Effect of Friction on Unbonded Elastomeric Bearings

Effect of Friction on Unbonded Elastomeric Bearings James M. Kelly, M.ASCE1; and Dimitrios Konstantinidis2 Abstract: This paper describes a theoretical analysis of a type of thermal expansion bridge bearing which could be used as a lightweight low-cost elastomeric seismic isolator for application to housing, schools, and other public buildings in earthquake-prone areas of the developing world. The analysis covers the effect of the frictional resistance of the supports on the vertical stiffness of this type of isolator. The most important aspect of these bearings is that they do not have end plates, which reduces their weight, but also means that they are not bonded to the upper and lower support surfaces and are held in place only by friction. This at first sight might seem to be a deficiency of this design, but it has the advantage that it eliminates the presence of tensile stresses in the bearings. It is these tensile stresses and the bonding requirements that arise from them that lead to the high costs of conventional isolation bearings. A theoretical analysis of the response of these bearings to vertical load shows that slip between the unbonded surfaces and rigid supports above and below can have a significant influence on the vertical stiffness and the internal pressure distribution. DOI: 10.1061/共ASCE兲EM.1943-7889.0000019 CE Database subject headings: Load bearing capacity; Bridges; Friction; Seismic effects; Stiffness; Earthquakes.

Introduction Many building and bridge structures are provided with seismic isolation bearings specifically designed for that purpose. These bearings are considerably more expensive than unbonded bridge bearings, which are designed to accommodate smaller lateral displacements, such as those due to thermal expansions. Seismic isolation bearings are much more difficult to manufacture since they are made to more stringent requirements, and they are considerably heavier. A recent experimental study by the writers 共Konstantinidis et al. 2008兲, which examined the behavior of unbonded thermal expansion bridge bearings when subjected to horizontal displacements much larger than what they are designed for but which they are expected to experience during an earthquake, showed that these bearings can accommodate shear strains up to 200% 共or more, depending on their height兲 without damage. The primary reason for this is the fact that the top and bottom surfaces can roll off the supports and no tension stresses are produced. The findings of this study motivated an investigation into the possibility of using unbonded thermal expansion bridge bearings as low-cost seismic isolators for application to housing, schools, and other public buildings in earthquake-prone areas of the developing world 共Kelly and Konstantinidis 2007兲. Motivated by the possibility of this application, the present paper looks into the compressive behavior of these types of bearings. The primary weight in a seismic isolator is due to the steel 1

Professor Emeritus, Pacific Earthquake Engineering Research Center, Univ. of California at Berkeley, 1301 S. 46th St., Richmond, CA 948044698. E-mail: [email protected] 2 Postdoctoral Scholar, Pacific Earthquake Engineering Research Center, Univ. of California at Berkeley, 1301 S. 46th St., Richmond, CA 94804-4698 共corresponding author兲. E-mail: [email protected] Note. This manuscript was submitted on August 4, 2008; approved on January 7, 2009; published online on March 6, 2009. Discussion period open until February 1, 2010; separate discussions must be submitted for individual papers. This paper is part of the Journal of Engineering Mechanics, Vol. 135, No. 9, September 1, 2009. ©ASCE, ISSN 0733-9399/ 2009/9-953–960/$25.00.

reinforcing plates, which are used to provide the vertical stiffness of the rubber-steel composite element. A typical rubber isolator has two large end plates and 20 or more thin reinforcing plates. The high cost of producing the isolators results from the labor involved in preparing the steel plates and laying-up of the rubber sheets and steel plates for vulcanization bonding in a mold. The steel plates are cut, sand blasted, acid cleaned, and then coated with bonding compound. Next, the compounded rubber sheets with the interleaved steel plates are put into a mold and heated under pressure for several hours to complete the manufacturing process. The manufacturing process for conventional isolators has to be done very carefully to satisfy the testing requirements in current codes. The bond between the rubber and the steel reinforcement and between the rubber and the end plates must be very good to survive the large tensile stresses that develop during these tests. The weight and cost of an unbonded bridge bearing are considerably lower than those of a seismic isolator because a bridge bearing uses thinner steel reinforcing plates, no end plates and no bonding to the top and bottom support surfaces. Of course, the performance objectives that unbonded bridge bearings are expected to satisfy are lower than those of seismic isolators, yet they offer a very appealing low-cost alternative in developing countries where conventional seismic isolators are not affordable. Friction in rubber is relatively high, but there is always the possibility that some level of lubrication can be introduced either intentionally or by accident. This results in a reduction of frictional resistance that develops in the support-rubber interface, and thus it is important to be able to predict the effect of slip in these bearings. An analysis of the mechanics of the bearing with slip will be given, and this will include the effect on the internal pressure in the bearing and the reduction of the vertical stiffness caused by slip. It will also be used to estimate the maximum amount of slip at the unbonded surfaces. In addition to the unbonded surfaces, another aspect of these bearings that distinguishes them from seismic isolation bearings is that the outer layers are only half the thickness of the inner layers. This means that if the friction is high enough to simulate the fully bonded situation at the top and bottom surfaces, these two outer layers

