mohr’s circle: plane stress

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• Equations for plane stress transformation have a graphical solution that is easy to remember and use.

• This approach will help you to “visualize” how the normal and shear stress components vary as the plane acted on is oriented in different directions.

MOHR’S CIRCLE: PLANE STRESS

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• Eqns discussed earlier are rewritten as

• Parameter can be eliminated by squaring each eqn and adding them together.


22cos2sin2

''

xy

yx

yx

' cos2 sin 2 12 2

x y x y

x xy

xyyx

yxyx

x2

2

''2

2

'22

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3

• If x, y, xy are known constants, thus we compact the Eqn as,


42

2

where

3

2

2

2''

22

'

xyyx

yx

avg

yxavgx

R

R

4

• Establish coordinate axes; positive to the right and positive downward, Eqn 3 represents a circle having radius R and center on the axis at pt C (avg, 0). This is called the Mohr’s Circle.


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• To draw the Mohr’s circle, we must establish the and axes.

• Center of circle C (avg, 0) is plotted from the known stress components (x, y, xy).

• We need to know at least one pt on the circle to get the radius of circle.


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Case 1 (x’ axis coincident with x axis)1. = 02. x’ = x

3. x’y’ = xy.

• Consider this as reference pt A, and plot its coordinates A (x, xy).

• Apply Pythagoras theorem to shaded triangle to determine radius R.

• Using pts C and A, the circle can now be drawn.


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Case 2 (x’ axis rotated 90 counterclockwise)

1. = 90

2. x’ = y

3. x’y’ = xy.

• Its coordinates are G (y, xy).

• Hence radial line CGis 180counterclockwise from “reference line” CA.


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Procedure for Analysis

Construction of the circle

1. Establish coordinate system where abscissa represents the normal stress , (+ve to the right), and the ordinate represents shear stress , (-ve downward).

2. Use positive sign convention for x, y, xy, plot the center of the circle C, located on the axis at a distance avg = (x + y)/2 from the origin.


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3. Plot reference pt A (x, xy). This pt represents the normal and shear stress components on the element’s right-hand vertical face. Since x’ axis coincides with x axis, = 0.


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4. Connect pt A with center C of the circle and determine CA by trigonometry. The distance represents the radius R of the circle.

5. Once R has been determined, sketch the circle.


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Principal stress

• Principal stresses 1 and 2 (1 2) are represented by two pts B and Dwhere the circle intersects the -axis.


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Principal stress

• These stresses act on planes defined by angles p1 and p2. They are represented on the circle by angles 2p1 and 2p2

and measured from radial reference line CA to lines CBand CD respectively.


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Principal stress

• Using trigonometry, only one of these angles needs to be calculated from the circle, since p1 and p2 are 90 apart. Remember that direction of rotation 2p on the circle represents the same direction of rotation p from reference axis (+x) to principal plane (+x’).


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Maximum in-plane shear stress

• The average normal stress and maximum in-plane shear stress components are determined from the circle as the coordinates of either pt Eor F.


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• The angles s1 and s2 give the orientation of the planes that contain these components. The angle 2s

can be determined using trigonometry. Here rotation is clockwise, and so s1 must be clockwise on the element.


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Stresses on arbitrary plane

• Normal and shear stress components x’ and x’y’

acting on a specified plane defined by the angle , can be obtained from the circle by using trigonometry to determine the coordinates of pt P.


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Stresses on arbitrary plane

• To locate pt P, known angle for the plane (in this case counterclockwise) must be measured on the circle in the same direction 2(counterclockwise), from the radial reference line CA to the radial line CP.


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EXAMPLE 1Due to applied loading, element at pt A on solid cylinder as shown is subjected to the state of stress. Determine the principal stresses acting at this pt.

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EXAMPLE 1 (SOLN)Construction of the circle

• Center of the circle is at

• Initial pt A (12, 6) and the center C (6, 0) are plotted as shown. The circle having a radius of

MPa60MPa12 xyyx

MPa62

012

avg

MPa49.86612 22 R

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EXAMPLE 1 (SOLN)Principal stresses

• Principal stresses indicated at pts B and D. For 1 > 2,

• Obtain orientation of element by calculating counterclockwise angle 2p2, which defines the direction of p2 and 2 and its associated principal plane.

