MEI solutions to exercise 4 1
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1 MEI-55 solutions to exercise Problem Solve the stationary two-dimensional heat transfer problem shown in the figure below by using linear elements Use symmetry to reduce the problem size The material is assumed to be homogeneous and isotropic with thermal conductivity k The loading is given with prescribed heat flux on the boundary x = L as q s = q c (y/l)( y/l) i O K " $ K G! "! # K N Solution The weak form is simply (now f ) ( w) T k u da = wq n da Using the elementwise interpolation u (e) = Nu (e) and w (e) = Nw (e), the element stiffness matrix has the form kb T B da, (e) where B is the discrete gradient operator matrix In the case of linear elements, it has the form [ ] b b B = b 3 c c c 3 The elements and 3 are equal as well as elements and The local nodes are numerated as shown in the table below thus for the elements and 3 the coefficients which are needed are: b = y y 3 =, b = y 3 y = L, b 3 = y y = L, c = x 3 x = L, c = x x 3 =, c 3 = x x = L node elem
2 MEI-55 solutions to exercise The area of all elements is A = L, thus K () = K (3) = k + k For elements and : K () = K () = k b = y y 3 = L, b = y 3 y = L, b 3 = y y =, + k = k c = x 3 x =, c = x x 3 = L, c 3 = x x = L = k 5 5 Since the teperature is prescribed for nodes,,3 and 5 there are only two active unknowns, ie u and u 6 The local-global degrees of freedom are related as shown in the table below (same table as before, but now the essential boundary nodes are not shown) node elem The global stiffness matrix can be assembled from the element contributions as thus K = K () + K() 33 + K(3) 33, K 6 = K (3) 3, K 66 = K (3) + K() 33, [ ] K K K = 6 = k K 6 K 66 [ 5 In the load vector there is only one non-zero component, which is f 6 = N 3 q nds = N 3 q c (y/l) ( (y/l)) ds = L (q c + 3 q c + ) = 5 q cl The global balance equations are k which has the solution { u = 9 u 6 [ 5 [ 5 ] { u u 6 ] { 5 = { 5 ] q c L, qc L k = 5 { 9 qc L k
3 MEI-55 solutions to exercise 3 The heat flux is constant in each element q (e) = kbu (e) = k A (e) [ b b b 3 c c c 3 ] u (e) u (e) u (e) 3 For element, u () = u () =, u () 3 = u 6, thus q (e) = k { { b3 u 6 = A (e) c 3 5/7 q c Calculate the heat flux for other elements and draw the vectors!
4 MEI-55 solutions to exercise Problem Compute the St Venant s torsion constant I t for a beam having a square cross-section (side length L) and made of a homogeneous isotropic material, by using the finite element method and a triangular mesh as shown in the figure below The problem can be formulated with St Venant s stress function Φ as Φ,xx Φ,yy =, with boundary conditions Φ = The torsional constant is obtained from equation I t = Φ(x, y)da Determine also the shear stress distribution from a twist θ = /L The shear stresses can be computed from τ zx = Φ,y, τ zy = Φ,x y u 3 u u 3 L L x Solution The weak form of the PDE is ˆΦ Φ da = The first integral can be transformed as ˆΦ Φ da = ˆΦ Φ da = ˆΦ da ˆΦ Φ n ds+ ˆΦ Φ da = ˆΦ Φ da The stiffness matris is thus similar to the heat transfer problem when the thermal conductivity k = The element matrices from elements, 3 and are identical Thus it is necessary to form only the element matrices for elements and The constants b i and c i are calculated in the table below b = y y 3, c = x 3 x b = y 3 y, c = x x 3 () b 3 = y y, c 3 = x x
5 MEI-55 solutions to exercise 5 K () = elements,3, element i b i c i b i c i -L/ -L/ -L/ L/ L/ L/ 3 L/ -L/ The area of all elements is A (e) = L /3 and the element matrices are + = K () = =, () + The local-global numbering is shown in the table below node elem The assembly of the global stiffness matrix is below The global stiffness matrix is thus K = K () 33 + K() 33 + K(), (3) K = K () 3 + K(), K 3 = K () 3, () K = K () + K(3) 33 + K(), K 3 = K () 3, K 33 = K () 33, K = The inverse of the stiffness matrix is K = 3 3, (5) (6)
6 MEI-55 solutions to exercise 6 The load vector is f = f () 3 + f () 3 + f () = L = 6 L, f = 6 L, (7) f 3 = 8 L Finally the nodal values of the stress function Φ i can be computed as Φ = 96 GL θ 6GL θ, Φ = 7 9 GL θ 885GL θ, (8) Φ 3 = 5 3 GL θ 563GL θ Let s compute the integral I t = ΦdA = 8 e= (e) ΦdA Since the stress function Φ is linear in each element, the element integrals are easy to compute 3 ΦdA = N i Φ i da = 3 (e) (e) 3 A(e) Φ i, and (A (e) = L /3) i= () ΦdA = 3 ΦdA = () 3 ΦdA = (3) 3 () ΦdA = 3 The value for the torsion constant is thus I t = 6 ( 3 3 L L 3 96, L 3 L 3 L ( ), i= 7 9, ( ) ) = 7 5 L 76L The exact solution with three significant digits is I t = L let s finally compute the shear stresses, which are constants in each element since we have used linear elements: τ zx = Φ,y, τ zy = Φ,x Element : Φ = N 3 Φ, Φ,x, Φ,y = c 3 A Φ = GLθ
7 MEI-55 solutions to exercise 7 Element : Element 3: Φ = N Φ + N 3 Φ, Φ,x = b A Φ + b 3 A Φ = 5 8 GLθ, Φ,y = c A Φ + c 3 A Φ = 7 8 GLθ The element correspondingly Φ = N 3 Φ, Φ,x, Φ,y = c 3 A Φ = 7 8 GLθ
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