Lecture 8  Viscoelasticity and Deformation

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Lecture 8 – Viscoelasticity and Deformation. Deformation due to applied forces varies widely among different biomaterials Depends on many factors Rate of applied force Previous loading Moisture content Biomaterial composition. Lecture 8 – Viscoelasticity and Deformation.

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Lecture 8 – Viscoelasticity andDeformation
• Deformation due to applied forces varies widely among different biomaterials
• Depends on many factors
• Rate of applied force
• Moisture content
• Biomaterial composition

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Normal stress: Force per unit area applied perpendicular to the plane

Normal strain: Change in length per unit of length in the direction of the applied normal stress

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Modulus of elasticity

Linear region of stress strain curve

E = σ/ε

For biomaterials: apparent E = σ/ε at any given point (secant method)

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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.

BAE2023 Physical Properties of Biological Materials

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/E

εy = -μ·εz

εx = -μ·εz

BAE2023 Physical Properties of Biological Materials

Poisson’s Ratio

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 - 0.21 – 0.29
• Wood – 0.3 to 0.5

More porous means smallerPoisson’s Ratio

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Shear Modulus:

Ratio of shear stress to shear strain

G = τ/γ

Measured with parallel

plate shear test

(pg. 119)

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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 lateral movement of the upper

surface w/respect to the lower surface

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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

dilitationalstresses

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Dilatation: (Vf – V0)/V0

Δ V = Vf – V0

K = ΔP/(Δ V / V0)

ΔP = Average normal stress, uniform hydrostatic gauge

pressure

K = average normal stress/dilatation

V is negative, so K is negative

•Example of importance: K (Soybean oil) > K (diesel)

•Will effect the timing in an engine burning biodiesel

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Apples compress easier than potatoes so they have a smaller bulk modulus, K (pg. 120) but larger bulk compressibility

K-1=bulk compressibility

• Sharp drop in curve = failure

BAE2023 Physical Properties of Biological Materials

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Stress-Strain Diagram, pg. 122

Toughness: Area under curve until it fails

Bio yield point: Failure point

• Resilient materials “spring back”…all energy is

Hysteresis: strain density – resilience

Figure 4.6, page 124

Figure 4.7, page 125

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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

Fig. 4.9

BAE2023 Physical Properties of Biological Materials

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

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.

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Creep: Figure 4.11 pg. 130

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Tensile Testing

• Not as common

as compression testing

• Harder to do

See figure 4.12 page 132

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Bending

E=modulus of elasticity

D=deflection

F=force

I = moment of inertia

E=L3(48DI)-1

I=bh3/12

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Bending

• Can be used for testing critical tensile stress at failure
• Max tensile stress occurs at bottom surface of beam

σmax=3FL/(2bh2)

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Contact Stresses (handout from Mohseninbook)

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”

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Contact Stresses

Assumptions:

• Material is homogeneous
• 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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

Contact Stresses

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

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

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

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

BAE2023 Physical Properties of Biological Materials

Lecture 8 – Viscoelasticity andDeformation

BAE2023 Physical Properties of Biological Materials

HW Assignment Due 2/11

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 425.5 N. Calculate the stress εz , and strain σz on the apple cylinder.

BAE2023 Physical Properties of Biological Materials

HW Assignment Due 2/11

Problem 2:

A sample of freshly harvested miscanthusis 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?

BAE2023 Physical Properties of Biological Materials

HW Assignment Due 2/11

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.

BAE2023 Physical Properties of Biological Materials

HW Assignment Due 2/11

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)

BAE2023 Physical Properties of Biological Materials