Behaviour of Materials

1. Introduction

When a force is applied on a body it suffers a change in shape, that is, it deforms. A force to resist the deformation is also set up simultaneously within the body and it increases as the deformation continues. The process of deformation stops when the internal resisting force equals the externally applied force. If the body is unable to put up full resistance to external action, the process of deformation continues until failure takes place. The deformation of a body under external action and accompanying resistance to deform are referred to by the terms strain and stress respectively.

2. Stresses

Stress is defined as the internal resistance set up by a body when it  is deformed. It is measured in N/m² and this unit is specifically called Pascal (Pa). A bigger unit of stress is the mega Pascal (MPa).

1 Pa = 1N/m²,

1MPa = 10⁶ N/m² =1N/mm².

2.1. Three Basic Types of Stresses

Basically three different types of stresses can be identified. These are related to the nature of the deforming force applied on the body. That is, whether they are tensile, compressive or shearing.

2.1.1. Tensile Stress

Consider a uniform bar of cross sectional area A subjected to an axial tensile force P. The stress at any section x-x normal to the line of action of the tensile force P is specifically called tensile stress p_t . Since internal resistance R at x-x is equal to the applied force P, we have,

p_t         = (internal resistance at x-x)/(resisting area at x-x)

=R/A

=P/A.

Under tensile stress the bar suffers stretching or elongation.

2.1.2. Compressive Stress

If the bar is subjected to axial compression instead of axial tension, the stress developed at x-x is specifically called compressive stress p_c.

p_c         =R/A

= P/A.

Under compressive stress the bar suffers shortening.

2.1.3. Shear Stress

Consider the section x-x of the rivet forming joint between two plates subjected to a tensile force P as shown in figure.

The stresses set up at the section x-x acts along the surface of the section, that is, along a direction tangential to the section. It is specifically called shear or tangential stress at the section and is denoted by q.

q          =R/A

=P/A.

2.2. Normal or Direct Stresses

When the stress acts at a section or normal to the plane of the section, it is called a normal stress or a direct stress. It is a term used to mean both the tensile stress and the compressive stress.

2.3. Simple and Pure Stresses

The three basic types of stresses are tensile, compressive and shear stresses. The stress developed in a body is said to be simple tension, simple compression and simple shear when the stress induced in the body is (a) single and (b) uniform. If the condition (a) alone is satisfied, the stress is called pure tension or pure compression or pure shear, as the case may be.

2.4. Volumetric Stress

Three mutually perpendicular like direct stresses of same intensity produced in a body constitute a volumetric stress. For example consider a body in the shape of a cube subjected equal normal pushes on all its six faces. It is now subjected to equal compressive stresses p in all the three mutually perpendicular directions. The body is now said to be subjected to a volumetric compressive stress p.

Volumetric stress produces a change in volume of the body without producing any distortion to the shape of the body.

3. Strains

Strain is defined a the ratio of change in dimension to original dimension of a body when it is deformed. It is a dimensionless quantity as it is a ratio between two quantities of same dimension.

3.1. Linear Strain

Linear strain of a deformed body is defined as the ratio of the change in length of the body due to the deformation to its original length in the direction of the force. If l is the original length and δl the change in length occurred due to the deformation, the linear strain e induced is given by

e=δl/l.

Linear strain may be a tensile strain , e_t or a compressive strain e_c according as δl refers to an increase in length or a decrease in length of the body. If we consider one of these as +ve then the other should be considered as –ve, as these are opposite in nature.

3.2.Lateral Strain

Lateral strain of a deformed body is defined as the ratio of the change in length (breadth of a rectangular bar or diameter of a circular bar) of the body due to the deformation to its original length (breadth of a rectangular bar or diameter of a circular bar) in the direction perpendicular to the force.

3.3. Volumetric Strain

Volumetric strain of a deformed body is defined as the ratio of the change in volume of the body to the deformation to its original volume. If V is the original volume and δV the change in volume occurred due to the deformation, the volumetric strain e_v induced is given by e_v =δV/V

Consider a uniform rectangular bar of length l, breadth b and depth d as shown in figure. Its volume V is given by,

V   = lbd

δV = δl bd + δb ld + δd lb

δV /V = (δl / l) + (δb / b) + (δd / d)

e_v = e_x +e_y +e_z

This means that volumetric strain of a deformed body is the sum of the linear strains in three mutually perpendicular directions.

