Principal stress| It’s important facts and 3D state of stress

Principal Stress

How to define principal stress ? | What is principal stress explain with an example ?

Principal Stress Definition:

Principal Stress is the maximum and minimum stresses derived from normal stress at an angle on a plane where shear stress is zero.

How to calculate principal stress ?

Principal stress equation | Principal stress formula:
Maximum and Minimum principal stress equations:

\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

Principal stress derivation | Determine the principal planes and the principal stresses

Normal stresses:

\sigma x'=\frac{\sigma x+\sigma y}{2}+\frac{(\sigma x-\sigma y)(cos2\Theta )}{2}+\sigma xysin2\Theta

\sigma y'=\frac{\sigma x+\sigma y}{2}-\frac{(\sigma x-\sigma y)(sin2\Theta )}{2}+\sigma xycos2\Theta

-\frac{(\sigma x-\sigma y)(cos2\Theta )}{2}+\sigma xysin2\Theta


\frac{dx'}{d\Theta }=0

tan2\Theta =\frac{\sigma xy}{\frac{(\sigma x-\sigma y)}{2}}

tan2\Theta_{p} =\frac{\sigma xy}{\frac{(\sigma x-\sigma y)}{2}}

“p” represents the principal plane.

There are two principal stresses,
one at angle 2\Theta
and other at 2\Theta+180
Maximum and Minimum principal stresses:

R=\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

cos2\Theta =\frac{\left ( \sigma x-\sigma y \right )}{2R}

sin2\Theta =\frac{\sigma xy}{R}

substitute in equation 1:

\sigma x'=\frac{\left ( \sigma x+\sigma y \right )}{2}+\frac{1}{R}[\left ( \frac{\sigma x-\sigma y}{2} \right )^{2}+\sigma xy^{2}]

substitute value of R

Maximum and minimum normal stresses are the principal stresses:

\sigma max=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

\sigma min=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

The state of Stress:

The principal stress is the reference co-ordinate axes to the representation of the stress matrix and this stress components are the significance of the state of stress could be represented as,

Stress tensor:

\tau ij=\begin{bmatrix}\sigma 1 & 0 & 0 \\0 & \sigma 2 & 0 \\0 &0 &\sigma 3\end{bmatrix}

Principal stresses from stress tensor and stress invariants |principal stress invariants

There are three principal planes at any stressed body, with normal vectors n, called principal directions where the stress vector is in the same direction as normal vector n with no shear stresses and these components depend on the alignment of the co-ordinate system.

A Stress vector parallel to the normal unit vector n is specified as,

\tau ^{\left ( n \right )}=\lambda n=\sigma _{n}n

\lambda represents constant of proportionality.

The principal stress vectors represented as,

\sigma ij nj=\lambda ni

\sigma ij nj-\lambda nij=0

The magnitude of three principal stresses gives three linear equations.
The determinant of the coefficient matrix is equal to zero and represented as,

\begin{vmatrix}\sigma ij-\lambda \delta ij\end{vmatrix}=\begin{bmatrix}\sigma 11-\lambda &\sigma 12 &\sigma 13 \\\sigma 21 & \sigma 22-\lambda &\sigma 23 \\\sigma 31 & \sigma 32 & \sigma 33-\lambda\end{bmatrix}

Principal stresses are the form of normal stresses, and the stress vector in the coordinate system is represented in the matrix form as follows:

\sigma ij=\begin{bmatrix}\sigma 1 & 0 & 0\\0 & \sigma 2 & 0\\0 &0 &\sigma 3\end{bmatrix}

I1, I2, I3 are the stress invariants of the principal stresses,
The stress invariants are dependent on the principal stresses and are calculated as follows,

I1=\sigma 1+\sigma 2+\sigma 3

I2=\sigma 1\sigma 2+\sigma 2\sigma 3+\sigma 3\sigma 1

I3=\sigma 1.\sigma 2.\sigma 3

The principal stresses equation for stress invariants:

\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

Principal stress trajectories | Principal directions of stress

Stress trajectories show the principal stress directions and their varying magnitude of the principal stresses.

Von mises stress vs principal stress

Von mises principal stress equation

Von Mises is the theoretical measure of the stress yield failure criterion in ductile materials.
The positive or negative sign depends on the principal stresses.
Principal stresses Boundary conditions:

\sigma 12=\sigma 23=\sigma 31=0

Theories of failure give the yield stresses of the components subjected to multiaxial loading. Further, when it is compared with the yield point of the components shows the margin of the safety of the component.

