Complete Soil Mech. Undestanding Pakage: ABHAY)

1. 1
Copyright© 2001
Content
 Stress Transformation
 A Mini Quiz
 Strain Transformation
CClliicckk hheerree
CClliicckk hheerree
CClliicckk hheerree
Approximate Duration: 20 minutes

2. N. Sivakugan
Copyright© 2001
Plane Stress Transformation

3. 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)
sz = 0; txz = 0; tzy = 0

4. 4
Copyright© 2001
Plane Stress Loading
x
y
Therefore, the state of stress at a
point can be defined by the three
independent stresses:
sx; sy; and txy
sx
sy txy
A
A

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Objective
A
x
y
sx
sy txy
A
State of Stress at A
If sx, sy, and txy
are known, …

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Objective
A
x
y
s’x
s’y t’xy
A
State of Stress at A
…what would be
s’x, s’y, and t’xy?
x’
y’
q

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Transformation
x
y
x’
y’
q
sy txy
q sx
A
State of Stress at A
txy
s’x t’xy=? =?

8. 8
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Transformation
Solving equilibrium equations for the wedge…
æ -
+ ÷ ÷ø
æ +
=
x + ÷ ÷ø
q t q
s s s s
x y x y
s cos 2 sin 2
' xy
2 2
ö
ç çè
ö
ç çè
æ -
x y
' s c xy
xy + ÷ ÷ø
q t q
s s
t in 2 os 2
2
ö
ç çè
= -

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Principal Planes & Principal Stresses
Principal
Plan~ earse the two planes where the normal stress (s) is
the maximum or minimum
~ there are no shear stresses on principal planes
~ these two planes are mutually perpendicular
~ the orientations of the planes (qp) are given by:
ö
÷ ÷ø
æ
ç çè
t
xy
1 1
p s -
s
= -
x y
q
2
tan
2
gives two values (qp1 and qp2)

10. 10
Copyright© 2001
Principal Planes & Principal
Stresses
x
qp1
Orientation of Principal
Planes
qp2
90°

11. æ -
s s
ö
R x y t
+ ÷ ÷ø
11
Copyright© 2001
Principal Planes & Principal Stresses
Principal
Stres~s aeres the normal stresses (s) acting on the principal planes
ö
æ +
s s
R y x + ÷ ÷ø
ç çè
s s
= =
max 1 2
ö
æ +
s s
R y x - ÷ ÷ø
ç çè
s s
= =
min 2 2
2
2
2 xy
ç çè
=

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Maximum Shear (tmax)
~ maximum shear stress occurs on two mutually
perpendicular planes
~ orientations of the two planes (qs) are given by:
ö
÷ ÷ø
æ -
ç çè
1 1
= - -
s s
x y
s t
xy
q
2
tan
2
gives two values (qs1 and qs2)
tmax = R
2
2
æ -
s s
ö
R x y t
+ ÷ ÷ø
2 xy
ç çè
=

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Maximum Shear
x
Orientation of Maximum Shear Planes
qs1
qs2
90°

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Copyright© 2001
Principal Planes & Maximum Shear
Planes
x
Principal plane
Maximum shear plane
qp = qs ±
45°
45°

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Copyright© 2001
Mohr Circles
From the stress-transformation equations (slide 7),
2 2
2
æ +
- x y
ö
t
' R xy
x 2
úû
+ '
= ù
é
êë
÷ ÷ø
ç çè
s s
s
Equation of a circle, with
variables being sx’ and txy’

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Mohr Circles
sx’
txy’
(sx + sy)/2
R

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Mohr Circles
A point on the Mohr circle represents the sx’
and txy’ values on a specific plane.
 q 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
sx’
txy’
q = 90°
q = 0
q
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
sx’
txy’
q
…..when we rotate the
plane by q°, we go 2q°
on the Mohr circle.
2q

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Mohr Circles
sx’
txy’
s2
s1
tmax

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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
QQuuiitt CCoonnttiinnuuee

22. 200 kPa
60 kPa
22
Copyright© 2001
A Mohr Circle Problem
A 40 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. 200 kPa
60 kPa
A 40 kPa
s (kPa)
Drawing Mohr Circle
t (kPa)
R = 100
120

