1. CHAPTER OBJECTIVES
Apply the stress transformation methods derived in Chapter 9 to similarly transform strain
Discuss various ways of measuring strain
Develop important material-property relationships; including generalized form of Hooke’s law

2. CHAPTER OUTLINE
Plane Strain
General Equations of Plane-Strain Transformation
Strain Rosettes
Material-Property Relationships

3. 10.1 PLANE STRAIN
As explained in Chapter 2.2, general state of strain in a body is represented by a combination of 3 components of normal strain (x, y, z), and 3 components of shear strain (xy, xz, yz).
Strain components at a pt determined by using strain gauges, which is measured in specified directions.
A plane-strained element is subjected to two components of normal strain (x, y) and one component of shear strain, xy.

4. The deformations are shown graphically below.
10.1 PLANE STRAIN
The deformations are shown graphically below.
Note that the normal strains are produced by changes in length of the element in the x and y directions, while shear strain is produced by the relative rotation of two adjacent sides of the element.

5. Note that plane stress does not always cause plane strain.
In general, unless  = 0, the Poisson effect will prevent the simultaneous occurrence of plane strain and plane stress.
Since shear stress and shear strain not affected by Poisson’s ratio, condition of xz = yz = 0 requires xz = yz = 0.

6. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Sign Convention
To use the same convention as defined in Chapter 2.2.
With reference to differential element shown, normal strains x and y are positive if they cause elongation along the x and y axes
Shear strain xy is positive if the interior angle AOB becomes smaller than 90.

7. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Sign Convention
Similar to plane stress, when measuring the normal and shear strains relative to the x’ and y’ axes, the angle  will be positive provided it follows the curling of the right-hand fingers, counterclockwise.
Normal and shear strains
Before we develop the strain-transformation eqn for determining x;, we must determine the elongation of a line segment dx’ that lies along the x’ axis and subjected to strain components.

8. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
Components of line dx and dx’ are elongated and we add all elongations together.
From Eqn 2.2, the normal strain along the line dx’ is x’ =x’/dx’. Using Eqn 10-1,

9. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
To get the transformation equation for x’y’, consider amount of rotation of each of the line segments dx’ and dy’ when subjected to strain components. Thus,

10. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
Using Eqn 10-1 with  = y’/x’,
As shown, dy’ rotates by an amount .

11. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
Using identities sin ( + 90) = cos , cos ( + 90) =  sin ,
Thus we get

12. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
Using trigonometric identities sin 2 = 2 sin cos, cos2 = (1 + cos2 )/2 and sin2 + cos2 = 1, we rewrite Eqns 10-2 and 10-4 as

13. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Normal and shear strains
If normal strain in the y direction is required, it can be obtained from Eqn 10-5 by substituting ( + 90) for . The result is

14. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Principal strains
We can orientate an element at a pt such that the element’s deformation is only represented by normal strains, with no shear strains.
The material must be isotropic, and the axes along which the strains occur must coincide with the axes that define the principal axes.
Thus from Eqns 9-4 and 9-5,

15. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Principal strains
Maximum in-plane shear strain
Using Eqns 9-6, 9-7 and 9-8, we get

16. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
Maximum in-plane shear strain
Using Eqns 9-6, 9-7 and 9-8, we get

17. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
IMPORTANT
Due to Poisson effect, the state of plane strain is not a state of plane stress, and vice versa.
A pt on a body is subjected to plane stress when the surface of the body is stress-free.
Plane strain analysis may be used within the plane of the stresses to analyze the results from the gauges. Remember though, there is normal strain that is perpendicular to the gauges.
When the state of strain is represented by the principal strains, no shear strain will act on the element.

18. 10.2 GENERAL EQNS OF PLANE-STRAIN TRANSFORMATION
IMPORTANT
The state of strain at the pt 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 strains is 45 from the element representing the principal strains.

19. EXAMPLE 10.1
A differential element of material at a pt is subjected to a state of plane strain x = 500(10-6), y =  300(10-6), which tends to distort the element as shown. Determine the equivalent strains acting on an element oriented at the pt, clockwise 30 from the original position.

