Mohr circle (Complete Soil Mech. Undestanding Pakage: ABHAY)

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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)

σz = 0; τxz = 0; τzy = 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:

σx; σy; and τxy

σx

σy τxy

A

A

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Objective

A

x

y

σx

σy τxy

A

State of Stress at A

If σx, σy, and τxy are known, …

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Objective

A

x

y

σ’x

σ’y τ’xy

A


…what would be σ’x, σ’y, and τ’xy?

x’y’

θ

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Transformation

x

yx’

y’θ

A


θ σx

σy τxyτxy

σ’x=?τ’xy=?

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Transformation

θτθσσσσ

σ 2 sin2 cos22

' xyyxyx

x +

−+

+=

θτθσσ

τ 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 (σ) is

the maximum or minimum

~ the orientations of the planes (θp) are given by:

−

= −

yx

xyp σσ

τθ

2tan

2

1 1

gives two values (θp1 and θp2)

~ there are no shear stresses on principal planes

~ these two planes are mutually perpendicular

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xθp1

θp2

90°

Orientation of Principal Planes

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Principal Stresses

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

Ryx +

+==

21max

σσσσ

Ryx −

+

==22min

σσσσ

2

2

2 xyyxR τ

σσ+

−=

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Maximum Shear (τmax)~ maximum shear stress occurs on two mutually perpendicular planes

−−= −

xy

yxs τ

σσθ

2tan

2

1 1

gives two values (θs1 and θs2)

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

τmax = R2

2

2 xyyxR τ

σσ+

−=

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Maximum Shear

xθs1

θs2

90°

Orientation of Maximum Shear Planes

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Principal Planes & Maximum Shear Planes

x

Principal plane

Maximum shear plane

θp = θs ± 45 °

45°

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Mohr CirclesFrom the stress-transformation equations (slide 7),

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2

'2

' Rxyyx

x =+

+− τ

σσσ

Equation of a circle, with variables being σx’ and τxy’

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Mohr Circles

σx’

τxy’

(σx + σy)/2

R

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Mohr Circles

A point on the Mohr circle represents the σ x’ and τ xy’ values on a specific plane.

θ 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

σx’

τxy’

θ = 90°

θ = 0

θ

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

σx’

τxy’

…..when we rotate the plane by θ°, we go 2θ°

on the Mohr circle.

θ2θ

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Mohr Circles

σx’

τxy’

σ2

σ1

τmax

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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

σ (kPa)

τ (kPa)

R = 100

Drawing Mohr Circle

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σ (kPa)

τ (kPa)

σ1= 220

σ2= 20

Principal Stresses

R = 100

σ (kPa)

τ (kPa)

τmax = 100

Maximum Shear Stresses

A40 kPa

200 kPa

60 kPa

σ (kPa)

τ (kPa)

R = 100

120

Positions of x & y Planeson Mohr Circle

60

40

60

θ

tan θ = 60/80

θ = 36.87°

σ (kPa)

τ (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

σ (kPa)

τ (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


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).


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.


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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)

εz = 0; γxz = 0; γzy = 0

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Plane Strain TransformationSign Convention:

Shear strain (γ ): decreasing angle positive

e.g.,

Normal strains (εx and εy): extension positive

x

y

beforex

y

after

εx positiveεy negative

γ positive

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Principal Strains

−

= −

yx

xyp εε

γθ 1tan

2

1Gives two values (θp1 and θp2)

~ maximum (ε1) and minimum (ε2) principal strains

~ occur along two mutually perpendicular directions, given by:

Ryx +

+=

21

εεε

Ryx −

+=

21

εεε

22

22

+

−= xyyxR

γεε

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Strain Rosettes~ measure normal strain (ε) in three directions; use

these to find εx, εy, and γxy

e.g., 45° Strain Rosette

x

45°

45°

ε0

ε90

ε45

εx = ε0

εy = ε90

γxy = 2 ε45 – (ε0 + ε90)

measured

Mohr circle (Complete Soil Mech. Undestanding Pakage: ABHAY)

Engineering

Transcript of Mohr circle (Complete Soil Mech. Undestanding Pakage: ABHAY)