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Concept of Strain Energy

Hemant More — Sat, 23 Nov 2019 11:46:05 +0000

Science > Physics > Elasticity > Concept of Strain Energy

In this article, we shall study, work done in stretching wire and the concept of strain energy.

Work done in Stretching a Wire:

Consider a wire of length ‘L’ and area of cross-section ‘A’ be fixed at one end and stretched by suspending a load ‘M’ from the other end. The extension in the wire takes place so slowly that it can be treated as quasi-static change; because internal elastic force in the wire is balanced by the external applied force and hence acceleration is zero.

Let at some instant during stretching the internal elastic force be ‘f’ and the extension produced be ‘x’. Then,

Since at any instant, the external applied force is equal and opposite to the internal elastic force, we can say that the work done by the external applied force in producing a further infinitesimal dx is

Let ‘ l ‘ be the total extension produced in the wire, and work done during the total extension can be found by integrating the above equation.

This is an expression for the work done in stretching wire.

Strain Energy:

The work done by the external applied force during stretching is stored as potential energy (U) in the wire and is called as strain energy in the wire. Thus the strain energy is given by

Its S.I. unit is J (joule) and its dimensions are [L²M¹T ^-2].

Strain Energy Per Unit Volume of a Wire:

The work done by external applied force during stretching is stored as potential energy (U) in the wire and is called as strain energy in the wire. Dividing both sides above equation by AL, the volume of the wire.

This is an expression for strain energy or potential energy per unit volume of stretched wire. This is also called as the energy density of the strained wire. Its S.I. unit is J m^-3 and its dimensions are [L^-1M¹T ^-2].

Different Forms of Expression of Strain Energy per Unit Volume:

By definition of Young’s modulus of elasticity

Now. Young’s modulus of elasticity for a material of a wire is constant.

Thus, strain energy per unit volume ∝ (stress)² i.e. strain energy per unit volume is directly proportional to the square of the stress.

Note:

More work is to be done for stretching a steel wire than stretching a copper wire because steel is more elastic than copper. Due to which more restoring force is produced in the steel, hence we have to do more work to overcome these larger restoring forces.

Numerical Problems:

Example – 1:

Find the work done in stretching a wire of length 2 m and of sectional area 1 mm² through 1 mm if Young’s modulus of the material of the wire is 2 × 10¹¹ N/m².

Given: Area = A = 1 mm² = 1 × 10^-6 m², Length of wire = L = 2m, Extension in wire = l = 1mm = 1 × 10^-3 m, Young’s modulus = Y =2 × 10¹¹ N/m².

To Find: Work done = W =?

Solution:

Young’s modulus of elasticity = Y = FL/Al

∴ F = YAl/L

∴ F = (2 × 10¹¹× 1 × 10^-6× 1 × 10^-3)/2

∴ F = 100 N

Now Work done in stretching wire = ½ Load × Extension

∴ Work done = ½ × 100 × 1 × 10^-3

∴ Work done = 0.05 J

Ans: Work done in stretching wire is 0.05 J

Example – 2:

Calculate the work done in stretching a wire of length 3 m and cross-sectional area 4 mm² when it is suspended from rigid support at one end and a load of 8 kg is attached at the free end. Y = 12 × 10¹⁰ N/m² and g = 9.8 m/s².

Given: Area = A = 4 mm² = 4 × 10^-6 m², Length of wire = L = 3m, Load = 8 kg-wt = 8 × 9.8 N, Young’s modulus = Y = 12 × 10¹⁰ N/m².

To Find: Work done = W =?

Solution:

Young’s modulus of elasticity = Y = FL/Al

∴ l = FL/AY

∴ l = (8 × 9.8 × 3) / (4 × 10^-6 × 12 × 10¹⁰)

∴ l = 4.9 × 10^-4 m

Now Work done in stretching wire = ½ Load ×Extension

∴ Work done = ½ × 8 × 9.8 × 4.9 × 10^-4

∴ Work done = 1.921 × 10^-2 J = 0.0192 J

Ans: Work done in stretching wire is 0.0192 J

Example – 3:

When the load on a wire is increased slowly from 3 to 5 kg wt, the elongation increases from 0.6 to 1 mm. How much work is done during the extension? g = 9.8 m/s².