3.4. Shear Strain

Shear strain is defined as the strain accompanying a shearing action. It is the angle in radian measure through which the body gets distorted when subjected to an external shearing action. It is denoted by Ф.

Consider a cube ABCD subjected to equal and opposite forces Q across the top and bottom forces AB and CD. If the bottom face is taken fixed, the cube gets distorted through angle f to the shape ABC’D’. Now strain or deformation per unit length is

Shear strain of cube = CC’ / CD = CC’ / BC = f radian

4. Relationship between Stress and Strain

Relationship between Stress and Strain are derived on the basis of the elastic behaviour of material bodies.

            A standard mild steel specimen is subjected to a gradually increasing pull by Universal Testing Machine. The stress-strain curve obtained is as shown below.

A         -Elastic Limit

B          - Upper Yield Stress

C         - Lower Yield Stress

D         -Ultimate Stress

E          -Breaking Stress

4.1. Elasticity and Elastic Limit

Elasticity of a body is the property of the body by virtue of which the body regains its original size and shape when the deformation force is removed. Most materials are elastic in nature to a lesser or greater extend, even though perfectly elastic materials are very rare.

The maximum stress upto which a material can exhibit the property of elasticity is called the elastic limit. If the deformation forces applied causes the stress in the material to exceed the elastic limit, there will be a permanent set in it. That is the body will not regain its original shape and size even after the removal of the deforming force completely. There will be some residual strain left in it.

Yield stress

When a specimen is loaded beyond the elastic limit the stress increases and reach a point at which the material starts yielding this stress is called yield stress.

Ultimate stress

Ultimate load is defined as maximum load which can be placed prior to the breaking of the specimen. Stress corresponding to the ultimate load is known as ultimate stress.

Working stress

Working stress= Yield stress/Factor of safety.

4.2. Hooke’s Law

Hooke’s law states that stress is proportional to strain upto elastic limit. If I is the stress induced in a material and e the corresponding strain, then according to Hooke’s law,

p / e = E, a constant.

This constant E is called the modulus of elasticity or Young’s Modulus, (named after the English scientist Thomas Young).

It has later been established that Hooke’s law is valid only upto a stress called the limit of proportionality which is slightly less than the elastic limit.

4.3. Elastic Constants

Elastic constants are used to express the relationship between stresses and strains. Hooke’s law , is stress/strain = a constant, within a certain limit. This means that any stress/corresponding strain = a constant, within certain limit. It follows that there can be three different types of such constants. (which we may call the elastic constants or elastic modulae) corresponding to three distinct types of stresses and strains. These are given below.

(i)Modulus of Elasticity or Young’s Modulus (E)

Modulus of Elasticity is the ratio of direct stress to corresponding linear strain within elastic limit. If p is any direct stress below the elastic limit and e the corresponding linear strain, then E = p / e.

(ii)Modulus of Rigidity or Shear Modulus (G)

Modulus of Rigidity is the ratio of shear stress to shear strain within elastic limit. It is denoted by N,C or G. if q is the shear stress within elastic limit and f the corresponding shear strain, then G = q / f.

(iii) Bulk Modulus (K)

Bulk Modulus is the ratio of volumetric stress to volumetric strain within the elastic limit. If p_v is the volumetric stress within elastic limit and e_v the corresponding volumetric strain, we have K = p_v / e_v.

5. Poisson’s Ratio

Any direct stress is accompanied by a strain in its own direction and called linear strain and an opposite kind strain in every direction at right angles to it, lateral strain. This lateral strain bears a constant ratio with the linear strain. This ratio is called the Poisson’s ratio and is denoted by (1/m) or µ.

Poisson’s Ratio = Lateral Strain / Linear Strain.

Value of the Poisson’s ratio for most materials lies between 0.25 and 0.33.

6. Complementary Strain

Consider a rectangular element ABCD of a body subjected to simple shear of intensity q as shown. Let t be the thickness of the element.

Total force on face AB is , F_AB = stress X area = q X AB X t.

Total force on face CD is, F_CD = q X CD Xt = q XAB Xt.

FAB and FCD being equal and opposite, constitute a couple whose moment is given by,

M =F_AB X BC = q XAB X BC X t

Since the element is in equilibrium within the body, there must be a balancing couple which can be formed only by another shear stress of some intensity q’ on the faces BC and DA. This shear stress is called the complementary stress.