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Table 1. Geometric Characteristics of the Steel-Laminated Elastomeric Pads Bearing designation

Rubber height Number of Total height Width Depth 共mm兲 steel shims 共mm兲共mm兲共mm兲

S-48 S-120

48 120

4 10

55.6 139.0

375 375

575 575

1. Fig. 1. 120-mm-tall unbonded laminated elastomeric bearing 共plan dimensions: 375 mm by 575 mm兲

will each be four times stiffer than each inner layer. On the other hand if the top and bottom surfaces are fully lubricated to the extent that there is no shear stress on each surface, the two outer layers will act as if they were a single inner layer since the analysis for the fully bonded inner layer has zero shear stress on the center of the layer. Thus, the two outer layers together will have the stiffness of an inner layer. Frictional slip will cause the stiffness of these two outer layers to vary between these two extreme values.

points on a vertical line before deformation lie on a parabola after loading, and 2. horizontal planes remain horizontal. In this paper we are only interested in the theory for a bearing in the form of a long strip when the effects of the ends can be neglected and the strip is taken to be infinite in the y direction, thus the problem is two dimensional, and the deformation is plane strain. The theory for an arbitrarily shaped pad is given in Kelly 共1997兲. We consider a pad of thickness t and width 2b and locate a rectangular Cartesian coordinate system, 共x , y , z兲, in the middle surface of the pad, as shown in Fig. 3共a兲. Fig. 3共b兲 shows the displacements, u , w, in the coordinate directions under assumptions 1 and 2

冉

u共x,z兲 = u0共x兲 1 −

4z2 t2

冊共1兲

w共x,z兲 = w共z兲

Bearing Designs Fig. 1 is a photograph of a typical unbonded elastomeric bearing, and Fig. 2 shows its cross section. The rubber compound is Neoprene 共polychloroprene兲 with a specified hardness of 55 on the Shore A scale. Each intermediate rubber layer is 12 mm thick while the top and bottom rubber layers are 6 mm thick. A 3-mm protective cover surrounds the bearing on the sides. The bearings are laminated with 1.9-mm-thick 共14-ga兲 A1011 steel shims. Table 1 shows detailed geometric properties for two typical-size elastomeric bridge bearings.

This displacement field satisfies the constraint that the top and bottom surfaces of the pad are bonded to rigid substrates. The assumption of incompressibility produces a further constraint on the normal components of strain, ⑀xx, ⑀zz, in the form ⑀xx + ⑀zz =

⳵u ⳵w + =0 ⳵x ⳵z

共2兲

and this leads to

冉

冊

4z2 du0 dw =0 1− 2 + dx t dz When integrated through the thickness, this gives

Compression of Fully Bonded Single Pad A linear elastic theory is the most common method used to predict the compression and the bending stiffness of a thin elastomeric pad. The first analysis of the compression stiffness was done using an energy approach by Rocard 共1937兲. Further developments were made by Gent and Lindley 共1959兲 and Gent and Meinecke 共1970兲. A very detailed description of the theory is given by Kelly 共1997兲 and need not be repeated here. The analysis is an approximate one based on the kinematic assumptions that

Z

dy =1

Y

X

t

3 mm

3 mm

6 mm

2b

(a) Z, w

12 mm 1.9 mm steel plates

6 mm

t (b)

375 mm

Fig. 2. Cross section of 120-mm-tall unbonded laminated elastomeric bearing

X,u

Fig. 3. 共a兲 Coordinate system for an infinite strip pad of width 2b; 共b兲 displacement field

