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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