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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.

Shear Stress, Shear Strain, and Modulus of Rigidity

Hemant More — Thu, 14 Nov 2019 09:17:25 +0000

Science > Physics > Elasticity > Shear Stress, Shear Strain, and Modulus of Rigidity

In this article, we shall study the concept of shear stress, shear strain, and modulus of rigidity.

Shear Stress:

When the deforming forces are such that there is a change in the shape of the body, then the stress produced is called shearing stress. Shear stress is also called as tangential stress.

Mathematically,

Shear stress = Shearing force (F) / Area under shear

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

Shear Strain:

When the deforming forces are such that there is a change in the shape of the body, then the strain produced in the body is called shear strain.

Shearing strain is defined as the ratio of relative displacement of any layer to its perpendicular distance from the fixed layer.

Mathematically,

tan θ = x/h

Modulus of Rigidity:

Within the elastic limit, the ratio of the shear stress to the corresponding shear strain in the body is always constant, which is called modulus of rigidity.

It is denoted by the letter ‘η’. Its S.I. unit of stress is N m^-2 or Pa (pascal) and its dimensions are [L^-1M¹T^-2].

Consider a rigid body as shown in the figure which is fixed along the surface CD. Let it be acted upon by tangential force F along surface AB as shown. Let lateral surface AD get deflected through angle θ as shown. The tangential force F per unit area of surface AB is called shear stress.

Characteristics of Modulus of Rigidity or Shear Modulus:

Within the elastic limit, it is the ratio of shear stress to shear strain
It is associated with the change in the shape of a body.
It exists in solids only.
It describes an object’s tendency to shear
The shear modulus of a material of a body is given by

Characteristics of Moduli of Elasticity:

Modulus of elasticity is the property of the material of a body and is independent of the stress and strain on the body.
a material is said to be elastic if it has a greater value of modulus of elasticity.
The modulus of elasticity for rigid bodies is infinity.
Young’s modulus is the property of solids only. While bulk modulus exists for all the three states of matter.
Gases possess two bulk moduli of elasticity. (i) Isothermal bulk modulus K_iso = P and (ii) Adiabatic bulk modulus K_adia = γP
The elasticity of a substance decreases with the increase in the temperature.

Distinguishing Between Young’s Modulus, Bulk Modulus and Modulus of Rigidity:

Young’s Modulus of Elasticity	Bulk Modulus of Elasticity	Modulus of Rigidity
Within the elastic limit, it is the ratio of longitudinal stress to longitudinal strain	Within the elastic limit, it is the ratio of volumetric stress to volumetric strain	Within theelastic limit, it is the ratio of shear stress to shear strain
It is associated with the change in the length of a body.	It is associated with the change in the volume of a body.	It is associated with the change in the shape of a body.
It exists in solid material bodies	It exists in solids, liquids, and gases.	It exists in solids only.
It is a measure of the stiffness of a solid material	It determines how much the body will compress under a given amount of external pressure.	It describes an object’s tendency to shear
Young’s modulus of the material of a wire is given by	The bulk modulus of the material of a body is given by	Shear modulus of the material of a body is given by

Relation Between the Moduli of Elasticity:

Numerical Problems:

Example – 1:

The area of the upper face of a rectangular block is 0.5 m x 0.5 m and the lower face is fixed. The height of the block is 1 cm. a shearing force applied to the top face produces a displacement of 0.015 mm. Find the strain, stress and the shearing force. Modulus of rigidity = η = 4.5 × 10¹⁰ N/m².

Given: Area under shear = A = 0.5 m x 0.5 m = 0.25 m², Height of the block = h = 1 cm = 1 × 10^-2 m, Displacement of top face = x = 0.015 mm = 0.015 × 10^-3 m = 1.5 × 10^-5 m, Modulus of rigidity = η = 4.5 × 10¹⁰ N/m².

Solution:

To Find: Shear strain =? Shear stress =? Shearing force = F =?

Shear strain = tanθ = x/h = (1.5 × 10^-5) / (1 × 10^-2) = 1.5 × 10^-3Modulus of rigidity = η = Shear stress / Shear strain
∴ Shear stress = η × Shear strain = 4.5 × 10¹⁰ × 1.5 × 10^-3∴ Shear stress = 6.75 × 10⁷ N/m².
Shear stress = F/A
∴ F = Shear stress × Area
∴ F = 6.75 × 10⁷ × 0.25
∴ F = 1.69 × 10⁷N
Ans: Shear strain = 1.5 × 10^-3, Shear stress = 6.75 × 10⁷N/m²,

Shearing force = 1.69 × 10⁷N.

Example – 2:

A metallic cube of side 5 cm, has its lower surface fixed rigidly. When a tangential force of 10⁴kg. wt. is applied to the upper surface, it is displaced through 0.03 mm. Calculate (1) the shearing stress (2) the shearing strain and (3) the modulus of rigidity of the metal.

Given: Area under shear = A = 5 cm x 5 cm = 25 cm² = 25 × 10^-4 m², Height of the block = h = 5 cm = 5 × 10^-2 m, Displacement of top face = x = 0.03 mm = 0.03 × 10^-3 m = 3 × 10^-5 m, Shearing force = 10⁴kg-wt = 10⁴ × 9.8 N.

To Find: Shear strain =? Shear stress =? Modulus of rigidity = η =?

Solution:

Shear stress = F/A
∴ Shear stress = (10⁴ × 9.8)/( 25 × 10^-4)
∴ Shear stress = 3.92 × 10⁷ N
Shear strain = tanθ = x/h = (3 × 10^-5 ) / (5 × 10^-2 ) = 6 × 10^-4Modulus of rigidity = η = Shear stress / Shear strain
η = (3.92 × 10⁷) / (6 × 10^-4) = 6.53 × 10¹⁰N/m²
Ans: Shear stress = 3.92 × 10⁷ N Shear strain = 6 × 10^-4,

Modulus of rigidity = 6.53 × 10¹⁰N/m²

Example – 3:

A 5 cm cube of substance has its upper face displaced by 0.65 cm by a tangential force of 0.25 N. Calculate the modulus of rigidity of the substance.

Given: Area under shear = A = 5 cm x 5 cm = 25 cm² = 25 × 10^-4 m², Height of the block = h = 5 cm = 5 × 10^-2 m, Displacement of top face = x = 0.65 cm = 0.65 × 10^-2 m = 6.5 × 10^-3 m, Shearing force = 0.25 N.

To Find: Modulus of rigidity = η =?

Solution:

Modulus of rigidity = η = Fh/Ax
∴ η = (0.25 × 5 × 10^-2) / (25 × 10^-4 × 6.5 × 10^-3)
∴ η =769 N/m²
Ans: Modulus of rigidity = 769N/m²

Example – 4:

A tangential force of 2100 N is applied on a surface area 3 × 10^-6m² which is 0.1 m from a fixed face of a block of material. The force produces a shift of 7 mm of the upper surface with respect to the bottom. Calculate the modulus of rigidity of the material.

Given: Area under shear = A = 3 × 10^-6m², Height of the block = h = 0.1 m, Displacement of top face = x = 7 mm = 7 × 10^-3 m, Shearing force = 2100 N.

To Find: Modulus of rigidity = η =?

Solution:

Modulus of rigidity = γ = Fh/Ax
∴ η = (2100 × 0.1) / (3 × 10^-6 × 7 × 10^-3)
∴ η =10¹⁰ N/m²
Ans: Modulus of rigidity = 10¹⁰ N/m²

Example – 5:

A metal plate has an area of face 1m x 1m and thickness of 1 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress, strain and magnitude of the tangential force applied. Modulus of rigidity of metal is ϒ = 8.4 × 10¹⁰ N/m²

Given: Area under shear = A = 1 m x 1 cm = 1 m², Thickness of plate = h = 1 cm = 1 × 10^-2 m, Displacement of top face = x = 0.005 cm = 0.005 × 10^-2 m = 5 × 10^-5 m, Modulus of rigidity = η = 8.4 × 10¹⁰ N/m²

To Find: Shear strain =? Shear stress =? Shearing force = F =?

Solution:

Shear strain = tanθ = x/h = (5 × 10^-5) / (1 × 10^-2) = 5 × 10^-3Modulus of rigidity = η = Shear stress / Shear strain
∴ Shear stress = η × Shear strain = 8.4 × 10¹⁰ × 5 × 10^-3∴ Shear stress = 4.2 × 10⁸ N/m².
Shear stress = F/A
∴ F = Shear stress × Area
∴ F = 4.2 × 10⁸ ×1
∴ F = 4.2 × 10⁸N
Ans: Shear strain = 5 × 10^-3, Shear stress = 4.2 × 10⁸N/m²,

Shearing force = 4.2 × 10⁸ N.

Example – 6:

A metal plate has an area of face 1m x 1m and thickness of 5 cm. One face of a larger area is fixed and a tangential force is applied to the opposite face. The displacement of the edge produced thereby is 0.005 cm. Find the shearing stress and shear strain. Modulus of rigidity of metal is η = 4.2 × 10⁶ N/m²

Given: Area under shear = A = 1 m x 1 cm = 1 m², Thickness of plate = h = 5 cm = 5 × 10^-2 m, Displacement of top face = x = 0.005 cm = 0.005 × 10^-2 m = 5 × 10^-5 m, Modulus of rigidity = η = 4.2 × 10⁶ N/m²

To Find: Shear strain =? Shear stress =? Shearing force = F =?