Maximum principal stress is considered for brittle elements such as casting components (i.e., clutch housing, gearbox, etc.)
Von-mises stress theory is based on shear strain energy theory is suggested for ductile materials like aluminum, steel components.

Why von mises stress is recommended for ductile and Principal Stress for brittle materials?

Failure of brittle materials used to uni-axial test is along a plane vertical to the axis of loading. So, the failure is because of normal stress generally. Out of all theories of failure, principal stress theory is based on normal Stress. Hence for brittle materials, principal stress theory is recommended,

Ductile materials fail at 45 degrees inclined at the plane of loading. So, the failure is due to shear stress. Out of all theories of failure shear strain energy or von-mises theory and maximum shear stress theory is based on shear stress. By comparison, von mises gives better results. Hence for ductile materials, von mises theory is recommended.

Different types of stress

Absolute principal Stress | Effective principal Stress:

The principal stresses are based on maximum Stress and minimum Stress. So, the range of the Stress is between the maximum and minimum Stress, (stress range is limited and less) and might lead to higher fatigue life. So, it is important to find out the effective principal Stress that gives the maximum value out of the two over the given period of time.

What is Maximum Normal Stress theory?

This states that brittle failure occurs when the maximum principal Stress exceeds the compression or the tensile strength of the material. Suppose that a factor of safety n is considered in the design. The safe design conditions require that.

Maximum principal stress equation

-\frac{Suc}{n}<{\sigma 1,\sigma 2,\sigma 3}<\frac{Sut}{n}

Where σ1, σ2, σ3 are three principal stresses, maximum, minimum, and intermediate, in the three directions, Sut and Suc are the ultimate tensile strength and the ultimate compressive strength, respectively.

To avoid brittle failure, the principal stresses at any point in a structure should lie within the square failure envelope based on the maximum normal stress theory.

Maximum principal stress theory |Maximum principal stress definition

consider two-dimensional stress state and the corresponding principal stresses such as σ1 >σ2 >σ3
Where σ3=0, σ2 may be compressive or tensile depending on the loading conditions where σ2 may be less or greater than σ3.

According to maximum principal stress theory, failure will occur when
σ1 or σ2 =σy or σt
The conditions are represented graphically with coordinates σ1,σ2. If the state of Stress with coordinates (σ1,σ2 ) falls outside of the rectangular region, failure will occur as per the maximum principal stress theory.

Mohr’s circle principal stresses

Explain the Mohr’s circles for the three-dimensional state of stress:

  • Consider a plane with a reference point as P. Sigma is represented as normal Stress and tau by the shearing stress on the same plane.
  • Take another plane with reference point Q representing sigma and tau as normal stress and shear stress, respectively. Different planes are passing through point p, different values of principal and shear stress.
  • For each plane n, a point Q with coordinates as shear stress and principal Stress can be located.
  • Determine the normal and shear stresses for point Q in all possible directions of n.
  • Obtain three principal stresses as maximum principal Stress, minimum principal Stress, and intermediate principal Stress and represent them in ascending order of the values of the stresses.
  • Draw three circles with diameters as the difference between the principal stresses.
moh's circle:principal stress
Image credit:SanpazMohr Circle, marked as public domain, more details on Wikimedia Commons
  • The shaded area region is the Mohr’s circle plane region.
  • The circles represent the Mohr’s circles.
  • (σ1-σ3) and the associated normal stress is (σ1+σ3)
  • There are three normal stresses, so are three shear stresses.
  • The principal shear planes are the planes where shear stresses act and principal normal stress acts at a plane where shear stress is ‘0’ and shear stress act at a plane where normal principal stress is zero. The principal shear stress act at 45° to the normal planes.

The shear stresses are denoted by \tau 1,\tau 2,\tau 3
And the principal stresses are denoted by \sigma 1,\sigma 2,\sigma 3

Third principal stress

3rd principal Stress is relative to the maximum compressive stress due to the loading conditions.

3D principal stress examples:

For three-dimensional case, all three planes have zero shear stresses, and these planes are mutually perpendicular, and normal stresses have maximum and minimum stress values and these are the normal stresses that represent the principal maximal and minimal stress.

These principal stresses are denoted by,
σ1,σ2, σ3.
3D Stress in hub-a a steel shaft is force-fitted into the hub.
3D Stress in machine component.

Principal deviatoric stress:

Principal deviatoric stresses are obtained by subtracting mean Stress from each principal Stress.

Intermediate principal Stress:

The principal Stress, which is neither maximum nor minimum, is called intermediate Stress.