MPa

MPa

5.1449.86

49.2649.8

2

1

5.22

0.45612

6tan2 1

2

2

p

p

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• The element is orientated such that x’ axis or 2

is directed 22.5 counterclockwise from the horizontal x-axis.

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EXAMPLE 2State of plane stress at a pt is shown on the element. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act.

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EXAMPLE 2 (SOLN)Construction of circle

• Establish the , axes as shown below. Center of circle C located on the -axis, at the pt:

MPa60MPa90MPa20 xyyx

MPa352

9020

avg

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• Pt C and reference pt A (20, - 60) are plotted. Apply Pythagoras theorem to shaded triangle to get circle’s radius CA,

MPa4.81

5560 22

R

R

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EXAMPLE 2 (SOLN)


• Maximum in-plane shear stress and average normal stress are identified by pt E or Fon the circle. In particular, coordinates of pt E (35, 81.4) gives

MPa

MPaplane-in

max

35

4.81

avg

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EXAMPLE 2 (SOLN)


• Counterclockwise angle s1 can be found from the circle, identified as 2s1.

3.21

5.4260

3520tan2

1

11

s

s

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EXAMPLE 2 (SOLN)


• This counterclockwise angle defines the direction of the x’ axis. Since pt E has positive coordinates, then the average normal stress and maximum in-plane shear stress both act in the positive x’ and y’ directions as shown.

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EXAMPLE 3State of plane stress at a pt is shown on the element. Represent this state of stress on an element oriented 30 counterclockwise from position shown.

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• Establish the , axes as shown. Center of circle Clocated on the -axis, at the pt:

MPa6MPa12MPa8 xyyx

MPa22

128

avg

30


• Initial pt for = 0 has coordinates A (8, 6) are plotted. Apply Pythagoras theorem to shaded triangle to get circle’s radius CA,

MPa66.11

610 22

R

R

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EXAMPLE 3 (SOLN)Stresses on 30 element

• Since element is rotated 30 counterclockwise, we must construct a radial line CP, 2(30) = 60counterclockwise, measured from CA ( = 0).

• Coordinates of pt P (x’, x’y’) must be obtained. From geometry of circle,

04.2996.3060

96.3010

6tan 1

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EXAMPLE 3 (SOLN)Stresses on 30 element

• The two stress components act on face BD of element shown, since the x’ axis for this face if oriented 30counterclockwise from the x-axis.

• Stress components acting on adjacent face DE of element, which is 60 clockwise from +x-axis, are represented by the coordinates of pt Q on the circle.

• This pt lies on the radial line CQ, which is 180 from CP.

MPa

MPa

66.504.29sin66.11

20.804.29cos66.112

''

'

yx

x

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EXAMPLE 3 (SOLN)

Stresses on 30 element

• The coordinates of pt Q are

• Note that here x’y’ acts in the y’ direction.

)(Check!MPa

MPa

66.504.29sin66.11

2.1204.29cos66.112

''

'

yx

x

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• Occasionally, circular shafts are subjected to combined effects of both an axial load and torsion.

• Provided materials remain linear elastic, and subjected to small deformations, we use principle of superposition to obtain resultant stress in shaft due to both loadings.

• Principal stress can be determined using either stress transformation equations or Mohr’s circle.

STRESS IN SHAFTS DUE TO AXIAL LOAD AND TORSION

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EXAMPLE 4Axial force of 900 N and torque of 2.50 Nm are applied to shaft. If shaft has a diameter of 40 mm, determine the principal stresses at a pt P on its surface.

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EXAMPLE 4 (SOLN)Internal loadings

• Consist of torque of 2.50 Nm and axial load of 900 N.

