CHAPTER 5

Transformations of Stress and Strain

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•General Equations for Plane Stress Transformation •Mohr’s Circle for Plane Stress •Absolute Maximum Shear Stress •Transformation of Plane Strain •Mohr’s Circle for Plane Strain •Introduction to Measurements of Strain: Strain Rosettes

•Determination of Stresses from Strains

Reference: R. C. Hibbeler, Mechanics of Materials, 7th Edition, Prentice Hall, 2008 F. P. Beer, E. R. Johnston Jr. J. T. DeWolf and D. F. Mazurex, Mechanics of Materials, 5th Edition in SI Units, McGraw-Hill, 2009

Plane-Stress Transformation

General state of stress at a point is characterized by 6 independent normal and shear stress components.

It can be analyzed in a single plane of a body, the material can said to be subjected to plane stress.

2

Stress components from one orientation of an element can transform to an element having a different orientation.

=

3

Example 9.1

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

Solution:

The element is sectioned by the line a–a.

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

4

(Ans) MPa 8.68

030sin30cos2530cos30sin80

30cos30cos2530sin30cos50 ;0'

x'y'

x'y'y

AA

AAAF

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

(Ans) MPa 15.4

030cos30sin2530sin30sin80

30sin30cos2530cos30cos50 ;0

'

''

x

xx

AA

AAAF

5

(Ans) MPa 8.68

030cos30sin5030sin30sin25

30sin30cos8030cos30cos25- ;0'

x'y'

x'y'y

AA

AAAF

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

(Ans) MPa 8.25

030sin30sin5030cos30cos25

30cos30cos8030sin30cos25 ;0

'

''

x

xx

AA

AAAF

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

6

General Equations of Plane-Stress Transformation

Positive normal stress acts outward from all faces and positive shear stress acts upward on the right-hand face of the element.

2cos2sin2

2sin2cos22

''

'

xy

yx

yx

xy

yxyx

x

7

Example 9.2 The state of plane stress at a point is represented by the element. Determine the state of stress at the point on another element oriented 30° clockwise from the position shown.

Solution: From the sign convention we have,

To obtain the stress components on plane CD,

30 MPa 25 MPa 50 MPa 80 xyyx

(Ans) MPa 8.682cos2sin2

(Ans) MPa 8.252sin2cos22

''

'

xy

yx

yx

xy

yxyx

x

8

To obtain the stress components on plane BC,

60 MPa 25 MPa 50 MPa 80 xyyx

(Ans) MPa 8.682cos2sin2

(Ans) MPa 15.42sin2cos22

''

'

xy

yx

yx

xy

yxyx

x

The results are shown on the element as shown.

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Principal Stresses and Maximum In-Plane Shear Stress

In-Plane Principal Stresses

Orientation of the planes will determine the maximum and minimum normal stress.

The solution has two roots, thus we obtain the following principle stress.

2/2tan

yx

xy

p

21

2

2

2,1 where22

xy

yxyx

10

Maximum In-Plane Shear Stress

Orientation of an element will determine the maximum and minimum shear stress.

The solution has two roots, thus we obtain the maximum in-plane shear stress and averaged normal stress.

xy

yx

s

2/2tan

2

2

plane-inmax 2

xy

yx

2

yx

avg

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Example 9.3

When the torsional loading T is applied to the bar 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.

Solution: From the sign convention we have,

a) Maximum in-plane shear stress is

xyyx 0 0

(Ans) 02

2

2

2

plane-inmax

yx

avgxy

yx

b) For principal stress,

(Ans) 22

135,452/

2tan

2

2

2,1

12

xy

yxyx

pp

yx

xy

p

12

Example 9.6

The state of plane stress at a point on a body is represented on the element. Represent this stress state in terms of the maximum in-plane shear stress and associated average normal stress.

Solution:

Since we have 60,90 ,20 xyyx

(Ans) MPa 352

(Ans) MPa 4.812

2

2

plane-inmax

yx

avg

xy

yx

The maximum in-plane shear stress and average normal stress is

3.111,3.21

2/2tan 12 ss

xy

yx

s

13

Mohr’s Circle—Plane Stress Plane stress transformation is able to have a graphical solution that is easy to

remember.

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Example 9.8

The torsional loading T produces the state of stress in the shaft as shown. Draw Mohr’s circle for this case.

Solution:

We first construct of the circle, xyyx and 0,0

0 , plane-inmax avg

The center of the circle C is on the axis at 02

yx

avg

Point A is the average normal stress and maximum in-plane shear stress,

Principal stresses are identified as points B and D on the circle.

21 ,

15

Example 9.10

The state of plane stress at a point is shown on the element. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act.

Solution:

We first construct of the circle, 60 and 90,20 xyyx

MPa 4.815560 22 R

The center of the circle C is on the axis at

MPa 352

9020

avg

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

Max in-plane shear stress and average normal stress are

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

The counter-clockwise angle is

(Ans) 3.2160

3520tan2 1

1

s

16

Example 9.12

An axial force of 900 N and a torque 2.5 Nm of are applied to the shaft. The shaft diameter is 40 mm, find the principal stresses at a point P on its surface.

Solution:

The stresses produced at point P is

kPa 2.71602.0

900 kPa, 9.198

02.0

02.05.224

2

A

P

J

Tc

kPa 1.3582

2.7160

avg

The principal stresses can be found using Mohr’s circle,

Principal stresses are represented by points B and D,

(Ans) kPa 5.517.4091.358

(Ans) kPa 7.7677.4091.358

2

1

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Example 9.13

The beam is subjected to the distributed loading of w = 120 kN/m. Determine the principal stresses in the beam at point P, which lies at the top of the web. Neglect the size of the fillets and stress concentrations at this point. I = 67.4(10-6) m4.

