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what is strain energy

Concept of Strain Energy

By Hemant More
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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 [L 2 M 1 T -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 -1 M 1 T -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 × 10 11 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 × 10 11 N/m².

To Find: Work done = W =?

Solution:

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

∴ F = (2 × 10 11 × 1 × 10 -6 × 1 × 10 -3 )/2

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 × 10 10 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 × 10 10 N/m².

To Find: Work done = W =?

Solution:

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

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

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 × (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

∴ 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 × (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 × 10 4 J/m³

Ans : The strain energy per unit volume of the wire 2.08 × 10 4 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 = 10 11 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 = 10 11 N/m².

To Find: Energy stored = dU =?

Solution:

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

∴ l = (2 × 9.8 × 1) / (1 × 10 -6 × 10 11 )

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 × 10 11 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 × 10 11 N/m².

To Find: Energy stored = dU =?

Solution:

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

∴ F = (1.25 × 10 11 × 1.5 × 10 -6 × 2 × 10 -3 )/2.5

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 × 10 11 N/m².

Given: Strain = l/L = 0.5 % = 0.5 × 10 -2 = 5 × 10 -3 , Young’s modulus = Y = 1.2 × 10 11 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 × 10 11

∴ dU/V = ½ × 25 × 10 -6 × 1.2 × 10 11

∴ dU/V = 1.5 × 10 6 J/m³

Ans: The strain energy per unit volume of the wire 1.5 × 10 6 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 × 10 4 J/m³

Ans: The strain energy per unit volume of the wire 4.9 × 10 4 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 × 10 10 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 × 10 10 N/m².

To Find: Work done = W =?

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

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

∴ 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 × 10 10

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 × 10 10 N/m².

To Find: Energy stored = dU =?

Solution:

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

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

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 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.

What is strain energy

by Marina Gandelsman

Strain energy is one of fundamental concepts in mechanics and its principles are widely used in practical applications to determine the response of a structure to loads.

Strain Energy in Uniaxial Loads

Consider a prismatic bar of length L subjected to a tensile force P. The load is applied slowly, so there are no effects due to motion. Such loads are called static loads. As the load reaches its full value P, the bar gradually elongates to L + d .

During this process, the load P gradually moves over the length d and does a certain amount of work. From physics we recall that

W = F * d

However, in this case, the force varies in magnitude (from F=0 to F=P). To find the value of work done under these conditions, we look at a load-displacement diagram to determine the manner in which the force varies.

The work done by the load is equal to the area under the curve. As the load is applied, strains are produced and their presence increases the energy of our bar. This strain energy is the energy absorbed by the bar as a result of its deformation under load. From the principle of conservation of energy we know that this energy is equal to the work done by the load, assuming no other energy transfer (such as heat) occurred.

U = W = S L P(x) dx

Sometimes this energy is referred to as internal work, to distinguish it from work done by the load. The unit of strain energy is the same as work – J (SI) and ft-lb (British).
If the force P is gradually removed, the bar will shorten and at least a portion of the strain energy will be recovered in the form of work. If the material has not exceeded its elastic limit, the bar will return to its original length L, otherwise a permanent set will remain.

If the material of the bar follows Hooke’s Law, the load-displacement curve is a straight line, P = k d , and the strain energy stored in the bar is:

U = W = k d 2 /2 = P d /2

The complimentary energy in the preceding picture is used with Castigliano’s theorems which apply to linearly elastic systems for small deformations as well as Complementary Strain Energy theorems.

The total strain energy in a bar composed of several sections is equal to the sum of strain energies in each section; however, it is important to realize that the total strain energy for a bar with several loads is not equal to the sum of strain energies for each load.

The total strain energy determined from a load-deformation curve is not really indicative of material since the results will depend on the size of the test specimen. In order to eliminate size as a factor, we consider the strain energy per unit volume (also known as strain energy density). Since P = s A and d = e L, (2) can be rewritten as follows:

U = ( se /2)*AL
u = U/AL
u = se /2 = s 2 /2E = E e 2 /2 (3)

The unit of strain energy density are J/m3 (SI) and in-lb/in3 (British).

