BMES 345 CH03 Problem Set 20240402(1)

BMES 345 CHAPTER 3 PROBLEMS: LINEAR VISCOELASTICITY Contents Problem 1: Spring-Dashpot Models ............................................................................................................................. 2 Problem 2: Spring-Dashpot Models ............................................................................................................................. 3 Problem 3: Viscoelasticity ............................................................................................................................................. 4 Problem 4: Viscoelasticity ............................................................................................................................................. 5 Problem 5: Spring-Dashpot Models ............................................................................................................................. 6 Problem 6: Spring-Dashpot Models ............................................................................................................................. 7 Problem 7: Spring-Dashpot Models ............................................................................................................................. 8 Problem 8: Spring-Dashpot Models ............................................................................................................................. 9 Problem 9: Spring-Dashpot Models ........................................................................................................................... 10 Problem 10: Spring-Dashpot Models ......................................................................................................................... 11 [Solution] Problem 1 ...................................................................................................................................................... 12 [Solution] Problem 2 ...................................................................................................................................................... 13 [Solution] Problem 3 ...................................................................................................................................................... 16 [Solution] Problem 4 ...................................................................................................................................................... 18 [Solution] Problem 5 ...................................................................................................................................................... 20 [Solution] Problem 6 ...................................................................................................................................................... 22 [Solution] Problem 7 ...................................................................................................................................................... 23 [Solution] Problem 8 ...................................................................................................................................................... 25 [Solution] Problem 9 ...................................................................................................................................................... 27 [Solution] Problem 10 .................................................................................................................................................... 28 1

Problem 1: Spring-Dashpot Models Match the creep curve with the corresponding spring-dashpot model. 2

Problem 3: Viscoelasticity Consider the following stress relaxation curve for an osteoblast (bone-producing cell) subjected to a constant strain ε 0 = 0.03: 0 50 100 150 200 250 300 350 400 0 5 10 15 20 25 30 35 40 45 50 Stress (Pa) Time (s) Based on the stress vs. time graph, determine the following: a) Instantaneous modulus b) Equilibrium modulus c) Relaxation time constant d) If the same osteoblast was subjected to a creep test at a constant stress σ 0 = 600 Pa, what would the instantaneous and equilibrium (asymptotic) strains be? Sketch the creep curve with appropriately labeled and scaled axes. 4

Problem 4: Viscoelasticity In a study of how botox-induced unloading of the Achilles tendon altered the biomechanical properties of the tissue (Khayyeri et al, Scientific Reports 2017), researchers measured the viscoelastic properties of rat Achilles tendons using a creep test. Using a constant applied stress of σ 0 = 16 MPa , the researchers obtained the following creep curve: 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0 30 60 90 120 150 Normal Strain Time (s) Find the following parameters (remember to show your work): a) Instantaneous modulus E 0 b) Equilibrium modulus E ∞ c) Creep time constant τ c 5

Problem 6: Spring-Dashpot Models Derive the general constitutive equations for the (a) Maxwell model and (b) Voigt model. Do this WITHOUT copying from the course notes – this is to help you understand where the model equations come from. 7

Problem 7: Spring-Dashpot Models Developing strategies to regenerate peripheral nerves is an ongoing area of research. Understanding the mechanical properties of human peripheral nerves is one component in developing functional replacement tissues. In a previous study (Xu et al., Comparison of viscoelasticity between normal human sciatic nerve and amniotic membrane, Neural Regeneration Research , 2013), the stress relaxation behavior of human sciatic nerves was measured. The average stress relaxation curve from that study is shown below (for an applied constant strain ε 0 = 0.0863): 0.15 0.20 0.25 0.30 0.35 0.40 0 1000 2000 3000 4000 5000 6000 7000 8000 Stress (MPa) Time (s) Using these data, and assuming the human sciatic nerve can be idealized by the standard linear solid spring-dashpot model, estimate the model parameters E 1 , E 2 , and η (the constants for the two springs and the dashpot in the standard linear solid model). 8

Problem 9: Spring-Dashpot Models Derive the stress relaxation response of the Generalized Maxwell Model: 10

