Mohr s Circle By drawing Mohr s circle, the stress transformation in -D can be done graphically. σ = σ x + σ y τ = σ x σ y + σ x σ y cos θ + τ xy sin θ, 1 sin θ + τ xy cos θ. Note that the angle of rotation, θ, is defined by comparing the two normals on the surfaces. σ \ τ [\ τ σ σ [ θ θ How to draw a Mohr circle Each point on a Mohr circle represents the orientation of surface. Therefore, one can start by mapping each side of the element to a corresponding point on the Mohr circle, i.e. 1. Normal Stress Horizontal axis a Tensile stress Right side b Compressive Stress Left side. Shear Stress Vertical axis a Clockwise rotation Upper side b Counterclockwise rotation Lower side 3. Draw a circle by connecting the four corners. 4. Rotation of θ in the physical elements is translated into θ on the Mohr circle space. Note that we avoided using terms such as positive rotation etc.. on purpose. Example: Draw Mohr s circle for σ x, σ y, τ xy = 3, 4,. 1
FZURWDWLRQ % $ FRPSUHVVLRQ % $ WHQVLRQ FFZURWDWLRQ Example 1: Using Mohr s circle, one can explain why the angle of the failure surface of a metal bar is 45 degrees. The maximum shear stress theory in failure of metals postulates that Failure yielding of metals is predicted when the maximum shear stress reaches a critical value. When a metal bar is subject to uniaxial tension, the corresponding Mohr s circle is the one centered at σ/, 0 with a radius of σ/. So the maximum shear stress is at the top of the circle σ/, σ/ which is rotation from the surface subject to the uniaxial tension by 90 degrees which is translated into rotation of the physical element by 45 degrees. Example : Drawing Mohr s circle can also explain why hydrostatic stresses never cause failure in metals. If an element is subject to hydrostatic stresses σ x = σ y = σ, the corresponding Mohr s circle is degenerated to a single point whose maximum shear stress remains 0. Strains It should be noted that the stress is a physical quantity while the strain is a geometrical quantity. O O If one wants to measure the rate of deformation based on the original length, one can define E l l o l o. 3 On the other hand, the rate of deformation can be also defined based on the present length as e l l o. l 4 Since the definitions are different, their values are different naturally. However, if one assumes that the deformation is infinitesimal, i.e. where ϵ is an infinitesimal quantity, l = l o 1 + ϵ, 5 E = l o1 + ϵ l o l o 6 = ϵ. 7 e = l o1 + ϵ l o l 8
= ϵ 1 + ϵ 9 = ϵ1 ϵ + ϵ ϵ 3 +... 10 ϵ, 11 the two definitions coincide. When a point in a deformable continua moves from a to x as shown in the figure, the rate of deformation can be measured by comparing the length of two infinitesimal vectors at the both points as da u dx a x ds = dx da 1 = dx i dx i da i da i. 13 If this quantity is to be expressed based on the original point, a, the position, x, can be regarded as a function of a as Substituting this to eq.13, one obtains x i = x i a j, 14 dx i = a j da j. 15 ds = dx i dx i da i da i 16 = a j da j a k da k da i da i 17 = da j da k δ jk da j da k 18 a j a k xi = δ jk da j da k 19 a j a k E jk da j da k, 0 where E jk 1 xi δ jk. 1 a j a k 3
