Content Curator
Strain Energy is the energy stored in a body because of deformation or strain. It is denoted by ‘U’. It is equal to the work done during the deformation process. It is also known as Resilience. The unit of strain energy is N-m or Joules. Here, we will discuss the derivation and expression of strain energy under different loads applied.
Read More: Mechanical Properties of Solids
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Key Terms:- Stress, Strain energy, Force, Resilience, stretch, Joules, work done.
Strain Energy
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When a load is applied to a body or material, it deforms and work is done upon it. If the applied load or force is removed, the body comes back to its original form. For example, when we stretch a spring, it always comes back to its earlier form.
Strain Energy
Read More: Stress and Strain Curve
So, we can say that,
Strain Energy = Work done
But strain energy is stored in a body only if,
- The load is in the elastic limit of the deformable material.
- Permanent deformation doesn’t occur.
- The amount of load and type of load also changes the amount of strain energy.
- These are the ideal conditions for the derivation of strain energy.
If strain energy is distributed inside the body uniformly, then strain energy per unit volume is known as Strain Energy Density.
Read More: Elastic behaviour of solids
Derivation of Strain Energy
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Consider a body of original length ‘L’ and cross-sectional area ‘A’ stretched by a force ‘F’ acting along its length. It gets stretched and an elongation ‘I’ is produced in it.
Strain in a Body
If the body is perfectly elastic, then,
Longitudinal stress = F/A
and
Longitudinal strain = l/L
We know that,
∴ Young’s modulus (E) = Longitudinal stress/ Longitudinal strain
E = F/A X l/L
∴ F = YAl/L … (i)
The magnitude of stretching force increases from zero to F during the elongation of the body. Let ‘f’ be the restoring force and ‘x’ be its corresponding extension at a certain instant during the process of extension.
∴ f = EAx/L … (ii)
Let ‘dW’ be the work done for the further small extension ‘dx’.
Work = force X displacement
∴ dW = fdx
∴ dW = EAx/L dx … (iii) [from (ii)]
Read More: Beam Deflection Formula
The total amount of work done in stretching the body from x = 0 to x = l, can be found out by integrating the equation (iii).
W = ∫0l dW
Substituting from equation (iii)
= ∫0lEAx/L dx
= EA/L ∫0lxdx
On Integrating,
W = EA/L [x2/2]0l
On removing the limits,
= EA/L [l2/2 - 02/2]
= EAl/L x l/2
But, EAl/L = F … (i)
W = ½ x F x l
∴ Work done by stretching a wire,
W = ½ x load x extension
Work done by the stretching force is equal to strain energy gained by the body.
∴ Strain energy = ½ x load x extension
Work done per unit volume = work done in stretching wire/volume of the body
= ½ x F x l /V
= ½ F x l / Ax L
=½ x F/A x l/L
=½ x stress x strain
Strain energy per unit volume = ½ x stress x strain
Read More: Poisson's Ratio
Strain Energy based on Load
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Strain energy may change depending on the load applied into two categories as,
- Strain Energy when a load is applied gradually
- Strain energy when a load is applied suddenly
Strain Energy When Load is Applied Gradually
Strain energy stored in a body is equal to the work done by an applied load in stretching the body.
Applying load gradually
Let,
Gradually applied load = P
Extension of the body = x
Cross-sectional area = A
Length of the body = L
Volume of the body = V
Young’s Modulus = E
Strain energy stored in the body = U
Stress induced in the body = σ
Work done = Area of load extension curve
Work done = Area of OBC
→ ½ P x … (i) [Area = ½ height base]
Load, P = Stress Area
P = σ x A
Extension, x = Strain length
Extension, x = Stress/E L [as strain = Stress/E, Hooke’s Law]
Extension, x = σ/E L
Substituting values of P & x in eq. (i),
Work done by load = ½ × σ A/E L
U = σ2 / 2E × V [as A x l = V]
The maximum energy stored up to the elastic limit is Proof Resilience. If σ is taken as elastic limit stress, we get Proof Resilience.
Proof Resilience = σ2 / 2E × V
The proof resilience per unit volume is known as Modulus of Resilience ( M.o.R).
M.o.R = (σE)2 / 2E
Strain Energy when a load is applied suddenly
When a load is applied suddenly to a body, the load is constant throughout the process of deformation of the body.
