KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS

Chapter 8 KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS Figure 8.1: 195

196 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS 8.1 Motivation In Chapter 3, the conservation of linear momentum for a static, stead-state, solid continuum bod resulted in three equations relating the Cauch stress components: -component: -component: z-component: 0 = ρg + σ 0 = ρg + σ 0 = ρg z + σ z + σ + σ + σ z + σ z + σ z + σ zz For a solid, these equations are often referred to as the equilibrium equations since the provide a statement of equilibrium that the internal forces (which produce internal stresses) within a continuum bod must satisf. We note again that equation provides 3 equilibrium equations that are in terms of 6 unknown stress components σ, σ, σ zz, σ = σ, σ z = σ z, and σ z = σ z, provided that ρ and g, g, g z are known. Therefore, in the same manner as for fluids and heat transfer, the differential equations of equilibrium alone cannot be used to solve for the unknown stresses; in fact, additional equations are required to affect a solution. These additional equations (kinematic equations and constitutive equations) will be considered in this and the following chapter. When the process of developing these additional sets of equations (kinematics and constitution) has been completed, we will have the equilibrium equations in terms of stresses, the constitutive equations relating stresses to strains (deformation gradients) and the kinematic equations that relate strains to displacements. As a result of finall combining the 3 equation sets, we will be able to determine the displacements of a solid continuum bod subjected to loading on the bod. 8. Kinematics In this chapter, we wish to describe the kinematic behavior of a solid continuum bod b defining quantities called strains in terms of the gradients of displacement components. Kinematic relations are epressions that define the motion of a bod. For a rigid bod, kinematic relations are epressions that can be written b considering the geometr of the bod and its motion. In ENGR 11, kinematic relations were utilized to relate various variables; for eample, the tangential and angular velocities were related b v = rω in the stud of a rotating rigid bod about a fied point. For rope-pulle sstems, epressions were found which related the velocit of one portion of the sstem to another; these relations were a result of the geometr of the pulle sstem and the assumption that the rope was inetensible. In this chapter, we are interested in the continuum bod that is deformable. In the eample below, we consider an elastic bar as shown in Figure 8.a. If the bar is subjected b an aial force F, it will stretch an amount δ as shown in Figure 8.b. The quantit δ L is a measure of the change in length relative to the original length and is defined to be the aial strain for the bar. In Figure 8.d, a shear load is applied that is parallel to the top surface as shown. The angle θ measures the amount the original angle in Figure 8.c has changed from a right angle. This angle θ is related to the shear strain. In this chapter we will mathematicall formalize these simple ideas to develop epressions for strains in terms of displacement components. 8.3 Kinematics of a Deformable Solid The construction of kinematic relations for a deformable bod will involve the mathematical description of how two points within the bod move relative to each other when eternal loads are applied to the bod. In order to describe the motion of a point within the bod, we will define the initial position of a material point in the undeformed bod and the final position of the same point in the deformed bod. The relative position of the material point from initial to final position will (8.1)

8.3. KINEMATICS OF A DEFORMABLE SOLID 197 L a) undeformed 90 c) undeformed F F θ F L + δ b) stretched (deformed) d) sheared (deformed) Figure 8.: Eample of Aial and Shear Deformation define its displacement. B defining two material points on the initial bod that are separated b some distance and comparing to the same two material points in the deformed bod, we will be able to define how much the vector connecting the two points has changed in magnitude and direction. In the figure below ever point in the bod during its initial undeformed state can be located relative to some coordinate sstem. Consider two points P and Q as shown below where P is located at position r and Q is located at r+ r. The position vector for P is given in terms of components b r = i+j+zk, and the position vector for Q is given b r+ r =(+ )i+( + )j+(z + z)k. As forces are applied to the bod, the bod reaches the final deformed state where these two points Undeformed Bod u (r + r) Deformed Bod Q Q P r u (r) P r r r z Figure 8.3: Position and Displacements Vectors for Deformable Bod are now located at P and Q. The relative position between Q and P is given b r, and the relative position between P and Q is r. The new position vectors are r = i + j + z k and