Solution:

Shear strain = tanθ = x/h = (5 × 10^-5) / (5 × 10^-2) = 10^-3Modulus of rigidity = η = Shear stress / Shear strain
∴ Shear stress = η × Shear strain = 4.2 × 10⁶ × 10^-3∴ Shear stress = 4.2 × 10³ N/m².
Ans: Shear strain = 10^-3, Shear stress = 4.2 × 10³N/m².

Example – 7:

A copper metal cube has each side of length 1m. The bottom edge of a cube is fixed and a tangential force of 4.2 × 10⁸ N is applied to the top surface. Calculate the lateral displacement of the surface, if the modulus of rigidity of copper is 14 × 10¹⁰ N/m².

Given: Area under shear = A = 1 m x 1 cm = 1 m², Height of cube = h =1 m, Modulus of rigidity = η = 14 × 10¹⁰ N/m², Shearing force = F = 4.2 × 10⁸ N

To Find: Displacement of top face = x =?

Solution:

Modulus of rigidity = η = Fh/Ax
∴ x = Fh/Aη
∴ x = ( 4.2 × 10⁸× 1)/(1 ×14 × 10¹⁰ )
∴ x = ( 4.2 × 10⁸× 1)/(1 ×14 × 10¹⁰ )
∴ x = 3 × 10^-3 m = 3mm
Ans: Displacement of top face is3mm

Example – 8:

The frame of a brass plate of an outer door design has area 1.60 m² and thickness 1cm. The brass plate experiences a shear force due to the earthquake. How large parallel force must be exerted on each of the edges if the lateral displacement is 0.32 mm. Modulus of rigidity for brass is 3.5 × 10¹⁰ N/m².

Given: Area under shear = A = 1.60 m², Thickness = h =1 cm =1 × 10^-2m, Modulus of rigidity = η = 3.5 × 10¹⁰ N/m², Displacement of top face = x = 0.32 mm = 0.32 × 10^-3m =3.2 × 10^-4m

To Find: Shearing force = F =?

Solution:

Modulus of rigidity = η = Fh/Ax
∴ F = Aηx /h
∴ x = ( 1.60 × 3.5 × 10¹⁰ × 3.2 × 10^-4)/(1 × 10^-2 )
∴ x = 1.792 × 10⁹ N
Ans: Shearing force is 1.792 × 10⁹ N

The post Shear Stress, Shear Strain, and Modulus of Rigidity appeared first on The Fact Factor.

Behaviour of Ductile Material With Increasing Load

Hemant More — Thu, 14 Nov 2019 07:39:40 +0000

Science > Physics > Elasticity > Behaviour of Ductile Material With Increasing Load

In this article, we shall study the construction and use of Searle’s apparatus to find Young’s modulus of wire and behaviour of ductile material of wire under increasing load.

Searle’s Experiment:

Apparatus :

Two identical wires A and B are suspended from rigid support so that the points of suspension are very close to each other. Searle’s apparatus blocks are attached to the lower ends of the wires by means of chucks F₁ and F₂.

Searle’s apparatus block consists of two metal frames P and Q. The two frames are loosely connected by cross strips in such a way that the frame Q can move relatively with respect to frame P. A spirit level S is hinged to the frame P and is rested on the tip of a micrometer screw M which can work in a nut fixed in the frame Q. At the lower end, each frame carries a hanger from which slotted weights can be suspended. Wire A is a dummy wire from which a fixed load of about 1 kg (deadweight) is suspended.

Procedure:

Initial readings and settings: Initially, the length (L) of wire B is measured. Its mean radius (r) is found with the help of a micrometer screw gauge. The wire A is experimental wire, it is initially subjected to a sufficient load called ‘zero load’ (about 1 kg) to avoid kinks in the wire. Micrometer screw is adjusted to bring the bubble in the spirit level at the centre and the reading is noted. This is called ‘zero reading’.
Loading the wire: The load suspended from wire B is then increased in equal steps of about 0.5 kg-wt. let ‘m’ be the mass in the hanger. Each time, after waiting for about two minutes, the bubble is brought to the centre by rotating the screw and micrometer reading is noted. This is extension or elongation (l₁) in the wire. This way five to six readings are taken.
Unloading the wire: After loading procedure is complete the wire is unloaded in the same steps of 0.5 kg-wt and the readings ( l₂) are noted again for each previous step.

Calculations:

The mean of the readings for loading ( l₁) and unloading ( l₂) is calculated and represented as (l) for each step. Then Young’s modulus of the material is calculated in each step using formula,

The average value of Young’s Modulus (Y) is calculated. Care should be taken to avoid possible errors.

Sources of Errors and Their Elimination:

Error due to kinks in the wire: Initially, there may be kinks in the wire, if a load is attached at the free end the kinks will get straightened and observed elongation will be much greater than the actual elongation. To avoid such error sufficient weights are attached to remove all the kinks in the wire. This sufficient or adequate weight is called zero load.
Errors due to a backlash of the screw: As we are using a screw gauge, there is a possibility of error due to backlash. This error can be eliminated by rotating the screw in one direction only when the load is increased and in the opposite direction only when the load is decreased.
Error due to bending (yielding) of the support: If a single wire is used, and if the support from which the wire suspended bends, then the measured extension will be much greater than the actual extension in the wire. This error is eliminated by using dummy wire A. As both the wires are suspended from the same support, if the bending of support occurs, both wires will be lowered to the same extent and there will be no shift in the position of the bubble in the spirit level.
Error due to thermal expansion or contraction: Since a long wire is used, a small change in temperature during the course of the experiment will produce a measurable change in length of the wire due to thermal expansion, then the measured extension will be greater than the actual extension in the wire. This error is eliminated by using a dummy wire. As both experimental wire and dummy wire are of the same material and same original length, the change in length due to change in temperature will be the same for both the wires and thus there will be no shift in the position of the bubble in spirit level.
Error due to the crossing of the elastic limit and/or slipping of the wire from the chucks: If anyone or both these errors are present, the readings of the micrometer screw S while unloading will be different from the corresponding readings while loading. If the two sets of reading do not agree, then the experiment has to be repeated after tightening the chucks. Also, the maximum load to which the wire is subjected must be reduced.

Behaviour (Stress Strain Curve) of Wire of Ductile Material Under Steadily Increasing Load:

The behaviour of wire under increasing load can be studied using Searle’s apparatus. The wire whose behaviour is to be studied is used in the apparatus, at the free end, increasing loads are applied. For each load, stress and strain are calculated. Then the behaviour of wire is studied by plotting a graph, stress versus strain.

For ductile material, the graph is as shown. From O to A graph is a straight line which clearly indicates that the stress is directly proportional to strain, which indicates that Hooke’s Law is obeyed in this region. Point A is called the limit of proportionality.

The elastic limit is the point up to which Hooke’s law is applicable. Stress corresponding to this is called the elastic limit. If the load is removed before the elastic limit is crossed, then the wire will be able to recover its original length completely.

If the load is further increased, we get curve AA’ which indicates that Hooke’s law is not obeyed. The extension starts increasing faster than the load, and the graph bends towards the strain axis. If the wire is strained up to a point A’ and then if the load is removed, the wire is not able to recover its original length. However, the wire still retains its elastic properties. We can see it by the fact, that when the load is steadily reduced, a new straight-line graph such as A’O’ is obtained. In this case, the wire undergoes permanent deformation. The corresponding permanent strain OO’ is called a permanent set or permanent strain or residual strain.

If the load is increased further, a point B is reached, at which the tangent to the curve becomes parallel to the strain axis. It indicates that there is an extension in the wire without an increase in the load. Here wire exhibits plastic flow. The point B is called a yield point and corresponding stress is called yield stress.

Initially, as wire elongates area of cross-section decreases uniformly, but if the wire is loaded beyond point B, stress at some local point starts increasing rapidly due to neck formation in that region and ultimately wire breaks. This point is called the breaking point, and corresponding stress is called breaking stress or ultimate stress or ultimate strength.

For ductile material, there is neck formation at breaking point C. Before breaking ductile material always show plastic flow. For obtaining an appreciable extension of wire in Serle’s experiment, the specimen wire should be long and thin.

Important Points on the Stress-Strain Curve:

Elastic limit: It is the maximum stress to which a body can be subjected without permanent deformation.
Breaking Point: It is the point at which a body subjected to stress breaks (fails).
Breaking Stress: The stress required to cause actual fracture of a material is called the breaking stress or the ultimate strength.
Yield Point: It is the point at which the extension in a wire begins to increase even without any increase in load.
Set: It is the permanent strain produced in a wire when it is stretched beyond the elastic limit.

Types of Material on Their Elastic Behaviour:

Ductile Materials:

Materials which have a great plastic range get stretched too long thin wires before they break are called ductile materials. Hence thin wires can be formed using ductile materials. e.g. steel, aluminium, gold, copper, silver, etc.