Given: Initial Load = F₁ = 3 kg wt = 3 × 9.8 N, Final load =F₂ = 5 kg-wt = 5 × 9.8 N, Initial extension l₁ = 0. 6 mm = 0.6 × 10^-3 m = 6 × 10^-4 m, Final extension = l₂ = 1mm = 1 × 10^-3 m = 10 × 10^-4 m, g = 9.8 m/s² .

To Find: Work done = W =?

Solution:

Work done = W = W2 – W1

∴ Work done = ½ × F₂× l₂ – ½ × F₁ × l₁

∴ Work done = ½ × (F₂ × l₂ – F₁ × l₁)

∴ Work done = ½ × (5 × 9.8 × 10 × 10^-4 – 3 × 9.8 × 6 × 10^-4)

∴ Work done = ½ × 9.8 × 10^-4(50 – 18)

∴ Work done = ½ × 9.8 × 10^-4× 32

∴ Work done =1.568 × 10^-2= 0.01568 J

Ans: Work done is 0.01568 J

Example – 4:

A spring is compressed by 1 cm by a force of 3.92 N. What force is required to compress it by 5 cm? What is the work done in this case? Assume the Hooke’s Law.

Given: Initial Load = F₁ = 3.92 N, Initial extension l₁ = 1 cm = 1 × 10^-2 m, Final extension = l₂ = 5 cm = 5 × 10^-2 m.

To Find: Final Load = F₂ =? Work done = W =?

Solution:

We have Force constant = K = F/l

Hence F₁/l₁ = F₂/l₂

Hence F₂ = (F₁× l₂)/ l₁ = (3.92× 5 × 10^-2) /(1 × 10^-2)

∴ F₂ = (3.92× 5 × 10^-2) / (1 × 10^-2)

Ans: (9.8 N; 0.49)

Example – 5:

A wire 4m long and 0.3 mm in diameter is stretched by a load of 0.8 kg. If the extension caused in the wire is 1.5 mm, find the strain energy per unit volume of the wire.g = 9.8 m/s²

Given: Length of wire = L = 4m, Diameter = 0.3 mm, Radius of wire = r = 0.3/2 = 0.15 mm = 015 × 10^-3 m = 1.5 × 10^-4 m, Area = Load applied = F = 0.8 kg-wt = 0.8 × 9.8 N, Extension in wire = l = 1.5 mm = 1.5 × 10^-3 m, .g = 9.8 m/s².

To Find: Strain energy per unit volume = dU/V =?

Solution:

Strain energy per unit volume =½ × Stress × Strain

∴ dU/V =½ × (F/A) × (l/L)

∴ dU/V =½ × (Fl/AL)

∴ dU/V =½ × (Fl/πr²L)

∴ dU/V =½ × (0.8 × 9.8 × 1.5 × 10^-3) / (3.142 × (1.5 × 10^-4)² × 4)

∴ dU/V =½ × (0.8 × 9.8 × 1.5 × 10^-3) / (3.142 × 2.25 × 10^-8 × 4)

∴ dU/V = 2.08 × 10⁴ J/m³

Ans : The strain energy per unit volume of the wire 2.08 × 10⁴ J/m³

Example – 6:

Find the energy stored in a stretched brass wire of 1 mm² cross-section and of an unstretched length 1 m when loaded by 2 kg wt. What happens to this energy when the load is removed? Y = 10¹¹N/m².

Given: Area = A = 1 mm² = 1 × 10^-6 m², Length of wire = L = 1 m, Load = 2 kg-wt = 2 × 9.8 N, Young’s modulus = Y = 10¹¹ N/m².

To Find: Energy stored = dU =?

