Kul Finite element method I, Exercise 08/2016

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1 Kul Finite element metod I, Eercise 08/016 Demo problems 1. A square tin slab (1) is loaded by a po force () as sown in te figure. Derive te relationsip between te force magnitude F and displacement u X 4 = a. Young s modulus E, Poisson s ratio ν, and tickness of te slab t are constants. e ernal distributed forces are zeros. Assume plane-stress conditions, start wit te virtual work density and use bilinear approimation. L Y,y 3 4 F 1 1 L X, Answer u X 4 F 1 ν = 6 Et 3 ν Y,y A,E 4. A structure, consisting of a tin slab and a bar, is loaded by a orizontal force F acting on node 1. Material properties are E and ν, tickness of te slab is t and te cross-sectional area of te bar is A. Determine displacement of node 1 u X1 = a1 and u = a by using a linear bar element and a linear plane-stress Y1 element. F 1 t,e,ν 3 X, (1 + ν ) F Answer ux1 = 4 t + 4 A(1 + ν ) E u Y1 = 0 3. Po force F is acting on node 1 of te tetraedron element of te figure. Nodes, 3 and 4 are fied so tat displacement components are zeros. Determine displacement u Z1 = a of node 1 if ux1 = uy1 = 0. Material properties E and ν are constants. Use linear approimation. Answer u Z1 = 6 ( 1+ ν)( 1 ν) 1 ν F E y z F e demo problems are publised in te course omepage on Fridays. e problems are related to te topic of te n weeks lecture (ue all K1 15). Solutions to te problems are eplained in te weekly eercise sessions (u all K3 118) and will also be available in te ome page of te course. Please, notice tat te problems of te midterms and te final eam are of tis type.

2 Lecture problem Given te virtual work epression of a tin slab model and approimations for te displacement components, be prepared to write te virtual work epressions. Lecture problems are specified and solved during te lecture (ue all K1 15). e time allocated for tis is 30 min.

3 Home problem y,y A long wall aving triangular cross-section of te figure, and made of omogeneous, isotropic, linearly elastic material, is subjected to its own weigt. Material properties E, ν, ρ are constants. Determine te displacement components u X 3 = a1 and u Y 3 = a of node 3. Nodes 1 and are fied. Use one tree-node element and assume plane strain conditions. L 3 E,ν,ρ 1 L g,x Solution template Under te plane strain conditions, te virtual work densities of tin slab are δu 1 ν ν 0 u Et δw = δv, y ν 1 ν 0 v, y, (1 ν)(1 ν) + δu 0 0 (1 ) /, y δv ν + u, y + v, u f = δ v f. y 1. Only te sape function associated wit node 3 is needed as te oter nodes are fied (displacement vanises) N 3 =. In terms of te displacement components u X 3 = a 1 and u Y 3 = a of node 3, element approimations of te displacement components and teir derivatives are u = v u, = v, and u, y =. v, y 3. Wen te approimation is substituted tere, te virtual work densities simplify to δu 1 ν ν 0 u Et δw = δv, y ν 1 ν 0 v, y (1 ν)(1 ν) + δu 0 0 (1 ) /, y δv ν + u, y + v,

4 1 ν ν 0 Et δw = ν 1 ν 0 (1 ν)(1 ν) (1 ν ) / a1 a1 = δa a and u f a1 = δ v f = y δ a. 4. Integrations over te element give te virtual work epressions (note tat te virtual work density of ernal forces is constant) δa = δw d = δa a and 1 a1 L ( L y)/ = = 0 ( y L)/ δw d ( δw d ) dy a1 L a1 = dy δa = 0 δa. 5. Principle of virtual work = + = 0 δa gives a1 = δ a 1 ( a a a1 ) = 0 δ a

5 a1 = 0 a a1 = a e compulsory ome problems are publised in te course omepage on Fridays and te deadline for answers is te n weeks Friday Return your omework answers o te green course mailbo tat can be found from te corridor of te K3 building lobby (Puumieenkuja 5A). Please, use te solution templates given.

