Chapter Objectives To develop the transformation of stress components from one orientation of the coordinate system to another orientation. To determine

Chapter Objectives

To develop the transformation of stress components from one orientation of the coordinate system to another orientation.

To determine the principal stress and maximum in-plane shear stress at a point.

To develop the transformation of plane strain components from one orientation of the coordinate system to another orientation.

Copyright © 2011 Pearson Education South Asia Pte Ltd

Chapter Objectives (cont)

To determine the principal strain and maximum in-plane shear strain at a point.

To develop Mohr’s circle for analyzing stress and strain components.

To discuss strain rosettes for strain components. To present the relationships between material

properties, such as the elastic modulus, shear modulus, and Poisson’s ratio.


1. Reading Quiz

2. Applications

3. General equations of plane-stress transformation

4. Principal stresses and maximum in-plane shear stress

5. Mohr’s circle for plane stress

6. Absolute maximum shear stress

In-class Activities


7. Equations of plane-strain transformation

8. Principal and maximum in-plane shear strain

9. Mohr’s circle for plane strain

10. Measurement of strains

11. Stress-strain relationship

12. Concept Quiz

READING QUIZ

1) Which of one the following statements is incorrect?

a) The principal stresses represent the maximum and minimum normal stress at the point

b) When the state of stress is represented by the principal stresses, no shear stress will act on the element

c) When the state of stress is represented in terms of the maximum in-plane shear stress, no normal stress will act on the element

d) For the state of stress at a point, the maximum in-plane shear stress usually associated with the average normal stress.


READING QUIZ

2) Which one of the following statements is incorrect for plane-strain?

a) σz = γyz = γxz = 0 while the plane-strain has 3 components σx, σy and γxy.

b) Always identical to the state of plane stress

c) Identical to the state of plane stress only when the Poisson’s ratio is zero.

d) When the state of strain is represented by the principal strains, no shear strain will act on the element.


APPLICATIONS




APPLICATIONS


APPLICATIONS (cont)


APPLICATIONS (cont)


APPLICATIONS (cont)


GENERAL EQUATIONS OF PLANE-STRESS TRANSFORMATION


• The state of plane stress at a point is uniquely represented by three components acting on an element that has a specific orientation at the point.

• Sign Convention:– Positive normal stress acts outward

from all faces– Positive shear stress acts upwards

on the right-hand face of the element


GENERAL EQUATIONS OF PLANE-STRESS TRANSFORMATION (cont)



• Sign Convention (continued)

– Both the x-y and x’-y’ system follow the right-hand rule

– The orientation of an inclined plane (on which the normal and shear stress components are to be determined) will be defined using the angle θ. The angle θ is measured from the positive x to the positive x’-axis. It is positive if it follows the curl of the right-hand fingers.




• Normal and shear stress components:– Consider the free-body diagram of the segment




+ΣFx’ = 0; σx’ ∆A – (τxy ∆A sin θ) cos θ – (σy ∆A sin θ) sin θ – ( τxy ∆A cos θ) sin θ – (σx ∆A cos θ) cos θ = 0 σx’ = σx cos2 θ + σy sin2 θ + τxy (2 sin θ cos θ)

+ΣFy’ = 0; τx’y’ ∆A + (τxy ∆A sin θ) sin θ – (σy ∆A sin θ) cos θ – ( τxy ∆A cos θ) cos θ + (σx ∆A cos θ) sin θ = 0 τx’y’ = (σy – σx) sin θ cos θ + τxy (cos2 θ – sin2 θ)

σx’ = σx + σy 2

σx – σy 2 cos 2θ + τxy sin 2 θ +

τx’y’ = – σx + σy

2 sin 2θ + τxy cos 2 θ

σy’ = σx + σy 2

σx – σy 2 cos 2θ – τxy sin 2 θ –

VARIABLE SOLUTIONS


Please click the appropriate icon for your computer to access the variable solutions

EXAMPLE 1


The state of plane stress at a point on the surface of the airplane fuselage is represented on the element oriented as shown in Fig. 9–4a. Represent the state of stress at the point on an element that is oriented 30° clockwise from the position shown.

