Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 1 Stress Strain An understing of stress strain is essential for the analsis of metal forming operations. Often the words stress strain are used snonmousl b the nonscientific public. In engineering usage, however, stress is the intensit of force strain is a measure of the amount of deformation. 1.1 STRESS Stress σ is defined as the intensit of force at a point. where F is the force acting on a plane of area, A. If the stress is the same everwhere in a bod, σ = F/ A as A 0, (1.1) σ = F/A. (1.2) There are nine components of stress as shown in Figure 1.1. A normal stress component is one in which the force is acting normal to the plane. It ma be tensile or compressive. A shear stress component is one in which the force acts parallel to the plane. Stress components are defined with two subscripts. The first denotes the normal to the plane on which the force acts the second is the direction of the force. For eample, σ is a tensile stress in the -direction. A shear stress acting on the -plane in the -direction is denoted b σ. Repeated subscripts (e.g., σ, σ, σ zz ) indicate normal stresses. The are tensile if both the plane direction are positive or both are negative. If one is positive the other is negative the are compressive. Mied subscripts (e.g., σ z, σ, σ z ) denote shear stresses. A state of stress in tensor notation is epressed as σ σ σ z σ ij = σ σ σ z σ z σ z σ, (1.3) zz The use of the opposite convention should cause no problem because σ ij = σ ji. 1 in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 2 STRESS AND STRAIN z σ zz σ z σ z σ z σ z σ σ σ σ Figure 1.1. Nine components of stress acting on an infinitesimal element. where i j are iterated over,, z. Ecept where tensor notation is required, it is simpler to use a single subscript for a normal stress denote a shear stress b τ. For eample, σ σ τ σ. 1.2 STRESS TRANSFORMATION Stress components epressed along one set of orthogonal aes ma be epressed along an other set of aes. Consider resolving the stress component, σ = F /A, onto the aes as shown in Figure 1.2. The force, F, acts in the direction is F = F cos θ the area normal to is A = A / cos θ, so σ = F /A = F cos θ/(a / cos θ) = σ cos 2 θ. (1.4a) Similarl τ = F /A = F sin θ/(a / cos θ) = σ cos θ sin θ. (1.4b) Note that transformation of stresses requires two sine /or cosine terms. Pairs of shear stresses with the same subscripts that are in reverse order are alwas equal (e.g., τ ij = τ ji ). This is illustrated in Figure 1.3 b a simple moment balance F F A θ F Figure 1.2. The stresses acting on a plane, A, under a normal stress, σ. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 1.2 STRESS TRANSFORMATION 3 τ Figure 1.3. Unless τ = τ, there would not be a moment balance. τ τ τ on an infinitesimal element. Unless τ ij = τ ji, there would be an infinite rotational acceleration. Therefore τ ij = τ ji. (1.5) The general equation for transforming the stresses from one set of orthogonal aes (e.g., n, m, p) to another set of aes (e.g., i, j, k), is 3 3 σ ij = l im l jn σ mn. (1.6) n=1 m=1 Here, the term l im is the cosine of the angle between the i the m aes the term l jn is the cosine of the angle between the j n aes. This is often written more simpl as σ ij = l in l jn σ mn, (1.7) with the summation implied. Consider transforming stresses from the,, z ais sstem to the,, z sstem shown in Figure 1.4. Using equation 1.6, σ = l l σ + l l σ + l l zσ z + l l zσ z + l l zσ z + l zl zσ zz (1.8a) σ = l l σ + l l σ + l zl zσ z + l l σ + l l σ + l zl σ z + l l zσ z + l l zσ z + l zl zσ zz. (1.8b) z z Figure 1.4. Two orthogonal coordinate sstems. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 4 STRESS AND STRAIN These can be simplified to σ = l 2 σ + l 2 σ + l 2 z σ z + 2l l zτ z + 2l zl τ z + 2l l τ (1.9a) τ = l l σ + l l σ + l zl zσ z + (l l z + l zl )τ z + (l zl + l l z)τ z + (l l + l l )τ (1.9b) 1.3 PRINCIPAL STRESSES It is alwas possible to find a set of aes along which the shear stress terms vanish. In this case σ 1, σ 2 σ 3 are called the principal stresses. The magnitudes of the principal stresses, σ p, are the roots of σ 3 p I 1σ 2 p I 2σ p I 3 = 0, (1.10) where I 1, I 2 I 3 are called the invariants of the stress tensor. The are I 1 = σ + σ + σ zz, I 2 = σz 2 + σ z 2 + σ 2 σ σ zz σ zz σ σ σ (1.11) I 3 = σ σ σ zz + 2σ z σ z σ σ σ 2 z σ σ 2 z σ zzσ 2. The first invariant, I 1 = p/3 where p is the pressure. I 1, I 2 I 3 are independent of the orientation of the aes. Epressed in terms of the principal stresses the are I 1 = σ 1 + σ 2 + σ 3, I 2 = σ 2 σ 3 σ 3 σ 1 σ 1 σ 2 (1.12) I 3 = σ 1 σ 2 σ 3. EXAMPLE 1.1: Consider a stress state with σ = 70 MPa, σ = 35 MPa, τ = 20, σ z = τ z = τ z = 0. Find the principal stresses using equations 1.10 1.11. SOLUTION: Using equations 1.11, I 1 = 105 MPa, I 2 = 2,050 MPa, I 3 = 0. From equation 1.10, σ 3 p 105σ 2 p + 2,050σ p + 0 = 0, σ 2 p 105σ p + 2,050 = 0. The principal stresses are the roots, σ 1 = 79.1 MPa, σ 2 = 25.9 MPa σ 3 = σ z = 0. EXAMPLE 1.2: Repeat Eample 1.1, with I 3 = 170,700. SOLUTION: The principal stresses are the roots of σ 3 p 65σ 2 p + 1750σ p + 170,700 = 0. Since one of the roots is σ z = σ 3 = 40, σ p + 40 = 0 can be factored out. This gives σ 2 p 105σ p + 2,050 = 0, so the other two principal stresses are σ 1 = 79.1 MPa, σ 2 = 25.9 MPa. This shows that when σ z is one of the principal stresses, the other two principal stresses are independent of σ z. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 1.4 MOHR S CIRCLE EQUATIONS 5 1.4 MOHR S CIRCLE EQUATIONS In the special cases where two of the three shear stress terms vanish (e.g., τ = τ z = 0), the stress, σ z, normal to the plane is a principal stress the other two principal stresses lie in the plane. This is illustrated in Figure 1.5. For these conditions l z = l z = 0, τ z = τ z = 0, l = l = cos φ l = l = sin φ. Substituting these relations into equations 1.9 results in τ = cos φ sin φ( σ + σ ) + (cos 2 φ sin 2 φ)τ, σ = cos 2 φσ + sin 2 φσ + 2 cos φ sin φτ, (1.13) σ = sin 2 φσ + cos 2 φσ + 2 cos φ sin φτ. These can be simplified with the trigonometric relations, sin 2φ = 2 sin φ cos φ cos 2 φ = cos 2 φ sin 2 φ to obtain τ = sin 2φ(σ σ )/2 + cos 2φτ, (1.14a) σ = (σ + σ )/2 + cos 2φ(σ σ ) + τ sin 2φ, (1.14b) σ = (σ + σ )/2 cos 2φ(σ σ ) + τ sin 2φ. (1.14c) If τ is set to zero in equation 1.14a, φ becomes the angle θ between the principal aes the aes. Then tan 2θ = τ /[(σ σ )/2]. (1.15) The principal stresses, σ 1 σ 2, are then the values of σ σ, σ 1,2 = (σ + σ )/2 ± [(σ σ )/ cos 2θ] + τ sin 2θ or [ 1/2 σ 1,2 = (σ + σ )/2 ± (1/2) (σ σ ) 2 + 4τ] 2. (1.16) σ z τ σ τ σ σ τ τ φ Figure 1.5. Stress state for which the Mohr s circle equations appl. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 6 STRESS AND STRAIN τ σ (σ 1 -σ 2 )/2 θ σ 2 2θ (σ -σ )/2 τ σ 1 σ (σ +σ )/2 σ Figure 1.6. Mohr s circle diagram for stress. τ σ σ 3 σ 2 τ 2θ σ 1 σ σ Figure 1.7. Three-dimensional Mohr s circles for stresses. A Mohr s circle diagram is a graphical representation of equations 1.15 1.16. The form a circle of radius (σ 1 σ 2 )/2 with the center at (σ 1 + σ 2 )/2 as shown in Figure 1.6. The normal stress components are plotted on the ordinate the shear stress components are plotted on the abscissa. Using the Pthagorean theorem on the triangle in Figure 1.6, (σ 1 σ 2 )/2 = { [(σ + σ )/2] 2 + τ 2 } 1/2 (1.17) tan(2θ) = τ /[(σ + σ )/2]. (1.18) A three-dimensional stress state can be represented b three Mohr s circles as shown in Figure 1.7. The three principal stresses σ 1, σ 2 σ 3 are plotted on the ordinate. The circles represent the stress state in the 1 2, 2 3 3 1 planes. EXAMPLE 1.3: Construct the Mohr s circle for the stress state in Eample 1.2 determine the largest shear stress. O. Mohr, Zivilingeneur (1882), p. 113. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 1.5 STRAIN 7 τ τ ma = 59.6 Figure 1.8. Mohr s circle for stress state in Eample 1.2. σ 3 = -40 σ = 35 σ 2 = 25.9 τ = 20 σ = 70 σ σ 1 = 79.1 A Figure 1.9. Deformation, translation, rotation of a line in a material. A l 0 l B B SOLUTION: The Mohr s circle is plotted in Figure 1.8. The largest shear stress is τ ma = (σ 1 σ 3 )/2 = [79.1 ( 40)]/2 = 59.6 MPa. 1.5 STRAIN Strain describes the amount of deformation in a bod. When a bod is deformed, points in that bod are displaced. Strain must be defined in such a wa that it ecludes effects of rotation translation. Figure 1.9 shows a line in a material that has been deformed. The line has been translated, rotated, deformed. The deformation is characterized b the engineering or nominal strain, e e = (l l 0 )/l 0 = l/l 0. (1.19) An alternative definition is that of true or logarithmic strain, ε, defined b which on integrating gives ε = ln(l/l 0 ) = ln(1 + e) dε = dl/l, (1.20) ε = ln(l/l 0 ) = ln(1 + e). (1.21) The true engineering strains are almost equal when the are small. Epressing ε as ε = ln(l/l 0 ) = ln(1 + e) eping, so as e 0,ε e. There are several reasons wh true strains are more convenient than engineering strains. The following eamples indicate wh. True strain was first defined b P. Ludwig, Elemente der Technishe Mechanik, Springer, 1909. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 8 STRESS AND STRAIN EXAMPLE 1.4: (a) A bar of length, l 0, is uniforml etended until its length, l = 2 l 0. Compute the values of the engineering true strains. (b) What final length must a bar of length l 0, be compressed if the strains are the same (ecept sign) as in part (a)? SOLUTION: (a) e = l/l 0 = 1.0, ε= ln(l/l 0 ) = ln 2 = 0.693 (b) e = 1 = (l l 0 )/l 0,sol = 0. This is clearl impossible to achieve. ε = 0.693 = ln(l/l 0 ), so l = l 0 ep(0.693) = l 0 /2. EXAMPLE 1.5: A bar 10 cm long is elongated to 20 cm b rolling in three steps: 10 cm to 12 cm, 12 cm to 15 cm, 15 cm to 20 cm. (a) Calculate the engineering strain for each step compare the sum of these with the overall engineering strain. (b) Repeat for true strains. SOLUTION: (a) e 1 = 2/10 = 0.20, e 2 = 3/12 = 0.25, e 3 = 5/15 = 0.333, e tot = 0.20 +.25 +.333 = 0.833, e overall = 10/10 = 1. (b) ε 1 = ln(12/10) = 0.182, ε 2 = ln(15/12) = 0.223, ε 3 = ln(20/15) = 0.288, ε tot = 0.693, ε overall = ln(20/10) = 0.693. With true strains, the sum of the increments equals the overall strain, but this is not so with engineering strains. EXAMPLE 1.6: A block of initial dimensions, l 0,w 0, t 0, is deformed to dimensions of l, w, t. (a) Calculate the volume strain, ε v = ln(v/v 0 ) in terms of the three normal strains, ε l,ε w ε t. (b) Plastic deformation causes no volume change. With no volume change, what is the sum of the three normal strains? SOLUTION: (a) ε v = ln[(lwt)/(l 0 w 0 t 0 )] = ln(l/l 0 ) + ln(w/w 0 ) + ln(t/t 0 ) = ε l + ε w + ε t. (b) If ε v = 0, ε l + ε w + ε t = 0. Eamples 1.4, 1.5 1.6 illustrate wh true strains are more convenient than engineering strains. 