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Mathematical analogy of a beam on elastic supports as a beam on elastic foundation
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Applied Mathematical Modelling 32 (2008) 688–699 www.elsevier.com/locate/apm
Mathematical analogy of a beam on elastic supports as a beam on elastic foundation Motohiro Sato *, Shunji Kanie, Takashi Mikami Graduate School of Engineering, Hokkaido University, Kita 13 Nishi 8, Kita-ku, Sapporo, Japan Received 1 January 2006; received in revised form 1 December 2006; accepted 6 February 2007 Available online 20 February 2007
Abstract This paper presents the mathematical hypothesis that a beam on equidistant elastic supports (BOES) can be considered as a beam on an elastic foundation (BOEF) in static and free vibration problems. This modeling of BOES as BOEF is presumed to be applicable to a limited range of support stiffness, spacing and flexural rigidity of the beam. The authors investigate the applicability of the modeling of BOEF from the property of characteristic solutions obtained from governing equations of both BOES and BOEF. In this study, the formulation of BOES leads to governing difference equations, and the motions of BOEF are expressed by differential equations. This is because exact solutions must be employed in order to verify their analogy accurately. The characteristic solutions obtained from these two governing equations are compared to each other in order to investigate the relationship between them. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Submerged floating tunnel; Beams on elastic foundation; Beams on elastic supports; Dynamic analysis
1. Introduction There are many civil engineering structures that can be idealized as beams on elastic supports (from now on abbreviated to ‘BOES’). As an example, submerged floating tunnel (SFT) is shown in Fig. 1. It is a whole new structure for strait crossings, which is stabilized in the water by the balance between tunnel buoyancy and the mooring force of the tension legs. This class of floating structure moored to the sea bed by tethers is referred to variously as tension leg platforms, tethered production platforms, and so on (for example, see [1]). If the tunnels length is very great compared with its cross-sectional diameter, the beam theory is applicable when calculating its deformations and responses. Moreover, SFT can be regarded as a beam on discrete elastic supports if its tension legs are separately spaced along its length and the elasticity of the tension legs is taken into consideration. It is well known that in many engineering cases, if separate elastic supports are equally spaced at close enough intervals along the beam, it can be modeled by a beam on an equivalent imaginary elastic foundation *
Corresponding author. E-mail address: [email protected] (M. Sato).
0307-904X/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2007.02.002
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
h
(a) Cross sectional shape
h
689
h
(b) Side view
Fig. 1. Submerged floating tunnel.
(here abbreviated to ‘BOEF’). The major advantage of applying the theory of BOEF is the resulting ability to evaluate the basic characteristics quickly and easily. The beam on elastic foundation model is suitable for simplified analysis of the global behavior of the structure under consideration and is therefore very effective in enabling designers to assume structural dimensions. There have been many studies on BOEF theory; for example Hetenyi [2] gives a comprehensive analysis of BOEF. Therefore, if it is possible to replace separate elastic supports with a continuous elastic foundation, designers can easily investigate behavioral characteristics by using previously discovered analytical results such as these. In the case of BOES, the hypothesis that the approximation of a uniform elastic foundation is applicable was investigated by Ellington [3]. However, those studies were confined within the limits of static problems. If this modeling of BOEF is used for a basic structural design of a submerged floating tunnel as shown in Fig. 1, it is also necessary to investigate the analogy between BOES and BOEF with consideration given to dynamic effects. The purpose of this paper is to make the applicability of this modeling clear while showing the conditions in which BOES can be considered as BOEF in static and free vibration problems. To confirm the validity and accuracy of the analogy between them, these solutions must be exact because approximate solutions have some inherent errors. Therefore, in this paper exact solutions are developed for the deflections and motions of both BOES and BOEF. In the case of beams on elastic supports, the method of finite difference [4] can be applied for determination of the governing equation and its solutions. Similarly, motions of beams on elastic foundation are expressed by differential equations. The characteristic solutions are obtained from these two equations and those solutions are compared to each other to investigate their equivalence. In order to verify the results of comparing the characteristic solutions, actual static and dynamic characteristics should also be considered. Subsequently, static deformations, bending moments, natural frequencies and modal shapes of BOES and BOEF are compared to each other. 2. Governing equations and its solutions The deflections and motions of the beam described in this paper are based on the Bernoulli–Euler beam theory. The beam is assumed to have flexural rigidity EI and mass per unit length m. 2.1. Beam on elastic supports (BOES) As mentioned previously, if the beam is uniformly supported the method of finite difference is applicable. Fig. 2 shows the coordinate system used in this paper for introducing the governing difference equations of BOES. All elastic supports are assumed to have an axial rigidity kv and be spaced at a regular interval h along the beam (x-) axis. The governing difference equation of motion can be obtained by the relationship between the end forces and end displacements for a typical beam segment extending from support r to r + 1 as shown
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r+1
r
y
h
ϕ
r
vr
ϕ
vr+1
r+1
x
M r+1,r
M r,r+1
S r,r+1
S r+1,r
Fig. 2. Notations and positive directions.