EXAMPLE 1 (cont)


• The free-body diagram of the segment is as shown.

• The element is sectioned by the line a-a.

Solution

EXAMPLE 1 (cont)

Applying the equations of force equilibrium in the x’ and y’ direction,

Solution

(Ans) MPa 15.4

030cos30sin2530sin30sin80

30sin30cos2530cos30cos50 ;0

'

''

x

xx

AA

AAAF

(Ans) MPa 8.68

030sin30cos2530cos30sin80

30cos30cos2530sin30cos50 ;0'

x'y'

x'y'y

AA

AAAF


EXAMPLE 1 (cont)

Repeat the procedure to obtain the stress on the perpendicular plane b–b.

Solution


(Ans) MPa 8.25

030sin30sin5030cos30cos25

30cos30cos8030sin30cos25 ;0

'

''

x

xx

AA

AAAF

(Ans) MPa 8.68

030cos30sin5030sin30sin25

30sin30cos8030cos30cos25- ;0'

x'y'

x'y'y

AA

AAAF

The state of stress at the point can be represented by choosing an element oriented.

EXAMPLE 2


The state of plane stress at a point is represented by the element shown in Fig. 9–7a. Determine the state of stress at the point on another element oriented 30° clockwise from the position shown.

EXAMPLE 2 (cont)


From the sign convention we have,

To obtain the stress components on plane CD,

Solution

30 MPa 25 MPa 50 MPa 80 xyyx

(Ans) MPa 8.682cos2sin2

(Ans) MPa 8.252sin2cos22

''

'

xyyx

yx

xyyxyx

x

EXAMPLE 2 (cont)


To obtain the stress components on plane BC,

The results are shown on the element as shown.

Solution

60 MPa 25 MPa 50 MPa 80 xyyx

(Ans) MPa 8.682cos2sin2

(Ans) MPa 15.42sin2cos22

''

'

xyyx

yx

xyyxyx

x

IN-PLANE PRINCIPAL STRESS


• The principal stresses represent the maximum and minimum normal stress at the point.

• When the state of stress is represented by the principal stresses, no shear stress will act on the element.

Solving this equation leads to θ = θp

2

2

2,1 22

2/2tan

xyyxyx

yx

xyp

2cos22sin22

'xy

yxx

d

d

IN-PLANE PRINCIPAL STRESS (cont)


2cos22sin22

'xy

yxx

d

d

IN-PLANE PRINCIPAL STRESS (cont)


Solving this equation leads to θ = θp; i.e 2/2tan

yx

xyp

2

2

2,1 22 xyyxyx

MAXIMUM IN-PLANE PRINCIPAL STRESS


• The state of stress can also be represented in terms of the maximum in-plane shear stress. In this case, an average stress will also act on the element.

• Solving this equation leads to θ = θs; i.e

• And there is a normal stress on the plane of maximum in-plane shear stress

2

2

plane-inmax 2 xyyx

02sin2cos22

''

xyyxyx

d

d

xy

yxs

2/2tan

2yx

avg

EXAMPLE 3


When the torsional loading T is applied to the bar in Fig. 9–13a, it produces a state of pure shear stress in the material. Determine (a) the maximum in-plane shear stress and the associated average normal stress, and (b) the principal stress.

EXAMPLE 3 (cont)


From the sign convention we have,

a) Maximum in-plane shear stress is

b) For principal stress,

Solution xyyx 0 0

(Ans) 02

2

2

2

plane-inmax

yx

avgxyyx

(Ans) 22

135,452/

2tan

2

2

2,1

12

xyyxyx

ppyx

xyp

EXAMPLE 3 (cont)


If we use

Thus, acts at as shown in Fig. 9–13b, and acts on the other face

Solution45

2p

90sin00

2sin2cos22' xy

yxyxx

2 452p 1

451p

EXAMPLE 4


When the axial loading P is applied to the bar in Fig. 9–14a, it produces a tensile stress in the material. Determine (a) the principal stress and (b) the maximum in-plane shear stress and associated average normal stress.