954 / JOURNAL OF ENGINEERING MECHANICS © ASCE / SEPTEMBER 2009


du0 3⌬ 3 = = ⑀c dx 2t 2

where ⌬ = the change of thickness of the pad 共⌬ ⬎ 0 in compression兲 and ⑀c = ⌬ / t = compression strain. The other assumptions of the theory are that the material is incompressible and that the stress state is dominated by the pressure, p, in the sense that the normal stress components can be taken as −p. The vertical shear stress components are included but the in-plane shear stress is assumed to be negligible. The only equation of stress equilibrium in this case is d␴xx d␶xz + =0 dx dz

共4兲

共5兲

The assumption of linear elastic behavior means that

冊

⳵u ⳵w 8G + = − 2 zu0 ⳵z ⳵x t

12G⑀c d2 p =− dx2 t2

共7兲

The boundary condition, p = 0, at the edges of the pad completes the system for the pressure distribution, p共x兲, in the pad. The effective compression modulus, Ec, of the pad is obtained by integrating p共x兲 over the area of the pad to determine the resultant load P. Ec is then given by P A⑀c

共8兲

The value of Ec for a single rubber layer is controlled by the shape factor, S, defined as S=

loaded area free area

which is a dimensionless measure of the aspect ratio of a single layer of the elastomer. For an infinite strip of width 2b, with a single layer thickness of t, we have S = 共2b · 1兲 / 共2t · 1兲 = b / t, as can be seen from Fig. 3共a兲.

Effect of Frictional Resistance on Vertical Stiffness of Fully Bonded Single Pad Before developing the solution for stiffness of the upper and lower layers of the bearing which are not bonded to steel endplates but held in place only by friction, it is useful to review the analysis of a single pad under pressure and restrained only by friction between the elastomer and the upper and lower rigid surfaces. This is a relatively simple analysis and is an extension of the pressure solution for a fully bonded pad. It will be used to motivate the much more complicated analysis for the pad which is bonded on one side and frictionally constrained on the other. For an infinite strip of width 2b, as shown in Fig. 3, the pressure is governed by Eq. 共7兲, which with p = 0 at x = ⫾ b gives

steel shim l

p(x +dx)

p(x) x

p(x)

x +dx

x

p(x +dx)

x +dx

s

(b)

(a)

Fig. 4. Definitions of surface shears

6G⑀c 2 2 共b − x 兲 t2

共9兲

where G = shear modulus of the elastomer. In this case, the load per unit length of the strip, P, is given by

冕

b

p共x兲dx =

−b

8G⑀cb3 t2

Since the shape factor is S = b / t and the area per unit length is A = 2b Ec =

共6兲

which together with Eqs. 共3兲 and 共5兲 leads to the pressure solution

Ec =

t

P=

d␶xz dp = dz dx

冉

t/2

p共x兲 =

which with the assumption that ␴xx = ␴zz = −p provides the sole equation of equilibrium as

␶xz = G␥xz = G

s

s

共3兲

P = 4GS2 A⑀c

共10兲

The only equation of stress equilibrium in this case remains Eq. 共5兲. Integration of this through the thickness of the pad with the definition of the surface shear stresses as shown in Fig. 4共a兲 ␶xz兩z=t/2 = − ␶s ;

␶xz兩z=−t/2 = ␶s

共11兲

leads, using Eq. 共9兲, to ␶s = −

t dp 6G⑀cx = 2 dx t

共12兲

It is clear that the shear stress increases toward the edges, whereas the pressure decreases, and if the constraint is controlled only by friction, that is by ␶s ⱕ ␮p where ␮ = friction coefficient, at some point slip must happen. In fact, if the pressure was given by Eq. 共9兲, the equality of ␶s and ␮p would be at x = b

冑

1+

1 1 − 4␮2S2 2␮S

共13兲

For example, if ␮ = 1, then slip occurs for x = b − t / 2, showing that for the high level of friction typical of rubber against steel, very little slip would take place. On the other hand, if the surface is fully lubricated, such that ␮ = 0, there will be no shear stress at any point in the pad, and the pressure is given by E0⑀c, where E0 = 3G is the Young’s modulus of the material. In most cases of thin bonded pads, this is completely negligible since the Ec of a single bonded layer can be two orders of magnitude larger than this. In the case of a pad with partial slip, E0 is required as a nonzero boundary condition at the edge of the pad for the homogeneous equation for the pressure.