in Fig. 2, and the equations of equilibrium at a point on an arbitrary elastic support r. First, the dynamic-stiffness expressions are derived by taking account of the actual distribution of mass and stiffness for a beam segment, as 9 9 8 9 8 8 vr =h > > hS r;rþ1 > hS r;rþ1 > > > > > > > > > > > > > = = > = < hS < v =h > < hS rþ1;r rþ1 rþ1;r ¼ KðXÞ þ ; ð1Þ > > > > > u M M r;rþ1 > > > > > > > r;rþ1 r > > : > > > > ; ; ; : : M rþ1;r urþ1 M rþ1;r where K(X) is the dynamic-stiffness matrix [5], v, u, M and S denote the deflection, slope, bending moment and shearing force, respectively; the subscript r and r + 1 of v, u and r, r + 1 and r + 1, r of M and S denote the support point under consideration as shown in Fig. 2. M and S show the fixed-end moment and shearing force respectively. In the case of bending vibration analyses, K(X) contains the non-dimensional frequency parameter X which is a function of the circular frequency x, as 2 3 6c4 12c5 12c6 6c3 7 EI 6 6 12c6 12c5 6c4 6c3 7 KðXÞ ¼ 6 ð2aÞ 7; h 4 6c3 6c4 4c1 2c2 5 6c4
6c3
2c2
4c1
where mx2 h4 ; 24EI sin a cosh a cos a sinh a a; c1 ¼ 4ð1 cos a cosh aÞ sinh a sin a a; c2 ¼ 2ð1 cos a cosh aÞ sin a sinh a a2 ; c3 ¼ 6ð1 cos a cosh aÞ X¼
ð2bÞ ð2cÞ ð2dÞ ð2eÞ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
691
c4 ¼
cosh a cos a a2 ; 6ð1 cos a cosh aÞ
ð2fÞ
c5 ¼
sin a cosh a þ cos a sinh a 3 a; 12ð1 cos a cosh aÞ
ð2gÞ
sinh a þ sin a a3 ; 12ð1 cos a cosh aÞ ffiffiffiffiffiffiffiffiffi p 4 a ¼ 24X:
c6 ¼
ð2hÞ ð2iÞ
Next, the equilibrium of a typical elastic support r requires that S r;rþ1 þ S r;r1 ¼ k v vr ; M r;rþ1 þ M r;r1 ¼ 0:
ð3aÞ ð3bÞ
Substituting Eq. (1) into S in Eq. (3a) and M in Eq. (3b) leads to c2 ðurþ1 þ ur1 Þ þ 4c1 ur þ 3c4 ðV r1 V rþ1 Þ ¼
T rh ; 2EI
ð4aÞ 2
c4 ðurþ1 ur1 Þ þ 4ðc5 þ K v ÞV r 2c6 ðV r1 þ V rþ1 Þ ¼
P rh ; 6EI
ð4bÞ
where Vr = vr/h, Tr = Mr,r+1 + Mr,r1, Pr = Sr,r+1 + Sr,r1 and Kv ¼
k v h3 24EI
ð5Þ
is the so-called relative stiffness of the supports. Elimination of u in Eqs. (4a) and (4b) gives the next difference equation ( ) 2 3c24 ðE E1 Þ h c4 ðE1 EÞ P rh 1 þ 4ðc5 þ K v Þ 2c6 ðE þ E Þ V r ¼ Tr þ ; ð6Þ 2EI c2 ðE þ E1 Þ þ 4c1 3 c2 ðE þ E1 Þ þ 4c1 where E is the shifting operator. Assuming for the complementary solution of Eq. (6) that V r ¼ D expðkD rÞ;
ð7Þ
where D is an arbitrary constant and kD is to be determined. Substituting this in the homogeneous part of Eq. (6) leads to 6c22 sinh2 kD D þ 4ðc5 þ K v c6 cosh kD Þ expðkD rÞ ¼ 0: ð8Þ c2 cosh kD þ 2c1 By setting the expression in brackets to zero, we obtain a quadratic for coshkD that is given by ð3c24 2c2 c6 Þcosh2 kD þ f2c2 ðc5 þ K v Þ 4c1 c6 g cosh kD þ 4c1 ðc5 þ K v Þ 3c24 ¼ 0:
ð9Þ
This equation, in general, yields two distinct values of cosh kD. Corresponding to each root of cosh kD, there are two possible value of kD, namely ±k. In static problems, the basic concept is the same as for dynamic problems. The static beam stiffness can be considered as representing a special case of the dynamic stiffness where the applied frequency is zero, that is 2 3 12 12 6 6 7 EI 6 6 12 12 6 6 7 ð10Þ Kð0Þ ¼ 6 7: h 4 6 6 4 2 5 6
6
2
4
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Applying Eq. (10) as a substitution for Eq. (2a) yields the following governing equation: ( ) 3ðE E1 Þ2 1 2ðE þ E 2Þ þ 4K v V r ¼ 0: E þ E1 þ 4
ð11Þ
Therefore the characteristic equation is derived by substituting Eq. (7) into Eq. (11) as follows: cosh2 kD 2ð1 K v Þ cosh kD þ 1 þ 4K v ¼ 0: The solution of Eq. (12) is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh kD ¼ 1 K v K 2v 6K v :
ð12Þ
ð13Þ
The values of kD and the general solutions Vr are determined in three cases corresponding to the range of Kv as (i) 0 < Kv < 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh kD ¼ 1 K v i 6K v K 2v ;
ð14Þ
kD ¼ ðaD bD iÞ; V r ¼ D1 sin bD r sinh aD r þ D2 sin bD r cosh aD r þ D3 cos bD r sinh aD r þ D4 cos bD r cosh aD r;
ð15Þ ð16Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 4 þ 2K v þ 6K v aD ¼ cosh ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 pffiffiffiffiffiffiffiffi 4 þ 2K v 6K v bD ¼ cos1 ; 2 1
ð17Þ ð18Þ
(ii) Kv = 6 kD ¼ ðaD þ ipÞ; r V r ¼ ð1Þ ðD1 þ D2 rÞðD3 sinh aD r þ D4 cosh aD rÞ;
ð19Þ ð20Þ
where aD ¼ cosh1 5:
ð21Þ
(iii) Kv > 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 kD ¼ cosh K v 1 K v 6K v þ ip ;
ð22Þ
V r ¼ ð1Þr ðD1 sinh aD1 r þ D2 cosh aD1 r þ D3 sinh aD2 r þ D4 cosh aD2 rÞ;
ð23Þ
where aD1 aD2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ cosh K v 1 K 2v 6K v ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ cosh1 K v 1 þ K 2v 6K v : 1
ð24Þ ð25Þ
2.2. Beam on elastic foundation Next, we consider the beam on an elastic foundation equivalent to equidistant elastic supports, as shown in the previous section. Neglecting the term of damping force, the governing differential equation of motion for the deflection curve v(x, t) of BOEF is known to be m
o2 vðx; tÞ o4 vðx; tÞ þ EI þ k 0v vðx; tÞ ¼ 0; ot2 ox4
ð26Þ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
693
where k 0v is foundation modulus. If elastic supports are equivalent to an elastic foundation, they can be replaced by a continuously distributed imaginary foundation. In this case, this modulus k 0v is written as kv ð27Þ k 0v ¼ : h Introducing the following non-dimensional parameters of x k 0 h4 mx2 h4 n ¼ ; Kv ¼ v ; X ¼ ; h 24EI 24EI into Eq. (26) and assuming that the solution of Eq. (26) has the form vðx; tÞ ¼ V ðnÞðA1 sin xt þ A2 cos xtÞ:
ð28Þ
It should be noted that V(n) is dimensionless displacement divided by h. Substituting Eq. (28) into Eq. (26), the ordinary differential equation with respect to n may be written as d4 V ðnÞ þ 24ðK v XÞV ðnÞ ¼ 0: dn4
ð29Þ
Eq. (29) can be solved in the same way as the difference equation by assuming a solution in the form V ðnÞ ¼ C expðkc nÞ;
ð30Þ
where C is an arbitrary constant and kc is to be determined. Substituting this into Eq. (15), we obtain the characteristic equation k4c þ 24ðK v XÞ ¼ 0;
ð31Þ
which has the solutions (i) X > Kv kc ¼ aC ; aC i;
ð32Þ
V ðnÞ ¼ C 1 cos aC n þ C 2 sin aC n þ C 3 cosh aC n þ C 4 sinh aC n;
ð33Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aC ¼ 4 24ðX K v Þ:
ð34Þ
(ii) X < Kv kc ¼ ð1 þ iÞaC ; ð1 iÞaC ; V ðnÞ ¼ C 5 sin aC n sinh aC n þ C 6 sin aC n cosh aC n þ C 7 cos aC n sinh aC n þ C 8 cos aC n cosh aC n;
ð35Þ ð36Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aC ¼ 4 6ðK v XÞ:
ð37Þ
In static problems, the solution can be obtained by setting X, which is the parameter in terms of inertia force, pffiffiffiffiffiffiffiffi to zero, that is aC ¼ 4 6K v . It is clear that the form of V(n) corresponds to Eq. (36). 3. Analogy between BOES and BOEF For static problems, the forms of both characteristic solutions correspond with each other when 0 < Kv < 6; however in the case of Kv P 6, the two forms are quite different. Consequently, discussions of analogy between BOES and BOEF should be undertaken when 0 < Kv < 6 for static problems. We can use the solutions of Eqs. (9) and (17), (18) to estimate their analogy because these solutions are both exact. Fig. 3 shows the comparisons of arguments of two characteristic solutions, BOES and BOEF. For BOEF, the real part of the characteristic root is equal to the imaginary one independent of Kv as indicated in Eq. (35). Therefore, the arguments are p/4(0.7854) for any value of Kv that satisfies 0 < Kv < 6. On the other hand, the arguments for BOES are not necessarily p/4; however as shown in Fig. 3, there is little
694
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699 0.79
arg.