6 Kul Finite element metod I; Formulae collection GENERAL i ix iy iz I I j jx jy j = Z J = i j k J k kx ky k Z K K Coordinate systems: { } X 1 i = Y Z Strain-stress: ε 1 ν ν σ 1 ε yy ν 1 ν = σyy E ε ν ν 1 zz σzz γy σy 1 γ yz = σyz G γz σz E G = (1 + ν ) or σ 1 ν ν ν ε ε E σ yy ν 1 ν ν = εyy [ E] εyy (1 ν)(1 ν) + σ ν ν 1 ν zz εzz εzz σy γy σyz = G γ yz σ γ z z 1 ν 0 E [ E] σ = ν ν 0 0 (1 ν ) / 1 ν ν 0 E [ E] ε = ν 1 ν 0 (1 + ν)(1 ν) 0 0 (1 ν ) / Strain-displacement: ε u, ε yy = uy, y ε u zz z, z γ y u, y + uy, γ yz = uy, z + uz, y γ u + u z z, z ELEMEN CONRIBUION (constant load) Bar (aial): F EA 1 1 u f = F u 1 EA a1 = R ii ii f i, in wic R ii ii a i X 1 i = Y Z Bar (torsion): M GI 1 1 θ m 1 = rr 1 M θ Beam (z): Fz uz 1 6 M y1 EI yy y1 z θ f = F 3 z uz 1 6 M y y θ

7 FX ux1 FX Po loads: F Y = uy1 F Y F Z u Z1 F Z PRINCIPLE OF VIRUAL WORK MX θ X1 MX M Y = θy1 M Y M M θ Z1 Z1 Z e = + = = 0 δa = δwd e E Bar: δw = δu EAu δw = δuf Bar (torsion): = δφ GI φ = δφm rr Beam (z): Beam (y): δw = EI w δw = δwf z yy δw = δ veizzv, δw = δvf y Beam (Bernoulli): A S z Sy u, u = δ v E S I I v δφ GI φ w S w δ y Iyz Iyy z zz zy rr u f δφ Sy fy + Sz fz δw = δv fy+ δw, Sy f w f v δ δ S f Plane-stress (y): z z z δu u δw = δv, y te [ ] σ v, y δu, y + δv u, y + v, u f = δ v f y Plane-strain (y): δu u δw = δv, y te [ ] ε v, y δu, y + δv u, y + v, u f = δ v f y Kircoff-plate (y):

8 w w 3 t δw = δw, yy [ E] σ w, yy 1 δ wy w, y Reissner-Mindlin plate (y): δw = δwf z δθ θ 3 t w, y δφ w, y φ = δφ, y [ E] σ φ, y tg 1 + δθ w, + θ δφ δθ, y φ θ, y δw = δwf z Body (yz): ε σ γ y σ y yy yy yz yz = δε σ δγ σ δε σ δγ σ zz zz z z u f δ y δw = v f δ w f z or u,, y + δ, y + u u v u v δw = δv, y [ E] v, y δv, z + δw, y G v, z + w, y δw w δw + δu w + u, z, z, z, z APPROXIMAIONS (some) u = N a ξ = Quadratic (line): N1 1 3ξ + ξ N = N = 4 ξ(1 ξ) N ξ(ξ 1) 3 u1 a = u (bar) u 3 Cubic (line): N10 (1 ξ) (1 + ξ) N 11 (1 ξ) ξ N 0 N = = (3 ξξ ) N 1 ξ ( ξ 1) u10 uz1 u θ 11 y1 a = ( = ) (beam bending) u u 0 z u 1 θ y Linear (triangle): N = 1 y1 y y 3 y VIRUAL WORK EXPRESSIONS uxi FXi θ Xi MXi Rigid body (force): = δuyi FYi + δθyi MYi δu Zi F Zi δθ Zi M Zi

9 Bar (aial): 1 EA 1 1 u1 u = δu 1 1 u u1 f 1 = δ u 1 Bar (torsion): θ1 GIrr 1 1 θ 1 δθ 1 1 θ = θ1 m 1 = ) δθ 1 Beam (z): z uz1 y1 EI yy θ y1 3 z z y θ y u δθ = δu u δθ uz1 6 δθ y1 z f = δuz 1 6 δθ y Beam (y): y uy1 z1 EIzz θz1 3 y uy z θz u δθ = δ u δθ uy1 6 δθz1 fy = ) δ uy 1 6 δθ z CONSRAINS Frictionless contact: n u A = 0 Jo: ub = ua Rigid body (link): ub = ua + θa ρab. θb = θa

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