Compression of Single Pad with Slip on Both Surfaces To accommodate the slip at the surface due to the shear stresses overcoming the frictional resistance between the rubber and the



Z

p共x兲 = − constrained

slipping u1

u0

t

u 1(b) u 0(b)

u0

It is convenient to absorb the unknown constant into the other constants and write the pressure as

X

x1 b

p共x兲 = 6G⑀c

Fig. 5. Coordinate system and displacement fields for a single infinitely long strip pad with slip on both surfaces

steel or concrete, we need to modify the original kinematic assumptions. In this approach, as before, the rubber is assumed to be incompressible and the pressure is assumed to be the dominant stress component. The kinematic assumption of quadratically variable displacement is supplemented by an additional displacement that is constant through the thickness and is intended to accommodate the squeezing out of the rubber where slip occurs, as shown in Fig. 5. Thus, in this case the displacement pattern that leads to the pressure solution of Eq. 共9兲 is replaced by

冉

冊

4z2 u共x,z兲 = u0共x兲 1 − 2 + u1共x兲 t w共x,z兲 = w共z兲

␶xz = G␥xz

共16兲

which with ␥xz =

8z ⳵u ⳵w + = − 2 u0 ⳵z ⳵x t

共17兲

8z dp = − 2 u0 dx t

p共x兲 = 3G⑀ce共2␮/t兲共b−x兲

Symmetry implies that A = 0, giving

共22兲

and at x = x1 we have continuity in p共x兲 and dp / dx 共since ␶s must be continuous across x = x1兲. From Eqs. 共20兲 and 共22兲, we have 12G⑀c dp =− x, dx t2

0 ⱕ x ⱕ x1

2␮ dp = − 3G⑀c e共2␮/t兲共b−x兲, t dx

x1 ⱕ x ⱕ b

which in turn gives us the equation for x1 2␮x1 共2␮/t兲x 1 = ␮ 2e 共2␮/t兲b e t

共23兲

The procedure is now to solve for 2␮x1 / t knowing ␮ and b and then to evaluate p共x1兲 from which B can be determined. From this, we calculate ␶s and then u0 from ␶s = 4Gu0 / t. Finally, the extent of the slip given by u1共x兲 for x1 ⱕ x ⱕ b is calculated from the equation of incompressibility, Eq. 共15兲. We denote ␥ = 2␮x1 / t and ␭ = ␮2e2␮S and solve y = ␭e−y. Since ␭ ⬎ 0, and the right hand side always decreases from ␭ at y = 0, and the left side increases from zero, there is always a solution. From the root y, we readily obtain x1. The two expressions for the pressure at x = x1

冉冊

共18兲

x21 b2

p共x+1 兲 = 3G⑀ce2␮S关1−共x1/b兲兴 are equal because of continuity, giving B=

共19兲

In the region where no slip has occurred, i.e., 0 ⱕ x ⱕ x1, the slip displacement is zero, and Eq. 共19兲 can be inverted to give u0 in terms of pressure and inserted into the incompressibility constraint, Eq. 共15兲, to give 6G⑀c p共x兲 = − 2 x2 + Ax + B t

共21兲

p共x−1 兲 = 6GS2⑀c B −

which with the assumption that ␴xx = ␴zz = −p provides the sole equation of equilibrium as

共20兲

which has the solution p共x兲 = Ce−2␮bx/t, where C is a constant of integration. To determine it, we use as boundary condition at x = b the fact that p共b兲 = 3G⑀c. Thus, for x1 ⱕ x ⱕ b, the pressure is given by

from Eq. 共14兲, gives

⳵ ␴xx 8G = 2 u0 ⳵x t

冉冊

dp 2␮ + p=0 dx t

共15兲

The only equation of stress equilibrium in this case remains Eq. 共4兲, and the assumption of elastic behavior means that

冉冊

b2 x2 x2 2 2 B − 2 = 6GS ⑀c B − 2 t b b

For x1 ⱕ x ⱕ b, we need to determine u1共x兲 subject to the simultaneous requirements that ␶s = ␮p and ␶s = −共t / 2兲dp / dx. This leads to the equation

共14兲

where u1共x兲 = 0 for 0 ⱕ x ⱕ x1, with x1 being the location where slip starts. The constraint of incompressibility consistent with the displacement pattern of Eq. 共14兲 leads to du0 3 du1 3 ⌬ 3 + = = ⑀c dx 2 dx 2 t 2

6G⑀c 2 x +B t2

1 2␮S关1−共x /b兲兴 x21 1 e + 2 2S2 b

共24兲

The pressure distribution is then given by

冉冊

x2 ; b2

0 ⱕ x ⱕ x1

p共x兲 = 3G⑀ce2␮S关1−共x/b兲兴 ;

x1 ⱕ x ⱕ b

p共x兲 = 6GS2⑀c B −

共25兲

Let us take, for example, the case where b / t = S = 10 and ␮ = 0.3. We get ␭ = 36.3 and the result for y is 2.63, from which we have x1 / b = 0.438. Eq. 共24兲 then gives B = 0.3375, and the pressure dis-



