Transcript Lecture 8 * Viscoelasticity and Deformation

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
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Lecture 8 – Viscoelasticity and
Deformation
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
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Lecture 8 – Viscoelasticity and
Deformation
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
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Lecture 8 – Viscoelasticity and
Deformation
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/E
εy = -μ·εz
εx = -μ·εz
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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 smaller Poisson’s Ratio
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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
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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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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 lateral movement of the upper
surface w/respect to the lower surface
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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
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Lecture 8 – Viscoelasticity and
Deformation
Dilatation: (Vf – V0)/V0
Δ V = Vf – V 0
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
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Lecture 8 – Viscoelasticity and
Deformation
Apples compress easier than potatoes so they
have a smaller bulk modulus, K (pg. 120) but
larger bulk compressibility
-1
K =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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Lecture 8 – Viscoelasticity and
Deformation
Stress-Strain Diagram, pg. 122
Toughness: Area under curve until it fails
Bio yield point: Failure 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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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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Effect of water potential and strain rate on stressstrain curve of cylindrical Ida Red apple tissue
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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.
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Lecture 8 – Viscoelasticity and
Deformation
Creep: Figure 4.11 pg. 130
A continual increase in deformation (strain) with time
with constant load
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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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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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Lecture 8 – Viscoelasticity and
Deformation
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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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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Lecture 8 – Viscoelasticity and
Deformation
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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Lecture 8 – Viscoelasticity and
Deformation
Contact Stresses
𝑆𝑚𝑎𝑥 =
3
𝐹
2 Π𝑎𝑏
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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Lecture 8 – Viscoelasticity and
Deformation
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Lecture 8 – Viscoelasticity and
Deformation
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Lecture 8 – Viscoelasticity and
Deformation
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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.
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HW Assignment Due 2/11
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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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.
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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)
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