system.(ii)
An equivalent model is exclusively applicable to a certain excitation, which may be a translational excitation in a certain direction or a rotational excitation about a certain axis. The excitation may be an arbitrary time function and need not be a sinusoidal time function.(iii)
The equivalent model parameters generally cannot satisfy all the above constraint equations (Equations (8)∼(12a) and (12b)) at the same time.Based on the above equivalence concept, the modified equivalence step is summarized as follows.(i)
The force Equation (11a) (or Equation (12a)) is applied to acquire the sloshing masses Mn (n = 1, 2, 3, …) and fixed mass M0. The obtained equivalent masses automatically satisfy the law of mass conservation (Equation (8)).(ii)
The spring constantsKn (n = 1, 2, 3, …) are computed using Equation (10).(iii)
The location expressionsh0 and hn (n = 1, 2, 3, …) of the equivalent masses are solved by using the corresponding moment Equation (11b) (or Equation (12b)).In the modified method, the excitation in the modified method can be an arbitrary time function. The equivalent parameters are uniquely determined by Equations (11a), (10), and (11b) (or Equations (12a), (10), and (12b)); therefore, the equivalent parameters by the modified method generally do not satisfy Equation (9) (the mass-center conservation). However, the equivalent model can simulate the entire fluid-sloshing effect on the associated structure in coupling analysis of fluid and structure, and the correct structural responses can still be obtained, whether the mass center of the equivalent system is preserved or not.Based on the modified equivalence method, Li et al. [24] proposed a seminumerical and semianalytical method for constructing liquid equivalent models in arbitrarily shaped aqueducts. The numerical examples in [24] showed that the liquid equivalent model of the U-section aqueduct can provide approximately the same force and moment as those of the original fluid, and the seismic responses of the liquid equivalent system and the original fluid-structure system agree well.The above discussion is on the spring-mass equivalent model. The modeling problem of the pendulum equivalent model of sloshing fluid is similar to the one above.


