Akhtar Kamal

AKHTAR KAMAL (120450119156)

Advanced Strength Of MaterialBranch Mechanical-4C(1)Collage SVMIT

Akhtar Kamal

INTRODUCTION1) Elastic Strain Energy due to

Gradual Loading2) Elastic Strain Energy due to

Sudden Loading3) Elastic Strain energy due to

impact loading4) Elastic Strain Energy due to

Principal Stresses5) Energy of Dilation And

Distortion

Akhtar Kamal

What is strain energy

When the body is subjected to gradual sudden or impact load the body deforms and work done upon it

The material behave like a perfect spring and oscillates about its mean position

If the elastic limit is not exceeded this work is stored in the body This work done or energy stored in the body is called strain energy

Akhtar Kamal

Some Important Definition And Question

(1)Resilience Total strain energy stored in body is called resilience It is denoted as lsquorsquo

Wherehellip

(2)Proof ResilienceMaximum strain energy which can be stored in a body at elastic limit is called proof resilience It is denoted as lsquorsquo

Wherehellip at elastic limit(3)Modulus of resilienceMaximum strain energy which can be stored in a body per unit volume at elastic limit is called Modulus of resilience It is denoted as lsquorsquo

Akhtar Kamal

Strain Energy Due to Gradual Loading

Considr a bar of length is and uniform section area subjected to gradual load

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

INTRODUCTION1) Elastic Strain Energy due to

Gradual Loading2) Elastic Strain Energy due to

Sudden Loading3) Elastic Strain energy due to

impact loading4) Elastic Strain Energy due to

Principal Stresses5) Energy of Dilation And

Distortion

Akhtar Kamal

What is strain energy

When the body is subjected to gradual sudden or impact load the body deforms and work done upon it

The material behave like a perfect spring and oscillates about its mean position

If the elastic limit is not exceeded this work is stored in the body This work done or energy stored in the body is called strain energy

Akhtar Kamal

Some Important Definition And Question

(1)Resilience Total strain energy stored in body is called resilience It is denoted as lsquorsquo

Wherehellip

(2)Proof ResilienceMaximum strain energy which can be stored in a body at elastic limit is called proof resilience It is denoted as lsquorsquo

Wherehellip at elastic limit(3)Modulus of resilienceMaximum strain energy which can be stored in a body per unit volume at elastic limit is called Modulus of resilience It is denoted as lsquorsquo

Akhtar Kamal

Strain Energy Due to Gradual Loading

Considr a bar of length is and uniform section area subjected to gradual load

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

What is strain energy

When the body is subjected to gradual sudden or impact load the body deforms and work done upon it

The material behave like a perfect spring and oscillates about its mean position

If the elastic limit is not exceeded this work is stored in the body This work done or energy stored in the body is called strain energy

Akhtar Kamal

Some Important Definition And Question

(1)Resilience Total strain energy stored in body is called resilience It is denoted as lsquorsquo

Wherehellip

(2)Proof ResilienceMaximum strain energy which can be stored in a body at elastic limit is called proof resilience It is denoted as lsquorsquo

Wherehellip at elastic limit(3)Modulus of resilienceMaximum strain energy which can be stored in a body per unit volume at elastic limit is called Modulus of resilience It is denoted as lsquorsquo

Akhtar Kamal

Strain Energy Due to Gradual Loading

Considr a bar of length is and uniform section area subjected to gradual load

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Some Important Definition And Question

(1)Resilience Total strain energy stored in body is called resilience It is denoted as lsquorsquo

Wherehellip

(2)Proof ResilienceMaximum strain energy which can be stored in a body at elastic limit is called proof resilience It is denoted as lsquorsquo

Wherehellip at elastic limit(3)Modulus of resilienceMaximum strain energy which can be stored in a body per unit volume at elastic limit is called Modulus of resilience It is denoted as lsquorsquo

Akhtar Kamal

Strain Energy Due to Gradual Loading

Considr a bar of length is and uniform section area subjected to gradual load

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Strain Energy Due to Gradual Loading

Considr a bar of length is and uniform section area subjected to gradual load

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

S tress DueiquestGradual Load Since the load is applied gradually(ie it increases from 0

to P) average load is considered Work done on the bar = Area of the load ndash

Deformation diagram

Work stored in the bar = Area of the resistance ndash Deformation diagram

Work done = Work stored

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Strain Energy Due to Gradual Loading

Strain energy = = = = =

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Elastic Strain Energy due to Sudden Loading

When the load is applied suddenly the value of the load is P throughout the deformation

But Resistance R increases from 0 to R

Work done on the bar

Work store in the bar= =

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Work done = work store =

Hence the maximum stress intensity due to a suddenly applied load twice the stress intensity produced by the load of the same magnitude applied gradually

Akhtar Kamal

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

THANK YOU

By-akhtar

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

STARIN ENERGY DUE TO IMPACT LOADING Work done on the bar= = =

Akhtar Kamal

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Work stored in the bar = strain energy

Work done = Work stored

Akhtar Kamal

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

When is very small as compared tothen work done

Impact factor the ratio of maximum dynamic deformation to the static

deformation is called the impact factor But The ratio is sometimes known as load factor

Akhtar Kamal

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Strain energy due to bending

consider two transverse section 1-1 and 2-2 of a beam distant apart as showing

strain energy stored in small strip pf area da

strain energy stored in entire section of a beam =

Akhtar Kamal

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Strain energy due to torsion consider a small elements ring of thickness at radius

Akhtar Kamal

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

For hollow shaft For a hollow shaft the range of integrating will be

Akhtar Kamal

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

THANK YOU

Akhtar Kamal

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Energy Of Dilation and DistortionTotal strain energy given by equation (1) of article 15 can be separated into the following two strain energies

a) Strain energy of dilatation (dilation) or volume metric strain energy (strain energy of uniform compression or tension)

b) Strain Energy of distortion(shear strain energy)

