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Friday, 18 August 2017

MAXIMUM STRAIN ENERGY THEORY OF FAILURE


Today we will understand here the theories of failure, in strength of material, with the help of this post.

As we know very well that when a body or component or material will be subjected with an external load, there will be developed stresses and strains in the body or component.

As per hook’s law, stress will be directionally proportional to the strain within the elastic limit or we can say in simple words that if an external force is applied over the object, there will be some deformation or changes in the shape and size of the object. Body will secure its original shape and size after removal of external force.

Within the elastic limit, there will be no permanent deformation in the body i.e. deformation will be disappeared after removal of load.

If external load is applied beyond the elastic limit, there will be a permanent deformation in the body i.e. deformation will not be disappeared after removal of load. Component or material or body will be said to be failed, if there will be developed permanent deformation in the body due to external applied load.

Theories of failure help us in order to calculate the safe size and dimensions of a machine component when it will be subjected with combined stresses developed due to various loads acting on it during its functionality.

There are following theories as listed here for explaining the causes of failure of a component or body subjected with external loads.

4. The maximum strain energy theory
5. The maximum shear strain energy theory

We will now understand here the maximum strain energy theory with the help of this article. The maximum strain energy theory is also termed as Haigh’s theory and this theory is the best theory for failure of ductile material. The maximum strain energy theory is not suitable under hydrostatic stress situation.

According to the theory of maximum strain energy, “The failure of a material or component will occur when the maximum value of strain energy per unit volume exceeds the limiting value of strain energy per unit volume i.e. value of strain energy per unit volume corresponding to the yield point of the material under tension test”.

Therefore in order to avoid the condition of failure of the component, maximum value of strain energy per unit volume must be below than the value of strain energy per unit volume corresponding to the yield point of the material under tension test.  

Let us consider a body which is subjected with tensile load which is increasing gradually up to its elastic limit from value 0 to value P and therefore deformation or extension of the body is also increasing from 0 to x and we can see it in following load extension diagram as displayed here.
We have following information from above load extension diagram for body which is subjected with tensile load up to its elastic limit.

σ = Stress developed in the body
E = Young’s Modulus of elasticity of the material of the body
A= Cross sectional area of the body
P = gradually applied load which is increasing gradually up to its elastic limit from value 0 to value P
P = σ. A
x = Deformation or extension of the body which is also increasing from 0 to x
L = Length of the body
V= Volume of the body = L.A
U = Strain energy stored in the body or component

As we have already discussed that when a body will be loaded within its elastic limit, the work done by the load in deforming the body will be equal to the strain energy stored in the body.

Strain energy stored in the body = Work done by the load in deforming the body
Strain energy stored in the body = Area of the load extension curve
Strain energy stored in the body = Area of the triangle ABC

U = (1/2). AB. BC
U = (1/2) x. P
U = (1/2) σ. A. e. L
U = (1/2) σ. e. V

Therefore, strain energy per unit volume = (1/2) σ. e

Let us consider the condition of three dimensional stress systems. Where σ1, σ2 and σ3 are the principal stresses and e1, e2 and e3 are the corresponding strains.

e1= (1/E) x [σ1- μ (σ2 + σ3)]
e2= (1/E) x [σ2- μ (σ3 + σ1)]
e3= (1/E) x [σ3- μ (σ1 + σ2)]

Let us determine the value of strain energy per unit volume for a three dimensional stress system
U = (1/2E) x [σ12 + σ22 + σ32- 2μ (σ1 σ2 + σ2 σ3 + σ3 σ1)]

Value of strain energy per unit volume corresponding to the yield point of the material under tension test will be given by following equation as mentioned here.
Uy = (1/2E) x σt2
Where σt is the principle stress at elastic limit under tension test

Condition of failure

Maximum value of strain energy per unit volume > Value of strain energy per unit volume corresponding to the yield point of the material under tension test
(1/2E) x [σ12 + σ22 + σ32- 2μ (σ1 σ2 + σ2 σ3 + σ3 σ1)] > (1/2E) x σt2
[σ12 + σ22 + σ32- 2μ (σ1 σ2 + σ2 σ3 + σ3 σ1)] > σt2

If we consider the case of two dimensional stress systems, we will have σ3= 0 and therefore we can write here the condition of failure for two dimensional stress system and here it is.
[σ12 + σ22 - 2μ σ1 σ2] > σt2

Condition for safe design

Maximum value of strain energy per unit volume Permissible strain energy per unit volume
Permissible principle stress = Principle stress at elastic limit / F.O.S
Permissible principle stress = σt / F.O.S

For three dimensional stress systems
For tri-axial state of stress, we will have following equation
[σ12 + σ22 + σ32- 2μ (σ1 σ2 + σ2 σ3 + σ3 σ1)]  (σt / F.O.S) 2

For two dimensional stress systems
[σ12 + σ22 - 2μ σ1 σ2 (σt / F.O.S) 2

Do you have suggestions? Please write in comment box.

We will now discuss the maximum shear strain energy theory, in the category of strength of material, in our next post.

Reference:

Strength of material, By R. K. Bansal
Image Courtesy: Google

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