Brittle Materials:

Few materials break quite suddenly as soon as the stress-strain curve starts deviating from the straight line after the elastic limit. They are called brittle materials. Hence thin wires cannot be formed using brittle materials. e.g. glass, ceramics, cast iron, etc.

The post Behaviour of Ductile Material With Increasing Load appeared first on The Fact Factor.

Stresses and Strains in Compound Wires

Hemant More — Thu, 14 Nov 2019 07:05:49 +0000

Science > Physics > Elasticity > Stresses and Strains in Compound Wires

Problem – 1:

For two wires of the same material, both the radii and lengths are in the ratio 1:2. What should be the ratio of stretching forces on the wires if equal extensions are to be produced in the two?

Given: Ratio of radii = r₁ / r₂ = 1/2, Ratio of length = L₁ / L₂ = 1/2, Extension equal l₁ = l₂ hence l₁ / l₂ = 1, Material is same hence ratio of Young’s moduli Y₁ / Y₂ = 1,

To Find: Ratio of stretching force = F₁ / F₂ =?

Solution:

Ans: The ratio of stretching forces is 1:2.

Problem – 2:

Two wires of different material, but of the same cross-section and length are stretched by the same force. Find the ratio of Young’s moduli are of their material if the elongations produced in them are in the ratio 3:1.

Given: Same Cross-section, hence the ratio of area = A₁ /A₂ = 1, Lengths are same hence the ratio of length = L₁ / L₂ = 1, Same stretching force hence F₁ / F₂ = 1, Ratio of elongation l₁ / l₂ = 3:1,

To Find: Ratio of Young’s moduli = Y₁ / Y₂ =?

Solution:

Ans: The ratio of Young’s moduli is 1:3.

Problem – 3:

Two wires are of the same material. Wire 1 if four times longer than the wire 2 but wire 1 has a diameter double that of wire 2. Compare stresses and elongations produced in the wires when under the same load.

Given: Material is same hence ratio of Young’s moduli Y₁ / Y₂ = 1, Length L₁ = 4 L₂ hence L₁ / L₂ = 4, Diameter d₁ = 2 d₂ hence d₁/d₂ = 2, hence ratio of radii = r₁/r₂ = 2, Load is same F₁ / F₂ = 1

To Find: Ratio of stresses =? Ratio of elongations =?

Solution:

To find the ratio of stresses

To find ratio of elongations

Ans: The ratio of stresses is 1:4

The ratio of elongation is 1:1. Hence elongations are equal

Example – 4:

Two wires of the same material have lengths in the ratio 2:1 and diameters are in the ratio 2:1 Find the ratio of extensions produced in the wires when the stretching forces acting on them are in the ratio 2:1

Given: Material is same hence ratio of Young’s moduli Y₁ / Y₂ = 1, Ratio of lengths = 2: 1 i.e. L₁ / L₂ = 2, Ratio of diameters = 2:1 hence d₁/d₂ = 2, hence ratio of radii = r₁/r₂ = 2, Load is same F₁ / F₂ = 2:1

To Find: Ratio of extensions =?

Solution:

Ans: The ratio of extension is 1: 1.

Hence the extensions in two wires are equal

Problems Based on Compound wire:

Example – 5:

A brass wire (Y = 11 × 10¹⁰ N/m²) and a steel wire (22 × 10¹⁰ N/m²) of the same length and cross-section are joined end to end. The composite wire is hung from a rigid support and weight is suspended from the free end. Find the extension in each wire if the increase in the length of the composite wire is 0.279 cm.

Given: Young’s modulus for brass Y_b = 11 × 10¹⁰ N/m², Young’s modulus for steel Y_s = 22 × 10¹⁰ N/m², Lengths of wire are same i.e L_b / L_s = 1, Crosssection are same i.e. A_b/A_s = 1, Load is same F_b / F_s = 1, Total extension in composite wire = l_b + l_s = 0.279 cm

To Find: extension in brass wire and steel wire =?

Solution:

l_b = 2 l_s

l_b + l_s = 0.279 cm

∴ 2 l_s + l_s = 0.279 cm

∴ 3 l_s = 0.279 cm

∴ l_s = 0.093 cm

Now l_b = 2 l_s = 2 × 0.093 = 0.186 cm

Ans: Extension in brass wire = 0.186 cm and

extension in steel wire = 0.o93 cm

Example – 6:

A uniform brass wire (Y = 10 × 10¹⁰ N/m²) and a uniform steel wire (Y = 20 × 10¹⁰ N/m²) each of length 3.14 m and diameter 2 × 10^-3 m are joined end to end to form a composite wire is hung from a rigid support and a load suspended from the free end. If the increase in the length of the composite wire is 6 × 10^-3 m, find the increase in the length of each wire.

Given: Young’s modulus for brass Y_b = 10 × 10¹⁰ N/m², Young’s modulus for steel Y_s = 20 × 10¹⁰ N/m², Lengths of wire are same each 3.14 m i.e L_b / L_s = 1, Diameter is same each 2 × 10^-3 m, Hence cross-section are same i.e. A_b/A_s = 1, Load is same F_b / F_s = 1, Total extension in composite wire = l_b + l_s = 6 × 10^-3 m

To Find: extension in brass wire and steel wire =?

Solution:

l_b = 2 l_s

l_b + l_s = 6 × 10^-3 m = 6 mm

∴ 2 l_s + l_s = 6 mm

∴ 3 l_s = 6 mm

∴ l_s = 2 mm

Now l_b = 2 l_s = 2 × 2 = 4 mm

Ans: Extension in brass wire = 4 mm and

extension in steel wire = 2 mm

Example – 7:

A light rod 1 m long is suspended horizontally by two wires of the same length and of the same cross-section but of different materials. The Young’s modulus of material of one wire is 30 x 1010N/m2 and that of the other is 20 x 1010N/m2. At what point should a weight W be hung from the rod so that it still remains horizontal? Ans: (0.4 m from the first wire)

Example – 8:

A weightless rod 105 cm long is suspended horizontally by two wires P and Q of equal length. The crosssection of P is 1 mm2 and that of Q is 2 mm2. From what point on the rod should a weight be suspended in order to produce equal strains in P and Q? Yp= 2 x 1011N/m2; YQ = 1011 N/m2. Ans: (Midpoint)

Example – 9:

A brass wire of length 5 m and of sectional area 1 mm2 is hung from a rigid support with a brass weight of volume 1000 cc hanging from the other end. Find the decrease in length of the wire when the brass weight is completely immersed in water. Y = 1011N/m2. Take g = 10 m/s2. Density of brass = 8400 kg/m3 and of water = 1000 kg/m3. Ans: (0.5 mm)

The post Stresses and Strains in Compound Wires appeared first on The Fact Factor.

Bulk Modulus of Elasticity

Hemant More — Thu, 14 Nov 2019 02:49:27 +0000

Science > Physics > Elasticity > Volumetric Stress, Bulk Modulus of Elasticity

In this article, we shall study the concept of volumetric stress, volumetric strain, and bulk modulus of elasticity.

Volumetric stress:

When the deforming forces are such that there is a change in the volume of the body, then the stress produced in the body is called volume stress. e.g. Solid sphere placed in a fluid under high pressure. Mathematically,

Volumetric Stress = Load / Area = Pressure Intensity = dP

S.I. Unit of stress is N m^-2 or Pa (pascal) and its dimensions are [L^-1M¹T^-2]. Units and dimensions of stress are the same as that of pressure.

The internal restoring force per unit area developed in a body when the body is compressed uniformly from all sides is called hydrostatic stress or hydraulic stress.

Volumetric strain:

When the deforming forces are such that there is a change in the volume of the body, then the strain produced in the body is called volume strain.

Mathematically

Volumetric strain = – Change in volume (dV)/ Original Volume (V)

The negative sign indicates the decrease in the volume

The volumetric strain has no unit and no dimensions.

Bulk Modulus of Elasticity:

Within the elastic limit, the ratio of volumetric stress to the corresponding volumetric strain in a body is always constant, which is called as Bulk modulus of elasticity.

It is denoted by the letter ‘K’. Its S.I. Unit of stress is N m^-2 or Pa (pascal) and its dimensions are [L^-1M¹T^-2].

Mathematically,

Characteristics of Bulk Modulus of Elasticity:

Within the elastic limit, it is the ratio of volumetric stress to volumetric strain.
It is associated with the change in the volume of a body.
It exists in solids, liquids, and gases.
It determines how much the body will compress under a given amount of external pressure.
The bulk modulus of a material of a body is given by

Compressibility:

The reciprocal of bulk modulus of elasticity is called as compressibility. Mathematically

Compressibility = 1 / K

Its S.I. unit is m² N^-1 or Pa-1 and its dimensions are [L^-1M-¹T²].

Numerical Problems:

Example – 1:

A solid rubber ball has its volume reduced by 14.5% when subjected to uniform stress of 1.45 × 10⁴ N/m². Find the bulk modulus for rubber.