EXAMPLE 4 (cont)


From the established sign convention,

Principal Stress

Since no shear stress acts on this element,

Solution

(Ans) 0 21

0 0 0 xyyx

EXAMPLE 4 (cont)


Maximum In-Plane Shear Stress

To determine the proper orientation of the element,

Solution

2090sin

2

02cos2sin

2''

xyyx

yx

(Ans) 22

0

2

(Ans) 2

02

0

2

45,45;0

2/02/2tan

avg

22

2

2

planein max

21

yx

xxy

yx

ssxy

yxs

MOHR’S CIRCLE OF PLANE STRESS


• A geometrical representation of equations 9.1 and 9.2; i.e.

• Sign Convention:

σ is positive to the right, and τ is positive downward.

2sin2sin2

2sin2cos22

''

'

xyyx

yx

xyyxyx

x

Please refer to the website for the animation: Mohr’s Circle

EXAMPLE 5


Due to the applied loading, the element at point A on the solid shaft in Fig. 9–18a is subjected to the state of stress shown. Determine the principal stresses acting at this point.

EXAMPLE 5 (cont)


Construction of the CircleFrom Fig. 9–18a,

The center of the circle is at

The reference point A(-12,-6) and the center C(-6, 0) are plotted in Fig. 9–18b.The circle is constructed having a radius of

Solution

MPa 62

012

avg

MPa 60MPa,12 xyyx , τ σ σ

MPa 49.86612 22 R

EXAMPLE 5 (cont)


Principal Stress

The principal stresses are indicated by the coordinates of points B and D.

The orientation of the element can be determined by calculating the angle

Solution

(Ans) MPa 5.1449.86

(Ans) MPa 49.2649.8

, have We

2

1

21

5.22

0.45612

6tan2

2

2

1

p

p

EXAMPLE 6


The state of plane stress at a point is shown on the element in Fig. 9–19a. Determine the maximum in-plane shear stress at this point.

EXAMPLE 6 (cont)


Construction of the Circle

We first construct of the circle,

The center of the circle C is on the axis at

From point C and the A(-20, 60) are plotted, we have

Solution

60 and 90,20 xyyx

MPa 4.815560 22 R

MPa 352

9020

avg

EXAMPLE 6 (cont)


Maximum In-Plane Shear Stress.

Max in-plane shear stress and average normal stress are

The counter-clockwise angle is

Solution

(Ans) MPa 35 , MPa 81.4plane-inmax avg

(Ans) 3.2160

3520tan2 1

1

s

EXAMPLE 7


The state of plane stress at a point is shown on the element in Fig. 9–20a. Represent this state of stress on an element oriented 30°counterclockwise from the position shown.

EXAMPLE 7 (cont)


Construction of the Circle

We first construct of the circle,

The center of the circle C is on the axis at

From point C and the A(-8, -6) are plotted, we have

Solution

6 and 12,8 xyyx

66.11610 22 R

MPa 22

128

avg

EXAMPLE 7 (cont)


Stresses on 30° Element

From the geometry of the circle,

The stress components acting on the adjacent face DE of the element, which is 60° clockwise from the positive x axis, Fig. 9–20c, are represented by the coordinates of point Q on the circle.

Solution

(Ans) MPa 66.504.29cos66.11

(Ans) MPa 20.804.29cos66.112

04.2996.3060 96.3010

6tan

''

'

1

yx

x

(Ans) (check) MPa 66.504.29sin66.11

(Ans) MPa 22.1204.29cos66.112

y'x'

'

x

ABSOLUTE MAXIMUM SHEAR STRESS


• State of stress in 3-dimensional space:

2

2

minmaxavg

minmaxmax abs

ABSOLUTE MAXIMUM SHEAR STRESS (cont)



ABSOLUTE MAXIMUM SHEAR STRESS (cont)



EXAMPLE 8


The point on the surface of the cylindrical pressure vessel in Fig. 9–24a is subjected to the state of plane stress. Determine the absolute maximum shear stress at this point.

