Mechanical properties of Materials

SESSION 1

Mechanical Properties of Materials Mechanical properties describe the response of the material to the applied force or torque. The mechanical behavior of a material reflects the relationship between its response or deformation to an applied load or force. Important mechanical properties are….. Elasticity Plasticity Ductility Strength Tensile strength Ultimate strength Hardness Creep Fracture Fatigue

Mechanical Properties of Materials Elasticity: It is the property of the material by virtue of which the material regains its original dimensions after the stress is removed. Ex: Steel, Spring, Plasticity: It is the property of the material, where the material has a permanent deformation after the stress is removed. Ex: Rubber band.

Mechanical Properties of Materials Ductility: It is the ease with which the materials can be drawn in the form of thin wires and sheets. Ex: Platinum, Iron, Copper, Aluminum and Zinc. Brittleness: It is the property of the material by virtue of which it will fracture without any appreciable deformation. Thus it is the opposite of ductility. Ex: glass

Mechanical Properties of Materials Strength: The ability of a material to sustain loads without failure. It refers to the stress up to which the body obeys Hooke’s law. It is load/unit area. There are two types of strengths. Yield strength: It refers to the strength beyond which it exhibits plasticity. Ultimate strength: It refers to the strength at which the material breaks or fractures.

Mechanical Properties of Materials Hardness: It refers to the ability of the material to withstand plastic deformation or indentations produced in the material. Creep: It refers to the changes that occur in the dimensions of the material as a function of time, when subjected to a constant load. Fracture: It refers to the breakage of a material into separate parts under the action of stress. Fatigue: It deals with the changes in the mechanical properties of the material under the action of a cyclical or repeated stress.

Mechanical Properties of Materials To understand and describe how materials deform (elongate, compress, twist) or break as a function of applied load, time, temperature and other conditions. We need first to discuss standard test methods and standard language for mechanical properties of materials.

Mechanical Properties of Materials Basic property of materials Ability to with stand loads with out failure Expressed in terms of load per unit area N/m2 of kN/m2 Stress(): Applied force per unit area. Stress = F/A, Units: Pa or N/m2 Strain(): The fractional change in the dimensions of the body. It is a ratio of change in length to original length.

CONCEPTS OF STRESS AND STRAIN If a load is static or changes relatively with time and is applied over a cross sectional surface of a material, the mechanical behavior can change due to following principles …… Tension Compression Shear OR Torsional These are also called “simple stress–strain test”

CONCEPTS OF STRESS AND STRAIN Tension Tests: One of the most common mechanical stress–strain tests is performed in tension. A tensile load produces an elongation and positive linear strain. Dashed lines represent the shape before deformation; solid lines, after deformation. A tensile stress occurs when equal and opposite forces are directed away from each other. F W Tension

Test conducted on Universal Testing Machine (UTM). Longitudinal Strain:-It is defined as the increase in length (Δl) per unit original length (l) when deformed by the external force. Longitudinal Strain= Δl l Test conducted on Universal Testing Machine (UTM). Also used for compression and shear stress. gauge length specimen extensometer

Tensile Strength, TS TS engineering stress strain engineering strain • Maximum stress on engineering stress-strain curve. y strain Typical response of a metal F = fracture or ultimate strength Neck – acts as stress concentrator engineering TS stress engineering strain • Metals: occurs when noticeable necking starts. • Polymers: occurs when polymer chains are aligned and about to break.

Tensile Strength If stress maintained specimen will break Fracture Strength

Compression Tests: A compressive load produces contraction and a negative linear strain. A compressive stress occurs when equal and opposite forces are directed toward each other. F W Compression Volumetric Strain:-It is defined as change in volume (ΔV) per unit original volume (v), when deformed by external force. Volume Strain= -ΔV v

Mechanical Properties of Materials Shear Tests: Shear force is applied parallel to the upper face, reactive force acts on the lower face in the opp. direction. This produces a change in the shape without any change of size. Torsional deformation (i.e angle of twist ϕ) produced by an applied torque T

