CH1.PDF

11

CHAP 1. STRESS-STRAIN ANALYSIS

Stress and strain are fundamental concepts in mechanical analysis and design. The safety of a structure is often evaluated based on stress that the structure carries on. If stress is great than a limit value, then the structure fails and, possibly, collapses. Thus, stress is often used as criteria for mechanical design. When loads are applied to the deformable structure, it generates internal forces (stress) against the loads. In order to generate the internal force the structure must deform its shape, which is related to strain. Consider a simple spring example, as shown in Figure 1.1. If a force F is applied to a spring, it generates a counterforce (or internal force) by deforming its length to ∆l. In this case, the internal force corresponds to stress, and the deformation corresponds to strain. The internal force and deformation are related by

F K l= ∆

where K is called the spring constant. The relation between the internal force and deformation is linear, which is a simple case of linear elasticity.

F

∆l l

Figure 1.1 Deformation of a spring under axial force

1.1. Stress

Surface Traction (Stress Vector) Surface traction or stress vector is defined as a force intensity acting on a plane. Consider a deformable body that is in equilibrium under external forces (see Figure 1.2(a)). To see the effect of the internal surface traction, we imaginarily cut the structure

12 Finite Element Analysis and Design

by a plane whose normal vector is n through a point P, as shown in Figure 1.2(b). In order for the left half of the structure to be in equilibrium, there must be internal forces transmitted across the cutting plane. Such a distributed force is called the surface traction or stress vector. The stress vector T(n) acting at the point P on a plane whose normal is n is defined as

( )

0limA A∆ →

∆=∆

n FT (1.1)

In Eq. (1.1), the right superscript (n) is used to denote that this stress vector is defined on a plane whose normal is n. In addition, the stress vector is defined at a point by reducing the surface area. It is clear from Eq. (1.1) that the dimension of stress is force per unit area (pressure). In addition to the information of a force, the stress vector contains the information of a surface on which it is defined.

∆F

n

f1

f2

f3

f1

f2

f3

f4

f6

f5

(a) (b)x y

z

P ∆A

Figure 1.2 Surface traction acting on a plane

Example 1.1 Consider a truss component with the cross-sectional area A = 0.2 m2, as shown in Figure 1.3. If an axial force F = 100 N is applied to the truss, determine the stress vector on the cross-section.

Chap. 1 Stress-Strain Analysis 13

FF

F

A

n T(n)

Figure 1.3 Equilibrium of the truss component under the axial force

To simplify analysis, let us assume that the surface traction is equally distributed on the cross-section of the truss. In fact, this is the fundamental assumption of truss. The force on the cross-section A is obtained by integrating the constant stress vector T(n) over the cross-sectional area. In this simple example, the stress vector T(n) is in the direction of n. Thus, the magnitude of stress can be obtained by T = T(n)·n. Since the truss component is in the equilibrium status, the internal force must be the same with F in magnitude, i.e.,

( )

A AF dA T dA TA= ⋅ = =∫∫ ∫∫nT n

2 2

100 N N500 500Pa

0.2m m

FT

A∴ = = = =

Note that Pascal (Pa) = N/m2 is used for the stress unit, which is the same as the unit for the pressure. In fact, the pressure is a special situation of stress in which the magnitude of stress is the same in every direction.

Normal Stress and Shear Stress The stress vector T(n) does not act in general in the direction of n. Thus, we can separate the stress vector into the one that is parallel to n and the one that is orthogonal to n.

σn normal stress stress component parallel to n

τn shear stress stress component orthogonal to n

From Figure 1.4, normal stress can be calculated by σn = T(n)·n, and shear stress can be calculated from the following relations:

2( ) 2 2

n nσ τ= +nT


2( ) 2

n nτ σ= −nT (1.2)

σn

τn

T(n)

n

P

Figure 1.4 Normal and shear stresses at a point P

Example 1.2 When the stress vector at a point is ( ) {3, 4, 5}T=nT on the plane whole normal is parallel to the z-coordinate, calculate the normal and shear stresses.

( )

2( ) 2 2 2

2

3 0 4 0 5 1 5

3 4 5 50

50 5 5

n

n

σ

τ

= ⋅ = × + × + × =

= + + =

= − =

n

n

T n

T

Rectangular or Cartesian Stress Components Decomposing the stress vector into the normal and shear components can be further generalized to a three-dimensional cube whose sides are parallel to the coordinate system. Once we have found the components of the stress vectors acting on the coordinate planes we will relate these to the components of T(n). In order to represent stress components in the Cartesian coordinate system, an infinitesimal cube whose edges are parallel to the x, y, and z axes is taken from inside of the structure in Figure 1.2, as shown in Figure 1.5. Three sides of the edge have a length of ∆x, ∆y, and ∆z. In order for the cube to be in equilibrium, each side of the cube has internal forces on the plane. For example, consider a plane in the positive x-direction whose area is ∆Ax = ∆y×∆z. The internal force on the plane ∆Ax can be represented using x-y-z coordinate system, as


x y zF F F∆ = ∆ + ∆ + ∆F i j k (1.3)

where i, j, and k are unit vectors in the x, y, and z directions, respectively. Using the decomposition in Eq. (1.3) and the definition of the stress vector in Eq. (1.1), the stress vector on the plane ∆Ax can also be decomposed in three Cartesian components as

( )xx xy xzσ τ τ= + +iT i j k

where in this case i is the normal vector to the plane, and

0

0

0

lim

lim

lim

x

x

x

xxx

Ax

yxy

Ax

zxz

Ax

F

A

F

A

F

A

σ

τ

τ

∆ →

∆ →

∆ →

∆ = ∆∆ = ∆

∆=∆

(1.4)

are three components of the stress vector. A special notation is used in representing the components of stress vector. In σxx, for example, the first subscript indicates the face on which the stress is acting and the second subscript indicates its direction. When two subscripts are the same it is normal stress, otherwise it is shear stress. Note that in each plane one normal stress and two shear stresses exist. If the same procedure is carried out for the planes in the positive y- and z-axis, then a total of nine stress components can be defined, whose explanation is summarized in Table 1.1. The same stress components are defined for the planes in the negative coordinate directions.

