*INTRODUCTION:*

*INTRODUCTION:*

**There are three types of static stresses to which materials can be subjected: tensile, compressive and shear.**

Tensile stresses will in general stretch the material, compressive stresses tend to press it, and shear includes stresses that will in general cause adjacent segment of the material to slide against one another.

The stress–strain relationship is the essential relationship that explains the mechanical properties of materials for all three types.

*TENSILE PROPERTIES:*

*TENSILE PROPERTIES:*

The tensile test is the most all known method for examining the stress–strain relationship, particularly for metals.

In the test, a force is applied that pulls the material, tending to prolong it and diminish its breadth.

Steps in tensile test

Norms by ASTM (American Society for Testing and Materials) indicate the preparation of the test specimen and the conduct of the test itself.

The common specimen and general arrangement of the tensile test.

The begining test specimen has an original length Lo and area Ao.

The length is estimated as the distance between the gage marks, and the zone is estimated as the (usually round) cross part of the specimen.

During the testing of a metal, the specimen extends, then necks, and at the end fractures.

The load and the adjustment in length of the specimen are recorded as testing continues, to give the data needed to decide common progress of a tensile test.

Setup of tensile test

There are two different types of stress–strain relationship:

(1) engineering stress–strain

(2) true stress–strain.

The first is more significant in design, and the second is more significant in assembling.

*ENGINEERING STRESS–STRAIN:*

*ENGINEERING STRESS–STRAIN:*

The engineering stress and strain in a tensile test are characterised comparative to the first region and length of the test specimen.

These quality are of interest in design because the designer expects that the strains experienced by any part of the product won’t altogether change its shape.

The components are intended to withstand the expected stresses experience in service.

The engineering stress at any point on the curve is characterised as the force divided by the original area.

s = F / Ao

where,

s = engineering stress, MPa (lb/in2),

F = applied force in the test, N (lb), and

Ao = unique area of the test specimen, mm2 (in2).

The engineering strain at any point in the test is given by,

e = L – Lo / Lo

where,

e = engineering strain, mm/mm (in/in)

L = length at any point during the

elongation, mm (in)

Lo = original gage length, mm (in).

The units of engineering strain are given as mm/mm (in/in), however think of it as

representing elongation per unit length, without units.

The stress–strain relationship has two regions, indicating two distinct forms of behavior:

(1) Elastic.

(2) Plastic.

In the elastic region, the relationship between stress and strain is linear, and the material exhibits elastic behavior by returning to its original length when the load (stress) is released.

The relationship is defined by Hooke’s law,

s = Ee

where,

E = modulus of elasticity, MPa (lb/in2 ), a measure of the inherent stiffness of a material.

Elongation is defined as,

EL = Lf – Lo / Lo

where,

EL = elongation, often expressed in percent.

Lf = specimen length at fracture, mm (in), measured as the distance between gage marks after the two parts of the specimen have been put back together.

Lo = original specimen length, mm (in).

*TRUE STRESS-STRAIN:*

*TRUE STRESS-STRAIN:*

Thoughtful readers may be troubled by the use of the original area of the test specimen to calculate engineering stress, rather than the actual (instantaneous) area that becomes increasingly smaller as the test proceeds.

If the actual area were used, the calculated stress value would be higher.

The stress value obtained by dividing the instantaneous value of area into the applied load is defined as the true stress:

s = F /A

where,

s = true stress, MPa (lb/in2)

F = force, N (lb); and

A = actual (instantaneous) area resisting the load, mm2 (in2).

Similarly, true strain provides a more realistic assessment of the ‘‘instantaneous’’ elongation per unit length of the material.

**The value of true strain in a tensile test can be estimated by dividing the total elongation into small increments, calculating the engineering strain for each increment on the basis of its starting length, and then adding up the strain values. **

At the end of the test (or other deformation), the final strain value can be calculated using,

L = Lf.

When the engineering stress–strain data are plotted using the true stress and strain values, the resulting curve would appear.

In the elastic region, the plot is virtually the same as before.

Strain values are small, and true strain is nearly equal to engineering strain for most metals of interest.

The respective stress values are also very close to each other.

The reason for these near equalities is that the cross-sectional area of the test specimen is not significantly reduced in the elastic region.