## 4. Comparison among the Equivalent Models Using Different Equivalence Methods

### 4.1. Different Equivalent Models in a Rectangular Tank

The equivalent model parameters in a rectangular tank derived by the traditional, modified, and Housner equivalence methods are presented in Table1. Figures 3–7 show the variations of M1/M, h1/H, M0/M, h0/H, and ω21a/g with H/a, respectively. Table 1 shows that except for the position ratio expressions hn/H (n = 1, 2, 3, …) of the sloshing masses, the model parameters by the traditional and modified equivalence methods are completely identical. Although the two position ratio (h0/H) expressions of the fixed mass appear different from each other, they are actually the same (as seen in Figure 6).Table 1
Traditional model [1–3]
Modified model [22]
Housner model [4]
M
0
/
M
=
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
0
/
M
=
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
0
/
M
=
tanh
3
a
/
H
/
3
a
/
H
h
0
/
H
=
1
/
2
−
∑
n
=
1
∞
16
⋅
tanh
2
n
−
1
π
H
/
2
a
/
π
3
2
n
−
1
3
H
/
a
⋅
1
−
tanh
2
n
−
1
π
H
/
4
a
/
2
n
−
1
π
H
/
4
a
/
1
−
∑
n
=
1
∞
16
⋅
tanh
2
n
−
1
π
H
/
4
a
/
π
3
2
n
−
1
3
H
/
a
h
0
/
H
=
1
/
2
+
1
/
3
⋅
a
/
H
2
−
2
H
/
a
2
∑
n
=
1
∞
2
+
2
n
−
1
π
H
/
2
a
⋅
sinh
2
n
−
1
π
H
/
2
a
−
cosh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
4
⋅
cosh
2
n
−
1
π
H
/
2
a
/
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
h
0
/
H
=
3
a
/
H
/
2
tanh
3
a
/
H
−
1
/
8
M
n
/
M
=
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
n
/
M
=
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
1
/
M
=
2.5
/
3
⋅
tanh
2.5
H
/
a
/
H
/
a
h
n
/
H
=
1
−
tanh
2
n
−
1
π
H
/
4
a
/
2
n
−
1
π
H
/
4
a
h
n
/
H
=
1
+
2
−
cosh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
⋅
sinh
2
n
−
1
π
H
/
2
a
h
1
/
H
=
1
+
2
−
cosh
2.5
H
/
a
/
2.5
H
/
a
⋅
sinh
2.5
H
/
a
n
=
1
ω
n
2
=
g
⋅
2
n
−
1
π
/
2
a
⋅
tanh
2
n
−
1
π
H
/
2
a
ω
n
2
=
g
⋅
2
n
−
1
π
/
2
a
⋅
tanh
2
n
−
1
π
H
/
2
a
ω
1
2
=
2.5
g
/
a
⋅
tanh
2.5
H
/
a
n
=
1
K
n
=
2
M
g
/
H
⋅
H
/
a
2
⋅
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
2
K
n
=
2
M
g
/
H
⋅
H
/
a
2
⋅
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
2
K
1
=
2.5
/
3
⋅
g
/
H
⋅
tanh
2
2.5
H
/
a
n
=
1Figure 3
Mass ratioM1/M vs. the fluid depth ratio H/a (rectangular tank).Figure 4
Position ratioh1/H vs. the fluid depth ratio H/a (rectangular tank).Figure 5
Mass ratioM0/M vs. the fluid depth ratio H/a (rectangular tank).Figure 6
Position ratioh0/H vs. the fluid depth ratio H/a (rectangular tank).Figure 7
The first nondimensional sloshing frequencyω12a/g vs. the fluid depth ratio H/a (rectangular tank).Note from Figure4 that the traditional position ratio h1/H is reduced with the decrease of the ratio H/a. The tendency of variation is apparently irrational. The resultant moment acting on the tank includes the contributions of hydrodynamic pressures from the sidewalls and the bottom of the rectangular tank. When the depth of fluid decreases, the relative pressure contribution from the bottom will increase, resulting in an increase of the height h1 of sloshing mass M1. However the traditional result is just the opposite. So the position ratio h1/H by the traditional equivalence step is not correct [22]. Under a translational excitation, the equivalent system will not offer the same moment as the original system because of the incorrect h1. The cause of the mistake is that the position ratio h1/H was determined by the rotational excitation (Equation (12a)) instead of the translational excitation (Equation (11b)).Figures3 and 4 show that the mass ratio M1/M and its position ratio h1/H given by Housner agree well with those obtained using the modified equivalence method. Figures 5 and 6 show that the mass ratio M0/M and its position ratio h0/H given by Housner are somewhat different from the exact solutions obtained using the traditional (or modified) equivalence method. The maximum relative differences of M0/M and h0/H are approximately 10.5% and 14.8%, respectively. The calculation precision of Housner’s solution is not satisfactory for engineering designs. From Figure 7, the first nondimensional frequency ω12a/g by Housner agrees well with that by the traditional (or modified) equivalence method.


