1. Lecture 8 – Viscoelasticity and Deformation
HW4 due today
Lab #1 due Wednesday, 2/6
Study Abroad Deposit due today by 5:00 pm
Lecture Wednesday 2/6 at 3: Ag Hall
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

2. Lecture 8 – Viscoelasticity and Deformation
Poisson’s Ratio
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
2/04/08
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
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

4. Lecture 8 – Viscoelasticity and Deformation
Apples compress easier than potatoes so they have a smaller bulk modulus, K
-78 MPa vs MPa
K-1 =bulk compressibility
Strain energy density: area under the loading curve of stress-strain diagram
Sharp drop in curve = failure
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

5. Lecture 8 – Viscoelasticity and Deformation
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 4.5, page 122
Figure 4.7, page 125
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

6. 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
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

7. Lecture 8 – Viscoelasticity and Deformation
Tensile testing
Not as common as compression testing
Harder to do
See figure 4.12 page 132
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

8. Lecture 8 – Viscoelasticity and Deformation
Tensile testing
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

9. 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
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

10. 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)
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

11. Lecture 8 – Viscoelasticity and Deformation
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
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

12. Lecture 8 – Viscoelasticity and Deformation
Shear modulus: ratio of shear stress to shear strain, G = τ/γ
Measured with parallel plate shear test
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

13. Lecture 8 – Viscoelasticity and Deformation
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

14. Lecture 8 – Viscoelasticity and Deformation
Dilitational stress or strain causes a change in volume (usually normal stress or strain)
Deviatoric stress or strain causes a change in shape (usually shear stress or strain)
Bulk modulus K=avg. normal stress/dilatation
Dilatation = (Vf-V0)/V0
Average stress: uniform hydrostatic gauge pressure ΔP SO….
Bulk modulus K= ΔP / (ΔV/V0) (K will be negative under compression!)
Interpretation: with a change in pressure, how much is the volume going to change
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

15. 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”
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

16. 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
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

17. 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

18. 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

19. 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

20. Lecture 8 – Viscoelasticity and Deformation
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

21. BAE2023 Physical Properties of Biological Materials Lecture 8
HW#5 (continued)
Problem 3:
One half of a soybean seed is shaped into a beam with a square cross section of 1.15 mm by 1.15 mm. The sample beam is supported at two points 0.5 mm apart and a load is applied halfway between the support points. If the ultimate tensile strength is 5.23 MPa, what would be the force F (newtons) required to cause the sample to fail?
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

22. BAE2023 Physical Properties of Biological Materials Lecture 8
HW#5 Assignment
Problem 4:
A block of bologna has a bottom surface of 3 cm x 4 cm. The block is held securely in a meat slicing machine. The blade that moves across the top surface of the salami applies a uniform lateral force of 16 N. The shear modulus, G, of the bologna is 1.7 kPa. Estimate the how far the top surface of the salami will move compared to the bottom surface.
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

23. BAE2023 Physical Properties of Biological Materials Lecture 8
HW#5 Assignment
Problem 5:
An hydrostatic bulk compression test on a sample of oranges indicates an average bulk modulus of 325 psi. Testing of specimens from the same group of oranges shows a compression modulus E of 412 psi. Some of the whole oranges are subjected to a parallel plate test. The average orange diameter is inches and the axial deformation for the whole oranges was inches for a force of 4.25 lbs. Estimate the modulus of elasticity for the whole oranges using the Hertz formula for contact stresses.
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8

24. BAE2023 Physical Properties of Biological Materials Lecture 8
HW#5 Assignment
Problem 6:
A 8 lb. crate of apples is placed on top of an open box of strawberries (single layer). A previous test of a similar box of strawberries indicated a compression modulus of 312 psi. The average diameter of a strawberry is inches. The axial deformation at 8 lb is inches. Estimate the modulus of elasticity for the strawberries using the Hertz formula. (Bulk modulus = 218 psi)
2/04/08
BAE2023 Physical Properties of Biological Materials Lecture 8