σxx σyy

σzz

τxy

τxz

τyx

τyz

τzx τzy

x y

z

∆x

∆y

∆z∆F

Figure 1.5 Stress components in Cartesian coordinate system


Table 1.1 Description of stress components

Stress Component Description σxx Normal stress on the x face in the x direction σyy Normal stress on the y face in the y direction σzz Normal stress on the z face in the z direction τxy Shear stress on the x face in the y direction τyx Shear stress on the y face in the x direction τyz Shear stress on the y face in the z direction τzy Shear stress on the z face in the y direction τxz Shear stress on the x face in the z direction τzx Shear stress on the z face in the x direction

As shown in Table 1.1, a knowledge of the nine components is necessary in order to determine the components of the stress vector T(n) acting on an arbitrary plane with normal n. Sign convention of stress components is quite different from that of regular force vectors because stress components contain information of the surface on which they are defined. A stress component is positive when both the surface normal and the stress component are either in the positive or in the negative coordinate direction. If the surface normal is in the positive direction and the stress component is in the negative direction, then the stress component has a negative sign. For example, the signs of σxx and τxy can be determined by

sgn( ) sgn( ) sgn( )

sgn( ) sgn( ) sgn( )xx x

xy y

F

F

στ

= × ∆= × ∆

n

n

Normal stress is positive when it is a tensile stress and negative when it is compressive. Shear stress acting on the positive face is positive when it is acting in the positive coordinate direction. The positive directions of all the stress components are shown in Figure 1.5.

Stress Components on an Arbitrary Plane through a Point If the components of stress at a point are known, it is possible to determine the traction or stress vector acting on any plane passing through that point. Let n be the unit normal to the plane on which we want to determine the stress vector. The normal vector can be represented in a similar way in Eq. (1.3), as

x

x y z y

z

n

n n n n

n

= + + =

n i j k


Figure 1.6 shows a coordinate system at a point P and a tetrahedron PABC where the normal to the face ABC is n. If A is the area of the triangle ABC, then the areas of triangles PAB, PBC, and PAC are given by Anz, Anx, Any, respectively. Let the perpendicular distance from point P to the plane ABC be h and let the surface traction acting on the face ABC be given by ( ) ( ) ( ) ( )

x y zT T T= + +n n n nT i j k .

x

y

z

B

A

C

σzz τzx

τzy

σyy

τyz τyx

σxx

τxz

τxy

T(n) n

P

Figure 1.6 Stress vector and stress components acting on faces of an infinitesimal tetrahedron, cutting from the body in Figure 1.2

From the definition of stress vector in Eq. (1.1), the force on the surface can be calculated by multiplying stress with the surface area. Since the whole structure is in equilibrium, the forces applied to the tetrahedron in Figure 1.6 are also balanced. Force balance in the x-direction yields

( ) 0x x xx x yx y zx zF T A An An Anσ τ τ= − − − =∑ n

If the size of the tetrahedron is not small enough, then the stress component on the plane is not constant. Since we are interested in the stress component at a point, the size of the tetrahedron is infinitesimal (i.e. h → 0) and, thus, the above equation holds true. After dividing the above equation with the area A, we obtain the following relation:

( )x xx x yx y zx zT n n nσ τ τ= + +n (1.5)

This is nothing but a summation of x-directional stress components, and their contribution is proportional to area ratio. Similarly, force balance in the y- and z-directions yields

( )

( )y xy x yy y zy z

z xz x yz y zz z

T n n n

T n n n

τ σ ττ τ σ

= + += + +

n

n (1.6)


From Eqs. (1.5) and (1.6), it is clear that the stress vector acting on the surface whose normal is n can be determined if nine stress components are available. Using matrix notation, we can write these equations, as

( ) [ ]= ⋅nT σ n (1.7)

where

[ ]xx yx zx

xy yy zy

xz yz zz

σ τ ττ σ ττ τ σ

=

σ (1.8)

is the stress matrix. We use [•] symbol to denote a matrix. [σ] is often called a stress tensor. From Eq. (1.2), the normal stress and shear stress on plane ABC can therefore be expressed as

( )

2( ) 2

[ ]n

n n

σ

τ σ

= ⋅ = ⋅ ⋅

= −

n

n

T n n σ n

T (1.9)

Example 1.3 The state of stress at a particular point relative to xyz coordinates is given by the following stress matrix:

3 7 7

[ ] 7 4 0

7 0 2

− = −

σ

Determine the normal and shear stresses on a surface intersecting the point and parallel to the plane given by the equation of 4 4 2 2x y z− + = . In order to determine the surface traction T(n), it is necessary to determine the unit normal vector to the plane. From the elementary geometry, the plane normal is in the direction of d = {4, –4, 2}T. and 6=d . Thus, the unit normal vector is

2 2 1

{ , , }3 3 3

T= −n

The surface traction can be obtained as


( )

3 7 7 2 51

[ ] 7 4 0 2 23

7 0 2 1 4

− = ⋅ = ⋅ − = − −

nT σ n

Since the surface traction and the unit normal vector is available, the remaining calculation is similar to that of Example 1.2, 12

( )

2( ) 2 2 2

2( )

2 2 1 25 2 4

3 3 3 3

5 2 4 45

4 401

9 3

n

n

σ

τ

= ⋅ = × − × − × =

= + + =

= − =

n

n

n

T n

T

T

Symmetry of Stress Matrix and Vector Notation The nine components of the stress matrix can be reduced to six using the symmetric property of the stress matrix. Figure 1.7 shows an infinitesimal component (∆l×∆l) of the structure with shear stresses. The stress matrix at a point can be obtained as ∆l approaches to zero. The positive direction of the shear stress τxy in surface BD is in the positive y-direction, while in surface AC it is in the negative y-direction. When the structure is in the equilibrium status, the forces and moments are

