Stagnation and rotating-disk flows over a compliant surface

Z. angew. Math. Phys. 52 (2001) 770–7920010-2571/01/050770-23 $ 1.50+0.20/0c© 2001 Birkhauser Verlag, Basel

Zeitschrift fur angewandteMathematik und Physik ZAMP

Stagnation and rotating-disk flows over a compliant surface

N. Phan-Thien and K. S. Yeo

Abstract. In this paper, exact solutions to the stagnation flow over a compliant surface andthe flow about a rotating disk coated by a compliant layer are discussed. The compliant layeris modelled by the Mooney rubberlike material. In the stagnation flow, the deformation ofthe rubber layer depends on two dimensionless groups: one is the ratio of pressure to the shearmodulus of the material, and the other is the ratio of the boundary layer thickness to the thicknessof the layer. It is shown that this deformation has a limit point at a critical value of the pressureforce, beyond which the solution does not exist. In the rotating-disk flow, the deformation of therubber layer depends on two dimensionless groups, one is a material parameter, and the otheris a flow loading parameter, the ratio of the flow-induced viscous stress to the shear modulusmultiplied by the ratio of the thickness of the layer to the boundary layer thickness. At largevalue of a material parameter, it is found that this deformation has a limit point at a criticalvalue of the flow loading parameter.

Keywords. Compliant surface, similarity solution, rubberlike elasticity, stagnation flow, rotat-ing flow.

1. Introduction

The flow-compliant wall interactions result in a set of very non-linear problems, andare, therefore, not at all well understood. Yeo et al (1994) considered the effects ofthe compliant wall on the stability of the boundary layer flow in which the wall wasmodelled as a linear compressible elastic body, which in turn was allowed to deformfinitely. It was found that the mean loading state, as induced by the hydrostaticpressure, could greatly reduce the effective shear modulus of the wall therebyleading to significant changes in the flow stability characteristics. The study wasnot entirely satisfactory in the sense that the constitutive equation chosen wasappropriate for a compressible elastic body undergoing small deformation, whereasthe actual deformation was allowed to be finite. (The fact that the (arbitrary)hydrostatic pressure can directly influence the effective shear modulus may bea consequence of the constitutive assumption.) Recently, Cooper and Carpenter(1997) considered the linear stability of the boundary-layer coaxial-disk flow overa compliant layer, modelled as a linear viscoelastic material. The problem reducedto solving a sixth-order eigenvalue problem and it was found that the compliant

Vol. 52 (2001) Stagnation and rotating-disk flows over a compliant surface 771

layer promoted a strong instability which is associated with the horizontal motionof the layer.

Although the literature on this area is quite extensive (e.g., see the Introductionof Cooper and Carpenter 1997), the non-linear flow interactions with compliantwalls (modelled by a non-linear constitutive law) are not well understood, and itis desirable to single out a flow field which is not too complicated, and to assumea reasonable but non-linear constitutive equation for the compliant surface to il-lustrate the mechanisms of the interactions. This is done here for the plane andaxi-symmetric stagnation flow, and the flow about a rotating disk (see, for exampleSchlichting 1979), where the compliant surface is modelled by the incompressibleMooney material (Mooney 1940, Rivlin 1948). It will be shown that the similaritysolutions of the stagnation and the coaxial-disk flows at steady state generate trac-tion fields which are compatible with those generated by the compression and tor-sional deformations of a Mooney material (Klingbeil and Shield 1966, Phan-Thien1988b). Thus, they constitute exact solutions to the flow-induced deformationproblems. The simplicity of these similarity solutions allows the fluid flow to beuncoupled from the deformation, and hence the problems are tractable and solv-able to any arbitrary degree of accuracy. For the stagnation flow, it is found thatthe deformation of the compliant layer is governed by two dimensionless groups:one is the pressure force normalised by the shear modulus of the material, andthe other is the ratio of the boundary layer thickness to the thickness of the layer.Keeping the latter ratio of thicknesses constant, the deformation is found to loseits stability at a limit point which corresponds to a critical value of the pressureforce. The cause of this behaviour may be traced to the softening effect of the ma-terial due to the flow (pressure) loading. For the flow between coaxial disks, it isfound that the deformation of the layer is governed by a shear loading parameter.However, whether or not the deformation has a limit point depends on a materialparameter, which is the ratio of the second to the first normal stress differences inshear deformation. At a critical value of this parameter, the deformation loses itsstability at a limit point. A motivation for this problem arises from our desire tounderstand some of the issues in the instabilities which arise with soft coatings onunderwater structures.

2. Stagnation flow

2.1. Two-dimensional flow

Figure 1 illustrates the stagnation flow geometry considered here. The plane stag-nation flow (Hiemenz 1911, Howarth 1935) is impinging on a compliant layerrigidly attached to a plane surface. Two Cartesian coordinate systems are used,x, y, z for the stagnation flow and x1, x2, x3 for the deformation of the com-pliant layer. The plane z = 0 represents the interface between the fluid and the

772 N. Phan-Thien and K. S. Yeo ZAMP

x3, z

x1

h

x

Figure 1. Flow geometry

compliant layer, and the plane x3 = 0 is the rigid plane surface to which thelayer is rigidly attached. The boundary conditions at z → ∞ correspond to thefrictionless potential flow

u = γ0x, v = 0, w = −γ0z, (1)

where γ0 is the imposed strain rate at infinity and u = u, v, w is the velocityvector.

