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Constant strain triangular
- 1. MANE 4240 & CIVL 4240 Introduction to Finite Elements Constant Strain Triangle (CST) Prof. Suvranu De
- 2. Reading assignment: Logan 6.2-6.5 + Lecture notes Summary: • Computation of shape functions for constant strain triangle • Properties of the shape functions • Computation of strain-displacement matrix • Computation of element stiffness matrix • Computation of nodal loads due to body forces • Computation of nodal loads due to traction • Recommendations for use • Example problems
- 3. Finite element formulation for 2D: Step 1: Divide the body into finite elements connected to each other through special points (“nodes”) x y Su ST u v x px py Element ‘e’ 3 2 1 4 y x v u 1 2 3 4 u1 u2 u3 u4 v4 v3 v2 v1 = 4 4 3 3 2 2 1 1 v u v u v u v u d
- 5. TASK 2: APPROXIMATE THE STRAIN and STRESS WITHIN EACH ELEMENT ......v y)(x,N u y)(x,Ny)(x,vy)(x,u v y)(x,N v y)(x,N v y)(x,N v y)(x,Ny)(x,v u y)(x,N u y)(x,N u y)(x,N u y)(x,Ny)(x,u 1 1 1 1 xy 4 4 3 3 2 2 1 1 y 4 4 3 3 2 2 1 1 x + ∂ ∂ + ∂ ∂ ≈ ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ ≈ ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ ≈ ∂ ∂ = xyxy yyyyy xxxxx γ ε ε Approximation of the strain in element ‘e’
- 7. Displacement approximation in terms of shape functions Strain approximation in terms of strain-displacement matrix Stress approximation Summary: For each element Element stiffness matrix Element nodal load vector dNu = dBD=σ dBε = ∫= e V k dVBDB T S e T b e f S S T f V T dSTdVXf ∫∫ += NN
- 8. Constant Strain Triangle (CST) : Simplest 2D finite element • 3 nodes per element • 2 dofs per node (each node can move in x- and y- directions) • Hence 6 dofs per element x y u3 v3 v1 u1 u2 v2 2 3 1 (x,y) v u (x1,y1) (x2,y2) (x3,y3)
- 9. 166212 dNu ××× = = = 3 3 2 2 1 1 321 321 v u v u v u N0N0N0 0N0N0N y)(x,v y)(x,u u The displacement approximation in terms of shape functions is = 321 321 N0N0N0 0N0N0N N 1 1 2 2 3 3u (x,y) u u uN N N≈ + + 1 1 2 2 3 3v(x,y) v v vN N N≈ + +
- 10. Formula for the shape functions are A ycxba N A ycxba N A ycxba N 2 2 2 333 3 222 2 111 1 ++ = ++ = ++ = 12321312213 31213231132 23132123321 33 22 11 x1 x1 x1 det 2 1 xxcyybyxyxa xxcyybyxyxa xxcyybyxyxa y y y triangleofareaA −=−=−= −=−=−= −=−=−= == where x y u3 v3 v1 u1 u2 v2 2 3 1 (x,y) v u (x1,y1) (x2,y2) (x3,y3)
- 11. Properties of the shape functions: 1. The shape functions N1, N2 and N3 are linear functions of x and y x y 2 3 1 1 N1 2 3 1 N2 1 2 3 1 1 N3 = nodesotherat inodeat 0 ''1 Ni
- 12. 2. At every point in the domain yy xx i i = = = ∑ ∑ ∑ = = = 3 1i i 3 1i i 3 1i i N N 1N
- 13. 3. Geometric interpretation of the shape functions At any point P(x,y) that the shape functions are evaluated, x y 2 3 1 P (x,y) A1 A3 A2 A A A A A A 3 3 2 2 1 1 N N N = = =
- 14. Approximation of the strains xy u v u v x y x Bd y y x ε εε γ ∂ ∂ ∂ = = ≈ ∂ ∂ ∂ + ∂ ∂ = ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = 332211 321 321 332211 321 321 000 000 2 1 y)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,Ny)(x,N y)(x,N 0 y)(x,N 0 y)(x,N 0 0 y)(x,N 0 y)(x,N 0 y)(x,N bcbcbc ccc bbb A xyxyxy yyy xxx B
- 15. Element stresses (constant inside each element) dBD=σ Inside each element, all components of strain are constant: hence the name Constant Strain Triangle
- 16. IMPORTANT NOTE: 1. The displacement field is continuous across element boundaries 2. The strains and stresses are NOT continuous across element boundaries
- 17. Element stiffness matrix ∫= e V k dVBDB T Atk e V BDBdVBDB TT == ∫ t=thickness of the element A=surface area of the element Since B is constant t A
- 18. S e T b e f S S T f V T dSTdVXf ∫∫ += NN Element nodal load vector
