MOHR’S STRESS CIRCLE CLASS NOTES FOR MECHANICAL ENGINEERING
MOHR’S STRESS CIRCLE
CLASS NOTES FOR MECHANICAL
Every item in nature is under complex
loading. Hence, it is under complex
stresses. But a material fails in a simple
manner. It may fail in simple tension/
strain/shear strain energy. Mohr’s Stress
Circle is a graphical method to determine
simple stresses and strains from complex
stresses and strains. Mohr’s circle is a graphical transformation of two dimensional plane stresses. It is extremely simple and useful.
MOHR’S STRESS CIRCLE
Since it is a graphical method, this post deals with drawing of the Mohr’s Circle from the given 2-dimensional complex stresses.
We know from the analytical method
σθ = (σx + σy)/2 + ((σx — σy)/2) (Cos2θ) + τ Sin2θ
τθ = (( σx — σy)/2) (Sin2θ) — τ Cos2θ
Square and add, we get the equation of a circle as given below.
[σθ – (σx — σy)/2)]2 +τ2 = [(1/2) (σx — σy) + 4τ2 ]1/2
Thus it was concluded that the complex system can be analyzed graphically by a circle.
This circle will represent the state of stress on any plane in the stressed element.
MOST IMPORTANT THREE THINGS TO REMEMBER FOR MOHR’S CIRCLE METHOD
(i) All measurement of stresses are made from the pole.
(ii) Every radius (radial line) of the circle is a plane in the stressed element.
(iii) θ on stressed element is 2θ on Mohr’s circle or vice versa.
PROCEDURE FOR DRAWING MOHR’S CIRCLE
VARIOUS STEPS FOR DRAWING MOHR’S STRESS CIRCLE
1. Draw horizontal axis as σ axis (tensile or compressive axis)
2. + σx, or +σy (tensile stress) in the + x direction (right side)
3. – σx, or -σy (compressive stress) in the – x direction (left side)
4. Draw vertical axis as τ axis (shear stress axis).
5. These two axis intersect at origin ‘ O’ called the POLE.
6. Shear stress + τ (anticlockwise) in the + y direction (upwards from Pole)
7. Shear stress – τ (clockwise shear stress) in the – y direction(downwards)
8. ASSUME σX AND σY TO BE TENSILE.
9. ASSUME τ WITH σX AS ANTICLOCKWISE (+)
10. ASSUME τ WITH σY AS CLOCKWISE (-)
11. Mark OM = + σx using some scale as say 10 N/mm2 = 1 cm
12. Mark MN = Shear Stress (+ τ) vertically upwards at point ‘M’
13. Now point ‘N’ will lie on the circle as it represents the state of stress on the given vertical plane AB.
14. Corresponding radius with respect to N will represent the reference plane AB since every radius is a plane.
15. Mark OL = +σy
16. Mark LQ = Shear stress (- τ ) vertically downwards at point ‘L’
17. Point ‘Q’ will also lie on the circle as it represents the state of stress on the given horizontal plane BC.
18. Radius with respect to Q will represent the plane BC
19. Join points ‘N’ and ‘Q’ to intersect at point ‘P’ the center of the circle.
20. Draw Mohr’s circle taking ‘P’ as center and PN=PQ=Radius
HOW TO FIND THE VARIOUS QUANTITIES FROM MOHR’S CIRCLE
(i) To find principal stresses σ1 and σ2—–points of zero shear stress
Mark two extreme ends on the horizontal diameter as S and R. These are the points of zero shear stress. Therefore measure OS = σ1
OR = σ2
(ii) To find principal planes θpp1 and θpp2
Since every radial line is a plane.
Point S joined to center P becomes a radial line. Similarly lines PN,PQ, PS, PN, PT and PV are radial lines and hence are planes. Hence Radial line PS is the principal plane because of zero shear stress on it.
But the given reference plane is AB on stressed element and PN on Mohr’s circle. Measure angle NPS between the given plane AB and the principal plane PS
Angle NPS = 2θpp1
Since angle on Mohr’s circle is twice of angle on the stressed element, angle θpp1 can be found.
Angle NPR will be 2θpp2.
It is angle between the given plane AB and the other principal plane PR.
NOTE: Both the angles 2θpp1 and2θpp2 are measured clockwise or anticlockwise.
Angle NPR = 2θpp2= 1800 + 2θpp1
θpp2= 900 + θpp1
Hence Angle between principal planes will be 900 on the stressed element. It is 1800 on Mohr’s circle.
OR Principal planes at right angles to each other.
(iii) Determination of planes of maximum shear and maximum shear stress
The point of maximum shear stress will be exactly above as well as exactly below the center of the circle.
The two ends of the vertical diameter will be points of maximum shear.
Let the point on top be T and at the bottom be V.
Measure PT = + Maximum shear stress
Measure PV= – maximum shear stress
Since TP is a radius, becomes the plane of maximum shear
Measure angle TPN = 2θms1
It is the angle between the given plane AB and the plane of maximum shear PT. Another plane of maximum shear will be radius PV.
Measure angle NPV = 2θms2
2θms2 = 2θms1 + 1800
θms2 = θms1 + 900
Thus planes of maximum shear will be at right angles.
(iv) To Find angle between a principal plane and a plane of maximum shear i.e. between θpp1 and θms1
Angle between θpp1 and θms1 on Mohr’s circle will be 900.
But actually angle between θpp1 and θms1 will be 450because angle on Mohr’s circle is twice of angle on stressed element.
HENCE ANGLE BETWEEN A PRINCIPAL PLANE AND A PLANE OF MAXIMUM SHEAR IS 450
Mohr’ circle represents normal (σ) and shear stresses ( τ ) on a plane inclined to the principal plane. Resultant stress will be (σ2 +τ2)0.5. Hence there will be an angle between the resultant stress and the principle plane, called the angle of Obliquity. For maximum angle of obliquity, ß, Tan ß =Sin– [(σx – σy)/(σx – σy)]. The shear stress on the plane of maximum obliquity will be less than the maximum shear stress.
There is only normal stress on a principal plane. A principal plane is a plane of zero shear stress.
There are three planes, major principal plane, intermediate principal plane and minor principal plane.
Fig. ELEMENTS OF MOHR’S STRESS CIRCLE
Fig. 3- DIMENSIONAL STRESS SYSTEM
THREE- DIMENSIONAL STRESS SYSTEM
3 Mohr’s Stress Circles (Circle 1, Circle 2 and Circle 3)
Zero shear stress on principal planes
There is normal stress on the plane of maximum shear
|Principal stresses||Shear stress on principal planes|
|1.||C1A, C1C||OA=σ1, OC=σ3||Zero|
|2.||C2A, C2B||OA=σ1, OB=σ2||Zero|
|3.||C3B, C3C||OB=σ2, OC=σ3||Zero|
|Sr. No.||Planes of maximum shear||Maximum shear stress||Normal stress on the plane of maximum shear|
|2.||C2E, C2H||C3F= C3I= (σ1-σ2)/2||OC2=(σ1-σ2)/2|