Principal Planes & Stresses-Part I

The concept of principal stresses and strains is an important factor in designing of everyday machine elements or components. It is an important factor for an engineering which help in manifest an element.

The true understanding of this concept will come to you by imaging the following situation:

Normal force applied on a cube

Suppose we have a cube in space under Cartesian co-ordinates . Now you apply some force in a particular direction say in positive x-direction. The force we applied is of finite magnitude and for a significant amount of time so the the cube  is taken from its stable elastic region to plastic region. Now the cube is deformed in the direction of the applied force,i.e, the force was tensile in nature. In our case the only stress generated throughout the volume of cube is normal stress.

Traction force applied on a cube

Now suppose you have a cube fixed at one face and we are applying force parallel(traction force)on the face opposite to the fixed side. Now the force acting on the plane will have a shear effect.

But in real-life application we hardly find a system in which either tensile or compressive force is acting or shear force is acting. The resultant of these two forces are taken in account. To any extent the complexity of object may become there always exist a set of three orthogonal planes on which shear stresses are zero and there are only the normal stresses on these planes. These planes are called pr

incipal planes and normal stresses on theses planes are called principal stress. Because of  this property of orthogonal plane the exercise of  finding the principal stresses and its direction has became common.

Stresses on Inclined plane-Part I

A cubic element

Consider a cube in space.The cube is having the following stresses definition:
On the lower surface of the cube, normal stress acting is σ(1)

On the face adjacent to it, normal stress acting is σ(2)

Now a plane approaches the cube at an angle θ with respect to the lower face of the cube and gives a wedge shaped element. Now we confine our analysis on this element.

Since the stresses σ(1) and σ(2) are along the standard co-ordinate axis, we need the resolve these stresses to get the stresses along and perpendicular to the inclined plane.

Stress generated on inclined plane

Let, the cause of generation of stress on lower face be P(1)

The cause of generation of stress on face adjacent to lower face be P(2)

Component of P(1) along and perpendicular to inclined plane AB: P(1)sinθ and P(1)cosθ respectively
Component of P(2) along and perpendicular to inclined plane AB: P(2)cosθ and P(2)sinθ respectively.

Resolution of applied stresses

Therefore, normal resisting force acting on the inclined plane AB: P(1)cosθ + P(2)sinθ
Shear resisting force acting on the inclined plane AB: P(1)sinθ-P(2)cosθ
Let σ(θ) and 𝛕(θ) be the normal and shear stress on inclined plane AB and t be the thickness of this wedge shaped element.

Then,

ABσ(θ)t=P(1)cos(θ)+P(2)cos(θ)

=σ(1)BCtcos(θ)+σ(2)ACt

σ(θ)=σ(1)(BC/AB)cos(θ)+σ(2)(AC/AB)sin(θ)

BC/AC=cos(θ) and BC/AB=sin(θ)

σ(θ)=σ(1)cos(θ)cos(θ)+σ(2)sin(θ)sin(θ)

=σ(1)(1+cos(2θ))/2+σ(2)(1-cos(2θ))/2

σ(θ)=(σ(1)+σ(2))/2+((σ(1)-σ(2))/2)cos(2θ)...Normal stress on inclined plane

In the same way,

𝛕(θ)=((σ(1)-σ(2))/2)sin(2θ)...Shear stress on inclined plane