Given: Volumetric strain = 14.5 % = 14.5 × 10^-2, Volumetric stress = 1.45 × 10⁴ N/m²,

To Find: Bulk modulus of elasticity =?

Solution:

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴ K = (1.45 × 10⁴) / (14.5 × 10^-2) = 10⁵ N/m²

Ans: Bulk modulus of elasticity of rubber is 10⁵ N/m²

Example -2:

What pressure should be applied to a lead block to reduce its volume by 10% Bulk modulus for lead = 6 × 10⁹ N/m²?

Given: Volumetric strain = 10 % = 10 × 10^-2 , Bulk modulus of elasticity = 6 × 10⁹ N/m².

To Find: Pressure intensity =?

Solution:

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴ Volumetric stress = K ×Volumetric strain

∴ Pressure intensity = K ×Volumetric strain

∴ Pressure intensity = 6 × 10⁹ ×10 × 10^-2

∴ Pressure intensity = 6 × 10⁸ N/m²

Ans: Pressure intensity is 6 × 10⁸ N/m²

Example – 3:

A volume of 5 litres of water is compressed by a pressure of 20 atmospheres. If the bulk modulus of water is 20 × 10⁸ N/m². , find the change produced in the volume of water. Density of Mercury = 13,600 kg/m³; g = 9.8 m/s². Normal atmospheric pressure = 75 cm of mercury.

Given: Original Volume = 5 L = 5 × 10^-3 m³, Pressure = dP = 20 atm = 20 × 75 × 10^-2 × 13600 × 9.8 N/m², Bulk modulus of elasticity of water = 20 × 10⁸ N/m².

To Find: Change in volume = dV =?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴ Change in volume = dV = (dP × V)/ K

∴ dV = 5 × 10^-6 m³ = 5 cc

Ans: The change produced in the volume is 5 cc.

Example – 4:

A volume of 10^-3 m³ of water is subjected to a pressure of 10 atmospheres. The change in volume is 10^-6 m³. Find the bulk modulus of water. Atm. pressure = 10⁵N/m².

Given: Original Volume = 10^-3 m³, Pressure = dP = 10 atm = 10 × 76 × 10^-2 × 13600 × 9.8 N/m², Change in volume = dV =10^-6 m³,

To Find: Bulk modulus of elasticity of water =?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴ K = (10 × 76 × 10^-2 × 13600 × 9.8 × 10^-3)/ 10^-6

∴ K = 1.01 × 10⁹ N/m²

Ans: Bulk modulus of elasticity of water is 1.01 × 10⁹ N/m²

Example – 5:

Two litres of water, when subjected to a pressure of 10 atmospheres, are compressed by 1.013 cc. Find the compressibility of water.

Given: Original Volume = 2 L = 2 × 10^-3 m³, Pressure = dP = 10 atm = 10 × 76 × 10^-2 × 13600 × 9.8 N/m², Change in volume = dV =1.013 cc = 1.013 × 10^-6 m³,

To Find: Compressibility of water =?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴ K = (10 × 76 × 10^-2 × 13600 × 9.8 × 2 × 10^-3)/(1.013 × 10^-6)

∴ K = 2 × 10⁹ N/m²

Compressibility = 1/K = 1/ (2 × 10⁹)

Compressibility = 5 × 10^-10 m²/N

Ans: Compressibility of water is 5 × 10^-10 m²/N

Example – 6:

Bulk modulus of water is 2.05 × 10⁹ N/m². What change of pressure will compress a given quantity of water by 0.5%?

Given: Bulk modulus of water = K = 2.05 × 10⁹ N/m², Volumetric strain = 0.5 % = 0.5 × 10^-2= 5 × 10^-3

To Find: Change in pressure = dP =?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = dP / Volumetric strain

∴ dP = K × Volumetric strain

∴ dP = 2.05 × 10⁹ × 5 × 10^-3

∴ dP = 1.025 × 10⁷ N/m²

Ans: Change in pressure is 1.025 × 10⁷ N/m²

Example – 7:

Calculate the change in volume of a lead block of volume 1 m³ subjected to pressure of 10 atmospheres. Also calculate compressibility of lead. 1 atm = 1.013 × 10⁵ N/m², K = 8 × 10⁵ N/m².

Given: Original Volume = 1 m³, Pressure = dP = 10 atm = 10 × 1.013 × 10⁵ N/m², Bulk modulus of elasticity = K = 8 × 10⁹ N/m².

To Find: Change in volume = dV =? Compressibility = ?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴ Change in volume = dV = (dP × V)/ K

∴ Change in volume = dV = (10 × 1.013 × 10⁵ × 1)/ 8 × 10⁹

∴ dV = 1.27 × 10^-4m³

Compressibility = 1/K = 1/ (8 × 10⁹)

Compressibility = 1.25 × 10^-10 m²/N

Ans: Change in volume is 1.27 × 10^-4m³ and

compressibility of lead is 1.25 × 10^-10 m²/N

Example – 8:

Find the increase in the pressure required to decrease volume of mercury by 0.001%. Bulk modulus of mercury = 2.8 × 10¹⁰ N/m².

Given: Volumetric strain = 0.001% = 0.001 × 10^-2 = 10^-5, Bulk modulus of elasticity = 2.8 × 10¹⁰ N/m².

To Find: Pressure intensity =?

Solution:

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴ Volumetric stress = K ×Volumetric strain

∴ Pressure intensity = K ×Volumetric strain

∴ Pressure intensity = 2.8 × 10¹⁰ × 10^-5

∴ Pressure intensity = 2.8 × 10⁵ N/m²

Ans: Pressure intensity is 2.8 × 10⁵ N/m²

Example – 9:

A solid brass sphere of volume 0.305 m³ is dropped in an ocean, where water pressure is 2 × 10⁷ N/m². The bulk modulus of water is 6.1 × 10¹⁰ N/m². What is the change in volume of the sphere?

Given: Original Volume = 0.305 m³, Pressure = dP = 2 × 10⁷ N/m²², Bulk modulus of elasticity = K =6.1 × 10¹⁰ N/m²

To Find: Change in volume = dV =?

Solution:

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴ Change in volume = dV = (dP × V)/ K

∴ Change in volume = dV = (2 × 10⁷ × 0.305)/ (6.1 × 10¹⁰)

∴ dV = 10^-4m³

Ans: Change in volume = 10^-4m³

The post Bulk Modulus of Elasticity appeared first on The Fact Factor.

Numerical Problems on Poisson’s Ratio

Hemant More — Tue, 12 Nov 2019 04:47:57 +0000

Science > Physics > Elasticity > Numerical Problems on Poisson’s Ratio

In this article, we shall study the concept of poisson’s ratio and numerical problems on it. The concept of this constant (Poisson,s ratio) 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.

Numerical Problems:

Example – 1:

When a brass rod of diameter 6 mm is subjected to a tension of 5 × 10³ N, the diameter changes by 3.6 × 10^-4 cm. Calculate the longitudinal strain and Poisson’s ratio for brass given that Y for the brass is 9 × 10¹⁰ N/m².

Given: Diameter of rod = D = 6 mm, Radius of wire = 6/2 = 3 mm = 3 × 10^-3 m, Load F = 5 × 10³ N, Change in diameter = d = 3.6 × 10^-4 cm = 3.6 × 10^-6 m, Y for the brass is 9 × 10¹⁰ N/m².

To Find: Longitudinal strain =? Poisson’s ratio = ?,

Solution:

Y = Longitudinal Stress /Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Longitudinal Strain = F / (A × Y)

∴ Longitudinal Strain = F / (π r² × Y)

∴ Longitudinal Strain = 5 × 10³ / (3.142 × (3 × 10^-3)² × 9 × 10¹⁰)

∴ Longitudinal Strain = 5 × 10³ / (3.142 × 9 × 10^-6 × 9 × 10¹⁰)

∴ Longitudinal Strain = 1.96 × 10^-3

Now, Lateral strain = d /D = (3.6 × 10^-6)/ (6 × 10^-3) = 6 × 10^-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Poisson’s ratio = (6 × 10^-4) / (1.96 × 10^-3) = 0.31

Ans: Longitudinal strain is 1.96 × 10^-3 and Poisson’s ratio is 0.31.

Example – 2:

A metal wire of length 1.5 m is loaded and an elongation of 2 mm is produced. If the diameter of the wire is 1 mm, find the change in the diameter of the wire when elongated. σ = 0.24.

Given: Original length of wire = L = 1.5 m, Elongation in wire = 2 mm = 2 × 10^-3 m, Diameter of wire = D = 1 mm, Poisson’s ratio = σ = 0.24.

To Find: Change in diameter = d =?

Solution:

Longitudinal strain = l/L = (2 × 10^-3)/1.5 = 1.33 × 10^-3

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.24 × 1.33 × 10^-3= 3.2 × 10^-4

Lateral strain = d / D

∴ d = Lateral strain × D = 3.2 × 10^-4× 1 × 10^-3 = 3.2 × 10^-7m

Ans: The change in diameter is 3.2 × 10^-7m

Example – 3:

A metallic wire (Y = 20 × 10¹⁰ N/m². and σ = 0.26) of length 3 m and diameter 0.1 cm is stretched by a load of 10 kg. Calculate the decrease in diameter of the wire.

