In this article, you will learn a complete overview of principal stress such as its definition, formula with derivation, calculation, and much more.
Principal stresses play the very most important role to find out theories of failure such as maximum principal stress theory, maximum shear stress theory, etc.
But before knowing the principal stress first of all we have to know what is stress, normal stress, shear stress, principal plane, etc. Such that we can better understand the principal stresses.
So without wasting time let's get started.
What is Stress?
It can be defined as an externally applied load that develops an internal resistance per unit area of a body called stress.
Mathematically stress is defined as the load per unit area of the body's cross-section (or) the force per unit area.
Normal Stress
The stress on a body or plane in the normal direction is called normal stress.
The normal stress on an oblique plane inclined at an angle is given by,
σₙ = (σx+σy)/2 + {(σx−σy)/2}cos2θ + τsin2θ
Shear Stress
It is defined as the force acting parallel to the area is called shear stress or tangential stress.
Shear stress on an oblique plane is given by,
σₜ = {(σx−σy)/2}sin2θ - τcos2θ
Principal Plane
When only Normal Stress is applied to a body or plane, no shear stress is acting on that plane, i.e. shear stress is zero, such the plane is called the principal plane.
Principal Stress
A body or a plane may even be subjected to a combination of normal stress as well as tangential stress (shear stress) called principal stress.
Principal stress can be divided into two types
- Maximum Principal Stress
- Minimum Principal Stress
Maximum Principal Stress
The maximum amount of normal stress acting on the main plane is called maximum principal stress.
The formula of maximum principal stress is given by,
σ₁ = {(σx +σy)/2} + √[{(σx −σy)/2} ² - τ ²]
Minimum Principal Stress
The minimum amount of normal stress acting on the main plane is called the minimum principal stress.
The formula of minimum principal stress is given by,
σ₂ = {(σx +σy)/2} - √[{(σx −σy)/2} ² - τ ²]
Derivation of Principal Stress
As we know, on the principal plane shear or tangential stress is zero.
So,
σₜ = 0
{(σx−σy)/2}sin2θ - τcos2θ = 0
sin2θ/cos2θ = 2τ/(σx−σy)
tan2θ = 2τ/(σx−σy)
Now, we will draw a geometric representation of the above equation.
So,
The diagonal of the right-angled triangle will be,
± √{ ( σx−σy ) ² + 2τ ²}
± √{ ( σx−σy ) ² + 4τ ²}
Then,
Sin2θ = Height/Diagonal
Sin2θ = ± 2τ/√{ ( σx−σy )² + 4τ²}
Cos2θ = ± ( σx−σy )/√{ ( σx−σy )² + 4τ²}
As we know that the normal stress on an oblique plane inclined at an angle is,
σₙ = (σx+σy)/2 + {(σx−σy)/2}cos2θ + τsin2θ
So the values of principal stresses are obtained by substituting the values of sin2θ and cos2θ in this equation.
σₙ = (σx+σy)/2 ± [{(σx−σy)/2} ×( σx−σy )/√{ ( σx−σy ) ² + 4τ²}] ±τ[2τ/√{ ( σx−σy )² + 4τ²}]
After calculating these values then,
σₙ = {(σx +σy)/2} ± √[{(σx −σy)/2}² - τ²]
So,
The maximum principal stress,
σ₁ = {(σx +σy)/2} + √[{(σx −σy)/2} ² - τ²]
And,
The minimum principal stress,
σ₂ = {(σx +σy)/2} - √[{(σx −σy)/2} ² - τ²]
Sign Conventions For Principal Stresses
There are the following sign conventions to take in mind while calculating the principal stresses:
- All the tensile stresses are taken as positive while all the compressive stresses are taken as negative.
- The shear stress which tends to rotate the element in the clockwise direction ( vertical faces or x - x-axis ) is taken as positive while the shear stress which tends to rotate the element anticlockwise direction ( horizontal faces or y - y-axis ) is taken as negative.
Calculation of Principal Stress
Question
In a strained material, two direct stress of 860 N / mm² ( tensile ) and 440 N / mm ² ( compressive ) are acting on two mutually perpendicular planes respectively with a shear stress of 300 N / mm² in a clockwise direction. Find maximum and minimum principal stress?
Solution
Given Data,
σx = 860 N / mm²
σy = - 440 N / mm ²
τ = 300 N / mm²
As we know,
Maximum principal stress,
σ₁ = {(σx +σy)/2} + √[{(σx −σy)/2} ² - τ²]
σ₁ = {(860 - 440)/2} + √[{(860 + 440)/2} ² - 300²]
= 210 + 715.89
σ₁ = 925.89 N / mm²
Minimum principal stress,
σ₂ = {(σx +σy)/2} - √[{(σx −σy)/2} ² - τ ²]
σ₂ = {(860 - 440)/2} - √[{(860 + 440)/2} ² - 300 ²]
= 210 - 715.89
σ₂ = -505.89 N / mm²
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