Mechanical Properties

Transcription

1 Stress-strain behavior of metals Elastic Deformation Plastic Deformation Ductility, Resilience and Toughness Hardness 108 Elastic Deformation bonds stretch δ return to initial Elastic means reversible! Linearelastic δ Non-Linearelastic 109

2 Plastic Deformation bonds stretch & planes shear planes still sheared δelastic + plastic δplastic Plastic means permanent! linear elastic δplastic linear elastic δ 110 Engineering Stress Tensile stress, σ: Shear stress, τ: t t Area, A Area, A s σ= t A o original area before loading t τ= s A o Stress has units: N/m 2 or lb/in 2 s t 111

3 Common States of Stress Simple tension: cable Ao = cross sectional Area (when unloaded) σ = A o σ σ Simple shear: drive shaft A c M s A o M 2R Ski lift (photo courtesy P.M. Anderson) τ = τ s Ao Note: τ = M/A c R here. 112 Common States of Stress (con.) Simple compression: A o Canyon Bridge, Los Alamos, NM (photo courtesy P.M. Anderson) Balanced Rock, Arches National Park (photo courtesy P.M. Anderson) σ = Ao Note: compressive structure member (σ < 0 here). 113

4 Common States of Stress (con.) Bi-axial tension: Hydrostatic compression: Pressurized tank (photo courtesy P.M. Anderson) σθ > 0 σz > 0 ish under water (photo courtesy P.M. Anderson) σ h < Engineering Strain Tensile strain: δ/2 Lateral strain: ε= δ L o ε L = δ L w o L o δ/2 Shear strain: θ/2 δ L /2 δ L /2 w o π/2 - θ γ = tan θ Strain is always dimensionless. π/2 θ/2 115

5 Stress-Strain Testing 116 Typical Engineering Stress-Strain Behavior of Metal 117

6 True Stress and Strain σ ε T T = A i l = ln l i o σ T = σ(1+ε) ε T = ln(1+ε) 118 Linear Elastic Properties Modulus of Elasticity, E: (also known as Young's modulus) Hooke's Law : Poisson's ratio, ν: metals: n ~ 0.33 ceramics: ~0.25 polymers: ~0.40 Units: E: [GPa] or [psi] ν: dimensionless σ = E ε σ ν = εl ε ε L ε -ν 1 E 1 Linearelastic ε simple tension test 119

7 Linear Elastic Properties (con.) Elastic Shear modulus, G: τ = G γ τ 1 G γ M M simple torsion test 120 Simple tension test: tensile stress, σ Plastic Deformation (at lower temperatures, T < T melt /3) Elastic+Plastic at larger stress Elastic initially permanent (plastic) after load is removed εp engineering strain, ε plastic strain 121

8 Yield Strength, σ y Stress at which noticeable plastic deformation has occurred. σy tensile stress, σ when εp = εp = engineering strain, ε 122 Tensile Strength, TS Maximum possible engineering stress in tension. Necking formed TS engineering stress Typical response of a metal strain 123

9 Ductility Plastic tensile strain at failure: Engineering tensile stress, σ smaller %EL (brittle if %EL<5%) larger %EL (ductile if %EL>5%) L f Lo % EL = 100 L o Ao Lo A f Lf Engineering tensile strain, ε Another ductility measure: Ao Af % RA = 100 Ao Note: %RA and %EL are often comparable. --Reason: crystal slip does not change material volume. --%RA > %EL possible if internal voids form in neck. 124 Resilience Energy to deform elastically and recover when unloading. Approximate by the area under the stress-strain curve up to the point of yielding. Modulus of Resilience: U r = ½ σ y ε y 125

10 Toughness Energy to break a unit volume of material. Approximate by the area under the stress-strain curve. Engineering tensile stress, σ smaller toughness (ceramics) larger toughness (metals, PMCs) smaller toughnessunreinforced polymers Engineering tensile strain, ε 126 Hardness Resistance to permanently indenting the surface. Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. e.g., 10mm sphere apply known force (1 to 1000g) measure size of indent after removing load most plastics D brasses Al alloys d easy to machine steels file hard Smaller indents mean larger hardness. cutting tools nitrided steels diamond increasing hardness 127

11 Hardening An increase in σy due to plastic deformation. σ σ y 1 σ y 0 large hardening small hardening unload reload ε 128 Design/Safety actors Design uncertainties mean we do not push the limit. actor of safety, N σ working = σ y N Often N is between 1.2 and 4 Ex: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of ,000N π d 2 /4 σ working 5 = σ y N 1045 plain carbon steel: σy=310mpa TS=565MPa = 220,000N d Lo 129

12 Summary Stress and strain: These are size-independent measures of load and displacement, respectively. Elastic behavior: This reversible behavior often shows a linear relation between stress and strain. To minimize deformation, select a material with a large elastic modulus (E or G). Plastic behavior: This permanent deformation behavior occurs when the tensile (or compressive) uniaxial stress reaches sy. Toughness: The energy needed to break a unit volume of material. Ductility: The plastic strain at failure. 130 Example: Problem 6.24 A cylindrical rod 380 mm long having a diameter of 10 mm is to be subjected to a tensile load. If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm when the applied load is 24,500 N which of the four metals or alloys listed below are possible candidates? Justify your choice(s). Material Modulus of Elasticity (GPa) Yield Strength (MPa) Tensile Strength (MPa) Aluminum Alloy Brass Copper Steel alloy

13 irst to compute the stress: σ = 24,500 N = = 312 MPa A o π 10 x m 2 Of the metal alloys listed, only brass and steel have yield strengths greater than this stress. Next to compute the elongation produced in brass and steel: or brass, or steel, l = σl o E l = σ l o E (312 MPa)(380 mm) = 100 x 10 3 = 1.19 mm MPa (312 MPa)(380 mm) = 207 x 10 3 = 0.57 mm MPa Therefore, of these four alloys, only steel satisfies the stipulated criteria. 132