### 4.2. Different Equivalent Models in an Upright Cylindrical Tank

For the equivalent model in a cylindrical tank under the translational excitation, the traditional model [3, 25] and Housner’s model [4] are commonly used. The (new) supplementary solution of the equivalent model by the modified equivalence method is presented in Appendix. The respective parameters of the three equivalent models are listed in Table 2. The variations of model parameters with the liquid-depth ratio H/R are shown in Figures 8–12, respectively.Table 2
Traditional model [3, 25]
Modified model (see Appendix)
Housner’s model [4]
M
0
/
M
=
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
0
/
M
=
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
0
/
M
=
tanh
3
R
/
H
/
3
R
/
H
h
0
/
H
=
1
/
2
−
1
/
2
⋅
R
/
H
2
+
∑
n
=
1
∞
4
R
2
/
H
2
ξ
1
n
2
ξ
1
n
2
−
1
⋅
cosh
ξ
1
n
H
/
R
/
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
H
ξ
1
n
ξ
1
n
2
−
1
h
0
/
H
=
1
/
2
+
∑
n
=
1
∞
F
n
J
2
ξ
1
n
R
2
/
ξ
1
n
H
2
⋅
1
−
1
/
cosh
ξ
1
n
H
/
R
−
F
n
J
1
ξ
1
n
I
1
n
/
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
H
ξ
1
n
ξ
1
n
2
−
1
h
0
/
H
=
3
/
8
⋅
1
+
4
/
3
⋅
3
⋅
R
/
H
tanh
3
⋅
R
/
H
−
1
H
/
R
≤
1.5
M
n
/
M
=
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
n
/
M
=
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
1
/
M
=
0.3859
R
/
H
⋅
tanh
27
/
8
⋅
H
/
R
n
=
1
h
n
/
H
=
1
−
tanh
ξ
1
n
H
/
2
R
/
ξ
1
n
H
/
2
R
h
n
/
H
=
1
−
R
/
ξ
1
n
H
⋅
1
/
tanh
ξ
1
n
H
/
R
−
1
/
sinh
ξ
1
n
H
/
R
−
ξ
1
n
J
2
ξ
1
n
/
J
1
ξ
1
n
sinh
ξ
1
n
H
/
R
h
1
/
H
=
1
−
R
/
H
⋅
0.544
/
tanh
1.837
⋅
H
/
R
−
0.835
/
sinh
1.837
⋅
H
/
R
n
=
1
ω
1
n
2
=
g
ξ
1
n
/
R
⋅
tanh
ξ
1
n
H
/
R
ω
1
n
2
=
g
ξ
1
n
/
R
⋅
tanh
ξ
1
n
H
/
R
ω
1
2
=
27
/
8
⋅
g
/
R
⋅
tanh
27
/
8
⋅
H
/
R
n
=
1
K
n
=
2
M
g
tanh
2
ξ
1
n
H
/
R
/
ξ
1
n
2
−
1
H
K
n
=
2
M
g
tanh
2
ξ
1
n
H
/
R
/
ξ
1
n
2
−
1
H
K
1
=
0.7889
M
g
/
H
⋅
tanh
2
27
/
8
⋅
H
/
R
n
=
1Figure 8
The first nondimensional sloshing frequencyω12R/g vs. the fluid depth ratio H/R (cylindrical tank).Figure 9
Mass ratiosM1/M and M0/M vs. the fluid depth ratio H/R (cylindrical tank).Figure 10
Condition of mass conservation (cylindrical tank).Figure 11
Position ratioh1/H vs. the fluid depth ratio H/R (cylindrical tank).Figure 12
Position ratioh0/H vs. the fluid depth ratio H/R (cylindrical tank).Expression (A22a) in Appendix is the first natural frequency that is identical to the traditional solution. Figure8 shows the variation of ω12R/g with H/R. From Figure 8, the first nondimensional frequency of Housner’s model agrees well with Equation (A22a).The expressions of mass ratiosMn/M (Equation (A17)) and M0/M (Equation (A18)) in Appendix are the same as the formulae obtained using the traditional method. Figure 9 shows the variations of mass ratios M0/M and M1/M with the liquid-depth ratio H/R. The mass ratios M1/M and M0/M given by Housner are somewhat different from Equations (A17) and (A18), respectively. The maximum errors of M0/M and M1/M (by Housner) relative to the (present) exact solutions (Equations (A17) and (A18)) reach approximately 7% and 15%, respectively. The calculation precision of Housner’s solution is not satisfactory.Figure10 shows the condition of mass conservation (Equation (8)). The equivalent masses given by traditional and modified methods strictly satisfy the condition of mass conservation when using the equation SI=M0/M+∑n=1∞Mn/M=1 and approximately satisfy the conservation condition when using the equation SI≈M0+M1/M≈1. The expression M0+M1/M≈1 indicates that the higher-mode masses are very small and can be ignored. However, the equivalent masses by Housner are not the satisfactory results as far as mass conservation is concerned.Figure11 shows the position ratio h1/H of the first convective mass M1 versus the liquid-depth ratio H/R. The ratio h1/H by Housner is somewhat different from the exact solution (Equation (A22d)) when the ratio H/R is smaller, and is notably consistent with the exact solution when the ratio H/R is higher. The position ratio h1/H given by the traditional method is reduced with the decrease of the ratio H/R, similar to the behavior shown in Figure 4. The tendency of h1/H is unreasonable, similar to the rectangular tank case. The results indicate that the traditional solution of h1/H is not correct.Figure12 shows the position ratio h0/H of fixed mass M0 versus the liquid-depth ratio H/R. The ratio h0/H by Housner agrees with the exact solution (Equation (A20)). However, the ratio h0/H obtained using the traditional method is quite different from the exact solution, especially when H/R<1.5. The negative position ratio h0/H appears in the traditional solution, which is contrary to common sense. The traditional solution of h0/H is apparently incorrect.It is found from Figures11 and 12 that the correct location formulae of h1 and h0 cannot be obtained by the traditional equivalence method. Thus, the equivalent model cannot provide the same dynamic moment as the original fluid system under the translational excitation. Because h1/H and h0/H were determined by the rotational excitation (Equation (12a)) instead of the translational excitation (Equation (11b)). It is not rational to use the two types of different excitations to build one equivalent model.