( ) 0xy yx

xy yx

M l τ ττ τ

= ∆ − =⇒ =

∑

In order to satisfy the requirement of ΣM = 0 the shear stress components acting on perpendicular faces and in directions perpendicular to the line of intersection of the faces must be equal in magnitude and directed relative to each other. Thus, in a similar way the following relations can be derived:

xy yx

yz zy

xz zx

τ ττ ττ τ

===

(1.10)

Equation (1.10) implies that the stress matrix is symmetric and that only six components are enough to fully represent the stress at a point. Thus, instead of the 3×3 stress matrix the following 6×1 stress vector notation is often used:


{ }

xx

yy

zz

xy

yz

zx

σσστττ

=

σ (1.11)

τxy

τyx

x

y

τxy

τyx

O

∆l

∆l

A B

C D

Figure 1.7 Moment equilibrium of shear stress at a point

Principal Stresses As shown in the previous section, the stress vector at a point has different components in different planes. If we find a plane on which the stress vector only has normal stress, then the normal stress is called the principal stress and the normal direction of the plane is called the principal direction. It was stated earlier that the stress vector acting on different planes through a point can be different. In other words, the stress vector acting on this plane is along the normal vector to the plane. It will be shown that there are three such planes and three corresponding principal stresses. These stresses are the extreme values (maximum or minimum) of stress at the point. When the normal to a plane is the principal direction, the surface normal and the stress vector are in the same direction (T(n) // n). Thus, the stress vector on a plane can be represented by the production of a scalar and the normal vector, as

( ) λ=nT n (1.12)

Combining Eq. (1.12) with Eq. (1.7) for the stress vector, we obtain

[ ] λ⋅ =σ n n (1.13)


Equation (1.13) represents the eigenvalue problem, where λ is an eigenvalue and n is an eigenvector. Equation (1.13) can be rearranged as

( )[ ] [ ]λ− ⋅ =σ I n 0

where [I] is a 3×3 identity matrix. In component form, the above equation can be written in the following form:

0

0

0

xx yx zx x

xy yy zy y

xz yz zz z

n

n

n

σ λ τ ττ σ λ ττ τ σ λ

− − = −

(1.14)

Note that n = 0 satisfies the above equation, which yields a trivial solution. The above set of linear simultaneous equations will have a non-trivial solution only if the determinant of the coefficient matrix is zero, i.e.,

0xx yx zx

xy yy zy

xz yz zz

σ λ τ ττ σ λ ττ τ σ λ

−− =

− (1.15)

Expanding this determinant, we obtain the following cubic equation:

3 21 2 3 0I I Iλ λ λ− + − = (1.16)

where

1

2 2 22

2 2 23 2

xx yy zz

xx yy yy zz zz xx xy yz zx

xx yy zz xy yz zx xx yz yy zx zz xy

I

I

I

σ σ σσ σ σ σ σ σ τ τ τσ σ σ τ τ τ σ τ σ τ σ τ

= + += + + − − −= + − − −

are three invariants of the stress matrix [σ], which is independent of coordinate systems. Three roots of cubic Eq. (1.16) correspond to the three principal stresses. We will denote them by σ1, σ2, and σ3. Such a cubic equation can be solved using numerical methods such as the Newton-Raphson method or an analytical method. The numerical method for solving the cubic equation (1.16) can be found in the literature [3]. The analytical method is explained as follows. Consider a cubic equation of the form:

3 2 0y py qy r+ + + =


A general analytical solution for the above cubic equation can be written as

1

2

3

cos3 3

2cos( )

3 34

cos( )3 3

py g

py g

py g

φ

φ π

φ π

= −

+= −

+= −

where

3

2

3

cos2 27

2 3

1(3 )

31

(2 9 27 )27

b

a

g a

a q p

b p pq r

φ = − − = − = −

= − +

It can be shown that eigenvalues of a real symmetric matrix is always real (not imaginary). Since the stress matrix is symmetric it implies that the principal stress values calculated will be real or in other words, one can always compute three real (not always unique) principal stresses at any point in a solid.

Example 1.4 For the stress matrix give below, determine the principal stresses.

3 1 1

[ ] 1 0 2

1 2 0

=

σ

From Eq. (1.15), the determinant of the coefficient matrix becomes

3 1 1

1 2 0

1 2

λλ

λ

−− =

−

Even if the above cubic equation can be solved using the method explained above, here we solve the cubic equation using factor analysis. By expanding the determinant, we obtain


2

2

(3 )( 4) ( 2) (2 )

( 3)( 2)( 2) 2( 2)

( 2)[ ( 3)( 2) 2]

( 2)( 5 4)

( 2)( 1)( 4) 0

λ λ λ λλ λ λ λ

λ λ λλ λ λλ λ λ

− − − − − + += − − + − + += + − − − += − + − += − + − − =

Three solutions of the above equation are three principal stresses. Thus,

1 2 34, 1, 2σ σ σ= = = −

Principal Directions Once the principal stresses have been computed, we can substitute them one at a time into Eq. (1.14) to obtain the linear system of matrix equation that have three unknowns nx, ny, and nz. For example if the first principal stress σ1 is replaced by λ, then we obtain

11

11

11

0

0

0

xx yx zx x

xy yy zy y

xz yz zz z

n

n

n

σ σ τ ττ σ σ ττ τ σ σ

− − = −

(1.17)

Corresponding to each principal value, we will get a principal direction that will be denoted as n1, n2, and n3. In Eq. (1.17), 1 1 1, , andx y zn n n are components of n1. The above

equations are three linear simultaneous equations with three unknowns. However, since the determinant of the matrix is zero (i.e., the matrix is singular), they are not independent. Thus, an infinite number of solutions exist. We need one more equation to find a unique value for the principal directions ni. We can find a unique values for these vectors by assuming that they are unit vectors so that

2 2 2 2( ) ( ) ( ) 1, 1,2,3i i i i

x y zn n n i= + + = =n (1.18)

