slide cơ học vật chất rắn chapter 8 2 new two dimensional problem solution

.c om ng co an cu u du o ng th Chapter 8: Two-dimensional problem solution (Part 2) TDT  University  -­‐  2015 CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   8.4 Polar Coordinate Formulation cu u du o ng th an co 8.6 Example Polar Coordinate Solutions ng 8.5 General Solutions in Polar Coordinates c om Institute for computational science CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.4 Polar Coordinate Formulation ng 8.5 General Solutions in Polar Coordinates cu u du o ng th an co 8.6 Example Polar Coordinate Solutions CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.4 Polar Coordinate Formulation c om Airy Stress Function Approach φ = φ(r,θ) Biharmonic  Governing  Equa-on   ng Airy  Representa-on   ⎧ ∂ϕ ∂ 2ϕ ⎪σ r = r ∂r + r ∂θ ⎪ ∂ 2ϕ ⎪ ⎨σ θ = co ⎛ ∂ ∂ ∂ ⎞⎛ ∂ ∂ ∂ ⎞ ∇ ϕ =⎜ + + + + ϕ =0 ⎟⎜ 2 ⎟ r ∂r r ∂θ ⎠⎝ ∂r r ∂r r ∂θ ⎠ ⎝ ∂r th an ∂r ⎪ ⎪ ∂ ⎛ ∂ϕ ⎞ τ = − ⎜ ⎟ ⎪ rθ ∂r ⎝ r ∂θ ⎠ ⎩ du o ng   Trac-on  Boundary  Condi-ons   cu u Tr = f r (r , θ ) , Tθ = fθ (r , θ ) σr   τrθ   R y σθ   r r θ CuuDuongThanCong.com S https://fb.com/tailieudientucntt x h"p://incos.tdt.edu.vn   Institute for computational science 8.4 Polar Coordinate Formulation Strain-­‐Displacement   an du o ng th σ r = λ (er + eθ ) + 2µ er σ θ = λ (er + eθ ) + 2µ eθ σ z = λ (er + eθ ) = ν (σ r + σ θ ) τ rθ = 2µ erθ , τ θ z = τ rz = er = 1 (σ r −νσ θ ) , eθ = (σ θ −νσ r ) E E ez = − ν (σ r + σ θ ) = − cu CuuDuongThanCong.com ν E −ν +ν erθ = τ rθ , eθ z = erz = E u ⎧ ∂ur ⎪er = ∂r ⎪ ⎪ ∂uθ ⎞ 1⎛ ⎨eθ = ⎜ ur + ⎟ r ∂ θ ⎝ ⎠ ⎪ ⎪ ⎛ ∂ur ∂uθ uθ ⎞ + − ⎟ ⎪erθ = ⎜ r ∂ θ ∂ r r ⎠ ⎝ ⎩ Plane  stress   co Plane  strain   ng c om Plane Elasticity Problem Hooke’s  Law   https://fb.com/tailieudientucntt (er + eθ ) h"p://incos.tdt.edu.vn   Institute for computational science c om 8.4 Polar Coordinate Formulation ng 8.5 General Solutions in Polar Coordinates cu u du o ng th an co 8.6 Example Polar Coordinate Solutions CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.5 General Solutions in Polar Coordinates ϕ (r ,θ ) = f (r )ebθ c om 8.3.1 General Michell Solution ⎛ ∂ ∂ ∂ ⎞⎛ ∂ ∂ ∂ ⎞ ∇ ϕ =⎜ + + + + ϕ =0 ⎟⎜ 2 ⎟ r ∂r r ∂θ ⎠⎝ ∂r r ∂r r ∂θ ⎠ ⎝ ∂r co ng − 2b2 − 2b2 b2 (4 + b2 ) f ′′′′ + f ′′′ − f ′′ + f ′+ f =0 r r r r4 ng + (a4 + a5 log r + a6 r + a7 r log r )θ th ϕ = a0 + a1 log r + a2 r + a3r log r an Solving the equation gives the general Michell solution (restricted to the periodic case) ∞ cu u du o a + (a11r + a12 r log r + 13 + a14 r + a15 rθ + a16 rθ log r ) cos θ r b + (b11r + b12 r log r + 13 + b14 r + b15 rθ + b16 rθ log r ) sin θ r + ∑ (an1r n + an r 2+ n + an 3r − n + an r 2− n ) cos nθ We will use various terms from this general solution to solve several plane problems in polar coordinates n=2 ∞ + ∑ (bn1r n + bn r 2+ n + bn r − n + bn r 2− n ) sin nθ n=2 CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.5 General Solutions in Polar Coordinates Navier Equation Approach u=ur(r)er (Plane Stress or Plane Strain) c om 8.3.2 Axisymmetric Solutions