The viscous flow problem is solved by (Hiemenz 1911)

u = xf ′ (z) , v = 0, w = −f (z) , (2)

P0 − P =12ργ2

0

[x2 + F (z)

], (3)

where P is the hydrostatic pressure, P0 is a reference pressure, ρ is the fluiddensity, f and F are functions to be determined and the prime denotes a deriva-tive with respect to z. Re-scaling of the vertical coordinate ζ = z

√γ0/ν, where

ν is the kinematic viscosity, and re-writing the velocity as

f (z) =√

γ0νφ (ζ) , (4)

one finds that the Navier-Stokes equations reduce to a non-linear third-order equa-tion in φ (detailed in Schlichting 1979). This equation was solved numerically byHiemenz (1911) and the numerical solution was later refined by Howarth (1935).Their numerical solutions showed that φ′′ (0) ≈ 1.2326. On the interface be-tween the compliant wall and the fluid, we have, noting that f ′ (z) = γ0φ

′ (ζ) ,f ′′ (z) = γ0

√γ0/νφ′′ (ζ) ,

σ33 = −P = −P0 +12ργ2

0

(x2 + F

)∣∣∣∣z=0

= −P0 +12ργ2

0x2, (5)

σ13 = ηxf ′′ (0) = ηγ0x

√γ0

νφ′′ (0) . (6)


Note that these stress components are up to quadratic in x, and these need to bematched by the deformation of the compliant wall.

The compliant layer is modelled as an incompressible rubber-like material, withthe constitutive equation (Mooney 1940, Rivlin 1948) for the stress tensor S :

S = αB− βB−1, (7)

where α and β are material constants, B = FFT is a strain tensor (Fingerstrain tensor) in which

F =∂x∂X

(8)

is the deformation gradient, where X denotes the position of a particle in theundeformed configuration and x denotes the position of the same particle in thedeformed configuration. This constitutive model has been found to be adequatefor rubber under moderate deformation (Treloar 1976); at large deformation, αand β need to be dependent on either the strain or the stress invariants for agood fit to experimental data (Phan-Thien 1988a). In a simple shear deformationin the 13-plane with shear strain γ , the stress components are

S11 = −P0 + α(1 + γ2

)− β,

S22 = −P0 + α− β,

S33 = −P0 + α− β(γ2 + 1

), (9)

S13 = (α + β) γ,

and consequently the first and the second normal stress differences are N1 =S11 − S33 = (α + β) γ2, N2 = S33 − S22 = −βγ2, respectively, and the shearmodulus is S13/γ = α + β. Data on polymers seem to suggest that β is of theorder 0.1α.

In the present flow, and under the non-uniform pressure loading, the layer iscompressed from the initial thickness H to the current uniform thickness h. Thegeneral kinematics of the squeezing deformation have been considered before, seeKlingbeil and Shield (1966) and Phan-Thien (1988b). Here, we show that thisdeformation is admissible and satisfies the interfacial boundary conditions (5-6)and therefore is the solution of the present problem. This deformation is describedby

X1 = x1g (ξ) , X2 = x2, X3 = hG (ξ) , (10)

where X1, X2, X3 : −∞ < X1 < ∞,−∞ < X2 < ∞, 0 ≤ X3 ≤ H represents theundeformed configuration, x1, x2, x3 : −∞ < x1 < ∞,−∞ < x2 < ∞, 0 ≤ x3 ≤ hthe deformed configuration, and ξ = x3/h ∈ [0, 1] is the scaled vertical coordinate.The boundary conditions are

g (0) = 1, G (0) = 0. (11)


The deformation gradient and its inverse are given by

[F] =

g−1 0 −x1

h

g′

gG′0 1 0

0 01G′

,

[F−1

]=

g 0

x1

hg′

0 1 00 0 G′

,

respectively, where the prime denotes a derivative with respect to ξ. Incompress-ibility requires that det [F] = 1, or

gG′ = 1. (12)

The Finger strain tensor and its inverse are given by, respectively,

[B] =[FFT

]=

x21

h2g′2 + G′2 0 −x1

hgg′

0 1 0

−x1

hgg′ 0

1G′2

,

[B−1

]=

[F−TF−1

]=

g2 0x1

hgg′

0 1 0x1

hgg′ 0

x21

h2g′2 + G′2

.

