Question

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed time and the relaxation time; τ is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer with the stress relaxation that obeys equation above was suddenly pulled in a tension to a measured strain of 0.49; the stress necessary to maintain this constant strain was measured as a function of time. Determine Er(6) for this material if the initial stress level was 3.1 MPa (440 psi), which dropped to 0.41 MPa (59 psi) after 32 s.

Answer 1

`E_r(6)=4.35614\ MPa`

Step-by-step explanation:

`\epsilon` = Strain = 0.49

`\sigma _0` = 3.1 MPa

At t = Time = 32 s
`\sigma` = 0.41 MPa

`\tau` = Time-independent constant

Stress relation with time

`\sigma=\sigma _0exp\left(-(t)/(\tau)\right)`

at t = 32 s

`0.41=3.1exp\left(-(32)/(\tau)\right)\\\Rightarrow exp\left(-(32)/(\tau)\right)=(0.41)/(3)\\\Rightarrow -(32)/(\tau)=ln(0.41)/(3)\\\Rightarrow \tau=-(32)/(ln(0.41)/(3))\\\Rightarrow \tau=16.0787\ s`

The time independent constant is 16.0787 s

`E_(r)(t)=(\sigma(t))/(\epsilon_0)`

At t = 6

`\\\Rightarrow E_(r)(6)=(\sigma(6))/(\epsilon_0)`

From the first equation

`\sigma(t)=\sigma _0exp\left(-(t)/(\tau)\right)\\\Rightarrow \sigma(6)=3.1exp\left(-(6)/(16.0787)\right)\\\Rightarrow \sigma(6)=2.13451`

`E_r(6)=(2.13451)/(0.49)\\\Rightarrow E_r(6)=4.35614\ MPa`

`E_r(6)=4.35614\ MPa`