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Design Guide Article Series #1 - Selecting the Right Elastomer

Article: # 1
Author: Paul Herbst, Consulting Engineer, Corry Rubber Corporation

The "So What" and "Who Cares" of Engineering a Part With an Elastomer

Just what is different about this substance called “elastomer” or “rubber”? In discussions about the design of products with an elastomer, it is always necessary to establish a frame of reference for those who are unfamiliar with the material. It is even prudent to do the same when discussions take place between those who think they understand the product.

Where and why are Elastomer Products used?
With just a short thought process, a large number of applications of elastomeric products can be established. There are suspension systems for vehicles and engines, seals of many shapes and forms, pump impellers, shock and vibration protectors for many products, floor mats, tires, etc. Within each of these general categories are many specific applications. Some of these products are highly “engineered” and others are not.

So that is the where and now the why. Elastomers possess some unique properties that are used to advantage in these applications.

  • They can be deflected to large deformation. This permits their use in applications where they perform as springs or seals.
  • They can be molded to many different shapes with different spring parameters in different directions. A suspension element can carry load in three different directions, with different spring rates in each one.
  • They provide energy loss when deflected. The amount can be compounded into the product within limits to match application needs.
The ‘so what’ is that each property, with it’s unique characteristics, is used to ultimate advantage in many applications. The fact that a suspension element can be characterized with different spring rates in different directions, and can be molded in a shape to fit a particular available space, and can provide some damping, is a valuable tool to a designer of the suspension.The ‘who cares’ should be obvious when the options are considered. The designer of the product really cares, or should care, about each of the special parameters that allow the optimization of a design.

Some Elastomer Properties
Strength - The tensile strength of an elastomer is less than a metal and its capacity to store energy is greater than a metal. Some generalized comparative numbers of two different elastomer types are:

  • Natural rubber; Tensile strength is 3000psi at 600% elongation.
  • Silicone rubber; Tensile strength is 1200 psi at 400% elongation.
It is the large deformation that gains the energy storage capability.

Environment - Natural rubber is the highest strength and toughest elastomer. Its limitations are; the loss of strength with increased temperature, the increase of stiffness with lower temperature, and loss of strength with exposure to some fluids.

Silicone rubber has a relatively broad temperature capability and can be compounded for resistance to fluids that effect natural rubber.

Other elastomers have varying properties in this same range. Elastomer selection is based on meeting the design criteria at the least cost.

Sizing a Part for Spring Rate
Spring rate is one of the major design characteristics in the engineering application of an elastomer. The large deformation capability leads to applications where the spring rate is the determining parameter. Suspension systems, seals, and impellers are examples of elastomer spring systems. In motion accommodation devices, knowledge of the spring rate is needed to determine the loads generated due to deformation. In order to understand how the spring rate is calculated, a few examples are as follows.

There are two basic modes of deflection used for an elastomeric product. The first is simple shear and the second is compression.

In simple shear, the formula for spring rate is: K=GxA/t, where K is the spring rate, G is the shear modulus, A is the shear loaded area, and t is the thickness of the elastomer.

In compression, the formula for spring rate is: Kc=Ec*A/t, where Kc is the compression spring rate, Ec is the compression modulus, A is the compression loaded area, and t is the elastomer thickness.

The spring rate formula’s come directly from material property definitions.

  • Stress=Load/Area=P/A
  • Strain=Deflection/Thickness (or length)=x/t
  • Modulus=Stress/Strain=G=P*t/A*x
  • Spring Rate=Load/Deflection=P/x=G*A/t

Elastomer Modulus
Static Modulus - The term modulus needs to be explained. Since elastomers are non-linear, the term means a tensile stress at a certain extension. There are three moduli that are of interest, the shear modulus, the tensile modulus, and the compression modulus. The compression modulus is a function of the shear and tensile moduli. A standard test specimen is used to determine the material properties.

Dynamic Modulus - Due to the viscoelastic nature of the elastomer, the dynamic modulus is not the same value as the static modulus. The dynamic modulus is measured on a test machine that can input a sinusoidal deflection into a standard test sample. The output is then broken into components that are in phase with the input deflection and ninety degrees out of phase with the input.

  • G’ is the dynamic elastic modulus and is in phase with the input deflection.
  • G” is the dynamic damping modulus and is ninety degrees out of phase with the input deflection.
  • G* is the total dynamic modulus and lags the input deflection by the angle
  • Loss factor is defined as G”/G’ and the symbol is used.
  • Dynamic spring rate is defined as K*=(A/t)G’(1+). G’ varies with frequency, strain, and temperature. The product application determines the selection of the proper value from a known database.

Calculating Sprint Rate

Shear of a flat sandwich

  • Ks=G*A/t=F/
  • A=L*W
  • G has values of 70 to 200 psi.
Compression of a flat sandwich

  • Kc=Ec*A/t=F/
  • A=L*W
  • Ec is dependent on shape factor, which is a measure of how much restraint there is on the part to allow the elastomer to move or “bulge”.
  • Ec=Eo(1+2kS^2)
  • Eo=Youngs Modulus=4.5G-71 (Approximatly)
  • S=Shape
  • Factor=L*W/2(L+W)t
  • K=numerical factor=0.444+(23.3/G)
This calculation for compression spring rate is used up to 10% strain, at which time the spring rate becomes non-linear.

Non linearities can be approximated by : F=Ec*A*(/T)(1+/T)

Shear of a circular sandwich

  • Ks=F/=G*A/T
  • A=(Ro^2-Ri^2)
  • G – Shear Modulus
  • A – Load Area
  • T – Elastomer Section Thickness
Compression of circular flat sandwich

  • Kc=F/=Ec*A/T
  • A=(Ro^2-Ri^2)
  • Ec=Eo(1+2kS^2)
  • S=Shape Factor= Load Area / Bulge Area = (Ro^2-Ri^2)/2(Ro+Ri)T=(Ro-Ri)/2T
  • Eo=4.5G-71 Approximately
  • k=0.444+23.3/G
  • G – Shear Modulus
  • Ec – Compression Modulus
  • A – Load Area
  • Tubular mount in shear
  • Ks=F/=2.73*G*L/log10(D/d)
Tubular mount in compression

  • Kc=F/=Ec*A/T
  • A=L*Di
  • Ec=Eo(1+2kS^2)
  • S=2*L*Di/(Do+Di)T
  • Eo=4.5*G-71 Approximately
The calculation of tubular mount compression rate is very approximate since the tension side of the mount can be modified by mold pressure and post molding processing.

Sizing a Part
The shear stress under static load should be less than 50 psi. The shear strain under static load should be less than 50%.

The compression strain under static load should be less than 15%.

Dynamic loads can be 3 to 4 times the static loads.

Each application of an elastomeric part has allowable stress and strain values that vary from these general guidelines.

Design for Life
Drift - An elastomeric part will continue to deflect under load. This deflection is a Log decrement phenomenon This means that the amount of deflection varies linearly with the Log of time. The amount of deflection in 1 day is the same as that in 10 days, is the same as that in 100 days, etc. This deflection has to be accounted for in the design process.

Fatigue - An elastomeric part subjected to a dynamic load will fatigue. There will be a loss of spring rate due to the break down of the elastomer. This also has to be accounted for in the design

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