We were discussing formula for bending stress in beam, basic
concepts of strength of materials, strain energy, resilience, proof resilience and modulus of resilience in our previous posts.

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.

Ïƒ = 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 = Ïƒ. 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) Ïƒ. A. e. L

U = (1/2) Ïƒ. e. V

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

Let us consider the condition of three dimensional stress
systems. Where Ïƒ

_{1}, Ïƒ_{2}and Ïƒ_{3}are the principal stresses and e_{1}, e_{2}and e_{3}are the corresponding strains.
e

_{1}= (1/E) x [Ïƒ_{1}- Î¼ (Ïƒ_{2}+ Ïƒ_{3})]
e

_{2}= (1/E) x [Ïƒ_{2}- Î¼ (Ïƒ_{3}+ Ïƒ_{1})]
e

_{3}= (1/E) x [Ïƒ_{3}- Î¼ (Ïƒ_{1}+ Ïƒ_{2})]
Let us determine the value of strain
energy per unit volume for a three dimensional stress system

**U = (1/2E) x [**

**Ïƒ**

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }+ Ïƒ_{3}^{2}- 2Î¼ (Ïƒ_{1}Ïƒ_{2}+ Ïƒ_{2}Ïƒ_{3 }+ Ïƒ_{3}Ïƒ_{1})]
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**

**Ïƒ**

_{t}^{2}
Where Ïƒ

_{t}is the principle stress at elastic limit under tension test###
*Condition
of failure*

*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 [Ïƒ

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }+ Ïƒ_{3}^{2}- 2Î¼ (Ïƒ_{1}Ïƒ_{2}+ Ïƒ_{2}Ïƒ_{3 }+ Ïƒ_{3}Ïƒ_{1})] > (1/2E) x Ïƒ_{t}^{2}**[**

**Ïƒ**

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }+ Ïƒ_{3}^{2}- 2Î¼ (Ïƒ_{1}Ïƒ_{2}+ Ïƒ_{2}Ïƒ_{3 }+ Ïƒ_{3}Ïƒ_{1})] > Ïƒ_{t}^{2}
If we consider the case of two
dimensional stress systems, we will have Ïƒ

_{3}= 0 and therefore we can write here the condition of failure for two dimensional stress system and here it is.**[**

**Ïƒ**

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }- 2Î¼ Ïƒ_{1}Ïƒ_{2}] > Ïƒ_{t}^{2}###
*Condition
for safe design*

*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 = Ïƒ

_{t}/ F.O.S*For three dimensional stress systems*

For tri-axial state of stress, we will
have following equation

**[**

**Ïƒ**≤

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }+ Ïƒ_{3}^{2}- 2Î¼ (Ïƒ_{1}Ïƒ_{2}+ Ïƒ_{2}Ïƒ_{3 }+ Ïƒ_{3}Ïƒ_{1})]**(Ïƒ**

_{t}/ F.O.S)^{ 2}

*For two dimensional stress systems*

**[**

**Ïƒ**≤

_{1}^{2}_{ }+ Ïƒ_{2}^{2 }- 2Î¼ Ïƒ_{1}Ïƒ_{2}]**(Ïƒ**

_{t}/ 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

## No comments:

## Post a comment