Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống
1
/ 14 trang
THÔNG TIN TÀI LIỆU
Thông tin cơ bản
Định dạng
Số trang
14
Dung lượng
236,16 KB
Nội dung
1 Chapter III: Strain 1 Content • Engineering strain versus true strain • Strain increment • Strain rates • Equivalent strain rate and equivalent strain • Examples of application Chapter III: Strain 2 Engineering Strain & True Strain λ λλ λ 0 λ λλ λ λ λλ λ 1 intermediate configuration initial configuration final configuration F F d λ λλ λ 2 Chapter III: Strain 3 Mechanical engineering strain: 1 0 0 e − = l l l Elongation is based to the initial length. 1 0 1 0 ln d d d ε ε = → = = ∫ l l l l l l l l True (logarithmic, natural) strain: Forming technology is related to the actual length, useful to describe the large amounts of deformation. Chapter III: Strain 4 Remark on Plastic Strains elastic plastic ε ε ε = + The total strain can be splitted (phân tách) into: In metalforming, plastic strains are much larger than elastic ones: plastic ε ε ≈ since 3 Chapter III: Strain 5 Examples Example 3.1: a) A uniform bar of length λ 0 is uniformly extended until its final length is λ = 2 λ 0 . Compute the values of engineering strain and true strain for this extension. b) To what final length, λ, must a bar of initial length λ 0 , be compressed if the strains are to be the same (except for sign) as those in part (a)? Example 3.2: A uniform bar of 100 mm initial length is elongated to a length of 200 mm in three stages: Stage 1: 100 mm to 120 mm, Stage 2: 120 mm to 150 mm and Stage 3: 150 mm to 200 mm. Calculate the engineering and true strains for each stage and compare the sums of the three with the overall values of the strains. Chapter III: Strain 6 Engineering Strain & True Strain ( ) 01 0 0 0 ln ln ln 1 ln 1 e ε + ∆ ∆ = = = + = + l ll l l l l ( ) 2 3 ln 1 2 3 e e e e ε = + ≈ − + − K Series expansion: Hence: 0 e e ε → → e 0.001 0.010 0.02 0.05 0.10 0.20 0.50 1.0 ε εε ε 0.0009995 0.00995 0.0198 0.0487 0.0953 0.182 0.405 0.693 ε/ ε/ε/ ε/ e 0.9995 0.995 0.990 0.976 0.953 0.912 0.811 0.693 4 Chapter III: Strain 7 Advantages and disadvantages resulting from the calculation with e and ɛ Large strains lead to numerical values which are not easy to deal with. (True strain ɛ=1 equals percentage strain of about 173%.) The logarithmic strain is mechanical more sensible as the strain related to the initial length and present calculation advantages. So in procedures with several forming steps the total strain equals the summation of the single strains. In contrast to (ngược với) the strain the true strain take in consideration not only the initial and the end state but also all intermediate steps of the total forming. In the calculation with strains only, the single strains can not be added to the total strain, because they are related to the initial length. Chapter III: Strain 8 Definition: True Strain The point has initial coordinations: x, y, z, After forming it has current coordinations: x’, y’, z’. Displacements of point following x, y, z directions are: x’ - x = u x = u y’ - y = u y = v z’ - z = u z = w Elongations and Deformations angle (Shear Strain) on area z 5 Chapter III: Strain 9 True Strain Differentials Similarly we can define the true strain differentials: { } x y z xy i yz zx yx zy xz d d d d E E d d d d d ε