0.788 0.786 0.784
BOES BOEF
0.782 0.78 0.001
0.01
0.1
1
Kv Fig. 3. Comparison of arguments (static).
1
Ω = 0.1
Re./Im.
0.98
Ω = 4.0
0.96 0.94
BOES BOEF
0.92
Ω = 1.0 0.9 0.001
0.01
0.1
1
Kv Fig. 4. Ratio of the real part of solutions to the imaginary part.
difference between the two arguments over the range of about Kv 6 0.05. It can be said that this range is the condition in which the approximation of a uniform elastic foundation is applicable in static problems. Fig. 4 shows the comparison of the ratio of the real part of solutions for kD or kC to the imaginary part. The two solutions agree well with each other when Kv 6 0.05. There are no significant differences. This range, Kv 6 0.05, is the condition in which the approximation of a uniform elastic foundation is applicable in dynamic problems. This result shows that the applicable condition for dynamic problems is similar to that for static problems. 3.1. Static characteristics In order to verify the results of comparing the characteristic solutions in the preceding section, actual static and dynamic characteristics should also be considered. In this section, static deformations and bending moments of BOES and BOEF subjected to a single concentrated force P at the center of the simple beam are taken up, and the effects of Kv on the differences between BOES and BOEF are considered. In addition, comparisons of free vibration characteristics are studied. 4. Beam on elastic supports (BOES) Here two cases are considered as shown in Fig. 5. The three moment equation and equilibrium near the concentrated load can be written as h h ðM r1 þ 4M r þ M rþ1 Þ þ ðV r1 2V r þ V rþ1 Þ ¼ ðM r;r1 þ M r;rþ1 Þ; 6EI 3EI ðM r1 2M r þ M rþ1 Þ þ k v h2 V r P r h ¼ M r ;
ð38Þ ð39Þ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
695
P 1
0
0
1
R'
(a) Case-A
P 1
0
1
R'
(b) Case-B Fig. 5. Analytical cases for the static problem.
where M is the fixed-end moment and P is the fixed-end shearing force. From Eq. (38) the bending moment equation is given, containing shifting operators, by h E þ E1 2 Mr ¼ Vr 6EI E þ E1 þ 4 ¼ fð1 2j1 Þ sin bD r sinh aD r þ 2j2 cos bD r cosh aD rgD1 þ fð1 2j1 Þ sin bD r cosh aD r þ 2j2 cos bD r sinh aD rgD2 þ fð1 2j1 Þ cos bD r sinh aD r 2j2 sin bD r cosh aD rgD3 þ fð1 2j1 Þ cos bD r cosh aD r 2j2 sin bD r sinh aD rgD4 ;
ð40Þ
where D1–D4 are constants that can be determined from the boundary conditions at the end of the beam, and sin aD ðcosh aD þ 2 cos bD Þ ; ð2 þ cosh aD cos bD Þ2 þ ðsinh aD sin bD Þ2 sin bD ð2 cosh aD þ cos bD Þ g2 ¼ 2 2 2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g
g1 ¼
j1 ¼
ð41Þ
ð1 þ 2 cosh aD cos bD Þð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ2 2
2
2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g 3 sinh aD sin bD j2 ¼ : 2 2 2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g
; ð42Þ
Case-A r 1 ¼ r ¼ 0;
r þ 1 ¼ 1;
M r;r1 ¼ M 0;0 ¼
3 Ph; 16
M r;rþ1 ¼ M 0;1 ¼ 0;
P r ¼ 0;
1 M r ¼ Ph: 2
The boundary conditions are V R0 ¼ 0
and
M R0 ¼ 0
at R0 ¼
R1 2
ð43Þ
696
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
and the three moment equation and the equilibrium equation are h Ph2 ð5M 0 þ M 1 Þ þ V 1 V 0 ¼ ; 6EI 16EI h Ph2 ðM 0 M 1 Þ þ 4K v V 0 ¼ : 6EI 12EI
ð44Þ ð45Þ
Case-B r 1 ¼ r þ 1 ¼ 1;
r ¼ 0;
M r;r1 ¼ M r;rþ1 ¼ M 0;1 ¼ 0;
Pr ¼ P;
M r ¼ 0:
The boundary conditions are V R0 ¼ 0
and
M R0 ¼ 0
at R0 ¼
R 2
ð46Þ
and the three moment equation and the equilibrium equation are h ð2M 0 þ M 1 Þ þ V 1 V 0 ¼ 0; 6EI h Ph2 ðM 0 M 1 Þ þ 2K v V 0 ¼ : 6EI 12EI
ð47Þ ð48Þ
4.1. Beam on elastic foundation (BOEF) In the case of a simply supported BOEF with concentrated load P in the middle of the span, it is extremely easy to determine the deflection curve and bending moment by introducing the following conditions l l l ð49Þ V ¼ 0 and M ¼ 0 at n ¼ ; 2h 2h 2h uð0Þ ¼ 0 and
Sð0Þ ¼
P 2
at n ¼ 0:
ð50Þ
When this is done, V(n) and M(n) yield P aC 0 2 2k v h ðcosh aC R
V ðnÞ ¼
þ cos aC RÞ
fsin aC n cosh aC ðR nÞ sin aC ðR nÞ cosh aC n
þ cos aC n sinh aC ðR nÞ cos aC ðR nÞ sinh aC ng; MðnÞ ¼
ð51Þ
Ph f sin aC n cosh aC ðR nÞ þ sin aC ðR nÞ cosh aC n 4aC ðcosh aC R þ cos aC RÞ
þ cos aC n sinh aC ðR nÞ cos aC ðR nÞ sinh aC ng:
ð52Þ
4.2. Results and discussion Fig. 6 shows the deflections and Fig. 7 the bending moments of BOES and BOEF from the center to the right end of both beams (R 0 = 5). Compared to the differences of deflections, those of the bending moment are larger, however in all cases of Kv = 0.005 and Kv = 0.05, the differences are quite insignificant. Table 1 gives full details of the difference between BOES and BOEF at the point at which the concentrated force is acting (i.e. at the center of the beam). The above-mentioned results suggest that the approach of comparing the characteristic solutions of BOES with those of BOEF for static problems is valid.