Note that the stress components,

S11 = α

(x2

1

h2g′2 + g−2

)− βg2,

S22 = α− β,

S33 = αg2 − β

(x2

1

h2g′2 + g−2

),

S13 = − (α + β)x1

hgg′,

are quadratic at most in x1 (including the pressure, see below). This impliesthat they can be matched to the flow-induced stresses. From the balance of linearmomentum,

∂P∂x1

=x1

h2

[2αg′2 − (α + β)

(g′2 + gg′′

)], (13)

∂P∂x3

= − (α + β)12h

(g2

)′+

α

h

(g2

)′ − β

h

(x2

1

h2g′2 + g−2

)′, (14)

we find the solution

P =x2

1

2h2Q1 + Q2 + Q0, (15)


with

Q1 =[2αg′2 − (α + β)

(g′2 + gg′′

)],

Q2 = −12

(α + β) g2 + αg2 − βg−2.

Compatibility between (13-14) requires

Q′1 = −2β

(g′2

)′,

which can be simplified tog′g′′ − gg′′′ = 0, (16)

which has as the first integral

g (ξ)′′ − C2g (ξ) = 0, (17)

where C is an integration constant. Another integration leads to

g = e−Cξ + D[eCξ − e−Cξ

](18)

where the boundary condition g(0) = 1 has been incorporated; C and D areintegration constants. Denoting g1 = g (1) , g′1 = g′ (1) , g′′1 = g′′ (1) , theseconstants of integration are found by requiring the stresses are matched:

σ33 = −P0 +12ργ2

0x2 = −(

x21

2h2Q1 + Q2 + Q0

)+ αg2 − β

(x2

1

h2g′2 + g−2

)∣∣∣∣ξ=1

,

σ13 = ηγ0x

√γ0

νφ′′ (0) = − (α + β)

x1

hgg′

∣∣∣ξ=1

.

The only terms that need to be matched are terms of O(x1) and O(x21). Terms

of O(1) take care of themselves through the arbitrary pressure P0 and Q0.At O

(x2

1

):

12ργ2

0h2 = −12Q1 − βg′2

∣∣∣∣z=1

= −12

[2αg′21 − (α + β)

(g′21 + g1g

′′1

)]− βg′12,

which leads toΓ = −g′21 + g1g

′′1 = 4DC2 (1−D) , (19)

where

Γ =ργ2

0h2

α + β(20)

is a flow parameter, representing the pressure loading relative to the shear modulusof the material.At O (x1) :

ηγ0h

√γ0

νφ′′ (0) = − (α + β) g1g

′1, (21)


Flow parameter Γ

10-8 10-7 10-6 10-5 10-4 10-3 10-2 10-1 100

Ca

nd

D

10-11

10-10

10-9

10-8

10-7

10-6

10-5

10-4

10-3

10-2

10-1

100

101

102

δ = 0.1δ = 1δ = 10

stable

unstable

D

C

Figure 2. Graph of solution versus the flow parameter Γ.

or,1.2326Γδ = C

[e−2C − 2De−2C −D2

(e2C − e−2C

)], (22)

where

δ =1h

√ν

γ0(23)

is the ratio of the boundary layer thickness (√

ν/γ0 ) to the layer thickness. Tosum up, the deformation of the compliant layer requires the solution of the non-linear algebraic equations

4DC2 (1−D)− Γ = 0,

C[e−2C − 2De−2C −D2

(e2C − e−2C

)]− 1.2326Γδ = 0, (24)

which can be written symbolically as

G (χ; Γ, δ) = 0, (25)

where χ = (C,D) represents the solution.The set of equations is solved by the path tracking method by Keller (1977),

as implemented by Deuflhard et al (1987). The solution space for fixed values ofδ is shown in Figure 2. It is clear that the solution loses its stability at limitpoints that depend on δ; the critical (limiting) values of Γ as function of δ are


ξ

0.01 0.1 1

Ga

nd

g

0.01

0.1

1

10

unstable

stable

G

g

Figure 3. Two solutions at δ = 0.1 and Γ = 0.1. One is physical and the other unphysical.

δ Limit point, Γcr

0.01 0.4290.1 0.3491 0.11410 0.0145

Table 1. Critical values of the flow parameter Γcr

tabulated in Table 1. The solution curves below the limit points represent physicalsolutions, and above, non-physical solutions.

To decipher which branch is physical, we consider two possible solutions atδ = 0.1 and for the same loading Γ = 0.1, shown in Figure 3. The physicalsolution, which corresponds to the lower solution branch, has G(1) ≈ 3.67 andthe non-physical solution, which corresponds to the upper solution branch, hasG(1) ≈ 0.58 . Noting that G(1) = H/h, this then implies that the layer swellingup in response to a positive pressure loading, which is clearly non-physical. Thenon-physical nature of the upper branch solution is clearly seen by plotting the


Flow parameter Γ

10-5 10-4 10-3 10-2 10-1 100

No

rma

lise

dsh

ea

rstr

ess

at

the

wa

ll

0.001

0.01

0.1

1

10

δ =0.1

δ =1.0

δ =10

Figure 4. Normalised shear stress at the wall.

normalised shear stress at the wall,

hσ13

(α + β) x1= −gg′|ξ=0 = C(1− 2D) ,

against the flow loading, given in Figure 4. In the figure, the shear stress inthe unstable solution branch is seen increasing as the flow loading decreases tozero. In another words, the strain along this solution branch increases significantlyas Γ → 0. Given the same stress level this implies the effective shear modulusdecreases with the pressure loading along this unstable branch of solution. Strictlyspeaking, apart from the non-existence of solution for Γ > Γcr, there is nothingmuch that one can say about the nature of the solution on the non-physical branch.However, from the vast literature on instability, one can conclude that either theconstitutive relation is no longer adequate to cope with the deformation at hand,or that the kinematics chosen become unstable (exchange of stability, Keller 1977),and turn into three-dimensional.