Table 2. Compression Modulus of Single Pad for Different Friction Coefficients 共S = 10兲

1

p 6G S 2 c

bonded

µ =1

0.8

0.6

µ = 0.5

evolution of x1 /b 0.4

0.2 µ = 0.1

0

0

0.2

0.4

0.6

0.8

1

x/b

Fig. 6. Pressure distributions for a single pad 共S = 10兲 with slip on both surfaces and ␮ ranging from 0.1 to 1 and for a fully bonded single pad

tribution from Eq. 共25兲 is plotted in Fig. 6 together with pressure distributions for various values of ␮ between 0.1 and 1 and the pressure distribution for a fully bonded pad given by Eq. 共9兲. We note a considerable reduction in peak pressure with decreasing ␮. The shear stresses that result from this pressure distribution are x t dp = 6G⑀c ; t 2 dx

0 ⱕ x ⱕ x1

␶s = 3␮G⑀ce2␮S关1−共x/b兲兴 ;

x1 ⱕ x ⱕ b

␶s = − and

In the slipped region x1 ⱕ x ⱕ b, where u1 ⫽ 0, use of ␶s = 4Gu0 / t gives 3 u0 = ␮tG⑀ce2␮S关1−共x/b兲兴 ; 4

x1 ⱕ x ⱕ b

共26兲

We now calculate u1 from the equation of incompressibility, Eq. 共15兲, which when integrated from x1 to x ⱖ x1 with u0共x1兲 = 3x1⑀c / 2 and u1共x1兲 = 0 gives

冉

u1共x兲 = x −

x1

0

which leads to Ec =

冉

0.035 0.223 0.438 0.579 0.671 0.734 0.778 0.812 0.838 0.858

x21 2

x1 P = 6GS2 B− 2b⑀c b 3b

冊

+

b

x1

3Ge共2␮/t兲共b−x兲dx

0.036 0.162 0.338 0.480 0.585 0.660 0.717 0.761 0.795 0.822

0.024 0.089 0.215 0.346 0.457 0.544 0.613 0.668 0.712 0.748

Similar results for values of ␮ between 0.1 and 1.0 are given in Table 2 for S = 10, and plots of x1 / b and Ec / 4GS2 for these values and for S = 15 and S = 20 are shown in Fig. 7.

Effect of Frictional Resistance on Vertical Stiffness of Bearing with Slip on Top and Bottom Surfaces Friction in rubber is relatively high, but there is always the possibility that some level of lubrication can be introduced either intentionally or by accident. Thus, it is important to be able to predict the effect of slip in these bearings. An analysis of the mechanics of the bearing with slip will be given, and this will include the effect on the internal pressure in the bearing and the reduction of the vertical stiffness caused by slip. It will also be used to estimate the maximum amount of slip at the unbonded surfaces. In addition to the unbonded surfaces, another aspect of these bearings that distinguishes them from other isolation bearings is that the outer layers are only one-half the thickness of the inner

0.8

x 1 /b

冉冊

x2 dx + b2

B

Ec 4GS2

1

b ␮t 2− ⑀c 2 b

冉冊冕

6GS2 B −

0.069 0.894 2.627 4.634 6.710 8.803 10.898 12.990 15.076 17.158

共27兲

0.6 0.4 S=10 S=15 S=20

0.2

册

(a)

0 1 0.8

Ec 4G S 2

冋冕

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

冊

To calculate the value of Ec that is developed when the pad slips, we must integrate the pressure over the range −b ⱕ x ⱕ b to determine P and divide by 2b⑀c. We have P = 2⑀c

y

␮t 2␮S关1−共x/b兲兴 e ⑀c 2

Thus, the maximum amount of slip is u1共b兲 =

␮

x1 b

3G 2␮S关1−共x /b兲兴 1 关e − 1兴 2␮S

0.6 0.4 0.2

共28兲 In the case of ␮ = 0.3 and S = 10 with B = 0.3375 and x1 / b = 0.438, we have Ec / 4GS2 = 0.2149, which implies a very substantial reduction in the modulus for this value of the friction coefficient.

