1. Content Stress Transformation A Mini Quiz Strain Transformation
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Approximate Duration: 20 minutes

2. Plane Stress Transformation

3. Plane Stress Loading z = 0; xz = 0; zy = 0
~ 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 x y

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

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

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

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

8. Transformation
Solving equilibrium equations for the wedge…

9. Principal Planes & Principal Stresses
~ 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:
gives two values (p1 and p2)

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

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

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

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

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

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

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

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

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

20. Mohr Circles
xy’
x’ 2
max 1

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. A Mohr Circle Problem A The stresses at a point A are shown on right.
40 kPa
200 kPa
60 kPa
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. A  (kPa)  (kPa) Drawing Mohr Circle 200 kPa 60 kPa 40 kPa 120

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

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

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

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

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

29. Testing Times…
Do you want to try a mini quiz?
YES
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30. The state of stress at a point A is shown.
Question 1: A
30 kPa
90 kPa
40 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. At A, what would be the principal stresses?
Question 2: A
30 kPa
90 kPa
40 kPa
At A, what would be the principal stresses?
Answer 2:
10 kPa, 110 kPa
Press RETURN for the answer
Press RETURN to continue

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

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

34. Plane Strain Transformation

35. Plane Strain Loading z = 0; xz = 0; zy = 0
~ 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 x y

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

37. Plane Strain Transformation
Sign Convention:
Normal strains (x and y): extension positive
Shear strain ( ): decreasing angle positive x y
before x y
after
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. Mohr Circles
(x + y)/2
x’
xy’ 2 R

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

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

44. The End