198 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS r + r =( + )i +( + )j +(z + z )k. Wedescribe the displacement of point P in moving from position P to P b the vector u(r). From the figure we see that the displacement u is the result of subtracting the position vector of original point, r, from r. The displacement of Q can be similarl described. Utilizing the position and displacement vectors shown on the figure, we can write the following: displacement vector of P to P : u(r) =r r (8.) displacement vector of Q to Q : u(r + r) (r + r ) (r + r) =u(r)+ r r (8.3) Note that the displacement vector has three components that can be written as u (,, z )= (,, z ) u (,, z )= (,, z ) (8.4) u z (,, z )=z (,, z ) z From vector calculus we now use the following definition of the gradient of a vector function for r 0 u(r + r) =u(r)+ r u (8.5) where the displacement gradient u has the following matri representation. [ u] = u u u u u u u z u z u z Equation will then reduce to the following equation after substituting from (8.6) r = r + r u (8.7) which epresses the deformation of material elements r in terms of r (original element) and the displacement gradient. 8.4 Definition of Strain Strain is a measure of the relative change in length and rotation of material segments. Consider two material elements of length r 1 and r undergoing deformations that will bring them into new locations r 1 and r, respectivel, as shown in the figure below. A vector operation between these two neighboring material element vectors that measures changes in length and relative rotation is the dot product. Consider therefore taking the dot product of these two vectors in the deformed state to obtain the following scalar result: Thus, r 1 r = [ r 1 + r 1 u] [ r + r u] = r 1 r + r 1 [ r u]+[ r 1 u] r +[ r 1 u] [ r u] = r 1 r + r 1 [ u + u T + u u T ] r }{{} E r 1 r = r 1 r + r 1 E r

8.4. DEFINITION OF STRAIN 199 Undeformed State Deformed State r 1 r r 1 u (r,t) r r r We alread know from equation (8.7): r 1 = r 1 + r 1 ( u) r = r + r ( u) z Figure 8.4: Vector Definitions Used in Strain where the quantit E that measures the deviation of the dot product between two deformed material elements from the dot product between the same material element, in the undeformed state, is defined to be the finite strain measure: E = 1 ( u + u T + u u T ) (8.8) From our discussions of vector calculus, we note that the quantit u is represented b a 3 3 matri and the notation u T indicates the transpose of the 3 3 matri. In this tet, we will not deal with the finite strain measure but onl with the infinitesimal stress that will be introduced net. In looking at equation 8.8, we note that it contains two terms that are linear in the displacement gradient u and the third term that is quadratic in the displacement gradient vector. For small displacement gradients, it is reasonable to neglect the higher order term so that u u T 0 for small displacement gradients B neglecting the higher order terms from the finite strain measure, we obtain the infinitesimal strain tensor given b: ε = 1 ( u + u T ) (8.9) The matri representation of strain can be written in terms of its components in Cartesian coordinates: ( ) u 1 u ε ε ε z + ( u 1 uz + ) u ( ) ( ) [ε] = ε ε ε z 1 = u + u u 1 uz + u (8.10) ε z ε z ε zz ) ) 1 ( u + uz 1 ( u + uz We make the following comments regarding the finite and infinitesimal strain tensors: B observation, [ε] =[ε] T and [E] =[E] T, i.e. both the finite and infinitesimal strains are represented b smmetric matrices. u z

00 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS Strain defines the relative deformation of material elements. B definition, it accounts for the change in the relative angle between two material elements and the change in length from the initial to the deformed state. ( r 1 r = cos θ r 1 r ) ( r 1 r = cos θ r 1 r ) (8.11) r 1 θ r 1 θ r r Figure 8.5: Onl when E = 0, the two dot products remain the same, i.e. r 1 r = r 1 r and therefore the angle θ and the lengths of the material elements remain the same. Since the dot product between two vectors is invariant under rigid bod rotation or translation, the finite strain E is also invariant under rigid bod motion of the deformed state. (CAUTION : ε is NOT invariant under rigid bod motion of the deformed state.) The derivations above are for the deformation of a bod from a reference to a deformed configuration and the coordinate sstem is defined in the reference state. However b assuming infinitesimal deformations, the difference between reference and deformed coordinates is neglected. In this chapter we make no distinction between Eulerian and Lagrangian descriptions onl because we deal with infinitesimal strains. 8.5 A -D Approach to Strain Consider a more intuitive -D geometrical approach to define ε (defined to be the change in length of a line segment which is originall oriented in the direction and undergoes displacements u and u )asshown below: First, define the strain of a line segment in the direction to be: ε = = change in length of infinitesimal line segment original length of line segment (8.1) As before, point P is located at position and point Q is a position r + r. Define the displacement of point P to be u () and u () inthe and directions respectivel, and that of Q as u ( + ) and u ( + ). From these definitions and the sketch above, we can write the following: = = [ { + u ( + ) u ()} + {u ( + ) u ()} ] 1 [ { + u = [ { 1+ u } + } { + u } ] 1 [ { ( = 1+ u [ { } ] 1 u = 1+ u + )} { + u } ] 1 (8.13) ( ) u + ( ) ] 1 u

