Stresses on Inclined Planes-II

In our analytical approach, we take several assumptions, and each of these assumptions pushes our problem in hand towards the ideal region, which may help in easily solving our problem but it also fills our solution with many errors. Our whole study of solid mechanics is based on several assumptions, but in real-life problem solving cases, these assumptions cannot be applied. However for the higher level of study we first have to clear our basic fundamentals and then approach to the next level. Giving equal importance to these assumptions is an important practice cause it help to appreciate what is really going on inside our object under observation.

In this article we examine the stresses on a inclined plane.

Stresses on Inclined Plane- Part II

The basic element for analysing stress

In my previous article of "Principal Planes and Stresses-Part I", I have derived the expression for normal and shear stresses due to the effect of normal forces acting on any set of two orthogonal planes of a cuboid. However in real life problem, it is rare case to find only normal force acting on the surface of an object. The normal forces are often accompanied by traction forces which generate shear stress on our object.

If we go deeper in real life scenario, forces acting on the object is often inclined in nature. In this case, we resolve the forces along our standard Cartesian co-ordinate system and then we find the effect of these resolved forces on the inclined plane.

Imagine the following situation:

Normal and  Shear force acting simultaneously

Consider a cube on which a normal and traction force is being acted.
Now a cutting plane is approaching our cuboid and cuts our object in two pieces and this process generates a set of two orthogonal planes and one inclined plane.

Now we examine the stress generated on our inclined plane.


Expression

Triangular Element 

Consider the following parameters:
θ=Angle of inclination of plane AB
σ(1)=Normal stress acting plane BC
σ(2)=Normal stress acting on plane AC
𝛕=Shear stress acting on plane BC and AC
t=thickness of the triangular element
σ(1)*BC*t=Normal force acting on plane BC
σ(2)*AC*t=Normal force acting on plane AC
𝛕*BC*t=Shear force acting on plane BC
𝛕*AC*t=Shear force acting on plane AC

Total horizontal force=σ(2)*AC*t+𝛕*BC*t=F
Total vertical force=σ(1)*BC*t+𝛕*AC*t=F'

Consider the following resolution of forces F and F' along and perpendicular to the inclined plane AB:

Resolving of Stresses on Inclined Plane

Let σ(θ) be the normal stress developed on plane AB.

Then,

σ(θ)ABt=Fcos(θ)+F'sin(θ)

σ(θ)ABt=σ(1)BCtcos(θ)+𝛕ACtcos(θ)+σ(2)ACtsin(θ)+𝛕(BC)tsin(θ)

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

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

=σ(1)(1+cos(2θ))/2+2𝛕sin(θ)cos(θ)+σ(2)(1-cos(2θ))/2

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

Similarly,

𝛕(θ)=((σ(1)-σ(2))/2)sin(2θ)+𝛕cos(2θ)...Required Result for Shear stress on inclined plane