slide cơ học vật chất rắn chapter 8 1 new two dimensional problem solution

.c om ng co an cu u du o ng th Chapter 8: Two-dimensional problem solution (Part 1) TDT  University  -­‐  2015 CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.1 Two-dimensional problem solution ng 8.2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8.3 Cartesian Coordinate Solutions Using Fourier Methods CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.1 Two-dimensional problem solution ng 8.2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8.3 Cartesian Coordinate Solutions Using Fourier Methods CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.1 Two-dimensional problem solution c om Using the Airy Stress Function approach, it was shown that the plane elasticity formulation with zero body forces reduces to a single governing co ng biharmonic equation In Cartesian coordinates it is given by ng th an ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ + 2 + = ∇ 4ϕ = ∂x ∂x ∂y ∂y du o and the stresses are related to the stress function by cu u ∂ 2ϕ ∂ 2ϕ ∂ 2ϕ σ x = , σ y = , τ xy = − ∂y ∂x ∂x∂y We now explore solutions to several specific problems in both Cartesian and Polar coordinate systems CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.1 Two-dimensional problem solution ng 8.2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8.3 Cartesian Coordinate Solutions Using Fourier Methods CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ + 2 + = ∇ 4ϕ = ∂x ∂x ∂y ∂y ng c om The biharmonic equation co In Cartesian coordinates we choose Airy stress function solution of polynomial form ∞ ∞ m=0 n =0 ng th an ϕ ( x, y ) = ∑∑ Amn x m y n du o where Amn are constant coefficients to be determined This method produces polynomial stress distributions, and thus would not satisfy general boundary conditions However, we can cu u modify such boundary conditions using Saint-Venant’s principle and replace a non-polynomial condition with a statically equivalent loading This formulation is most useful for problems with rectangular domains, and is commonly based on the inverse solution concept where we assume a polynomial solution form and then try to find what problem it will solve CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science ∞ ∞ ϕ ( x, y ) = ∑∑ Amn x y m c om 8.2 Cartesian Coordinate Solutions Using Polynomials ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ + + = ∇ ϕ =0 2 ∂x ∂x ∂y ∂y n ng m=0 n =0 co Noted that the three lowest order terms with m + n ≤ not contribute to the stresses and an will therefore be dropped It should be noted that second order terms will produce a constant th stress field, third-order terms will give a linear distribution of stress, and so on for higher-order ng polynomials du o Terms with m + n ≤ will automatically satisfy the biharmonic equation for any choice of cu u constants Amn However, for higher order terms, constants Amn will have to be related in order to have the polynomial satisfy the biharmonic equation For example, the 4th-order polynomial terms A40x4+A22x2y2+A04y4 will not satisfy the biharmonic equation unless 3A40+A22+3A04=0 This condition specifies one constant in terms of the other two, thus leaving two