Read: pg 130 – 168 (rest of Chpt. 4) Lecture 8 – Viscoelasticity and Deformation Read: pg 130 – 168 (rest of Chpt. 4) 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 - 0.21 – 0.29 Wood – 0.3 to 0.5 More porous means smaller Poisson’s 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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) 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

Lecture 8 – Viscoelasticity and Deformation 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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. 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

Lecture 8 – Viscoelasticity and Deformation Tensile testing Not as common as compression testing Harder to do See figure 4.12 page 132 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

Lecture 8 – Viscoelasticity and Deformation Tensile testing 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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) 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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” 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

In the case of 2 contact spheres, pg 354: Lecture 8 – Viscoelasticity and Deformation In the case of 2 contact spheres, pg 354: 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

Lecture 8 – Viscoelasticity and Deformation To determine the elastic modulus, E… for steel flat plate: for steel spherical indentor: 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

Lecture 8 – Viscoelasticity and Deformation 12/7/2018 BAE2023 Physical Properties of Biological Materials Lecture 8

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 425.5 N. Calculate the stress εz , and strain σz on the apple cylinder. 12/7/2018 BAE2023 Physical Properties of Biological Materials

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? 12/7/2018 BAE2023 Physical Properties of Biological Materials

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. 12/7/2018 BAE2023 Physical Properties of Biological Materials

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