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Content. Stress Transformation A Mini Quiz Strain Transformation. Click here. Click here. Click here. Approximate Duration: 20 minutes. Plane Stress Transformation. y. x. Plane Stress Loading.

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Content
Content

  • Stress Transformation

  • AMini Quiz

  • Strain Transformation

Click here

Click here

Click here

Approximate Duration: 20 minutes



Plane stress loading

y

x

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 (z-direction)

z = 0; xz = 0; zy = 0


Plane stress loading1

y

xy

x

A

A

y

x

Plane Stress Loading

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

x; y; and xy


Objective

y

xy

x

A

A

y

x

Objective

State of Stress at A

If x, y, and xy are known, …


Objective1

’y

’xy

’x

A

y

y’

A

x’

x

Objective

State of Stress at A

…what would be ’x, ’y, and ’xy?


Transformation

y

xy

xy

’xy=?

’x=?

x

y

y’

x’

x

Transformation

A

State of Stress at A


Transformation1
Transformation

Solving equilibrium equations for the wedge…


Principal planes principal stresses

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:


Principal planes principal stresses1

p2

p1

x

90

Principal Planes & Principal Stresses

Orientation of Principal Planes


Principal planes principal stresses2
Principal Planes & Principal Stresses

Principal Stresses

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


Maximum shear max

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


Maximum shear

s2

s1

x

90

Maximum Shear

Orientation of Maximum Shear Planes


Principal planes maximum shear planes

45

Principal plane

x

Maximum shear plane

Principal Planes & Maximum Shear Planes

p = s± 45


Mohr circles

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

Mohr Circles

From the stress-transformation equations (slide 7),


Mohr circles1

(x + y)/2

R

x’

xy’

Mohr Circles


Mohr circles2
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 circles3

= 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….


Mohr circles4

2

x’

xy’

Mohr Circles

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


Mohr circles5

x’

2

max

1

xy’

Mohr Circles


From the three musketeers
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
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?

Answer 1: 50 kPa

Press RETURN for the answer

Press RETURN to continue


Question 2:

90 kPa

40 kPa

A

30 kPa

At A, what would be the principal stresses?

Answer 2:

10 kPa, 110 kPa

Press RETURN to continue

Press RETURN for the answer


Question 3:

90 kPa

40 kPa

A

30 kPa

At A, will there be any compressive stresses?

Answer 3:

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

Press RETURN to continue

Press RETURN for the answer


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?

Answer 4:

0

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

Press RETURN to continue

Press RETURN for the answer



Plane strain loading

y

x

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 (z-direction)

z = 0; xz = 0; zy = 0


Plane strain transformation1
Plane Strain Transformation

Similar to previous derivations. Just replace

 by , and

 by /2


Plane strain transformation2

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 transformation3
Plane Strain Transformation

Same format as the stress transformation equations


Principal strains
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
Maximum Shear Strain (max)

max/2 = R

p = s± 45


Mohr circles6

(x + y)/2

R

x’

xy’

2

Mohr Circles


Strain gauge

electrical resistance strain gauge

Strain Gauge

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


Strain rosettes

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)



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