20. EXAMPLE 10.1 (SOLN)
Since  is counterclockwise, then  = –30, use strain-transformation Eqns 10-5 and 10-6,

21. EXAMPLE 10.1 (SOLN)
Since  is counterclockwise, then  = –30, use strain-transformation Eqns 10-5 and 10-6,

22. EXAMPLE 10.1 (SOLN)
Strain in the y’ direction can be obtained from Eqn 10-7 with  = –30. However, we can also obtain y’ using Eqn 10-5 with  = 60 ( = –30 + 90), replacing x’ with y’

23. EXAMPLE 10.1 (SOLN)
The results obtained tend to deform the element as shown below.

24. EXAMPLE 10.2
A differential element of material at a pt is subjected to a state of plane strain defined by x = –350(10-6), y = 200(10-6), xy = 80(10-6), which tends to distort the element as shown. Determine the principal strains at the pt and associated orientation of the element.

25. EXAMPLE 10.2 (SOLN)
Orientation of the element
From Eqn 10-8, we have
Each of these angles is measured positive counterclockwise, from the x axis to the outward normals on each face of the element.

26. EXAMPLE 10.2 (SOLN)
Principal strains
From Eqn 10-9,

27. EXAMPLE 10.2 (SOLN)
Principal strains
We can determine which of these two strains deforms the element in the x’ direction by applying Eqn 10-5 with  = –4.14. Thus

28. EXAMPLE 10.2 (SOLN)
Principal strains
Hence x’ = 2. When subjected to the principal strains, the element is distorted as shown.

29. EXAMPLE 10.3
A differential element of material at a pt is subjected to a state of plane strain defined by x = –350(10-6), y = 200(10-6), xy = 80(10-6), which tends to distort the element as shown. Determine the maximum in-plane shear strain at the pt and associated orientation of the element.

30. EXAMPLE 10.3 (SOLN)
Orientation of the element
From Eqn 10-10,
Note that this orientation is 45 from that shown in Example 10.2 as expected.

31. EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
Applying Eqn 10-11,
The proper sign of can be obtained by applying Eqn 10-6 with s = 40.9.

32. EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
Thus tends to distort the element so that the right angle between dx’ and dy’ is decreased (positive sign convention).

33. EXAMPLE 10.3 (SOLN)
Maximum in-plane shear strain
There are associated average normal strains imposed on the element determined from Eqn 10-12:
These strains tend to cause the element to contract.

34. 10.5 STRAIN ROSETTES
We measure the normal strain in a tension-test specimen using an electrical-resistance strain gauge.
For general loading on a body, the normal strains at a pt are measured using a cluster of 3 electrical-resistance strain gauges.
Such strain gauges, arranged in a specific pattern are called strain rosettes.
Note that only the strains in the plane of the gauges are measured by the strain rosette. That is ,the normal strain on the surface is not measured.

35. 10.5 STRAIN ROSETTES
Apply strain transformation Eqn 10-2 to each gauge:
We determine the values of x, y xy by solving the three equations simultaneously.

36. 10.5 STRAIN ROSETTES
For rosettes arranged in the 45 pattern, Eqn becomes
For rosettes arranged in the 60 pattern, Eqn becomes

37. EXAMPLE 10.8
State of strain at pt A on bracket is measured using the strain rosette shown. Due to the loadings, the readings from the gauges give a = 60(10-6), b = 135(10-6), and c = 264(10-6). Determine the in-plane principal strains at the pt and the directions in which they act.

38. EXAMPLE 10.8 (SOLN)
Establish x axis as shown, measure the angles counterclockwise from the +x axis to center-lines of each gauge, we have a = 0, b = 60, and c = 120 Substitute into Eqn 10-16,

39. EXAMPLE 10.8 (SOLN)
Solving Eqns (1), (2) and (3) simultaneously, we get
The in-plane principal strains can also be obtained directly from Eqn Reference pt on Mohr’s circle is A [60(10-6), –74.5(10-6)] and center of circle, C is on the  axis at avg = 153(10-6). From shaded triangle, radius is

40. EXAMPLE 10.8 (SOLN)
The in-plane principal strains are thus
Deformed element is shown dashed. Due to Poisson effect, element also subjected to an out-of-plane strain, in the z direction, although this value does not influence the calculated results.

41. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hooke’s law
Material at a pt subjected to a state of triaxial stress, with associated strains.
We use principle of superposition, Poisson’s ratio (lat = long), and Hooke’s law ( =  E) to relate stresses to strains, in the uniaxial direction.
With x applied, element elongates in the x direction and strain is this direction is

42. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hooke’s law
With y applied, element contracts with a strain ‘’x in the x direction,
Likewise, With z applied, a contraction is caused in the z direction,

43. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hooke’s law
By using the principle of superposition,

44. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Generalized Hooke’s law
If we apply a shear stress xy to the element, experimental observations show that it will deform only due to shear strain xy. Similarly for xz and xy, yz and yz. Thus, Hooke’s law for shear stress and shear strain is written as

45. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Relationship involving E, , and G
We stated in chapter 3.7:
Relate principal strain to shear stress,
Note that since x = y = z = 0, then from Eqn , x = y = 0. Substitute into transformation Eqn 10-19,

46. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Relationship involving E, , and G
By Hooke’s law, xy = xy/G. So max = xy/2G.
Substitute into Eqn and rearrange to obtain Eqn
Dilatation and Bulk Modulus
Consider a volume element subjected to principal stresses x, y, z.
Sides of element are dx, dy and dz, and after stress application, they become (1 + x)dx, (1 + y)dy, (1 + z)dz, respectively.

47. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Dilatation and Bulk Modulus
Change in volume of element is
Change in volume per unit volume is the “volumetric strain” or dilatation e.
Using generalized Hooke’s law, we write the dilatation in terms of applied stress.

48. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Dilatation and Bulk Modulus
When volume element of material is subjected to uniform pressure p of a liquid, pressure is the same in all directions.
As shear resistance of a liquid is zero, we can ignore shear stresses.
Thus, an element of the body is subjected to principal stresses x = y = z = –p. Substituting into Eqn and rearranging,

49. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
Dilatation and Bulk Modulus
This ratio (p/e) is similar to the ratio of linear-elastic stress to strain, thus terms on the RHS are called the volume modulus of elasticity or the bulk modulus. Having same units as stress with symbol k,
For most metals,  ≈ ⅓ so k ≈ E.
From Eqn 10-25, theoretical maximum value of Poisson’s ratio is therefore  = 0.5.
When plastic yielding occurs,  = 0.5 is used.

50. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
IMPORTANT
When homogeneous and isotropic material is subjected to a state of triaxial stress, the strain in one of the stress directions is influence by the strains produced by all stresses. This is the result of the Poisson effect, and results in the form of a generalized Hooke’s law.
A shear stress applied to homogenous and isotropic material will only produce shear strain in the same plane.
Material constants, E, G and  are related mathematically.

51. 10.6 MATERIAL-PROPERTY RELATIONSHIPS
IMPORTANT
Dilatation, or volumetric strain, is caused by only by normal strain, not shear strain.
The bulk modulus is a measure of the stiffness of a volume of material. This material property provides an upper limit to Poisson’s ratio of  = 0.5, which remains at this value while plastic yielding occurs.

52. EXAMPLE 10.10
Copper bar is subjected to a uniform loading along its edges as shown. If it has a length a = 300 mm, width b = 50 mm, and thickness t = 20 mm before the load is applied, determine its new length, width, and thickness after application of the load. Take Ecu = 120 GPa, cu = 0.34.

53. EXAMPLE (SOLN)
By inspection, bar is subjected to a state of plane stress. From loading, we have
Associated strains are determined from generalized Hooke’s law, Eqn 10-18;

54. EXAMPLE (SOLN)
Associated strains are determined from generalized Hooke’s law, Eqn 10-18;

55. EXAMPLE (SOLN)
The new bar length, width, and thickness are

56. EXAMPLE 10.11
If rectangular block shown is subjected to a uniform pressure of p = 20 kPa, determine the dilatation and change in length of each side. Take E = 600 kPa,  = 0.45.

57. EXAMPLE (SOLN)
Dilatation
The dilatation can be determined using Eqn with x = y = z = –20 kPa. We have

58. EXAMPLE (SOLN)
Change in length
Normal strain on each side can be determined from Hooke’s law, Eqn 10-18;

59. EXAMPLE (SOLN)
Change in length
Thus, the change in length of each side is
The negative signs indicate that each dimension is decreased.

60. CHAPTER REVIEW
When element of material is subjected to deformations that only occur in a single plane, then it undergoes plain strain.
If the strain components x, y, and xy are known for a specified orientation of the element, then the strains acting for some other orientation of the element can be determined using the plane-strain transformation equations.
Likewise, principal normal strains and maximum in-plane shear strain can be determined using transformation equations.

61. CHAPTER REVIEW
Hooke’s law can be expressed in 3 dimensions, where each strain is related to the 3 normal stress components using the material properties E, and , as seen in Eqns
If E and  are known, then G can be determined using G = E/[2(1 + ].
Dilatation is a measure of volumetric strain, and the bulk modulus is used to measure the stiffness of a volume of material.