Given: Original length of wire = L = 3 m, Diameter of wire = D = 0.1 cm = 0.1 × 10^-2 m = 1 × 10^-3 m, Radius of wire = r = 0.1/2 = 0.05 cm = 0.05 × 10^-2 m = 5 × 10^-4 m,, Stretching load = 10 kg = 10 x 9.8 N, Young’s modulus of elasticity = Y = 20 × 10¹⁰ N/m², and Poisson’s ratio = σ = 0.26

To Find: Decrease in diameter = d =?

Solution:

Y = Longitudinal Stress /Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Longitudinal Strain = F / (A × Y)

∴ Longitudinal Strain = F / (π r² × Y)

∴ Longitudinal Strain = (10 x 9.8) / (3.142 × (5 × 10^-4)² × 20 × 10¹⁰)

∴ Longitudinal Strain = (10 x 9.8) / (3.142 × 25 × 10^-8 × 20 × 10¹⁰)

∴ Longitudinal Strain = 6.24 × 10^-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.26 × 6.24 × 10^-4= 1.62 × 10^-4

Lateral strain = d / D

∴ d = Lateral strain × D = 1.62 × 10^-4×1 × 10^-3 = 1.62 × 10^-7m

Ans: The decrease in diameter is 1.62 × 10^-7m

Example – 4:

A copper wire 3m long and 1 mm² in cross-section is fixed at one end and a weight of 10 kg is attached at the free end. If Y for copper is 12.5 × 10¹⁰ N/m² and σ = 0.25 find the extension, lateral strain and the lateral compression produced in the wire.

Given: Original length of wire = L = 3 m, Area of cross-section of wire = A = 1 mm² = 1 × 10^-6 m², Stretching load = 10 kg = 10 × 9.8 N, Young’s modulus of elasticity = Y = 12.5 × 10¹⁰ N/m², and Poisson’s ratio = σ = 0.25

To Find: Extension = l =? Lateral strain = ?, Lateral compression = ?

Solution:

Y = Longitudinal Stress /Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Longitudinal Strain = F / (A × Y)

∴ Longitudinal Strain = 10 × 9.8 / (1 × 10^-6 × 12.5 × 10¹⁰)

∴ Longitudinal Strain = 7.84 × 10^-4

Now, Longitudinal Strain = l/L

∴ l = Longitudinal strain × L

∴ l = 7.84 × 10^-4 × 3 =2.352 × 10^-3 m = 2.352 mm

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.25 × 7.84 × 10^-4= 1.96 × 10^-4

Area of cross-section = A = 1 × 10^-6 m²

∴ π r² = 1 × 10^-6

∴ r² = 1 × 10^-6/ π = = 1 × 10^-6/ 3.142

∴ r² = 0.3183 × 10^-6

∴ r = 5.64 × 10^-4 m

Diameter = D = 2r = 2 × 5.64 × 10^-4 m = 11.28 × 10^-4 m

Now, Lateral strain = d / D

∴ d = Lateral strain × D = 1.96 × 10^-4× 11.28 × 10^-4 = 2.21 × 10^-7m

Ans: Elongation = 2.352 mm, Lateral strain = 1.96 × 10^-4, Lateral compression = 2.21 × 10^-7m.

Example – 5:

A wire of diameter 2 mm and length 5 m is stretched by a load of 10 kg. Find the extension produced in the wire if Y = 12 × 10¹⁰ N/m². If σ = 0.35 for the material of the wire, find the lateral contraction.

Given: Original length of wire = L = 5 m, Diameter of wire = D = 2 mm = 2 × 10^-3m , Radius of wire = 2/2 = 1mm = 1 × 10^-3m, Stretching load = 10 kg = 10 × 9.8 N, Young’s modulus of elasticity = Y = 12 × 10¹⁰ N/m², and Poisson’s ratio = σ = 0.35

To Find: Lateral contraction =?

Solution:

Y = Longitudinal Stress /Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Longitudinal Strain = F / (A × Y)

∴ Longitudinal Strain = F / (π r² × Y)

∴ Longitudinal Strain = (10 x 9.8) / (3.142 × (1 × 10^-3)² × 12 × 10¹⁰)

∴ Longitudinal Strain = (10 x 9.8) / (3.142 × 1 × 10^-6 × 12 × 10¹⁰)

∴ Longitudinal Strain = 2.6 × 10^-4

Now, Longitudinal Strain = l/L

∴ l = Longitudinal strain × L

∴ l = 2.6 × 10^-4 × 5 = 1.3 × 10^-3 m = 1.3 mm

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.35 ×2.6 × 10^-4= 9.1 × 10^-5

Now, Lateral strain = d / D

∴ d = Lateral strain × D = 9.1 × 10^-5× 2 × 10^-3 = 1.82 × 10^-7m

Ans: Elongation = 1.3 mm, Lateral contraction = 1.82 × 10^-7m

Example – 6:

Find the longitudinal stress to be studied to a wire to decrease its diameter uniformly by 1%. Poisson’s ratio = 0.25, Young’s modulus = 2 × 10¹¹N/m².

Given: Lateral strain = 1 % = 1 × 10^-2, Young’s modulus of elasticity = Y = 2 × 10¹¹ N/m² . and Poisson’s ratio = σ = 0.25

To Find: Longitudinal stress =?

Solution:

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Longitudinal strain =Lateral strain / Poisson’s ratio

∴ Longitudinal strain = 1 × 10^-2 / 0.25 = 4 × 10^-2

Y = Longitudinal Stress /Longitudinal Strain

∴ Longitudinal Stress = Longitudinal Strain × Y

∴ Longitudinal Stress = 4 × 10^-2 × 2 × 10¹¹= 8 × 10⁹N/m² .

Ans: Longitudinal stress = 8 × 10⁹N/m²

Example – 7:

A copper wire 3 m long is stretched to increase its length by 0.3 cm. Find the lateral strain produced in the wire. If Poisson’s ratio for copper is 0.26.

Solution:

Given: Length of wire = L = 3m, Increase in length = l = 0.3 cm = 0.3 × 10^-2m = 3 × 10^-3m, Poisson’s ratio = σ = 0.26

To Find: Lateral strain =?

Longitudinal strain = l/L = (3 × 10^-3)/ 3 = 10^-3

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.36 × 10^-3= 3.6 × 10^-4

Ans: Lateral strain = 3.6 × 10^-4

Example – 8:

A steel wire having cross-sectional area 1 mm² is stretched by 10 N. Find the lateral strain produced in the wire. Young’s modulus for steel is 2 × 10¹¹ N/m² and Poisson’s ratio is 0.291.

Solution:

Given: Area of cross-section = 1 mm² = 1 × 10^-6m², Stretching Load = 10 N, Young’s modulus for steel= Y = 2 × 10¹¹ N/m², Poisson’s ratio = σ = 0.291

To Find: Lateral strain =?

Y = Longitudinal Stress / Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Longitudinal strain = F / (A × Y)

∴ Longitudinal strain = 10 / (1 × 10^-6× 2 × 10¹¹) = 5 × 10^-5

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 5 × 10^-5= 1.455 × 10^-5

Ans: Lateral strain =1.455 × 10^-5

Example – 9:

A load 1 kg produces a certain extension in the wire of length 3 m and radius 5 × 10^-4m. How much will be the lateral strain produced in the wire? Given Y = 7.48 × 10¹⁰ N/m², σ = 0.291.

Solution:

Given: Load attached = F = 1 kg = 1 × 9.8 N, Length of wire = L = 3 m, Radius of cross-section = r = 5 × 10^-4m cross-section = 1 mm² = 1 × 10^-6m², Stretching Load = 10 N, Young’s modulus = Y = 7.48 × 10¹⁰ N/m², Poisson’s ratio = σ = 0.291

To Find: Lateral strain =?

Y = Longitudinal Stress / Longitudinal Strain

∴ Y = F / (A × Longitudinal Strain)

∴ Y = F / (π r² × Longitudinal Strain)

∴ Longitudinal strain = (1 × 9.8) /(3.142 × (5 × 10^-4)² × 7.48 × 10¹⁰)

∴ Longitudinal strain = (1 × 9.8) /(3.142 × 25 × 10^-8 × 7.48 × 10¹⁰)

∴ Longitudinal strain = 1.67 × 10^-4

Poisson’s ratio = Lateral strain / Longitudinal strain

∴ Lateral strain =Poisson’s ratio × Longitudinal strain

∴ Lateral strain = 0.291 × 1.67 × 10^-4= 4.86 × 10^-5

Ans: Lateral strain = 4.86 × 10^-5

The post Numerical Problems on Poisson’s Ratio appeared first on The Fact Factor.

Numerical Problems on Stress, Strain, and Young’s Modulus

Hemant More — Tue, 12 Nov 2019 03:07:34 +0000

Science > Physics > Elasticity > Numerical Problems on Stress, Strain, and Young’s Modulus

In this article, we shall study concept application and numerical problems on longitudinal stress, longitudinal strain, Young’s modulus of elasticity.

