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Copyright© 2001
Content
Stress Transformation
A Mini Quiz
Strain Transformation
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Approximate Duration: 20 minutes
N. Sivakugan
Copyright© 2001
Plane Stress Transformation
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Plane Stress Loading
x
y
~ 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
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Plane Stress Loading
x
y
Therefore, the state of stress at a point can be defined by the three independent stresses:
x; y; and xy
x
y xy
A
A
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Objective
A
x
y
x
y xy
A
State of Stress at A
If x, y, and xy are known, …
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Objective
A
x
y
’x
’y ’xy
A
State of Stress at A
…what would be ’x, ’y, and ’xy?
x’y’
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Transformation
x
yx’
y’
A
State of Stress at A
x
y xy
xy
’x=?’xy=?
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Transformation
2 sin2 cos22
' xyyxyx
x
2 os2 in2
' cs xyyx
xy
Solving equilibrium equations for the wedge…
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Principal Planes & Principal Stresses
Principal Planes~ are the two planes where the normal stress () is
the maximum or minimum
~ the orientations of the planes (p) are given by:
yx
xyp
2tan
2
1 1
gives two values (p1 and p2)
~ there are no shear stresses on principal planes
~ these two planes are mutually perpendicular
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Principal Planes & Principal Stresses
xp1
p2
90
Orientation of Principal Planes
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Principal Planes & Principal Stresses
Principal Stresses
~ are the normal stresses () acting on the principal planes
Ryx
21max
Ryx
22min
2
2
2 xyyxR
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Maximum Shear (max)~ maximum shear stress occurs on two mutually perpendicular planes
xy
yxs
2tan
2
1 1
gives two values (s1 and s2)
~ orientations of the two planes (s) are given by:
max = R2
2
2 xyyxR
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Maximum Shear
xs1
s2
90
Orientation of Maximum Shear Planes
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Principal Planes & Maximum Shear Planes
x
Principal plane
Maximum shear plane
p = s ± 45
45
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Mohr CirclesFrom the stress-transformation equations (slide 7),
22
2
'2
' Rxyyx
x
Equation of a circle, with variables being x’ and xy’
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Mohr Circles
x’
xy’
(x + y)/2
R
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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 ….
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Mohr Circles
x’
xy’
= 90
= 0
When we rotate the plane by 180°, we go a full round (i.e., 360°, on the Mohr circle.
Therefore….
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Mohr Circles
x’
xy’
…..when we rotate the plane by °, we go 2°
on the Mohr circle.
2
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Mohr Circles
x’
xy’
2
1
max
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From the three Musketeers
Get the sign convention right
Mohr circle is a simple but powerful
technique
Mohr circle represents the state of stress at a point;
thus different Mohr circles for different points in the
body
QuitQuit ContinueContinue
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A40 kPa
200 kPa
60 kPaThe stresses at a point A are shown on right.
A Mohr Circle Problem
Find the following:
major and minor principal stresses,
orientations of principal planes,
maximum shear stress, and
orientations of maximum shear stress planes.
A40 kPa
200 kPa
60 kPa
(kPa)
(kPa)
R = 100
Drawing Mohr Circle
120
(kPa)
(kPa)
1= 220
2= 20
Principal Stresses
R = 100
(kPa)
(kPa)
max = 100
Maximum Shear Stresses
A40 kPa
200 kPa
60 kPa
(kPa)
(kPa)
R = 100
120
Positions of x & y Planeson Mohr Circle
60
40
60
tan = 60/80
= 36.87°
(kPa)
(kPa)
Orientations of Principal Planes
A40 kPa
200 kPa
60 kPa
36.9°
18.4°
major principal plane
71.6°
minor principal plane
Orientations of Max. Shear Stress Planes
(kPa)
(kPa)
A40 kPa
200 kPa
60 kPa
36.9°
53.1°
26.6°
116.6°
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Testing Times…
Do you want to try a mini quiz?
Oh, NO!YESYES
Question 1:
A30 kPa
90 kPa
40 kPaThe 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
Question 2:
A30 kPa
90 kPa
40 kPaAt A, what would be the principal stresses?
Answer 2: 10 kPa, 110 kPa
Press RETURN for the answer Press RETURN to continue
Question 3:
A30 kPa
90 kPa
40 kPaAt 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
Question 4:
B90 kPa
90 kPa
0 kPaThe 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 for the answer Press RETURN to continue
N. Sivakugan
Copyright© 2001
Plane Strain Transformation
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Plane Strain Loading
x
y
~ 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
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Plane Strain TransformationSimilar to previous derivations. Just replace
by , and by /2
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Plane Strain Transformation
Sign Convention:
Shear strain ( ): decreasing angle positive
e.g.,
Normal strains (x and y): extension positive
x
y
beforex
y
after
x positivey negative positive
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Plane Strain Transformation
2 sin2
2 cos22
' xyyxyxx
2 os2
2 in22
'cs xyyxxy
Same format as the stress transformation equations
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Principal Strains
yx
xyp
1tan
2
1Gives two values (p1 and p2)
~ maximum (1) and minimum (2) principal strains
~ occur along two mutually perpendicular directions, given by:
Ryx
21
Ryx
21
22
22
xyyxR
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Maximum Shear Strain (max)
max/2 = R22
22
xyyxR
p = s ± 45
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Mohr Circles(x + y)/2
R x’
xy’2
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Strain Gauge
electrical resistance strain gauge
~ measures normal strain (), from the change in electrical resistance during deformation
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Strain Rosettes~ measure normal strain () in three directions; use
these to find x, y, and xye.g., 45° Strain Rosette
x
45°
45°
0
90
45
x = 0
y = 90
xy = 2 45 – (0 + 90)
measured