ε ε ε ε ε ε ε ε = = where: ( ) , etc. x du d x ε ∂ = ∂ ( ) ( ) 1 , etc. 2 xy du dv d y x ε ∂ ∂ = + ∂ ∂ Here: u, v and w are displacements in x, y and z directions and x, y, z are current coordinates. Chapter III: Strain 10 True Strain Differentials T x xy xz yx y yz zx zy z ε ε γ γ γ ε γ γ γ ε = 1 2 1 2 1 2 1 2 1 2 1 2 ε ε ε ε ε ε ε ε ε ε ij xx xy xz yx yy yz zx zy zz = Tensor of strain: ε ∂ ∂ ε ∂ ∂ ε ∂ ∂ x x y y z z u x u y u z = = = ∂ ∂ + ∂ ∂ =γ ∂ ∂ + ∂ ∂ =γ ∂ ∂ + ∂ ∂ =γ z u x u y u z u x u y u xz zx z y yz y x xy Where: 6 Chapter III: Strain 11 Strain Rates ( ) x du d x ε ∂ = ∂ Sometimes it is more convenient (thuận tiện) to introduce strain rates: Consider the strain differential: Dividing both sides by the differential time dt : ( ) x du d dt dt x ε ∂ = ∂ Or: ( ) x x du dt d u dt x x ε ε ∂ ∂ = = = ∂ ∂ & & Sometimes the strain-rate tensor is called “Rate-of-Deformation” tensor ij ij D ε = & Chapter III: Strain 12 Strain Rates ( ) ( ) xx xx du dt d du x u dt dt x x ε ε ∂ ∂ ∂ ∂ = = = = ∂ ∂ & & ( ) ( ) yy yy d dv dt dv y v dt dt y y ε ε ∂ ∂ ∂ ∂ = = = = ∂ ∂ & & [ ] 1 ( ) ( ) 1 2 2 xy xy yx dv x du y d v u dt dt x y ε ε ε ∂ ∂ + ∂ ∂ ∂ = = = = + ∂ ∂ & & & & Similar to strain, we have terms of strain rates: 7 Chapter III: Strain 13 Strain Rate Tensor xx xy xz ij yx yy yz zx zy zz ε ε ε ε ε ε ε ε ε ε ≡ & & & & & & & & & & xx u x ε ∂ = ∂ & & yy v y ε ∂ = ∂ & & zz w z ε ∂ = ∂ & & 1 2 xz zx w u x z ε ε ∂ ∂ = = + ∂ ∂ & & & & 1 2 xy yx v u x y ε ε ∂ ∂ = = + ∂ ∂ & & & & 1 2 yz zy v w z y ε ε ∂ ∂ = = + ∂ ∂ & & & & Normal Strain Rate Components: Shear Strain Rate Components: The Strain Rate Tensor in 3-D: particle velocity in -direction particle velocity in -direction particle velocity in -direction u x v y w z & & & Chapter III: Strain 14 Examples Example 2.3: A uniform bar with current length λ is extended at its free end by a tool with velocity of v tool . Determine the strain rate in axial direction. λ λλ λ v tool Example 2.4: A sheet with dimensions shown in the figure is sheared by a tool with velocity v tool . Determine the shear strain rate. v tool y x h 8 Chapter III: Strain 15 Strain Rates in Cylindrical Coordinate System z r θ P (r, ,z) θ 1 1 2 1 1 2 1 2 r r r z z z r z zr rz u u u r r u u r z u u z r θ θ θ θ θ θ θ ε ε θ ε ε θ ε ε ∂ ∂ = = − + ∂ ∂ ∂ ∂ = = + ∂ ∂ ∂ ∂ = = + ∂ ∂ & & & & & & & & & & & & & , , 1 r rr r r r z zz z z u u u u r r u u z θ θθ ε ε θ ε ∂ ∂ = = = + ∂ ∂ ∂ = = ∂ & & & & & & & & & rr r rz ij r z zr z zz θ θ θθ θ θ ε ε ε ε ε ε ε ε ε ε ≡ & & & & & & & & & & Chapter III: Strain 16 Strain Rates for Axisymmetrical Problems Conditions of axial symmetry: 0 and 0 where stands for any quant ity u θ θ ∂ ⊗ = = ⊗ ∂ & Using these conditions results: , , 1 r z rr r r r zz z z u u u u u r r z θθ ε ε ε ∂ ∂ = = = = = ∂ ∂ & & & & & & & & 1 0 0 2 r z r r z z zr rz u u z r θ θ θ θ ε ε ε ε ε ε ∂ ∂ = = = = = = + ∂ ∂ & & & & & & & & 9 Chapter III: Strain 17 Strain Rates in Spherical Coordinate System { } ( ) , , , r, , , , , , 1 1 cot sin 1 1 1 1 2 sin 1 1 cot 2 sin 1 1 2 rr r r r r r r r r r r r u u u u r u u r u u u r u u u r u u u r ϕϕ ϕ ϕ θ θθ θ θ ϕ ϕ ϕ ϕ ϕ ϕθ θϕ θ ϕ θ ϕ ϕ θ θ θ θ θ ε ε θ θ ε ε ε θ ε ε θ θ ε ε = = + + = + = = − + = = + − = = + − & & & & & & & & & & & & & & & & & & & & & & & & r θ P (r, ) ϕ, θ ϕ Chapter III: Strain 18 Strain Rates in Spherical