(b)

0

0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

µ

1

Fig. 7. 共a兲 Location where slip initiates for a single pad; 共b兲 compression modulus as a fraction of the compression modulus of a fully bonded pad



Z

␥xz = slipping

constrained

u1

u0 t

x1 b

Fig. 8. Coordinate system and displacement fields for bearing with two slipping outer layers

layers. This means that if the friction is high enough to simulate the fully bonded situation at the top and bottom surfaces these two outer layers will each be four times stiffer than each inner layer. On the other hand, if the top and bottom surfaces are fully lubricated to the extent that there is no shear stress on either surface, the two outer layers will act as if they were a single inner layer since the analysis for the fully bonded inner layer has zero shear stress on the center of the layer. Thus, the two layers will have together the stiffness of an inner layer. Frictional slip will cause the stiffness of these two outer layers to vary between these two extreme values. To determine the effect of slip on the vertical stiffness of the two outer layers of the bearing, the inner layers are replaced by a central line since they will play no role in their deformation other than providing bonding of each layer at the inner surface. The deformation is divided into two parts, viz., the parabolic displacement field assumed for the fully bonded bearing and an additional linear displacement pattern that varies from zero at the inner surface to u1共x兲 at the outer edges to characterize the slip. Symmetry in both horizontal and vertical directions is assumed and deformation in the y direction is neglected. This displacement field is shown in Fig. 8 and takes the form

冉冊

2z 2z 2z 1− + u1共x兲 t t t

w共x,z兲 = w共z兲

共29兲

The normal strains are given by

␶s =

␶l =

2G 共u0 + u1兲 t

⳵ ␶xz dp = ⳵z dx

共32兲

共33兲

and integration of this through the upper half layer 0 ⱕ z ⱕ t / 2 gives

冕

t/2

0

⳵ ␶xz t dp dz = = − 共␶s + ␶l兲 ⳵z 2 dx

which is also shown in the diagram. From this equation and the fact that for both the nonslip and slipped regions, ␶s + ␶l = 4Gu0 / t, we have a relationship between the pressure and the displacement field valid everywhere in the form 8G dp = − 2 u0 dx t

共34兲

We assume that slip occurs at x = x1 and assume symmetry and continuity at x = 0 and x = x1. In the nonslip region 0 ⱕ x ⱕ x1, we have u1 = 0, and with the above equation we have u0 = −共t2 / 8G兲dp / dx. Substitution into the equation of incompressibility, which now takes the form du0 / dx = 6⑀c, gives

冉冊

48G⑀c d2 p 2 =− dx t2

共35兲

from which

⳵ w dw = ⑀zz = ⳵ z dz

p共x兲 = −

The material is assumed to be incompressible so that

冉冊

2z du0 2z du1 2z dw =0 + 1− + dx t t dx t dz

Integration of this through the half thickness 0 ⱕ z ⱕ t / 2 gives

24G⑀c 2 x + Ax + B t2

Symmetry dictates that A = 0, and, if there were no slip, we could use a boundary condition at x = b to determine B. In this case we cannot use such a condition and must leave B as an unknown. It is convenient to absorb it into the other constants and write

冉冊

冉冊

du0 t t ⌬ du1 t + w共0兲 = + =−w dx 12 dx 4 2 2 du0 du1 +3 = 6⑀c dx dx

2G 共u0 − u1兲; t

The normal stresses ␴xx, ␴yy, and ␴zz are all represented by the pressure p共x兲 = −␴xx = −␴yy = −␴zz. The surface shear stress is defined in the negative x direction on the top surface and in the positive direction on the bottom surface, as shown in Fig. 4共b兲, to allow us to make use of symmetry. The assumption that all the normal stress components are equal to the negative pressure reduces the equations of stress equilibrium to the single equation

⳵ u du0 2z 2z du1 2z = 1− + ⑀xx = t ⳵ x dx t dx t

⑀xx + ⑀zz =

共31兲

and the shear stress is ␶xz = G␥xz. The important shear stresses are the surface stresses which will be denoted by ␶s and defined by ␶s = −␶xz 兩z=t/2 and ␶s = ␶xz 兩z=−t/2 and the inner shear stresses at the bonded surface which will be denoted by ␶l = ␶xz 兩z=0. These two shear stresses can be expressed in terms of the displacement variables u0 and u1 by

u0

X

u共x,z兲 = u0共x兲

冉冊

⳵u ⳵w 2 8z 2 + = − u0 + u1 ⳵z ⳵x t t2 t

p共x兲 = 24GS2⑀c B −

48G⑀c dp =− x, dx t2

共30兲

where the compression strain ⑀c = ⌬ / t is positive in compression. The only significant shear strain is given by

x2 , b2

0 ⱕ x ⱕ x1

0 ⱕ x ⱕ x1

共36兲

共37兲

The displacement field in this region is simply u0 = 6x⑀c and u1 = 0. We note that at x = x1 the value of the surface shear stress is



related to the pressure through ␶s = ␮p, where ␮ is the coefficient of friction, so that from Eqs. 共30兲 and 共34兲 48Gx1 b ⑀c = 24␮G⑀c 2 t t or x1 =

