A short review of continuum mechanics


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1 A short review of continuum mechanics Professor Anette M. Karlsson, Department of Mechanical ngineering, UD September, 006 This is a short and arbitrary review of continuum mechanics. Most of this material can be found in a number of text books, including Fung (00).. Stress Figure. Forces on a solid Forces are vectors. The notation used is as follows: F, where F is the force vector, e i is a direction of a coordinate system (,, ) = Fie i i= force component in direction i. Newton s second law states that F = ma where a is the acceleration vector. The equivalent moment equilibrium is given by uler s rule M = mα where α is the angular acceleration vector. x x x, and F i is the Newton s third law: Forces on two opposite crosssections are equal in magnitude but opposite in direction. (Figure. B) Since we are using vectors, it is important to clearly define what coordinate system is used. Coordinate systems only exist in our minds, and are only there to help us solve a problem. Thus, we can select the coordinate system any way we like. In solid mechanics, it is common to select the coordinate system so that on axis is parallel to the surface of interest, and the second is normal to the surface... Definition of stress Consider a small, infinitesimal, material element in a coordinate system spanning the coordinate axes ( x, x, x x ), and where each side of the material element have the sides xi, i = (,,). ach side of the element has the area A i, where i denotes the normal direction of the surface. The cube is subjected to the
2 forces F ij as shown in Fig.., where i corresponds to the surface in where it acts and j to the direction of the force. Stress is defined as Fij σ ij = lim (no sum of i) Ai 0 A i If i = j normal stress and if i j shear stress. In this general case, it we have nine stress components: six normal stresses and six shear stresses. Traction is the stress vector on each surface, that is T = σ, σ, σ i ( ) Figure. : forces and stresses on a finite material element.. The equality of shear stress components Investigate the moment equilibrium of the cube. Start with the moment about the x axis: Σ M x = 0 Moment arm } σ σ + σ + Area + σ σ + x + σ + x = 0 Which can be simplified to
3 σ + x + σ + x = 0 We note that x = V is the volume of the element. Since the volume is finite, we can divide thought with x and get σ + + σ + x = 0 σ Next, let 0, 0, 0. We note that, for example lim = x 0 x σ then = 0. This is true for all mixed terms. Thus, we have: x { { finite 0 which is finite, σ + σ = 0 or σ = σ (.) Similarly, we can show that σ = σ and that σ = σ. It follows that each shear stress components has a companion shear stress. This companion shear stress is equal in magnitude and is acting on the surface which has its normal paralleling the shear stress considered. Try at home: verify that σ = σ and σ = σ.. Stress equilibrium Next, we will study the equation of equilibrium. First, sum all the forces in the x direction: Σ = 0 F x σ σ σ + σ + dx dx dx + σ + σ + dx dx dx x x σ σ + σ + dx dxdx = 0 x Simplifying, and again use that And similarly: σ σ σ x x x + + = σ σ σ x x x + + = 0 x = V 0 0 we get (.a) (.b)
4 σ σ σ + + = 0 x x x (.c) These are the equilibrium equation expressed stress. These can be written more compactly as σ ji = 0 for {i =, and } (.') j= x j Or alternatively, in short form σ ji, j = 0 (.'') Try at home: show that eq. (.b) and (.c) are true!. Strain The displacement of one material particle from its initial location to its current, can be described by the u = u, u, u, where u i is the displacement in the x i direction. displacement vector ( ) Green s Strain Tensor is commonly used in solid mechanics. In this formulation, the current position of a particle is expressed in the initial coordinate systems. The ijcomponent of the strain is expressed as i.e., u j ui uk u k εij = + + (.) xi xj k = xi xj and ε ε u u u u u u u u = x x x x x x x x u u u u u u u u = x x x x x x x x Try at home: write ε and ε. Start with eq. (.)! u u u u = x x x x The Green s strain tensor is used when it is important to update the analysis to incorporate the new position of the structure. Sometimes, this is referred to as large strains or large deformations. Note that large strains are not necessarily the same as large deformation, but in either case, eq. (.) must be used. Luckily, in many problems in solid mechanics the assumption small strains can be made. In this case, the strain and deformation is assumed to be small, so that it is not needed to update for the new position. Higher order terms in eq. (.) is ignored, and we only have: u j u i εij = + xi x j For example, and u u u ε = + = x x x (.4) 4