Mechanical Properties of Materials Force per unit area applied parallel to the surface of a body trying to displace the upper layers of the body is called shearing stress. Shearing Stress or Tangential Stress = Force/Area F  l d = F/A0 When change takes place in the shape of the body, the strain is called shear strain. The shear strain () is defined as the tangent of the strain angle  Strain = tanΘ = d / l

STRESS-STRAIN BEHAVIOR Elastic deformation Reversible: when the stress is removed, the material returns to the dimensions it had before the loading. Usually strains are small (except for the case of some plastics, e.g. rubber).  Plastic deformation Irreversible: when the stress is removed, the material does not return to its original dimensions.

STRESS-STRAIN BEHAVIOR Elastic Deformation: In tensile tests, if the deformation is elastic, the stress-strain relationship is called Hooke's law: σ = E ε   E is Young's modulus or modulus of elasticity, has the same units as σ, N/m2 or Pa     Higher E →higher “stiffness” Slope=modulus of elasticity E stress strain load unload

Elastic Deformation:  F F  Elastic means reversible! bonds stretch 1. Initial 2. Small load 3. Unload F  bonds stretch return to initial F  Linear- elastic Non-Linear- Elastic means reversible!

Elastic Deformation: Nonlinear Elastic Behavior In some materials (many polymers, concrete...), elastic deformation is not linear, but it is still reversible.

Elastic Moduli α Bond strength Elastic Moduli α Bond strength. Elastic Moduli α 1/interatomic distance ( r ). Temperature α r α 1/Elastic Moduli Deformation : Amount of force required to change the dimension of the body.

Poisson’s ratio is defined as the ratio of the ELASTIC ELASTIC PROPERTIES OF MATERIALS When a tensile stress is imposed on a metal specimen, an elastic elongation and accompanying strain result in the direction of the applied stress ( z direction), As a result of this elongation, there will be constrictions in the lateral (x and y) directions perpendicular to the applied stress; from these contractions, the compressive strains and may be determined. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then a parameter termed Poisson’s ratio is defined as the ratio of the lateral and axial strains X he applied stress is uniaxial (only in the z direction), and the material is isotropic, th

Poissons Ratio Poissons Ratio = Lateral Strian/Longitudional Strain (Areal/Length) Δd/d/Δl/l Isotropic – when we apply stress,strain is produced in the body along two axes are same. Anisotropic- when we apply stress, strain produced in the body along the two axes is not same.

Elastic Deformation : Poisson’s ratio loaded unloaded

Elastic Deformation : Poisson’s ratio Materials subject to tension shrink laterally. Those subject to compression, bulge. The ratio of lateral and axial strains is called the Poisson's ratio ν. Sign in the above equations shows that lateral strain is in opposite sense to longitudinal strain. ν is dimensionless. Theoretical value for isotropic material: 0.25 Maximum value: 0.50, Typical value: 0.24 - 0.30 ν = − ε x / ε z= − ε y /ε z

Plastic deformation: F F  Plastic means permanent! plastic 1. Initial 2. Small load 3. Unload p lanes still sheared F  elastic + plastic bonds stretch & planes shear plastic F  linear elastic plastic Plastic means permanent!

Plastic(permanent) deformation: (at lower temperatures, i.e. T < Tmelt/3) • Simple tension test: Elastic+Plastic at larger stress engineering stress,  Elastic initially permanent (plastic) after load is removed p plastic strain engineering strain, 

Atomic Perspective of Plastic Deformation. Atomic perspective, plastic deformation corresponds to the breaking of bonds with original atom neighbors and then reforming bonds with new neighbors as large numbers of atoms or molecules move relative to one another; upon removal of the stress they do not return to their original positions. The mechanism of this deformation is different for crystalline and amorphous materials. For crystalline solids, deformation is accomplished by means of a process called slip, which involves the motion of dislocations. Plastic deformation in noncrystalline solids (as well as liquids) occurs by a viscous flow mechanism,

SESSION 2

TYPES OF MODULUS Within the proportional limit, the strain produced in a body is directly proportional to the applied stress Stress α Strain Stress = E Strain Where ‘E’ is a constant called the modulus of elasticity of the material

Young’s Modulus Units: Pa or N/m2 For materials whose length is much greater than the width or thickness, we are concerned with the longitudinal modulus of elasticity, or Young’s Modulus (Y). Units: Pa or N/m2

The bulk modulus is negative because of decrease in V. Not all deformations are linear. Sometimes an applied stress F/A results in a decrease of volume. In such cases, there is a bulk modulus B of elasticity. The bulk modulus is negative because of decrease in V.

