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1. Content • Stress Transformation • AMini Quiz • Strain Transformation Click here Click here Click here Approximate Duration: 20 minutes

2. Plane Stress Transformation

3. y x Plane Stress Loading ~ where all elements of the body are subjected to normal and shear stresses acting along a plane (x-y); none perpendicular to the plane (z-direction) z = 0; xz = 0; zy = 0

4. y xy x A A y x Plane Stress Loading Therefore, the state of stress at a point can be defined by the three independent stresses: x; y; and xy

5. y xy x A A y x Objective State of Stress at A If x, y, and xy are known, …

6. ’y ’xy ’x A y y’ A x’  x Objective State of Stress at A …what would be ’x, ’y, and ’xy?

7. y xy xy ’xy=? ’x=? x  y y’ x’  x Transformation A State of Stress at A

8. Transformation Solving equilibrium equations for the wedge…

9. gives two values (p1 and p2) Principal Planes & Principal Stresses Principal Planes ~ are the two planes where the normal stress () is the maximum or minimum ~ there are no shear stresses on principal planes ~ these two planes are mutually perpendicular ~ the orientations of the planes (p) are given by:

10. p2 p1 x 90 Principal Planes & Principal Stresses Orientation of Principal Planes

11. Principal Planes & Principal Stresses Principal Stresses ~ are the normal stresses () acting on the principal planes

12. gives two values (s1 and s2) Maximum Shear (max) ~ maximum shear stress occurs on two mutually perpendicular planes ~ orientations of the two planes (s) are given by: max = R

13. s2 s1 x 90 Maximum Shear Orientation of Maximum Shear Planes

14. 45 Principal plane x Maximum shear plane Principal Planes & Maximum Shear Planes p = s± 45

15. Equation of a circle, with variables being x’ and xy’ Mohr Circles From the stress-transformation equations (slide 7),

16. (x + y)/2 R x’ xy’ Mohr Circles

17. Mohr Circles • A point on the Mohr circle represents the x’ and xy’values on a specific plane. •  is measured counterclockwise from the original x-axis. • Same sign convention for stresses as before. i.e., on positive planes, pointing positive directions positive, and ….

18.  = 0 x’   = 90 xy’ Mohr Circles When we rotate the plane by 180°, we go a full round (i.e., 360°, on the Mohr circle. Therefore….

19.  2 x’ xy’ Mohr Circles …..when we rotate the plane by °, we go 2° on the Mohr circle.

20. x’ 2 max 1 xy’ Mohr Circles

21. From the three Musketeers Mohr circle represents the state of stress at a point; thus different Mohr circles for different points in the body Mohr circle is a simple but powerful technique Get the sign convention right Quit Continue

22. 200 kPa 60 kPa A 40 kPa A Mohr Circle Problem The stresses at a point A are shown on right. Find the following: • major and minor principal stresses, • orientations of principal planes, • maximum shear stress, and • orientations of maximum shear stress planes.

23. 200 kPa 60 kPa A 40 kPa 120  (kPa) R = 100  (kPa) Drawing Mohr Circle

24. 1= 220  (kPa) 2= 20 R = 100  (kPa) Principal Stresses

25.  (kPa) max = 100  (kPa) Maximum Shear Stresses

26. 200 kPa 60 kPa A 40 kPa R = 100 60 120  40  (kPa) 60  (kPa) Positions of x & y Planeson Mohr Circle tan  = 60/80  = 36.87°

27. 200 kPa 60 kPa A 40 kPa 71.6°  (kPa) 36.9° major principal plane 18.4°  (kPa) Orientations of Principal Planes minor principal plane

28. 200 kPa 26.6° 60 kPa A 40 kPa 53.1°  (kPa) 36.9°  (kPa) 116.6° Orientations of Max. Shear Stress Planes

29. Testing Times… Do you want to try a mini quiz? YES Oh, NO!

30. Question 1: 90 kPa 40 kPa A 30 kPa The state of stress at a point A is shown. What would be the maximum shear stress at this point? Answer 1: 50 kPa Press RETURN for the answer Press RETURN to continue

31. Question 2: 90 kPa 40 kPa A 30 kPa At A, what would be the principal stresses? Answer 2: 10 kPa, 110 kPa Press RETURN to continue Press RETURN for the answer

32. Question 3: 90 kPa 40 kPa A 30 kPa At A, will there be any compressive stresses? Answer 3: No. The minimum normal stress is 10 kPa (tensile). Press RETURN to continue Press RETURN for the answer

33. Question 4: 90 kPa 0 kPa B 90 kPa The state of stress at a point B is shown. What would be the maximum shear stress at this point? Answer 4: 0 This is hydrostatic state of stress (same in all directions). No shear stresses. Press RETURN to continue Press RETURN for the answer

34. Plane Strain Transformation

35. y x Plane Strain Loading ~ where all elements of the body are subjected to normal and shear strains acting along a plane (x-y); none perpendicular to the plane (z-direction) z = 0; xz = 0; zy = 0

36. Plane Strain Transformation Similar to previous derivations. Just replace  by , and  by /2

37. y y x x before after Plane Strain Transformation Sign Convention: Normal strains (x andy): extension positive Shear strain ( ): decreasing angle positive e.g., x positive y negative  positive

38. Plane Strain Transformation Same format as the stress transformation equations

39. Principal Strains ~ maximum (1) and minimum (2) principal strains ~ occur along two mutually perpendicular directions, given by: Gives two values (p1 and p2)

40. Maximum Shear Strain (max) max/2 = R p = s± 45

41. (x + y)/2 R x’ xy’ 2 Mohr Circles

42. electrical resistance strain gauge Strain Gauge ~ measures normal strain (), from the change in electrical resistance during deformation

43. 90 measured 45 45° 0 45° x Strain Rosettes ~ measure normal strain () in three directions; use these to find x, y, and xy e.g., 45° Strain Rosette x = 0 y = 90 xy = 2 45– (0+ 90)

44. The End