It can be shown that the planes on which the principal stresses act are mutually perpendicular. Let us consider any two principal directions ni and nj, with i j. If σi and σj are the corresponding principal stresses, then they satisfy the following equations:

[ ]

[ ]

i ii

j jj

σσ

⋅ =⋅ =

σ n n

σ n n

Multiplying the first equation by nj and the second equation by ni, we obtain


[ ]

[ ]

j i j ii

i j i jj

σσ

⋅ ⋅ = ⋅⋅ ⋅ = ⋅

n σ n n n

n σ n n n

Subtracting the first equation from the second and considering [σ] = [σ]T, we obtain

( ) 0i ji jσ σ− ⋅ =n n

This implies that if the principal stresses are distinct, i.e., σi σj, then

0i j⋅ =n n

which means that ni and nj are perpendicular. The three planes on which the principal stresses act are mutually perpendicular. There are three cases for principal directions: a) σ1, σ2, and σ3 are distinct ⇒ principal directions are three unique mutually orthogonal unit vectors. b) σ1 = σ2 and σ3 are distinct ⇒ n3 is a unique principal direction, and any two orthogonal directions on the plane that is perpendicular to n3 are principal directions. c) σ1 = σ2 = σ3 ⇒ any three orthogonal directions are principal directions. This state of stress corresponds to a hydrostatic pressure.

Example 1.5 Calculate the principal direction corresponding to σ3 = –2 of Example 1.4. By substituting λ = –2 into eigenvalue problem in Eq. (1.14), we obtain the following simultaneous equations:

5 0

2 2 0

2 2 0

x y z

x y z

x y z

n n n

n n n

n n n

+ + =

+ + =

+ + =

Even if there are three equations and three unknowns, one cannot solve it because three equations are not independent. In fact, the second the third equations are identical. This situation happens because the principal stresses are chosen in such a way that the matrix equations are dependent each other. As a result, two independent equations exist with three unknowns. Thus, we cannot obtain a unique solution. Let solve the second equation for nx, yielding

2( )x y zn n n= − +


Substituting into the first equation yields

9( ) 0y zn n− + =

Thus, any ny and nz that satisfy ny = –nz can be a solution to the original eigenvalue problem. A unique solution can be obtained if the condition in Eq. (1.18) is imposed, as

0 01 1

1 , 12 2

1 1

= − = −

n or n

Transformation of Stress Let the state of stress at a point be given by the stress matrix [σ]xyz where the components are expressed in terms of the xyz coordinate as shown in Figure 1.8. What are the components of stress at the same point, when expressed in terms of a different coordinate system x′y′z′ (i.e., [σ]x′y′z′)? It will be shown that the stress matrix [σ]x′y′z′ can be obtained by rotating the stress matrix [σ]xyz using a transformation matrix that is obtained from direction cosines.

x

y

z

x′ y′

z′

d1d2

d3

Figure 1.8 Coordinate transformation of stress

Let us define x′y′z′ coordinates system whose unit vectors are given in the xyz coordinate system, as

1 2 31 1 1

1 1 2 2 3 32 2 21 2 33 3 3

, ,

d d d

d d d

d d d

= = =

d d d

For example, d1 = {1, 0, 0}T in x′y′z′ coordinates, while 1 1 1 11 2 3{ , , }d d d=d in xyz coordi-

nates. In fact, these three unit vectors are direction cosines of x′y′z′ coordinates with


respect to xyz coordinates. Using these vectors, let us define the rotational transformation matrix as

1 2 31 1 1

1 2 3 1 2 32 2 21 2 33 3 3

[ ] [ ]

b b b

b b b

b b b

= =

N b b b (1.19)

The matrix [N] transforms a vector in the x′y′z′ coordinates into a vector in the xyz coordinates, while its transpose [N]T transforms a vector in the xyz coordinates into a vector in the x′y′z′ coordinates. For example, a unit vector ax′y′z′ = {1, 0, 0}T in x′y′z′ co-ordinates can be represented in the xyz coordinates as

111 1213

[ ]xyz x y z

b

b

b′ ′ ′

= ⋅ = =

a N a b

Transformation of the stress matrix is more complicated than transformation of a vector. The representation of stress vector in Eq. (1.7) is

1 2 3( ) ( ) ( ) 1 2 3[ ] [ ] [ ] [ ] [ ]xyz xyz xyz= =b b bT T T σ b b b σ N (1.20)

Equation (1.20) represents three stress vectors on the planes perpendicular to b1, b2, and b3 in the xyz coordinates system. However, each stress vector is defined in the xyz coordinates system. Thus, by multiplying the transposed matrix [N]T the stress vectors can be represented in the x′y′z′ coordinates. The stress matrix in the new coordinate system can be obtained by

[ ] [ ] [ ] [ ]Tx y z xyz′ ′ ′ =σ N σ N (1.21)

In Eq. (1.21), matrix [N] is used twice to transform the stress matrix. The first one is used to transform plane information, while the second is used to transform force information.

Example 1.6 The state of stress at a point with respect to the xyz coordinates is

2 1 0

[ ] 1 2 0

0 0 2

=

σ


Determine the stress matrix relative to the x′y′z′ coordinates.

x

y

z, z′

x′ y′

45o

Figure 1.9 Coordinate transformation of Example 1.6

Since the z′-direction is the same as the z-direction, the coordinate transformation is

basically the same as two-dimensional rotation. Let cos45 sin45 1/ 2a = ° = ° = , then the transformation matrix becomes

0

[ ] 0

0 0 1

a a

a a

− =

N

Multiplying its transpose with the stress matrix yields

0 2 1 0 3 3 0

[ ] [ ] 0 1 2 0 0

0 0 1 0 0 2 0 0 2

T

a a a a

a a a a

= − = −

N σ

Finally, the stress matrix at the transformed coordinates becomes

2

2

3 3 0 0 6 0 0 3 0 0

[ ] [ ][ ] 0 0 0 2 0 0 1 0

0 0 2 0 0 1 0 0 2 0 0 2

T

a a a a a

a a a a a

= − − = =

N σ N

Note that the stress matrix is diagonal after transformation, which means that the x′y′z′ coordinates is the principal directions.

