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# Linear Viscoelasticity – Part 5 – Storage and Loss Modulus

## Introduction

In previous parts of this series I covered the theory of linear viscoelasticity based on stress relaxation and constant strain rate tension tests. In this article I will show how a linear viscoelastic material model can be used to predict the response due to a sinusoidal strain history. The derivation will show how the storage and loss moduli are defined, and how they can be determined from the stress response.

### Previous Part

• In many applications the load is cyclic.
• It is easy to measure the steady-state cyclic response using a Dynamic Mechanical Analysis (DMA) machine.
• We can calibrate a linear viscoelastic model to dynamic data (E’ and E’’).

Figure 1. DMA test machine.

## Cyclic Response - Full Solution

In this example I will determine the stress response of a linear viscoelastic material loaded with a sinusoidal strain history: $$\varepsilon(t) = \varepsilon_0 \sin(\omega t)$$, and I will only consider times $$t \ge 0$$. In part 1 of this series I showed that the stress for any strain history can be obtained from:

$\sigma(t) = \displaystyle \int_0^{t} E_R(t – \tau) \dot{\varepsilon}(\tau) d\tau.$

(1)

Inserting the sinusoidal strain history into this equation gives:

$\sigma(t) = \displaystyle \int_0^{t} E_R(t-\tau) \varepsilon_0 \omega \cos(\omega \tau) d\tau.$

(2)

Rewrite the integral based on the variable substitution: $$s = t – \tau$$:

$\sigma(t) = \displaystyle \int_0^t E_R(s) \varepsilon_0 \omega \cos[\omega (t-s)] ds.$

(3)

## Example - Full Solution

To evaluate the integral and calculate the actual stress response we need to specify the stress relaxation modulus. Let’s assume the following 1-term Prony expression: $$E_R(t) = E_0 e^{-\alpha t}$$. Inserting this into Equation (3), and evaluating the integrals gives:

$\sigma(t) = \displaystyle\frac{E_0 \varepsilon_0 \omega}{\alpha^2 + \omega^2} \left[ \alpha \cos(\omega t) + \omega \sin(\omega t) + e^{-\alpha t} \cdot f(\alpha, \omega, t) \right]$

(4)

Note 1: I did not write out the whole equation,  I left some terms in the function $$f(\alpha, \omega,t)$$. Note 2: In steady state, part of the stress is in-phase with the strain, and part of the stress is out-of phase with the strain.

The predicted stress-strain response due to a sinusoidal strain history can also be calculated using MCalibration (see Figure 2). In this case the steady-state response is reached after about one cycle.

Figure 2. Predicted cyclic response from a one-term Prony series model.

## Cyclic Response - Steady State

Equation (3) gives the full transient solution for the stress. If we are only interested in the steady state response then we can still use Equation (3) if we simply change the upper limit of the integral to infinity (this may not be obvious at first, but think about it for a while). Furthermore, if we recall that $$\cos(\alpha-\beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$$, then the stress is given by:

(5)

which can also be written:

$\sigma(t) = \varepsilon_0 \left[ E'(\omega) \sin(\omega t) + E”(\omega) \cos(\omega t)\right]$

(6)

## Tan Delta

From Equation (6) we know that the stress is given by a sin and a cos term, which can also be written: $$\sigma(t) = \sigma_0 \sin(\omega t + \delta)$$.

This equation can also be written: $$\sigma(t) =\sigma_0 \sin(\omega t) \cos\delta + \sigma_0 \cos(\omega t) \sin\delta$$.

Which gives:
$\tan\delta = \displaystyle \frac{E”}{E’}$

## Summary

• The storage (E’) and loss modulus (E’’) are often measured using DMA experiments.
• The Prony series terms can be determined from the dynamic data.

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