µ µ µ µ

3.5

冉冊 x21 B− 2 b

µ = 0.1

冉冊

x21 ␮b2 B− 2 2t b

2.5

µ = 0.01

1

which shows that the location of the slip is not affected by the level of the compression strain. For the region x1 ⱕ x ⱕ b, where u1 ⫽ 0, the equation of incompressibility becomes du0 du1 +3 = 6⑀c dx dx

共39兲

0

2 4G dp 2␮ + p = − ␶l = − 2 共u0 + u1兲 dx t t t We can replace u1 using the integrated form of the equation of incompressibility leading to

冉

冊

0

0.2

共4y − 1兲ey = 共3␮S − 1兲e3␮S where here y = 3␮Sx1 / b. Knowing y, we can then solve for x1 / b which depends only on the coefficient of friction and the shape factor S = b / t. Equating p共x−1 兲 and p共x+1 兲,

冋

冉冊

24GS2⑀c B −

冋

冉

x21 1 x1 + 共3␮S − 1兲e3␮S关1−共x1/b兲兴 − 3␮S − 1 b b2 18␮2S2

e

p共b兲 − e

3␮x/t

12G⑀c p共x兲 = − t2

冕

x ⬘e

3␮x⬘/t

dx⬘

x

If we assume that p共b兲 = 0 共in this case we do not need to use the Young’s modulus, E0, to provide the boundary condition兲 and carry out the integration, the result for p共x兲 takes the form p共x兲 =

冋

冉

4G⑀c x 共3␮S − 1兲e3␮S关1−共x/b兲兴 − 3␮S − 1 b 3␮2

冊册

,

x1 ⱕ x ⱕ b 共42兲

from which we have

冊册

and leads to the result for the pressure in the nonslip region in the from p共x兲 = 24GS2⑀c B −

b

冊册

共44兲

Integration from x to b gives 3␮b/t

冉

x21 4G⑀c x1 = 共3␮S − 1兲e3␮S关1−共x1/b兲兴 − 3␮S − 1 b b2 3␮2

冉冊

12G⑀c 3␮x/t d 3␮x/t 共e p兲 = − xe dx t2

1

Thus, we need to solve the identity

共41兲

Using the integrating factor e3␮x/t, this can be written in the form

0.8

4G⑀c dp + 共x 兲 = − 关1 + 共3␮S − 1兲e3␮S关1−共x1/b兲兴兴 dx 1 ␮t

B=

12G⑀c dp 3␮ + p=− x dx t t2

0.6

Fig. 9. Pressure distributions for bearing 共S = 10兲 with slip on the top and bottom surfaces

reduces to

which in turn, by replacing u0 from Eq. 共34兲, gives

0.4

x/b

共40兲

In addition, over the slipped region, we have ␶s = ␮p and also dp / dx = −2共␶s + ␶l兲 / t, which taken together give

µ =0

0.5

This can be integrated over the region x1 ⱕ x ⱕ b using the fact that continuity requires that u0共x1兲 = 6x1⑀c and u1共x1兲 = 0 giving u0共x兲 + 3u1共x兲 = 6x⑀c

evolution of x1 /b

µ = 0.05

2 1.5

共38兲

4G 2 dp 2␮ + p=− 2 u0 + 2x⑀c dx t t 3

=1 = 0.6 = 0.4 = 0.2

3 p 6G S 2 c

2

4

x2 , b2

0 ⱕ x ⱕ x1

共45兲

In Fig. 9, pressure distributions are shown for a bearing with S = 10 and slip on the top and bottom surfaces for various values of ␮. To calculate the value of Ec, we integrate the pressure over the range −b ⱕ x ⱕ b to determine P and then divide by 2b⑀c. From Eqs. 共42兲 and 共45兲, we have P = 2⑀c