Applying load suddenly
Read More: Thermal Stress
Load applied suddenly = P
Length of the body = L
Area of cross-section = A
Volume of the body = V
Young’s Modulus = E
Extension of the body = x
Stress induced in the body = σ
Strain energy stored in the body = U
Work done by load = load extension
Work done by load = P x … (i)
Maximum strain energy to the work done,
U = σ2/2E × V
U = σ2/2E × A × L … (ii)
Equating the strain energy to the work done,
σ2/2E × A × L = P × σ / E × L [as x = σ / E × L, Hooke’s Law]
Or
σ×A /2 = P
Or
σ = 2 × P/A
This shows that the maximum stress induced due to a suddenly applied load is twice the stress induced when the same load is applied gradually.
The value of strain energy stored in the body due to suddenly applied load can be determined by the value of the stress (σ) induced in the body due to suddenly applied load.
Read More:
Comparison of Strain Energy
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When we compare the strain energy stored in a body due to gradually applied load & due to suddenly applied load, we get,
σgradual = Pgradual/A
σsudden = 2Psudden/A
Or
σsudden = 2σgradual
And Strain energy, U = σ2 / 2E × V
Ugradual = σ2 /2E × V
Usuddenly = (2σgradual)2 /2E × V = 4σ2/2E × V
∴ Ugradual / Usuddenly = (4σ2/2E × V) × 2E/σ2V = 4
So, the strain energy stored in a body due to suddenly applied load is four times compared to the strain energy stored in a body due to gradually applied load.
Strain energy stored due to Shear Stress
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After straining or deformation of ABCD to ABC1D1,
Energy Stored Due to Shear Stress
Let,
Shear load applied gradually = P
Shear stress developed in the body due to shear force = σ
Shear strain = CC1/CB = DD1/DA = ɸ
Modulus of rigidity = G
Length = l
Breadth = b
Height = h
Cross-sectional area = A
Volume = V
Shear deformation = CC1 or DD1
Strain energy = U
Shear stress = shear force / Area
Area of top face = l × b
= P/A = P/ l × b
P = σ × l × b
Shear strain,ɸ = CC1/CB
CC1= CB × ɸ
If shear force P is applied gradually, the average load will be equal to P/2.
Work done by gradually applied shear force = Average load × distance
Work done by gradually applied shear force = P/2 × CC1
= ½ × (σ × l × b) × (CB ×ɸ)
= ½ × σ × l × b × h ×ɸ [as P = σ × l × b, & CB = h]
= ½ × σ × σ/G × V [as l×b×h = V, shear strain = σ/G]
= ½ × σ2/G × V
And work done is equal to strain energy,
Hence,
Strain energy stored, U = σ2V / 2G
Read More: Stress
Things to Remember
- Strain energy is equal to the work done.
- Strain energy is given by U = Fδ /2, where, δ is compression and F is the force applied.
- Strain energy is given by U = ½ Vσε, where, V is volume, σ is stress, ε is strain. (σ ∝ ε)
- Strain energy is given by, U = σ2/2E × V, where, σ is stress, E is Young’s modulus, V is the volume of a body.
- Resilience or the total strain energy stored in a body is given by the formula U = σ2/2E × V, where, σ is stress, E is Young’s modulus, V is volume of the body.
- Proof Resilience, Up = (σE)2/ 2E × V, where UE = stress at elastic limit.
- Modulus of Resilience, Um = (σE)2/2E
- For shear stress, strain energy, U = σ2V/2G, where σ = shear stress, G = Modulus of Rigidity or shear modulus, and V = volume.
Sample Questions
Ques: A bar having an area of 90 mm2 has a length of 3m. Determine the strain energy if the stress of 300MPa is applied when stretched. Young’s modulus is given as 200GPa. (3 marks)
Ans. Given,
Area of the bar, A = 90 mm2
Length of the bar, L= 3m
Stress applied in the bar, σ = 300 MPa.
Young’s modulus, E = 200 GPa.
Now, the volume of the bar is given by,
V=area × length
= (90 × 10-6) × 3
= 270 × 10-6 m3
Now, the strain energy formula is given as, U= σ2 / 2E× V
= (Stress)2 / 2E × volume of material
= (300×106)2 / 2 × (200×109) × 270 ×10-6
= 83.3 x 10-6 J
Therefore, the strain energy stored inside the rod is 83.3 x 10-6 J.
Ques: Calculate the Work Done in Stretching a Wire of Length 5m and its Cross-Sectional Area is Given by 1mm2 is Deformed by the Length of 1mm if Young’s Modulus is Given of the Wire is Given by 2 × 1011 N/m2. (3 Marks)
Ans: Given,
Length of the wire = l = 5m
Area of the wire = A = 1mm2 =10-6m2
Deformation due to stretching = δl =1mm=10-3mA
Young’s modulus of the wire = E = 2 × 1011 N/m2
Young’s Modulus of Elasticity is given by,
E = Fl / Aδl
∴ F = EAδl / l = (2 × 1011 N/m2) (10-6m2) (10-3m) / 5m
⇒ F = 40N
Now work done on stretching the wire is given by,
⇒ U = Fδl / 2
⇒ U = (40×10-3) / 2
⇒ U = 0.02J
Therefore, 0.02J of work is done in stretching 5m long wire.