This is called Green s strain Lagrangian strain in fluid mechanics. On the other hand, if one wants to measure the rate of deformation based on the current position, the point a can be regarded as a function of x as Substituting this to eq.13, one obtains a i = a i x j, da i = a i dx j. 3 where ds = dx i dx i da i da i 4 = dx i dx i a i dx j a i x k dx k 5 = δ jk dx j dx k a i a i dx j dx k 6 x k = δ jk a i a i dx j dx k 7 x k e jk dx j dx k, 8 e jk 1 This is called Cauchy s strain Euler s strain in fluid mechanics. δ jk a i a i. 9 x k In short, E ij and e ij are different in values as the definitions are different. Nevertheless, their combination with the infinitesimal elements are the same, i.e. ds = E ij da i da j 30 = e ij dx i dx j. 31 It is possible to express Green s strains and Cauchy s strains in terms of the displacement, i.e. For Green s strains, u i = x i a i. 3 x i a j = u i a j + a i, 33 a j = i a j + δ ij. 34 E jk = 1 xi δ jk 35 a j a k = 1 i + δ ij i + δ ik δ jk 36 a j a k = 1 i i + k + j + δ jk δ jk 37 a j a k a j a k = 1 k + j + i i. 38 a j a k a j a k 4
For Cauchy s strains, a k x i = x k u k x i, 39 a k = δ ki k. 40 e ij = 1 = 1 = 1 δ ij δ ki k δ ij δ ij i δ kj k i + j k k j + k k 41 4. 43 If the deformation is small such that the second order terms in eqs.38,43 are dropped, eq.38 and eq.43 become the same and the both can be expressed as ϵ ij 1 i + j. 44 Note that this definition differs from the definition of engineering strain by 1/ for the shear components. Problem A square plate is rotated from the first quarter to the third quarter as shown below. Compute the strains. a,b x,y Answer Since so x = a y = b, u = x x = x, v = y y = y, 45 46 ϵ 11 = x =, ϵ 1 = 1 y + x v = 0, 47 v ϵ = y =, 5
but the rotation above does not accompany any deformation rigid body rotation. What went wrong??? Solution The displacement in the rotation above is not small so one has to use the formula for finite deformation either E ij Green s strains or e ij Cauchy s strain. Let s compute the Green strain, E ij. so 1 E 11 = Other E ij s can be shown to be 0 verify!. u = a, v = b, 1 a 1 + 1 a 1 = 1 a + 1 = + k k a 1 a 1 a + a + v a + + = 0. 48 49 Shear deformation Pure shear Simple shear 1. Pure shear: x = 1 + ϵx, y y = 1+ϵ. 50 u = ϵx, 1 v = 1+ϵ 1 y, 51. Simple shear ϵ ij = ϵ 0 0 ϵ ϵ 0 0 ϵ 1+ϵ. 5 x = x + ϵy y = y. u = ϵy v = 0, 53 54 6
The strain components are ϵ ij = 0 ϵ ϵ 0. 55 As discussed in class, shear deformation is defined as the strain whose trace is 0, i.e. This definition is independent of the frame of reference. Compatibility Conditions Problem 1: Does the system ϵ ii = 0. 56 allow permissible consistent displacement fields? ϵ xx = xy ϵ xy = x ϵ yy = 0, 57 \ [ In other words, does the system have solutions? 1 x = xy = x y + v x v y = 0, 58 The answer to the question above is no as shown in class detailed omitted. Problem : Does the system have solutions? Answer: From eq.59, x = x + 3y y = x, 59 from which u = x + 3ydx + gy = x + 3xy + gy, 60 y = 3y + g y, 61 which is in contradiction with eq.59 so this system of PDEs does not have solutions. 7
