1. Read: pg 130 – 168 (rest of Chpt. 4)
Lecture 8 – Viscoelasticity and Deformation
Read: pg 130 – 168 (rest of Chpt. 4)
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BAE2023 Physical Properties of Biological Materials Lecture 8

2. Lecture 8 – Viscoelasticity and Deformation
Poisson’s Ratio, μ (pg. 115)
Ratio of the strain in the direction perpendicular to the applied force to the strain in the direction of the applied force.
For uniaxial compression:
εz = σz/E, εy = -μ·εz and εx = -μ·εy
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BAE2023 Physical Properties of Biological Materials Lecture 8

3. Lecture 8 – Viscoelasticity and Deformation
Poisson’s Ratio
For multi-axial compression
See equations in 4.2 page 117
Maximum Poisson’s = 0.5 for incompressible materials to 0.0 for easily compressed materials
Examples: gelatin gel – 0.50
Soft rubber – 0.49
Cork – 0.0
Potato flesh – 0.45 – 0.49
Apple flesh – 0.29
Wood – 0.3 to 0.5
More porous means smaller Poisson’s
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BAE2023 Physical Properties of Biological Materials Lecture 8

4. Lecture 8 – Viscoelasticity and Deformation
In addition to Normal stresses: Shearing Stresses
Shear stress: force per unit area acting in the direction parallel to the surface of the plane,τ
Shear strain: change in the angle formed between two planes that are orthogonal prior to deformation that results from application of sheer stress, γ
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BAE2023 Physical Properties of Biological Materials Lecture 8

5. Lecture 8 – Viscoelasticity and Deformation
Shear modulus: ratio of shear stress to shear strain, G = τ/γ
Measured with parallel plate shear test (pg. 119)
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BAE2023 Physical Properties of Biological Materials Lecture 8

6. Lecture 8 – Viscoelasticity and Deformation
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BAE2023 Physical Properties of Biological Materials Lecture 8

7. Lecture 8 – Viscoelasticity and Deformation
Example Problem
The bottom surface (8 cm x 12 cm) of a rectangular block of cheese (8 cm wide, 12 cm long, 3 cm thick) is clamped in a cheese grater.
The grating mechanism moving across the top surface of the cheese applies a lateral force of 20N.
The shear modulus, G, of the cheese is 3.7kPa.
Assuming the grater applies the force uniformly to the upper surface, estimate the latera movement of the upper surface w/respect to the lower surface.
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BAE2023 Physical Properties of Biological Materials Lecture 8

8. Lecture 8 – Viscoelasticity and Deformation
Stresses and Strains: described as deviatoric or dilitational
Dilitational: causes change in volume
Deviatoric: causes change in shape but negligible changes in volume
Bulk Modulus, K: describes response of solid to dilitational stresses
K = average normal stress/dilatation
Dilatation: (Vf – V0)/V0
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BAE2023 Physical Properties of Biological Materials Lecture 8

9. Lecture 8 – Viscoelasticity and Deformation
K = average normal stress/dilatation
Dilatation: (Vf – V0)/V0
Average normal stress = ΔP, uniform hydrostatic gauge pressure
ΔV = Vf – V0
So: K = ΔP/(ΔV / V0)
Δ V is negative, so K is negative
Example of importance: K (Soybean oil) > K (diesel)
Will effect the timing in an engine burning biodiesel
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BAE2023 Physical Properties of Biological Materials Lecture 8

10. Lecture 8 – Viscoelasticity and Deformation
Apples compress easier than potatoes so they have a smaller bulk modulus, K (pg. 120) but larger bulk compressibility
K-1 =bulk compressibility
Strain energy density: area under the loading curve of stress-strain diagram
Sharp drop in curve = failure
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BAE2023 Physical Properties of Biological Materials Lecture 8

11. Lecture 8 – Viscoelasticity and Deformation
Stress-Strain Diagram, pg. 122
Area under curve until it fails = toughness
Failure point = bioyield point
Resilience: area under the unloading curve
Resilient materials “spring back”…all energy is recovered upon unloading
Hysteresis = strain density – resilience
Figure 4.6, page 124
Figure 4.7, page 125
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BAE2023 Physical Properties of Biological Materials Lecture 8

12. Lecture 8 – Viscoelasticity and Deformation
Factors Affecting Force-Deformation Behavior
Moisture Content, Fig. 4.6b
Water Potential, Fig. 4.8
Strain Rate: More stress required for higher strain rate, Fig. 4.8
Repeated Loading, Fig. 4.9
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BAE2023 Physical Properties of Biological Materials Lecture 8

13. Lecture 8 – Viscoelasticity and Deformation
Stress Relaxation: Figure 4.10 pg 129
Material is deformed to a fixed strain and strain is held constant…stress required to hold strain constant decreases with time.
Creep: Figure 4.11 pg. 130
A continual increase in deformation (strain) with time with constant load
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BAE2023 Physical Properties of Biological Materials Lecture 8