5 ε u u = + x x Strain compatibility is a set of equations that can be used to verify that a given strain field is valid. It is given by εij εkl ε ik ε jl + + x x x x x x x x or more compact k l i j j l i k εij, kl εkl, ij εik, jl ε jl, ik = = 0 (.5) See for example Fung (00) for how to derive these compatibility equations... A note about strain Strain can be divided into several components. For example, in the one dimensional case, εtot = εf + εt (.6) where ε tot is the total strain, ε F is a strain due to a force, and ε T is a strain due to a temperature change. In this case, the total strain can be zero, resulting in ε F = ε T. This case is achieved when a thin rod is clamed between two ends and the temperature is changed from its initial position. Since the ends cannot move, the total deformation and total strain is zero. But to balance the thermally induced strain, the balancing strain which will be associated with an axial force and stress will be induced. We will revisit and extend this problem later. The thermal strain is sometimes refer to as a stress free strain since the strain itself does is not associated with a force or stress. In the simple rod case mentioned above, there would have the total strain would have equaled the thermal strain if the ends would have been free to move.. Mathematical Formulation of Problems in Thermoelasticity The general problem of thermoelasticity has 5 unknown variables: 6 stress components, 6 strain components and displacements. (We have not shown it here, but in a similar way that only six stress components exist, one can show that only six strain components exist.) To solve the general problem, 5 equations are needed. These are found from Three equilibrium equations: eq. (.) σ ji = 0 for {i =, and } j= x j u j u i Six straindisplacement relations: eq. (.) or (.4) εij = + (small strains) xi x j Six stressstrain relations (constitutive laws). For time independent, linearelastic materials Hook s law is most commonly used, which will be discussed in the following... Linear lastic StressStrain Relationships Hooke s law for a general anisotropic, linear elastic material is given by 5
6 C ( T ) σ = ε α δ (.7) ij ijkl kl ij ij k= l= where α ij is the coefficient of thermal expansion [/ o C], T [ o C] is the change in temperature from the reference configuration ( zero thermal strain: normally the stress free configuration), and δij is Kronecker s delta, defined by for i = j δij = 0 for i j Furthermore, in eq. (.7) C ijkl is an 4dimensional tensor, which include 8 constants. One can show that C = C = C = C, which reduces the number of constants to. Thus, the most general linear, ijkl jikl ijlk klij elastic material has material constants that must be determined through experimental investigations. A triclinic crystal is the crystal that has the lowest degree of symmetry, which requires all constants to be described. More common form of material anisotropy are found in the orthorhombic crystal that requires nine constants, the tetragonal crystal that requires six constants and the cubic crystal that requires three constants. For isotropic, linear elastic materials, only two constants are required. There are several ways to define these, but the most common nomenclature in engineering mechanics is by means of lastic modulus (Young s modulus), [Pa] and Poisson s ratio, ν [unit less]. Hooke s law for isotropic, linear elastic materials is written as: ij ij kk ij ij k = + ν ν ε = σ σ δ + α Tδ (.8) The shear modulus, G, is not an independent constant but can be related to and ν by G = ( ν ) Writing eq. (.8) explicitly we get for i = j = : ε = σ ν( σ + σ ) + α T and for i= ; j = σ ε = G To try at home: verify that the two expressions above are correct... Twodimensional state of stress The full three dimensional state of stress as discussed above can many times be simplified. Here, we will introduce two common approximations: plane stress and plane strain.... Plane Stress Plane stress is an approximation for thin structures. It should not be confused with plate theory. In the plane stress approximation, it is assumed that stress in one plane is observed and since the thickness 6
7 in the third direction is so small compared to all other dimensions, the stress is approximately zero. Thus, in this case σ σ σ 0. q. (.8) can then be rewritten as ε = [ σ νσ ] + α T (.9a) ε = [ σ νσ] + α T (.9b) σ G ε = (.9c)... Plane Strain Plane strain is an exact (not approximate) solution for many cases. It corresponds to the case when there is strain in two planes, and vanishing strain in the third direction. An example is a long and slender bar fixed between two fixed supports. ε = ε = ε = 0. In this case, σ 0, but can be determined from Hooke s law: ε = σ ν( σ + σ ) + α T 0 = Thus, ( ) σ = ν σ + σ α T. (.0) Hooke s law for plane strain is obtained by substituting eq. (.0) into eq (.8) for ε, ε and ε. A generalized form of plane strain is found when the ends are allowed to move, but no distortion or ε x, x, x = const. bending take place, i.e., ( ) Figure : Plane stress and Plane strain Try at home: Determine Hook s law for plane strain. Try at home: Problem Consider a rod, with its length axis in the direction, made from linearelastic material under uniform stress σ = p. Young s modulus is and the Poisson s ratio is ν. The rod is constrained so that no lateral contraction can develop in the direction. The rod is free to move in the direction. Assume that 7