constants to be determined by the boundary conditions CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials ∞ ∑∑ m(m − 1)(m − 2)(m − 3) A mn m= n =0 ∞ x m−4 ∞ ∞ y + ∑∑ m(m − 1)n(n − 1) Amn x m − y n − n m=2 n=2 ∞ co + ∑∑ n(n − 1)(n − 2)(n − 3) Amn x m y n − = ng ∞ c om Considering the general case, substituting the series into the governing biharmonic equation yields an m=0 n =4 ∞ th Collecting like powers of x and y, the preceding equation may be written as ∞ ng ∑∑ ⎡⎣(m + 2)(m + 1)m(m − 1) A m=2 n=2 m + 2, n − + 2m(m − 1)n(n − 1) Amn + du o +(n + 2)(n + 1)n(n − 1) Am − 2,n + ⎤⎦ x m − y n − = cu u Because this relation must be satisfied for all values of x and y, the coefficient in brackets must vanish, giving the result (m + 2)(m + 1)m(m − 1) Am + 2,n − + 2m(m − 1)n(n − 1) Amn + (n + 2)(n + 1)n(n − 1) Am − 2,n + = For each m, n pair, this equation is the general relation that must be satisfied to ensure that the polynomial grouping is biharmonic CuuDuongThanCong.com https://fb.com/tailieudientucntt   h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials y   T   c om Example 8.1 Uniaxial Tension of a Beam Stress  Field     th an ⎧⎪σ x (±l , y ) = T , σ y ( x, ±c) = Boundary Conditions: ⎨ ⎪⎩τ xy (±l , y ) = τ xy ( x, ±c) = co 2l   u du o ng Since the boundary conditions specify constant stresses on all boundaries, try a second-order stress function of the form σ x = A02 , σ y = τ xy = ϕ = A02 y T   x   ng 2c   Displacement  Field  (Plane  Stress)     ∂u T = ex = (σ x −νσ y ) = ∂x E E ∂v T = ey = (σ y −νσ x ) = −ν ∂y E E u= T T x + f ( y ) , v = −ν y + g ( x) E E τ ∂u ∂v + = 2exy = xy = ⇒ f ′( y ) + g ′( x) = ∂y ∂x µ cu The first boundary condition implies that A02 = T/2, and all other boundary conditions are f ( y ) = −ωo y + uo identically satisfied Therefore the stress field g ( x) = ωo x + vo Rigid-Body Motion solution is given by “Fixity conditions” needed to determine RBM terms σ x = T , σ y = τ xy = CuuDuongThanCong.com u (0, 0) = v(0, 0) = u (0, c) = ⇒ f ( y ) = g ( x) = https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials Example 8.2 Pure Bending of a Beam c om y   2l   Boundary Conditions: ∫ c −c σ x (±l , y ) ydy = − M ng −c σ x (±l , y )dy = , du o ∫ c th σ y ( x, ±c) = , τ xy ( x, ±c) = τ xy ( ±l , y ) = an co Stress Field   σ x = A03 y , σ y = τ xy = cu ϕ = A03 y u Expecting a linear bending stress distribution, try 2nd- stress function of the form Moment boundary condition implies that A03 = -M/4c3, and all other boundary conditions are identically satisfied Thus the stress field is 3M σ x = − y , σ y = τ xy = 2c CuuDuongThanCong.com M   x   ng 2c   M   Displacement Field (Plane Stress) ∂u 3M 3M =− y ⇒u=− xy + f ( y ) ∂x Ec Ec3 ∂v 3M 3Mν =ν y ⇒v= y + g ( x) ∂y Ec Ec3 ∂u ∂v 3M + =0⇒ − x + f ′( y ) + g ′( x) = ∂y ∂x Ec3 ⎧ f ( y ) = −ω0 y + u0 ⎪ ⎨ 3M g ( x ) = x + ω0 x + v0 ⎪⎩ Ec3 “Fixity conditions” to determine RBM terms: v(±l , 0) = and u (−l , 0) = u0 = ω0 = , v0 = −3Ml /16 Ec3 https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials Solution Comparison of Elasticity with Elementary Mechanics of Materials c om y   2c   M   x   co ng M   2l   an   th Elasticity Solution M y , σ y = τ xy = I Mxy M u=− ,v= [4ν y + x − l ] EI 