1. Week 4 MECHANICAL PROPERTIES AND TESTS
Materials Science

2. Stress
Stress is a measure of the intensity of the internal forces acting within a deformable body.
Mathematically, it is a measure of the average force per unit area of a surface within a the body on which internal forces act
The SI unit for stress is Pascal (symbol Pa), which is equivalent to one Newton (force) per square meter (unit area).
Three types of stresses -> Tensile; Compressive; Shear
These internal forces are produced between the particles in the body as a reaction to external forces applied on the body. … and result in deformation of the body's shape. Beyond certain limits of material strength, this can lead to a permanent change of shape or physical failure (yield strength and ultimate strength) .
The unit for stress is the same as that of pressure, which is also a measure of force per unit area
Tensile stress tends to stretch the material; Compressive stress tends to squeeze it; Shear stress tends to cause the adjacent portions of the material to slide against each other

3. Mechanism of Stress (Tensile)
Axial stress in a prismatic bar axially loaded

4. Strain
Strain is deformation of a physical body under the action of applied forces
It is the geometrical measure of deformation representing the relative displacement between particles in the material body
Strain is a dimensionless quantity
Strain accounts for elongation, shortening, or volume changes, or angular distortion
Normal stress causes normal strain (tensile or compressive)
Shear strain is defined as the change in angle between two originally orthogonal material lines
3. which can be expressed as a decimal fraction, a percentage or in parts-per notation
5. If there is an increase in length, the normal strain is called tensile strain, otherwise, if there is reduction in the length of the material, it is called compressive strain

5. Types of Strains tensile load produces an elongation and
4. Torsion is a variation of pure shear, wherein a structural member is twisted in the manner of the figure
tensile load produces
an elongation and
positive linear
strain.
compressive load
produces contraction
and a negative linear
strain.
torsional
deformation

6. Tensile Test and Stress-strain relationship

7. Tensile Test
Used for determining UTS, yield strength, %age elongation, and Young’s Modulus of Elasticity
The ends of a test piece are fixed into grips. The specimen is elongated by the moving crosshead; load cell and extensometer measure, respectively, the magnitude of the applied load and the elongation
0. The ability of a material to resist breaking under tensile stress is one of the most important and widely measured properties of materials used in structural applications. Tensile tests are simple, relatively inexpensive, and fully standardized. By pulling on something, you will very quickly determine how the material will react to forces being applied in tension
1. Tensile test gives valuable information on the tensile strength (TS=maximum load/original area) and ductility (percentage reduction in area or percentage elongation) of the material.
2. The strain is usually applied by means of a motor-driven crosshead and the elongation of the specimen is indicated by its relative movement

8. Stress-Strain Relationship
0. The stress-strain curve is the basic relationship that describes the mechanical properties of materials. During tensile testing of a material sample, the stress–strain curve is a graphical representation of the relationship between stress, derived from measuring the load applied on the sample, and strain, derived from measuring the deformation of the sample, i.e. elongation, compression, or distortion. The area under the elastic portion of the curve is known as the modulus of resilience (property of a material to absorb energy when it is deformed elastically and then, upon unloading to have this energy recovered)

9. Important Terms (Stress-Strain Rel.)
Elastic Limit -> Maximum amount of stress up to which the deformation is absolutely temporary
Proportionality Limit -> Maximum stress up to which the relationship between stress & strain is linear.
Hooke’s Law -> Within elastic limit, the strain produced in a body is directly proportional to the stress applied.
σ = E ε
0. This curve is for a ductile material

10. Important Terms (Stress-Strain Rel.)
Young’s Modulus of elasticity -> the ratio of the uniaxial stress over the uniaxial strain in the range of stress in which Hooke's Law holds
Elasticity -> the tendency of a body to return to its original shape after it has been stretched or compressed
Yield Point -> the stress at which a material begins to deform plastically
It is a measure of STIFFNESS for a material. E = Stress/Strain; Units for “E” are same as that for Stress. .
For pure metals it is yield point and for alloys it is yield strength (obtained by intersection of a line at 0.2% strain offset to the original stress-strain curve). Yield strength is also called as yield stress. Yield point marks the transition to the plastic region and start of plastic deformation

11. Important Terms (Stress-Strain Rel.)
Plasticity -> the deformation of a material undergoing non-reversible changes of shape in response to applied forces
Ultimate Strength -> It is the maxima of the stress-strain curve. It is the point at which necking will start.
Necking -> A mode of tensile deformation where relatively large amounts of strain localize disproportionately in a small region of the material
Plastic range of a ductile material is the area of interest for manufacturing ppl. The region is not governed by the Hooke’s Law and the deformation magnitude is exceedingly large as compared to that in elastic range for the same increment in the applied stress .
During necking, the stress begins to decline and a localized elongation sets in the test specimen. Instead of continuing strain uniformly, straining becomes concentrated in one small region of the specimen.