8.5. A -D APPROACH TO STRAIN 01 Q P u ( + ) u () u () u ( + ) P Q ( + ) Figure 8.6: Two-Dimensional Geometr for Strain We make use of the series approimation 1+a = 1+ 1 a (for small a). Thus, the last result is approimatel: The strain becomes or [ = ε = 1+ u + 1 [ 1+ u + 1 ε = u + 1 ( ) u + 1 ( u ) + 1 ( ) u + 1 ( ) ] u ( ) ] u For small gradients, the quadratic terms are neglected and we obtain u (8.14) (8.15) ( ) u (8.16) ε = u (8.17) Shear strain is a measure of the rotation of line segments that form a right angle in the undeformed bod as shown below. The shear strain is) defined geometricall as the average of the two rotations. ( u u and so that ε = 1 + u Geometricall, each of the two terms is the tangent of an angle as shown above. ε is called a shear strain and geometricall is 1 (average) of the angular rotation of line segments and which originall form a right angle. In contrast, the engineering shear strain γ is defined as the sum of these two angles, i.e., γ =ε = u + u. The definition of the engineering shear strain γ from a graphical viewpoint is an approimation (similar to the square root approimation made in ε. From the geometr above, d d = cos θ d d. Note that from the geometr in the above sketch, we can show that: cos θ = cos ( π γ ) = sin γ γ = u + u =ε.

0 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS u u θ = π θ u u Figure 8.7: Two-Dimensional Geometr for Shear Strain As in the square root approimation made for ε (for the geometrical interpretation of strain), an assumption of small rotations has been made in defining the shear strain ε. 8.6 Visualization of the Infinitesimal Strain in 3-D The infinitesimal strain tensor is given in terms of displacements b equation (8.9). Based on the derivation of strain ε and ε in the previous section, each term in the strain tensor represents either aial etension of a line segment (related to aial strain) or rotation of a line segment (related to shear strain). The three diagonal terms are referred to as aial or normal strains and represent the stretching of a material in each of the three coordinate directions. The three terms above the diagonal are referred to as shear strains and represent the shearing of a material in each of the three planes formed b the --z aes. To see this more clearl, consider the diagram below: The upper left-hand term in the matri (or the 1,1 term) represents aial etension of a line segment in the direction when subjected to an aial force in the direction, and is the aial strain ε. The (,3) term represents the change in the right angle, or shearing of a square ling in the -z plane, and is the shear strain ε z. The (,) term represents the stretching of a line originall in the direction when subjected to an aial force in the direction and is called the aial strain ε.together, the 6 independent components of the strain tensor represent all possible deformation states of a material point, i.e., aial strain in the, and z direction and shear strain in -, -z, and -z planes. 8.7 Strain Transformation In Chapter 5, a stress transformation was developed which allowed us to determine the normal and shear components of stress on a surface with unit normal n. The normal component is given b (see Equation (5.5)): (1 )( )( 1) σ n = σ = n σ n = [n] [σ] [n] = σ cos θ +σ sin θ cos θ + σ sin θ (1 )( )( 1) The epression σ n = σ = n σ n = [n] [σ] [n] gives the component of stress, σ n,inthe direction of the unit normal (or in the direction of the -ais which makes an angle θ to the -ais).

8.7. STRAIN TRANSFORMATION 03 d d γ γ z z d ε = d d d z ε = γ z ε z = γ z [ε] = d d d γ z z ε = d d d z ε z = γ z SYMMETRIC z dz dz dz ε zz = dz dz dz Figure 8.8: We can perform eactl the same set of transformations on strain to obtain the strain transformation given b: (1 )( )( 1) ε n = ε = n ε n = [n] [ε] [n] (8.18) = ε cos θ +ε sin θ cos θ + ε sin θ The quantit ε n is the component of strain in the direction of a unit normal n. ε n is often called the unit elongation in the n direction (just as ε is the unit elongation in the i or -coordinate direction). It should be noted that the strain transformation applies to the strain tensor [ε] and is not applicable to engineering shear strains unless the engineering strain components are first converted to tensor strain components. In a manner similar to that done in Chapter 5, we can also show that the shear compnent of (1 )( )( 1) strain is given b ε = n ε n = [n] [ε] [n ]= (ε ε ) sin θ cos θ + ε (cos θ sin θ). Since the tensor strain transformation equations are identical to the tensor stress transformation equations, we can also define a Mohr s circle for strains. In appling Mohr s circle for strains, we follow eactl the same steps as was done for stress. We leave this topic for the student s net course in structural mechanics. We note in passing the following regarding second-order tensors:

04 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS t (n) n n t (n) σ θ σ σ = τ s σ = σ n σ σ Figure 8.9: 1. Both [σ] and [ε] are second-order tensors.. All second-order tensors follow the same transformation form in transforming from (,, z ) to another Cartesian coordinate sstem (,,z ), i.e., σ n = σ = n σ n and ε n = ε = n ε n are identical transformations. 3. The same transformation applies to another second-order tensor, the moment of inertia of a cross-section A. For cross-section like that shown below, we can define moments of inertia as follows: I = da, I = da, I = da A A With respect to the - coordinate sstem at some angle θ, the moment of inertial about the -ais is defined b: I = ( ) da A A θ Figure 8.10: However, since moments of inertia are second-order tensors, we can also obtain I b appling the coordinate transformation to the - moments of inertia: (n is unit vector in direction).