- 19. Element nodal load vector due to body forces ∫∫ == ee A T V T b dAXtdVXf NN = = ∫ ∫ ∫ ∫ ∫ ∫ e e e e e e A b A a A b A a A b A a yb xb yb xb yb xb b dAXNt dAXNt dAXNt dAXNt dAXNt dAXNt f f f f f f f 3 3 2 2 1 1 3 3 2 2 1 1 x y fb3x fb3y fb1y fb1x fb2x fb2y 2 3 1 (x,y) Xb Xa
- 20. EXAMPLE: If Xa=1 and Xb=0 = = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ 0 3 0 3 0 3 0 0 0 3 2 1 3 3 2 2 1 1 3 3 2 2 1 1 tA tA tA dANt dANt dANt dAXNt dAXNt dAXNt dAXNt dAXNt dAXNt f f f f f f f e e e e e e e e e A A A A b A a A b A a A b A a yb xb yb xb yb xb b
- 21. Element nodal load vector due to traction ∫= e TS S T S dSTf N EXAMPLE: x y fS3x fS3y fS1y fS1x 2 3 1 ∫− − = e l S along T S dSTtf 31 31 N
- 22. Element nodal load vector due to traction EXAMPLE: x y fS3x 2 31 ∫ − − = e l S along T S dSTtf 32 32 N fS3y fS2x fS2y (2,0) (2,2) (0,0) = 0 1 ST tt dyNtf ex l alongS =×× = = ∫ − − 12 2 1 )1( 32 2 322 0 0 3 3 2 = = = y x y S S S f tf f Similarly, compute 1 2
- 23. Recommendations for use of CST 1. Use in areas where strain gradients are small 2. Use in mesh transition areas (fine mesh to coarse mesh) 3. Avoid CST in critical areas of structures (e.g., stress concentrations, edges of holes, corners) 4. In general CSTs are not recommended for general analysis purposes as a very large number of these elements are required for reasonable accuracy.
- 24. Example x y El 1 El 2 1 23 4 300 psi 1000 lb 3 in 2 in Thickness (t) = 0.5 in E= 30×106 psi ν=0.25 (a) Compute the unknown nodal displacements. (b) Compute the stresses in the two elements.
- 25. Realize that this is a plane stress problem and therefore we need to use psi E D 7 2 10 2.100 02.38.0 08.02.3 2 1 00 01 01 1 × = −− = ν ν ν ν Step 1: Node-element connectivity chart ELEMENT Node 1 Node 2 Node 3 Area (sqin) 1 1 2 4 3 2 3 4 2 3 Node x y 1 3 0 2 3 2 3 0 2 4 0 0 Nodal coordinates
- 26. Step 2: Compute strain-displacement matrices for the elements = 332211 321 321 000 000 2 1 bcbcbc ccc bbb A B Recall 123312231 213132321 xxcxxcxxc yybyybyyb −=−=−= −=−=−= with For Element #1: 1(1) 2(2) 4(3) (local numbers within brackets) 0;3;3 0;2;0 321 321 === === xxx yyy Hence 033 202 321 321 ==−= −=== ccc bbb −− − − = 200323 003030 020002 6 1)1( B Therefore For Element #2: −− − − = 200323 003030 020002 6 1)2( B
- 27. Step 3: Compute element stiffness matrices 7 )1(T)1()1(T)1()1( 10 2.0 05333.0 02.02.1 3.00045.0 2.02.02.13.04.1 3.05333.02.045.05.09833.0 BDB)5.0)(3(BDB × − − −− −−− = == Atk u1 u2 u4 v4v2 v1
- 29. Step 4: Assemble the global stiffness matrix corresponding to the nonzero degrees of freedom 014433 ===== vvuvu Notice that Hence we need to calculate only a small (3x3) stiffness matrix 7 10 4.102.0 0983.045.0 2.045.0983.0 × − − =K u1 u2 v2 u1 u2 v2
- 30. Step 5: Compute consistent nodal loads = y x x f f f f 2 2 1 = yf2 0 0 ySy ff 2 10002 +−= The consistent nodal load due to traction on the edge 3-2 lb x dx x dxN tdxNf x x x S y 225 2 9 50 2 50 3 150 )5.0)(300( )300( 3 0 2 3 0 3 0 233 3 0 2332 −= −= −= −= −= −= ∫ ∫ ∫ = = − = − 3 2 3232 x N =−
- 31. lb ff ySy 1225 1000 22 −= +−= Hence Step 6: Solve the system equations to obtain the unknown nodal loads fdK = − = − − × 1225 0 0 4.102.0 0983.045.0 2.045.0983.0 10 2 2 1 7 v u u Solve to get ×− × × = − − − in in in v u u 4 4 4 2 2 1 109084.0 101069.0 102337.0
- 32. Step 7: Compute the stresses in the elements )1()1()1( dBD=σ With [ ] [ ]00109084.0101069.00102337.0 d 444 442211 )1( −−− ×−××= = vuvuvu T Calculate psi − − − = 1.76 1.1391 1.114 )1( σ In Element #1
- 33. )2()2()2( dBD=σ With [ ] [ ]44 224433 )2( 109084.0101069.00000 d −− ×−×= = vuvuvu T Calculate psi − = 35.363 52.28 1.114 )2( σ In Element #2 Notice that the stresses are constant in each element