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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
3
Copyright© 2001
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)
σz = 0; τxz = 0; τzy = 0
4
Copyright© 2001
Plane Stress Loading
x
y
Therefore, the state of stress at a point can be defined by the three independent stresses:
σx; σy; and τxy
σx
σy τxy
A
A
5
Copyright© 2001
Objective
A
x
y
σx
σy τxy
A
State of Stress at A
If σx, σy, and τxy are known, …
6
Copyright© 2001
Objective
A
x
y
σ’x
σ’y τ’xy
A
State of Stress at A
…what would be σ’x, σ’y, and τ’xy?
x’y’
θ
7
Copyright© 2001
Transformation
x
yx’
y’θ
A
State of Stress at A
θ σx
σy τxyτxy
σ’x=?τ’xy=?
8
Copyright© 2001
Transformation
θτθσσσσ
σ 2 sin2 cos22
' xyyxyx
x +
−+
+=
θτθσσ
τ 2 os2 in2
' cs xyyx
xy +
−
−=
Solving equilibrium equations for the wedge…
9
Copyright© 2001
Principal Planes & Principal Stresses
Principal Planes~ are the two planes where the normal stress (σ) is
the maximum or minimum
~ the orientations of the planes (θp) are given by:
−
= −
yx
xyp σσ
τθ
2tan
2
1 1
gives two values (θp1 and θp2)
~ there are no shear stresses on principal planes
~ these two planes are mutually perpendicular
10
Copyright© 2001
Principal Planes & Principal Stresses
xθp1
θp2
90°
Orientation of Principal Planes
11
Copyright© 2001
Principal Planes & Principal Stresses
Principal Stresses
~ are the normal stresses (σ) acting on the principal planes
Ryx +
+==
21max
σσσσ
Ryx −
+
==22min
σσσσ
2
2
2 xyyxR τ
σσ+
−=
12
Copyright© 2001
Maximum Shear (τmax)~ maximum shear stress occurs on two mutually perpendicular planes
−−= −
xy
yxs τ
σσθ
2tan
2
1 1
gives two values (θs1 and θs2)
~ orientations of the two planes (θs) are given by:
τmax = R2
2
2 xyyxR τ
σσ+
−=
13
Copyright© 2001
Maximum Shear
xθs1
θs2
90°
Orientation of Maximum Shear Planes
14
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Principal Planes & Maximum Shear Planes
x
Principal plane
Maximum shear plane
θp = θs ± 45 °
45°
15
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Mohr CirclesFrom the stress-transformation equations (slide 7),
22
2
'2
' Rxyyx
x =+
+− τ
σσσ
Equation of a circle, with variables being σx’ and τxy’
16
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Mohr Circles
σx’
τxy’
(σx + σy)/2
R
17
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Mohr Circles
A point on the Mohr circle represents the σ x’ and τ xy’ values on a specific plane.
θ 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
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Mohr Circles
σx’
τxy’
θ = 90°
θ = 0
θ
When we rotate the plane by 180°, we go a full round (i.e., 360°, on the Mohr circle.
Therefore….
19
Copyright© 2001
Mohr Circles
σx’
τxy’
…..when we rotate the plane by θ°, we go 2θ°
on the Mohr circle.
θ2θ
20
Copyright© 2001
Mohr Circles
σx’
τxy’
σ2
σ1
τmax
21
Copyright© 2001
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
22
Copyright© 2001
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
σ (kPa)
τ (kPa)
R = 100
Drawing Mohr Circle
120
σ (kPa)
τ (kPa)
σ1= 220
σ2= 20
Principal Stresses
R = 100
σ (kPa)
τ (kPa)
τmax = 100
Maximum Shear Stresses
A40 kPa
200 kPa
60 kPa
σ (kPa)
τ (kPa)
R = 100
120
Positions of x & y Planeson Mohr Circle
60
40
60
θ
tan θ = 60/80
θ = 36.87°
σ (kPa)
τ (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
σ (kPa)
τ (kPa)
A40 kPa
200 kPa
60 kPa
36.9°
53.1°
26.6°
116.6°
29
Copyright© 2001
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
35
Copyright© 2001
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)
εz = 0; γxz = 0; γzy = 0
36
Copyright© 2001
Plane Strain Transformation
Similar to previous derivations. Just replace
σ by ε, and
τ by γ/2
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Copyright© 2001
Plane Strain TransformationSign Convention:
Shear strain (γ ): decreasing angle positive
e.g.,
Normal strains (εx and εy): extension positive
x
y
beforex
y
after
εx positiveεy negative
γ positive
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Copyright© 2001
Plane Strain Transformation
θγ
θεεεε
ε 2 sin2
2 cos22
' xyyxyxx +
−+
+=
θγ
θεεγ
2 os2
2 in22
'cs xyyxxy +
−−=
Same format as the stress transformation equations
39
Copyright© 2001
Principal Strains
−
= −
yx
xyp εε
γθ 1tan
2
1Gives two values (θp1 and θp2)
~ maximum (ε1) and minimum (ε2) principal strains
~ occur along two mutually perpendicular directions, given by:
Ryx +
+=
21
εεε
Ryx −
+=
21
εεε
22
22
+
−= xyyxR
γεε
40
Copyright© 2001
Maximum Shear Strain (γmax)
γmax/2 = R22
22
+
−= xyyxR
γεε
θp = θs ± 45 °
41
Copyright© 2001
Mohr Circles(εx + εy)/2
R εx’
γxy’2
42
Copyright© 2001
Strain Gauge
electrical resistance strain gauge
~ measures normal strain (ε), from the change in electrical resistance during deformation
43
Copyright© 2001
Strain Rosettes~ measure normal strain (ε) in three directions; use
these to find εx, εy, and γxy
e.g., 45° Strain Rosette
x
45°
45°
ε0
ε90
ε45
εx = ε0
εy = ε90
γxy = 2 ε45 – (ε0 + ε90)
measured