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699 0
kvvr /P ( or k'vhv(ξ )/P)
kvvr /P ( or k'vhv(ξ )/P)
0
697
K v = 0.005
0.2
K v = 0.05 BOES BOEF
0.4
K v = 0 .5
0.6 0
1
2 3 r (or ξ )
4
0.2
K v = 0.05 0.4
BOES BOEF
K v = 0 .5
0.6
5
0
1
(a) Case-A
2
3 r (or ξ )
4
5
(b) Case-B
0
K v = 0.05 0.2
K v = 0 .5
K v = 0.005 BOES BOEF
0.4
0.6
Mr /Ph ( or M(ξ)/Ph)
Mr /Ph ( or M(ξ)/Ph)
Fig. 6. Deflections.
0
1
2 3 r (or ξ )
4
5
0
K v = 0.05 K v = 0.5 0.2
K v = 0.005 BOES BOEF
0.4
0.6
0
1
(a) Case-A
2
3 r (or ξ )
4
5
(b) Case-B Fig. 7. Bending moments.
Table 1 Differences between BOES and BOEF at the center of the beam Kv
V(n)/Vr (Case-A)
V(n)/Vr (Case-B)
M(n)/Mr (Case-A)
M(n)/Mr (Case-B)
0.005 0.05 0.5
1.000 0.995 0.940
1.000 1.003 1.035
0.985 0.951 0.840
1.032 1.106 1.456
5. Free vibration For simply supported beams, modal shape functions Vr (BOES) and V(n) (BOES) can be expressed as follows: npr ; n ¼ 1; 2; . . . ; ð53Þ V r ¼ Ad sin R npn V ðnÞ ¼ Ac sin ; n ¼ 1; 2; . . . ð54Þ R These functions satisfy the boundary conditions at r (or n) = 0, R. Inserting Eq. (53) into the left-hand side of Eqs. (6) and (54) in the left-hand side of Eq. (11), we find that 6c24 sin2 np np R For BOES : þ 4 c þ K c cos ¼ 0; ð55Þ 5 v 6 R c2 cos np þ 2c1 R 1 np4 For BOEF : X ¼ þ K v: ð56Þ 24 R
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M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
The solutions of the resulting eigenvalue problems give the vibration modes and the corresponding natural frequencies. These are the characteristic equations for determining the natural frequencies. The solutions of Eqs. (55) and (56) then provide the values of X which represent the frequencies of free vibration. 5.1. Results and discussion Fig. 8 shows the comparison of natural frequencies. The presence of an elastic foundation influences only the natural frequencies [6]. It is found that this figure also indicates the equivalence when Kv 6 0.05. Fig. 9 shows the modal shapes of BOEF and Fig. 10 shows those of BOES. Some differences exist between the two models. In the case of BOEF, the vibration modes have the same shapes as those for beams without 1
0.99
4
ΩD ΩC
0.98 n/R=0.1 n/R=0.5 n/R=0.9
0.97
0.96 0.01
0.05
0.1
0.5
1
Kv Fig. 8. Comparison of natural frequencies.
(a)
n = 0.1 R
(b)
n = 0.5 R
(c)
n = 0.9 R
Fig. 9. Modal shape (beams on elastic foundation).
(a)
n = 0.1 R
(b)
n = 0.5 R
(c)
n = 0.9 R
Fig. 10. Modal shape (beams on elastic supports). Solid lines: Kv = 0.05, broken lines: Kv = 0.5, dotted lines: Kv = 5.0.
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
699
elastic foundations. However in beams on elastic supports, mode shapes change as Kv increases. This is caused by the increase in local stiffness of the support. However when Kv 6 0.05, the two modes of beams are very similar to each other. The differences are quite insignificant. These results suggest that the approach in the previous section is valid. 6. Conclusions In this study, fourth order difference equations dealing with the deflection of BOES, and their solutions, were introduced to be compared with those of BOEF. The results on the analogy for static problems were the same as that of Ellington, that is, BOES and BOEF were nearly equal with respect to the expressions of deflection when Kv 6 0.05. This investigation was also developed to cover dynamic problems in a similar manner in this study. Accordingly, at least both are equivalent under the condition of Kv 6 0.05 even for the dynamic problems. Next, in order to verify these considerations, the deflections and bending moments of the beams when concentrated load was acting at the center of these beams, as well as natural frequencies and modal shapes were compared with each other. In all cases of these variables, there were no significant differences between solutions of BOES and BOEF within the range of Kv 6 0.05. The above results will be very effective particularly for decision-making in SFT projects and in work with similar types of structures where dynamic effects must be taken into consideration during the structural design phase. References [1] [2] [3] [4] [5] [6]
M.H. Patel, J.A. Witz, Compliant Offshore Structures, Butterworth-Heinemann, 1991. M. Hetenyi, Beams on Elastic Foundation, The University of Michigan Press, 1974. J.P. Ellington, The beam on discrete elastic supports, Bull. Int. Railway Cong. Assoc. 34 (12) (1957) 933–941. T. Wah, L.A. Calcote, Structural Analysis by Finite Difference Calculus, Van Nostrand Reinhold Company, New York, 1970. R.W. Clough, J. Penzien, Dynamics of Structures, McGraw-Hill, 1975. S. Timoshenko, D.H. Young, W. Weaver, Vibration Problems in Engineering, John Wiley & Sons, 1974.
