Chapter 3 Mechanical Properties of Materials What do they tell us? Relationship between stress and strain. How do we find them? Uniaxial stress/strain testing. Stress-strain characteristic is inherent to the material

Test Specimens Need to follow some standard procedure Concept of engineering stress/true stress σengstress = P/A0 A0  Original undeformed area of cross section σtruestress = P/A A = deformed area of cross section

Testing set-up

Videos Tensile Test MaterialsScience2000 http://www.youtube.com/watch?v=hD_NJaZIpT0&feature=related http://www.youtube.com/watch?v=VgE7TaXuUqI&feature=related http://www.youtube.com/watch?v=rzgumWPB_zc

Stress-Strain Diagrams Nominal Stress: σ = P/A0 Where A0 is original cross section Nominal Strain: ε = δ/L0 Where L0 is the original gauge length and δ is the change in gauge length No two stress-strain diagrams for a particular material will be exactly the same since the results depend on: material’s composition microscopic imperfections, way material is manufactured rate of loading temperature

Stress/strain curve

Testing

Necking

Stress/strain diagram for mild steel

Ductile/brittle behaviour A ductile material is subjected to large strains before it ruptures Two measures for ductility Percent elongation =((Lf – L0)/ L0)) x (100%) For mild steel this value is 38% Percent reduction in area = ((A0 – Af)/ A0)) x(100%) For mild steel this value is 60%

Concept of offset method for yield strength When yield point is not well defined (☺ ?)  use 0.2% strain criterion

Yield Strength Yield Strength is not a physical property of the material. We will use approach that Yield strength Yield point Elastic limit Proportional limit All coincide unless otherwise stated

What about rubber? No yield point Non-linear elastic behavior

Exhibit no yielding before failure Brittle materials Exhibit no yielding before failure

Grey Cast Iron

Concrete Reinforced with steel to give the tensile resistance

Effect of temperature

3.4 Hooke’s Law Linear elastic behavior leads to: σ = E ε (Hooke’s law) E = Young’s modulus (Thomas Young-1807) (aka: modulus of elasticity) E = (σpl / εpl) Typically E = 210 GPa (steel); 70 GPa(aluminium)

Modulus of Elasticity Modulus of Elasticity (E) indicates stiffness of a material. Stiff material -> large E (e.g. steel: E = 200 GPa) Spongy material -> small E (for vulcanized rubber, E = 0.70 MPa)

Strain Hardening Strain hardening is used to establish a higher yield point for a material The modulus of elasticity stays the same. The ductility decreases.

Strain Energy Density (u) u = (1/2) (σ2 / E) (linear elastic) Modulus of Resilience (ur) ur = (1/2) σpl εpl = (1/2) (σ2pl / E) =ability to absorb energy w/o permanent damage Modulus of Toughness  entire area under the stress-strain curve =total energy absorbed before fracture

Strength, Toughness and Ductility Strength ~ related to the height of the curve Ductility ~ related to the width of the curve Toughness ~ related to area under the curve

Chapter 3 Lecture Example 1 A tension test for a steel alloy results in the stress-strain diagram shown. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify on the graph the ultimate stress and the fracture stress.

Chapter 3 Lecture Example 1

Chapter 3 Lecture Example 2 The stress-strain diagram for an aluminum alloy that is used for making aircraft parts is shown. If a specimen of this material is stressed to 600 MPa, determine the permanent strain that remains in the specimen when the load is released. Also, find the modulus of resilience both before and after the load application.

Don’t cover in detail in class Don’t cover in detail in class. Mention: same force, different A, different stress. AB elastic, BC yields !

Poisson’s Ratio  A tensile force causes a deformable body to longitudinally elongate and to laterally contract. εlong = ( δ/L) εlat = ( δ’/r)

Tension

Compression

Poisson’s Ratio

Poisson’s Ratio  Value of  is positive Value of  same in tension and compression Typical value of  ~ 0.25, Range is between 0 and 0.5  constant only in the elastic range Only Longitudinal force is acting to cause the lateral strain.

Lecture Example 3: A steel pipe (Stainless 304) has a length of 1 Lecture Example 3: A steel pipe (Stainless 304) has a length of 1.5 m and inside and outside diameters of di=12 cm and do=15cm. It is compressed by an axial force P=500 kN. Before you calculate, make predictions a. the pipe gets longer/shorter. b. The pipe’s outer diameter increases/decreases/stays the same. c. The pipe’s wall thickness increases/decreases/stays the same. Now find the change in length of the pipe, the lateral strain, the new inner and outer diameters and the change in wall thickness.

Shear stress-strain behavior Can be determined by subjecting circular tubes to torsional loading

Shear Stress –Strain Diagram

Modulus of Rigidity – linear region G – how much a material resists twisting

Lecture Example 4: Determine the shear modulus, G, the proportional limit, and the ultimate shear stress. Find the maximum d where the material behaves elastically. What is the magnitude of V to cause d?

Bending test

Bending test

Impact Test

Failure of turbine blade due to creep Failure of steam pipe due to creep

Creep Creep is the time-related deformation of a material for which temperature and stress play an important role. Creep results from sustained loading below the measured yield point. Members are designed to resist the effects of creep based on their creep strength, which is the highest amount of stress a member can withstand during a specified amount of time without experiencing creep strain.

Fatigue Fatigue occurs in metals when stress or strain is cycled. It causes a brittle fracture to occur. Engineering Design: stress in the member should not exceed the material’s endurance or fatigue limit. This the maximum stress that the member can resist when enduring a specified number of cycles.

Material Properties Strength – Capacity to resist loads – yield stress. Higher y, higher strength. Resilience – Measure of energy absorbed without permanent damage . Toughness – Measure of energy absorbed before fracturing. Ductility – Property of material which allows large deformation before fracture. Brittleness – Property of material which allows little or no yielding before fracture. Endurance – Ability to sustain cyclic loads. Stiffness – Mechanical property indicated by E. Higher E means stiff material. Steeper slope in the Stress – Strain diagram. Rigidity - Mechanical property of material indicated by G.

3.27 A short cylindrical block of 2014-T6 Aluminum having an original diameter of 0.5 in and an original length of 1.5 in is placed in the smooth jaws of a vise and a compressive force of 800 lb is applied. Determine (a) The decrease in length. (b) The new diameter.