Stress Function Approach a1 + a3 + 2a2 r a σ θ = 2a3 log r − 12 + 3a3 + 2a2 r τ rθ = ng ϕ = a0 + a1 log r + a2 r + a3r log r an th ng Displacements - Plane Stress Case co σ r = 2a3 log r + d 2ur dur + − ur = dr r dr r ur = C1r + C2 r Gives Stress Forms σr = A A + B , σ = − + B , τ rθ = θ r2 r2 cu u du o ⎡ (1 +ν ) ⎤ − a + 2(1 − ν ) a r log r − (1 + ν ) a r + a (1 − ν ) r 3 ⎥⎦ E ⎢⎣ r + A sin θ + B cos θ Underlined terms represent 4rθ uθ = a3 + A cos θ − B sin θ + Cr rigid-body motion E ur = •  a3 term leads to multivalued behavior, and is not found following the displacement formulation approach CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.4 Polar Coordinate Formulation cu u du o ng th an co 8.6 Example Polar Coordinate Solutions ng 8.5 General Solutions in Polar Coordinates CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions c om Example  8.6  Thick-­‐Walled  Cylinder  Under  Uniform  Boundary  Pressure  p2   A σr = + B r A σθ = − + B r r1   an co  p1   ng General  Axisymmetric   Stress  Solu-on   cu Using Strain Displacement Relations and Hooke’s Law for plane strain gives the radial displacement CuuDuongThanCong.com σ r (r1 ) = − p1 , σ r (r2 ) = − p2 r12 r22 ( p2 − p1 ) A= r22 − r12 r12 p1 − r22 p2 B= r22 − r12 r12 r22 ( p2 − p1 ) r12 p1 − r22 p2 σr = + r22 − r12 r2 r22 − r12 σθ = − u du o ng th r2   Boundary  Condi-ons     r12 r22 ( p2 − p1 ) r12 p1 − r22 p2 + r22 − r12 r2 r22 − r12 +ν A r[(1 − 2ν ) B − ] E r r12 p1 − r22 p2 +ν ⎡ r12 r22 ( p2 − p1 ) = + (1 − 2ν ) ⎢− E ⎣ r22 − r12 r r22 − r12 ur = https://fb.com/tailieudientucntt ⎤ r⎥ ⎦ h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions c om Comparison of Flamant Results with 3-D Theory-Boussinesq’s Problem Cartesian Solution P   Px ⎛ z − 2ν u= − ⎜ 4πµ R ⎝ R R + z ⎛z P ⎡ 3x z R x (2 R + z ) ⎞ ⎤ − (1 − ν ) − + ⎢ ⎜ ⎟⎥ 2π R ⎣ R ⎝ R R + z R( R + z ) ⎠ ⎦ ⎛z P ⎡ 3y2 z R y (2 R + z ) ⎞ ⎤ σy = − − (1 − ν ) − + ⎢ ⎜ ⎟⎥ 2π R ⎣ R ⎝ R R + z R( R + z ) ⎠⎦ 3Pz P ⎡ 3xyz (1 − 2ν )(2 R + z ) xy ⎤ σz = − , τ xy = − − ⎥ 2π R 2π R ⎢⎣ R R( R + z )2 ⎦ σx = − an z   P(1 −ν ) u z ( R, 0) = 2πµ R 3Pyz 3Pxz τ yz = − , τ xz = − 2π R 2π R du o Free Surface Displacements ng th y   P ⎛ z2 ⎞ ⎞ ⎜ 2(1 −ν ) + ⎟ ⎟ , w= 4πµ R ⎝ R ⎠ ⎠ co x   Py ⎛ z − 2ν ⎞ − ⎟,v= ⎜ 4πµ R ⎝ R R + z ⎠ ng   Cylindrical Solution cu u Corresponding 2-D Results P uθ ( r , ) = − ⎡(1 + ν ) + log r ⎤⎦ πE ⎣ 3-D Solution eliminates the unbounded far-field behavior CuuDuongThanCong.com P ⎡ 3r z (1 − 2ν ) R ⎤ rz (1 − ν ) r ⎡ ⎤ σr = − + ur = − 2π R ⎢⎣ R R + z ⎥⎦ ⎢ ⎥ 4πµ R ⎣ R R+z ⎦ (1 − 2ν ) P ⎡ z R ⎤ P ⎡ z2 ⎤ σθ = − ⎢ uz = 2(1 −ν ) + ⎥ 2π R ⎣ R R + z ⎥⎦ ⎢ 4πµ R ⎣ R ⎦ 3Pz 3P rz σz = − , τ rz = − uθ = 2π R 2π R P https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions Example 9: Half-Space Under Uniform Normal Loading a ≤ x ≤- a c om    p   2Y sin θ cos θ πr 2Y σ y = σ r sin θ = − sin θ πr 2Y τ xy = σ r sin θ cos θ = − sin θ cos θ πr σ x = σ r cos θ = − x   ng a   θ1   an co θ2   a    y   θ   ng dx   th   du o cu u dθ   θ   2p dσ y = − 2p dτ xy = − 2p p [2(θ − θ1 ) + (sin 2θ − sin 2θ1 )] π 2π p θ2 p σy = − sin θ dθ = − [2(θ − θ1 ) − (sin 2θ − sin 2θ1 )] ∫ π θ1 2π p θ2 p τ xy = − sin θ cos θ d θ = [cos 2θ − cos 2θ1 ] ∫θ1 π π CuuDuongThanCong.com https://fb.com/tailieudientucntt σx = − 2p dY = pdx = prdθ /sinθ r   dσ x = − θ2 ∫θ cos θ dθ = − π π π cos θ dθ sin θ dθ sin θ cos θ dθ h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions Example 9: Half-Space Under Uniform Normal Loading a ≤ x ≤- a 0.4 Concentrated  Loading                 τmax/(Y/a)   0.3 0.25 ng 0.35 co Dimensionless Maximum Shear Stress Distributed  Loading                       τmax/p   0.2 an σy/p     CuuDuongThanCong.com 0.15 0.1 0.05 0 Dimensionless Distance, y / a   10 A plot of this maximum shear stress versus depth below the surface is shown in Figure 8-27 u Along y-axis below the loading, τxy = 0, and the x- and y-axes are principal at these points and the maximum shear stress is given by τmax = ½|σx - σy| cu   du o Dimensionless Distance, x/a   ng th Dimensionless Stress   0.45   τxy  /p     c om 0.5 τmax  -­‐  Contours https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions c om Example 9: Half-Space Under Uniform Normal Loading a ≤ x ≤- a x ng Generalized Superposition Method a a t(s) th an co Half-Space Loading Problems p(s) ng y p( s )( x − s ) 2 a t ( s)( x − s)3 σx = − ∫ ds − ∫ ds 2 2 2 − a − a π [( x − s ) + y ] π [( x − s ) + y ] u p( s) y2 σy = − ds − 2 ∫ − a π [( x − s ) + y ] π a cu y3 a du o 2y t ( s)( x − s) ∫ −a [( x − s)2 + y ]2 ds a p( s )( x − s ) 2 y a t ( s )( x − s ) τ xy = − s− ds 2 2 2 ∫ ∫ − a − a π [( x − s ) + y ] π [( x − s ) + y ] y2 a CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.6 Example Polar Coordinate Solutions c om Example 9: Half-Space Under Uniform Normal Loading a ≤ x ≤- a   ng th an co ng Photoelastic Contact Stress Fields (Uniform     Loading)   cu u du o (Point     Loading)   (Flat  Punch  Loading)   CuuDuongThanCong.com (Cylinder  Contact  Loading)   https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science ng co y   r   θ   ϑ   α   x   Stress  Free  Faces   β  =  2π  -­‐  α   ng th an Example 10: Notch/Crack Problems - Consider the wedge problem for the case where angle α is small and β is 2π-α - We pursue the case where α ≈ 0, and the notch becomes a crack - The boundary surfaces of the notch are taken to be stress free, and thus the problem involves only far-field loadings - Start with Michell solution, we try the stress Function in generalized form: c om 8.6 Example Polar Coordinate Solutions   du o ϕ = r λ [ A sin λθ + B cos λθ + C sin(λ − 2)θ + D cos(λ − 2)θ ] where λ is allowed a non-integer σ θ = λ (λ − 1)r λ −2 [ A sin λθ + B cos λθ + C sin(λ − 2)θ + D cos(λ − 2)θ ] cu u τ rθ = −(λ − 1)r λ −2 [ Aλ cos λθ − Bλ sin λθ + C (λ − 2) cos(λ − 2)θ − D(λ − 2)sin(λ − 2)θ ] Boundary Conditions: σ θ (r ,0) = τ rθ (r,0) = σ θ (r, 2π ) = τ rθ (r, 2π ) = ⇒ sin 2π ( λ − 1) = ⇒ λ = At Crack Tip r è 0: n + , n = 0,1,2,! Stress = O(r λ −2 ) , Displacement = O(r λ −1 ) Finite Displacements and Singular Stresses at Crack Tip è 1< λ