24. s (kPa)
Principal Stresses
t (kPa)
s1= 220
s2= 20
R = 100

25. s (kPa)
t (kPa)
tmax = 100
Maximum Shear Stresses

26. 200 kPa
60 kPa
A 40 kPa
s (kPa)
Positions of x & y Planes
on Mohr Circle
t (kPa)
R = 100
120
60
40
60
q
tan q = 60/80
q = 36.87°

27. 200 kPa
60 kPa
s (kPa)
Orientations of Principal Planes
t (kPa)
A 40 kPa
36.9°
18.4°
major principal
plane
minor principal
plane
71.6°

28. Orientations of Max. Shear
Stress Planes
200 kPa
60 kPa
s (kPa)
t (kPa)
A 40 kPa
53.1°
26.6°
36.9°
116.6°

29. 29
Copyright© 2001
Testing Times…
Do you want to try a mini quiz?
YYEESS Oh, NO!

30. Question 1:
90 kPa
The state of stress at a point A 40 kPa
is shown.
What would be the maximum
shear stress at this point?
A 30 kPa
Answer 1: 50 kPa
Press RETURN for the answer Press RETURN to continue

31. Question 2:
90 kPa
40 kPa At A, what would be the
principal stresses?
A 30 kPa
Answer 2: 10 kPa, 110 kPa
Press RETURN for the answer Press RETURN to continue

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

33. Question 4:
90 kPa
The state of stress at a point B 0 kPa
is shown.
What would be the maximum
shear stress at this point?
B 90 kPa
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

34. N. Sivakugan
Copyright© 2001
Plane Strain Transformation

35. 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)
ez = 0; gxz = 0; gzy = 0

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Copyright© 2001
Plane Strain Transformation
Similar to previous derivations. Just replace
s by e, and
t by g/2

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Copyright© 2001
Plane Strain Transformation
Sign Convention:
Normal strains (ex and ey): extension positive
Shear strain (g ): decreasing angle positive
e.g.,
x
y
before
x
y
after
ex positive
ey negative
g positive

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Copyright© 2001
Plane Strain Transformation
q
g
æ -
+ ÷ ÷ø
æ +
=
' x y x y q
x ÷ø
xy
÷ + e e e e
e sin 2
2
cos 2
2 2
ö
ç çè
ö
ç çè
q
g
æ -
xy x y s q
+ xy c ÷ø
÷ g e e
os 2
2
in 2
'
2 2
ö
ç çè
= -
Same format as the stress
transformation equations

39. ~ maximum (e1) and minimum (e2) principal strains
~ occur along two mutually perpendicular directions, given by:
R = x y xy
39
Copyright© 2001
Principal Strains
ö
÷ ÷ø
æ
ç çè
g
xy
1
p e -
e
= -
x y
q tan 1
2
Gives two values (qp1 and qp2)
ö
ç çè æ +
e e
R y x + ÷ ÷ø
e
=
1 2
ö
æ +
=
1 2
R y x - ÷ ÷ø
ç çè
e e
e
2 2
ö
+ æ ÷ ÷ø
æ -
2 2 ÷ ÷ø
ç çè
ö
ç çè
e e g

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Copyright© 2001
Maximum Shear Strain (gmax)
gmax/2 = R
2 2
ö
+ æ ÷ ÷ø
æ -
R = x y xy
2 2 ÷ ÷ø
ç çè
ö
ç çè
e e g
qp = qs ±
45°

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Copyright© 2001
Mohr Circles
(ex + ey)/2
R ex’
gxy’
2

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Copyright© 2001
Strain Gauge
~ measures normal strain (e), from the change in
electrical resistance during deformation
electrical resistance
strain gauge

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Copyright© 2001
Strain Rosettes
~ measure normal strain (e) in three directions; use
these to find ex, ey, and gxy
e.g., 45° Strain Rosette
x
e90
45° e0
45°
e45
ex = e0
measured
ey = e90
gxy = 2 e45 – (e0 + e90)