*TYPES OF STRESS STRAIN RELATIONSHIPS:*

*TYPES OF STRESS STRAIN RELATIONSHIPS:*

Much information about elastic–plastic behavior is provided by the true stress–strain curve.

As indicated, Hooke’s law governs the metal’s behavior in the elastic region, and the flow curve determines the behavior in the plastic region.

Three basic forms of stress–strain relationship describe the behavior of nearly all types of solid materials.

*PERFECTLY ELASTIC: *

*PERFECTLY ELASTIC:*

**The behavior of this material is defined completely by its stiffness, indicated by the modulus of elasticity E.**

It fractures rather than yielding to plastic flow.

Brittle materials such as ceramics, many cast irons, and thermosetting polymers possess stress–strain curves that fall into this category.

These materials are not good candidates for forming operations.

*ELASTIC AND PERFECTLY PLASTIC:*

*ELASTIC AND PERFECTLY PLASTIC:*

This material has a stiffness defined by E. Once the yield strength Y is reached, the material deforms plastically at the same stress level.

The flow curve is given by K = Y and n = 0. Metals behave in this fashion when they have been heated to sufficiently high temperatures that they recrystallize rather than strain harden during deformation.

Lead exhibits this behavior at room temperature because room temperature is above the recrystallization point for lead.

*ELASTIC AND STRAIN HARDENING:*

*ELASTIC AND STRAIN HARDENING:*

This material obeys Hooke’s law in the elastic region.

It begins to flow at its yield strength Y.

Continued deformation requires an ever increasing stress, given by a flow curve whose strength coefficient K is greater than Y and whose strain-hardening exponent n is greater than zero.

The flow curve is generally represented as a linear function on a natural logarithmic plot. Most ductile metals behave this way when cold worked.

Manufacturing processes that deform materials through the application of tensile

stresses include wire and bar drawing and stretch forming.

*COMPRESSION PROPERTIES:*

*COMPRESSION PROPERTIES:*

A compression test applies a load that squeezes a cylindrical specimen between two platens.

Compression test

As the specimen is compressed, its height is reduced and its cross-sectional area is increased.

Engineering stress is defined as,

s = F / Ao

where,

Ao = original area of the specimen.

This is the same definition of engineering stress used in the tensile test.

The engineering strain is defined as,

e = h – ho / ho

where,

h = height of the specimen at a particular moment into the test, mm (in)

ho = starting height, mm (in).

Because the height is decreased during compression, the value of e will be negative.

The negative sign is usually ignored when expressing values of compression strain.

When engineering stress is plotted against engineering strain in a compression test, the

results appear.

The curve is divided into elastic and plastic regions, as before the compression test.

*SHEAR PROPERTIES:*

*SHEAR PROPERTIES:*

Shear involves application of stresses in opposite directions on either side of a thin element to deflect.

The shear stress is defined as,

t = F / A

where,

t = shear stress, lb/in2 (MPa)

F = applied force, N (lb)

A = area over which the force is applied, in2 (mm2).

Shear stress and strain are commonly tested in a torsion test, in which a thin walled

tubular specimen is subjected to a torque.

As torque is increased, the tube deflects by twisting, which is a shear strain for this geometry.

*FREQUENTLY ASKED QUESTIONS:*

*FREQUENTLY ASKED QUESTIONS:*

*What is the relationship between stress and strain?*

The stress–strain relationship is the essential relationship that explains the mechanical properties of materials for all three types: tensile, compressive and shear.

*What is stress and strain formula?*

In the elastic region, the relationship between stress and strain is linear, and the material exhibits elastic behavior by returning to its original length when the load (stress) is released.

The relationship is defined by Hooke’s law,

s = Ee

where,

E = modulus of elasticity, MPa (lb/in2 ), a measure of the inherent stiffness of a material.

*What is Hooke’s law for stress and strain?*

Hooke’s law for stress and strain is stated as follows:

s = Ee

where,

E = modulus of elasticity, MPa (lb/in2 ), a measure of the inherent stiffness of a material.

*Are stress and strain inversely related?*

As per Hooke’s law, stress and strain are directly proportional.

*Why is stress strain important?*

The stress–strain relationship is the essential relationship that explains the mechanical properties of materials.

*What is the symbol of strain?*

The symbol of strain is e.