2.2. Axi-symmetric flow

The axi-symmetric flow over a compliant wall, although is much more complicated,can be handled in the same manner. First, the flow kinematics take the form (incylindrical coordinates r, θ, z , where z = 0 denotes the interface between thefluid and the compliant wall, Schlichting 1979)

ur = rf ′ (z) , uθ = 0, uz = −2f (z) , (26)

P0 − P =12ργ2

0

[r2 + F (z)

], (27)

where the velocity at z → ∞ is given by ur = U = γ0r, γ0 is the strain rateimposed at infinity, and the prime denotes a derivative with respect to z. All thephysical parameters in the flow kinematics can be removed (i.e., normalised tounity) by defining

ζ = z

√γ0

ν, f (z) =

√γ0νφ (ζ) , f ′ (z) = γ0φ

′ (ζ) , f ′′ (z) = γ0

√γ0

νφ′′ (ζ) ,

(28)where ν is the kinematic viscosity. The Navier-Stokes equations lead to a third-order differential equation for φ . The solution by Froessling, as reported inSchlichting (1979), showed that φ′′ (0) ≈ 1.3120.

On the interface between the compliant wall and the fluid, we have,

σzz = −P0 +12ργ2

0

(r2 + F

)− 4ηf ′∣∣∣∣z=0

= −P0 +12ργ2

0r2, (29)

σrz = ηrf ′′ (0) = ηγ0r

√γ0

νφ′′ (0) ≈ 1.3120ηγ0r

√γ0

ν, (30)

and these stress levels (quadratic in r ) need to be matched by the deformation ofthe compliant wall.

The deformation takes the form (using cylindrical coordinates r, θ, x3 , wherex3 = 0 denotes the rigid wall, Phan-Thien, 1988b)

R = rg (ξ) , Θ = θ, X3 = hG (ξ) , (31)g (0) = 1, G (0) = 0

Again, ξ = x3/h ∈ [0, 1] is the normalised coordinate, and R,Θ, X3 representsthe undeformed configuration, r, θ, x3 the deformed configuration. The Fingerstrain tensor and its inverse are given by, respectively,

[B] =

G′ +r2

h2(gg′)2 0 − r

hg3g′

0 G′ 0

− r

hg3g′ 0

1G′2

, (32)


[B−1

]=

g2 0r

hgg′

0 g2 0r

hgg′ 0

r2

h2(g′)2 + (G′)2

. (33)

Incompressibility requires thatg2G′ = 1. (34)

The stress components are quadratic in r and therefore can be matched with(29-30):

Srr = α

(G′ +

r2

h2(gg′)2

)− βg2,

Sθθ = αG′ − βg2,

Szz = αg4 − β

(r2

h2(g′)2 + g−4

),

Srz = − r

h

(αg3g′ + βgg′

).

The equations of balance∂P∂r

=r

h2

[−α(g2g′2 + g3g′′

)− β(g′2 + gg′′

)], (35)

∂P∂z

=1h

[2αg3g′ − 2βgg′ − β

(2

r2

h2g′g′′ − 4g−5g′

)]. (36)

This is solved by

P =r2

2h2Q1 + Q2 + Q0, (37)

where

Q1 = −α(g2g′2 + g3g′′

)− β(g′2 + gg′′

),

Q′2 = 2αg3g′ − 2βgg′ + 4βg−5g′,

Q′1 = −4βg′g′′ = −2β

(g′2

)′.

The last of the preceding set of equations represents the compatibility between thetwo balance equations. It results in

g2g′2 + g3g′′ + ε(gg′′ − g′2

)= C, (38)

where C is a constant of integration, and

ε =β

α(39)

is a material parameter.The solution needs to be matched to the stresses generated by fluid force.

Matching σzz yields

−(

r2

2h2Q1 + Q2 + Q0

)+ αg4 − β

(r2

h2(g′)2 + g−4

)= −P0 +

12ργ2

0r2.