Since F/A is generally pressure P, we may write: Units remain in Pascals (Pa)

A shearing stress alters only the shape of the body, leaving the volume unchanged. For example, consider equal and opposite shearing forces F acting on the cube below: A F  l d The shearing force F produces a shearing angle . The angle  is the strain and the stress is given by F/A as before.

Calculating Shear Modulus F  l d A Stress is force per unit area: The strain is the angle expressed in radians: The shear modulus S is defined as the ratio of the shearing stress F/A to the shearing strain : The shear modulus: Units is in Pascal.

Engineering and True Stress-Strain Diagrams The Engineering stress σn = F/A0 where F is the force and A0 the original area of cross section The Engineering strain, εn = (L-L0)/L0 where L is the length of the original gauge length under force F, and L0 is the original gauge length.

Engineering and True Stress-Strain Diagrams The stress is calculated based on the instantaneous area at any instant of load, then it is the true stress. Where, Ai is the actual area of the cross-section corresponding to load P.

Engineering and True Stress-Strain Diagrams Similarly, True strain Where, dℓ is the infinitesimal elongation, ℓi is the instantaneous length and ℓ0 is the original length. StraiStrain hardening reduces ductility, which increases the chances of brittle failure. n hardening reduces ductility, which increases the chances of brittle failure.

slope =Young’s (elastic) modulus yield strength Plastic region ultimate tensile strength 3 necking Slope=E Strain Hardening yield strength Fracture 5 Strain hardening reduces ductility, which increases the chances of brittle failure. 2 Elastic region slope =Young’s (elastic) modulus yield strength Plastic region ultimate tensile strength strain hardening fracture Plastic Region Stress (F/A) Elastic Region 4 1 Strain ( ) (L/Lo)

FRACTURE

Tensile properties: Ductility Ductility is a measure of the deformation at fracture. It is defined by percent of elongation %EL = (lf - l 0 )X100/ l 0 The yield and tensile strengths and modulus of elasticity decrease with increasing temperature, ductility increases with temperature.   Percentage reduction in area %RA = ( A0 – Af )X100/ A0

Toughness Toughness = the ability to absorb energy up to fracture = the total area under the strain-stress curve up to fracture Units: the energy per unit volume, e.g. J/m3 It can be measured by an impact test. A “tough” material has strength and ductility.

hardness Hardness is defined as the resistance of a material to (plastic) deformation or indentation. Hardness tests may be classified as 3 types. Scratch hardness test. (Moh’s hardness scale) 2) Indentation hardness test Brinell’s hardness test Rockwell’s hardness test Vickers hardness test 3) Rebound or dynamic hardness test. (Shore scleroscope test)

hardness The depth or size of indentation is measured. In the indentation test, a small indenter (sphere, cone, or pyramid) is forced into the surface of a material under conditions of controlled magnitude and rate of loading. The depth or size of indentation is measured. The tests somewhat approximate, but popular because they are easy and non-destructive (except for the small dent).

hardness Brinell Hardness Test: Rockwell Hardness Test: 2F BH = D[D- √(D2-Di2)] Rockwell Hardness Test:

hardness VICKERS HARDNESS TEST: Also known as Diamond Pyramid Hardness (DPH) Test VH = 1.854F/d2 , d = (d1+d2)/2

fatigue Fatigue is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading. Fatigue occurs when a material is subjected to repeated loading and unloading or fluctuating loads. Occurs in bridges, aircraft and machine components.

