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Presentation Transcript
Content
• Stress Transformation
• AMini Quiz
• Strain Transformation

Approximate Duration: 20 minutes

y

x

~ 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

y

xy

x

A

A

y

x

Therefore, the state of stress at a point can be defined by the three independent stresses:

x; y; and xy

y

xy

x

A

A

y

x

Objective

State of Stress at A

If x, y, and xy are known, …

’y

’xy

’x

A

y

y’

A

x’

x

Objective

State of Stress at A

…what would be ’x, ’y, and ’xy?

y

xy

xy

’xy=?

’x=?

x

y

y’

x’

x

Transformation

A

State of Stress at A

Transformation

Solving equilibrium equations for the wedge…

gives two values (p1 and p2)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:

p2

p1

x

90

Principal Planes & Principal Stresses

Orientation of Principal Planes

Principal Planes & Principal Stresses

Principal Stresses

~ are the normal stresses () acting on the principal planes

gives two values (s1 and s2)Maximum Shear (max)

~ maximum shear stress occurs on two mutually perpendicular planes

~ orientations of the two planes (s) are given by:

max = R

s2

s1

x

90

Maximum Shear

Orientation of Maximum Shear Planes

45

Principal plane

x

Maximum shear plane

Principal Planes & Maximum Shear Planes

p = s± 45

Equation of a circle, with variables being x’ and xy’Mohr Circles

From the stress-transformation equations (slide 7),

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 ….
 = 0

x’

 = 90

xy’

Mohr Circles

When we rotate the plane by 180°, we go a full round (i.e., 360°, on the Mohr circle. Therefore….

2

x’

xy’

Mohr Circles

…..when we rotate the plane by °, we go 2° on the Mohr circle.

x’

2

max

1

xy’

Mohr Circles
From the three Musketeers

Mohr circle represents the state of stress at a point; thus different Mohr circles for different points in the body

Mohr circle is a simple but powerful technique

Get the sign convention right

Quit

Continue

200 kPa

60 kPa

A

40 kPa

A Mohr Circle Problem

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.
200 kPa

60 kPa

A

40 kPa

120

 (kPa)

R = 100

 (kPa)

Drawing Mohr Circle

1= 220

 (kPa)

2= 20

R = 100

 (kPa)

Principal Stresses

 (kPa)

max = 100

 (kPa)

Maximum Shear Stresses

200 kPa

60 kPa

A

40 kPa

R = 100

60

120

40

 (kPa)

60

 (kPa)

Positions of x & y Planeson Mohr Circle

tan  = 60/80

 = 36.87°

200 kPa

60 kPa

A

40 kPa

71.6°

 (kPa)

36.9°

major principal plane

18.4°

 (kPa)

Orientations of Principal Planes

minor principal plane

200 kPa

26.6°

60 kPa

A

40 kPa

53.1°

 (kPa)

36.9°

 (kPa)

116.6°

Orientations of Max. Shear Stress Planes

Testing Times…

Do you want to try a mini quiz?

YES

Oh, NO!

Question 1:

90 kPa

40 kPa

A

30 kPa

The state of stress at a point A is shown.

What would be the maximum shear stress at this point?

Question 2:

90 kPa

40 kPa

A

30 kPa

At A, what would be the principal stresses?

10 kPa, 110 kPa

Question 3:

90 kPa

40 kPa

A

30 kPa

At A, will there be any compressive stresses?

No. The minimum normal stress is 10 kPa (tensile).

Question 4:

90 kPa

0 kPa

B

90 kPa

The state of stress at a point B is shown.

What would be the maximum shear stress at this point?

0

This is hydrostatic state of stress (same in all directions). No shear stresses.

y

x

~ 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

Plane Strain Transformation

Similar to previous derivations. Just replace

 by , and

 by /2

y

y

x

x

before

after

Plane Strain Transformation

Sign Convention:

Normal strains (x andy): extension positive

Shear strain ( ): decreasing angle positive

e.g.,

x positive

y negative

 positive

Plane Strain Transformation

Same format as the stress transformation equations

Principal Strains

~ maximum (1) and minimum (2) principal strains

~ occur along two mutually perpendicular directions, given by:

Gives two values (p1 and p2)

Maximum Shear Strain (max)

max/2 = R

p = s± 45

electrical resistance strain gaugeStrain Gauge

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

90

measured

45

45°

0

45°

x

Strain Rosettes

~ measure normal strain () in three directions; use these to find x, y, and xy

e.g., 45° Strain Rosette

x = 0

y = 90

xy = 2 45– (0+ 90)