Stress on inclined planes under axial loading:

Stress on inclined planes under axial loading:


When a body is under an axial load, the plane normal to the axis contains only the normal
stress as discussed in section.

However, if we consider an oblique plane that forms an angle with normal plane, it
consists shear stress in addition to normal stress.
Consider such an oblique plane in a bar. The resultant force P acting on that plane will

keep the bar in equilibrium against the external load P'


The resultant force P on the oblique plane can be resolved into two components Fn and Fs
that are acting normal and tangent to that plane, respectively.
If A is the area of cross section of the bar, A/cos is the area of the oblique plane. Normal
and shear stresses acting on that plane can be obtained as follows.
Fn= Pcosθ
Fs = -Psinθ (Assuming shear causing clockwise rotation negative).

σ = Pcos

θ/A/cos

θ = P/A cos

2

θ

τ = -Psin

θ/A/cos

θ = -P/Asin

θcos

θ

 It define the normal and shear stress values on an inclined plane that
makes an angle θ with the vertical plane on which the axial load acts.
From above equations, it is understandable that the normal stress reaches its maximum
when θ = 0o and becomes zero when θ = 90o.


But, the shear stress assumes zero value at θ = 0o and θ = 90o and reaches its maximum
when θ = 45o.
The magnitude of maximum shear stress occurring at θ = 45o plane is half of the maximum
normal stress that occurs at θ = 0o for a material under a uniaxial loading.


  τ =

max

P/

2A =

σmax/2

Now consider a cubic element A in the rod which is represented in two dimension

 such that one of its sides makes an angle with the vertical plane

To determine the stresses acting on the plane mn, equations 1.6 and 1.7 are used as such
and to knows the stresses on plane om, θ is replaced by θ + 90o.
Maximum shear stress occurs on both om and mn planes with equal magnitude and
opposite signs, when mn forms 45o angle with vertical plane.