8EI cu u du o ng σx = − I = 2c / Mechanics of Materials Solution Uses Euler-Bernoulli beam theory to find bending stress and deflection of beam centerline M y , σ y = τ xy = I M v = v( x, 0) = [4 x − l ] EI σx = − Two solutions are identical, with the exception of the x-displacements CuuDuongThanCong.com https://fb.com/tailieudientucntt   Institute for computational science h"p://incos.tdt.edu.vn   8.2 Cartesian Coordinate Solutions Using Polynomials Example 8.3 Bending of a Beam by Uniform Transverse Loading co ng 2c   w   wl   y   2l   ng th an wl   Boundary Conditions: c ⎧τ (x,±c) = 0; ∫ −c τ xy (±l, y)d y = ∓wl ⎪ xy ⎪⎪ c σ (x,c) = 0; ⎨ y ∫ −cσ x (±l, y) y d y = ⎪ c ⎪σ (x,−c) = −w; σ (±l, y)d y = ∫ −c x ⎪⎩ y c om Stress Field du o ϕ = A20 x + A21 x y + A03 y + A23 x y − cu u ⎧ σ = A y + A ( x y − y ) 03 23 ⎪ x ⎪⎪ ⎨σ y = A20 + A21 y + A23 y ⎪ ⎪τ xy = −2 A21 x − A23 xy ⎪⎩ CuuDuongThanCong.com BC’s     A23 y ⎧ 3w ⎛ l 2 ⎞ 3w 2 ⎪σ x = ⎜ − ⎟ y − (x y − y ) 4c ⎝ c ⎠ 4c ⎪ ⎪⎪ w 3w w y − y3 ⎨σ y = − + 4c 4c ⎪ 3w 3w ⎪ τ = − x + xy ⎪ xy c c ⎪⎩ https://fb.com/tailieudientucntt x     h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials Example 8.3 Bending of a Beam by Uniform Transverse Loading c om w   wl   x   co y   ng 2c   wl   th Elasticity Solution an 2l   u du o ng w w y3 c2 y σ x = (l − x ) y + ( − ) 2I I w ⎛ y3 ⎞ σ y = − ⎜ − c y + c3 ⎟ 2I ⎝ 3 ⎠ cu w τ xy = − x(c − y ) 2I Mechanics of Materials Solution My w = (l − x ) y I 2I σy = σx = τ xy = VQ w = − x (c − y ) It 2I Shear stresses are identical, while normal stresses are not CuuDuongThanCong.com https://fb.com/tailieudientucntt   Institute for computational science h"p://incos.tdt.edu.vn   c om 8.2 Cartesian Coordinate Solutions Using Polynomials σx  –  Stress  at  x=0   co ng σy  -­‐  Stress   an l/c  =  2   ng σx/w  -­‐  Elasticity     σx/w  -­‐  Strength  of  Materials   du o l/c  =  4   th l/c  =  3   cu u Maximum differences between the two theories exist at top and bottom of beam, and actual difference in stress values is w/ For most beam problems where l >> c, the bending stresses will be much greater than w, and thus the differences between elasticity and strength of materials will be relatively small CuuDuongThanCong.com  σy/w  -­‐  Elasticity      σy/w  -­‐  Strength  of  Materials   Maximum difference between the two theories is w and this occurs at the top of the beam Again this difference will be negligibly small for most beam problems where l >> c These results are generally true for beam problems with other transverse loadings https://fb.com/tailieudientucntt   h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials Example 8.3 Bending of a Beam by Uniform Transverse Loading c om w   wl   wl   2c   2l   co y   ng x   du o ng th an ⎧ w x3 y 2c y y3 2c 2 u= [(l x − ) y + x( − ) +ν x( − c y + )] + f ( y ) Displacement  Field   ⎪⎪ EI 3 3 ⎨ 2 2  (Plane  Stress)     ⎪v = − w [( y − c y + 2c y ) + ν (l − x ) y + ν ( y − c y )] + g ( x) ⎪⎩ EI 12 cu u f ( y ) = ω0 y + u , g ( x ) = Choosing Fixity Conditions w w x − [l − ( + ν )c ]x − ω0 x + v0 24 EI EI u (0, y ) = v(±l , 0) = 5wl u0 = ω0 = 0, v0 = 24 EI CuuDuongThanCong.com ⎡ 12 ⎛ ν ⎞ c ⎤ ⎢1 + ⎜ + ⎟ l ⎥ ⎝ ⎠ ⎦ ⎣ https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.2 