ME 350 – Lecture 2 – Chapter 3

ME 350 – Lecture 2 – Chapter 3 Mechanical Properties of Materials • Stress‑Strain Relationships • Tensile, compressive, and shear • Hardness • Effect of Temperature on Properties • Fluid Properties • Pseudoplastic and Viscoelastic Behavior of Polymers

Tensile Test Figure 3.2 Example tensile test: (1) no load; (2) uniform elongation and reduction of cross‑sectional area; (3) continued elongation, maximum load reached; (4) necking begins, load begins to decrease; and (5) fracture. If pieces are put back together as in (6), final length can be measured.

Stress & Strain • Stress is defined as force divided by area: Where σ = stress, F = applied force, and A = instantaneous or initial cross-sectional area Engineering stress: A = True stress: A = • Strain is defined at any point in the test as where e = engineering strain; ε = true strain; L = instantaneous length at any point during elongation; and Lo = original gage length

Typical Stress-Strain Plot Stress, σ

True Stress-Strain • Stress increases until necking begins • Metal becomes stronger as strain increases, known as

True Stress-Strain in Log-Log Plot Flow Curve: Experimentally, a higher value of “n” means that the metal can be strained further before the onset of necking

Conservation of Volume • Assume volume is conserved and area uniformly increases or decreases before “necking” (in a tensile test) or “barreling” (in a compressive stress). LfAf = L0A0

Problem 3.2 in Text p64 A test specimen has a gage length of 2.0 in and an area = 0.5 in2. During the test the specimen yields (0.2% yield pt.) under a load of 32,000 lb at a gage length of 2.0083 in. The max load of 60,000 lb is reached at a gage length = 2.60 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) the tensile strength. (a) Y = (b) ε = E = • Af= TS =

Problem 3.5 in Text p64 In a tensile test on a metal specimen, strain = 0.08 at a stress = 265 MPa. When the stress = 325 MPa, the strain = 0.27. Determine (a) the strength coefficient and (b) the strain-hardening exponent in the flow curve equation. (b) (a)

Categories of Stress-Strain Relationship • Perfectly – fractures rather than yields • Elastic and perfectly – flow curve K = and n = (heated metals can behave like this) • Elastic and (K Y and n 0)

Cross section increases Instead of “necking” → “” wider in middle than top or bottom K, n, Y, and E values should be Stress-Strain Curve in Compression

Shear Stress and Strain L • Shear stress defined as where F = applied force; A = deflection area; T = applied torque; R = tube radius; and t = tube wall thickness • Shear strain defined as where δ = deflection; b = deflection distance; α = angular deflection (rad); and L = the gauge length of the tube R

Torsion Stress-Strain Curve In the elastic region where G = For most materials, G 0.4E, where E = elastic modulus In the plastic region, where S = coefficient For most materials S 0.7 TS (tensile strength) n = strain hardening exponent

Problem 3.24 in Text p65 In a torsion test, a torque of 5000 ft-lb is applied which causes an angular deflection = 1° on a thin-walled tubular specimen whose radius = 1.5 in, wall thickness = 0.10 in, and gage length = 2.0 in. Determine (a) the shear stress, (b) shear strain, and (c) shear modulus, assuming the specimen had not yet yielded. (a)

Problem 3.24 in Text p65 In Problem 3.24, the specimen fails at a torque = 8000 ft-lb and an angular deflection = 23°. Calculate the shear strength of the metal.

Hardness • Resistance to permanent indentation • High hardness generally means more resistance to scratching and wear (desirable for tooling, etc.) • Common test methods: • Metals and general purpose - and • Ceramics - Vickers & Knoop • Polymers -

Brinell Hardness Test • Widely used for testing metals and nonmetals of low to medium hardness • A hard ball is pressed into specimen surface with a load of 500, 1500, or 3000 kg • Load divided into indentation area = Brinell Hardness Number (BHN), F = indentation load, kg; Db = diameter of ball, mm, Di = diameter of indentation, mm

Effect of Temperature on Properties

Recrystallization in Metals • Most metals strain harden at room temperature according to the flow curve (n > 0) • But if heated to sufficiently high temperature (above the recrystallization temperature) and deformed, strain hardening does not occur • Instead, new grains are formed that are free of strain • The metal behaves as a material; that is, n = • Tr ≈ Tm (temp in Kelvin)

Recrystallization and Manufacturing • Recrystallization can be exploited in manufacturing • Heating a metal above its recrystallization temperature prior to deformation allows a greater amount of straining, and lower forces and power are required to perform the process • Forming metals at temperatures above recrystallization temperature is called

Fluid Properties and Manufacturing • Fluids flow - They take the shape of the container that holds them • Knowing fluidic properties are important • Examples: • Metals are cast in molten state • Glass is formed in a heated and fluid state • Polymers are almost always shaped as fluids

Viscosity in Fluids • Viscosity is the resistance to flow that is characteristic of a given fluid • Viscosity is a measure of the internal friction when velocity gradients are present in the fluid • The more viscous the fluid, the the internal friction and the greater the resistance to flow • Reciprocal of viscosity is ‑ the ease with which a fluid flows

Viscosity • Viscosity can be defined using two moving parallel plates • Viscosity is the ratio of shear stress to shear rate: • For Newtonian fluids, viscosity is a • For non-Newtonian fluids, it is not • 1 milli-Pascal-second (mPa∙s) = 1 (cP) Shear Stress: Shear Rate Viscosity:

Problem 3.28 in Text p66 Two flat plates, separated by a space of 4 mm, are moving relative to each other at a velocity of 5 m/sec. The space between them is occupied by a fluid of unknown viscosity. The motion of the plates is resisted by a shear stress of 10 Pa due to the viscosity of the fluid. Assuming that the velocity gradient of the fluid is constant, determine the coefficient of viscosity of the fluid.

Viscosity of Polymers and Flow Rate • Viscosity of a thermoplastic polymer melt is not constant • It is affected by flow rate • Its behavior is non-Newtonian • A fluid that exhibits this decreasing viscosity with increasing shear rate is called • This complicates analysis of polymer shaping processes such as injection molding

Newtonian versus Pseudoplastic Fluids • Figure 3.18 Viscous behaviors of Newtonian and pseudoplastic fluids. Polymer melts exhibit pseudoplastic behavior – with increased shear rate, viscosity decreases

Viscoelastic Behavior • A temperature and strain rate dependent material property that is a combination of viscosity and elasticity • Viscous – solid exhibits a plastic deformation to a long duration small stress where material “flows” or • Elastic –liquid exhibits an elastic response to a short duration high stress • Examples: aluminum, synthetic polymers, wood… • Viscous – permanent deformation of a cheap soccer ball sitting in the sun • Elastic – liquid polymer “swells” coming from extruder

Elastic versus Viscoelastic Behavior (a) perfectly elastic response of material to stress applied over time; (b) viscoelastic response as a function of time exhibiting material “creep” and stress–strain graph hysteresis

Quotes • Think big dreams, but relish small pleasures. • Listen, opportunity sometimes knocks very softly. • Strive for excellence, not perfection. • Be tough minded, but tenderhearted. • Never underestimate your power to change yourself.

ME 350 – Lecture 2 – Chapter 3