Question

Using Mohr’s circle, derive expression for normal and tangential stresses on a diagonal plane of a material subjected to pure shear. Also state and explain mohr’s theorem for slope and deflection.

Answer 1

Mohr circle is a graphical method to find the stress system on any inclined plane through the body. It is a circle drawn for the compound stress system. The centre of the circle has the coordinate (σ_x - σ_y)/2, 0) and radius of circle is √((σ_x - σ_y)/2)² + τ²

by drawing the mohr’s circle of stress the following three systems can be determined.

(a) The normal stress, shear stress, and resultant stress on any plane. 

(b) Principal stresses and principal planes. 

(c) Maximum shearing stresses and their planes along with the associated normal stress. Mohr’s circle can also be drawn for compound strain system. 

1. Mohr’s circle of stress with reference to two mutually perpendicular principal stresses acting on a body consider both dike stresses.

Both principal stresses may be considered as (a) Tensile and (b) Compressive. Let us for tensile i.e.; σ_x > σ_y 

Steps: 

1. Mark OA = S_x and OB = S_y, along x-axis (on +ve side if tensile and -ve side if compressive). 

2. Draw the circle with BA as diameter, called mohr’s circle of stress. 

3. To obtain stress on any plane ¸ as shown in fig (a) measure angle ACP = 2¸ in counter-clock wise direction. 

4. The normal stress on the plane is S_n = OQ 

Shear stress is τ = PQ 

Resultant stress = σ_r = OP 

And Angle POQ = ϕ is known as angle of obliquity. 

5. Mohr’s stress circle for a two dimensional compound stress condition shown in fig. 

Let for a tensile σ_x > σ _y

Steps: 

l. Mark OA = σx and OB = σy along x axis 

[on +ve side if tensile and –ve side if compessive.] 

2. Mark AC = τ_xy and BD = τ_xy

3. Join C and D which bisects AB at E the centre of Mohr’s circle. 

4. With E as centre, either EC or ED as a radius draw the circle called Mohr’s circle of stress. 

5. Point P and Q at which the circle cuts S axis gives principal planes OP = σ₁, and OQ = σ₂ gives the two principal stresses. 

6. ∠CEP = 2θ₁ and ∠CEQ = 2θ₂ is measured in anti-clockwise direction. 

θ₁ = (1/2) ∠CEP and θ₁ = (1/2) ∠CEQ indicates principal planes. 

7. ER = EC = τ_max is the maximum shearing stress. 

θ′₁ = (1/2) ∠CER and θ′₂ = (1/2) ∠CES 

is measured anti-clockwise direction gives maximum shearing planes. 

8. To obtain stress on any plane ‘θ’ measure ∠CEX = 2θ in anticlockwise direction. 

Normal stress on the plane is, 

σ_n = OY 

Shear Stress, τ = _XY 

Resultant Stress is σ_r = OX 

And ∠XOY = ϕ is known as angle of obliquity.