Cartesian Coordinate Solutions Using Polynomials c om Example 8.3 Bending of a Beam by Uniform Transverse Loading ⎡⎛ ⎛ y 2c y ⎞ ⎛ y3 x3 ⎞ 2c ⎞ ⎤ − ⎢⎜ l x − ⎟ y + x ⎜ ⎟ +ν x ⎜ − c y + ⎟⎥ 3 3 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎧ y c y 2c y ⎡ y4 c2 y ⎤ ⎫ y + + ν ⎢( l − x ) + − ⎪ − ⎥⎪ 12 ⎦⎪ w ⎪ ⎣ v=− ⎨ ⎬ EI ⎪ x ⎡ l ⎛ ν ⎞ ⎤ ⎪ − + + + c x ⎜ ⎟ ⎢ ⎥ ⎪ 12 ⎪ ⎣2 ⎝5 2⎠ ⎦ ⎩ ⎭ co ng th an Displacement  Field    (Plane  Stress)     ng w u= EI ⎡ 12 ⎛ ν ⎞ c ⎤ ⎢1 + ⎜ + ⎟ l ⎥ ⎝ ⎠ ⎦ ⎣ u du o 5wl + 24 EI cu v(0, 0) = vmax 5wl = 24 EI Strength of Materials: vmax CuuDuongThanCong.com ⎡ 12 ⎛ ν ⎞ c ⎤ ⎢1 + ⎜ + ⎟ l ⎥ ⎝ ⎠ ⎦ ⎣ 5wl = 24 EI Good match for beams where l >> c https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science c om 8.1 Two-dimensional problem solution ng 8.2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8.3 Cartesian Coordinate Solutions Using Fourier Methods CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   Institute for computational science 8.3 Cartesian Coordinate Solutions Using Fourier Methods c om A more general solution scheme for the biharmonic equation may be found using Fourier methods Such techniques generally use separation of variables along with Fourier series or Fourier integrals ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ +2 2 + =0 ∂x ∂x ∂y ∂y α = ±i β an αx βy Choosing X = e , Y = e co ng ϕ ( x, y ) = X ( x)Y ( y ) th φ = sin β x ⎡⎣( A + C β y ) sinh β y + ( B + Dβ y ) cosh β y ⎤⎦ ng + cos β x ⎡⎣( A′ + C ′β y ) sinh β y + ( B′ + D′β y ) cosh β y ⎤⎦ du o + sin α y ⎡⎣( E + Gα x ) sinh α x + ( F + H α x ) cosh α x ⎤⎦ cu + φα =0 + φβ =0 e x − e− x sinh( x) = = −i sin(ix) x e + e− x cosh( x) = = cos ix u + cos α y ⎡⎣( E ′ + G′α x ) sinh α x + ( F ′ + H ′α x ) cosh α x ⎤⎦ The general solution includes the superposition of the general roots plus the zero root cases (zero root solutions) ⎧⎪φβ =0 = C0 + C1 x + C2 x + C3 x where ⎨ 2 ⎪⎩φα =0 = C4 y + C5 y + C6 y + C7 xy + C8 x y + C9 xy CuuDuongThanCong.com https://fb.com/tailieudientucntt   h"p://incos.tdt.edu.vn   Institute for computational science 8.3 Cartesian Coordinate Solutions Using Fourier Methods qosinπx/l   qol/π   2c   th (2) σ y ( x, − c ) = (3) u cu ∫ −c c −c σ x = β sin β x ⎡⎣ A sinh β y + C ( β y sinh β y + cosh β y ) du o (1) ng σ x (0, y ) = σ x (l , y ) = τ xy ( x, ±c) = σ y ( x, c) = −qo sin(π x / l ) l   ϕ = sin β x ⎡⎣( A + C β y ) sinh β y + ( B + D β y ) cosh β y ⎤⎦ Boundary Conditions: ∫ x   an co Stress Field c qol/π   ng Example 8.4 Beam with Sinusoidal Loading c om y   (4) τ xy (0, y )dy = −qol / π (5) τ xy (l , y )dy = qol / π (6) CuuDuongThanCong.com + B cosh β y + D ( β y cosh β y + 2sinh β y )⎤⎦ σ y = − β sin β x ⎡⎣( A + C β y ) sinh β y + ( B + Dβ y ) cosh β y ⎤⎦ τ xy = − β cos β x ⎡⎣ A cosh β y + C ( β y cosh β y + 2sinh β y ) + B sinh β y + D ( β y sinh β y + cosh β y )⎤⎦ https://fb.com/tailieudientucntt   h"p://incos.tdt.edu.vn   Institute for computational science 8.3 Cartesian Coordinate Solutions Using Fourier Methods qosinπx/l   qol/π   Example 8.4 Beam with Sinusoidal Loading c om y   qol/π   2c   co ng x   an th du o ng Condition (2) gives ⎧⎪ A = − D ( β c β c + 1) ⎨ ⎪⎩ B = −C ( β c coth β c + 1) cu