Problem 3: Does the system have solutions? Answer: From eq.6, x = 3x + 3y y = 3x, 6 from which which is compared with eq.6 to yield so this system of PDEs does have solutions as u = 3x + 3ydx + gy = x 3 + 3xy + gy, 63 y = 3x + g y, 64 gy = const. ux, y = x 3 + 3xy + C, 65 The general question is under what conditions does a system of partial differential equations x = fx, y y = gx, y, have solutions? Obviously one can find a necessary condition easily, i.e. if the above system has solutions, it follows 66 u x y = f y u y x = g x, 67 f y = g x. 68 It can be shown that this condition is also a sufficient condition proof omitted eq.68 is a necessary and sufficient condition for the system to have solutions. Redo the above using index notation, i.e. a system of PDEs, have solutions if or u,i = f i, 69 f 1, f,1 = 0, 70 f i,j e ij3 = 0, 71 where e ijk is the permutation symbol. Equation 71 is called the compatibility condition. 8
Strain compatibility conditions The strain-displacement relationship is expressed as u i,j + u j,i = ϵ ij. 7 The compatibility condition for the above system of differential equations must be written for both i and j, i.e. which yields only one equation in -D as ϵ ij,kl e jk3 e il3 = 0, 73 or ϵ 11, ϵ 1,1 + ϵ,11 = 0, 74 ϵ xx y The following Mathematica command can derive eq.75 automatically: ϵ xy x y + ϵ yy x = 0. 75 strain={{e11,e1},{e1,e}} Sum[Dt[strain[[i,j]],x[k],x[l]] Signature[{j,k,3}] Signature[{i,l,3}],{i,},{k,},{j,},{l,}] Compatibility condition 3-D 1. Symmetrization operator a ij 1 a ij + a ji. 76. Anti-symmetrization operator Note that a [ij] 1 a ij a ji. 77 a ij = a ji a [ij] = a [ji]. Any quantity with two indices can be uniquely decomposed into its symmetrical part and anti-symmetrical part because 78 Example: For the displacement, u i, the decomposition is v ij = 1 v ij + v ji + 1 v ij v ji = v ij + v [ij]. 79 u i,j = 1 u i,j + u j,i + 1 u i,j u j,i = ϵ ij + ω ij, where ϵ ij is the strain tensor and ω ij is the rotation. This way, you can say that the strain is the symmetrical part of the displacement gradient and the rotation is the anti-symmetrical part of the displacement gradient. Compatibility conditions for a system of PDE, can be written as 80 u,i = g i, 81 9
the compatibility condition for the strain-displacement relation, g [i,j] = 0, 8 can be written as which is explicitly written as u i,j = ϵ ij, 83 ϵ [i[j,k]l] = 0, 84 ϵ [i[j,k]l] = 1 4 ϵ ij,kl + ϵ kl,ij ϵ lj,ki ϵ ik,jl = 0. Note that this condition yields only one independent equation in -D as 85 ϵ xx y and six independent conditions as shown in the textbook 6.3-4. ϵ xy x y + ϵ yy x = 0, 86 Example: Plastic deformation In theory of plasticity, when the body undergoes plastic deformation, the stress-strain relation is given incrementally. Consider the stress-strain relation for a fictitious plastic body given as where σ is the normal stress and τ is the shear stress. Problem Can you find the normal strain, ϵ, at σ, τ = 1, 1? 1. Apply σ normal stress first followed by τ shear stress. σ : 0 1 τ : 0 1 So dϵ = σ + τdσ + τdτ, 87 τ = 0, ϵ = 1 0 σ = 1, dσ = 0 ϵ = 1 0 dϵ = σdσ, σdσ = 1. dϵ = τdτ τdτ = 1. ϵ total = 1 + 1 = 3.. Apply τ shear stress first followed by σ normal stress. τ : 0 1 σ : 0 1 So σ = 0, ϵ = 1 0 τ = 1, dτ = 0 ϵ = 1 0 dϵ = τdτ, τdτ = 1. dϵ = 1 + σdσ, 1 + σdσ = 3. ϵ total = 1 + 3 + 1 = 5. 10
It was shown that the final total deformation depends on the history of loading. The fictitious stress-strain relation above implies that it is not possible to express the strain, ϵ, as a function of σ and τ uniquely. If the strain, ϵ, is expressed uniquely as a function of σ and τ, as it follows ϵ = ϵσ, τ, 88 dϵ = ϵ ϵ dσ + dτ σ τ 89 = P dσ + Qdτ, 90 where so the compatibility condition is P ϵ σ, Q ϵ τ, 91 P τ = Q σ. 9 P = σ + τ and Q = τ do not satisfy the compatibility condition. 11