Ques: When a force of 1000 N is applied to a body, it gets compressed by 1.2 mm. Determine the strain energy. (3 Marks)
Ans: Given,
Force F = 1000 N,
Compression δ = 1.2 mm
Strain energy formula is given by,
U = Fδ / 2
= 1000 × 1.2 × 10-3 / 2
Therefore, U = 0.6 J.
Ques: A load of 100N falls through a height of 2cm on the collar, rigidly attached to the lower end of a vertical bar 1.5m long and 1.5 cm2 cross-sectional area. Determine (3 Marks)
(i) Max. Inst. Stress Induced
(ii) Max. Inst. Elongation
(iii) Strain energy stored in vertical rod; Take E = 2 × 105 N/mm2.
Ans: Impact load P = 100N
Height, h = 2cm = 20mm
Length, L = 1.5m = 1500mm
Area, A = 1.5cm2 = 150 mm2
Volume, V = A × L = 1500 × 150 = 225000 mm3
E = 2 × 105 n/mm2
Let, δL = Max. elongation
σ = Max. Inst. Stress induced
U = Strain energy
(i) σ = P/A (1 + 1+2AEh/PL)
= 100/150 {1 + 1+ (2×150×2×105×20) /100×150}
= 100/150 (1 + 1+8000
= 60.23 N/mm2.
(ii) δl = σ/E × L
= 60.23×1500 / 2×105 = 0.452 mm
(iii) Strain Energy U = σ2/2E × V
= (60.23)2 × 225000 / 2×2×105 = 2045 N-mm = 2.045 N-m.
Ques: An unknown weight falls through a height of 10mm on a collar rigidly attached to the lower end of a vertical bar 500 cm long and 600 mm2. If the max extension of rod is to be 2mm, what is the corresponding stress & magnitude of unknown weight? [ E = 2 × 105 N/mm2] (3 Marks)
Ans: Height, h = 10mm
Length, L = 500cm = 5000mm
Area, A = 600mm2
Max. extension δl =2mm
E = 2 × 105N/mm2
σ = inst. Stress produced
P =weight falling on collar
E = stress/strain
σ = E × δL/L
= 2 × 105 × 2/5000 = 80 N/mm2
Value of falling weight,
σ = P/A (1 + 1+2AEh/PL)
80 = P/600 (1 + 1+2 × 600 × 2 × 105 × 10/P × 5000)
4800/P = 1 + 1+480000/P
4800/P -1 = 1+480000/P
Squaring on both sides,
(4800/P)2 + 12 - 2×4800×1/P = 1 + 480000/P [using (a-b)2 = a2 + b2 - 2ab]
Or, 2304000000/P2 = 480000/P + 9600/P = 576000/P
Or, 2304000000/P2 = 57600
Or, P = 2304000000/57600 = 4000N = 4kN.
Ques: The maximum stress produced by a pull in a bar of length 1m is 150N/mm2. The area of cross-sections and length are shown in the figure below. Calculate the strain energy stored in the bar if E = 2 × 105 N/mm2. (4 marks)
Ans: Given,
Length of bar, L = 1m = 1000mm
Max. stress, σ = 150 N/mm2
Part AB: Length, L1 = 475mm
Area, A1 = 200mm2
Part BC: Length, L2 = 50mm
Area, A2 = 100mm2
Part CD: Length, L3 = 475mm
Area, A3 = 200mm2
Value of E = 2 × 105 N/mm2
σ2 = stress in BC = 150 N/mm2
σ1 = stress in AB or CD
Load = σ1 × A1 = σ2 × A2
σ1 = σ2 × A2 /A1 = 150×100 /200 = 75 N/mm2
Strain energy in AB,
U1 = (σ1)2/2E × V1
V1 = A1 + L1 = 200 × 475 = 95000 mm3
∴ U1 = (σ1)2/2E × 95000
= (752/2×105) × 95000
= 1335.938 N-mm
Strain energy in part BC,
U2 = (σ2)2/2E × V2
= (1502/ 2 × 2×105) × A2 × L2 [ as V2 = A2 × L2]
= (1502/ 2 × 2×105) × 100 × 50
= 281.25 N-mm
Energy stored in part CD,
U3 = (σ3)2/2E × V3
= 1335.938 N-mm [ as V3=V1, σ3=σ1, U3 =U1]
∴ Total strain energy stored, U = U1 + U2 + U3 = 1335.938 + 281.25 + 1335.938 = 2953.126 N-mm.
