Plane Stress Transformation.pptx

Prof. Samirsinh Parmar
Asst. Professor, Dept. of Civil Engg.
Dharmasinh Desai University, Nadiad, Gujarat, INDIA
Mail: samirddu@gmail.com

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
Plane Stress Transformation- SPP, DoCL, DDU, Nadiad.

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

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

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

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

Transformation







 2
sin
2
cos
2
2
' xy
y
x
y
x
x 







 








 






 2
os
2
in
2
' c
s xy
y
x
xy 







 


Solving equilibrium equations for the wedge…

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:









 
y
x
xy
p




2
tan
2
1 1
gives two values (p1 and p2)
~ there are no shear stresses on principal planes
~ these two planes are mutually perpendicular

x
p1
p2
90
Orientation of Principal Planes

Principal Stresses
~ are the normal stresses () acting on the principal planes
R
y
x








 


2
1
max




R
y
x








 


2
2
min




2
2
2
xy
y
x
R 










 


Maximum Shear (max)
~ maximum shear stress occurs on two mutually
perpendicular planes







 

 
xy
y
x
s




2
tan
2
1 1
gives two values (s1 and s2)
~ orientations of the two planes (s) are given by:
max = R
2
2
2
xy
y
x
R 










 


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

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

Mohr Circles
From the stress-transformation equations (slide 7),
2
2
2
'
2
' R
xy
y
x
x 














 
 



Equation of a circle, with
variables being x’ and xy’

Mohr Circles
x’
xy’
(x + y)/2
R

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 ….

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….

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

Mohr Circles
x’
xy’
2
1
max

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
Quit Continue

A 40 kPa
200 kPa
60 kPa
The 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.

A 40 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

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

 (kPa)
 (kPa)
Orientations of Principal Planes
A 40 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)
A 40 kPa
200 kPa
60 kPa
36.9°
53.1°
26.6°
116.6°

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

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

Question 2:
A 30 kPa
90 kPa
40 kPa
At A, what would be the
principal stresses?
Answer 2: 10 kPa, 110 kPa

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).

Question 4:
B 90 kPa
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: 0
This is hydrostatic state of stress
(same in all directions). No shear
stresses.

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

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

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








 2
sin
2
2
cos
2
2
' xy
y
x
y
x
x 







 








 







2
os
2
2
in
2
2
'
c
s xy
y
x
xy








 


Same format as the stress
transformation equations

Principal Strains









 
y
x
xy
p



 1
tan
2
1
Gives two values (p1 and p2)
~ maximum (1) and minimum (2) principal strains
~ occur along two mutually perpendicular directions, given by:
R
y
x








 

2
1



R
y
x








 

2
1



2
2
2
2 















 
 xy
y
x
R




Maximum Shear Strain (max)
max/2 = R
2
2
2
2 















 
 xy
y
x
R



p = s ± 45

Mohr Circles
(x + y)/2
R x’
xy’
2

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

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

Plane Stress Transformation.pptx

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