Read: pg 130 168 (rest of Chpt. 4) 1
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: ε z = σ z /E, ε y = -µ ε z and ε x = -µ ε y 2
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 3
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 4
Shear modulus: ratio of shear stress to shear strain, G = τ/γ Measured with parallel plate shear test (pg. 119) 5
6
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. 7
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: (V f V 0 )/V 0 8
K = average normal stress/dilatation Dilatation: (V f V 0 )/V 0 Average normal stress = P, uniform hydrostatic gauge pressure V = V f V 0 So: K = P/( V / V 0 ) V is negative, so K is negative Example of importance: K (Soybean oil) > K (diesel) Will effect the timing in an engine burning biodiesel 9
Apples compress easier than potatoes so they have a smaller bulk modulus, K (pg. 120) but larger bulk compressibility K -1 Lecture 8 Viscoelasticity and =bulk compressibility Strain energy density: area under the loading curve of stress-strain diagram Sharp drop in curve = failure 10
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 11
Factors Affecting Force- 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
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 13
Tensile testing Not as common as compression testing Harder to do See figure 4.12 page 132 14
Tensile testing 15
Bending: E=modulus of elasticity D=deflection, F=force, I = moment of inertia E=L 3 (48DI) -1 I=bh 3 /12 16
Can be used for testing critical tensile stress at failure Max tensile stress occurs at bottom surface of beam σ max =3FL/(2bh 2 ) 17
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 18
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 19
S max 3 F = 2 π ab 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 20
In the case of 2 contact spheres, pg 354: 1 3 FA a = 0.721, a = contact area diameter 1 d1 + 1 d2 ( 1 + 1 ) 2 F d 1 d 2 Smax = 0.918, S 2 max = max contact stress A ( ) 1 3 2 2 1.04 1 1 1 2, combined deformation 1 3 D = F A d + d D = at point of contact 21
E Lecture 8 Viscoelasticity and To determine the elastic modulus, E for steel flat plate: ( µ ) 3 2 2 0.338K F 1 1 1 =, R radius of curvature 3 2 + = D R1 R 1 1 2 for steel spherical indentor: E ( µ ) 3 2 2 0.338K F 1 1 1 4 = 3 2 + + D R1 R 1 d2 1 2 22
1 E µ = 1, µ = 2 3k Poisson's Ratio (ratio of transverse contraction strain to longitudinal extension strain in the direction of stretching force), k=bulk modulus K=1.3514 for cost=0, and R 1 = R 1 23
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 ε z, and strain σ z on the apple cylinder. Biological Materials 24
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? Biological Materials 25
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. Biological Materials 26
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) Biological Materials 27