Mechanical Metallurgy : Response of metals to forces or loads

Transkript

1 Mechanical Metallurgy : Response of metals to forces or loads Mechanical assessment of Materials Structural materials Machine, aircraft, ship, car etc We need to know limiting values of which materials in service can withstand without failure. Forming of metals into useful shapes Forging, rolling, extrusion, drawing, machining, etc We need to know conditions of load and temperature to minimize the forces that are needed to deform metal without failure. What is failure?

2 Strength of materials Strength of materials deals with relationships between external loads which act on some part of a body (member) in equilibrium. internal resisting forces deformation In equilibrium condition, if there are external forces acting on the member, there will be internal forces resisting the action of the external forces. The internal resisting forces are usually expressed by the stress acting over a certain area, so that the internal force is the integral of the stress times the differential area over which it acts.

3 Basic Assumptions Continuous: Homogeneous: Isotropic: No voids or empty spaces. Has identical properties at all points. Has similar properties in all directions or orientation. Microscopic scale, metals are made up of an aggregate of crystal grains having different properties in different crystallographic directions. However, these crystal grains are very small, and therefore the properties are homogenous in the micrometers macroscopic scale. Macroscopic scale, engineering materials such as steel, cast iron, aluminum seems to be continuous, homogeneous and isotropic.

4 Stress - Strain (Average strain) (Average stress, Pa, MPa) True strain : True stress :

5 Bonding Forces and Energies

6 Elastik Modulus Elastic modulus depends on the microstructure and interatomic bonding forces. Modulus of Elasticity Values for Several Metals at Various CrystallographicOrientations (GPa) Metal [100] [110] [111] Aluminum Copper Iron Tungsten Copper

7 Elastik Modulus Hooke s Law : Measuring Young s modulus: F = k x ; k: spring constant E : modulus of elasticity or Young s modulus. E is resistance of a material to elastic deformation. E is a material property. (Pa, GPa) Ceramics, glasses,: (GPa) Melting Temp. (C) Diamond (C) 1000 Tungsten Carbide (WC) Silicon Carbide (SiC) 450 Aluminum Oxide (Al 2 O 3 ) Berylium Oxide (BeO) 380 Magnesium Oxide (MgO) 250 Zirconium Oxide (ZrO) Mullite (Al 6 Si 2 O 13 ) 145 Silicon (Si) 107 Silica glass (SiO 2 ) 94 Soda-lime glass (Na 2 O - SiO 2 ) 69 Metals: Tungsten (W) Chromium (Cr) Berylium (Be) Nickel (Ni) 214 Iron (Fe) Low Alloy Steels Stainless Steels Cast Irons Copper (Cu) Titanium (Ti) 116 Brasses and Bronzes Aluminum (Al) Acoustic methods, Resonant Frequency Method Polymers: Polyimides 3-5 Polyesters 1-5 Nylon 2-4 Polystryene Polyethylene Rubbers

8 Young s Moduli: Comparison E(GPa) 10 9 Pa Metals Alloys Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin Graphite Ceramics Semicond Diamond Si carbide Al oxide Si nitride <111> Si crystal <100> Glass - soda Concrete G raphite Polymers Polyester PET PS PC Composites /fibers Carbon fibers only C FRE( fibers)* A ramid fibers only A FRE( fibers)* Glass fibers only G FRE( fibers)* GFRE* CFRE * G FRE( fibers)* C FRE( fibers) * AFRE( fibers) * Epoxy only Based on data in Table B2, Callister 7e. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers PP HDP E PTF E Wood( grain) 0.2 LDPE

9 Deformations Under Axial Loading From Hooke s Law: E E P AE From the definition of strain: L Equating and solving for the deformation, PL AE With variations in loading, cross-section or material properties, PL i i i AE i i

10 Resonant Frequency Method (Hooke s Law) l A (m: mass) In our lab we use resonant frequency method in flexure (bending) mod.

11 Resilience, U r Ability of a material to store energy elastically U r 0 y d Energy per unit volume If we assume a linear stress-strain curve this simplifies to Schematic representation showing how modulus of resilience (corresponding to the shaded area) is determined from the tensile stress-strain behavior of a material. U 1 2 y y Adapted from Fig. 6.15, Callister 7e.

12 Resilience, U r Comparison of stress-strain curves for high and low resilient materials.

13 Example: A bar of a material with Young s modulus, E, length, L, and cross sectional area, A, is subjected to an axial load, P. Derive an expression for strain energy stored in the bar assuming linear elastic deformation. Solution 1: Energy stored per unit volume. Multiply by volume, AxL Replace = /E Notice =P/A Solution 2: Energy stored in a spring: Remember:

14 Multi axial stress

15 Multi axial stress Notice : magnitude of stress is negative

16 Multi axial stress Pressurized tank (photo courtesy P.M. Anderson) q > 0 z > 0

17 Multi axial stress

18 Multi axial stress

19 Multi axial stress

20 Plane Stress

21 Multiaxial Stress-Strain Nominal tensile strain Nominal lateral strain Poisson s ratio : Engineering shear strain for small strains G: Shear modulus, Pa, GPa

22 Poisson s Ratio In an isotropic material, ε x is equal to ε y. The value of n for two extreme cases: (1) when the volume remains constant the initial and final volumes, V 0 and V, are equal V = V 0 [(1 + ε x) (1 + ε z) (1 + ε z)] Neglecting the cross products of the strains, because they are orders of magnitude smaller than the strains themselves V = V 0 [1 + ε x + ε y + ε z] Since V=V 0, ε x + ε y + ε z= 0 For constant volume deformation.

23 Poisson s Ratio For the isotropic case, the two lateral contractions are the same (ε x =ε y ). Hence, 2ε x = - ε z Remember n = 0.5 For constant volume deformation. (2) when there is no lateral contraction n = 0 Teorik sınırlar

24 Multiaxial Stress-Strain

25 Elastic Constants For an isotropic material two elastic constant are required and enough to define elastic behaviour Hydrostatic or mean stress Volumetric strain (D=DV/ V 0 ) D = Bulk modulus