Ques: A steel rod is 2m long and 50mm in diameter. An axial pull of 199kN is suddenly applied. Calculate the instantaneous stress and also the instantaneous elongation produced in the rod. [ E = 200GN/m2]. (3 marks)
Ans: Length, L = 2m = 2 × 1000 = 2000mm
Diameter, d = 50mm
Area, A = Π/4 × d2 = Π/4 × 502 = 625Π mm2
Suddenly applied load, P = 100kN = 100 × 103N
E = 200GN/m2 = 2 × 109 N/m2
= 200 × 109 /106 N/mm2 = 200 × 103 N/mm2
Stress induced by suddenly applied load,
σ = 2 × P/A = 2 × 100×1000/625Π
= 101.86 N/mm2
dL = Elongation
= P/E × L = 101.86/200×103 × 2000
= 1.0186mm.
Ques: A tension bar 5m long is made up of two parts; 3m of its length has a cross-sectional area of 10cm2; the remaining 2m has a cross-sectional area of 20cm2. An axial load of 80kN is gradually applied. Find the total strain energy produced and compare this value with that obtained in a uniform bar of the same length and same volume under the same load. [ E = 2 × 105 N/mm2] (5 Marks)
Ans: Total length, L = 5m = 5000mm
Length of first part, L1 = 3m = 3000mm
Area of first part, A1 = 10cm2 = 1000mm2
Volume of first part, V1 = A1 × L1
= 1000×3000
= 3×106mm3
Length of second part, L2 = 2m = 2000mm
Area of second part, A2 = 20cm2
Volume of second part, V2 = A2 × L2
= 2000 × 2000
= 4×106 mm3
Axial gradual load, P = 80kN = 80 × 103 N
E = 2 × 105 N/mm2
Stress in first part, σ1 = load/A1 = 80 × 103 /1000 = 80 N/mm2
Stress in second part, σ2 = load/A2 = 80 × 103 /2000 = 40 N/mm2
Strain energy stored in first part, U1 = (σ1)2/2E × V1 = (802/2×2×105) × 3×106 = 48000 N-mm = 48 n-m
Stress energy in second part, U2 = (σ2)2/2E × V2 = (402/2×2×105) × 40×105 = 16000 N-mm = 16 N-m
∴ Total Strain energy, U = U1 + U2 = 48 + 16 = 64 N-m.
Strain energy stored in uniform bar,
Volume, V = V1 + V2 = 3×106 + 4×106 = 7×106 mm3
Length, L = 5m = 5000mm
Let A = Area of uniform bar,
V = A×L
Or, 7×106 = A × 5000
Or, A = 7×106 /5000
= 1400 mm2
Stress in uniform bar, σ = P/A = 80000/5000 = 57.143 N/mm2
Strain energy stored in uniform bar,
U = σ2/2E × V
= (57.143)2/2×2×105 × 7×106
= 57143 N-mm
= 57.143 N-m
∴ Strain energy in given bar/strain energy in uniform bar = 64/57.143 = 1.12.
Ques: The shear stress in a material at a point is given as 50 N/mm2. Determine local strain energy per unit volume stored in the material due to shear stress. [G = 8×104 N/mm2] (2 marks)
Ans: Shear stress, σ = 50 N/mm2
M.o.R. G = 8×104 N/mm2
Strain energy, U = σ2/2G × volume
= 502/ 2×8×104 × volume
= 0.015625 × volume
Strain energy per unit volume = 0.015625 × volume /volume
= 0.015625 N/mm.
Ques: A tensile load of 60kN is gradually applied to a circular bar of 4cm diameter and 5m length. If E = 2×105 N/mm2, determine: (3 marks)
(i) stretch in the rod,
(ii) stress in the rod,
(iii) strain energy.
Ans: Load, P = 60 kN = 6 ×103 N
Diameter, d = 4cm = 40mm
Area, A = π/4 × d2 = π/4 × 402 = 400π mm2
Length, L = 5m = 5000mm
Volume, V = A × L = 400π × 5000 = 2 × 106π mm3
E = 2×105 N/mm2
x = stretch or extension in the rod
σ = stress in the rod
U = strain energy
σ = P/A = 6000/400π = 47.746 N/mm2
Stretch, x = σ/E × L = 47.746/ 2 × 105 × 5000 = 1.19 mm
Strain energy absorbed, U = \(\sigma\)2/2E × V
= (47.746)2/ 2×2×105 × 2 × 106π = 35.81 N-m.
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