To accomplish this Let the principal strains be in the deration of principal stresses and respectively

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Energy Of Dilation and Distortion

But

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

Energy Of Dilation and Distortion From the above discussion following conclusions can be

made (a) If This means that there is no distortion (so that no

shearing stresses and shearing strains will be present anywhere in the block) but only volumetric change(dilation) occurs

(b) If This means that if the sum of three principal stress is

zero there is no volumetric change(dilation) but only the distortion occurs

The above to conclusion can be used to break the given three principal stresses into two sets of principal stresses such that one set produces dilation (volumetric change) only while the other produces distortion (shear stresses) only

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

STRAIN ENERGY DUE TO PRINCIPAL STRESSES Consider a small block of length δℓ width δb and

height δh subjected to three principal stresses σ1

σ2 and σ3 as shown in figure σ1 = Principal stress on face of area (δb times δh) σ2 = Principal stress on face of area (δℓ times δh)

σ3 = Principal stress on face of area (δℓ times δb) μ = Poissonrsquos ratio for the material

Akhtar Kamal

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

˙ Extention of the block in the direction of σ1

δℓı =εı δı

δℓı = [σ1 ndash μ (σ2+σ3 )] δℓ

Akhtar Kamal

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

˙ Strain energy due to σ1

= (Load due to σ1 in the direction of σ1) times δℓ1

= [σ1δbδh] x [σ1‒ μ (σ2+ σ3)] δℓ= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] (δbδhδℓ)]= [σ1sup2 ‒ μ (σ1 σ2 +σ1 σ3)] δV

Where δV= volume of block = δbδhδl

Akhtar Kamal

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Similarly

Strain energy due to σ2

= [σ2sup2 ‒ μ(σ2σ1 + σ2 σ3)]δV

Strain Energy due to σ3

= [σ3sup2 ‒ μ σ3 σ1 + σ3 σ2)]δV

Akhtar Kamal

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

˙ δu = Total Strain energy for volume δV = Sum of strain energies due to σ1σ2 and σ3 = [σ1sup2 ‒ μ(σ1 σ2 +σ1 σ3)]δV

+ [σ2 sup2 ‒ μ(σ2 σ1 +σ2 σ3)]δV

+ [σ3 sup2 ‒ μ(σ3 σ1 + σ3 σ2)]δV

˙ δu = [σ1sup2+ σ2 sup2+ σ3 sup2- 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] δV

Akhtar Kamal

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Thus for a body of Volume V Subjected to the principal Stresses σ1σ2 and σ3 total strain energy is given by

u= [σ1sup2+ σ2 sup2+ σ3 sup2 ‒ 2μ(σ1 σ2 + σ2 σ3 + σ3 σ1)] V

Sign for Principal StressesTension = + veCompression = -ve

Akhtar Kamal

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

DEDUCTION FOR THE SIMPLE CASES OF LOADING

The expression for the strain energy for the simple cases of stresses can be easily deducted from the general equation (1) for the strain energy

Akhtar Kamal

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

CASE 1 FOR BIAXIAL STRESS SYSTEM

˙ u= [σ1sup2+ σ2 sup2 ‒ 2μ(σ1 σ2)] V

Akhtar Kamal

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

CASE 2 - FOR ONLY ONE DIRECT STRESS σ1=σ σ2 =0 σ3 = 0 ˙ u= (σsup2 times V) ˙ u= V

Akhtar Kamal

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

CASE 3 - FOR SIMPLE SHEAR ONLYLet τ be simple shear in volume V

Then the principle stress will be

σ1= τ σ2 = ‒ τ σ3 =0

Substituting these values in (1) we get

u= [τ sup2(‒ τ )sup2+0 ‒ 2μ(τ)(‒ τ)] V

= [2 τ sup2+ 2 μ τ sup2]V

= (1+ μ) V

ButE =2G (1+ μ) G = Modulus of rigidity

Therefore =

u= VStrain energy per unit volume = u =

Akhtar Kamal

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

CASE 4 -DUE TO EQUAL TENSIONS APPLIED TO A VOLUMELet p= hydrostatic tension or hydrostatic pressure˙ Either σ1 =p σ2 =p and σ3=p

Or σ1= -p σ2 = -p and σ3 = -p

Substituting any one in equation (1) we get

u= [ psup2 + psup2 + psup2 ‒ 2μ( pp + pp + pp)] V = [3psup2 ‒ 2μ(3psup2)]V = (1- 2μ)V but E = 3k(1- 2μ)˙ = k = Bulk modulus

˙ u = V

Akhtar Kamal

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33

Akhtar Kamal

THANK YOU

• Akhtar kamal (120450119156)
• INTRODUCTION
• What is strain energy
• Some Important Definition And Question
• Strain Energy Due to Gradual Loading
• Slide 6
• Strain Energy Due to Gradual Loading (2)
• Elastic Strain Energy due to Sudden Loading
• Work done = work store =
• Slide 10
• STARIN ENERGY DUE TO IMPACT LOADING
• Slide 12
• Slide 13
• Strain energy due to bending
• Strain energy due to torsion
• For hollow shaft
• Slide 17
• Energy Of Dilation and Distortion
• Energy Of Dilation and Distortion (2)
• Energy Of Dilation and Distortion (3)
• Strain energy due to principal stresses
• Slide 22
• Slide 23
• Slide 24
• Slide 25
• Slide 26
• Slide 27
• Deduction for the simple cases of loading
• Case 1 For biaxial stress system
• Case 2 - For only one direct stress
• Case 3 - For simple shear only
• Case 4 -due to equal tensions applied to a volume
• Slide 33