Conversion Factors:

From	To	Factor
mm	m	x 10^-3
cm	m	x 10^-2
m	mm	x 10³
m	cm	x 10²
dyne	N	x 10^-5
N	dyne	x 10⁵
cm²	m²	x 10^-4
m²	cm²	x 10⁴
kPa	Pa	x 10³
MPa	Pa	x 10⁶
GPa	Pa	x 10⁹

Formulae:

Example – 1:

A wire 2 m long and 2 mm in diameter, when stretched by weight of 8 kg has its length increased by 0.24 mm. Find the stress, strain and Young’s modulus of the material of the wire. g = 9.8 m/s²

Given: Initial length of wire = L = 2 m, Diameter of wire = 2 mm, Radius of wire 2/2 = 1 mm = 1 × 10^-3 m, Weight attached = m = 2 kg, Increase in length = l = 0.24 mm = 0.24 × 10^-3 m, g = 9.8 m/s².

To Find: Stress =? Strain =? Young’s modulus of material = Y =?

Solution:

Stress = F / A = mg /π r²
∴ Stress = ( 8 × 9.8) /(3.142 ×(1 × 10^-3)²)
∴ Stress = ( 8 × 9.8) /(3.142 × 1 × 10^-6)
∴ Stress = 2.5× 10⁷N/m²
Strain = l / L = 0.24 × 10^-3 / 2
∴ Strain =0.12 × 10^-3 =1.2 × 10^-4Now, Young’s modulus of elasticity= Y = Stress / Strain

∴ Y = (2.5× 10⁷) / (1.2 × 10^-4)
∴ Y = 2.08 × 10¹¹ N/m²
Ans.: Stress = 2.5× 10⁷N/m², Strain =1.2 × 10^-4 , Yong’s modulus of elasticity= 2.08 × 10¹¹ N/m²

Example – 2:

A wire of length 2 m and cross-sectional area 10^-4 m² is stretched by a load 102 kg. The wire is stretched by 0.1 cm. Calculate longitudinal stress, longitudinal strain and Young’s modulus of the material of wire.

Given: Initial length of wire = L = 2 m, Cross-sectional area = A = 10^-4 m, Stretching weight = 102 kg wt = 102 × 9.8 N, Increase in length = l = 0.1 cm = 0.1 × 10^-2 m = 1 × 10^-3 m, g = 9.8 m/s².

To Find: Stress =? Strain = ?, Young’s modulus of material = Y = ?

Solution:

Stress = F / A = mg /A
∴ Stress = ( 102 × 9.8) /10^-4∴ Stress = 1 × 10⁷N/m²
Strain = l / L = 1 × 10^-3 / 2
∴ Strain = 0.5 × 10^-3 = 5 × 10^-4Now, Young’s modulus of elasticity= Y = Stress / Strain = (1 × 10⁷) / ( 5 × 10^-4)
∴ Y = 2 × 10¹⁰ N/m²
Ans.: Stress = 1 × 10⁷N/m², Strain = 5 × 10^-4 , Young’s modulus of elasticity= Y = 2 × 10¹⁰ N/m²

Example – 3:

A mild steel wire of radius 0.5 mm and length 3 m is stretched by a force of 49 N. calculate a) longitudinal stress, b) longitudinal strain c) elongation produced in the body if Y for steel is 2.1 × 10¹¹ N/m².

Given: Initial length of wire = L = 3 m, radius of wire = 0.5 mm = 0.5 × 10^-3 m = 5 × 10^-4 m,Force applied =49 N, Young’s modulus for steel = Y = 2.1 × 10¹¹ N/m².

To Find: Stress =? Strain =? elongation =?

Solution:

Stress = F / A = mg /π r²
∴ Stress = 49 /(3.142 ×(5 × 10^-4)²)
∴ Stress = 49 /(3.142 × 25 × 10^-8)
∴ Stress = 6.238 × 10⁷N/m²
Now, Y = Stress / Strain
∴ Strain = Stress / Y = (6.238 × 10⁷) / (2.1 × 10¹¹)
∴ Strain = 2.970 × 10^-4Now, Strain = l / L
∴ l = Strain × L
∴ l = 2.970 × 10^-4 × 3
∴ l = 8.91 × 10^-4 m = 0. 891 × 10^-3 m = 0.891 mm
Ans.: Stress = 6.238 × 10⁷N/m², Strain = 2.970 × 10^-4, Elongation = 0.891 mm.

Example – 4:

A metal wire 1 m long and of 2 mm diameter is stretched by a load of 40 kg. If Y = 7 × 10¹⁰ N/m² for the metal, find the (1) stress (2) strain and (3) force constant of the material of the wire.

Given: Initial length of wire = L = 1 m, Diameter of wire = 2 mm, Radius of wire = 2/2 = 1 mm = 1 × 10^-3 m, Load attached = m = 40 kg, Young’s modulus of material = Y = 7 × 10¹⁰ N/m².

To Find: Stress =? Strain = ?, Force constant = ?

Solution:

Stress = F / A = mg /π r²
∴ Stress = ( 40 × 9.8) /(3.142 ×(1 × 10^-3)²)
∴ Stress = ( 40 × 9.8) /(3.142 × 1 × 10^-6)
∴ Stress = 1.25 × 10⁸N/m²
Now, Y = Stress / Strain
∴ Strain = Stress / Y = 1.25 × 10⁸/ 7 × 10¹⁰∴ Strain = 1.78 × 10^-3Now, Strain = l /L
∴ extension = l = Strain × L
∴ l = 1.78 × 10^-3 × 1
∴ l = 1.78 × 10^-3 m
Now, force constant K = F/l = mg/l = ( 40 × 9.8) /(1.78 × 10^-3)
∴ Force constant K = 2.2 × 10⁵N/m
Ans.: Stress = 1.25 × 10⁸N/m², Strain = 1.78 × 10^-3, Force constant = 2.2 × 10⁵N/m

Example – 5:

What must be the elongation of a wire 5m long so that the strain is 1% of 0.1? If the wire has cross-selection of 1mm² and is stretched by 10 kg-wt, what is the stress?

Given: Initial length of wire = L = 5 m, Strain = 1% of 0.1 = 1 × 10^-2 × 0.1 = 1 × 10^-3, Area of cross-section = 1 mm² = 1 × 10^-6m², Load attached = F = 10 kg-wt = 10 × 9.8 N .

To Find: Elongation = l =? Stress = ?,

Solution:

Strain = l /L
∴ extension = l = Strain × L
∴ l = 1 × 10^-3 × 5
∴ l = 5 × 10^-3 m = 5 mm
Stress = F / A = mg /π r²
∴ Stress = ( 10 × 9.8) /( 1 × 10^-6)
∴ Stress = 9.8 × 10⁷N/m²
Ans.: Extension = 5 mm and Stress = 9.8 × 10⁷N/m²

Example-6:

A brass wire of length 2 m has its one end, fixed to a rigid support and from the other end a 4 kg wt is suspended. If the radius of the wire is 0.35 mm, find the extension produced in the wire. g = 9.8 m/s², Y = 11 × 10¹⁰ N/m²

Given: Initial length of wire = L = 2 m, Radius of wire = 0.35 mm = 0.35 × 10^-3 m = 3.5 × 10^-4 m, Load attached = F = 4 kg wt = 4 × 9.8 N, g = 9.81 m/s², Y = 11 × 10¹⁰ N/m².

To Find: Extension =?

Solution:

Y = FL /A l
∴ l = F L /π r² Y
∴ l = (4 × 9.8 × 2) /(3.142 × (3.5 × 10^-4)² × 11 × 10¹⁰ )
∴ l = (4 × 9.8 × 2) /(3.142 × 12.25 × 10^-8 × 11 × 10¹⁰ )
∴ l = 1.85 × 10^-3 m = 0.185 × 10^-2 m = 0.185 cm
Ans.: Extension of wire is 0.185 m

Example-7:

A wire of length 1.5 m and of radius 0.4 mm is stretched by 1.2 mm on loading. If the Young’s modulus of its material is 12.5 × 10¹⁰ N/m². , find the stretching force.

Given: Initial length of wire = L = 1.5 m, Radius of wire = 0.4 mm = 0.4 × 10^-3 m = 4 × 10^-4 m, Extension = l = 1.2 mm = 1.2 × 10^-3 m, g = 9.8 m/s², Young’s modulus = Y = 12.5 × 10¹⁰ N/m².

To Find: Stretching force = F =?

Solution:

Y = FL /A l
∴ F = AY l /L
∴ F = π r² Y l /L
∴ F = (3.142 × (4 × 10^-4)² × 12.5 × 10¹⁰ × 1.2 × 10^-3) /1.5
∴ F = (3.142 × 16 × 10^-8 × 12.5 × 10¹⁰ × 1.2 × 10^-3) /1.5
∴ F = 50.27 N
Ans.: Stretching force required = 50.27 N

Example – 8:

What force is required to stretch a steel wire 1 cm2 in cross-section to double its length? Y = 2× 10¹¹ N/m². Assume Hooke’s law.

