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## Q. 1.11

A bearing pad of the kind used to support machines and bridge girders consists of a linearly elastic material (usually an elastomer, such as rubber) capped by a steel plate (Fig. 1-53a). Assume that the thickness of the elastomer is h, the dimensions of the plate are a × b, and the pad is subjected to a horizontal shear force V.
Obtain formulas for the average shear stress $\tau_{\text {aver }}$in the elastomer and the horizontal displacement d of the plate (Fig. 1-53b).

## Verified Solution

Assume that the shear stresses in the elastomer are uniformly distributed throughout its entire volume. Then the shear stress on any horizontal plane through the elastomer equals the shear force V divided by the area ab of the plane (Fig. 1-53a):

$\tau_{ aver }=\frac{V}{a b}$    (1-23)

The corresponding shear strain [from Hooke’s law in shear; Eq. (1-21 $\tau=G \gamma$)] is

$\gamma=\frac{\tau_{\text {aver }}}{G_{e}}=\frac{V}{a b G_{e}}$    (1-24)

in which $G _{e}$ is the shear modulus of the elastomeric material. Finally, the horizontal displacement d is equal to htanγ (from Fig. 1-53b):

$d=h \tan \gamma=h \tan \left(\frac{V}{a b G_{e}}\right)$    (1-25)

In most practical situations the shear strain γ is a small angle, and in such cases we may replace tanγ by γ and obtain

$d=h \gamma=\frac{h V}{a b G_{e}}$      (1-26)

Equations (1-25) and (1-26) give approximate results for the horizontal displacement of the plate because they are based upon the assumption that the shear stress and strain are constant throughout the volume of the elastomeric material. In reality the shear stress is zero at the edges of the material (because there are no shear stresses on the free vertical faces), and therefore the deformation of the material is more complex than pictured in Fig. 1-53b. However, if the length a of the plate is large compared with the thickness h of the elastomer, the preceding results are satisfactory for design purposes.