β= π l πc l π ⎡π c πc πc⎤ 2 ⎢ + sinh cosh ⎥ l ⎣ l l l ⎦ u Condition (3) gives C = −qo sinh l   D= −qo sinh l π ⎡π c πc πc⎤ 2 ⎢ − sinh cosh ⎥ l ⎣ l l l ⎦ Condition (1) and condition (5,6) will be satisfied CuuDuongThanCong.com πc https://fb.com/tailieudientucntt   h"p://incos.tdt.edu.vn   Institute for computational science 8.3 Cartesian Coordinate Solutions Using Fourier Methods qosinπx/l   qol/π   Example 8.4 Beam with Sinusoidal Loading qol/π   x   ng 2c   c om y   co Displacement Field l   β β } th an cos β x { A (1 + ν ) sinh β y + B (1 + ν ) cosh β y v = − sin β x { A (1 + ν ) cosh β y + B (1 + ν ) sinh β y E E + C ⎡⎣(1 + ν ) β y sinh β y + cosh β y ⎤⎦ + C ⎡⎣(1 + ν ) β y cosh β y − (1 +ν ) sinh β y ⎤⎦ u=− ng + D ⎡⎣(1 +ν ) β y cosh β y + 2sinh β y ⎤⎦ − ωo y + uo du o u (0, 0) = v(0, 0) = v(l , 0) = } + D ⎡⎣(1 +ν ) β y sinh β y − (1 +ν ) cosh β y ⎤⎦ + ωo y + vo ω0 = v0 = , u0 = β ⎡ B (1 + ν ) + 2C ⎤⎦ E⎣ cu u Dβ sin β x ⎡⎣ + (1 +ν ) β c β c ⎤⎦ E 3q0l π x ⎡ +ν π c πc⎤ 3q0l v( x, 0) = − sin + D≈− For the case l >> c 2c π E l ⎢⎣ l l ⎥⎦ 4c π 3q0l πx v ( x , 0) = − sin Strength of Materials 2c3π E l v( x, 0) = CuuDuongThanCong.com https://fb.com/tailieudientucntt   Institute for computational science h"p://incos.tdt.edu.vn   Must use series representation for Airy stress function to handle general boundary loading βn = n =1 ∞ th m =1 ∞ du o ∞ ng σ x = ∑ β n2 cos β n x ⎡⎣ Bn cosh β n y + Cn ( β n y sinh β n y + cosh β n y )⎤⎦ n =1 − ∑ α m2 cos α m y [ Fm cosh α m x + Gmα m x sinh α m x ] m =1 ∞ x    p(x)   Boundary Conditions σ x ( ± a, y ) = τ xy (± a, y ) = τ xy ( x, ±b) = σ y ( x, ± b ) = − p ( x ) cu u σ y = −∑ β n2 cos β n x [ Bn cosh β n y + Cn β n y sinh β n y ] + 2C0 a   b   an + ∑ cos α m y [ Fm cosh α m x + Gmα m x sinh α m x ] + C0 x nπ l a   ng ϕ = ∑ cos β n x [ Bn cosh β n y + Cn β n y sinh β n y ] b    p(x)   co ∞  y   c om 8.3 Cartesian Coordinate Solutions Using Fourier Methods Example 8.5 Rectangular Domain with Arbitrary Boundary Loading Use Fourier series theory to handle + ∑ α m2 cos α m y ⎡⎣ Fm cosh α m x + Gm (α m x sinh α m x + cosh α m x )⎤⎦ general boundary conditions, and this m =1 generates a doubly infinite set of ∞ τ xy = ∑ β n sin β n x ⎡⎣ Bn sinh β n y + Cn ( β n y cosh β n y + sinh β n y )⎤⎦ equations to solve for unknown n =1 constants in stress function form See ∞ + ∑ α m sin α m y ⎡⎣ Fm sinh α m x + Gm (α m x cosh α m x + sinh α m x )⎤⎦ text for details m =1 n =1 ∞ CuuDuongThanCong.com https://fb.com/tailieudientucntt h"p://incos.tdt.edu.vn   cu u du o ng th an co ng c om Institute for computational science CuuDuongThanCong.com https://fb.com/tailieudientucntt ... computational science c om 8. 1 Two- dimensional problem solution ng 8. 2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8. 3 Cartesian Coordinate Solutions Using Fourier Methods... computational science c om 8. 1 Two- dimensional problem solution ng 8. 2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8. 3 Cartesian Coordinate Solutions Using Fourier Methods... computational science c om 8. 1 Two- dimensional problem solution ng 8. 2 Cartesian Coordinate Solutions Using Polynomials cu u du o ng th an co 8. 3 Cartesian Coordinate Solutions Using Fourier Methods