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**Stress**

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

StressIt is denoted by a symbol ‘σ’. And mathematically expressed as

σ = P/a

As particular stress generally holds true only at a point, therefore it is defined mathematically as

**Types of Stress**

Only two basic stresses exist.

- Normal stress and
- Shear stress.

**1. Normal Stresses**

If the force applied are perpendicular or normal to areas concerned, then these are termed as normal stresses.

The normal stresses are generally denoted by a Greek letter (σ).

Uniaxial Normal Stress

**Biaxial Normal Stress**

**Triaxial Normal Stress****(i) Tensile or compressive stresses: **The normal stresses can be either tensile or compressive depending upon the direction of the load.

**(ii) Sign convention: **The tensile forces are termed as (+ve ) while the compressive forces are termed as negative (-ve).

**2. Shear Stress**

When cross-sectional area of a block of material is subject to a distribution of forces which are parallel to the area concerned. Such forces are associated with a shearing of the material, are known as shear forces. The stress produced by these forces are known as shear stresses.

Shear stressIt is generally denoted by ‘τ’ and expressed as

τ = shear resistance/Shear area = P_{s}/A

The complementary shear stresses are equal in magnitude.

The same form of relationship can be obtained for the other two pair of shear stress components to arrive at the relations,

τ_{xy }= τ_{yx}

τ_{yz} = τ_{z}_{y}

**(i) Sign convections for shear stresses**

- Shear stress tending to turn the element Clockwise is taken as
**Positive**. - Shear stress tending to turn the element Anticlockwise is taken as
**Negative**.

**Strain**

When a prismatic bar is subjected to axial load, it undergoes a change in length, as indicated in Figure. This change in length is usually called deformation.

Deformation of bar under axial loadIf the axial force is tensile, the length of the bar is increased, while if the axial force is compressive, there is shortening of the length of the bar.

The deformation (i.e. elongation or shortening) per unit length of the bar is termed as strain and denoted by ε or e.

Classification of strain

**1. Longitudinal strain**

L = Length of the body,

P = Tensile force acting on the body,

δL = Increase in the length of the body in the direction of P.

Then,

longitudinal strain = δL/L

**2. Lateral strain**

δL = Increase in length,

δb = Decrease in breadth, and

δd = Decrease in depth.

StrainThe longitudinal strain = δL/L

**3. Shear Strain**

Change in initial right angle between two-line elements which are parallel to x and y-axis respectively.

Shear strain

Y = φ

tanФ = δ/L

For smaller angles, tanφ ≈ φ

Ф = δ/L ⇒ δ = Ф.L

τ = P/A

G = τ/y or y = τ/G

**Stress - Strain Diagram**

The mechanical properties of a material are determined in the laboratory by performing tests on small specimens of the material, in the materials testing laboratory. The most common materials test is the tension test performed on a cylindrical specimen of the material.

Stress strain Diagram

It is customary to base all the stress calculations on the original cross-sectional area of the specimen, and since the latter is not constant, the stresses so calculated are known as Nominal stresses.

**Linear Elasticity: Hooke's Law**

The slope of stress-strain curve is called the young’s modulus of elasticity (E):

Slope of stress-strain curve,

E = σ/∈

σ = ∈E

This equation is known as Hooke’s law.

**Type of Metal Behaviour**

**Stress strain Diagram**

**Elongation of Bar Under Different Conditions**

**Uniformly Tapering Circular Bar**

Let us now consider a uniformly tapering circular bar, subjected to an axial force P, as shown in Figure. The bar of length L has a diameter d_{1}at one end and d_{2}at the other end (d_{2}> d_{1}).

Uniformly tapering circular barΔL = 4PL / πEd_{1}d_{2}

**Principle of Superposition**

Bars In Series

A prismatic bar subjected to multiple loads

It stated that

"the resultant strain in bar will be equal to the algebraic sum of the strains caused by the individual forces acting along the length of the member.

Thus, if a member of uniform section is subjected to a number of forces, the resulting deformation (ΔL) is given by

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