Strain Definitions and Components

CHAPTER 1

Stress/Strain Definitions and Components 1.1 GENERAL CONCEPT Engineering systems must be designed to withstand the actual and probable loads that may be imposed on them. Hence the wall of a dam must be of adequate strength to hold out mainly the reservoir water pressure but also to withstand other loads, such as seismic occasional shocks, thermal expansions/contractions, and many others. A tennis racket is designed to take dynamic and impact loads imposed by a fast-moving flying tennis ball. It must also be adequately designed to withstand impact loads when incidentally hitting a hard ground. An oil drilling equipment must be designed to suitably and adequately drill through different types of rock materials, but at the same time ensuring that its imposing loads would not cause rock formation integrity affecting the stability of the drilled well. The concept of solid mechanics provides the analytical methods of designing solid engineering systems with adequate strength, stiffness, stability, and integrity. Although it is different but very much overlaps with the concept and analytical methods provided by continuum mechanics. Solid mechanics is used broadly across all branches of the engineering science including many applications, such as oil and gas exploration, drilling, completion, and production. In this concept the behavior of an engineering object, subjected to various forces and constraints (as shown in Fig. 1.1), is evaluated using the fundamental laws of Newtonian mechanics, that governs the balance of forces, and the mechanical properties or characteristics of the materials from which the object is made. The two key elements of solid mechanics are the internal resistance of a solid object to balance the effects of imposing external forces, represented by a term called stress, and the shape change and deformation of the solid object in response to external forces, denoted by strain. The next sections of this chapter are devoted to defining these two elements and their relevant components. Petroleum Rock Mechanics DOI: https://doi.org/10.1016/B978-0-12-815903-3.00001-7

© 2019 Elsevier Inc. All rights reserved.

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Heat flux

Z F1

F2

Solid core

Y

Fn

X

Outflow flux

F3

Figure 1.1 A solid object subjected to various forces and constraints.

1.2 DEFINITION OF STRESS In general, stress is defined as average force acting over area. This area may be a surface, or an imaginary plane inside a material. Since the stress is a force per unit area, as given in the equation below, it is independent of the size of the body. σ5

Force F 5 Area A

(1.1)

where σ is the stress (Pa or psi), F is the force (N or lbf), and A represents the surface area (m2 or in.2). Stress is also independent of the shape of the body. We will show later that the stress level depends on the orientation. The criterion that governs this is the force balance and the concept of Newton’s second law. Fig. 1.2 illustrates a simple one-dimensional stress state where a body is loaded to a uniform stress level of σaxial. Since the body is in equilibrium, an action stress from the left must be balanced by a reaction stress on the right. By defining an arbitrary imaginary plane inside the body the forces acting on this plane must balance as well, regardless of the orientation of the plane. Two types of stresses are therefore resulted from the equilibrium condition; these are normal stress σ, which acts normal to the plane, and shear stress τ, which acts along the plane. The normal stress may result to tensile or compressive failure and the shear stress to shear failure where the material is sheared or slipped along a plane.

Stress/Strain Definitions and Components

σ

σaxial

5

σaxial θ τ

σ θ

σaxial

τ

Figure 1.2 A stress component may result into normal and shear stresses acting on an imaginary plane.

σz τzy τzx

τyz

τxz

σy τxy

τyx

σx Figure 1.3 Three-dimensional stress state of a cube.

Note 1.1: Stress component may be transformed into other stress components by defining arbitrary planes inside the body. The law governing this is the balance of forces.