Only O(r2) needs to be matched, terms of O(1) will take care of themselves:

−Q1 − 2βg′2 = ργ20h2,

α(g2g′2 + g3g′′

)+ β

(−g′2 + gg′′)∣∣

ξ=1= ργ2

0h2,

org2g′2 + g3g′′ + ε

(gg′′ − g′2

)∣∣ξ=1

= Γ, (40)

where

Γ =ργ2

0h2

α(41)

is the flow loading parameter. Comparing (38) and (40), it is clear that C = Γ.Now, matching σrz yields

− (g3g′ + εgg′

)∣∣ξ=1

= 1.3120ργ2

0h2

α

1h

√ν

γ0(42)

In summary, the set of governing equations for this case are

g2g′2 + g3g′′ + ε(gg′′ − g′2

)= Γ (43)

g (0) = 1

− (g3g′ + εgg′

)∣∣ξ=1

= 1.3120Γδ

where δ =1h

√ν

γ0is the ratio between the boundary layer thickness and the

compliant layer thickness, as before.Since there is no closed form solution for (43) for the general case of ε 6= 0, ,

the solution procedure is more complex. Here, we use a combination of finitedifference and path tracking method, similar to that employed in Phan-Thien(1988b) to cover the whole solution space.

First, the interval ξ ∈ [0, 1] is uniformly divided into N sub-intervals, andthe solutions gi = g (ξi) are sought at the nodal points ξi = i/N, i = 1, . . . , N.Central second-order accurate finite difference approximations are used for thederivatives,

g′i = g′ (ξi) =gi+1 − gi−1

2∆, ∆ =

1N

g′′i = g′′ (ξi) =gi+1 − 2gi + gi−1

∆2.

Boundary conditions are satisfied by setting g0 = 1 and inventing the node gN+1

to satisfy the boundary condition at ξ = 1. The numerical equations to be solvedare

Gi (gj ; Γ, δ, ε) = g2i g′2i + g3

i g′′i + ε(gig

′′i − g′2i

)− Γ = 0, (44)

GN+1 (gj ; Γ, δ, ε) = g3Ng′N + εgNg′N + 1.3120Γδ = 0.

This set of N + 1 non-linear equations are solved by the path tracking method,as implemented by Deuflhard et al (1987).


Flow parameter Γ

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35

Norm

alis

ed

shear

str

ess

at

the

wall

0.01

0.1

1

δ = 0.001

δ = 0.01

δ = 0.1

δ = 1

ε = 0.1

Figure 5. The normalised shear stress at the wall plotted against the flow parameter.

δ ε = 0 ε = 0.1 ε = 1

0.001 0.292 0.335 0.7200.01 0.285 0.326 0.7020.1 0.230 0.262 0.565

1 0.0733 0.0836 0.179


A typical set of results for ε = 0.1 is shown in Figure 5, where the shear stressat the wall, Srzh/(αr) = −(1 + ε)g′ (0) is plotted against the flow parameter Γ.It is clear that the deformation loses its stability at the limit point. The upperbranch of the solution is un-physical, corresponding to a finite wall shear stress asloading tends to zero. Table 2 tabulates the limiting values of the flow parameterat criticality.

A plot of two possible solution at ε = 1, δ = 1 and Γ = 0.1 is given inFigure 6. With the un-physical solution the layer has been significantly compressed( H/h = G(1) is much larger for the un-physical solution), due to a much softeningof the compliant layer.


ξ

0.0 0.2 0.4 0.6 0.8 1.0

Ga

nd

g

0

1

2

3

4

stable

unstable

G

g

Figure 6. Graph of two possible solution at Γ = 0.1, ε = 1 and δ = 1.

x, z

rθ

ω

Figure 7. Flow geometry.

3. Rotating-disk flow

3.1. General equations

Figure 7 illustrates the flow geometry, where the axi-symmetric flow about a ro-tating disk (Schlichting 1979) coated by a compliant layer, to be being considerednext. Two cylindrical coordinate systems are used, r, θ, x for the rotating flow


and r, θ, z for the deformation of the compliant layer. The plane x = 0 repre-sents the interface between the fluid and the compliant layer, and the plane z = 0is the rigid plane surface to which the layer is rigidly attached. The boundaryconditions for the fluid problem are

x = 0 : u = 0, v = rω, w = 0,

x →∞ : u = 0, v = 0. (45)

where ω is the angular velocity of the disk, and u = u, v, w is the velocityvector.

The flow problem is solved by von Kßrmßn kinematics (Kßrmßn 1921)

u = rωF (ζ) , v = rωG (ζ) , w =√

νωH (ζ) , (46)

ζ = x

√ω

ν, (47)

where F, G and H are functions to be determined, ν is the kinematic viscosityand ζ denotes the normalised vertical coordinate. The pressure is given by

P = ρνωQ (ζ) , (48)

In compressibility and the momentum equations lead to

2F + H ′ = 0,F′′ − (

F2 −G2 + HF′)

= 0,G′′ − (2FG + HG′) = 0,

H ′′ −HH ′ = Q′. (49)

This set of equations is subjected to the boundary values

F (0) = 0, G (0) = 1, H (0) = 0, F (∞) = 0, G (∞) = 0. (50)

This system of equations has been solved by Sparrow and Gregg (1960), whoshowed that

F′(0) = 0.510, G′(0) = −0.6159, Q(0) = 0.