14. Lecture 8 – Viscoelasticity and Deformation
Tensile testing
Not as common as compression testing
Harder to do
See figure 4.12 page 132
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BAE2023 Physical Properties of Biological Materials Lecture 8

15. Lecture 8 – Viscoelasticity and Deformation
Tensile testing
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BAE2023 Physical Properties of Biological Materials Lecture 8

16. Lecture 8 – Viscoelasticity and Deformation
Bending: E=modulus of elasticity D=deflection, F=force, I = moment of inertia
E=L3(48DI)-1
I=bh3/12
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BAE2023 Physical Properties of Biological Materials Lecture 8

17. Lecture 8 – Viscoelasticity and Deformation
Can be used for testing critical tensile stress at failure
Max tensile stress occurs at bottom surface of beam
σmax=3FL/(2bh2)
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BAE2023 Physical Properties of Biological Materials Lecture 8

18. Lecture 8 – Viscoelasticity and Deformation
Contact Stresses (handout from Mohsenin book)
Hertz Problem of Contact Stresses
Importance:
“In ag products the Hertz method can be used to determine the contact forces and displacements of individual units”
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BAE2023 Physical Properties of Biological Materials Lecture 8

19. Lecture 8 – Viscoelasticity and Deformation
Assumptions:
Material is homogeneous
Loads applied are static
Hooke’s law holds
Contacting stresses vanish at the opposite ends
Radii of curvature of contacting solid are very large compared to radius of contact surface
Contact surface is smooth
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BAE2023 Physical Properties of Biological Materials Lecture 8

20. Lecture 8 – Viscoelasticity and Deformation
Maximum contact stress occurs at the center of the surface of contact
a and b are the major and minor semiaxes of the elliptic contact area
For ag. Products, consider bottom 2 figures in Figure 6.1
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BAE2023 Physical Properties of Biological Materials Lecture 8

21. In the case of 2 contact spheres, pg 354:
Lecture 8 – Viscoelasticity and Deformation
In the case of 2 contact spheres, pg 354:
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BAE2023 Physical Properties of Biological Materials Lecture 8

22. Lecture 8 – Viscoelasticity and Deformation
To determine the elastic modulus, E…
for steel flat plate:
for steel spherical indentor:
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BAE2023 Physical Properties of Biological Materials Lecture 8

23. Lecture 8 – Viscoelasticity and Deformation
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BAE2023 Physical Properties of Biological Materials Lecture 8

24. BAE2023 Physical Properties of Biological Materials
Lecture 9
HW Assignment Due 2/15
Problem 1:
An apple is cut in a cylindrical shape 28.7 mm in diameter and 22.3 mm in height. Using an Instron Universal Testing Machine, the apple cylinder is compressed. The travel distance of the compression head of the Instron is 3.9 mm. The load cell records a force of N. Calculate the stress εz , and strain σz on the apple cylinder.
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BAE2023 Physical Properties of Biological Materials

25. BAE2023 Physical Properties of Biological Materials
Lecture 9
HW Assignment Due 2/15
Problem 2:
A sample of freshly harvested miscanthus is shaped into a beam with a square cross section of 6.1 mm by 6.1 mm. Two supports placed 0.7 mm apart support the miscanthus sample and a load is applied halfway between the support points in order to test the Force required to fracture the sample. If ultimate tensile strength is 890 MPa, what would be the force F (newtons) required to cause this sample to fail?
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BAE2023 Physical Properties of Biological Materials

26. BAE2023 Physical Properties of Biological Materials
HW Assignment Due 2/15
Problem 3:
Ham is to be sliced for a deli tray. A prepared block of the ham has a bottom surface of 10 cm x 7 cm. The block is held securely in a meat slicing machine. A slicing blade moves across the top surface of the ham with a uniform lateral force of 27 N and slices a thin portion of meat from the block. The shear modulus, G, of the ham is 32.3 kPa. Estimate the deflection of the top surface with respect to the bottom surface of the block during slicing.
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BAE2023 Physical Properties of Biological Materials

27. BAE2023 Physical Properties of Biological Materials
HW Assignment Due 2/15
Problem 4:
Sam, the strawberry producer, has had complaints from the produce company that his strawberries are damaged during transit. Sam would like to know the force required to damage the strawberries if they are stacked three deep in their container.
The damage occurs on the bottom layer at the interface with the parallel surface of the container and also at the point of contact between the layers of strawberries.
An hydrostatic bulk compression test on a sample of Sam’s strawberries indicates an average bulk modulus of 225 psi. Testing of specimens from Sam’s strawberry crop shows a compression modulus E of 200 psi. The average strawberry diameter is 1.25 inches and the axial deformation due to the damage in transit averages 0.23 inches. The modulus of elasticity for Sam’s variety of strawberries is reported to be 130 psi.
Estimate the force Sam’s strawberries may be encountering during transit. (Hertz method)
12/7/2018
BAE2023 Physical Properties of Biological Materials