EXAMPLE 8 (cont)


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

Same result for can be obtained from direct application of

Solution

(Ans) MPa 16 , MPa 16 avgmax abs

(Ans) MPa 162

032

MPa 162

32

2

avg

1max abs

EXAMPLE 8 (cont)


By comparison, the maximum in-plane shear stress can be determined from the Mohr’s circle,

Solution

(Ans) MPa 242

1632

MPa 82

1632

avg

max abs

EQUATIONS OF PLANE-STRAIN TRANSFORMATION


• In 3D, the general state of strain at a point is represented by a combination of 3 components of normal strain σx, σy, σz, and 3 components of shear strain γxy, γyz, γxz.

• In plane-strain cases, σz, γxz and γyz are zero.

• The state of plane strain at a point is uniquely represented by 3 components (σx, σy and γxy) acting on an element that has a specific orientation at the point.

Please refer to the website for the animation: Strain Transformation

EQUATIONS OF PLANE-STRAIN TRANSFORMATION (cont)


Note: Plane-stress case ≠ plane-strain case



• Positive normal strain σx and σy cause elongation• Positive shear strain γxy causes small angle AOB• Both the x-y and x’-y’ system follow the right-hand rule• The orientation of an inclined plane (on which the

normal and shear strain components are to be determined) will be defined using the angle θ. The angle is measured from the positive x- to positive x’-axis. It is positive if it follows the curl of the right-hand fingers.



• Normal and shear strains– Consider the line segment dx’

sin'

cos'

dydy

dxdx

2sin2

2cos22

2sin2

2cos22

cossinsincos'

cossincos'

'

'

22'

xyyxyxy

xyyxyxx

xyxxx

xyyx

dx

x

dydydxx



• Similarly,

2sincossin

sincossin'

xyyx

xyyx dydydxdy

90cossincos

90sin90cos90sin2

2

xyyx

xyyx

y



2cos2

2sin22

sincoscossin

''

22''

xyyxyx

xyyxyx

EXAMPLE 9


A differential element of material at a point is subjected to a state of plane strain which tends to distort the element as shown in Fig. 10–5a. Determine the equivalent strains acting on an element of the material oriented at the point, clockwise 30° from the original position.

666 10200 , 10300 , 10500 xyyx

EXAMPLE 9 (cont)


• Since θ is positive counter-clockwise,

Solution

(Ans) 10213

302sin2

10200302cos10

2

30050010

2

300500

2sin2

2cos22

6'

666

'

x

xyyxyxx

(Ans) 10793

302cos2

10200302sin10

2

300500

2cos2

2sin22

6''

66

''

yx

xyyxyx

EXAMPLE 9 (cont)


• By replacement,

Solution

(Ans) 104.13

602sin2

10200602cos10

2

30050010

2

300500

2sin2

2cos22

6'

666

'

x

xyyxyxy

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN


• Similar to the deviations for principal stresses and the maximum in-plane shear stress, we have

• And,

22

2,1 222 2tan

xyyxyx

yx

xyp

2 ,

222

2tan

22

plane-inmax yxavg

xyyx

xy

yxS

PRINCIPLE AND MAXIMUM IN-PLANE SHEAR STRAIN (cont)


• When the state of strain is represented by the principal strains, no shear strain will act on the element.

• The state of strain at a point can also be represented in terms of the maximum in-plane shear strain. In this case an average normal strain will also act on the element.

• The element representing the maximum in-plane shear strain and its associated average normal strain is 45° from the element representing the principal strains.

EXAMPLE 10


A differential element of material at a point is subjected to a state of plane strain defined by which tends to distort the element as shown in Fig. 10–7a. Determine the maximum in-plane shear strain at the point and the associated orientation of the element.

666 1080 , 10200 , 10350 xyyx

EXAMPLE 10 (cont)


• Looking at the orientation of the element,

• For maximum in-plane shear strain,

Solution

311 and 9.40

80

2003502tan

s

xy

yxs

(Ans) 10556

2226

planein max

22

planein max

xyyx

MOHR’S CIRCLE FOR PLANE STRAIN


• A geometrical representation of Equations 10-5 and 10-6; i.e.,

• Sign convention: ε is positive to the right, and γ/2 is positive downwards.