Maximum Shear Stress Maximum shear stress plays an important role in the failure criteria of the material. Let σ1, σ2, and σ3 be the three principal stresses at a point such that σ1 7 σ2 7 σ3. The stress status at this point can be described using the Mohr’s circle as in Figure 1.10.


σ

τ

σ1 σ3 σ2

τmax

Figure 1.10 Maximum shear stress

In the above diagram, any point (σ, τ) located in the shaded area represents a plane through the point that has σ and τ as the normal and shear stress components, respectively. The maximum shear stress is

1 3max 2

σ στ −=

Note that σ1 is the maximum principal stress and σ3 is the minimum principal stress (σ1 7 σ2 7 σ3), and σ1 – σ3 is the diameter of the largest circle.

1.2. Strain

Tensile (Normal) Strain When a force is applied to the structure, it deforms its shape. The deformation of the structure is quantitatively measured by strain. Strain is defined as the elongation per unit length. For example, the length of the rectangular component in Figure 1.11(a) increases from ∆x to ∆x + ∆ux. Also, the height in Figure 1.11(b) increases from ∆y to ∆y + ∆uy. In both cases, the tensile (normal) strains in x- and y-directions are defined by

0

0

lim

lim

x xxx

x

y yyy

y

u u

x xu u

y y

ε

ε

∆ →

∆ →

∆ ∂= =∆ ∂∆ ∂

= =∆ ∂

(1.22)


P P P ∆x ∆ux

∆y

∆uy

∆ux

∆uy

θ2

θ1

(a) (b) (c)

π/2 – γ12

Figure 1.11 Definition of strain

Because normal strain is the ratio of two lengths, it is a dimensionless quantity; that is, it has no units. Thus, strain is expressed as a pure number, independent of any system of units.

Shear Strain In general, however, the deformation of the structure cannot fully be represented with two tensile strains. The distortion of the structure, as shown in Figure 1.11(c), can be represented by shear strain. Shear strain is defined as the tangent of the change in angle between two originally perpendicular axes. Since in linear elastic material the amount of deformation is very small, the angles in Figure 1.11(c) can be approximated by

1 1

2 2

tan

tan

y

x

u

xu

y

θ θ

θ θ

∆=

∆∆=∆

∼

∼

Using θ1 and θ2, the change of angle can be measured as

1 2 0 0lim limy yx x

xy x y

u uu u

x y x yγ θ θ

∆ → ∆ →

∆ ∂∆ ∂= + = + = +∆ ∆ ∂ ∂

(1.23)

The angle γxy is a measure of the distortion, or change in shape, of the component and is called the shear strain. The units of shear strain are radians. Instead of shear strain in Eq. (1.23), the following definition of shear strain is often used:

1 1

2 2y x

xy xy

u u

x yε γ

∂ ∂= = + ∂ ∂ (1.24)


Equations (1.22) and (1.24) indicate that the strain components depend linearly on the derivatives of the displacement components. It is important to emphasize that our derivation for the normal and shear strains is valid under the assumption of small displacement derivatives. Thus, the strain measure in this class is limited for small strains and small rotations. Shear strain in a component is positive when the angle between two positive (or two negative) faces is reduced. The strain is negative when the angle between two positive (or two negative) faces is increased. Thus, the shear strain in Figure 1.11(c) is positive and positive shear stresses produce positive shear strains. In general three-dimensional case, the strain matrix can be defined as

[ ]xx xy xz

yx yy yz

zx zy zz

ε ε εε ε εε ε ε

=

ε

As is clear from the definition in Eq. (1.24), the strain matrix is symmetric. Same as the stress vector, the symmetric strain matrix can be represented using a vector notation as

{ }

xx

yy

zz

xy

yz

zx

εεεγγγ

=

ε

where γxy, γyz, and γzx are used instead of εxy, εyz, and εzx.

Deformation (or Displacement) Field The deformation of a solid can be described by specifying the displacement at every point in the solid. Let u(x, y, z) be the displacement vector at a point P(x, y, z). Since this vector is a function of spatial coordinates x, y, and z, it represents a vector field that is referred to as the deformation field or the displacement field. The new position of the point after deformation can be computed as

′ = +r r u

where r is the position vector of the original location of point P and r′ id the location of the final position of the point after deformation has occurred (see Figure 1.12).


r′

r

u

y

z

n

P

x

Figure 1.12 Deformation and strain

Assuming n is an arbitrary unit vector, the component of displacement along this direction is

nu = ⋅u n

The normal strain in the direction of n can be defined as

nnn

u

nε ∂=

∂

If the state of strain is known at a point, the normal strain at that point in the direction of n can be computed as

[ ]nnε = ⋅ ⋅n ε n

Transformation of Strain Let the state of strain at a point be given by the strain matrix [ε]xyz where the components are expressed in terms of xyz coordinates shown in Figure 1.8. What are the components of strain at the same point, when expressed in terms of a different coordinates system x′y′z′ (i.e., [ε]x′y′z′)? Let b1, b2, and b3 be unit vectors along the axes x′, y′, and z′, respectively, represented in xyz coordinates. The strain matrix expressed with respect to the x′y′z′ coordinates system is

[ ] [ ] [ ] [ ]Tx y z xyz′ ′ ′ =ε N ε N

which is the same transformation with the stress matrix as in Eq. (1.21).


Principal Strains Principal strain is the extensional strain on a plane, where there is no shear strain. The principal strains and their directions can be computed as the eigenvalues and eigenvectors of the following eigenvalue problem:

[ ] λ⋅ =ε n n

where n is the eigenvector of matrix [ε]. Same as the stress matrix, the above eigenvalue problem has three solutions ε1, ε2, and ε3 in general, which are called the principal strains. If the principal strains ε1, ε2, and ε3 are known, then the maximum shear strain can be computed as

1 3max 2

ε εγ −=

where ε1 7 ε2 7 ε3 are the principal strains.