Applied Mathematical Modelling 32 (2008) 688–699 www.elsevier.com/locate/apm
Mathematical analogy of a beam on elastic supports as a beam on elastic foundation Motohiro Sato *, Shunji Kanie, Takashi Mikami Graduate School of Engineering, Hokkaido University, Kita 13 Nishi 8, Kita-ku, Sapporo, Japan Received 1 January 2006; received in revised form 1 December 2006; accepted 6 February 2007 Available online 20 February 2007
Abstract This paper presents the mathematical hypothesis that a beam on equidistant elastic supports (BOES) can be considered as a beam on an elastic foundation (BOEF) in static and free vibration problems. This modeling of BOES as BOEF is presumed to be applicable to a limited range of support stiffness, spacing and flexural rigidity of the beam. The authors investigate the applicability of the modeling of BOEF from the property of characteristic solutions obtained from governing equations of both BOES and BOEF. In this study, the formulation of BOES leads to governing difference equations, and the motions of BOEF are expressed by differential equations. This is because exact solutions must be employed in order to verify their analogy accurately. The characteristic solutions obtained from these two governing equations are compared to each other in order to investigate the relationship between them. Ó 2007 Elsevier Inc. All rights reserved. Keywords: Submerged floating tunnel; Beams on elastic foundation; Beams on elastic supports; Dynamic analysis
1. Introduction There are many civil engineering structures that can be idealized as beams on elastic supports (from now on abbreviated to ‘BOES’). As an example, submerged floating tunnel (SFT) is shown in Fig. 1. It is a whole new structure for strait crossings, which is stabilized in the water by the balance between tunnel buoyancy and the mooring force of the tension legs. This class of floating structure moored to the sea bed by tethers is referred to variously as tension leg platforms, tethered production platforms, and so on (for example, see [1]). If the tunnels length is very great compared with its cross-sectional diameter, the beam theory is applicable when calculating its deformations and responses. Moreover, SFT can be regarded as a beam on discrete elastic supports if its tension legs are separately spaced along its length and the elasticity of the tension legs is taken into consideration. It is well known that in many engineering cases, if separate elastic supports are equally spaced at close enough intervals along the beam, it can be modeled by a beam on an equivalent imaginary elastic foundation *
Corresponding author. E-mail address: [email protected] (M. Sato).
0307-904X/$ - see front matter Ó 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.apm.2007.02.002
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
h
(a) Cross sectional shape
h
689
h
(b) Side view
Fig. 1. Submerged floating tunnel.
(here abbreviated to ‘BOEF’). The major advantage of applying the theory of BOEF is the resulting ability to evaluate the basic characteristics quickly and easily. The beam on elastic foundation model is suitable for simplified analysis of the global behavior of the structure under consideration and is therefore very effective in enabling designers to assume structural dimensions. There have been many studies on BOEF theory; for example Hetenyi [2] gives a comprehensive analysis of BOEF. Therefore, if it is possible to replace separate elastic supports with a continuous elastic foundation, designers can easily investigate behavioral characteristics by using previously discovered analytical results such as these. In the case of BOES, the hypothesis that the approximation of a uniform elastic foundation is applicable was investigated by Ellington [3]. However, those studies were confined within the limits of static problems. If this modeling of BOEF is used for a basic structural design of a submerged floating tunnel as shown in Fig. 1, it is also necessary to investigate the analogy between BOES and BOEF with consideration given to dynamic effects. The purpose of this paper is to make the applicability of this modeling clear while showing the conditions in which BOES can be considered as BOEF in static and free vibration problems. To confirm the validity and accuracy of the analogy between them, these solutions must be exact because approximate solutions have some inherent errors. Therefore, in this paper exact solutions are developed for the deflections and motions of both BOES and BOEF. In the case of beams on elastic supports, the method of finite difference [4] can be applied for determination of the governing equation and its solutions. Similarly, motions of beams on elastic foundation are expressed by differential equations. The characteristic solutions are obtained from these two equations and those solutions are compared to each other to investigate their equivalence. In order to verify the results of comparing the characteristic solutions, actual static and dynamic characteristics should also be considered. Subsequently, static deformations, bending moments, natural frequencies and modal shapes of BOES and BOEF are compared to each other. 2. Governing equations and its solutions The deflections and motions of the beam described in this paper are based on the Bernoulli–Euler beam theory. The beam is assumed to have flexural rigidity EI and mass per unit length m. 2.1. Beam on elastic supports (BOES) As mentioned previously, if the beam is uniformly supported the method of finite difference is applicable. Fig. 2 shows the coordinate system used in this paper for introducing the governing difference equations of BOES. All elastic supports are assumed to have an axial rigidity kv and be spaced at a regular interval h along the beam (x-) axis. The governing difference equation of motion can be obtained by the relationship between the end forces and end displacements for a typical beam segment extending from support r to r + 1 as shown
690
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
r+1
r
y
h
ϕ
r
vr
ϕ
vr+1
r+1
x
M r+1,r
M r,r+1
S r,r+1
S r+1,r
Fig. 2. Notations and positive directions.