Solution:

kNm 6.30 kN 84 MV

The equilibrium of the sectioned beam is as shown where

(Ans) MPa 2.3501.0104.67

015.0175.01075.084

(Ans) 4.45104.67

1.0106.30

6

6

3

It

VQ

I

My

At point P,

18

MPa 6.649.417.22

MPa 2.197.229.41

2

1

7.222

04.45

Thus the results are as follows,

Thus the radius is calculated as 41.9, thus the principle stresses are

The centre of the circle is and point A is (-45.4, -32.5).

The counter-clockwise angle is

6.282.572 22 pp

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Absolute Maximum Shear Stress

The absolute maximum shear stress and associated average normal stress can also be found by using Mohr’s circle.

2

2

minmaxminmaxmax abs

avg

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Example 9.14

Due to the applied loading, the element at the point on the frame is subjected to the state of plane stress shown. Find the principal and absolute maximum shear stress at the point.

Solution:

The centre of the circle is kPa 102

020

avg

The controlling point is A (-20, -40).

The radius is kPa 2.41401020 22R

21

Thus the Mohr’s circle is drawn accordingly.

kPa 2.512.4110

kPa 2.312.4110

min

max

0.38

1020

40tan2 1

Principal stresses are at the points where the circle intersects the σ axis,

From the circle, the counter-clockwise angle is

Since there is no principal stress on the element in the z direction, we have

(Ans) kPa 2.51 ,0 kPa, 2.31 minintmax

22

For absolute maximum shear stress,

(Ans) kPa 102

2.512.31

2

(Ans) kPa 2.412

2.512.31

2

minmax

minmaxmax abs

avg

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Plane Strain General state of strain at a point in a body is represented by 3 normal strains

and 3 shear strains .

The normal and shear strain components will vary according to the orientation of the element.

zyx

zyx

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General Equations of Plane-Strain Transformation

Strain-transformation equations

2cos2

2sin22

2sin2

2cos22

2sin2

2cos22

''

'

'

xyyxyx

xyyxyx

y

xyyxyx

x

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Principal Strains

Maximum In-Plane Shear Strain

Maximum in-plane shear strain and associated average normal strain are as follow:

22

2,1222

2tan

xyyxyx

yx

xy

p

2 ,

222

2tan

22

plane-inmax yx

avg

xyyx

xy

yx

S

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Example 10.3

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

Solution:

Looking at the orientation of the element,

For maximum in-plane shear strain,

666 1080 , 10200 , 10350 xyyx

311 and 9.40

80

2003502tan

s

xy

yx

s

(Ans) 10556

222

6

planein max

22

planein max

xyyx

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Mohr’s Circle—Plane Strain

We can also solve problems involving the transformation of strain using Mohr’s circle.

It has a center on the ε axis at point C(εavg, 0) and a radius R.

22

22 and

2

where

xyyxyx

avg R

2

2

''2

'2

Ryx

avgx

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Example 10.5

The state of plane strain at a point is represented by the components Determine the maximum in-plane shear strains and the orientation of an element.

Solution:

From the coordinates of point E, we have

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

666 10120 , 10150 , 10250 xyyx

6

6

planein max ''

6planein max ''

1050

10418

108.2082

avg

yx

yx

(Ans) 7.36

35.82902

1

1

s

s

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Absolute Maximum Shear Strain

Absolute maximum shear strain is found from the circle having the largest radius.

It occurs on the element oriented 45° about the axis from the element shown in its original position.

2

minmax

minmaxmax abs

avg

30

Plane Strain

For plane strain we have,

This value represents the absolute maximum shear strain for the material.

maxmax''max abs zx minmaxmax''max abs yx

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Example 10.7

The state of plane strain at a point is represented by the strain components, Determine the maximum in-plane shear strain and the absolute maximum shear strain.

Solution:

From the strain components, the centre of the circle is on the ε axis at

Since , the reference point has coordinates

666 10150 , 10200 , 10400 xyyx

66 10100102

200400

avg

610752

xy 66 1075,10400 A

Thus the radius of the circle is 9622103091075100400

R

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Computing the in-plane principal strains, we have

From the circle, the maximum in-plane shear strain is

66

min

66

max

1040910309100

1020910309100

(Ans) 1061810409209 66

minmaxplanein max

10409 , 0 , 10209 6

minint

6

max

From the above results, we have Thus the Mohr’s circle is as follows,

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Strain Rosettes

Normal strain in a tension-test specimen can be measured using an electrical-resistance strain gauge.

The strain-transformation equation for each gauge is as follow:

ccxycycxc

bbxybybxb

aaxyayaxa

cossinsincos

cossinsincos

cossinsincos

22

22

22

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Material-Property Relationships

Generalized Hooke’s Law

For a triaxial state of stress, the general form for Hooke’s law is as follow:

They are valid only for a linear–elastic materials.

Hooke’s law for shear stress and shear strain is written as

yxzzzxyyzyxx vE

vE

vE

1

, 1

, 1

xzxzyzyzxyxyGGG

1

1

1

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Relationship Involving E, v, and G

Dilatation and Bulk Modulus

Dilatation, or volumetric strain, is caused only by normal strain, not shear strain.

Bulk modulus is a measure of the stiffness of a volume of material.

Plastic yielding occurs at v = 0.5.

v

EG

12

v

Ek

213

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