Stress vs. Strain Stress and strain in the previous two sections share many similar properties. Table 1.2 summarizes these common properties between stress and strain.

Table 1.2 Comparison of stress and strain

[σ] is a symmetric 3×3 matrix [ε] is a symmetric 3×3 matrix Normal stress in the direction n is

[ ]nnσ = ⋅ ⋅n σ n Normal strain in the direction n is

[ ]nnε = ⋅ ⋅n ε n

Transformation of stress [ ] [ ] [ ] [ ]T

x y z xyz′ ′ ′ =σ N σ N Transformation of strain [ ] [ ] [ ] [ ]T

x y z xyz′ ′ ′ =ε N ε N

Three mutually perpendicular principal directions and principal stresses can be computed as eigenvalues and eigenvectors of the stress matrix as [ ] λ⋅ =σ n n

Three mutually perpendicular principal directions and principal strains can be computed as eigenvalues and eigenvectors of the strain matrix as [ ] λ⋅ =ε n n

1.3. Stress-Strain Relationship An applied load to the structural component develops distributed stress within the component to satisfy force equilibrium. However, this stress can only be generated by changing the shape of the component, which is defined as strain. Thus, intuitively, there must be a relationship between stress and strain. Even if the actual relation between stress and strain is complicated, it is often simplified to meet engineering purposes. Linear


elastic relationship is the simplest and most dominantly used one in the engineering applications, which is the topic of this section.

Uni-axial Stress A cylindrical test specimen is loaded along its axis as shown Figure 1.13. This type of loading ensures that the specimen is subjected to uni-axial state of stress so that if the z-axis is along the axis of the specimen σzz = F/A, and all other components of stress are zero. Here F is the axial force applied to the specimen and A is the cross-sectional area of the specimen. For small values of the applied load F, the specimen deforms elastically so that the specimen returns to its original length after the load is released. In this range of deformation, the deformation of the specimen varies linearly with the load so that stress-strain relation is linear, i.e.,

zz zzEσ ε= (1.25)

where proportional constant E is the Young’s modulus of elasticity.

FF

Figure 1.13 Uniaxial Tension Test

For uni-axial loading (one-dimensional stress) when the specimen extended in the z-direction due to the applied load, it also shrinks in the x- and y-directions because the total volume of the structure remains constant. The ratio between the tensile strain in the z-direction and the shrinking in x- and y-directions is called the Poisson’s ratio. Since the dimensions in x- and y-axes decrease, εxx and εyy are negative and defined by

xx yy zzε ε νε= = −

where ν is the Poisson’s ratio. If the stress-strain relation of the uni-axial tension test in Figure 1.13 is plotted, then a typical ductile material may show a behavior as in Figure 1.14. The explanation of terms in the figure is summarized in Table 1.3.


Table 1.3 Explanations of uniaxial tension test

Terms Explanation Proportional limit

The greatest stress for which the stress is still proportional to the strain

Elastic limit The greatest stress without resulting in any permanent strain on release of stress

Yield stress The stress required to produce 0.2% plastic strain Strain hardening A region where more stress is required to deform the material Ultimate stress The maximum stress the material can resist Necking Cross section of the specimen reduces during deformation Fracture Material failure

Proportional limit

Yield stress

Ultimate stress

Strain hardening

Necking

Fracture

σ

ε

Figure 1.14 Stress-strain diagram for a typical ductile material in tension

Since after the material yields the shape of the structure permanently changes, many engineering applications are involved in designing the structure so that the maximum stress is smaller than the yield stress of the material. Under this range of the stress, the stress-strain relation can be approximated by linear relation.

Linear Elastic Relationship (Generalized Hooke’s Law) The one-dimensional stress-strain relation in the previous section can be extended to the three-dimensional state of stress. When the stress-strain relation is linear, the relationship between stress and strain can be written as

{ } [ ] { }= ⋅σ C ε (1.26)

where


11 12 13 14 15 16

21 22 23 24 25 26

31 32 33 34 35 36

41 42 43 44 45 46

51 52 53 54 55 56

61 62 63 64 65 66

{ } , [ ] , and { }

xx xx

yy yy

zz zz

xy xy

yz yz

xz

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

c c c c c c

σ εσ εσ εσ γσ γσ

= = =

σ C ε

xzγ

The [C] is called the stress-strain matrix. For isotropic materials the 36 constants in the matrix [C] is reduced to two. Thus, it can be shown that

1 0 0 01 1

1 0 0 01 1

1 0 0 01 1(1 )

[ ] 1 2(1 )(1 2 ) 0 0 0 0 02(1 )

1 20 0 0 0 0

2(1 )

1 20 0 0 0 0

2(1 )

E

ν νν ν

ν νν ν

ν νν νν

νν νν

νν

νν

− − − −

− − −= −+ − −

−

− −

−

C

where E is the Young’s modulus and ν is Poisson’s ratio. Alternately, strain can be written as a function of stresses as follows:

( )

( )

( )

xxxx yy zz

yyyy xx zz

zzzz xx yy

xyxy

yzyz

xzxz

E E

E E

E E

G

G

G

σ νε σ σ

σ νε σ σ

σ νε σ σ

τγ

τγ

τγ

= − + = − + = − + = = =

(1.27)


where G is the shear modulus defined by

2(1 )

EG

ν=

+

Example 1.7 The stress at a point in a body is

3

5 3 2

[ ] 3 1 0 10 psi

2 0 4

= − ×

σ

Determine the strain components in the x-, y-, and z-direction if E = 10×106 psi, and ν = 0.3. From Eq. (1.27),

3 46

3 46

3 46

46

6

46

1[5 0.3( 1 4)] 10 4.1 10

10 101

[ 1 0.3(5 4)] 10 3.7 1010 10

1[4 0.3(5 1)] 10 2.8 10

10 102(1 0.3)

3000 7.8 1010 102(1 0.3)

0 010 102(1 0.3)