Principal stress angle | Orientation of principal stress: θP

Orientation of the principal Stress is computed by equating shear stress to zero in x-y direction at the principle plane rotated through an angle theta. Solve θ to get θP, the principal stress angle.

Important Frequently Asked Questions (FAQs):

Maximum principal stress theory is applicable for which material?

Answer: Brittle materials.

What are the 3 principal stresses? | What is maximum and minimum principal stress ?

Maximum principal Stress | Major principal stress: Most tensile (σ1)
Minimum principal Stress | Minor Principal Stress: Most compressive (σ3)
Intermediate principal Stress (σ2)

Principal Stress vs normal Stress:

Normal Stress is the force applied to the body per unit area. Principal Stress is the stress applied to the body having zero shear stress principal Stress is in the form of normal Stress giving maximum and minimum stresses on the principal plane.

Principal Stress vs Bending Stress:

Bending Stress is the Stress that occurs in the body due to the application of a large amount of load that causes the object to bend.

Principal Stress vs axial Stress:

Axial Stress and principal stress are the parts of normal stress.

What is the significance of principal Stress?

Principal Stress shows the maximum and minimum normal stress. Maximum normal Stress shows the component’s ability to sustain the maximum amount of force.

What are the principal stresses in a shaft with torque applied ?

The shear stress due to torque has maximum magnitude at the outer fiber. The bending stress is due to horizontal loads (horizontal gear forces, belt, or chain forces) that induce bending stresses which are maximum at the outer fibers.

Why shear stress is zero on principal plane ?

The normal Stress is maximum or minimum, and shear stress is zero.

tan2\Theta _{\tau-max}=-(\frac{\sigma x-\sigma y}{2\tau xy})

\tau max=\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

when shear stress=0,

\tau max=\frac{\begin{vmatrix}\sigma x-\sigma y\end{vmatrix}}{2}

Important Principal stress problems:

1) A rectangular stress vector having shear stress in XY direction of 60Mpa and normal tensile Stresses of 40Mpa.How to find principal stresses ?

Given: \sigma x=\sigma y=40Mpa , \tau=60Mpa
Principal stresses are calculated as,

\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


2)What are the coordinates of the center of Mohr’s circle for an element subjected to two mutually perpendicular stresses, one tensile of magnitude 80MPa and other compressive of magnitude 50MPa?

σx = 80 MPa,
σy = -50 MPa
Co-ordinates of center of Mohr’s circle =[ ½( σx + σy),0]
= [(30/2),0]
= (15,0)

3)A body was subjected to two mutually perpendicular stresses of -4MPa and 20MPa, respectively. Calculate the shear stress on the plane of the shear.

σx+σy /2= -4+20/2 = 8Mpa
Radius= σ1-σ2/2 = 20-(-4)/2 = 12
where σx,σy are principal stresses
at pure shear stress,σn=0
shear stress= squareroot12^2-8^2= 8.94Mpa.

4) Application of principal stress | Find the principal stresses for the following cases.

i)σx=30 Mpa, σy=0, \tau=15Mpa.


\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

ii)σx=0,σy=80MMpa, \tau=60Mpa.

\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


iii)\tau=10Mpa, σx=50Mpa,σy=50Mpa.

\sigma 1=\frac{\sigma x+\sigma y}{2}+\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


\sigma 1=\frac{\sigma x+\sigma y}{2}-\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}


5) The maximum principal Stress is given 100 Mpa, and the Minimum principal Stress is 50 MPa. Calculate the maximum shear stress and the orientation of the principal plane using Mohr’s circle.

Maximum principal stress=100Mpa(compressive)
Minimum principal stress=50 Mpa(compressive)
Maximum shear stress is the radius of the Mohr’s circle, then we can write as follows.

R=\sqrt{(\frac{\sigma x-\sigma y}{2})^{2}+\tau xy^{2}}

\tau max=25Mpa

2θ = 90, from the maximum principal stress direction.
So, the orientation at that point is θ = 45 from the maximum principal stress direction.

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About Sulochana Dorve

I am Sulochana. I am a Mechanical Design Engineer— in design Engineering, in Mechanical Engineering. I have worked as an intern at Hindustan Aeronautics limited in the design of the armament department. I have experience working in R&D and design. I am skilled in CAD/CAM/CAE: CATIA | CREO | ANSYS Apdl | ANSYS Workbench | HYPER MESH | Nastran Patran as well as in Programming languages Python, MATLAB and SQL.
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