1.3 STRESS COMPONENTS We start with a general three-dimensional case as shown in Fig. 1.3. This figure shows a cube with the respective stresses. Only the stresses acting on the faces of the cube are shown. Balance of forces requires that equal stresses act in the opposite direction on the three sides of the cube. Nine different components of stress can be seen in Fig. 1.3. These are required to determine the state of stress at a point. The stress components

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can be grouped into two categories, that is, σx ; σy ; and σz , as normal stresses and τ xy ; τ yx ; τ xz ; τ zx ; τ yz ; and τ zy as shear stresses. The stress components have indices, which relate to the Cartesian coordinate system. The first index defines the axis normal to the plane on which the stress acts. The second index defines the direction of the stress component. Normal stresses with two identical indexes are given with one index, for example, σxx
σx . For the cube to be in the state of equilibrium, we verify the stress state about z-axis as shown in Fig. 1.4 and define a moment balance about the origin, which is resulted into P

Mo 5 2 ðσx dyÞ

     dx dy  dy 1 τ xy dy dx 2 τ yx dx dy 1 σy dx 1 ðσx dyÞ 2 2 2

  dx 2 σy dx 50 2     P or Mo 5 τ xy dy dx 2 τ yx dx dy 5 0 or τ xy 5 τ yx (1.2) By writing similar equations for x and y axes the stress state can now be defined by three normal and three shear stresses as given in the following equation: 2 3 σx τ xy τ xz ½σ 5 4 τ xy σy τ yz 5 (1.3) τ xz τ yz σz The stress matrix given above is symmetric about its diagonal. y

σy τyx y

σx

τxy

dx dy

σx

τxy

o

x

τyx σy

Figure 1.4 Stresses acting on xy plane.

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Stress/Strain Definitions and Components

Note 1.2: In the analysis of solid rocks, compressive stresses are usually defined as positive entities and tensile stresses as negative. This is opposite the sign convention used for the analysis of other engineering materials.

1.4 DEFINITION OF STRAIN When a body is subjected to loading, it will undergo displacement and/or deformation. This means that any point in/on the body will be shifted to another position. Deformation is normally quantified in terms of the original dimension, and it is represented by strain that is a dimensionless parameter. Strain is therefore defined as deformation divided by the original or nondeformed dimension and is simply expressed by ε5

Δl lo

(1.4)

where ε is the strain, Δl is the deformed dimension (m or in.), and lo is the initial dimension (m or in.). Strains are categorized into engineering strain and scientific strain. While the initial/original dimension is used throughout the analysis in the engineering strain, in the scientific strain, the actual dimension, which changes with time, is applied. Eq. (1.4) is derived using the concept of small deformation theory. If large deformations are involved, Eq. (1.4) is no longer valid, and other definitions are required. Two main large deformation formulas are introduced by Almansi and Green. These are expressed by ε5

l 2 2 lo2 2l 2

(1.5)

known as Almansi strain formula, and ε5

l 2 2 lo2 2lo2

(1.6)

known as Green strain formula, respectively. It can be shown that for small deformations, Eqs. (1.5) and (1.6) will be simplified to Eq. (1.4). The error of using Eq. (1.4) may be negligible for many cases compared to other assumptions.

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1.5 STRAIN COMPONENTS Assuming small deformation theory, we will study the deformation of a square under loading as shown in Fig. 1.5. It can be seen that the square is moved (translated) and has changed shape (deformed). The translation in space has actually no effect on stresses, whereas the deformation causes change of stress and is therefore of interest in failure analysis. Angle of deformation can be expressed by tan α 5

ð@v=@xÞdx dx 1 ð@u=@xÞdx

and approximated by tan α 

@v @x

(1.7)

using the small deformation theory. The strains in x and y directions are defined by εx 5

ð@u=@xÞdx @u 5 ; dx @x

εy 5

ð@v=@yÞdy @v 5 dy @y

εxy 5

ð@v=@xÞdx @v 5 ; dx @x

εyx 5

ð@u=@yÞdy @u 5 dy @y

and

y ∂u dy ∂y v+

∂v dy ∂y

dy

∂v dx ∂x

α

o’

x

o dx

∂u dx ∂x

u+

Figure 1.5 A square shape before and after loading.