12. Important Terms (Stress-Strain Rel.)
Fracture Point -> The stress calculated immediately before the fracture.
Ductility -> The amount of strain a material can endure before failure.
Ductility is measured by percentage elongation or area reduction
Or Fracture Strength -> Fracture stress is less than ultimate strength
Ductility is the ability of the material to plastically strain without failure. Ductility refers to workability of any material

13. Important Terms (Stress-Strain Rel.)
A knowledge of ductility is important for two reasons:
It indicates to a designer the degree to which a structure will deform plastically before fracture.
It specifies the degree of allowable deformation during fabrication
Figure -> Schematic representations of tensile stress–strain behavior for brittle and ductile materials loaded to fracture

14. Engineering stress– strain behavior for Iron at three temperatures

15. Resilience
Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered
Modulus of Resilience (Ur) is the strain energy per unit volume required to stress a material from an unloaded state up to the point of yielding.
Last -> Put σ = ЄE and then integrate

16. Resilience Assuming a linear elastic region
For SI units, this is joules per cubic meter (J/m3, equivalent to Pa)
Thus, resilient materials are those having high yield strengths and low moduli of elasticity; such alloys would be used in spring applications

17. Shear and Torsional Tests
Shear Stress:
The shear strain γ is defined as the tangent of the strain angle θ
Torsion is a variation of pure shear, wherein a structural member is twisted in the manner of the figure
Torsional stress -> The shear stress on a transverse cross section resulting from a twisting action
Torsional forces produce a rotational motion about the longitudinal axis of one end of the member relative to the other end
… where F is the load or force imposed parallel to the upper and lower faces .
Examples of torsion are found for machine axles and drive shafts, and also for twist drills. Torsional tests are normally performed on cylindrical solid shafts or tubes. A shear stress is a function of the applied torque T, whereas shear strain γ is related to the angle of twist θ. In circular sections, the resultant shearing stress is perpendicular to the radius

18. ANELASTICITY
In most engineering materials, there also exists a time-dependent elastic strain component.
This time-dependent elastic behavior is known as anelasticity
For metals the anelastic component is normally small and is often neglected
For some polymeric materials its magnitude is significant; in this case it is termed viscoelastic behavior
1. … That is, elastic deformation will continue after the stress application, and upon load release some finite time is required for complete recovery
2. … and it is due to time-dependent microscopic and atomistic processes that are attendant to the deformation
4. Viscoelasticity is the property of materials that exhibit both viscous and elastic characteristics when undergoing deformation

19. EXAMPLE PROBLEM 6.1
A piece of copper originally 305mm (12 in.) long is pulled in tension with a stress of 276MPa (40,000psi). If the deformation is entirely elastic, what will be the resultant elongation?
Magnitude of E for copper from Table 6.1 is 110GPa

20. Poisson’s Ratio
Poisson’s ratio is defined as the ratio of the lateral and axial strains
Theoretically, Poisson’s ratio for isotropic materials should be 1/4; furthermore, the maximum value for ν is 0.50
0. When a tensile stress is imposed on a metal specimen, an elastic elongation and accompanying strain result in the direction of the applied stress (arbitrarily taken to be the 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,
2. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then εx = εy. The negative sign is included in the expression so that ν will always be positive
3. For many metals and other alloys, values of Poisson’s ratio range between 0.25 and 0.35
4. In most metals G is about 0.4E

21. EXAMPLE PROBLEM 6.2
A tensile stress is to be applied along the long axis of a cylindrical brass rod that has a diameter of 10mm. Determine the magnitude of the load required to produce a mm change in diameter if the deformation is entirely elastic.
For the strain in the x direction:
1. When the force F is applied, the specimen will elongate in the z direction and at the same time experience a reduction in diameter of mm in x-direction