冉

再冕

x1

冉冊冕冋

24GS2 B −

0

x − 3␮S − 1 b

冊册

dx

冎

x2 dx + b2

b

x1

4G 共3␮S − 1兲e3␮S关1−共x/b兲兴 3␮2

which leads to

4G⑀c dp =− 关1 + 共3␮S − 1兲e3␮S关1−共x/b兲兴兴 dx ␮t

共43兲

At this point we can use continuity across x = x1 of the pressure and its derivative to determine the two unknowns B and x1. At x = x1 we have 48G⑀cx1 dp − 共x 兲 = − dx 1 t2

冉冊冋冉冊

Ec = 24GS2 −

x21 4G x1 共1 − 3␮S兲共1 − e3␮S关1−共x1/b兲兴兲 B− 2 + b 3b 9␮3S

2G x1 x1 − 共2 − 3␮S兲 2 − 3␮S b 3␮2 b

册

共46兲

Table 3 lists values of x1 / b and Ec / 4GS2 for S = 10 and various values of ␮ and Fig. 10 shows these quantities as a function of ␮ for S = 10, 15, and 20.



1

Table 3. Compression Modulus for Bearing with Slip on the Top and Bottom Surfaces 共S = 10兲 B

0.018 0.568 1.842 4.725 7.687 10.668 13.657 16.650 19.645 22.641 25.638 28.635

0.060 0.378 0.614 0.788 0.854 0.889 0.911 0.925 0.936 0.943 0.950 0.955

0.303 0.522 0.684 0.817 0.872 0.902 0.920 0.933 0.942 0.949 0.954 0.959

1.122 1.733 2.359 2.985 3.271 3.433 3.536 3.607 3.660 3.700 3.731 3.757

0.8

x 1 /b

0.01 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

y

Ec 4GS2

0.6 0.4

(a)

0 4 3.5 3 2.5 2 1.5 (b)

Conclusions This paper has provided an analysis of a type of elastomeric bearing that is not bonded to the supports above and below it but held in place solely by friction. These bearings have been previously tested and survived very large shear strains 共in the order of 200% or larger兲共Konstantinidis et al. 2008兲. While the seismic performance objectives that unbonded bearings are expected to achieve are not as high as those of isolation bearings, their low cost and light weight makes them very appealing for use as a low-cost alternative to seismic isolators for schools, housing, and other public buildings in highly seismic areas of the developing world where conventional seismic isolators are not affordable 共Konstantinidis et al. 2008兲. The mechanics of this frictional restraint do not appear to have been treated in the literature before. In this paper, the effect of the friction on the pressure distribution and the compression modulus of an unbonded bearing under compressive load are examined. Two cases are analyzed: the first case is for a single pad held in place by friction on both top and bottom surfaces and the second is for when one surface is able to slip and the other is bonded. This second case is an effort to predict the behavior of a type of elastomeric bridge bearing permitted in California for use as a thermal expansion bearing. The analysis shows that friction has a

S=10 S=15 S=20

0.2

Ec 4G S 2

µ

x1 b

1

0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

µ

1

Fig. 10. 共a兲 Location where slip initiates for a bearing with slip on the top and bottom surfaces; 共b兲 compression modulus as a fraction of the compression modulus of a fully bonded pad

very pronounced effect on both the maximum pressure and the compression modulus.

References Gent, A. N., and Lindley, P. B. 共1959兲. “The compression of bonded rubber blocks.” Proc. Inst. Mech. Eng., 173共3兲, 111–122. Gent, A. N., and Meinecke, E. A. 共1970兲. “Compression, bending and shear of bonded rubber blocks.” Polym. Eng. Sci., 10共1兲, 48–53. Kelly, J. M. 共1997兲. Earthquake-resistant design with rubber, 2nd Ed., Springer, London. Kelly, J. M., and Konstantinidis, D. 共2007兲. “Low-cost seismic isolators for housing in highly-seismic developing countries.” Proc., ASSISi 10th World Conf. on Seismic Isolation, Energy Dissipation and Active Vibrations Control of Structures 共CD-ROM兲, Istanbul, Turkey. Konstantinidis, D., Kelly, J. M., and Makris, N. 共2008兲. “Experimental investigation on the seismic response of bridge bearings.” Technical Rep. No. EERC 2008–02, Earthquake Engineering Research Center, Univ. of California, Berkeley, Calif. Rocard, Y. 共1937兲. “Note sur le calcul des propriétés élastique des supports en caoutchouc adhérent.” J. Phys. Radium, 8, 197–203.