Solution:

Young’s modulus of elasticity = Y = FL/Al

∴ l = FL/AY

∴ l = (2 × 9.8 × 1) / (1 × 10^-6 × 10¹¹)

∴ l = 1.96 × 10^-4 m

Now Work done in stretching wire = ½ Load ×Extension

∴ Work done = ½ × 2 × 9.8 × 1.96 × 10^-4

∴ Work done = 1.921 × 10^-3 J

Now energy stored = Work done in stretching wire

Ans: Energy stored is 1.921 × 10^-3 J

Example – 7:

A metal wire of length 2.5 m and are of cross section 1.5 × 10^-6 m² is stretched through 2 mm. Calculate work done during stretching. Y = 1.25 × 10¹¹ N/m².

Given: Area = A = 1.5 × 10^-6 m², Length of wire = L = 2.5 m, Extension = l = 2mm = 2 × 10^-3m, Young’s modulus = Y = 1.25 × 10¹¹ N/m².

To Find: Energy stored = dU =?

Solution:

Young’s modulus of elasticity = Y = FL/Al

∴ F = YAl/L

∴ F = (1.25 × 10¹¹ × 1.5 × 10^-6× 2 × 10^-3)/2.5

∴ F = 150 N

Now Work done in stretching wire = ½ Load ×Extension

∴ Work done = ½ × 150 × 2 × 10^-3

∴ Work done = 0.150 J

Now energy stored = Work done in stretching wire

Ans: Energy stored is 0.150 J

Example – 8:

A copper wire is stretched by 0.5% of its length. Calculate the energy stored per unit volume in the wire. Y = 1.2 × 10¹¹ N/m².

Given: Strain = l/L = 0.5 % = 0.5 × 10^-2 = 5 × 10^-3, Young’s modulus = Y = 1.2 × 10¹¹ N/m².

To Find: Strain energy per unit volume = dU/V =?

Solution:

Strain energy per unit volume = dU/V = ½ × (Strain)² × Y

∴ dU/V = ½ × (5 × 10^-3)² × 1.2 × 10¹¹

∴ dU/V = ½ × 25 × 10^-6 × 1.2 × 10¹¹

∴ dU/V = 1.5 × 10⁶J/m³

Ans: The strain energy per unit volume of the wire 1.5 × 10⁶J/m³

Example – 9:

Calculate the strain energy per unit volume in a brass wire of length 2.0 m and cross-sectional area 0.5 mm2, when it is stretched by 2mm and a force of 5 kg-wt is applied to its free end.

Given: Area = A = 0.5 mm² = 0.5 × 10^-6 m² = 5 × 10^-7 m², Length of wire = L = 2.0 m, Extension in wire = l = 2 mm = 2 × 10^-3 m, Load applied = F = 5 kg-wt = 5 × 9.8 N

To Find: Strain energy per unit volume = dU/V =?

Solution:

Strain energy per unit volume = dU/V = ½ × Stress × Strain

∴ Strain energy per unit volume = ½ × (F/A) × (l/L)

∴ Strain energy per unit volume = ½ × (Fl/AL)

∴ Strain energy per unit volume = ½ × (5 × 9.8 × 2 × 10^-3) / (5 × 10^-7 × 2)

∴ Strain energy per unit volume = 4.9 × 10⁴J/m³

Ans: The strain energy per unit volume of the wire 4.9 × 10⁴J/m³

Example – 10:

Calculate the work done in stretching a wire of length 2 m and cross-sectional area 0.0225 mm² when a load of 100 N is applied slowly to its free end. Young’s modulus of elasticity = 20 × 10¹⁰N/m².

Solution:

Given: Area = A =0.0225 mm² =0.0225 × 10^-6 m² = 2.25 × 10^-8 m², Length of wire = L = 2 m, Load applied = F = 100 N, Young’s modulus of elasticity = Y = 20 × 10¹⁰N/m².

To Find: Work done = W =?