Stress components

• Stresses produced at pt P are therefore

kPam

mmN9.198

02.02

02.050.24

J

Tc

kPa

m

N2.716

02.0

9004

A

P

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• Using Mohr’s circle, center of circle Cat the pt is

• Plotting C (358.1, 0) and reference pt A (0, 198.9), the radius found was R = 409.7 kPA. Principal stresses represented by pts B and D.

kPaavg 1.3582

2.7160

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• Clockwise angle 2p2 can be determined from the circle. It is 2p2 = 29.1. The element is oriented such that the x’ axis or 2 is directed clockwise p1 = 14.5 with the x axis as shown.

kPa

kPa

2

1

6.517.4091.358

8.7677.4091.358

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• The shear and flexure formulas are applied to a cantilevered beam that has a rectangular x-section and supports a load P at its end.

• At arbitrary section a-a along beam’s axis, internal shear Vand moment M are developed from a parabolic shear-stress distribution, and a linear normal-stress distribution.

STRESS VARIATIONS THROUGHOUT A PRISMATIC BEAM

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• The stresses acting on elements at pts 1 through 5 along the section.

• In each case, the state of stress can be transformed into principal stresses, using either stress-transformation equations or Mohr’s circle.

• Maximum tensile stress acting on vertical faces of element 1 becomes smaller on corresponding faces of successive elements, until it’s zero on element 5.

9.6 STRESS VARIATIONS THROUGHOUT A PRISMATIC BEAM

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• Similarly, maximum compressive stress of vertical faces of element 5 reduces to zero on that of element 1.

• By extending this analysis to many vertical sections along the beam, a profile of the results can be represented by curves called stress trajectories.

• Each curve indicate the direction of a principal stress having a constant magnitude.

9.6 STRESS VARIATIONS THROUGHOUT A PRISMATIC BEAM

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EXAMPLE 5Beam is subjected to the distributed loading of = 120kN/m. Determine the principal stresses in the beam at pt P, which lies at the top of the web. Neglect the size of the fillets and stress concentrations at this pt. I = 67.1(10-6) m4.

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EXAMPLE 5 (SOLN)

Internal loadings

• Support reaction on the beam B is determined, and equilibrium of sectioned beam yields

Stress components

• At pt P,

mkNkN 6.3084 MV

MPa

m

m

4.45

1074.6

100.0106.3046

3

mN

I

My

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EXAMPLE 5 (SOLN)

Stress components

• At pt P,

Principal stresses

• Using Mohr’s circle, the principal stresses at P can be determined.

MPa

mm

mmmN

2.35

010.01074.6

015.0175.01075.0108446

3

It

VQ

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45


• As shown, the center of the circle is at (45.4 + 0)/2 = 22.7, and pt A (45.4, 35.2). We find that radius R = 41.9, therefore

• The counterclockwise angle 2p2 = 57.2, so that

MPa

MPa

2

1

6.649.417.22

2.197.229.41

6.282p

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• A pt in a body subjected to a general 3-D state of stress will have a normal stress and 2 shear-stress components acting on each of its faces.

• We can develop stress-transformation equations to determine the normal and shear stress components acting on ANY skewed plane of the element.

ABSOLUTE MAXIMUM SHEAR STRESS

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• These principal stresses are assumed to have maximum, intermediate and minimum intensity: max int min.

• Assume that orientation of the element and principal stress are known, thus we have a condition known as triaxialstress.


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• Viewing the element in 2D (y’-z’, x’-z’,x’-y’) we then use Mohr’s circle to determine the maximum in-plane shear stress for each case.

9.7 ABSOLUTE MAXIMUM SHEAR STRESS

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• As shown, the element have a45 orientation and is subjected to maximum in-plane shear and average normal stress components.


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• Comparing the 3 circles, we see that the absolute maximum shear stress is defined by the circle having the largest radius.

• This condition can also be determined directly by choosing the maximum and minimum principal stresses:

max

abs

52

minmax

maxabs

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• Associated average normal stress

• We can show that regardless of the orientation of the plane, specific values of shear stress on the plane is always less than absolute maximum shear stress found from Eqn 5.

• The normal stress acting on any plane will have a value lying between maximum and minimum principal stresses, max min.


