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Https _www.iit.edu_arc_workshops_pdfs_mohr_circle

1. 1. Mohr’s Circle Academic Resource Center
2. 2. Introduction • The transformation equations for plane stress can be represented in graphical form by a plot known as Mohr’s Circle. • This graphical representation is extremely useful because it enables you to visualize the relationships between the normal and shear stresses acting on various inclined planes at a point in a stressed body. • Using Mohr’s Circle you can also calculate principal stresses, maximum shear stresses and stresses on inclined planes.
3. 3. Stress Transformation Equations sx 1 s x +s y = s x -s y cos2q + t xy sin 2q 2 2 s x +s y t x1y1 = sin 2q + t xy cos2q 2 1 2
4. 4. Derivation of Mohr’s Circle • If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1 for a given stress state. • Eliminate θ by squaring both sides of 1 and 2 equation and adding the two equations together.
5. 5. Derivation of Mohr’s Circle (cont’d)
6. 6. Mohr’s Circle Equation • The circle with that equation is called a Mohr’s Circle, named after the German Civil Engineer Otto Mohr. He also developed the graphical technique for drawing the circle in 1882. • The graphical method is a simple & clear approach to an otherwise complicated analysis.
7. 7. Sign Convention for Mohr’s Circle • Shear Stress is plotted as positive downward • θ on the stress element = 2θ in Mohr’s circle
8. 8. Constructing Mohr’s Circle: Procedure 1. Draw a set of coordinate axes with σx1 as positive to the right and τx1y1 as positive downward. 2. Locate point A, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note that point A on the circle corresponds to θ = 0°. 3. Locate point B, representing the stress conditions on the y face of the element by plotting its coordinates σx1 = σy and τx1y1 = -τxy. Note that point B on the circle corresponds to θ = 90°.
9. 9. Procedure (cont’d) 4. Draw a line from point A to point B, a diameter of the circle passing through point c (center of circle). Points A and B are at opposite ends of the diameter (and therefore 180° apart on the circle). 5. Using point c as the center, draw Mohr’s circle through points A and B. This circle has radius R. The center of the circle c at the point having coordinates σx1 = σavg and τx1y1 = 0.
10. 10. Stress Transformation: Graphical Illustration
11. 11. Explanation • On Mohr’s circle, point A corresponds to θ = 0. Thus it’s the reference point from which angles are measured. • The angle 2θ locates the point D on the circle, which has coordinates σx1 and τx1y1. D represents the stresses on the x1 face of the inclined element. • Point E, which is diametrically opposite point D is located 180° from cD. Thus point E gives the stress on the y1 face of the inclined element. • Thus, as we rotate the x1y1 axes counterclockwise by an angle θ, the point on Mohr’s circle corresponding to the x1 face moves ccw by an angle of 2θ.
12. 12. Explanation • Principle stresses are stresses that act on a principle surface. This surface has no shear force components (that means τx1y1=0) • This can be easily done by rotating A and B to the σx1 axis. • σ1= stress on x1 surface, σ2= stress on y1 surface. • The object in reality has to be rotated at an angle θp to experience no shear stress.
13. 13. Explanation • The same method to calculate principle stresses is used to find maximum shear stress. • Points A and B are rotated to the point of maximum τx1y1value. This is the maximum shear stress value τmax. • Uniform planar stress (σs) and shear stress (τmax) will be experienced by both x1 and y1 surfaces. • The object in reality has to be rotated at an angle θs to experience maximum shear stress.
14. 14. Example 1 Draw the Mohr’s Circle of the stress element shown below. Determine the principle stresses and the maximum shear stresses. What we know: σx = -80 MPa σy = +50 MPa τxy = 25 MPa Coordinates of Points A: (-80,25) B: (50,-25)
15. 15. Example 1 (cont’d)
16. 16. Example 1 (cont’d) Principle Stress:
17. 17. Example 1 (cont’d) Maximum Shear Stress:
18. 18. Example 2 Given the same stress element (shown below), find the stress components when it is inclined at 30° clockwise. Draw the corresponding stress elements. What we know: σx = -80 MPa σy = +50 MPa τxy = 25 MPa Coordinates of Points A: (-80,25) B: (50,-25)
19. 19. Example 2 (cont’d) Using stress transformation equation (θ=30°): sx - s x +s y 1 = s x -s y cos2q + t xy sin 2q 2 2 s x +s y t x1y1 = sin 2q + t xy cos2q 2 σx = -25.8 MPa σy = -4.15 MPa τxy = 68.8 MPa
20. 20. Example 2 (cont’d) Graphical approach using Mohr’s Circle (and trigonometry)
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