## 4.1. Different Equivalent Models in a Rectangular Tank

The equivalent model parameters in a rectangular tank derived by the traditional, modified, and Housner equivalence methods are presented in Table1. Figures 3–7 show the variations of M1/M, h1/H, M0/M, h0/H, and ω21a/g with H/a, respectively. Table 1 shows that except for the position ratio expressions hn/H (n = 1, 2, 3, …) of the sloshing masses, the model parameters by the traditional and modified equivalence methods are completely identical. Although the two position ratio (h0/H) expressions of the fixed mass appear different from each other, they are actually the same (as seen in Figure 6).Table 1
Traditional model [1–3]
Modified model [22]
Housner model [4]
M
0
/
M
=
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
0
/
M
=
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
0
/
M
=
tanh
3
a
/
H
/
3
a
/
H
h
0
/
H
=
1
/
2
−
∑
n
=
1
∞
16
⋅
tanh
2
n
−
1
π
H
/
2
a
/
π
3
2
n
−
1
3
H
/
a
⋅
1
−
tanh
2
n
−
1
π
H
/
4
a
/
2
n
−
1
π
H
/
4
a
/
1
−
∑
n
=
1
∞
16
⋅
tanh
2
n
−
1
π
H
/
4
a
/
π
3
2
n
−
1
3
H
/
a
h
0
/
H
=
1
/
2
+
1
/
3
⋅
a
/
H
2
−
2
H
/
a
2
∑
n
=
1
∞
2
+
2
n
−
1
π
H
/
2
a
⋅
sinh
2
n
−
1
π
H
/
2
a
−
cosh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
4
⋅
cosh
2
n
−
1
π
H
/
2
a
/
1
−
∑
n
=
1
∞
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
h
0
/
H
=
3
a
/
H
/
2
tanh
3
a
/
H
−
1
/
8
M
n
/
M
=
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
n
/
M
=
2
H
/
a
2
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
3
M
1
/
M
=
2.5
/
3
⋅
tanh
2.5
H
/
a
/
H
/
a
h
n
/
H
=
1
−
tanh
2
n
−
1
π
H
/
4
a
/
2
n
−
1
π
H
/
4
a
h
n
/
H
=
1
+
2
−
cosh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
⋅
sinh
2
n
−
1
π
H
/
2
a
h
1
/
H
=
1
+
2
−
cosh
2.5
H
/
a
/
2.5
H
/
a
⋅
sinh
2.5
H
/
a
n
=
1
ω
n
2
=
g
⋅
2
n
−
1
π
/
2
a
⋅
tanh
2
n
−
1
π
H
/
2
a
ω
n
2
=
g
⋅
2
n
−
1
π
/
2
a
⋅
tanh
2
n
−
1
π
H
/
2
a
ω
1
2
=
2.5
g
/
a
⋅
tanh
2.5
H
/
a
n
=
1
K
n
=
2
M
g
/
H
⋅
H
/
a
2
⋅
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
2
K
n
=
2
M
g
/
H
⋅
H
/
a
2
⋅
tanh
2
n
−
1
π
H
/
2
a
/
2
n
−
1
π
H
/
2
a
2
K
1
=
2.5
/
3
⋅
g
/
H
⋅
tanh
2
2.5
H
/
a
n
=
1Figure 3
Mass ratioM1/M vs. the fluid depth ratio H/a (rectangular tank).Figure 4
Position ratioh1/H vs. the fluid depth ratio H/a (rectangular tank).Figure 5
Mass ratioM0/M vs. the fluid depth ratio H/a (rectangular tank).Figure 6
Position ratioh0/H vs. the fluid depth ratio H/a (rectangular tank).Figure 7
The first nondimensional sloshing frequencyω12a/g vs. the fluid depth ratio H/a (rectangular tank).Note from Figure4 that the traditional position ratio h1/H is reduced with the decrease of the ratio H/a. The tendency of variation is apparently irrational. The resultant moment acting on the tank includes the contributions of hydrodynamic pressures from the sidewalls and the bottom of the rectangular tank. When the depth of fluid decreases, the relative pressure contribution from the bottom will increase, resulting in an increase of the height h1 of sloshing mass M1. However the traditional result is just the opposite. So the position ratio h1/H by the traditional equivalence step is not correct [22]. Under a translational excitation, the equivalent system will not offer the same moment as the original system because of the incorrect h1. The cause of the mistake is that the position ratio h1/H was determined by the rotational excitation (Equation (12a)) instead of the translational excitation (Equation (11b)).Figures3 and 4 show that the mass ratio M1/M and its position ratio h1/H given by Housner agree well with those obtained using the modified equivalence method. Figures 5 and 6 show that the mass ratio M0/M and its position ratio h0/H given by Housner are somewhat different from the exact solutions obtained using the traditional (or modified) equivalence method. The maximum relative differences of M0/M and h0/H are approximately 10.5% and 14.8%, respectively. The calculation precision of Housner’s solution is not satisfactory for engineering designs. From Figure 7, the first nondimensional frequency ω12a/g by Housner agrees well with that by the traditional (or modified) equivalence method.