Coordinate System: Symmetry For symmetry with: 0 and u u ϕ θ ϕ ∂ ⊗ = = ∂ & & , , , 0 2 r rr r r r r r r r u u r u r ϕϕ θθ ϕ ϕ ϕθ θϕ θ θ θ ε ε ε ε ε ε ε ε ε = = = = = = = = = & & & & & & & & & & & & 10 Chapter III: Strain 19 Total Finite Strains Total finite strains have physical meaning if and only if: 1. All shear strain rates are zero, 2. The straining path is straight, 0 ij i j ε = ∀ ≠ & constant, etc. xx yy ε ε = & & λ λλ λ v tool d x x Assuming uniform straining: tool x u v = & l ( ) { tool xx u x v u du x dx ε ∂ = = = ∂ & & & & l tool d v dt = l but so that xx d dt ε = l & l Integration: 0 0 0 t t xx xx d dt d dt dt ε ε = = = ⇒ ∫ ∫ ∫ l l l l & l l 0 ln xx ε = l l Chapter III: Strain 20 Principal Strains x y z 1 2 3 element before deformation element after deformation ( ) ( ) ( ) 1 1 0 2 1 0 3 1 0 ln ln ln a a b b c c ε ε ε = = = principal strain directions [...]... plastic strain for the process by assuming that there is no bulging and the cylinder is compressed from height h0 to h1 upper die (moving) ur h z r D lower die (fixed) Chapter III: Strain 24 12 Equivalent Strain Rate in 1/s Equivalent Strain Rate in Upsetting 4 0% 20% 40% 60% 3 2 1 vtool = 100 mm/s, h0 = 100 mm 0 0 10 20 30 40 50 60 70 Height Reduction in % Chapter III: Strain 25 Equivalent Plastic Strain. .. Plastic Strain The hardening of metals in multidimensional strain states is represented by the equivalent strain and its rate The von Mises equivalent plastic strain rate can be derived as: & ε = 2 (ε xx + ε yy + ε zz ) + 2 (ε xy + ε yz + ε xz ) &2 &2 &2 &2 &2 &2 3 The total equivalent strain is simply: & ε = ∫ ε dt t In terms of principal strains: ε = 2 2 (ε1 + ε 22 + ε 32 ) 3 Chapter III: Strain. .. 24 24 4 3 initial final Chapter III: Strain ∴ dV =0 dt 22 11 Condition of Volume Constancy (2) Expand (khai tri n) dV d = 0 = ( a1 ⋅ b1 ⋅ c1 ) dt dt dV da db dc = 0 = 1 ⋅ b1 ⋅ c1 + a1 ⋅ 1 ⋅ c1 + a1 ⋅ b1 ⋅ 1 dt dt dt dt 0= da1 db1 dc1 + + a1dt b1dt c1dt a1b1c1 ⇒ & & & & & & { 0 = ε1 + ε 2 + ε 3 ⇒ 0 = ε xx + ε yy + ε zz tensor ! Chapter III: Strain 23 Examples (1) F,... Equivalent Strain 3 2 0% 20% 40% 60% 80% 1 vtool = 100 mm/s, h0 = 100 mm 0 0 20 40 60 Height Reduction in % Chapter III: Strain 80 100 ∆h / h0 26 13 Examples (2) Example 2.6: Consider the frictionless axisymmetrical drawing process as shown in the figure Determine the equivalent plastic strain rate for the process by assuming that the axial velocity component is a function of the axial coordinate only Chapter. .. axisymmetrical drawing process as shown in the figure Determine the equivalent plastic strain rate for the process by assuming that the axial velocity component is a function of the axial coordinate only Chapter III: Strain 27 14 . 1 Chapter III: Strain 1 Content • Engineering strain versus true strain • Strain increment • Strain rates • Equivalent strain rate and equivalent strain • Examples of application Chapter III: Strain 2 Engineering. engineering and true strains for each stage and compare the sums of the three with the overall values of the strains. Chapter III: Strain 6 Engineering Strain & True Strain ( ) 01 0 0 0 ln. calculation with strains only, the single strains can not be added to the total strain, because they are related to the initial length. Chapter III: Strain 8 Definition: True Strain The point