The stresses at the interface are given by

σrx = ηω

√ω

νrF′ ≈ 0.510ηω

√ω

νr,

σθx = ηω

√ω

νrG′ ≈ −0.6159ηω

√ω

νr, (51)

σxx = −P + 2ηωH ′ = 0,

which are at most quadratic in r (the quadratic terms are all zero) and need tobe matched by the deformation of the compliant wall.

The compliant layer is again modelled as a Mooney material (Mooney 1940,Rivlin 1948), eq. (7). In the present flow, and under the (non-uniform) flow loading,


the layer is compressed from the initial thickness ∆ to the final thickness δ. Thegeneral kinematics of the torsional deformation have been considered before, seePhan-Thien (1988b):

R = rf (ξ) , Θ = θ − g (ξ) , Z = δh (ξ) , (52)f (0) = 1, g (0) = 0, h(0) = 0,

where ξ = z/δ ∈ [0, 1] is the normalised coordinate, and R,Θ,Z representsthe undeformed configuration, r, θ, z the deformed configuration. The inversedisplacement gradient is

[F−1

]=

f 0r

δf ′

0 f −r

δfg′

0 0 h′

. (53)

Incompressibility requires that det [F] = 1, which leads to f2h′ = 1. The Fingerstrain tensor and its inverse are given by, respectively,

[B] =

1f2

(r2

δ2(f ′)2f4 + 1

)−r2

δ2f3g′f ′ −r

δf3f ′

−r2

δ2f3g′f ′

1f2

(r2

δ2f6(g′)2 + 1

)r

δf4g′

−r

δf3f ′

r

δf4g′ f4

, (54)

[B−1

]=

f2 0r

δff ′

0 f2 −r

δf2g′

r

δff ′ −r

δf2g′

r2

δ2(f ′)2 +

r2

δ2f2(g′)2 + f−4

(55)

The stress components are quadratic in r and can be matched to the fluid-inducedstresses:

Srr = α1f2

(r2

δ2(f ′)2f4 + 1

)− βf2,

Sθθ = α1f2

(r2

δ2f6(g′)2 + 1

)− βf2,

Szz = αf4 − β

(r2

δ2

[(f ′)2 + f2(g′)2

]+ f−4

), (56)

Srz = −r

δ

(αf2 + β

)ff ′,

Srθ = −r2

δ2αf3g′f ′,

Sθz =r

δ

(αf2 + β

)f2g′


The equations of balance are

∂P∂r

=r

δ2

3α(f ′)2f2 − αf4(g′)2 − (

αf3f ′ + βff ′)′

(57)

0 = −4αf3g′f ′ +[(

αf4g′ + βf2g′)]′

(58)

∂P∂z

=1δ

−2

(αf3f ′ + βff ′

)+

[αf4 − β

(r2

δ2

[(f ′)2 + f2(g′)2

]+ f−4

)]′.

(59)

Equation (58) is uncoupled from the rest(f2 + ε

)fg′′ + 2εf ′g′ = 0, (60)

which has the solution

g′ = C2

(1 + εf−2

)= C2 (1 + εh′) ,

g = C2 (ξ + εh) ,

where C2 is an integration constant.The pressure field

P =r2

δ2Q2 + Q1 + Q0, (61)

where

2Q2 = 3α(f ′)2f2 − αf4(g′)2 − (αf3f ′ + βff ′

)′Q′

2 = −β((f ′)2 + f2(g′)2

)′(62)

Q′1 = −2

(αf3f ′ + βff ′

)+

[αf4 − βf−4

]′These compatibility conditions (62) can be integrated to yield

Q1 = −(

12αf4 + βf2

)+

[αf4 − βf−4

](63)

and (αf2 + β

)ff ′′ +

(αf2 − 2β

)f2g′2 − β (f ′)2 = C3 (64)

Matching the stress σzz at the interface ξ = 1

−(

r2

δ2Q2 + Q1 + Q0

)+ αf4 − β

(r2

δ2

[(f ′)2 + f2(g′)2

]+ f−4

)∣∣∣∣ξ=1

= 0

Terms of O(r2

)need only to be matched, terms of O(1) will take care of them-

selves through the arbitrary setting of Q0 :(αf2 + β

)ff ′′ +

(αf2 − 2β

)f2g′2 − β (f ′)2

∣∣∣ξ=1

= 0. (65)


Comparing (64) to (65) we find C3 = 0.

Matching σrz = 0.510ηω

√ω

νr :

(f2 + ε

)ff ′

∣∣ξ=1

+ 0.510ηω

α

√ω

νδ = 0 (66)

Matching σθz = −0.6159ηω

√ω

νr :

(f2 + ε

)f2g′

∣∣ξ=1

+ 0.6159ηω

α

√ω

νδ = 0 (67)

To sum up, the governing equations are(f2 + ε

)ff ′′ +

(f2 − 2ε

)f2g′2 − ε (f ′)2 = 0, (68)(

f2 + ε)fg′′ + 2εf ′g′ = 0,(

f2 + ε)ff ′

∣∣ξ=1

+ 0.510Γ = 0,(f2 + ε

)f2g′

∣∣ξ=1

+ 0.6159Γ = 0,

f (0) = 1, g(0) = 0,

where Γ =ηω

α

√ω

νδ is the flow loading parameter. This parameter is a ratio of

the viscous stress ( ηω ) to a shear modulus ( α ) multiplied by the ratio of thelayer thickness ( δ ) to a boundary layer thickness (

√ν/ω ).

3.2. Linearised equations

In the limit of small loading Γ ¿ 1 , the equations can be linearised. In effect, weseek a solution of the form

g = g1 + HOT, f = 1 + f1 + HOT

and neglect all higher-order terms ( HOT ). The linearised equation for g is

g′1 = C2 (1 + ε) , g′1(1) (1 + ε) = −0.6159Γ

giving

g1 = −0.6159Γ

1 + εξ. (69)

The linearised equation for f is

f ′′1 = 0, f ′1 (1) = −0.510Γ

1 + ε

giving

f1 = −0.510Γ

1 + εξ.


The top angular displacement of the layer is

Ω = θ −Θ|ξ=1 = −0.6159Γ

1 + ε,

i.e., equals to the flow loading divided by the shear modulus of the materials.This result may be arrived at simply by equating the flow induced shear stress,

σθz = −0.6159ηω

√ω

νr, to the elastic stress, σθz = (α + β) Ωr/δ. The vertical

displacement can be obtained from the incompressibility condition:

h′ = (1 + f1 + HOT)−2 ≈ 1− 2f1 = 1 + 0.5102Γ

1 + εξ,

which yields

h = ξ + 0.510Γ

1 + εξ2,

and consequently the initial thickness of the layer is (c.f., (52)),

∆δ

= 1 + 0.510Γ

1 + ε. (70)

The effect of the flow loading is to compress the layer, and at the same time inducesa negative angular displacement on the top surface which is proportional to theamount of compression:

Ω = −0.61590.510

(∆δ− 1

). (71)

As the full non-linear torsional deformation can be unstable at large angular dis-placement, we expect a loss of stability here, as the loading parameter increases.

3.3. Numerical solution

Since there is no closed form solution for (68), we use a combination of finitedifference and path tracking method, similar to that employed in Phan-Thien(1988b) to cover the whole solution space.

First, the interval ξ ∈ [0, 1] is uniformly divided into N sub-intervals, andthe solutions fi = f (ξi) , gi = g (ξi) are sought at the nodal points ξi = i/N,i = 1, . . . , N. Central second-order accurate finite difference approximations areused for the derivatives, e.g.,

f ′i = f ′ (ξi) =fi+1 − fi−1

2∆, ∆ =

1N

f ′′i = f ′′ (ξi) =fi+1 − 2fi + fi−1

∆2.

Boundary conditions are satisfied by setting f0 = 1, g0 = 0 and inventing thenodes fN+1 and gN+1 to satisfy the boundary conditions at ξ = 1. The numer-


Γ

0.01 0.1 1 10

Ω

-2

-1

0

ε = 0.0ε = 0.1ε = 0.13ε = 0.2ε = 0.3ε = 0.5ε = 1.0

Figure 8. The angular displacement of the top of the compliant layer, plotted against the flowloading parameter.

ical equations to be solved are

Gi (fj , gj ; Γ, ε) =(f2

i + ε)fif

′′i +

(f2

i − 2ε)f2

i g′i2 − εf ′2i = 0, (72)

GN+1 (fj , gj ; Γ, ε) =(f2

N + ε)fNf ′N + 0.510Γ = 0,

GN+1+i (fj , gj ; Γ, ε) =(f2

i + ε)fig

′′i + 2εf ′ig

′i = 0, (73)

G2N+2 (fj , gj ; Γ, ε) =(f2

N + ε)f2

Ng′N + 0.6159Γ = 0,

This set of 2N +2 non-linear equations is solved by the path tracking method, asimplemented by Deuflhard et al (1987). In all solutions, 50 intervals were usedalthough N = 20 were adequate.

Some typical sets of results for different values of ε are shown in Figure 8, wherethe angular displacement of the top of the compliant layer, Ω = θ−Θ|ξ=1 = g(1),is plotted against the flow parameter Γ. It is clear that the deformation losesits stability at the limit point. The lower branch of the solution is un-physical,corresponding to a finite angular displacement as the flow loading tends to zero.Table 3 tabulates the limiting values of the flow parameter at criticality.

A plot of two possible solutions at ε = 0.2 and Γ = 0.1 is given in Figure9. With the un-physical solution (broken lines) the layer has been significantlycompressed ( h(1) =

∫ 1

0dξ/f (ξ)2 ≈ 3.07 ), compared to the same value for the


ξ

0.0 0.2 0.4 0.6 0.8 1.0

Ra

dia

ld

isp

lace

me

nt

f

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

An

gu

lar

dis

pla

ce

me

nt

g

-0.8

-0.6

-0.4

-0.2

0.2

-1.0

0.0

non-physical

physical

Figure 9. Two possible solutions at Γ = 0.1 and ε = 0.2. The physical solutions arerepresented by the solid lines.