2000 5.2 1010 10

xx

yy

zz

xy

yz

xz

ε

ε

ε

γ

γ

γ

−

−

−

−

−

= − − + × = ××

= − − + × = − ××

= − − × = ××+= = ××+= =×+= = ××

Simplified Laws for Two-Dimensional Analysis The general three-dimensional stress-strain relation in Eq. (1.26) can be simplified when the stress or strain is limited. One-dimensional stress-strain relation has been studied in Eq. (1.25). Two-dimensional stress-strain relation can be categorized into two cases: plane stress and plane strain. Plane stress problem can be applied to a structure whose dimension of one axis is very small compared to other two dimensions, for example, a thin sheet metal. In such a structure, the load and stress in z-direction are zero. Thus, the following constraints are applied to the general stress-strain relation:

0, 0, 0zz xz yzσ τ τ= = =


From the relation in Eq. (1.26), it is clear that γxz = γyz = 0. In addition, from the relation of σzz = 0 the normal strain εzz can be calculated. Thus, a simplified relation for stress-strain can be obtained for the plane stress problem as

2

1 0

{ } 1 01

0 0 (1 ) / 2

xx xx

yy yy

xy xy

Eσ ν εσ ν ε

ντ ν γ

= = − −

σ (1.28)

Plane strain problem is similar to the plane stress problem. However in this case, the strain in the z-direction is assumed to be zero. This situation corresponds to a structure whose deformation in the z-direction is constrained (i.e., uz = 0), so that the following relation holds:

0, 0, 0zz xz yzε ε ε= = = (1.29)

In the plane strain problem, a different stress-strain relation can be obtained by applying the constraints in Eq. (1.29) into Eq. (1.26), to obtain

1 01

(1 ){ } 1 0

(1 )(1 2 ) 11 2

0 02(1 )

xx xx

yy yy

xy xy

E

ννσ ε

ν νσ εν ν ν

τ γνν

−

− = = + − − − −

σ

Note that the normal stress σzz is not zero in the plan strain problem, but can be calculated from εxx and εyy.

Equilibrium Equations Consider the equilibrium of a differential element represented in two-dimension by a square (see Figure 1.15) whose center is (x, y) and its sides are dx and dy, respectively.


x

y

2dxxx x

σ+

2dxxx x

σ−

2dyyy y

σ+

2dyyy y

σ−

2dyyx y

τ+

2dyyx y

τ−

2dxxy x

τ+

2dxxy x

τ−

Figure 1.15 Stress variations in infinitesimal components

Equilibrium in the x-direction yields the following equation:

2 2 2 2

0dx dx dy dyxx xx yx yxx x y ydy dy dx dxσ σ τ τ

+ − + −

− + − = (1.30)

If the first-order Taylor series expansion is used to represent stresses on the surface in terms of stresses at the center, the first two terms in Eq. (1.30) can be approximated by

2 2 2 2

xx xx xxdx dxxx xx xx xxx x x x

dx dxdy dy dy dy dxdy

x x x

σ σ σσ σ σ σ+ −

∂ ∂ ∂ − = + − − = ∂ ∂ ∂

Similarly, the last two terms can be approximated by

2 2 2 2

yx yx yxdy dyyx yx yx yxy y y y

dy dydx dx dx dx dxdy

y y y

τ τ ττ τ τ τ

+ −

∂ ∂ ∂ − = + − − = ∂ ∂ ∂

By substituting these two equations into Eq. (1.30), we obtain an equilibrium equation in the x-direction as

0yxxx

x y

τσ ∂∂ + =∂ ∂

(1.31)

Similarly, equilibrium in the y-direction yields the following equation:

0xy yy

x y

τ σ∂ ∂+ =

∂ ∂ (1.32)


Equations (1.31) and (1.32) are the equilibrium equations for a solid subjected to two-dimensional state of stress. We can similarly derive the equations for a three-dimensional state of stress, by considering the equilibrium of a three-dimensional differential element to obtain,

0

0

0

yxxx zx

xy yy zy

yzxz zz

x y z

x y z

x y z

τσ τ

τ σ τ

ττ σ

∂∂ ∂+ + = ∂ ∂ ∂∂ ∂ ∂ + + = ∂ ∂ ∂ ∂∂ ∂+ + =

∂ ∂ ∂

(1.33)

Equation (1.33) is obtained by considering force equilibrium in the x-, y-, and z-directions. As has been shown in Eq. (1.10), moment equilibrium yields symmetry of stress matrix.

Boundary Value Problem Consider an arbitrary body subjected to the certain forces and constraints are shown in Figure 1.16. The deformation of this object under applied loads can be fully determined by solving the deformation field.

( , , ) ( , , ) ( , , ) ( , , )x y zx y z u x y z u x y z u x y z= + +u i j k

Boundary condition

Applied loads

{σ} = [C] {ε}

Figure 1.16 Boundary value problem

The deformation behavior of a solid is governed by the equations that were described in the previous sections and the knowledge of the applied forces and boundary conditions. Such a problem, defined by partial differential equations and boundary conditions is called a boundary value problem.


a) Equilibrium equation in Eq. (1.33) b) Constitutive equations (stress-strain relation) in Eq. (1.26) c) Strain definitions (Section 1.2) d) Loads and boundary conditions (applied forces and displacement constraints)

Example 1.8 The displacement field for the thin beam shown in Figure 1.17 considering bending only is

23

2 32

( , ) ( )2 6

( , ) ( ) ( )2 2 6

x

y

P x Pu x y Lx y y

EI EIP P Lx x

u x y L x yEI EI

ν

ν

= − −

−= − − −

where P is the applied force at the tip, I is the area moment of inertia about bending axis, and L is the length of the beam. Determine the entire stress field.

x

y L P

Figure 1.17 Cantilever beam bending problem

Since the thickness of the beam is small, we can assume the plane stress condition along the z-direction. From the definition of strain,

2 2 2 2

( )

( )