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minmaxavg

52

Plane stress

• Consider a material subjected to plane stress such that the in-plane principal stresses are represented as max

and int, in the x’ and y’ directions respectively; while the out-of-plane principal stress in the z’ direction is min = 0.

• By Mohr’s circle and Eqn. 5,


72max

max''maxabs

zx

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Plane stress

• If one of the principal stresses has an opposite sign of the other, then these stresses are represented as max and min, and out-of-plane principal stress int = 0.

• By Mohr’s circle and Eqn. 5,


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minmax

max''maxabs

yx

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ABSOLUTE MAXIMUM SHEAR STRESSIMPORTANT• The general 3-D state of stress at a pt can be

represented by an element oriented so that only three principal stresses act on it.

• From this orientation, orientation of element representing the absolute maximum shear stress can be obtained by rotating element 45 about the axis defining the direction of int.

• If in-plane principal stresses both have the same sign, the absolute maximum shear stress occurs out of the plane, and has a value of

2max max

abs

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ABSOLUTE MAXIMUM SHEAR STRESSIMPORTANT

• If in-plane principal stresses are of opposite signs, the absolute maximum shear stress equals the maximum in-plane shear stress; that is

2minmax max

abs

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EXAMPLE 6Due to applied loading, element at the pt on the frame is subjected to the state of plane stress shown.

Determine the principal stresses and absolute maximum shear stress at the pt.

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The in-plane principal stresses can be determined from Mohr’s circle. Center of circle is on the axis at avg = (20 + 20)/2 = 10 kPa. Plotting controlling ptA (20, 40), circle can be drawn as shown. The radius is

kPa2.41

401020 22

R

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EXAMPLE 6 (SOLN)

Principal stresses

The principal stresses at the pt where the circle intersects the -axis:

From the circle, counterclockwise angle 2, measured from the CA to the axis is,

kPa

kPa

2.512.4110

2.312.4110

min

max

0.38

0.761020

40tan2 1

Thus,

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EXAMPLE 6 (SOLN)

Principal stresses

This counterclockwise rotation defines the direction of the x’ axis or min and its associated principal plane. Since there is no principal stress on the element in the z direction, we have

kPa

kPa

2.51

0

2.31

min

int

max

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EXAMPLE 6 (SOLN)Absolute maximum shear stress

Applying Eqns. 5 and 6,

kPa

avg

102

2.512.31

2minmax

kPa

maxabs

2.412

)2.512.31

2minmax

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EXAMPLE 6 (SOLN)Absolute maximum shear stress

These same results can be obtained by drawing Mohr’s circle for each orientation of an element about the x’, y’, and z’ axes. Since max and min are of opposite signs, then the absolute maximum shear stress equals the maximum in-plane shear stress. This results from a 45rotation of the element about the z’axis, so that the properly oriented element is shown.

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EXAMPLE 7The pt on the surface of the cylindrical pressure vessel is subjected to the state of plane stress. Determine the absolute maximum shear stress at this pt.

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EXAMPLE 7 (SOLN)Principal stresses are max = 32 MPa, int = 16 MPa, and min = 0. If these stresses are plotted along the axis, the 3 Mohr’s circles can be constructed that describe the stress state viewed in each of the three perpendicular planes.

The largest circle has a radius of 16 MPa and describes the state of stress in the plane containing max = 32 MPa and min = 0.

An orientation of an element 45 within this plane yields the state of absolute maximum shear stress and the associated average normal stress, namely,

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EXAMPLE 7 (SOLN)

An orientation of an element 45 within this plane yields the state of absolute maximum shear stress and the associated average normal stress, namely,

Or we can apply Eqns 5 and 6:

MPaMPa avgmaxabs 1616 σ

MPa

MPa

avg

maxabs

162

032

2

162

032

2

minmax

minmax

σ

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EXAMPLE 7 (SOLN)

By comparison, maximum in-plane shear stress can be determined from the Mohr’s circle drawn between max = 32 MPa and int = 16 MPa, which gives a value of

MPa

MPa

avg

maxabs

242

163216

82

1632

σ

mohr’s circle: plane stress

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