1. True strains for an equivalent amount of tensile compressive deformation are equal ecept for sign. 2. True strains are additive. 3. The volume strain is the sum of the three normal strains. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 1.6 SMALL STRAINS 9 If strains are small, true engineering strains are nearl equal. Epressing true strain as ε = ln( l 0 + l l 0 ) = ln(1 + l/l 0 ) = ln(1 + e) taking the series epansion, ε = e e 2 /2 + e 3 /3!...,itcanbeseen that as e 0,ε e. EXAMPLE 1.7: Calculate the ratio of ε/e for e = 0.001, 0.01, 0.02, 0.05, 0.1 0.2. SOLUTION: For e = 0.001, ε = ln(1.001) = 0.0009995; ε/e = 0.9995. For e = 0.01, ε = ln(1.01) =0.00995, ε/e = 0.995. For e = 0.02, ε = ln(1.02) =0.0198, ε/e = 0.99. For e = 0.05, ε = ln(1.05) = 0.0488, ε/e = 0.975. For e = 0.1, ε = ln(1.1) = 0.095, ε/e = 0.95. For e = 0.2, ε = ln(1.2) = 0.182, ε/e = 0.912. As e gets larger the difference between ε e become greater. 1.6 SMALL STRAINS Figure 1.10 shows a small two-dimensional element, ABCD, deformed into A B C D where the displacements are u v. The normal strain, e, is defined as e = (A D AD)/AD = A D /AD 1. (1.22) Neglecting the rotation e = A D d u + u + ( u/ )d /AD 1 = 1 or d e = u/. (1.23) Similarl, e = v/ e zz = w/ z for a three-dimensional case. ( u/ )d P B C v + ( v/ )d B C D d v A Q ( v/ )d A u d D u + ( u/ )d Figure 1.10. Distortion of a two-dimensional element. in this web service Cambridge Universit Press
Cambridge Universit Press 978-1-107-00452-8 - Metal Forming: Mechanics Metallurg, Fourth Edition Ecerpt 10 STRESS AND STRAIN The shear strain are associated with the angles between AD A D between AB A B. For small deformations AD A D A B v/ AB u/ (1.24) The total shear strain is the sum of these two angles, γ = γ = u + v. Similarl, (1.25a) γ z = γ z = v z + w (1.25b) γ z = γ z = w + u z. (1.25c) This definition of shear strain, γ, is equivalent to the simple shear measured in a torsion of shear test. 1.7 THE STRAIN TENSOR If tensor shear strains, ε ij, are defined as ε ij = (1/2)γ ij, (1.26) small shear strains form a tensor, ε ε ε z ε ij = ε ε ε z ε z ε z ε. (1.27) zz Because small strains form a tensor, the can be transformed from one set of aes to another in a wa identical to the transformation of stresses. Mohr s circle relations can be used. It must be remembered, however, that ε ij = γ ij /2 that the transformations hold onl for small strains. If γ z = γ z = 0, ε = ε l 2 + ε l 2 + γ l l (1.28) γ = 2ε l l + 2ε l l + γ (l l + l l ). (1.29) The principal strains can be found from the Mohr s circle equations for strains, ε 1,2 = ε + ε ± (1/2) [ (ε ε ) 2 + γ 2 1/2. 2 ] (1.30) Strains on other planes are given b ε, = (1/2)(ε 1 + ε 2 ) ± (1/2)(ε 1 ε 2 ) cos 2θ (1.31) γ = (ε 1 ε 2 ) sin 2θ. (1.32) 1.8 ISOTROPIC ELASTICITY Although the thrust of this book is on plastic deformation, a short treatment of elasticit is necessar to underst springback residual stresses in forming processes. in this web service Cambridge Universit Press