## 4.2. Different Equivalent Models in an Upright Cylindrical Tank

For the equivalent model in a cylindrical tank under the translational excitation, the traditional model [3, 25] and Housner’s model [4] are commonly used. The (new) supplementary solution of the equivalent model by the modified equivalence method is presented in Appendix. The respective parameters of the three equivalent models are listed in Table 2. The variations of model parameters with the liquid-depth ratio H/R are shown in Figures 8–12, respectively.Table 2
Traditional model [3, 25]
Modified model (see Appendix)
Housner’s model [4]
M
0
/
M
=
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
0
/
M
=
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
0
/
M
=
tanh
3
R
/
H
/
3
R
/
H
h
0
/
H
=
1
/
2
−
1
/
2
⋅
R
/
H
2
+
∑
n
=
1
∞
4
R
2
/
H
2
ξ
1
n
2
ξ
1
n
2
−
1
⋅
cosh
ξ
1
n
H
/
R
/
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
H
ξ
1
n
ξ
1
n
2
−
1
h
0
/
H
=
1
/
2
+
∑
n
=
1
∞
F
n
J
2
ξ
1
n
R
2
/
ξ
1
n
H
2
⋅
1
−
1
/
cosh
ξ
1
n
H
/
R
−
F
n
J
1
ξ
1
n
I
1
n
/
1
−
∑
n
=
1
∞
2
R
tanh
ξ
1
n
H
/
R
/
H
ξ
1
n
ξ
1
n
2
−
1
h
0
/
H
=
3
/
8
⋅
1
+
4
/
3
⋅
3
⋅
R
/
H
tanh
3
⋅
R
/
H
−
1
H
/
R
≤
1.5
M
n
/
M
=
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
n
/
M
=
2
R
tanh
ξ
1
n
H
/
R
/
ξ
1
n
ξ
1
n
2
−
1
H
M
1
/
M
=
0.3859
R
/
H
⋅
tanh
27
/
8
⋅
H
/
R
n
=
1
h
n
/
H
=
1
−
tanh
ξ
1
n
H
/
2
R
/
ξ
1
n
H
/
2
R
h
n
/
H
=
1
−
R
/
ξ
1
n
H
⋅
1
/
tanh
ξ
1
n
H
/
R
−
1
/
sinh
ξ
1
n
H
/
R
−
ξ
1
n
J
2
ξ
1
n
/
J
1
ξ
1
n
sinh
ξ
1
n
H
/
R
h
1
/
H
=
1
−
R
/
H
⋅
0.544
/
tanh
1.837
⋅
H
/
R
−
0.835
/
sinh
1.837
⋅
H
/
R
n
=
1
ω
1
n
2
=
g
ξ
1
n
/
R
⋅
tanh
ξ
1
n
H
/
R
ω
1
n
2
=
g
ξ
1
n
/
R
⋅
tanh
ξ
1
n
H
/
R
ω
1
2
=
27
/
8
⋅
g
/
R
⋅
tanh
27
/
8
⋅
H
/
R
n
=
1
K
n
=
2
M
g
tanh
2
ξ
1
n
H
/
R
/
ξ
1
n
2
−
1
H
K
n
=
2
M
g
tanh
2
ξ
1
n
H
/
R
/
ξ
1
n
2
−
1
H
K
1
=
0.7889
M
g
/
H
⋅
tanh
2
27
/
8
⋅
H
/
R
n
=
1Figure 8
The first nondimensional sloshing frequencyω12R/g vs. the fluid depth ratio H/R (cylindrical tank).Figure 9
Mass ratiosM1/M and M0/M vs. the fluid depth ratio H/R (cylindrical tank).Figure 10
Condition of mass conservation (cylindrical tank).Figure 11
Position ratioh1/H vs. the fluid depth ratio H/R (cylindrical tank).Figure 12
Position ratioh0/H vs. the fluid depth ratio H/R (cylindrical tank).Expression (A22a) in Appendix is the first natural frequency that is identical to the traditional solution. Figure8 shows the variation of ω12R/g with H/R. From Figure 8, the first nondimensional frequency of Housner’s model agrees well with Equation (A22a).The expressions of mass ratiosMn/M (Equation (A17)) and M0/M (Equation (A18)) in Appendix are the same as the formulae obtained using the traditional method. Figure 9 shows the variations of mass ratios M0/M and M1/M with the liquid-depth ratio H/R. The mass ratios M1/M and M0/M given by Housner are somewhat different from Equations (A17) and (A18), respectively. The maximum errors of M0/M and M1/M (by Housner) relative to the (present) exact solutions (Equations (A17) and (A18)) reach approximately 7% and 15%, respectively. The calculation precision of Housner’s solution is not satisfactory.Figure10 shows the condition of mass conservation (Equation (8)). The equivalent masses given by traditional and modified methods strictly satisfy the condition of mass conservation when using the equation SI=M0/M+∑n=1∞Mn/M=1 and approximately satisfy the conservation condition when using the equation SI≈M0+M1/M≈1. The expression M0+M1/M≈1 indicates that the higher-mode masses are very small and can be ignored. However, the equivalent masses by Housner are not the satisfactory results as far as mass conservation is concerned.Figure11 shows the position ratio h1/H of the first convective mass M1 versus the liquid-depth ratio H/R. The ratio h1/H by Housner is somewhat different from the exact solution (Equation (A22d)) when the ratio H/R is smaller, and is notably consistent with the exact solution when the ratio H/R is higher. The position ratio h1/H given by the traditional method is reduced with the decrease of the ratio H/R, similar to the behavior shown in Figure 4. The tendency of h1/H is unreasonable, similar to the rectangular tank case. The results indicate that the traditional solution of h1/H is not correct.Figure12 shows the position ratio h0/H of fixed mass M0 versus the liquid-depth ratio H/R. The ratio h0/H by Housner agrees with the exact solution (Equation (A20)). However, the ratio h0/H obtained using the traditional method is quite different from the exact solution, especially when H/R<1.5. The negative position ratio h0/H appears in the traditional solution, which is contrary to common sense. The traditional solution of h0/H is apparently incorrect.It is found from Figures11 and 12 that the correct location formulae of h1 and h0 cannot be obtained by the traditional equivalence method. Thus, the equivalent model cannot provide the same dynamic moment as the original fluid system under the translational excitation. Because h1/H and h0/H were determined by the rotational excitation (Equation (12a)) instead of the translational excitation (Equation (11b)). It is not rational to use the two types of different excitations to build one equivalent model.