The area under a complete stress-strain diagram gives a measure of a material’s ability to absorb energy up to fracture and is called toughness. The larger the area under the diagram, the tougher the material. A high modulus of toughness is important when a material is subject to impact loads. In the inelastic range, only a small part of the energy absorbed by the material is recoverable. Most of the energy is dissipated in the form of heat. The energy that may be recovered when a specimen has been stressed to point A is represented by triangle ABC. AB is parallel to OD since all materials essentially behave elastically upon the release of stress. The area OABO represents the inelastic strain energy (dissipated). The strain-energy density of the material when it is stressed to the proportional limit (D on the diagram) is called modulus of resilience. It is found by substituting the proportional limit s pl into one of elastic energy equations:

Resilience represents the ability of the material to absorb and release energy within the elastic range.

Strain Energy in Torsion

Consider a prismatic bar AB in pure torsion under the action of torque T. When the load is applied statically, the bar twists and the free end rotates through an angle f. Again, assume the material is linearly elastic and follows Hooke’s Law. The relationship between T and f will also be linear.
From this, we determine that

U = W = T f /2

Using the equation f = TL/GI p , we can express strain energy as

U = S L T(x) 2 /2GI p dx = T 2 L/2GI p = GI p f 2 /2L

If the bar is subjected to non-uniform torsion, the total strain energy is equal to the sum of strain energies of each segment with constant torque, but again, the total strain energy of a structure supporting several loads is not equal to the sum of strain energies from each load.

Strain Energy in Pure Shear

Consider an element of dimensions x, y, and z subjected to shear load t . As this element is deformed, the force on top plane reaches a final value of t xz. The total displacement of this force for a small deformation of the element is g y. Therefore,

W = U = 1/2 t xz * g y = 1/2 tg V = 1/2 t 2 V/G

The strain-energy density in this case is

u = 1/2 tg = 1/2 t 2 /G

Strain Energy in Bending

Consider a beam in pure bending by couples of moment M. Its material follows Hooke’s Law and rotations are small. The normal stress varies linearly from the neutral axis and s = -My/I. >From U = s І/2E and the stress equation, we get

U = M 2 L/2EI

Substituting q = ML/EI, we get

U = EI q 2 /2L

If the bending moment in a beam varies along its length, we can obtain the total strain energy by applying one of the preceding equations to an element of the beam and integrating along its length. From d q = 1/ r dx = d 2 v/dx 2 dx, we get

U = S L M(x) 2 /2EI dx = S L EI/2 (d 2 v/dx 2 ) 2 dx

The previous equations only consider the effect on bending on the beam. If shear forces are also present, additional strain energy will be stored in the beam, however, this energy is negligible in beams where L >> t. If a beam supports a single load (either point load P or moment M 0 , we can determine either deflection d (for P) or angle of rotation q (for M 0 ) from strain energy. The deflection is measured along the line of action of the load and is positive in the direction of the load. The angle of rotation is the angle of rotation of the beam axis at the point where the moment is applied. We can obtain the following equations:

U = W = P d /2
U = W = M 0 q /2

This method is limited in its applications because only one deflection (or angle) can be found and that deflection (or angle) must correspond to the load (or couple).
The equations demonstrated here are fairly basic and simplified in several ways. Only the perfectly elastic bodies do not dissipate any energy and thus can store all of the work as recoverable energy. Most real-life materials are not elastic. A lot of research is concerned with what happens in the postelastic regions, however, this discussion is beyond the scope of this paper.

Bibliography

Mechanics of Materials, Gere & Timoshenko, 4th edition, PWS Publishing
Engineering Mechanics of Materials, B.B. Muvdi and J.W. McNabb, Macmillan Publishing Co.
Engineering Mechanics of Solids, E. Popov, 2nd edition, Prentice Hall
Engineering Mechanics of Deformable Bodies, E.F. Byars and R.D. Snyder, 3rd edition, Intext Educational Publishers

© 1999 by Marina Gandelsman. Distribute freely (I borrowed quite liberally from my sources, obviously 🙂

What is strain energy by Marina Gandelsman Strain energy is one of fundamental concepts in mechanics and its principles are widely used in practical applications to determine the response of a