fatigue Reversed Stress cycle Repeated Stress cycle Irregular Stress cycle

R R Moore Reversed – Bending Fatigue Machine

fracture Separation of a body into pieces due to stress (i.e. Static or slowly changing with time), as temperatures below the melting point. There are four fractures. Brittle fracture(Ceramics ice, cold metals) Ductile fracture(most metals not too cold) Fatigue fracture Creep fracture

fracture Brittle fracture: A brittle fracture is the fracture which takes place by the rapid propagation of crack with a quite negligible plastic deformation. Ex: ceramics, ice, cold metals. Ductile fracture: Ductile fracture is the fracture which takes place by a slow propagation of crack with appreciable plastic deformation. Ex: most metals (not too cold). Fatigue fracture: It is the fracture which takes place under repeatedly applied fatigue stresses. Creep fracture: It is the fracture which takes place due to excessive creeping of materials under steady loading.

Problems 1.A specimen of copper having a rectangular cross section 15.2 mm 19.1 mm (0.60 in.0.75 in.) is pulled in tension with 44,500 N (10,000 lb ) force, producing only elastic deformation. Calculate the resulting strain. 2. A cylindrical specimen of a nickel alloy having an elastic modulus of 207 GPa (psi) and an original diameter of 10.2 mm (0.40 in.) will experience only elastic deformation when a tensile load of 8900 N (2000 lb) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm (0.010 in.). 3. An aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in) on an edge is pulled in tension with a load of 66,700 N (15,000 lb), and experiences an elongation of 0.43 mm (in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

Problems 4. For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa( 15X106psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 130 mm2 (0.2 in2) without plastic deformation? (b) If the original specimen length is 76 mm (3.0 in.), what is the maximum length to which it may be stretched without causing plastic deformation? 5. A cylindrical rod of steel ( E=207GPa, 30X106 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a load of 11,100 N (2500 lb ). If the length of the rod is 500 mm (20.0 in.), what must be the diameter to allow an elongation of 0.38 mm (0.015 in.)? 6. Consider a cylindrical nickel wire 2.0 mm (0.08 in.) in diameter and 3x104mm (1200 in.) long. Calculate its elongation when a load of 300 N (67 lb) is applied. Assume that the deformations totally elastic.  

SHORT QUESTIONS 1.The stress-strain graphs for two materials A and B are shown in the figure. Which material has greater Young’s modulus and larger elasticity? 2. A piece of copper originally 305 mm long is pulled in a tension with a stress of 276 M Pa. If the deformation is entirely elastic, what will be resultant elongation? Given 1 M Pa is 106 N/m2 3.Define brittleness. 4. What is the effect on hardness of the material, when it is heated up to a melting temperature at a slow rate and cooled suddenly. 5. A specimen of aluminum having a rectangular cross section 10 mm X 12.7 mm is pulled in tension with 35,000 N force, producing only elastic deformation. Calculate the resulting strain.

6. Identify the type of material whose stress-strain graph is as shown below. 7.What do you mean by Toughness? 8. Which property of a material makes them tough and hard. 9. Draw the stress strain curve for brittle material? 10. Differentiate the ductile and brittle nature of a material with an example. 11. Define Poisson’s ratio. 12. By what process maximum hardness is obtained for a steel part. 13. What do you understand by toughness, ductility, creep and fatigue 14. Explain stress and strain relationship using Hook's law?

LONG QUESTIONS 1. Explain the terms Yield strength, Ultimate tensile strength, necking and fracture in the Stress-Strain behaviour of ductile materials. 2.Compare and explain the Stress- Strain behaviour of ductile and brittle materials. 3.State Hooke’s law and distinguish True stress-strain and Engineering stress-strain. 4.Define Hardness and illustrate the procedure to measure the hardness number through Vicker’s hardness test. 5.Explain Brinell’s hardness test to measure hardness number of a material 6.Illustrate Rockwell’s hardness test and mention the advantages of it. Compare the three hardness tests in detail.

The End