## 5. Conclusions

This paper discussed three different modeling methods for constructing the liquid equivalent models. The equivalent models by the three methods were presented and compared with each other for a liquid in the rectangular and cylindrical tanks.The traditional equivalent model can provide correct sloshing frequencies, equivalent masses, and spring constants. However, the traditional equivalence method generally cannot give the correct location expressions of the equivalent masses. Thus, the traditional equivalent system cannot provide the same dynamic moment as that of the original fluid system. The reason is that two types of different excitations (translational and rotational excitations) were simultaneously used in one equivalent model. An equivalent model is exclusively applicable to a certain excitation (a translational excitation in a certain direction or a rotational excitation about a certain axis).Housner’s method is only suitable for vertical tanks with a flat bottom subjected to a horizontal translational excitation. Housner’s method is based on physical intuition, instead of fluid dynamics theory; therefore it is difficult to ensure the calculation precision of Housner’s solution.The present results showed that the rational and correct equivalent parameters can be acquired by using the modified equivalence method. The modified equivalence method is applicable to arbitrary-shape tanks. The obtained equivalent parameters generally do not satisfy the mass-center conservation (Equation (9)). Because the equivalent system can give the same dynamic effects (force and moment) as those of the original fluid system, the equivalent model can simulate the entire fluid-sloshing effect on the associated structure in coupling analysis of fluid and structure. The correct structural responses can still be obtained even if the equivalent parameters do not satisfy the mass-center conservation (Equation (9)). Therefore, whether the mass center of the equivalent system is preserved or not does not affect the dynamic response results of the structural system.Because of the defects in the existing liquid equivalent models in the rectangular and upright cylindrical tanks, the modified equivalent models in the two tanks are recommended for the dynamic response analysis of the structural system. The modified equivalent models in the two tanks in this paper are only applicable to the translational excitation.



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