Lecture 8 – Viscoelasticity and Deformation 2/4/2011 BAE2023 Physical Properties of Biological Materials 1 Deformation due to applied forces varies widely

BAE2023 Physical Properties of Biological Materials

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Lecture 8 – Viscoelasticity andDeformation

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Deformation due to applied forces varies widely among different biomaterials

Depends on many factors• Rate of applied force• Previous loading• Moisture content• Biomaterial composition


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Normal stress: Force per unit area applied perpendicular to the planeNormal strain: Change in length per unit of length in the direction of the applied normal stress


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Modulus of elasticityLinear region of stress strain curve E = σ/εFor biomaterials: apparent E = σ/ε at any given point (secant method)

Tangent method: slope of stress/strain curve at any point



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Poisson’s Ratio, μ• When a material is compressed in one direction, it usually

tends to expand in the other two directions perpendicular to the direction of compression

• The Poisson ratio is the ratio of the fraction (or percent) of expansion divided by the fraction (or percent) of compression, for small values of these changes.

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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 in Z direction:εz = σz/Eεy = -μ·εz εx = -μ·εz

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Poisson’s RatioMulti-axial Compression

See equations in 4.2 page 117Maximum Poisson’s = 0.5 for incompressible materials to 0.0 for easily compressed materialsExamples:• Gelatin gel – 0.50• Soft rubber – 0.49• Cork – 0.0• Potato flesh – 0.45 – 0.49• Apple flesh - 0.21 – 0.29• Wood – 0.3 to 0.5More porous means smaller Poisson’s Ratio2/4/2011


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Shearing StressesShear Stress: Force per unit area acting in the direction parallel to the surface of the plane, τShear 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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Shear Modulus:Ratio of shear stress to shear strain G = τ/γ

Measured with parallel plate shear test

(pg. 119)

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Example Problem

The bottom surface (8 cm x 12 cm) of a rectangular blockof cheese (8 cm wide, 12 cm long, 3 cm thick) is clampedin a cheese grater.• The grating mechanism moving across the top surface ofthe 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 theupper surface, estimate the lateral movement of the uppersurface w/respect to the lower surface

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Stresses and Strains: Described as Deviatoric or Dilitational

Dilitational: Causes change in volume

Deviatoric: Causes change in shape but negligiblechanges in volume

Bulk Modulus, K: describes response of solid todilitational stresses2/4/2011


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Dilatation: (Vf – V0)/V0

Δ V = Vf – V0

K = ΔP/(Δ V / V0)ΔP = Average normal stress, uniform hydrostatic gaugepressureK = average normal stress/dilatationV is negative, so K is negative

•Example of importance: K (Soybean oil) > K (diesel)•Will effect the timing in an engine burning biodiesel2/4/2011


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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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Stress strain curve for uniaxial compression of cylindrical sample of food product

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Stress-Strain Diagram, pg. 122Toughness: Area under curve until it fails Bio yield point: Failure point Resilience: Area under the unloading curve• Resilient materials “spring back”…all energy isrecovered upon unloadingHysteresis: strain density – resilienceFigure 4.6, page 124Figure 4.7, page 1252/4/2011


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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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Effect of water potential and strain rate on stress-strain curve of cylindrical Ida Red apple tissue

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Stress Relaxation: Figure 4.10 pg 129Material is deformed to a fixed strain and strain is held constant…stress required to hold strain constant decreases with time.

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Creep: Figure 4.11 pg. 130

A continual increase in deformation (strain) with time with constant load

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Tensile Testing• Not as common as compression testing• Harder to do

See figure 4.12 page 132

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Bending E=modulus of elasticityD=deflection F=forceI = moment of inertia E=L3(48DI)-1

I=bh3/12

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Bending• 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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Contact Stresses (handout from Mohsenin book)

Hertz Problem of Contact StressesImportance:“In ag. products the Hertz method can beused to determine the contact forces anddisplacements of individual units”

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Contact Stresses 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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Contact Stresses

Maximum contact stress occurs at the center of the surface of contact

a and b are the major and minor semi axes the elliptic contact area

For ag. Products, consider bottom 2 figures in Figure 6.1

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HW Assignment Due 2/11

Problem 1:An apple is cut in a cylindrical shape 28.7 mm indiameter and 22.3 mm in height. Using an InstronUniversal Testing Machine, the apple cylinder iscompressed. The travel distance of the compression head of the Instron is 3.9 mm. The load cell records a force of 425.5 N. Calculate the stress εz , and strain σz on the apple cylinder.

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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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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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Problem 4:•Sam, the strawberry producer, has had complaints from the produce companythat his strawberries are damaged during transit. Sam would like to know theforce required to damage the strawberries if they are stacked three deep in theircontainer.•The damage occurs on the bottom layer at the interface with the parallel surfaceof the container and also at the point of contact between the layers ofstrawberries.•An hydrostatic bulk compression test on a sample of Sam’s strawberries indicatesan average bulk modulus of 225 psi. Testing of specimens from Sam’sstrawberry crop shows a compression modulus E of 200 psi. The averagestrawberry diameter is 1.25 inches and the axial deformation due to the damagein transit averages 0.23 inches. The modulus of elasticity for Sam’s variety ofstrawberries is reported to be 130 psi.•Estimate the force Sam’s strawberries may be encountering during transit. (Hertzmethod)

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Documents

Lecture 8 – Viscoelasticity and Deformation 2/4/2011 BAE2023 Physical Properties of Biological Materials 1 Deformation due to applied forces varies widely