Given: Initial length of wire = L, Final length = 2L, Hence extension of wire = l = 2L – L = L, Area of cross-section = 1 cm² = 1 × 10^-4m², Young’s modulus of elasticity = Y = 2× 10¹¹ N/m².

To Find: Stretching force = F =?

Solution:

Y = FL /A l
∴ F = AY l /L
∴ F = (1 × 10^-4× 2× 10¹¹× L) /L
∴ F =2 × 10⁷Ans.: Stretching force required = 2 × 10⁷ N

Example – 9:

Find the maximum load which may be placed on a tungsten wire of diameter 2 mm so that the permitted strain not exceed 1/1000. Young’s modulus for tungsten = Y = 35 × 10¹⁰ N/m².

Given: Strain = 1/1000 = 10^-3, Young’s modulus of elasticity = Y = 35 × 10¹⁰ N/m², Diameter of wire = 2 mm, Radius of wire = 2/2 = 1 mm = 1 × 10^-3 m,

To Find: Maximum load = F =?

Solution:

Y = Stress /Strain = (F/A)/Strain
Y = F/(A × strain)
∴ F = π r² × Y× strain
∴ F = 3.142 × (1 × 10^-3)² × 35 × 10¹⁰× 10^-3∴ F = 3.142 × 1 × 10^-6× 35 × 10¹⁰× 10^-3∴ F = 1100 N
Ans.: Maximum load can be placed is 1100 N

Problem – 10:

A mass of 2kg is hung from a steel wire of radius 0.5 mm and length 3m. Compute the extension produced. What should be the minimum radius of wire so that elastic limit is not exceeded? Elastic limit for steel is 2.4 × 10⁸ N/m², Y for steel = Y = 20 × 10¹⁰ N/m²

Given: Radius of wire = 0.5 mm = 0.5 × 10^-3 m = 5 × 10^-4 m. Initial length of wire = L = 3m, Mass attached = m = 2 kg, Y for steel = Y = 20 × 10¹⁰ N/m²

To Find: Extension = l =?, Minimum radius of wire = r =?

Solution:

Part – I:

Y= F L /A l
∴ l = F L /A Y
∴ l = m g L /π r² Y
∴ l = (2 × 9.8 × 3) /(3.142 1 × (5× 10^-4)²× 20 × 10¹⁰ )
∴ l = (2 × 9.8 × 3) /(3.142 × 25× 10^-8× 20 × 10¹⁰ )
∴ l = 3.743 × 10^-4 m = 0.3743 mm

Part – II:

Given: Elastic limit for steel = Stress = 2.4 × 10⁸ N/m², Mass attached = m = 2 kg,

To Find: Radius of wire at elastic limit = r =?

Stress = F /A = F /π r²
∴ r² = mg / (π × Stress)
∴ r² = (2 × 9.8) / (3.142 × 2.4 × 10⁸)
∴ r² = 2.599× 10^-8∴ r = 1.612× 10^-4m = 0.1612× 10^-3m = 0.1612 mm

Ans.: Part – I:Change in length of wire is 0.3743 mm
Part – II: Radius of wire at elastic limit = 0.1612 mm

Example – 11:

A wire is stretched by the application of a force of 50 kg wt/sq. cm. What is the percentage increase in the length of the wire? Y = 7 × 10¹⁰ N/m², g = 9.8 m/s²

Given: Stress = 50 kg wt/sq. cm = 50 × 9.8 N / 10^-4 m² = 50 × 9.8 × 10⁴N/m², Young’s modulus of elasticity = Y = 7 × 10¹⁰ N/m². g = 9.8 m/s²

To Find: % elongation = % l/L =?

Solution:

Now, Y = Stress / Strain
∴ Strain = Stress / Y = (50 × 9.8 × 10⁴)/ (7 × 10¹⁰)
∴ Strain = 7 × 10^-5% elongation = Strain × 100 = 7 × 10^-5 × 100
% elongation = Strain × 100 = 0.007
Ans.: Elongation is 0.007 percent

Problem – 12:

A compressive force of 4 × 10⁴ N is exerted at the end of a bone of length 30 cm and 4 cm² square cross-sectional area. What will happen to the bone? Calculate the change in length of a bone. Compressive strength of bone is 7.7 × 10⁸ N/m² and Young’s modulus of bone is 1.5 × 10¹⁰ N/m²

Given: Initial length of wire = L = 30 cm = 0.30 m, Area of cross-section = 4 cm² = 4× 10^-4m², Load attached = F = 4 × 10⁴ N . Y = 1.5 × 10¹⁰ N/m². Maximum Stress = 7.7 × 10⁸ N/m².

To Find: Effect of loading =? Change in length = l = ?,

Solution:

Applied Stress = Applied force / Area of cross-section
Applied Stress = (4 × 10⁴ )/ (4× 10^-4) = 1 × 10⁸ N/m²
This stress is less than the maximum allowable stress (7.7 × 10⁸ N/m²)
Hence the bone will not break but will get compressed and its length decreases
Y= F L /A l
∴ l = (4 × 10⁴ × 0.3) /(4× 10^-4 × 1.5 × 10¹⁰)
∴ l = 2 × 10^-3 m = 2 mm
Ans.: The length of bone decreases by 2 mm

Example – 13:

The radius of a copper bar is 4 mm. What force is required to stretch the rod by 20% of its length assuming that the elastic limit is not exceeded? Y = 12 × 10¹⁰ N/m².

Given: Radius of wire = r = 4 mm = 4 × 10^-3 m, % elongation = Strain = 20% = 20 × 10^-2, Young’s modulus of elasticity = Y = 12× 10¹⁰ N/m².

To Find: Stretching force = F =?

Solution:

Y = Stress /Strain = (F/A)/Strain
Y = F/(A × strain)
∴ F = AY× strain
∴ F = π r² × Y × strain
∴ F = 3.142 × (4 × 10^-3)² × 12× 10¹⁰ × 20 × 10^-2∴ F = 3.142 × 16 × 10^-6 × 12× 10¹⁰ × 20 × 10^-2∴ F = 1.207× 10⁶ N
Ans.: Stretching force required = 1.207× 10⁶ N

Example – 14:

Find the change in length of a wire 5m long and 1 mm² in cross-section when the stretching force is 10 kg-wt. Y = 4.9 × 10¹¹ N/m², and g=9.8 m/s².

Solution:
Given: Initial length of wire = L = 5 m, Area of cross-section = 1 mm² = 1 × 10^-6m², Load attached = F = 10 kg-wt = 10 × 9.8 N . Y = 4.9 × 10¹¹ N/m², and g=9.8 m/s².
To Find: Change in length = l =?

Y = FL /A l
∴ l = F L /A Y
∴ l = (10 × 9.8 × 5) / (1 × 10^-6 × 4.9 × 10¹¹)
∴ l = 1 × 10^-3 m = 1 mm
Ans.: Change in length of wire is 1 mm

Example – 15:

Elastic limit is exceeded when the strain in a wire (Y=14 × 10¹¹ N/m²) exceeds 1/2000. If the area of the cross-section of the wire is 0.02 cm², find the maximum load that can be used for stretching the wire without causing a permanent set.

Given: Strain = 1/2000 = 5 × 10^-4, Young’s modulus of elasticity = Y = 14 × 10¹¹ N/m², Area of cross section = A = 0.02 cm² = 0.02 × 10^-4m² = 2 × 10^-6m²

To Find: Stretching force = F =?

Solution:

Y = Stress /Strain = (F/A)/Strain
Y = F/(A × strain)
∴ F = AY× strain
∴ F = 2 × 10^-6 × 14 × 10¹¹ × 5 × 10^-4∴ F = 1400 N
Ans.: Stretching force required = 1400 N

Example – 16:

Elastic limit of steel is exceeded when the stress on given steel wire exceeds 8.26 × 10⁸ N/m². Can a steel wire (Y = 2 × 10¹¹ N/m²) 2m long be stretched by 10 mm without exceeding the elastic limit?

Given: Initial length of wire = L = 2 m, Elastic limit = stress = 8.26 × 10⁸ N/m², Young’s modulus of elasticity = Y = 2 × 10¹¹ N/m².

To Find: To find whether wire can be stretched by 10 mm.

Solution:

Y = Stress /Strain = Stress/(l/L)
∴ Y = (Stress × L) / Y
∴ Y = (8.26 × 10⁸ × 2) / 2 × 10¹¹∴ Y = 8.26 × 10^-3 m = 8.26 mm
Ans.: Wire cannot be stretched up to 10 mm because the elastic limit will get crossed at extension of 8.26 mm.

Example – 17:

Young’s modulus of the material of a wire is 9.68 × 10¹⁰ N/m². A wire of this material of diameter 0.95 mm is stretched by applying a certain force. What should be the limit of this force if the strain is not to exceed 1 in 1000?

Given: Strain = 1/1000 = 10^-3 , Young’s modulus of elasticity = Y = 9.68 × 10¹⁰ N/m², Diameter of wire = 0.95 mm, Radius of wire = 0.95/2 = 0.475 mm = 0.475 × 10^-3 m = 4.75 × 10^-4 m

To Find: Stretching force = F =?

