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 3:30 122 Ag Hall 2/04/08 BAE2023 Physical Properties of Biological Materials Lecture 8

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

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 2/04/08 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 -78 MPa vs. -3.6 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 2.375 inches and the axial deformation for the whole oranges was 0.019 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

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 1.125 inches. The axial deformation at 8 lb is 0.025 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