Posted On:February 2, 2019, Posted By: Latest Interview Questions,
Views: 961, Rating :

Dear Readers, Welcome to __ Strength of Materials Interview Questions and Answers__ have been designed specially to get you acquainted with the nature of questions you may encounter during your Job interview for the subject of

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

The ratio of increase in length to th original length is known as tensile strain.

The stress induced in a body, when subjected to two equal and opposite pushes, as a result of which there is an decrease in length, is known as compressive stress.

The ratio of increase in length to th original length is known as compressive strain.

The stress induced in a body, when subjected to two equal and opposite forces, which are acting tangentially across the resisting section as a result of which the body tends to shear off across the section is known as shear stress and corresponding strain is known as shear strain.

A bar made up of two or more bars of equal length but of different materials rigidly fixed with each other behaving as one unit for extension or for compression when subjected to an axial tensile or compressive load is called as a composite bar.

Ductile materials ------------> steel, copper

Brittle materials ------------> wrought iron

Malleable materials ------------> cast iron

The ratio of lateral strain to the linear strain is a constant for a given material, when the material is stressed within the elastic limit. This ratio is Poisson’s ratio and it is generally denoted by 1/m or µ.

**Poisson’s ratio 1/m = µ = linear starin/ lateral starin.**

The relationship between modulus of elasticity, modulus of rigidity and Poisson’s ratio is given by

**E = 2C (1+ 1/m)**

E=Modulus of elasticity

C=Modulus of rigidity

1/m = Poisson’s ratio

Hooke’s law is stated as when a material is loaded within elastic limit, the stress is proportional to the strain produced by stress, or Stress/strain=constant. This constant is termed as modulus of elasticity.

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

Mathematically stress is written as

f=P/A

Where f= stress

P=external force or load

A=cross-sectional area.

Unit of stress: stress is represented in N/m2

**Strain:**

The ratio of change in dimension to the orginal dimension when subjected to an external load is termed as strain and is denoted by e. It has no unit.

The ratio of shear stress to the corresponding shear strain when the stress is within the elastic limit is known as modulus of rigidity or shear modulus and is denoted by C or Gor N

Shear stress / Shear strain = q / ?

The ratio of tensile stress or compressive stress to the corresponding strain is known as modulus of elasticity or young’s modulus and is denoted by E.

**stress / strain = E**

When a body is subjected to an uniform direct stress in all the three mutually perpendicular directions, the ratio of the direct stress to the corresponding volumetric strain is found to be a constant is called as the bulk modulus of the material and is denoted by K.

** K = direct stress / volumetric strain = P/A / dv/v**

Longitudinal strain: longitudinal strain is defined as the deformation of the body per unit length in the direction of the applied load.

Longitudinal strain= dL/L

Where L= length of the body.

P= tensile force acting on the body

dL= increase in the length of the body in the direction of P

1. Ductile material

2. Brittle material

It is defined as the of ultimate stress to the working stress or permissible stress.

** Factor of safety = ultimate stress / permissible stress**

**Two conditions employed in solving a composite bar are:**

1. P=f1A1 + f2A2

2. e= f1/E1 + f2/E2

Where P =total load

e= strain

f1 and f2 = stress set up in the respective materials.

E1 and E2 = modulus of elasticity of the respective materials.

Give the relationship between modulus of elasticity, modulus of rigidity and bulk modulus.

** E = 9KC / 3K+C**

Where

E= young’s modulus

K= bulk modulus

C= rigidity modulus

** E = 3K( 1 - 2/m )**

Where

E = Young’s modulus

K = Bulk modulus

1 / m = Rigidity modulus

The stability may be defined as an ability of a material to withstand high load without deformation.

**Example for gradually applied load:**

When we lower a body with the help of a crane, the body first touches the platform on which it is to be placed. On further releasing the chain, the platform goes on loading till it is fully loaded by the body. This is the case of gradually applied load.

**Example for suddenly applied load:**

When we lower a body with the help of a crane, the body is first of all, just above the platform on which it is to be placed. If the chain breaks at once at this moment the whole load of the body begins to act on the platform. This is the case of suddenly applied load.