Χσ

Procedure for drawing Mohr’s circle SIGN CONVENTIONS :

• Positive normal stresses (Tensile) are to the right.

• Negative normal stresses (Compressive) are to the left.

• Shear stressese that tend to rotate the stress element clockwise (CW) are plotted upward on

the ττττ - axis

• Shear stressese that tend to rotate the stress element counter clockwise (CCW) are plotted

downward on the ττττ - axis

Procedure

1. Identify the stress condition at the point of interest and represent it as the initial stress element the manner shown in Figure 1

2. The combination σσσσx and ττττxy is plotted as point 1 on the σσσσ−−−−ττττ plane.

3. The combination σσσσY and ττττyx is then plotted as point 2. Note that ττττxy and ττττyx always act in

opposite directions; Therefore, one point will be plotted above the σσσσ -axis and one will be below.

4. Draw a straight line between the two points.

5. This line crosses the σσσσ - axis at the centre of Mohr’s circle, which is also the value of the average normal stress applied to the initial stress element. The location of the centre can be observed from the data used to plot the points or computed from Equation

For convenience, label the centre O.

6. Identify the line from O through point 1 (σσσσx, ττττxy) the x-axis. This line corresponds to the original x-axis and is essential to correlating the data from Mohr’s circle to the original x and y directions.

7. The points O , σσσσx and point 1 form and important right triangle because the distance from O

to point 1, the hypotenuse of the triangle, is equal to the radius of the circle, R. Calling the other

two sides “a” and “b”; the following calculations can be made:

b = ττττxy

Note that the equation for R is identical to for the maximum shear stress on the element. Thus The length of the radius of Mohr’s circle is equal to the magnitude of the maximum shear stress.

8. Draw the complete circle with the centre at o and the radius R

[ ]yxavg σσσ +=2

1

[ ]yxa σσ −=2

1

xyyxbaR222

)(2

1τσσ +

−=+=

9. Draw the vertical diameter of the circle. The point at the top of the circle has the coordinates

(σσσσavg, ττττmax) where the shear stress has the clockwise (cw) direction .The point at the bottom of

the circle represents(σσσσavg, ττττmax) Where the shear stress is counter clockwise (ccw).

10. Identify the points on the σσσσ - axis at the ends of the horizontal diameter as σσσσP1 at the right

(the maximum principal stress) and σσσσP2( the minimum principlal stress). Note that the shear stress is zero at these points.

11. Determine the values for σσσσP1 and σσσσP2 from

σσσσP1 = “O” + R and σσσσP2 = “O” - R

where “O” represents the coordinate of the centre of the circle, σσσσavg and R is the radius of the circle. This Equation are identical to equations for the principal stresses. The following steps determine the angles of orientation of the principal stress element and maximum shear stress element. An important concept to remember is that angles obtained from Mohr’s circle are double the true angles.

12. The orientation of the principal stress element is determined by finding the angle from the x-

axis to the σσσσP1 - axis, labelled 2ϕ in Figure 2. From the data on the circle you can see that The argument of this inverse tangent function is the same as the absolute value of the argument. Problems with signs for the resulting angle are avoided by noting the direction from the x – axis

to the σσσσP1 - axis on the circle, clockwise for the present example. The principal stress element

is rotated in the same direction from the x- axis by an amount φφφφ to locate the face on which the

maximum principal stress σσσσP1 acts. 13. Draw the principal stress element in its proper orientation as determined from step 12 with

the two principal stresses σσσσP1 and σσσσP2 . 14. The orientation of the maximum shear stress element is determined by finding the angle

from the x-axis to the ττττmax -axis, labelled 2φφφφ ‘ in Figure 2. From trigonometry it can be shown that this is equivalent to finding the inverse tangent of a/b,

the reciprocal of the argument used to find 2φφφφ Again problems with signs for the resulting angle are avoided by noting the direction from

the x-axis to the ττττmax -axis on the circle, counter clockwise for the present example. Then the maximum shear stress element is rotated in the same direction from the x-axis by and amount to locate the face on which the clockwise maximum shear stress acts. 15. Draw the maximum shear stress element in its proper orientation as determined from step 14 with the shear stresses on all four faces and the average normal stress acting on each face.

a

b1tan2

−=φ