Problem 10: Spring-Dashpot Models Articular cartilage exhibits stress relaxation behavior that can be modeled using spring-dashpot models of viscoelasticity. Below is a plot of stress versus time for a sample of tibial articular cartilage obtained from a turkey knee (these are actual data obtained by students in BMES 301 Experimental Biomechanics): 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 10 20 30 40 50 60 Stress (MPa) Time (s) These data were obtained by performing an unconfined compression stress relaxation test at a fixed strain of ε = 0.10. Using your knowledge of spring-dashpot models and the accompanying data (in the Excel spreadsheet cartilage.xlsx), please complete the following: a) Fit the standard linear solid stress relaxation response to the above data. Determine E 1 , E 2 , and η. b) Fit the stress relaxation response to the 5 parameter version of the generalized Maxwell model. Determine E ∞ , E 1 , E 2 , η 1 , and η 2 . c) Which model fits the response better? 11

[Solution] Problem 2 In a recent study by Rubiano et al ( J Mech Behav Biomed Mater 2016), researchers injected adipose- derived stem cells into the damaged hearts of hypertensive rats, and observed significant improvements in mechanical properties and clinical outcomes. One of the ways they measured improvements was via viscoelastic testing of the heart tissue. In this study, they modeled the viscoelasticity of the heart using the standard linear solid. Rubiano et al determined that the standard linear solid parameters for the normal, healthy heart tissue were E 1 = 13 kPa, E 2 = 15 kPa, and η = 2,380 kPa-s. Using this information, sketch the following (make sure the curves and axes are clearly and properly labeled): a) A creep curve for the heart tissue, using an applied stress of σ 0 = 2 kPa. To plot the creep (or stress relaxation) curve, we will need to know the curve’s initial, equilibrium, and transient characteristics. These are best exemplified by the instantaneous modulus, equilibrium modulus, and relevant time constant. We can either recall how these constants relate to the standard linear solid parameters, or use the standard linear solid equations to find them: E 0 = E 1 + E 2 = 28 kPa E ∞ = E 1 = 13 kPa τ creep = η ( E 1 + E 2 ) E 1 E 2 = ( 2380 kPa − s )( 28 kPa ) ( 13 kPa )( 15 kPa ) = 342 seconds Now we can use the instantaneous modulus and equilibrium modulus to find the strain at t = 0 and at long time: ε ( 0 ) = σ 0 E 0 = 2 kPa 28 kPa = 0.071 ε t→ ∞ = σ 0 E ∞ = 2 kPa 13 kPa = 0.154 We can also use our “rule of thumb” to estimate the strain when t = 342 seconds (the creep time constant): ∆ ε = 0.154 − 0.071 = 0.083 0.071 + ( 0.667 ) ( 0.083 ) = 0.126 We can now plot the creep curve: 13

b) A stress relaxation curve for the heart tissue, using an applied strain of ε 0 = 0.05. For the stress relaxation curve, we first need to find the relaxation time constant: τ relaxation = η E 2 = 2380 kPa − s 15 kPa = 159 seconds Now we can use the instantaneous modulus and equilibrium modulus to find the stress at t = 0 and at long time: σ ( 0 ) = E 0 ε 0 = ( 28 kPa ) ( 0.05 ) = 1.4 kPa σ t→∞ = E ∞ ε 0 = ( 13 kPa ) ( 0.05 ) = 0.65 kPa We can also use our “rule of thumb” to estimate the stress when t = 159 seconds (the relaxation time constant): ∆ σ = 1.4 kPa − 0.65 kPa = 0.75 kPa 1.4 kPa − ( 0.667 ) ( 0.75 kPa ) = 0.90 kPa We can now plot the stress relaxation curve: 14

[Solution] Problem 3 Consider the following stress relaxation curve for an osteoblast (bone-producing cell) subjected to a constant strain ε 0 = 0.03: 0 50 100 150 200 250 300 350 400 0 5 10 15 20 25 30 35 40 45 50 Stress (Pa) Time (s) Based on the stress vs. time graph, determine the following: a) Instantaneous modulus The instantaneous modulus is the effective elastic modulus of the material at very short times, effectively t = 0. In the stress relaxation curve above, we see that the stress at t = 0 is ~390 Pa, so we can use that info and the constant applied strain = 0.03 to find the instantaneous modulus: E 0 = σ ( 0 ) ε 0 = 390 Pa 0.03 = 13,000 Pa = 13 kPa b) Equilibrium modulus The equilibrium modulus is the effective elastic modulus of the material as it approaches its equilibrium (fully relaxed) stress. Looking at the stress relaxation curve above, we estimate the equilibrium stress to be ~90 Pa. E ∞ = σ t→∞ ε 0 = 90 Pa 0.03 = 3,000 Pa = 3 kPa c) Relaxation time constant 16