22

22 and

2

where

xyyxyx

avg R

2

2

''2' 2

Ryxavgx

MOHR’S CIRCLE FOR PLANE STRAIN (cont)


EXAMPLE 11


The state of plane strain at a point is represented by the components:

Determine the principal strains and the orientation of the element.

666 10120 , 10150 , 10250 xyyx

EXAMPLE 11 (cont)


• From the coordinates of point E, we have

• To orient the element, we can determine the clockwise angle.

Solution

6

6

planein max ''

6planein max ''

1050

10418

108.2082

avg

yx

yx

(Ans) 7.36

35.82902

1

1

s

s

MEASUREMENT OF STRAINS BY STRAIN ROSETTES


• Ways of arranging 3 electrical-resistance strain gauges

• In general case (a):

ccxycycxc

bbxybybxb

aaxyayaxa

cossinsincos

cossinsincos

cossinsincos

22

22

22

VARIABLE SOLUTIONS


Please click the appropriate icon for your computer to access the variable solutions

MEASUREMENT OF STRAINS BY STRAIN ROSETTES (cont)


• In 45° strain rosette [case (b)],

• In 60° strain rosette [case (c)],

cabxy

cy

ax

2

cbxy

acby

ax

3

2

222

1

EXAMPLE 12


The state of strain at point A on the bracket in Fig. 10–17a is measured using the strain rosette shown in Fig. 10–17b. Due to the loadings, the readings from the gauges give

Determine the in-plane principal strains at the point and the directions in which they act.

666 10264 , 10135 , 1060 cba

EXAMPLE 12 (cont)


• Measuring the angles counter-clockwise,

• By substituting the values into the 3 strain-transformation equations, we have

• Using Mohr’s circle, we have A(60(10-6), 60(10-6)) and center C (153(10-6), 0).

Solution

120 and 60 ,0 cba

666 10149 , 10246 , 1060 zyx

(Ans) 3.19

, 109.33

, 10272

101.119105.7460153

p2

61

61

6622

R

STRESS-STRAIN RELATIONSHIP


• Use the principle of superposition

• Use Poisson’s ratio,

• Use Hooke’s Law (as it applies in the uniaxial direction),

allongitudinlateral

E

yxzzzxyyzyxx vE

vE

vE

1

, 1

, 1

STRESS-STRAIN RELATIONSHIP (cont)


• Use Hooke’s Law for shear stress and shear strain

• Note:

xzxzyzyzxyxy GGG 1

1

1

v

EG

12



• Dilatation (i.e. volumetric strain )zyxV

ve

zyxE

ve

21



• For special case of “hydrostatic” loading,

• Where the right-hand side is defined as bulk modulus R, i.e.

v

E

e

P

pzyx

213

v

Ek

213

EXAMPLE 13


The copper bar in Fig. 10–24 is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, b = 500 mm, and t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take 34.0 , GPa 120 cucu vE

EXAMPLE 13 (cont)


• From the loading we have

• The associated normal strains are determined from the generalized Hooke’s law,

• The new bar length, width, and thickness are therefore

Solution

0 , 80 , MPa 500 , MPa 800 zxyx

000850.0 , 00643.0 , 00808.0 yxz

zzxy

yzyx

x E

v

EE

v

EE

v

E

(Ans) mm 98.1920000850.020'

(Ans) mm 68.495000643.050'

(Ans) mm 4.30230000808.0300'

t

b

a

CONCEPT QUIZ

1) Which one of the following statements is untrue?

a) In 2-D state of stress, the orientation of the element representing the maximum in-plane shear stress can be obtained by rotating the element 45° from the element representing the principle stresses.

b) In 3-D state of stress, the orientation of the element representing the absolute maximum shear stress can be obtained by rotating the element 45° about the axis defining the direction of σint.

c) If the in-plane principal stresses are of opposite sign, then the absolute maximum shear stress equals the maximum in-plane stress, that is, τabs max = (σmax – σmin)/2

d) Same as (c) but the principal stresses are of the same sign.


Documents

Chapter Objectives To develop the transformation of stress components from one orientation of the coordinate system to another orientation. To determine