ε Γcrit

0 no limit0.10 no limit0.13 0.3060.20 0.2740.30 0.2950.50 0.3381.0 0.458


physical solution ( h(1) =∫ 1

0dξ/f (ξ)2 ≈ 1.051 ) due to a considerable softening of

the compliant layer.

4. A physical explanation

The physical explanation of this softening is much simpler if we consider a homo-geneous deformation

X1 = λ−1x1 − γx3,X2 = x2,X3 = λx3,


mapping the deformed configuration (−∞ < x1 < ∞, −∞ ≤ x2 ≤ ∞, 0 ≤x3 ≤ h ) back to the undeformed configuration (−∞ < X1 < ∞, −∞ ≤ X2 ≤∞, 0 ≤ X3 ≤ H ). This mapping represents a compression along the x3− axis forλ = H/h > 1 and a shearing in the 13− plane (with a shear strain of γ ). Thestress components in this homogeneous deformation are given by

S11 = α(λ2 + γ2

)− βλ−2,

S22 = α− β,

S33 = αλ−2 − β(γ2 + λ2

), (74)

S13 = (α + β) λ−1γ.

If there is no net force acting in the 1− direction,

σ11 = −P0 + α(λ2 + γ2

)− βλ−2 = 0

giving a pressure ofP0 = α

(λ2 + γ2

)− βλ−2. (75)

The force (per unit width) required to maintain no deformation in the 2− directionis

F2 = − [α

(λ2 + γ2

)− βλ−2 − α + β]h,

= − [α

(λ2 + γ2 − 1

)+ β

(1− λ−2

)]h. (76)

The squeezing force (per unit area) is

F3

A= − [

α(λ2 + γ2

)− βλ−2 − αλ−2 + β(γ2 + λ2

)],

= − (α + β)(λ2 + γ2 − λ−2

). (77)

The shearing force (per unit area) is

Fs

A= (α + β) λ−1γ. (78)

Thus, a shear deformation always enhances the squeezing force, whereas acompression always reduces the shearing force, i.e., reducing the “effective” shearmodulus, which can be defined as

Ge =Fs

γA= (α + β) λ−1. (79)

Denoting the elongational strain by e = (H − h) /H, one finds the work doneby the vertical loading is

12

(F3

A

)e =

12

(α + β) γ2e +12

(α + β)(λ2 − λ−2

)e,

whereas the work done by the shear loading is

12

(Fs

A

)γ =

12

(α + β) λ−1γ2 =12

(α + β) γ2 − 12

(α + β) γ2e.


The terms underlined, representing a reduction in the work done by shear loading,is precisely the contribution to the work done by the vertical loading. Thus the re-duction in the effective shear modulus may be attributed to the work contributionby the vertical loading to the shearing process.

We expect the same mechanism comes into play here in the flow loading case,which ultimately gives rise to the loss of stability through the limit point. This isin agreement with the finding of Yeo et al (1994) for compressible compliant wallthat flow loading reduces the effective modulus leading to changes in flow stability.

References

[1] Cooper, A.J. and Carpenter, P.W., J. Fluid Mech., 350 (1997), 231.[2] Deuflhard, P., Fiedler, B. and Kunkel, P., SIAM J. Numer. Anal., 24 (1987), 912.[3] Hiemenz, K., Die Grenzschicht an einem in den gleichformigen Flussigkeitsstrom einge-

tauchten geraden Kreiszylinder. Thesis Gottingen (1911).[4] Howarth, L., On the calculation of the steady flow in the boundary layer near the surface of

a cylinder in a stream. ARC RM 1632 (1935).[5] Keller, H., In Applications of Bifurcation Theory, Ed. P. Rabinowitz, Academic Press, New

York (1977).[6] Klingbeil, W.W. and Shield, R.T., ZAMP, 17 (1966), 281.[7] Mooney, M., J. Appl. Phys., 11 (1940), 582.[8] Phan-Thien, N., Rheol. Acta, 27 (1988a), 230.[9] Phan-Thien, N., ZAMM, 68 (1988b), 8.

[10] Rivlin, R.S., Philos. Trans. Roy. Soc. A, 241 (1948), 379.[11] Schlichting, H., Boundary-Layer Theory, 7th Edition, translated by J. Kestin, McGraw-Hill,

New York (1979).[12] Sparrow, E.M. and Gregg, J.L, ASME Transaction, J. Heat Transfer 82 (1960), 294.[13] Treloar, L.R.G., Proc. Roy. Soc. Lond. A, 351 (1976), 301.[14] von Kßrmßn, T., ZAMM, 1 (1921), 232.[15] Yeo, K.S., Khoo, B.C. and Chong, W.K., J. Fluids & Struct., 8 (1994), 529.

N. Phan-ThienDepartment of Mechanical EngineeringNational University of SingaporeSingapore 119260e-mail: [email protected]

K. S. YeoInstitute of High Performance ComputingNational University of SingaporeSingapore 119260

(Received: February 25, 1999; November 18, 1999)

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Stagnation and rotating-disk flows over a compliant surface

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