Per our rule of thumb, the relaxation time constant is equal to the time it takes for the material to undergo ~66% of its total relaxation. In this case, the amount of relaxation is given by: ∆ = 390 Pa − 90 Pa = 300 Pa Therefore, 66% relaxation occurs at: 390 Pa − ( 2 3 ) ( 300 Pa ) = 190 Pa At σ(t) = 190 Pa, t is approximately 5 to 7 seconds. Therefore, our relaxation time constant is around 5 – 7 seconds . d) If the same osteoblast was subjected to a creep test at a constant stress σ 0 = 600 Pa, what would the instantaneous and equilibrium (asymptotic) strains be? Sketch the creep curve with appropriately labeled and scaled axes. We can use our instantaneous and equilibrium modulus values to predict the instantaneous and equilibrium strain values for our osteoblast under a constant stress = 600 Pa: ε ( 0 ) = σ 0 E 0 = 600 Pa 13,000 Pa = 0.046 ε t→ ∞ = σ 0 E ∞ = 600 Pa 3,000 Pa = 0.2 The instantaneous strain value is the strain at t = 0. The equilibrium strain value is the asymptote that our creep curve should approach. In terms of time scale, creep and stress relaxation occur over similar (but not the same) time scales, so plotting 0 – 50 s is a reasonable guess: 0.00 0.05 0.10 0.15 0.20 0.25 0 10 20 30 40 50 Strain Time (s) 17

E ∞ = σ 0 lim t→∞ ε ( t ) Looking at the graph, we can approximate the asymptotic strain at some value around 0.20 – 0.21. I will use 0.20: E ∞ = σ 0 lim t→ ∞ ε ( t ) = 16 MPa 0.20 = 80 MPa c) (10 points) Creep time constant τ c The creep time constant is a characteristic time that describes how quickly the material creeps, in other words how quickly it approaches the asymptotic strain value. A good approximation is that the time constant is the time it takes for two-thirds or 66% of the creep deformation to occur (it is actually the time it takes for 1-1/ e or 63.2% of the creep to occur, but two-thirds is close enough). First we determine what the total creep deformation is: lim t→ ∞ ε ( t ) − ε ( 0 ) ≈ 0.20 − 0.05 = 0.15 Then we take 66% of that deformation and add it to the initial strain, to find the target strain value: 0.05 + ( 0.667 ) ( 0.15 ) = 0.15 So, if we look at our graph and find the time where ε(t) = 0.15, that should be approximately the time constant: ε ( t ) = 0.15 →τ ≈ 10 s 19

[Solution] Problem 5 Download the Excel file containing stress relaxation data (stress_relax.xlsx) from BB Learn. In that file, you have the stress relaxation response of a breast cancer cell subjected to a normal strain ε = 0.05. You also have a curve fit (which matches the general form of the stress relaxation response of a standard linear solid): σ ( t )= E 2 ε 0 exp ( − t τ ) + E 1 ε 0 Determine the parameters E 1 , E 2 , and τ of the breast cancer cell using the standard linear solid model. a) Using the equation above and the definitions of instantaneous modulus, equilibrium modulus, and the characteristic time constant, estimate the parameters. If you look at the form of the stress relaxation function, you can note that the instantaneous modulus = E 1 + E 2 , the equilibrium modulus = E 1 , and the relaxation time constant = τ . σ ( 0 ) = E 2 ε 0 + E 1 ε 0 → σ ( 0 ) ε 0 = E 0 = E 1 + E 2 lim t→ ∞ σ ( t )= σ t →∞ = E 1 ε 0 → σ t →∞ ε 0 = E ∞ = E 1 With this, you can make some good guesses at what they should be. E ∞ = σ t→∞ ε 0 = 10 Pa 0.05 = 200 Pa→ E 1 = 200 Pa E 1 + E 2 = E 0 = σ ( 0 ) ε 0 = 35 Pa 0.05 = 700 Pa→E 2 = 500 Pa Finally, if you look at the stress relaxation data, the time constant is on the order of 10 seconds, i.e., τ = 10 s. Inputting these values into the Excel spreadsheet, we see that the fit is decent, but the equilibrium modulus is too high (the asymptote is closer to 8 Pa, rather than 10 Pa) and the time constant should be lower. We can still get better estimated values. First, we can note that the equilibrium stress is closer to 8 Pa, giving us: E 1 = E ∞ = σ t→ ∞ ε 0 = 8 Pa 0.05 = 160 Pa E 1 + E 2 = 700 Pa→E 2 = 540 Pa To get closer to the actual time constant, we can calculate the change in stress (the total amount of stress relaxation). My rule of thumb is that the time constant is the time it takes for 66% of the relaxation to occur: 20