Question

Another model of viscoelastic behavior is the Kelvin model, in which the stress \(\sigma\) in the medium for a given strain \(\varepsilon\) and strain rate \(\dot{\varepsilon}\) is the superposition of linear elastic and linear viscous stresses, \(\sigma_{e}\) and \(\sigma_{f}\). Show that the rheological law for the Kelvin viscoelastic material is $$ \sigma=\varepsilon E+2 \mu \frac{d \varepsilon}{d t} $$ Show also that the response of the Kelvin viscoelastic material to the sudden application of a stress \(\sigma_{0}\) at time \(t=0\) is $$ \varepsilon=\frac{\sigma_{0}}{E}\left(1-e^{-t / \tau_{w}}\right) $$ Assume that \(\sigma=\sigma_{0}\) for \(t>0 .\) While stresses decay exponentially with time in a Maxwell material subjected to constant strain, strain relaxes in the same way in a Kelvin material subjected to constant stress.

Short answer

The rheological law is \( \sigma = E \varepsilon + 2 \mu \dot{\varepsilon} \) and the strain response is \( \varepsilon = \frac{\sigma_0}{E}(1-e^{-t/\tau_w}) \).

Step-by-step solution

Understand the Kelvin model

The Kelvin model, also known as the Kelvin-Voigt model, describes viscoelastic behavior by combining a spring (elastic element) and a dashpot (viscous element) in parallel. In this model, the total stress \( \sigma \) is the sum of elastic stress \( \sigma_e \) and viscous stress \( \sigma_f \). The elastic stress is proportional to strain \( \varepsilon \) and the viscous stress is proportional to the strain rate \( \dot{\varepsilon} \).

Express the stresses

For the Kelvin model:- The elastic stress \( \sigma_e = E \varepsilon \), where \( E \) is the modulus of elasticity.- The viscous stress \( \sigma_f = 2 \mu \dot{\varepsilon} \), where \( \mu \) is the viscosity coefficient.Thus, the total stress is given by the equation \( \sigma = E \varepsilon + 2 \mu \dot{\varepsilon} \).

Derive the strain response to constant stress

For a sudden application of stress \( \sigma_0 \) at \( t=0 \), the total stress \( \sigma = E \varepsilon + 2 \mu \dot{\varepsilon} = \sigma_0 \) for \( t>0 \). The differential equation describing the strain \( \varepsilon(t) \) under constant stress has the form:\[ E \varepsilon + 2 \mu \frac{d \varepsilon}{d t} = \sigma_0 \].

Solve the differential equation

The equation \( E \varepsilon + 2 \mu \frac{d \varepsilon}{d t} = \sigma_0 \) is linear with constant coefficients. Rearranging terms:\[ \frac{d \varepsilon}{d t} = \frac{\sigma_0}{2 \mu} - \frac{E}{2 \mu} \varepsilon \]This can be solved using an integrating factor or recognizing it as a first-order linear differential equation that has an exponential solution \( \varepsilon(t) = A e^{-t/\tau_w} + B \).

Apply initial conditions

Initially, when \( t=0 \), the strain is zero due to the sudden stress application. Therefore, \( \varepsilon(0) = 0 \). Substituting this into the solution form, we solve for constants \( A \) and \( B \), providing us the solution:\[ \varepsilon(t) = \frac{\sigma_0}{E} \left(1 - e^{-t/\tau_w}\right) \]Where \( \tau_w = \frac{2 \mu}{E} \) is the relaxation time for the Kelvin material.

Key concepts

Viscoelastic Behavior

In the study of materials, viscoelastic behavior is a fascinating area bridging between purely elastic and purely viscous responses. Imagine a material that acts a bit like a rubber band and a bit like honey. That's what viscoelasticity is all about. It describes materials that have properties of both solids and liquids to varying degrees.

When we apply stress to a viscoelastic material, part of its response is instant (elastic), while another part is time-dependent (viscous). This dual nature makes them unique and versatile in various applications, like damping vibrations or adsorbing shocks.

Viscoelastic materials have the ability to "remember" their previous states somewhat, meaning they can 'relax' back to their original shape after being deformed, but it takes some time due to their internal resistance.

Elastic Stress

Elastic stress refers to the immediate and recoverable response of a material when stress is applied. In the context of the Kelvin model, elastic stress is represented by a spring, a perfect metaphor for how it works.

When you stretch a spring, it stores energy and exerts an opposing force proportional to its deformation. Mathematically, this relationship is expressed as \(\sigma_e = E \varepsilon\), where \(E\) is the modulus of elasticity, and \(\varepsilon\) is the strain or deformation. The modulus of elasticity is a measure of a material's ability to withstand changes in length when under lengthwise tension or compression.

In practical terms, elastic stress is straightforward—it's the quick, reversible part of a material's response to stress, bouncing back immediately once the stress is relieved.

Viscous Stress

Unlike elastic stress, viscous stress is associated with the rate of deformation rather than the deformation itself. Think of squeezing a bottle of ketchup. The resistance you feel is due to its viscosity—a measure of the material's resistance to flow.

In the Kelvin model, this behavior is modeled by a dashpot. Viscous stress is proportional to the strain rate, expressed as \(\sigma_f = 2 \mu \dot{\varepsilon}\), where \(\mu\) is the viscosity coefficient, and \(\dot{\varepsilon}\) is the strain rate. This means it acts more prominently as the rate of strain changes over time.

Importantly, viscous stress causes energy dissipation in a system. This dampening effect is crucial in applications requiring shock absorption and noise reduction, helping to protect structures and improve comfort.

Strain Response

In the Kelvin model, the strain response describes how the material deforms over time when exposed to stress. If we suddenly apply stress to a viscoelastic material, how it "settles" or changes shape characterizes its strain response.

For instance, when a constant stress \(\sigma_0\) is applied suddenly, the strain \(\varepsilon(t)\) evolves over time. Initially, the strain is zero, indicating no deformation at the exact moment of stress application. Over time it follows the equation \(\varepsilon(t) = \frac{\sigma_0}{E} (1 - e^{-t/\tau_w})\), with \(\tau_w\) being the relaxation time.

This exponential relationship reflects the gradual adjustment within the material—a quick initial response due to the elastic component, followed by a slower approach to equilibrium as the viscous components come into play. This behavior highlights how viscoelastic materials deform over time, adapting to sustained forces in a manner akin to a slow-moving liquid.