( ) ( ) 02 2 2 2

xxx

xyy

y xxy

u PL x y

x EIu P

L x yx EIu u Py P x P x Py

Lx Lx yx y EI EI EI EI

ε

νε

ν νγ

∂= = −∂∂ −= = −∂∂ ∂= + = − − + − − = ∂ ∂

Substituting into Eq. (1.28) yields the stress field:


2

2

2

( ) ( ) ( )1

( ) ( ) 01

0

xx

yy

xy

E P P PL x y L x y L x y

EI EI I

E P PL x y L x y

EI EI

νσν

ν νσν

τ

= − − − = − −

= − − + − = − =

1.4. Exercise 1.1 A vertical force F is applied to a two-bar truss as shown in the figure. Let cross- sectional areas of trusses 1 and 2 be A1 and A2, respectively. Determine the area ratio between trusses 1 and 2 in order to have the same magnitude of stress.

lC B

A

45°

F

1.2 Find the principal stresses and the orientation of the principal axes of stresses for the following cases of plane stress. (a) σxx = 40 MN/m2, σyy = 0, τxy = 80 MN/m2 (b) σxx = 140 MN/m2, σyy = 20 MN/m2, τxy = −60 MN/m2 (c) σxx = −120 MN/m2, σyy = 50 MN/m2, τxy = 100 MN/m2 1.3 If the minimum principal stress is −7 MN/m2, find σxx and the angle which the principal stress axes make with the xy axes for the case of plane stress illustrated


x

y

21 MN/m2

56 MN/m2

σxx

1.4 At a point P in a body, Cartesian stress components are given by σxx = 8,000 N/cm2, σyy = −4,000 N/cm2, σzz = −4,000 N/cm2 and τxy = τyz = τzx = 8,000 N/cm2. Determine the traction vector, its normal component and shear component on a plane that is equally inclined to all three axes. 1.5 Determine the principal stresses and their associated directions, when the stress matrix at a point is given by

1 1 1

[ ] 1 1 2 MPa

1 2 1

=

σ

1.6 Let x′y′z′ coordinate system is defined using the three principal directions obtained from Exercise 1.5. Determine the transformed stress matrix [σ]x′y′z′ in the new coordinates system. 1.7 If the displacement field is given by

2 2

2

2

2

2 ( )

2

x

y

z

u x y

u y x y z

u z xy

= + = − − − = − −

(a) Write down 3×3 strain matrix. (b) What is the normal strain component in the direction of (1,1,1) at point (1,–3,1)? 1.8 For steel, the following material data are applicable: Young’s modulus E = 207 GPa and shear modulus G = 80 GPa. For the strain matrix at a point shown below, determine the symmetric 3×3 stress matrix.


0.003 0 0.006

[ ] 0 0.001 0.003

0.006 0.003 0.0015

− = − −

ε

1.9 A long, thin plate of width b, thickness t, and length L is placed between two frictionless rigid walls a distance b apart and is acted on by an axial force P. The material properties are Young’s modulus E and Poisson’s ratio ν. (a) Find the stress vector in the xyz coordinate system (b) Find the strain vector in the xyz coordinate system (c) Find the displacement field

L

P P

y

x

y

z b t

1.10 Draw nine stress components with their positive directions in the figure below.

x y

z


1.11 For the stress matrix below, the two principal stresses are given as σ3 = –3 and σ1 = 2, respectively. In addition, two principal directions corresponding to the two principal stresses are also given below. The yield stress of the structure is given as σY = 4.5.

1 0 2

[ ] 0 1 0

2 0 2

= −

σ , 1

1

50

2

5

= −

n and 3

2

50

1

5

=

n

(a) What is the normal and shear stress in the plane whose normal vector is parallel to (2, 1, 2)? (b) Calculate the missing principal stress σ2 and the principal direction n2. (c) Write stress matrix in the new coordinates system that is aligned with n1, n2, and n3. 1.12 With respect to the coordinate system xyz, the state of stress at a point P in a solid is:

[ ]20 0 0

0 50 0

0 0 50

MPaσ− =

x

y

z

m1

m2

m3

P

(a) m1, m2 and m3 are three mutually perpendicular vectors such that m1 makes 45º with both x- and y-axes and m3 is aligned with the z-axis. Compute the normal stresses on planes normal to m1, m2, and m3.


(b) Compute two components of shear stress on the plane normal to m1 in the directions m2 and m3. (c) Is the vector n = {0, 1, 1}T a principal direction of stress? Explain. What is the normal stress in the direction n? (d) Draw an infinitesimal cube with faces normal to m1, m2 and m3 and display the stresses on the positive faces of this cube. (e) Express the state of stress at the point P with respect to the x′y′z′ coordinates system that is aligned with the vectors m1, m2 and m3? (f) What are the principal stress and principal directions of stress at the point P with respect to the x′y′z′ coordinates system? Explain. (g) Compute the maximum shear stress at the point P. Which plane(s) does this maximum shear stress act on? 1.13 Figure 1.18 illustrates a thin plate of thickness t. An approximate displacement field which accounts for displacements due to the weight of the plate is given by

2 2( , ) (2 )2

( , ) ( )

x

y

u x y bx x yE

u x y y b xE

ρ ν

νρ

= − −

= − −

(a) Determine the corresponding plane stress field. (b) Qualitatively draw the deformed shape of the plate. (c) Determine the rotation of the plate at points A and B.

A B

y

x a a

b


Figure 1.18 Thin plate under self weight

1.14 The stress matrix at a particular point in a body is

7

2 1 3

[ ] 1 0 4 10 Pa

3 4 5

− − = × −

σ

Determine the corresponding strain if E = 20×1010 Pa and ν = 0.3. 1.15 Carry out the tutorial in I-DEAS S/W to create a part shown below. The tutorial can be found in I-DEAS menu: Help-Help Library-Tutorials. In the tutorial browser, choose “Fundamentals” in “Design, Part Modeling”. In the Fundamental Skills browser, you have to carry out items 18 and 19 to create a part shown below. Submit a picture of the part (Color or Black&White) as a proof of the tutorial.

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