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• Post author:Shibashis Ghosh
• Post published:January 25, 2009
• Post category:Engineering design
• Post comments:0 Comments

The green dots of the mohr’s circle represent the maximum and minimum principal stresses (σ₁ & σ₂₎ and the red dots represent maximum and minimum shear stresses (T_max & T_min ). The stress system of the Mohr’s circle will be:

The maximum and minimum shear stresses will be acting at 45⁰ away from the principal stresses.

How to draw a Mohr’s circle if normal stresses and shear stresses are given?

We will take a practical example to digest the process of drawing Mohr’s circle, but before going to that let’s see the sign conventions for structural elements:

• Tensile stress will always be considered as positive and compressive as negative.
• If a pair of shear stress create counterclockwise torque to an element, then it will be considered as positive or else negative
• Angle of an inclined plane will be measured from vertical and if it goes counterclockwise from vertical, then it will be positive or else negative.

Now, lets see an example of plotting a Mohr’s circle for the normal-X stress=4000 Mpa, normal-Y stress=3000 Mpa and shear stress= 1000 Mpa. Means the situation is like below:

Following steps need to be followed for creating Mohr’s circle:

1.  Plot point A (4000, 1000) and B (3000,-1000) with respect to a coordinate system.

2. Get the mid point of A and B and create a circle taking the mid point as center and joining A and B. The Mohr’s circle is  ready like below:

3.  Radius of circle will be calculated using a formula,

R=√ {[(σ_x– σ_y)/2]^{2 +} [T_xy ]²} ⁼1118 Mpa; the value is same as we get from Mohr’s circle diagram. This is also the value of maximum shear stress.

• Coordinate of the centre of Mohr  circle diagram = [(σ_x + σ_y )/2, 0 ]= [3500,0]. 
• X Coordinate of the point C will be the value of maximum principal stress= (X coordinate of centre+ radius) = (3500+1118) =4618.
• X coordinate of the point E will be the value of minimum principal stress= (X coordinate of centre – radius) = (3500-1118) = 2382.
• The angle between the line AB and EC is twice the angle between the principal and normal plane. From the graph it has been found as 63.43⁰. This angle could be calculated from the geometry of the circle as well, from the equation below:

Tan (2*θ) = (1000)/ (4000-3500)

θ= 31.7⁰ , 121.7⁰

So, the orientation of principal plane will be like below:

Another principal plane will be at 90⁰ to the plane shown.

Shibashis Ghosh

Hi, I am Shibashis, a blogger by passion and an engineer by profession. I have written most of the articles for mechGuru.com. For more than a decades i am closely associated with the engineering design/manufacturing simulation technologies. I am a self taught code hobbyist, presently in love with Python (Open CV / ML / Data Science /AWS -3000+ lines, 400+ hrs. )