Young’s modulus of elasticity = Y = FL/Al

∴ l = FL/AY

∴ l = (100 × 2) / (2.25 × 10^-8 × 20 × 10¹⁰)

∴ l = 4.444 × 10^-2 m

Now Work done in stretching wire = ½ Load ×Extension

∴ Work done = ½ × 100 × 4.444 × 10^-2

∴ Work done = 2.222 J

Ans: Work done in stretching wire is 2.222 J

Example – 11:

A uniform steel wire of length 3 m and area of cross-section 2 mm² is extended through 3mm. Calculate the energy stored in the wire, if the elastic limit is not exceeded. Young’s modulus of elasticity = Y = 20 × 10¹⁰

Given: Area = A =2 mm² =2 × 10^-6 m², Length of wire = L = 3 m, Extension = l = 3 mm = 3 × 10^-3 m, Young’s modulus of elasticity = Y = 20 × 10¹⁰N/m².

To Find: Energy stored = dU =?

Solution:

Young’s modulus of elasticity = Y = FL/Al

∴ F = YAl/L

∴ F = (20 × 10¹⁰× 2 × 10^-6× 3 × 10^-3)/3

∴ F = 400 N

Now Work done in stretching wire = ½ Load ×Extension

∴ Work done = ½ × 400 × 3 × 10^-3

∴ Work done = 0.6 J

Energy stored = work done in stretching wire = 0.6 J

Ans: Energy stored is 0.6 J

The post Concept of Strain Energy appeared first on The Fact Factor.

Longitudinal Stress and Strain

Hemant More — Fri, 25 Oct 2019 07:05:20 +0000

Science > Physics > Elasticity > Longitudinal Stress and Strain

In this article, we shall study the concept of stress, strain, and modulus of elasticity. Main focus will be on longitudinal stress, longitudinal strain, and Young’s modulus of elasticity.

Concept of Stress:

The net internal elastic force (restoring force) acting per unit area of the surface which is subjected to deformation is called stress.

Stress = F / A

Stress is denoted by the letter ‘σ’. S.I. unit of stress is N m^-2 (newton per square meter) or Pa (pascal) and its dimensions are [L^-1M¹T^-2]. Units and dimensions of stress are the same as that of pressure.

Characteristics of Stress:

Stress is an internal restoring force per unit area.
It opposes the change in the size, shape or both of the body. i.e. it opposes deformation.
It is a tensor quantity.
Its S.I. unit is Nm^-2 and its dimensions are [L^-1M¹T^-2].

Concept of Strain:

The change in dimension per unit original dimension of a body subjected to deforming forces is called Strain.

Strain = Change in dimension / Original dimension

A strain is denoted by letter ‘e’. It is a pure ratio, (Ratio of two similar quantities) hence it is unitless, dimensionless quantity [L⁰M⁰T⁰].

Characteristics of Strain:

The strain is defined as the ratio of change in dimensions of a body to its original dimensions when subjected to deformation.
For longitudinal loading, both the longitudinal and lateral strain are produced.
It is a scalar quantity.
it is unit less, dimension less quantity.

Hooke’s Law of Elasticity:

Statement: Within the elastic limit, the stress developed in the body is directly proportional to the strain produced in the body.

By Hooke’s Law,

Stress ∝ Strain

∴ Stress = Constant x Strain

∴ Stress / Strain = Constant

This constant of proportionality is called the modulus of elasticity or coefficient of elasticity. Its units and dimensions are the same as that of stress. Its S.I. unit is N m^-2 (newton per square meter) or Pa (pascal) and its dimensions are [L^-1M¹T^-2]. Depending upon the nature of stress and strain these constants are further classified as (1) Young’s modulus of elasticity (2) Bulk modulus of elasticity and (3) Modulus of rigidity.

Graphical Representation:

Characteristics of Modulus of Elasticity:

The ratio of the stress produced in a body to corresponding stress produced in it is called the modulus of elasticity of the material of the body.
It is the characteristic property of the material of the body.
Its S.I. unit is Nm-2 and its dimensions are [L-1M1T-2].
Depending upon loading there are three types of elasticity constant. If there is a change in the length of a body then the corresponding elastic constant is called Young’s modulus of elasticity. If there is a change in the volume of a body then the corresponding elastic constant is called the bulk modulus of elasticity. If there is a change in the shape of a body then the corresponding elastic constant is called modulus of rigidity.