8 the temperature is held constant. Define the effective Young s modulus by = σ / ε and the effective Poisson s ratio by ν = ε / ε. Show that owing to the lateral constraint, one has ν = and ν = ν ν... Transformation of Stresses and Strain (D) Consider the transformation of stresses from one coordinate system ( x, x, x ) to another ( x, x, x ), where the new coordinate system is obtained by a counter clockwise rotation θ about the x axis (see fig..4). Thus, x x. If the stresses in the original coordinate system are defined by ( σ, σ, σ ), the stresses in the rotated coordinate systems, ( σ, σ, σ ) are given by σ + σ σ σ σ = + cos θ + σ sin θ (.) σ σ σ = sin θ + σ cos θ (.b) σ is obtain by letting θ = θ + π /. A more rigorous method for rotating stresses between coordinate systems can be conducted utilizing tensor multiplication. This is a trivial exercise for the reader who is familiar with tensor formalisms, and is left for the interested reader to explore, for example in Fung (00). Here, we will continue to consider the two dimensional state of stress for enhancing the understanding of the dimensional state of stress. One useful way of visualizing the D state of stress is using Mohr s circle. ven though it can be somewhat involved to draw correctly, it is an efficient tool to visualize a state of stress. Figure 4. Transformation of stress from the coordinate system ( x, x, x ) to ( x, x, x ) 8
9 Figure 5. General Definition of DMohr s circle Mohr s circle for a general D application. The construction of Mohr s circle is outlined in Fig. 5A. ven for the D state of stress, a D state exists: If plane stress is considered, one principal stress is always zero. If plane strain is consider, the third stress is given by eq. (.0). It is possible to rotate the coordinate system so that all shear stresses vanish. This state of stress is defined as the principal stresses, where the stresses are denoted as σ, σ, σ. By definition σ > σ > σ (.) The absolute maximum shear stress is given by σ σ = = (.) ( σij ) τmax max It is evident that this maximum shear stress is located 90 o (π/) from the orientation of the principal stresses. Recall that an alternative way of obtaining the principal stresses is by finding the eigenvalues to the stress tensor: To try at home Problem : σ σ σ S (.4) = σ σ σ σ σ σ Consider the cases of (a) plane stress and (b) plane strain when σ = 0 MPa, σ = 5 MPa and σ = 0 xx yy xy and determine the absolute maximum shear stress. Assume ν = 0. and constant temperature. 9
10 4. lastic Strain nergy Work, is defined as W = F d δ, where F is the force vector and δ is the displacement. For example, if a force is exerted on an object, where the force is linearly increased from zero to a maximum force, F, the work W is given by the area under the force displacement curve, i.e., W = Fδ. Similarly, if a linear spring is stretched from its stress free point to a displacement δ, the work conducted on the spring is W = Fδ, where F is the force at δ. In the case of the spring, once the force is released it will move back to its original position. Thus, the work has been stored in the spring as potential energy, or elastic strain energy. We have "stored energy" work = Fδ = "Internal nergy" = U "Strain nergy" One can show that the total stored strain energy is U = σε ij ij dv (.5) i= j= V Figure 6 Work and Strain nergy XAMPL Determine the elastic strain energy in a beam segment under pure bending. Assume isotropic, linearelastic materials with elastic modulus,. Solution: The basic kinematics assumption for beam bending (according to the uler assumption ) ε ( x ) o ε x = + κ (i) o where ε ( x ) is the extensional strain in the x direction, as a function of x. ε is the extensional strain at the centroid κ is the change in curvature of the centroid, with κ = / R where R is the radius the beam segment during loading. For pure bending, no shear force is present, and we have σ = σ = σ = σ = σ. Thus, eq. (.5) simplifies to 0
11 U = σ ε dv (ii) V Furthermore, applying Hooke s law: σ = ε (iii) Next, substitute (iii) and ii) into (ii) L ( ) o U = ε x κ da + dx 0 A o If there is no normal force applied, the strain at the centroid, ε, vanishes. Furthermore, since pure bending, κ is constant and since there are no material variations is constant. Thus, U = κ ( x) da dx Ix Lκ = 0 A L where we used the definition of area moment of inertia I x A = x da Figure 7. the exaggerated deformation of a beam segment subjected to pure bending the general beam section with coordinate system based on the centroid. References YC Fung and P Tong, Classical and computational solid mechanics, Word Scientific 00
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