22. EXAMPLE PROBLEM 6.2

23. True Stress and Strain
The decline in the stress necessary to continue deformation past the point M, indicates that the metal is becoming weaker.
Material is increasing in strength.
True stress σT is defined as the load F divided by the instantaneous cross-sectional area Ai over which deformation is occurring
True strain ЄT is defined as:
2. On the other hand, … because of strain hardening. True stress equation applies from onset of yielding to onset of necking

24. True Stress and Strain
If no volume change occurs during deformation—that is, if Aili = A0l0
Then true and engineering stress and strain are related according to
The equations are valid only to the onset of necking; beyond this point true stress and strain should be computed from actual load, cross-sectional area, and gauge length measurements
Engineering stress is the one that we measure during experiment within plastic range. This is not true because we are measuring force only and we are dividing it with the initial cross sectional area, which is continuously decreasing

25. Assignment
(a) Completely describe “Compression Test”. (b) How is it different from Tensile test? (c) What are the effects of Friction and Workpiece’s height-to-diameter ratio on the test? (d) Derive relationship between true stress/strain and engineering stress/strain for compression test (also show by stress-strain curve)
Submission on or before 18-Oct-2012

26. EXAMPLE PROBLEM 6.4
A cylindrical specimen of steel having an original diameter of 12.8mm is tensile tested to fracture and found to have an engineering fracture strength σf of 460MPa. If its cross-sectional diameter at fracture is 10.7mm, determine:
(a) The ductility in terms of percent reduction in area
(b) The true stress at fracture
Ductility is computed as

27. EXAMPLE PROBLEM 6.4
True stress is defined by Equation where the area is taken as the fracture area Af However, the load at fracture must first be computed from the fracture strength as And the true stress is calculated as
Fracture strength is calculated based on the initial cross sectional area

28. Elastic Recovery after Plastic Deformation
Upon release of the load during the course of a stress–strain test, some fraction of the total deformation is recovered as elastic strain
During the unloading cycle, the curve traces a near straight-line path from the point of unloading (point D), and its slope is virtually parallel to the initial elastic portion of the curve
The magnitude of this elastic strain, which is regained during unloading, corresponds to the strain recovery
3. …. If the load is reapplied, the curve will traverse essentially the same linear portion in the direction opposite to unloading; yielding will again occur at the unloading stress level where the unloading began

29. Hardness
Hardness is the property of material by virtue of which it resists against surface indentation and scratches.
Macroscopic hardness is generally characterized by strong intermolecular bonds
Hardness is dependent upon strength and ductility
Common examples of hard matter are diamond, ceramics, concrete, certain metals, and superhard materials (PcBN, PcD, etc)
Good hardness means the material is resistant to abrasion and wear
however the behavior of solid materials under force is complex, therefore there are different measurements of hardness: scratch hardness, indentation hardness, and rebound hardness (Rebound hardness, also known as dynamic hardness, measures the height of the "bounce" of a diamond-tipped hammer dropped from a fixed height onto a material.)
Higher is the strength higher is the hardness. Higher is the ductility lesser is the hardness

30. Hardness Tests (BRINELL HARDNESS TEST)
Used for testing metals and nonmetals of low to medium hardness
The Brinell scale characterizes the indentation hardness of materials through the scale of penetration of an indenter, loaded on a material test-piece
A hardened steel (or cemented carbide) ball of 10mm diameter is pressed into the surface of a specimen using load of 500, 1500, or 3000 kg.
0. Swedish Engineer (John August Brinell) in year 1900.
3. For softer materials, a smaller force is used; for harder materials, a tungsten carbide ball is substituted for the steel ball

31. BRINELL HARDNESS TEST
where:
P = applied force (kgf)
D = diameter of indenter (mm)
d = diameter of indentation (mm)
The resulting BHN has units of kg/mm2, but the units are usually omitted in expressing the numbers
The d is average of d1 and d2
The Brinell hardness is represented in HB units

32. Rockwell Hardness Test
Rockwell test determines the hardness by measuring the depth of penetration of an indenter under a large load compared to the penetration made by a preload
A cone shaped indenter or small diameter ball (D = 1.6 or 3.2mm) is pressed into a specimen using a minor load of 10kg
Then, a major load of 150kg is applied
The additional penetration distance d is converted to a Rockwell hardness reading by the testing machine.
2. … , thus seating the indenter in the material.
3. …, causing the indenter to penetrate a certain distance beyond its initial position