Elastic Limit:

It is the upper limit of deforming force up to which if the deforming force is removed, the body regains its original shape and size completely. If the deforming force is increased beyond this limit, there is permanent deformation in the body called a permanent set. Elastic limit is a property of a body, while elasticity is the property of the material of the body.

Longitudinal Loading (Along the Length):

Longitudinal stress:

When the deforming forces are such that there is a change in the length of the body, then the stress produced in the body is called longitudinal stress. Longitudinal stress is further classified into two types. Tensile stress and compressive stress.

Longitudinal stress = F / A

Tensile stress:

When the deforming force is such that there is the increase in the length of the body, then the stress produced in the body is called tensile stress.

Compressive stress:

When the deforming force is such that there is a decrease in the length of the body, then the stress produced in the body is called compressive stress.

Longitudinal strain:

When the deforming forces are such that there is a change in the length of the body, then the strain produced in the body is called longitudinal strain. The longitudinal strain is further classified into two types. Tensile strain and tensile strain. Mathematically the longitudinal strain is given by

Longitudinal strain = Change in length(l) / Original length (L)

The longitudinal strain has no unit and no dimensions.

Tensile strain:

When the deforming force is such that there is an increase in the length of the body, then the strain produced in the body is called tensile strain.

Compressive strain:

When the deforming force is such that there is a decrease in the length of the body, then the strain produced in the body is called compressive strain.

Click the Following Link for Video Lecture

Young’s Modulus of Elasticity:

Within the elastic limit, the ratio of the longitudinal stress to the corresponding longitudinal strain in the body is always constant, which is called Young’s modulus of elasticity. It is denoted by letter “Y” or “E”. Young’s modulus of elasticity is not defined for liquids and gases. The international standard symbols for Young’s modulus E is derived from word élasticité (French for elasticity), while some authors use Y as it is the first letter of the expression Young’s modulus of elasticity.

Its S.I. unit of stress is N m^-2 (newton per square meter) or Pa (pascal) and its dimensions are [L^-1M¹T^-2]. Mathematically,

Youn’s modulus of elasticity = Longitudinal stress / Longitudinal strain

Young’s Modulus of Elasticity of the Material of a Wire:

Consider a wire of length ‘L’ and radius of cross-section ‘r’ is fixed at one end and stretched by suspending a load of ‘mg’ from the other end. Let ‘l‘ be the extension produced in the wire when it is fully stretched.

Now, by the definition of Yong’s modulus of elasticity we have This is an expression for Young’s modulus of elasticity of a material of a wire.

This is an expression for Young’s modulus of elasticity of a material of a wire.

Characteristics of Young’s Modulus of Elasticity:

Within the elastic limit, it is the ratio of longitudinal stress to longitudinal strain
It is associated with the change in the length of a body.
It exists in solid material bodies
It is a measure of the stiffness of a solid material.
Young’s modulus of the material of a wire is given by

Poisson’s Ratio:

The concept of this constant was introduced by physicist Simeon Poisson. When a rod or wire is subjected to tensile stress, its length increases in the direction of stress, but its transverse (lateral) dimensions decrease and vice-versa. i.e. when the length increase, the thickness decreases and vice-versa. In other words, we can say that the longitudinal strain is always accompanied by a transverse (lateral) strain.

The ratio of transverse strain to the corresponding longitudinal strain is called Poisson’s ratio. It is denoted by letter ‘m’. It has no unit. It is a dimensionless quantity.

Poisson’s Ratio = Lateral strain / Longitudinal strain

For homogeneous isotropic medium -1 ≤ m ≤ 0.5. In actual practice, Poisson’s ratio is always positive. There are some materials with a negative Poisson’s ratio. Poisson’s ratio of cork is zero, that of metal is 0.3 and that of rubber is 0.5.

Materials with a negative value of Poisson’s ratio are said to be auxetic. They grow larger in the transverse direction when stretched and smaller when compressed. Most auxetic materials are polymers with a crumpled, foamy structure. Pulling the foam causes the crumples to unfold and the whole network expands in the transverse direction.

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