8.7. STRAIN TRANSFORMATION 05 The moment of inertia tensor is defined b [ I I [I] = I I ] and the tensor transformation leads to I =[n][i][n] Eample 8-1 Given: u = k( + ) u = k( ) u z = 0 d d 1 k d 1 k 1 d k 1+k Figure 8.11: Required : Evaluate [ u] and [ε]. First we have the displacement gradient: [ε] = 1 ( u + u T ) = 1 [ u] = k k 0 k k 0 0 0 0 Solution k k 0 k k 0 0 0 0 + Eample 8- k k 0 k k 0 0 0 0 = k k( + ) 0 k( + ) k 0 0 0 0

06 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS Given: = k where: k =5 10 5 = + kz + k z = (1+k)z Required : (a) Determine the components of the infinitesimal strain tensor, ε. (b) Find the elongation per unit length of an element initiall in the direction of i +j. Solution a.) [ε] = 1 ( u + u T ) = k 1 1 0 k 1 k 1 k 0 k b.) [ ε n = n ε n n = 1 5 5 0 ] ε n = 0.

8.7. STRAIN TRANSFORMATION 07 Deep Thought Strain and Stress are two universal comcepts - everbod understands them, but nobod knows which one comes first.

08 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS 8.8 Problems 8.1 GIVEN : = k where: k =7 10 5 = 3 +4kz +k z = (5+k)z REQUIRED: a) Determine the components of the infinitesimal strain tensor, ε. b) Find the elongation per unit length of an element initiall in the direction of i, j and k. 8. GIVEN : u = + where: k =1 10 4 u = k( + ) u z = z REQUIRED: a) Determine the components of the infinitesimal strain tensor, ε. 8.3 GIVEN : [ε] = 4 3 3 1 0 0 1 10 5 ksi REQUIRED: a) Determine the elongations per unit length of two elements initiall in the directions of i and j. b) Determine the change in angle between the two material elements in a). 8.4 GIVEN : The unit elongations at a certain point on the surface of a bod are measured eperimentall b means of strain gages that are arranged to be 45 apart (referred to as a 45 strain rosette gauge) in the directions i, (i + j) and j where the measured elongations are designated b a, b and c, respectivel. REQUIRED: a) In terms of a, b, and c, determine the strain components ε,ε,ε b) If a =50 10 5,75b = 10 5, c = 100 10 5, determine the numerical values of ε,ε,ε 8.5 GIVEN : REQUIRED: u = 5 u = sin u z = cos z a) Determine the components of the infinitesimal strain tensor, ε.

8.8. PROBLEMS 09 b) How will the infinitesimal strain tensor components var on the surface of a clinder given b + z = a, where a is the radius. Assume the clinder is centered at the origin. 8.6 For an isotropic plate, the displacement components are found to be u (,, z ) = 0.00004 0.000 u (,, z ) = 0.00035 3 0.0006 u z (,, z ) = 0 a) Determine ε, ε and ε zz. b) Evaluate ε at the coordinates ( 0.6, 0.5, 0.3). c) Is this problem plane stress? Wh? 8.7 A rectangular piece of rubber has been deformed into the shape shown below b the dashed lines. Determine the average normal strain along the diagonal CB and side AC. Average normal strain is given b: ε ave = l l o where l is the length of a line. 3mm C D 00 mm 4mm A 150 mm B Problem 8.7 8.8 A particle in a elastic bod is originall located at position (,, z )inacartesian coordinate sstem. The final position (,,z )ofthe point (after deformation of the bod) is given b (assume units of inches): = k where: k =7 10 5 = 3 +4kz +k z = (5+k)z Note that if ou know the final position and initial position of a particle, then ou know its displacement. Refer to the notes to refresh our memor. a) Determine the components of the infinitesimal strain tensor, ε. b) Find the elongation per unit length of an element initiall in the direction of i, j, and k. 8.9 For an isotropic plate, the displacement components are found to be: u (,, z ) = 0.00004 0.000 u (,, z ) = 0.00035 3 0.0006 u z (,, z ) = 0 a) Determine the components of the infinitesimal strain tensor, ε. b) Evaluate ε at the coordinates ( 0.6, 0.5, 0.3). c) Is this problem plane stress? Wh?

10 CHAPTER 8. KINEMATIC RELATIONS IN DEFORMATION OF SOLIDS