HW#5 Due 2/13 (Friday) Lab #1 Due 2/18 (Next Wednesday) For Friday 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 = ΔP, uniform hydrostatic gauge pressure ΔV = V f V 0 So: K = ΔP/(ΔV / V 0 ) 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 9
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 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: 13 FA a = 0.721, a = contact area diameter 1 d1+ 1 d2 ( 1 + 1 ) 2 F d1 d 2 Smax = 0.918, S 2 max = max contactstress A ( ) 13 2 2 1.04 1 1 1 2, combined deformation 13 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 ) 32 2 0.338K F 1 1 1 =, R radius of curvature 32 + = D R1 R 1 12 for steel spherical indentor: E ( μ ) 32 2 0.338K F 1 1 1 4 = 32 + + D R1 R 1 d2 12 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