in Fig. 2, and the equations of equilibrium at a point on an arbitrary elastic support r. First, the dynamic-stiffness expressions are derived by taking account of the actual distribution of mass and stiffness for a beam segment, as 9 9 8 9 8 8 vr =h > > hS r;rþ1 > hS r;rþ1 > > > > > > > > > > > > > = = > = < hS < v =h > < hS rþ1;r rþ1 rþ1;r ¼ KðXÞ þ ; ð1Þ > > > > > u M M r;rþ1 > > > > > > > r;rþ1 r > > : > > > > ; ; ; : : M rþ1;r urþ1 M rþ1;r where K(X) is the dynamic-stiffness matrix [5], v, u, M and S denote the deflection, slope, bending moment and shearing force, respectively; the subscript r and r + 1 of v, u and r, r + 1 and r + 1, r of M and S denote the support point under consideration as shown in Fig. 2. M and S show the fixed-end moment and shearing force respectively. In the case of bending vibration analyses, K(X) contains the non-dimensional frequency parameter X which is a function of the circular frequency x, as 2 3 6c4 12c5 12c6 6c3 7 EI 6 6 12c6 12c5 6c4 6c3 7 KðXÞ ¼ 6 ð2aÞ 7; h 4 6c3 6c4 4c1 2c2 5 6c4
6c3
2c2
4c1
where mx2 h4 ; 24EI sin a cosh a cos a sinh a a; c1 ¼ 4ð1 cos a cosh aÞ sinh a sin a a; c2 ¼ 2ð1 cos a cosh aÞ sin a sinh a a2 ; c3 ¼ 6ð1 cos a cosh aÞ X¼
ð2bÞ ð2cÞ ð2dÞ ð2eÞ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
691
c4 ¼
cosh a cos a a2 ; 6ð1 cos a cosh aÞ
ð2fÞ
c5 ¼
sin a cosh a þ cos a sinh a 3 a; 12ð1 cos a cosh aÞ
ð2gÞ
sinh a þ sin a a3 ; 12ð1 cos a cosh aÞ ffiffiffiffiffiffiffiffiffi p 4 a ¼ 24X:
c6 ¼
ð2hÞ ð2iÞ
Next, the equilibrium of a typical elastic support r requires that S r;rþ1 þ S r;r1 ¼ k v vr ; M r;rþ1 þ M r;r1 ¼ 0:
ð3aÞ ð3bÞ
Substituting Eq. (1) into S in Eq. (3a) and M in Eq. (3b) leads to c2 ðurþ1 þ ur1 Þ þ 4c1 ur þ 3c4 ðV r1 V rþ1 Þ ¼
T rh ; 2EI
ð4aÞ 2
c4 ðurþ1 ur1 Þ þ 4ðc5 þ K v ÞV r 2c6 ðV r1 þ V rþ1 Þ ¼
P rh ; 6EI
ð4bÞ
where Vr = vr/h, Tr = Mr,r+1 + Mr,r1, Pr = Sr,r+1 + Sr,r1 and Kv ¼
k v h3 24EI
ð5Þ
is the so-called relative stiffness of the supports. Elimination of u in Eqs. (4a) and (4b) gives the next difference equation ( ) 2 3c24 ðE E1 Þ h c4 ðE1 EÞ P rh 1 þ 4ðc5 þ K v Þ 2c6 ðE þ E Þ V r ¼ Tr þ ; ð6Þ 2EI c2 ðE þ E1 Þ þ 4c1 3 c2 ðE þ E1 Þ þ 4c1 where E is the shifting operator. Assuming for the complementary solution of Eq. (6) that V r ¼ D expðkD rÞ;
ð7Þ
where D is an arbitrary constant and kD is to be determined. Substituting this in the homogeneous part of Eq. (6) leads to 6c22 sinh2 kD D þ 4ðc5 þ K v c6 cosh kD Þ expðkD rÞ ¼ 0: ð8Þ c2 cosh kD þ 2c1 By setting the expression in brackets to zero, we obtain a quadratic for coshkD that is given by ð3c24 2c2 c6 Þcosh2 kD þ f2c2 ðc5 þ K v Þ 4c1 c6 g cosh kD þ 4c1 ðc5 þ K v Þ 3c24 ¼ 0:
ð9Þ
This equation, in general, yields two distinct values of cosh kD. Corresponding to each root of cosh kD, there are two possible value of kD, namely ±k. In static problems, the basic concept is the same as for dynamic problems. The static beam stiffness can be considered as representing a special case of the dynamic stiffness where the applied frequency is zero, that is 2 3 12 12 6 6 7 EI 6 6 12 12 6 6 7 ð10Þ Kð0Þ ¼ 6 7: h 4 6 6 4 2 5 6
6
2
4
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Applying Eq. (10) as a substitution for Eq. (2a) yields the following governing equation: ( ) 3ðE E1 Þ2 1 2ðE þ E 2Þ þ 4K v V r ¼ 0: E þ E1 þ 4
ð11Þ
Therefore the characteristic equation is derived by substituting Eq. (7) into Eq. (11) as follows: cosh2 kD 2ð1 K v Þ cosh kD þ 1 þ 4K v ¼ 0: The solution of Eq. (12) is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh kD ¼ 1 K v K 2v 6K v :
ð12Þ
ð13Þ
The values of kD and the general solutions Vr are determined in three cases corresponding to the range of Kv as (i) 0 < Kv < 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi cosh kD ¼ 1 K v i 6K v K 2v ;
ð14Þ
kD ¼ ðaD bD iÞ; V r ¼ D1 sin bD r sinh aD r þ D2 sin bD r cosh aD r þ D3 cos bD r sinh aD r þ D4 cos bD r cosh aD r;
ð15Þ ð16Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffi 4 þ 2K v þ 6K v aD ¼ cosh ; pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 pffiffiffiffiffiffiffiffi 4 þ 2K v 6K v bD ¼ cos1 ; 2 1
ð17Þ ð18Þ
(ii) Kv = 6 kD ¼ ðaD þ ipÞ; r V r ¼ ð1Þ ðD1 þ D2 rÞðD3 sinh aD r þ D4 cosh aD rÞ;
ð19Þ ð20Þ
where aD ¼ cosh1 5:
ð21Þ
(iii) Kv > 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 2 kD ¼ cosh K v 1 K v 6K v þ ip ;
ð22Þ
V r ¼ ð1Þr ðD1 sinh aD1 r þ D2 cosh aD1 r þ D3 sinh aD2 r þ D4 cosh aD2 rÞ;
ð23Þ
where aD1 aD2
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ cosh K v 1 K 2v 6K v ; qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ cosh1 K v 1 þ K 2v 6K v : 1
ð24Þ ð25Þ
2.2. Beam on elastic foundation Next, we consider the beam on an elastic foundation equivalent to equidistant elastic supports, as shown in the previous section. Neglecting the term of damping force, the governing differential equation of motion for the deflection curve v(x, t) of BOEF is known to be m
o2 vðx; tÞ o4 vðx; tÞ þ EI þ k 0v vðx; tÞ ¼ 0; ot2 ox4
ð26Þ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
693
where k 0v is foundation modulus. If elastic supports are equivalent to an elastic foundation, they can be replaced by a continuously distributed imaginary foundation. In this case, this modulus k 0v is written as kv ð27Þ k 0v ¼ : h Introducing the following non-dimensional parameters of x k 0 h4 mx2 h4 n ¼ ; Kv ¼ v ; X ¼ ; h 24EI 24EI into Eq. (26) and assuming that the solution of Eq. (26) has the form vðx; tÞ ¼ V ðnÞðA1 sin xt þ A2 cos xtÞ:
ð28Þ
It should be noted that V(n) is dimensionless displacement divided by h. Substituting Eq. (28) into Eq. (26), the ordinary differential equation with respect to n may be written as d4 V ðnÞ þ 24ðK v XÞV ðnÞ ¼ 0: dn4
ð29Þ
Eq. (29) can be solved in the same way as the difference equation by assuming a solution in the form V ðnÞ ¼ C expðkc nÞ;
ð30Þ
where C is an arbitrary constant and kc is to be determined. Substituting this into Eq. (15), we obtain the characteristic equation k4c þ 24ðK v XÞ ¼ 0;
ð31Þ
which has the solutions (i) X > Kv kc ¼ aC ; aC i;
ð32Þ
V ðnÞ ¼ C 1 cos aC n þ C 2 sin aC n þ C 3 cosh aC n þ C 4 sinh aC n;
ð33Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aC ¼ 4 24ðX K v Þ:
ð34Þ
(ii) X < Kv kc ¼ ð1 þ iÞaC ; ð1 iÞaC ; V ðnÞ ¼ C 5 sin aC n sinh aC n þ C 6 sin aC n cosh aC n þ C 7 cos aC n sinh aC n þ C 8 cos aC n cosh aC n;
ð35Þ ð36Þ
where pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi aC ¼ 4 6ðK v XÞ:
ð37Þ
In static problems, the solution can be obtained by setting X, which is the parameter in terms of inertia force, pffiffiffiffiffiffiffiffi to zero, that is aC ¼ 4 6K v . It is clear that the form of V(n) corresponds to Eq. (36). 3. Analogy between BOES and BOEF For static problems, the forms of both characteristic solutions correspond with each other when 0 < Kv < 6; however in the case of Kv P 6, the two forms are quite different. Consequently, discussions of analogy between BOES and BOEF should be undertaken when 0 < Kv < 6 for static problems. We can use the solutions of Eqs. (9) and (17), (18) to estimate their analogy because these solutions are both exact. Fig. 3 shows the comparisons of arguments of two characteristic solutions, BOES and BOEF. For BOEF, the real part of the characteristic root is equal to the imaginary one independent of Kv as indicated in Eq. (35). Therefore, the arguments are p/4(0.7854) for any value of Kv that satisfies 0 < Kv < 6. On the other hand, the arguments for BOES are not necessarily p/4; however as shown in Fig. 3, there is little
694
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699 0.79
arg.
