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 = 10¹¹ N/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^-3 m, 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 × 10⁶    J/m³

Ans: The strain energy per unit volume of the wire 1.5 × 10⁶    J/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 × 10⁴ J/m³

Ans: The strain energy per unit volume of the wire 4.9 × 10⁴ J/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 × 10¹⁰ N/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 × 10¹⁰ N/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 × 10¹⁰ N/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

Related Topics:

• Classification of Materials
• Longitudinal Stress, Strain, and Young’s Modulus of Elasticity
• Numerical Problems on Stress, Strain, and Young’s Modulus of Elasticity
• Numerical Problems on Poisson’s ratio
• Numerical Problems on Compound Wires
• Behaviour of Ductile Material Under Increasing Load
• Volumetric Stress, Volumetric Strain, and Bulk Modulus of Elasticity
• Shear Stress, Shear Strain, and Modulus of Rigidity

For More Topics of Physics Click Here

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

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:

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^-3
Modulus 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 × 10⁷ N
Ans: Shear strain = 1.5 × 10^-3, Shear stress = 6.75 × 10⁷ N/m²,

Shearing force = 1.69 × 10⁷ N.

Example – 2:

A metallic cube of side 5 cm, has its lower surface fixed rigidly. When a tangential force of 10⁴ kg. 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  = 10⁴ kg-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^-4
Modulus of rigidity = η = Shear stress / Shear strain
η = (3.92 × 10⁷) / (6 × 10^-4) = 6.53 × 10¹⁰   N/m²
Ans: Shear stress = 3.92 × 10⁷ N Shear strain = 6 × 10^-4,

Modulus of rigidity = 6.53 × 10¹⁰   N/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^-6 m² 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^-6 m², 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^-3
Modulus 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 × 10⁸ N
Ans: Shear strain = 5 × 10^-3, Shear stress = 4.2 × 10⁸ N/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^-3
Modulus 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 × 10³ N/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^-2 m, Modulus of rigidity = η = 3.5 × 10¹⁰ N/m², Displacement of top face = x = 0.32 mm = 0.32 × 10^-3 m =3.2 × 10^-4 m

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

Related Topics:

• Classification of Materials
• Longitudinal Stress, Strain, and Young’s Modulus of Elasticity
• Numerical Problems on Stress, Strain, and Young’s Modulus of Elasticity
• Numerical Problems on Poisson’s ratio
• Numerical Problems on Compound Wires
• Behaviour of Ductile Material Under Increasing Load
• Volumetric Stress, Volumetric Strain, and Bulk Modulus of Elasticity
• Strain Energy

For More Topics of Physics Click Here

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