33. Rockwell Hardness Test
2. Difference in load and indenter geometry provides Rockwell Scales for different materials

34. Vickers Hardness Test Uses a pyramid shaped indenter made of diamond.
It is based on the principle that impressions made by this indenter are geometrically similar regardless of load.
The basic principle, as with all common measures of hardness, is to observe the questioned material's ability to resist plastic deformation from a standard source.
Accordingly, loads of various sizes are applied, depending on the hardness of the material to be measured
0. The Vickers hardness test was developed in 1924 by Smith and Sandland at Vickers Ltd (Sheffield UK) as an alternative to the Brinell method.
2. The Vickers test is often easier to use than other hardness tests since the required calculations are independent of the size of the indenter, and the indenter can be used for all materials irrespective of hardness
3. The Vickers test can be used for all metals and has one of the widest scales among hardness tests

35. Vickers Hardness Test Where: F = applied load (kg)
D = Diagonal of the impression made the indenter (mm)
The hardness number is determined by the load over the surface area of the indentation and not the area normal to the force

36. Knoop Hardness Test
It is a microhardness test - a test for mechanical hardness used particularly for very brittle materials or thin sheets
A pyramidal diamond point is pressed into the polished surface of the test material with a known force, for a specified dwell time, and the resulting indentation is measured using a microscope
Length-to-width ratio of the pyramid is 7:1
0. developed in 1939 (pronounced Knoop -> kuh-nūp)
… where only a small indentation may be made for testing purposes. These materials/parts may fracture if heavy loads are applied .
Applied loads are lighter than in Vickers test

37. Knoop Hardness Test (contd…)
The indenter shape facilitates reading the impressions at lighter loads
HK = Knoop hardness value; F = load (kg); D = long diagonal of the impression (mm)

38. Hardness of Metals and Ceramics
0. (See Section 3.2.2)
1, 2. When Brinell and Rockwell tests were developed, metals were the only principal engineering materials. For metals, hardness is related to strength. Brinell hardness is not suitable for ceramics (ceramic material is harder than the indentation ball)
WC -> Ceramic

39. Hardness of Polymers
Polymers have the lowest hardness values among the three categories.

40. TOUGHNESS
It is a property of material by virtue of which it resists against impact loads.
Toughness is the resistance to fracture of a material when stressed
Mathematically, it is defined as the amount of energy per volume that a material can absorb before rupturing
Toughness can be determined by measuring the area (i.e., by taking the integral) underneath the stress-strain curve

41. Toughness (contd…) Toughness = Where ε is strain
εf is the strain upon failure
σ is stress
The Area covered under stress strain curve is called toughness
The explicit mathematical description is “It's energy of mechanical deformation per unit volume prior to fracture” … .
If we restrict the upper limit of integration up to the yield point, than the energy absorbed per unit volume is known as modulus of resilience

42. Toughness (contd…)
Toughness is measured in units of joules per cubic meter (J/m3) in the SI system
Toughness and Strength -> A material may be strong and tough if it ruptures under high forces, exhibiting high strains
Brittle materials may be strong but with limited strain values, so that they are not tough
Generally, strength indicates how much force the material can support, while toughness indicates how much energy a material can absorb before rupture
2. Strength and toughness are related.

43. Effect of Temperature on Properties
Generally speaking, materials are lower in strength and higher in ductility, at elevated temperatures
0. Temperature has effect on all the mechanical properties
1,2. Thus, most of the materials can be formed easily at high temperatures

44. Hot Hardness
A property used to characterize strength and hardness at elevated temperatures is Hot Hardness
It is the ability of a material to retain its hardness at elevated temperatures
3. Steels can be alloyed to achieve significant improvements in hot hardness. Ceramics exhibit superior properties at high temperatures, that’s why these are used for high-temp applications like, turbine blades, cutting tools, outside skin of shuttles, and refractory materials (for furnace linings). Good hot hardness is also desirable in tooling materials used in many manufacturing operations.

45. Numerical Problems Problems 6.3 to 6.9; 6.14 to 6.23; 6.25 to 6.33;