4 Node Quad

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MANE 4240 & CIVL 4240 Introduction to Finite Elements Prof. Suvranu De Reading assignment: Logan 10.2 + Lecture notes Summary: ã Computation of shape functions for 4-noded quad ã Special case: rectangular element ã Properties of shape functions ã Computation of strain-displacement matrix ã Example problem ãHint at how to generate shape functions of higher order (Lagrange) elements Four-noded rectangular element Finite element formulation for 2D: Step 1: Divide the body into finite elements co
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  1 MANE 4240 & CIVL 4240Introduction to Finite Elements Four-nodedrectangular element Prof. Suvranu De Reading assignment:Logan 10.2 + Lecture notesSummary: ãComputation of shape functions for 4-noded quadãSpecial case: rectangular elementãProperties of shape functionsãComputation of strain-displacement matrixãExample problemãHint at how to generate shape functions of higher order (Lagrange) elements Finite element formulation for 2D:Step 1: Divide the body into finite elements connected to eachother through special points (“ nodes ”)xyS u S T uvx p x  p y Element ‘e’   3214yxvu1234u 1 u 2 u 3 u 4 v 4 v 3 v 2 v 1 ⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎨⎧= 44332211 vuvuvuvud Displacement approximation in terms of shape functions Strain approximation in terms of strain-displacement matrix Stress approximationSummary: For each elementElement stiffness matrixElement nodal load vector d Nu = dBD = σ  dB ε = ∫ = e V  k  dVBDB T     SeT be  f SST  f V T  dST dV  X  f  ∫∫ +=  N N  2 Constant Strain Triangle (CST) : Simplest 2D finite elementã3 nodes per elementã2 dofsper node (each node can move in x-and y-directions)ãHence 6 dofs per element xy u 3 v 3 v 1 u 1 u 2 v 2 231 (x,y) vu (x 1 ,y 1   )(x 2 ,y 2 )(x 3 ,y 3 ) 332211 332211 vy)(x, Nvy)(x, Nvy)(x, Ny)(x,v uy)(x, Nuy)(x, Nuy)(x, Ny)(x,u ++≈++≈ Formula for the shape functions are  A yc xba  N  A yc xba  N  A yc xba  N  222 333322221111 ++=++=++= 12321312213 31213231132 23132123321 332211 x1x1x1det21  x xc y yb y x y xa  x xc y yb y x y xa  x xc y yb y x y xa  y y ytriangleof area A −=−=−= −=−=−= −=−=−= ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡== where xy u 3 v 3 v 1 u 1 u 2 v 2 231 (x,y) vu (x 1 ,y 1 )(x 2 ,y 2 )(x 3 ,y 3 ) Approximation of the strains xy uvuv  x y  x B d  y y x ε ε ε γ   ∂⎧ ⎫⎪ ⎪∂⎧ ⎫⎪ ⎪∂⎪ ⎪⎪ ⎪= = ≈⎨ ⎬ ⎨ ⎬∂⎪ ⎪ ⎪ ⎪⎩ ⎭⎪ ∂ ∂ ⎪+⎪ ⎪∂ ∂⎩ ⎭ ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎣⎡∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂∂= 332211 321321332211 321321 00000021y)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, Ny)(x, N y)(x, N0y)(x, N0y)(x, N00y)(x, N0y)(x, N0y)(x, N bcbcbc cccbbb A x y x y x y  y y y x x x B ⎪⎪⎪⎪⎭⎪⎪⎪⎪⎬⎫⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧⎥⎦⎤⎢⎣⎡=⎭⎬⎫⎩⎨⎧= 332211321321 vuvuvu N0 N0 N0 0 N0 N0 N y)(x,vy)(x,uu d Nu = Approximation of  displacements   Element stiffness matrix ∫ = e V  k  dVBDB T  At k  e V  BDBdVBDB TT == ∫ t=thickness of the elementA=surface area of the elementSince Bis constanttA Element nodal load vector     SeT be  f SST  f V T  dST dV  X  f  ∫∫ +=  N N  3 Class exercise ∫ = eT  SST S dST  f   N For the CST shown below, compute the vector of nodal loads due to surface traction xy f  S3x f  S3y f  S2y f  S2x 231 ∫ − − = e lSalongT S dST t  f  31 32  N p y =-1(0,0)(1,0) Class exercise ∫ − − = e lSalongT S dST t  f  31 32  N xy f  S3x f  S3y f  S2y f  S2x 231p y =-1 ⎭⎬⎫⎩⎨⎧−= 10 S T  The only nonzero nodal loads are ∫ =− = 10322 2  xyalongS dx p N t  f   y ∫ =− = 10323 3  xyalongS dx p N t  f   y ( ) ( )  x x x y x y y x y y y y x y y A x y y y x y x  A xba A yc xba  N   yalong −=−−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡−=−+−=+=⎥⎦⎤⎢⎣⎡++= =− 1)()1(0x10x1x1det)1(x1x1x1det222 2311321113322111113311322 0222322 (can you derive this simpler?) 2)1)(1( 1010322 2 t dx xt dx p N t  f   x xyalongS  y −=−−==⇒ ∫∫ ==− Now compute ∫ =− = 10323 3  xyalongS dx p N t  f   y 4-noded rectangular element with edges parallel to thecoordinate axes: xy (x,y) vu (x 1 ,y 1 )(x 2 ,y 2 )(x 3 ,y 3 ) 1234 (x 4 ,y 4 ) 2b2a ∑ = ≈ 41i y)u(x,y)(x,u ii  N  ∑ = ≈ 41i y)v(x,y)(x,v ii  N  ã4 nodes per elementã2 dofsper node (each node can move in x-and y-directions)ã8 dofs per element  4 Generation of N 1 : xy 12342b2a N 1 l 1 (y) l 1 (x) 11 2121 )(  x x x x xl −−= At node 1has the property 0)(1)( 2111 ==  xl xl Similarly 4141 )(  y y y y yl −−= has the property 0)(1)( 4111 ==  yl yl Hence choose the shape function at node 1 as ( )( ) 42414212111 41)()( y y x x ab y y y y x x x x yl xl N  −−=⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ −−⎟⎟ ⎠ ⎞⎜⎜⎝ ⎛ −−==   ( )( )( )( )( )( )( )( ) 134243312421 41414141  y y x x ab N  y y x x ab N  y y x x ab N  y y x x ab N  −−−= −−=−−−= −−= Using similar arguments, choose Properties of the shape functions:1. The shape functions N 1 , N 2 , N 3 and N 4 are bilinear functionsof x and y ⎩⎨⎧= nodesother at inodeat  y x 0''1),( N i 3. Completeness  y y x x ii === ∑∑∑ === 41ii41ii41ii  N N1 N 2. Kronecker delta property   3. Along lines parallel to the x-or y-axes, the shape functionsare linear. But along any other line they are nonlinear.4. An element shape function related to a specific nodal point iszero along element boundaries not containing the nodal point.5. The displacement field is continuous across elements6. The strains and stresses are not constant within an elementnor are they continuous across element boundaries.
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