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Stress/Strain Definitions and Components

εxy 1 εyx 5

@v @u 1 5 2εxy 5 γ xy @x @y

(1.8)

where ε is known as normal strain and γ as the shear strain. The three-dimensional strain state can be derived, in the same way as the three-dimensional stress state [Eq. (1.3)], in a matrix form, as 3 2 εx 12γxy 12γxz 7 6 6 1γ εy 12γ yz 7 7 ½ε 5 6 2 xy 7 6 5 4 1 1 εz 2γ xz 2γ yz 2 ! !3 @u 1 @u @v 1 @u @w 1 1 7 6 6 @x 2 @y @x 2 @z @x 7 7 6 ! !7 6 6 1 @u @v @v 1 @v @w 7 7 6 1 1 (1.9) 56 7 @x @y 2 @z @y 7 6 2 @y 7 6 ! ! 7 6 7 6 1 @u @w 1 @v @w @w 5 4 1 1 2 @z @x 2 @z @y @z It is seen that the effects of second-order terms have been neglected by performing linearization. The equations derived are therefore valid for small deformations, which can be applied to most of engineering materials. If a material exhibits large deformations, the second-order terms become significant.

Example 1.1.

A circular solid piece of rock is tested in a compression testing rig to examine its stress/strain behavior. The sample is 6 in. in diameter and 12 in. in length with the compression load cell imposing a constant load of 10,000 lbf equally at both top and bottom of the rock sample. Assuming a measured reduction in length of 0.02 in., find the compressive stress and strain of the rock. Solution: We use σ 5 F/A, as defined by Eq. (1.1), where P 5 10,000 lb and A is A5

π 2 π d 5 3 62 5 28:27 in:2 4 4

Therefore the compressive stress in the rock piece is (Continued)

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(Continued)

σ5

F 10; 000 5 5 353:7 lbf=in:2 or psi A 28:27

The compressive strain, as defined in Eq. (1.4), is calculated as ε5

Δl 0:02 5 1:667 3 1023 in:=in: 5 1667 μin:=in: 5 lo 12

Problems 1.1.

1.2.

Assuming lo 5 200 mm and l 5 220 mm for a metallic rod under tension, determine strain using the three methods defined by Eqs. (1.4)(1.6) and compare the results. A plane strain test is being performed in a soil-testing apparatus. Before the test, needles are inserted at distances of 25 mm 3 25 mm as shown in Fig. 1.6. After deformation, the measured distances are given by

G

H

I

D

E

F

A

B

C

Figure 1.6 Plane strain test.

DF 5 48:2 mm AI 5 70 mm BH 5 49:1 mm GC 5 67:6 mm Make a plot of the test results on a millimeter paper assuming that line ABC remains unchanged during deformation and its corresponding points remain fixed. The final shape is a parallelogram with GHI parallel to ABC. Determine the deformations and the strains. (Continued)

Stress/Strain Definitions and Components

(Continued)

24.5 Tonnes

100 cm

Figure 1.7 Concrete post.

1.3.

1.4.

1.5.

A short post, constructed from a tube of concrete, supports a compressive load of 24.5 metric tonnes as shown in Fig. 1.7. The inner and outer diameters of the tube are 91 and 127 cm, respectively, and its length is 100 cm. The shortening of the post is measured as 0.056 cm. Determine the axial compressive stress and strain in the post. The effect of post’s weight is neglected. It is also assumed that the post does not buckle under the load. Three different solid steel balls are suspended by three wires of different lengths as shown in Fig. 1.8 and different diameters, that is, 0.12, 0.08, and 0.05 in., from the top to the bottom, respectively. Calculate a. Stresses in the wires and compare the results; b. Total elongation of the wires; and c. Strain in every wire and the total strain. Fig. 1.9 shows a circular steel rod of length 40 m and diameter 8 mm hangs in a mine shaft and holds an ore bucket of weight 1.5 kN at its lower end. Calculate the maximum axial stress taking account of the rod weight assuming the weight density of rod as 77.0 kN/m3. (Continued)

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(Continued)

2.2 ft 3.5 lb 2.8 ft 1 lb 2.5 ft 2 lb Figure 1.8 Balls suspended in wires.

40 m

8 mm

1.5 kN Figure 1.9 Ore bucket suspended by rod.