F_BC = q’ X BC X t

F_DA = q’ X DA X t = q’ X BC Xt

The couple formed by these two forces is M’ = F_BC X AB = q’ X BC X t

For equilibrium, M’ = M.

Therefore q’ = q

This enables us to make the following statement.

In a state of simple shear a  shear stress of any intensity along a plane is always accompanied by a complementary shear stress of same intensity along a plane at right angles to the plane.

7. Direct Stresses Developed Due to Simple Shear.

Consider a square element of side a and thickness t in a state of simple shear as shown in figure. It is clear that the shear stress on the forces of element causes it to elongate in the direction of the diagonal BD. Therefore a tensile stress of same intensity pt is induced in the elements along BD. ie, across the plane of the diagonal AC. The triangular portion ABC of the element is in equilibrium under the action of the following.

(1)F_AC = Normal force on face AC = pt X AC X t = pt X √2 aXt

(2)F_AB = Tangential force on face AB = q X BC X t = q aXt

(3)FBC = Tangential force on face BC = q X BC X t = q aXt

For equilibrium in the direction normal to AC,

FAC – F_AB cos45 – F_BC cos45 = 0

Pt X √2 at – q at X 1/√2 - q at X 1/√2 = 0

√2 pt – 2 q /√2 = 0

pt = q

It can also be seen that the shear stress on the faces of the element causes it to foreshorten in the direction of the diagonal BD. Therefore a compressive stress pc is induced in the element in the direction AC, ie across the plane of the diagonal BD. It can also be shown that pc = q.

It can thus be concluded that simple shear of any intensity gives rise to direct stresses of same intensity along the two planes inclined at 45º to the shearing plane. The stress along one of these planes  being tensile and that along the other being compressive.

8. Relationship among the elastic constants

8.1. Relationship between modulus of elasticity and modulus of rigidity

Consider a square element ABCD of side ‘a’ subjected to simple shear of intensity q as shown in figure.

It is deformed to the shape AB’C’D under the shear stress. Drop perpendicular BE to the diagonal DB’. Let Ф be the shear strain induced and let N be the modulus of rigidity.

The diagonal DB gets elongated to DB’. Hence there is tensile strain et in the diagonal.

et = (DB’ – DB) / DB = EB’ / DB

since this deformation is very small we can take L BB’E = 45º

EB’ = BB’ / √2 = AB tan Ф / √2 = a tan Ф / √2

DB = √2 a

et = (a tan Ф / √2)/ √2 a = tan Ф /2 = = Ф / 2 since Ф is small

ie et = ½ X q/N        ------------------ (1)

We know that stress along the diagonal DB is a pure tensile stress pt = q and that along the diagonal AC is a pure compressive stress pc also equal to q. hence the strain along the diagonal DB is et = q/E + 1/m X q/E

Ie et = q/E (1+1/m) ------------------ (2)

From (1) and (2) we have,

E = 2N(1+1/m)

This is the required relationship between E and N.

8.2. Relationship between Modulus of Elasticity E and Bulk Modulus K

Consider a cube element subjected to volumetric tensile stress pυ in X,Y and Z directions. Stress in each direction is equal to pυ. ie px = py = pz =pυ

Consider strains induced in X-direction by these stresses. px induces tensile strain , while py and pz induces compressive strains. Therefore,

ex  = px/E – 1/m[py/E + pz/E] = pυ/E[1-2/m]

due to the perfect symmetry in geometry and stresses

ey = pυ/E[1-2/m]

ez = pυ/E[1-2/m]

K = p υ / e υ =p υ/(ex+ey+ez) = pυ/[3pυ/E(1-2/m)]

ie E = 3K(1-2/m ) is the required relationship.

8.3. Relationship among the constants

From above,

E = 2N[1+(1/m)] and E = 3K[1-(2/m )]

E = 3K[1-2(E/2N -1)] = 3K[1-E/N +2]

9K = E[1+(3K/N)] = E[(N+3K)/N]

E = 9NK/(N+3K)

9. Bars of uniform section

Consider a bar of length l and Cross sectional area A. Let P be the axial pull on the bar,p the stresss induced ,e the strain in the bar and δl is the elongation.

Then   p= P/A

            e= p/E =P/(AE)       -------------(1)

e= δl/l                       -------------(2)

equating (1) and (2)

δl = Pl / (AE)

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