Learn Stresses and strain, unit of stress,Type of stresses,Tensile stress,Compressive stress,Tensile strain,compressive strain,Shear stress and shear strain

Table of Contents

## Stress:

The force of resistance per unit area, offered by a body against deformation is known as stress.

The external force acting on the body is called the load or force. The load is applied on the body while the stress is induced in the material of the body. A loaded member remains in equilibrium when the resistance offered by the member against the deformation and the applied load are equal.

Mathematically stress is written as,

\[σ = \frac{P}{ A }\]

where σ = Stress (also called intensity of stress),

P = External force or load,

and A = Cross-sectional area.

### Strain

When a body is subjected to some external force, there is some change of dimension of the body. The ratio of change of dimension of the body to the original dimension is known as strain. Strain is dimensionless. Strain may be:

1. Tensile strain,

2. Compressive strain,

3. Volumetric strain, and

4. Shear strain.

If there is some increase in length of a body due to external force, then the ratio of increase of length to the original length of the body is known as tensile strain. But if there is some decrease in length of the body, then the ratio of decrease of the length of the body to the original length is known as compressive strain. The ratio of change of volume of the body to the original volume is known as volumetric strain. The strain produced by shear stress is known as shear strain.

### Units of Stress.

The unit of stress depends upon the unit of load (or force) and unit of area. In M.K.S. units, the force is expressed in kgf and area in metre square (i.e., m^{2}). Hence unit of stress becomes as kgf/m^{2}.

If area is expressed in centimetre square (i.e., cm^{2}), the stress is expressed as kgf/cm^{2}.

In the S.I. units, the force is expressed in newtons (written as N) and area is expressed as m^{2}. Hence unit of stress becomes as N/m^{2}.

The area is also expressed in millimetre square then unit of force becomes as N/mm^{2}

1 N/m^{2} = 1 N/(100 cm)^{2} = 1 N/10^{4} cm^{2} = 10^{–4} N/cm^{2} or 10^{–6} N/mm^{2}

∴ 1 N/mm^{2} = 10^{6 }N/m^{2}.

Also 1 N/m^{2} = 1 Pascal = 1 Pa.

The large quantities are represented by kilo, mega, giga and terra. They stand for:

Kilo = 10^{3} and represented by …… k

Mega = 10^{6} and represented by …… M

Giga = 10^{9} and represented by …… G

Terra = 10^{12} and represented by …… T.

Thus mega newton means 10^{6} newtons and is represented by MN. T

he symbol 1 MPa stands for 1 mega pascal which is equal to 10^{6 pascal (or 106 N/m2).}

The small quantities are represented by milli, micro, nano and pico.

They are equal to

Milli = 10^{–3} and represented by …… m

Micro = 10^{–6} and represented by …… µ

Nano = 10^{–9} and represented by …… η

Pico = 10^{–12} and represented by …… p.

Notes.

1. Newton is a force acting on a mass of one kg and produces an acceleration of 1 m/s^{2 }i.e., 1 N = 1 (kg) × 1 m/s^{2}.

2. The stress in S.I. units is expressed in N/m^{2} or N/mm^{2}.

3. The stress 1 N/mm^{2 }= 10^{6} N/m^{2} = MN/m^{2}. Thus one N/mm^{2} is equal to one MN/m^{2}.

4. One pascal is written as 1 Pa and is equal to 1 N/m^{2}.

### Types of stresses and strain

The stress may be normal stress or a shear stress. Normal stress is the stress which acts in a direction perpendicular to the area. It is represented by σ (sigma). The normal stress is further divided into tensile stress and compressive stress.

#### Tensile Stress.

The stress induced in a body, when subjected to two equal and opposite pulls as shown in Fig. (a) as a result of which there is an increase in length, is known as tensile stress.

#### Tensile strain

The ratio of increase in length to the original length is known as tensile strain. The tensile stress acts normal to the area and it pulls on the area.

Let

P = Pull (or force) acting on the body,

A = Cross-sectional area of the body,

L = Original length of the body,

dL = Increase in length due to pull P acting on the body,

σ = Stress induced in the body,

and e = Strain (i.e., tensile strain).

Fig. (a) shows a bar subjected to a tensile force P at its ends. Consider a section x-x, which divides the bar into two parts. The part left to the section x-x, will be in equilibrium if P = Resisting force (R). This is shown in Fig. (b). Similarly the part right to the section x-x, will be in equilibrium if P = Resisting force as shown in Fig. (c). This resisting force per unit area is known as stress or intensity of stress.

∴ \[Tensile stress = σ = \frac{Resisting forceR}{( Cross-sectional area A)}=\frac{ Tensile loadP}{ A}\]

(∵ P = R)

or \[σ = \frac{P} {A}\]

And tensile strain is given by,\[ e =\frac{ Increase in length dL}{Original lengthL}\]

#### Compressive Stress.

The stress induced in a body, when subjected to two equal and opposite pushes as shown in Fig. (a) as a result of which there is a decrease in length of the body, is known as compressive stress.The compressive stress acts normal to the area and it pushes on the area. Let an axial push P is acting on a body in cross-sectional area A.

Compressive stress is given by

\[σ = \frac{Resisting ForceR} { Area A} =\frac{ Push P}{ Area A}=\frac{P}{A}\]

#### Compressive strain

the ratio of decrease in length to the original length is known as compressive strain. Due to external push P, let the original length L of the body decreases by dL.

Compressive stress is given by

\[e=\frac{Decrease in length }{Orginal Length}=\frac{dL}{L}\]

#### Shear Stress.

The stress induced in a body, when subjected to two equal and opposite forces which are acting tangentially across the resisting section as shown in Fig. as a result of which the body tends to shear off across the section, is known as shear stress. The corresponding strain is known as shear strain. The shear stress is the stress which acts tangential to the area. It is represented by τ.

Consider a rectangular block of height h, length L and width unity. Let the bottom face AB of the block be fixed to the surface as shown in Fig.(a). Let a force P be applied tangentially along the top face CD of the block. Such a force acting tangentially along a surface is known as shear force. For the equilibrium of the block, the surface AB will offer a tangential reaction P equal and opposite to the applied tangential force P.

Consider a section x-x (parallel to the applied force), which divides the block into two parts. The upper part will be in equilibrium if P = Resistance (R). This is shown in Fig. (b). Similarly the lower part will be in equilibrium if P = Resistance (R) as shown in Fig. (c). This resistance is known as shear resistance. And the shear resistance per unit area is known as shear stress which is represented by τ.

\[Shear stress τ = \frac{Shear resistance}{ Shear Area}=\frac{R}{A}=\frac{P}{LX1}\]

Note that shear stress is tangential to the area over which it acts.

As the bottom face of the block is fixed, the face ABCD will be distorted to ABC1D1 through an angle φ as a result of force P as shown in Fig.(d). And

#### Shear strain (φ)

\[φ = \frac{Transversal displacement} {Distance AD}\]

\[φ = \frac{DD}{ AD}=\frac{dl}{ h}\]