Solution:

Y = Stress /Strain = (F/A)/Strain
Y = F / (A × strain)
∴ F = π r² × Y× strain
∴ F = 3.142 × (4.75 × 10^-4)² × 9.68 × 10¹⁰× 10^-3∴ F = 68.62
Ans.: Limit of the stretching force required = 68.62 N

Example – 18:

The elastic limit of copper is 1.5 × 10⁸ N/m². A copper wire is to be stretched by a load of 10 kg. Find the minimum diameter the wire must have if the elastic limit is not to be exceeded.

Given: Elastic Limit = Stress = 1.5 × 10⁸ N/m², Load = F = 10 kg wt = 10 × 9.8.

To Find: Minimum diameter of the wire.

Solution:

Stress = F / A
∴ Stress = F / π r²
∴ r² = F / (π × Stress )
∴ r² = (10 × 9,8) / (3.142 × 1.5 × 10⁸ )
∴ r² = 20.79 × 10^-8∴ r = 4.56 × 10^-4m = 0.456 × 10^-3m = 0.456 mm
Diameter of wire = 2 × r = 2 × 0.456 mm =0.912 mm
Ans.: Diameter of wire is 0.912 mm

Example – 19:

What would be the greatest length of a steel wire which when fixed at one end can hang freely without breaking? Density of steel = 7800 kg/m³. Breaking stress for steel = 7.8 × 10⁸ N/m².

Given: Density of steel = ρ = 7800 kg/m³. Stress = 7.8 × 10⁸ N/m².

To Find: Greatest length of the wire =?.

Solution:

Stress = F / A = mg / A = V ρ g /A
∴ Stress = A L ρ g / A
∴ Stress = L ρ g
∴ L = Stress / ρ g
∴ L = 7.8 × 10⁸ / (7800 × 9.8)
∴ L = 7.8 × 10⁸ / (7800 × 9.8)
∴ L = 1.021 × 10⁴ m
Ans.: Maximum length of copper wire is 1.021 × 10⁴ m

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

The post Longitudinal Stress and Strain appeared first on The Fact Factor.

Classification of Materials

Hemant More — Fri, 25 Oct 2019 05:29:04 +0000

Science > Physics > Elasticity > Classification of Materials

In this article, we shall study the deformation in the body due to the application of deforming force and classification of materials on this behaviour.

Effects of Applied Force on a Body:

If the forces acting on a body are unbalanced i.e. the resultant force acting on the body is not zero, then the body sets into translational motion.
If the forces acting on a body are balanced but the net moment is not zero i.e. the resultant force acting on the body is zero but the resultant moment is not zero, then the body sets into rotational motion.
If the forces on the body are balanced and there is no net moment for the body then the force may produce a change in size, shape or both in the body.

Deformation and Deforming Force:

The forces applied to a body can produce a change in shape or size or both the shape and size of the body. Such a change is called deformation. The forces which produce deformation in the body are called deforming forces.

Characteristics of Deformation:

The extent of deformation is always directly proportional to the deforming force.
Due to deforming force the length of wire (chord) or volume of a body or shape of body changes.
When deformation takes place, the body is said to be strained or in a deformed state.
In the deformed state, the applied force is numerically equal to internal elastic restoring force within a body but in the opposite direction.
Deformation produced in a body is due to change in relative positions of the molecules within a body, due to applied deforming force.

Classification of Material on the Basis of Deformation:

Every engineering structure is subjected to external loading of various kinds (normal load, shear load or mixed load such as bending, torsion, etc.). When a substantial load is applied on the material, it automatically tends to deform. Depending upon the response of the material towards the deforming force, the materials are classified into three types: elastic material, plastic material, and rigid material. Corresponding material properties are elasticity, plasticity, and rigidity respectively.

Elasticity:

The property by virtue of which material bodies regain their original dimensions (size, shape or both) after removal of deforming force is called elasticity. The material exhibiting elasticity is called elastic material and the body is called the elastic body. e.g. Rubber, Steel, Aluminum, Sponge etc.

As solids have a definite shape and definite volume, therefore, they alone possess elasticity of shape as well as volume. Liquids have a definite volume but an indefinite shape. Hence, they possess volume elasticity. Gases possess volume elasticity.

When the body is stretched the inter-atomic spacing increases and when it is compressed the inter-atomic spacing decreases. In both cases, internal forces are created in the body which tend to restore the atoms back to their original positions. Such internal forces are called internal elastic forces or restoring forces. The magnitude of restoring or internal elastic force is the same as the applied force. These restoring forces are responsible for the elastic property of the body.

Characteristics of Elasticity:

Elasticity is the property by virtue of which material bodies regain their original dimensions (size, shape or both) after removal of deforming force (external force).
When the body is stretched the inter-atomic spacing increases and when it is compressed the inter-atomic spacing decreases. In both cases, internal restoring forces are created in the body which tend to restore the atoms back to their original positions.
The deforming force required is small.
The amount of elastic deformation is very small.
Within elastic limit the Hooke’s law is applicable.
Within the elastic limit, the stress-strain graph is a straight line.
Energy absorbed by the material during elastic deformation is called a module of resilience.
Elastic deformation Occurs before plastic deformation.
The mechanical and metallurgical properties of the solid material remain unaltered even after deformation.
It is important concept in designing mechanical and civil structures, frames and equipment.

Click the Following Link for Video Lecture

Plasticity:

When a body is acted upon by deforming forces the shape and/or size of the body changes. But, if the deforming forces are removed, the body retains its new shape and size. Such body is called a plastic body and the property is called plasticity. e.g. Plaster of Paris, Clay, Mud, Plastic, etc. show plasticity.

In plastic bodies, no internal elastic force or restoring force is produced. If produced they are negligible. Hence they cannot regain original shape but in the absence of restoring force, they retain their new shape.

A body which can easily be deformed when a very small deforming force is applied and which does not regains its original shape or size or both but retains its new shape is called a perfectly plastic body. No body in the universe is a perfectly plastic body. Wall putty can be considered as a perfectly plastic body.

Characteristics of Plasticity:

Plasticity is the property by virtue of which material bodies retain their new dimensions (size, shape or both) even after removal of deforming force (external force).
Due to no or negligible internal elastic force or restoring force they cannot regain original shape but retain their new shape.
The deforming force required is large.
The amount of elastic deformation is very large.
Hooke’s law is not applicable.
In the plastic region, the stress-strain graph is non-linear.
Total energy absorbed by the material during the elastic and plastic deformation region is called module of toughness.
Plastic deformation occurs only after the elastic deformation.
Many properties of the solid material change after plastic deformation.
It is important concept in sheet metal working, moulding, rolling, forging, riveting, etc.

Rigidity:

When a body is acted upon by deforming forces, the shape and size of the body do not get altered, whatever may be the magnitude of deforming forces. Such body is called a rigid body and the property is known as rigidity.

In rigid bodies, the internal force of attraction is so high that there is no relative motion between two particles of the body. Hence there is no change in the shape of the body.

In practice, it is not possible for us to get a perfectly rigid body. But large blocks of metal, stones can be considered rigid bodies.

Characteristics of Rigidity:

Rigidity is the property by virtue of which material bodies do not undergo a change in their original dimensions (size, shape or both) whatever be the magnitude of the deforming force (external force).
In this case, the distances between the constituent particles do not change.
However, maybe the large force is applied there is no change in the shape of the body.
Hooke’s law is not applicable.
As there is no change in the shape of a rigid body, no energy of deformation is in it.
There is a stress in rigid material but no strain.

To prove that within the elastic limit, Young’s modulus of the material of wire is the stress required to double the length of wire.

As the length is doubled, change in length = l = 2L – L = L

Strain = l / L = L/L = 1

Now, Young’s modulus of elasticity is given by

Y = Stress / Strain = Stress/1 = Stress

Hence within the elastic limit, Young’s modulus of the material of wire is the stress required to double the length of wire.

Distinguishing Between Elasticity and Plasticity:

Elasticity	Plasticity
The body exhibiting elasticity regains its shape or size after the removal of the external force.	The body exhibiting plasticity retains its new shape or size after the removal of the external force.
There is a temporary change in dimensions of the body on the application of the deforming force.	There is a permanent change in dimensions of the body on the application of the deforming force.
The internal restoring force is set up inside the body.	No Internal restoring force is set up inside the body. Or they are very negligible.
The ratio of stress to corresponding strain produced is constant.	The ratio of stress to the corresponding strain produced is not constant.
materials exhibiting elasticity: Steel, rubber, etc	Materials exhibiting plasticity: PVC, plaster of Paris, wax, etc.

Scientific Reasons:

Pressure and stress have the same dimensions but are two different physical quantities.

Both the pressure and the stress have the same dimensions [L-1M1T-2], but the pressure is a scalar quantity while stress is a tensor quantity. Hence they are two different physical quantities.

Steel is more elastic than a rubber.

The elastic properties depend on Young’s modulus of elasticity. Greater the value of Young’s modulus, the more the elastic material is. The value of Young’s modulus of elasticity is more for steel than that for rubber. Hence steel is more elastic than rubber.

The post Classification of Materials appeared first on The Fact Factor.

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