0.788 0.786 0.784
BOES BOEF
0.782 0.78 0.001
0.01
0.1
1
Kv Fig. 3. Comparison of arguments (static).
1
Ω = 0.1
Re./Im.
0.98
Ω = 4.0
0.96 0.94
BOES BOEF
0.92
Ω = 1.0 0.9 0.001
0.01
0.1
1
Kv Fig. 4. Ratio of the real part of solutions to the imaginary part.
difference between the two arguments over the range of about Kv 6 0.05. It can be said that this range is the condition in which the approximation of a uniform elastic foundation is applicable in static problems. Fig. 4 shows the comparison of the ratio of the real part of solutions for kD or kC to the imaginary part. The two solutions agree well with each other when Kv 6 0.05. There are no significant differences. This range, Kv 6 0.05, is the condition in which the approximation of a uniform elastic foundation is applicable in dynamic problems. This result shows that the applicable condition for dynamic problems is similar to that for static problems. 3.1. Static characteristics In order to verify the results of comparing the characteristic solutions in the preceding section, actual static and dynamic characteristics should also be considered. In this section, static deformations and bending moments of BOES and BOEF subjected to a single concentrated force P at the center of the simple beam are taken up, and the effects of Kv on the differences between BOES and BOEF are considered. In addition, comparisons of free vibration characteristics are studied. 4. Beam on elastic supports (BOES) Here two cases are considered as shown in Fig. 5. The three moment equation and equilibrium near the concentrated load can be written as h h ðM r1 þ 4M r þ M rþ1 Þ þ ðV r1 2V r þ V rþ1 Þ ¼ ðM r;r1 þ M r;rþ1 Þ; 6EI 3EI ðM r1 2M r þ M rþ1 Þ þ k v h2 V r P r h ¼ M r ;
ð38Þ ð39Þ
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
695
P 1
0
0
1
R'
(a) Case-A
P 1
0
1
R'
(b) Case-B Fig. 5. Analytical cases for the static problem.
where M is the fixed-end moment and P is the fixed-end shearing force. From Eq. (38) the bending moment equation is given, containing shifting operators, by h E þ E1 2 Mr ¼ Vr 6EI E þ E1 þ 4 ¼ fð1 2j1 Þ sin bD r sinh aD r þ 2j2 cos bD r cosh aD rgD1 þ fð1 2j1 Þ sin bD r cosh aD r þ 2j2 cos bD r sinh aD rgD2 þ fð1 2j1 Þ cos bD r sinh aD r 2j2 sin bD r cosh aD rgD3 þ fð1 2j1 Þ cos bD r cosh aD r 2j2 sin bD r sinh aD rgD4 ;
ð40Þ
where D1–D4 are constants that can be determined from the boundary conditions at the end of the beam, and sin aD ðcosh aD þ 2 cos bD Þ ; ð2 þ cosh aD cos bD Þ2 þ ðsinh aD sin bD Þ2 sin bD ð2 cosh aD þ cos bD Þ g2 ¼ 2 2 2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g
g1 ¼
j1 ¼
ð41Þ
ð1 þ 2 cosh aD cos bD Þð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ2 2
2
2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g 3 sinh aD sin bD j2 ¼ : 2 2 2fð2 þ cosh aD cos bD Þ þ ðsinh aD sin bD Þ g
; ð42Þ
Case-A r 1 ¼ r ¼ 0;
r þ 1 ¼ 1;
M r;r1 ¼ M 0;0 ¼
3 Ph; 16
M r;rþ1 ¼ M 0;1 ¼ 0;
P r ¼ 0;
1 M r ¼ Ph: 2
The boundary conditions are V R0 ¼ 0
and
M R0 ¼ 0
at R0 ¼
R1 2
ð43Þ
696
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
and the three moment equation and the equilibrium equation are h Ph2 ð5M 0 þ M 1 Þ þ V 1 V 0 ¼ ; 6EI 16EI h Ph2 ðM 0 M 1 Þ þ 4K v V 0 ¼ : 6EI 12EI
ð44Þ ð45Þ
Case-B r 1 ¼ r þ 1 ¼ 1;
r ¼ 0;
M r;r1 ¼ M r;rþ1 ¼ M 0;1 ¼ 0;
Pr ¼ P;
M r ¼ 0:
The boundary conditions are V R0 ¼ 0
and
M R0 ¼ 0
at R0 ¼
R 2
ð46Þ
and the three moment equation and the equilibrium equation are h ð2M 0 þ M 1 Þ þ V 1 V 0 ¼ 0; 6EI h Ph2 ðM 0 M 1 Þ þ 2K v V 0 ¼ : 6EI 12EI
ð47Þ ð48Þ
4.1. Beam on elastic foundation (BOEF) In the case of a simply supported BOEF with concentrated load P in the middle of the span, it is extremely easy to determine the deflection curve and bending moment by introducing the following conditions l l l ð49Þ V ¼ 0 and M ¼ 0 at n ¼ ; 2h 2h 2h uð0Þ ¼ 0 and
Sð0Þ ¼
P 2
at n ¼ 0:
ð50Þ
When this is done, V(n) and M(n) yield P aC 0 2 2k v h ðcosh aC R
V ðnÞ ¼
þ cos aC RÞ
fsin aC n cosh aC ðR nÞ sin aC ðR nÞ cosh aC n
þ cos aC n sinh aC ðR nÞ cos aC ðR nÞ sinh aC ng; MðnÞ ¼
ð51Þ
Ph f sin aC n cosh aC ðR nÞ þ sin aC ðR nÞ cosh aC n 4aC ðcosh aC R þ cos aC RÞ
þ cos aC n sinh aC ðR nÞ cos aC ðR nÞ sinh aC ng:
ð52Þ
4.2. Results and discussion Fig. 6 shows the deflections and Fig. 7 the bending moments of BOES and BOEF from the center to the right end of both beams (R 0 = 5). Compared to the differences of deflections, those of the bending moment are larger, however in all cases of Kv = 0.005 and Kv = 0.05, the differences are quite insignificant. Table 1 gives full details of the difference between BOES and BOEF at the point at which the concentrated force is acting (i.e. at the center of the beam). The above-mentioned results suggest that the approach of comparing the characteristic solutions of BOES with those of BOEF for static problems is valid.
