Transcript Document

Copyright© 2001
Content
 Stress Transformation Click here
 A Mini Quiz
Click here
 Strain Transformation Click here
Approximate Duration: 20 minutes
1
Copyright© 2001
Plane Stress Transformation
N. Sivakugan
Copyright© 2001
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 (zdirection)
z = 0; xz = 0; zy = 0
y
x
3
Copyright© 2001
Plane Stress Loading
Therefore, the state of stress at a
point can be defined by the three
independent stresses:
x; y; and xy
A
y
y
x
A
xy
x
4
Copyright© 2001
Objective
y
A
xy
A
x
y
x
State of Stress at A
If x, y, and xy
are known, …
5
Copyright© 2001
Objective
A
y’
State of Stress at A
y

x’
x
…what would be
’x, ’y, and ’xy?
6
Copyright© 2001
Transformation
y
’xy=?

y’
xy
’x=?
A
xy
x
y

x’
x
State of Stress at A
7
Copyright© 2001
Transformation
Solving equilibrium equations for the wedge…
 x  y   x  y 
 x '  
  
 cos 2   xy sin 2
 2   2 
 x  y 
 xy '  
 sin 2   xycos 2
 2 
8
Copyright© 2001
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:
 2 xy 
1
1

 p  tan 

2



y 
 x
gives two values (p1 and p2)
9
Copyright© 2001
Principal Planes & Principal Stresses
Orientation of Principal Planes
p2
p1
x
90
10
Copyright© 2001
Principal Planes & Principal Stresses
Principal Stresses
~ are the normal stresses () acting on the principal planes
 x  y 
 max   1  
  R
 2 
 x  y 
 min   2  
  R

2

 x  y 
2
R  
   xy
11
 2 
2
Copyright© 2001
Maximum Shear (max)
~ maximum shear stress occurs on two mutually
perpendicular planes
~ orientations of the two planes (s) are given by:
  x  y
1
1
 s  t an  
2
2 xy





gives two values (s1 and s2)
max = R
 x  y 
2
R  
   xy
 2 
2
12
Copyright© 2001
Maximum Shear
Orientation of Maximum Shear Planes
s2
s1
x
90
13
Copyright© 2001
Principal Planes & Maximum Shear Planes
45
Principal plane
x
Maximum shear plane
p = s ± 45
14
Copyright© 2001
Mohr Circles
From the stress-transformation equations (slide 7),
2

  x   y 
2
2

'



'

R


xy
 x 


 2 

Equation of a circle, with
variables being x’ and xy’
15
Copyright© 2001
Mohr Circles
(x + y)/2
R
xy’
x’
16
Copyright© 2001
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 ….
17
Copyright© 2001
Mohr Circles

=0
When we rotate the
plane by 180°, we go a
full round (i.e., 360°,
on the Mohr circle.
Therefore….
x’
 = 90
xy’
18
Copyright© 2001
Mohr Circles

2
…..when we rotate the
plane by °, we go 2°
on the Mohr circle.
x’
xy’
19
Copyright© 2001
Mohr Circles
x’
2
max
xy’
1
20
Copyright© 2001
From the three Musketeers
Mohr circle represents the
state of stress at a point; thus
different Mohr circles for
different points in the body
Get the sign
convention right
Quit
Mohr circle is a
simple but powerful
technique
Continue
21
Copyright© 2001
A Mohr Circle Problem
200 kPa
The stresses at a point A are
shown on right.
60 kPa
A
Find the following:
40 kPa
 major and minor principal stresses,
 orientations of principal planes,
 maximum shear stress, and
 orientations of maximum shear stress planes.
22
Drawing Mohr Circle
200 kPa
60 kPa
A
40 kPa
120
 (kPa)
R = 100
 (kPa)
Principal Stresses
1= 220
2= 20
 (kPa)
 (kPa)
R = 100
Maximum Shear Stresses
 (kPa)
max =
100
 (kPa)
200 kPa
Positions of x & y Planes
on Mohr Circle
60 kPa
A
40 kPa
R = 100
120
40
60
60

 (kPa)
tan  = 60/80
 (kPa)
 = 36.87°
Orientations of Principal Planes
200 kPa
60 kPa
A
40 kPa
minor principal
plane
71.6°
36.9°
 (kPa)
18.4°
major principal
plane
 (kPa)
Orientations of Max. Shear
Stress Planes
200 kPa
60 kPa
26.6°
A
53.1°
36.9°
 (kPa)
116.6°
40 kPa
 (kPa)
Copyright© 2001
Testing Times…
Do you want to try a mini quiz?
YES
Oh, NO!
29
Question 1:
The state of stress at a point A
is shown.
What would be the maximum
shear stress at this point?
90 kPa
40 kPa
A
30 kPa
Answer 1: 50 kPa
Press RETURN for the answer
Press RETURN to continue
Question 2:
90 kPa
At A, what would be the
principal stresses?
Answer 2:
40 kPa
A
30 kPa
10 kPa, 110 kPa
Press RETURN for the answer
Press RETURN to continue
Question 3:
90 kPa
At A, will there be any
compressive stresses?
Answer 3:
40 kPa
A
30 kPa
No. The minimum normal stress is 10 kPa (tensile).
Press RETURN for the answer
Press RETURN to continue
Question 4:
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 kPa
B
90 kPa
0
This is hydrostatic state of
stress (same in all directions).
No shear stresses.
Press RETURN for the answer
Press RETURN to continue
Copyright© 2001
Plane Strain Transformation
N. Sivakugan
Copyright© 2001
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 (zdirection)
z = 0; xz = 0; zy = 0
y
x
35
Copyright© 2001
Plane Strain Transformation
Similar to previous derivations. Just replace
 by , and
 by /2
36
Copyright© 2001
Plane Strain Transformation
Sign Convention:
Normal strains (x and y): extension positive
Shear strain ( ): decreasing angle positive
e.g.,
y
y
before
x
x positive
y negative
 positive
after
x
37
Copyright© 2001
Plane Strain Transformation
 xy
 x   y   x  y 
 x '  
sin 2
  
 cos 2 
2
 2   2 
 xy '
 xy
 x  y 
 
cos 2
sin 2 
2
2
 2 
Same format as the stress
transformation equations
38
Copyright© 2001
Principal Strains
~ maximum (1) and minimum (2) principal strains
~ occur along two mutually perpendicular directions, given by:
  xy 
1
1
 Gives two values (p1 and p2)
 p  tan 

2



y 
 x
 x   y 
1  
  R
 2 
 x   y 
1  
  R
 2 
  x   y    xy 
R  
  

 2   2 
2
2
39
Copyright© 2001
Maximum Shear Strain (max)
max/2 = R
  x   y    xy 
R  
  

 2   2 
2
2
p = s ± 45
40
Copyright© 2001
Mohr Circles
(x + y)/2
R
x’
xy’
2
41
Copyright© 2001
Strain Gauge
~ measures normal strain (), from the change in
electrical resistance during deformation
electrical resistance
strain gauge
42
Copyright© 2001
Strain Rosettes
~ measure normal strain () in three directions; use
these to find x, y, and xy
e.g., 45° Strain Rosette
x = 0
90
y = 90
45
45°
45°
measured
0
x
xy = 2 45 – (0 +
90)
43