M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699 0
kvvr /P ( or k'vhv(ξ )/P)
kvvr /P ( or k'vhv(ξ )/P)
0
697
K v = 0.005
0.2
K v = 0.05 BOES BOEF
0.4
K v = 0 .5
0.6 0
1
2 3 r (or ξ )
4
0.2
K v = 0.05 0.4
BOES BOEF
K v = 0 .5
0.6
5
0
1
(a) Case-A
2
3 r (or ξ )
4
5
(b) Case-B
0
K v = 0.05 0.2
K v = 0 .5
K v = 0.005 BOES BOEF
0.4
0.6
Mr /Ph ( or M(ξ)/Ph)
Mr /Ph ( or M(ξ)/Ph)
Fig. 6. Deflections.
0
1
2 3 r (or ξ )
4
5
0
K v = 0.05 K v = 0.5 0.2
K v = 0.005 BOES BOEF
0.4
0.6
0
1
(a) Case-A
2
3 r (or ξ )
4
5
(b) Case-B Fig. 7. Bending moments.
Table 1 Differences between BOES and BOEF at the center of the beam Kv
V(n)/Vr (Case-A)
V(n)/Vr (Case-B)
M(n)/Mr (Case-A)
M(n)/Mr (Case-B)
0.005 0.05 0.5
1.000 0.995 0.940
1.000 1.003 1.035
0.985 0.951 0.840
1.032 1.106 1.456
5. Free vibration For simply supported beams, modal shape functions Vr (BOES) and V(n) (BOES) can be expressed as follows: npr ; n ¼ 1; 2; . . . ; ð53Þ V r ¼ Ad sin R npn V ðnÞ ¼ Ac sin ; n ¼ 1; 2; . . . ð54Þ R These functions satisfy the boundary conditions at r (or n) = 0, R. Inserting Eq. (53) into the left-hand side of Eqs. (6) and (54) in the left-hand side of Eq. (11), we find that 6c24 sin2 np np R For BOES : þ 4 c þ K c cos ¼ 0; ð55Þ 5 v 6 R c2 cos np þ 2c1 R 1 np4 For BOEF : X ¼ þ K v: ð56Þ 24 R
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M. Sato et al. / Applied Mathematical Modelling 32 (2008) 688–699
The solutions of the resulting eigenvalue problems give the vibration modes and the corresponding natural frequencies. These are the characteristic equations for determining the natural frequencies. The solutions of Eqs. (55) and (56) then provide the values of X which represent the frequencies of free vibration. 5.1. Results and discussion Fig. 8 shows the comparison of natural frequencies. The presence of an elastic foundation influences only the natural frequencies [6]. It is found that this figure also indicates the equivalence when Kv 6 0.05. Fig. 9 shows the modal shapes of BOEF and Fig. 10 shows those of BOES. Some differences exist between the two models. In the case of BOEF, the vibration modes have the same shapes as those for beams without 1
0.99
4
ΩD ΩC
0.98 n/R=0.1 n/R=0.5 n/R=0.9
0.97
0.96 0.01
0.05
0.1
0.5
1
Kv Fig. 8. Comparison of natural frequencies.
(a)
n = 0.1 R
(b)
n = 0.5 R
(c)
n = 0.9 R
Fig. 9. Modal shape (beams on elastic foundation).
(a)
n = 0.1 R
(b)
n = 0.5 R
(c)
n = 0.9 R
Fig. 10. Modal shape (beams on elastic supports). Solid lines: Kv = 0.05, broken lines: Kv = 0.5, dotted lines: Kv = 5.0.
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elastic foundations. However in beams on elastic supports, mode shapes change as Kv increases. This is caused by the increase in local stiffness of the support. However when Kv 6 0.05, the two modes of beams are very similar to each other. The differences are quite insignificant. These results suggest that the approach in the previous section is valid. 6. Conclusions In this study, fourth order difference equations dealing with the deflection of BOES, and their solutions, were introduced to be compared with those of BOEF. The results on the analogy for static problems were the same as that of Ellington, that is, BOES and BOEF were nearly equal with respect to the expressions of deflection when Kv 6 0.05. This investigation was also developed to cover dynamic problems in a similar manner in this study. Accordingly, at least both are equivalent under the condition of Kv 6 0.05 even for the dynamic problems. Next, in order to verify these considerations, the deflections and bending moments of the beams when concentrated load was acting at the center of these beams, as well as natural frequencies and modal shapes were compared with each other. In all cases of these variables, there were no significant differences between solutions of BOES and BOEF within the range of Kv 6 0.05. The above results will be very effective particularly for decision-making in SFT projects and in work with similar types of structures where dynamic effects must be taken into consideration during the structural design phase. References [1] [2] [3] [4] [5] [6]
M.H. Patel, J.A. Witz, Compliant Offshore Structures, Butterworth-Heinemann, 1991. M. Hetenyi, Beams on Elastic Foundation, The University of Michigan Press, 1974. J.P. Ellington, The beam on discrete elastic supports, Bull. Int. Railway Cong. Assoc. 34 (12) (1957) 933–941. T. Wah, L.A. Calcote, Structural Analysis by Finite Difference Calculus, Van Nostrand Reinhold Company, New York, 1970. R.W. Clough, J. Penzien, Dynamics of Structures, McGraw-Hill, 1975. S. Timoshenko, D.H. Young, W. Weaver, Vibration Problems in Engineering, John Wiley & Sons, 1974.