Kul Finite element method I, Exercise 07/2016

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1 Kul Finite element metod I, Eercise 07/016 Demo problems y 1. Determine stress components at te midpo of element sown if u y = a and te oter nodal displacements are zeros. e approimations to te displacement components uv, are bi-linear. e material parameters E, ν and tickness t are constants. Use te strain-displacement and stress-strain relationsip of linearly elastic isotropic material and assume plane-stress conditions nswer ν a E yy = 1 1 ν (1 ν ) / y. e kinematic assumptions of Bernoulli beam model are u = u( ) zw ( ), u y = 0 and uz = w ( ). e kinetic assumptions are yy = zz = 0. Derive te epressions of te normal force N( ), bending moment M( ), and sear force Q ( ) by using definitions N = d, M = zd and Q = zd, in wic te egrals are over te cross-section. Moments of te cross-section are = 1d, S = zd, and I = z d. Use te stress-strain and strain-displacement relationsips of a omogeneous, isotropic and linearly elastic material. nswer N = Eu ESw M = ESu EIw Q = 0 3. e kinematic assumptions of imosenko (z-plane) beam model are u = u ( ) + zθ ( ), u y = 0 and uz = w ( ). e kinetic assumptions concerning stress are yy = zz = 0. Derive te epressions of te engineering strain and stress components of te model starting from te generic epressions 1 ν ν γy y 1 yy ν 1 ν 1 = yy E, yz = yz, ν ν 1 G zz zz γz z u, yy = uy, y u zz z, z and γ y u, y + uy, yz = uy, z + uz, y γ u + u z z,, z nswer = u + zθ γz = w + θ = Eu ( + zθ ) z = Gw ( + θ)

2 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.

3 y Determine stress components at te midpo of element sown if u y = a and te oter nodal displacements are zeros. e approimations to te displacement components uv, are bi-linear. e material parameters E, ν and tickness t are constants. Use te strain-displacement and stress-strain relationsip of linearly elastic isotropic material and assume plane-stress conditions. Solution Under te plane-stress condition, te stress-strain and strain-displacement relationsips of isotropic linearly elastic material are 1 ν 0 E yy ν 1 0 = yy 1 ν 0 0 (1 ν ) / y γy u, yy = uy, y γ y u, y + uy,. e material parameters are Young s modulus E and Poisson s ratio ν. e relationsips can be used to calculate stress out of te given displacement obtained e.g. from displacement analysis of a structure. Element approimation of te present case simplifies to (sape functions can be deduced from te figure) u (1 / )(1 y/ ) 0 ( / )(1 y/ ) 0 = = 0 (1 / )( y/ ) 0 ( / )( y/ ) 0 and u y (1 / )(1 y/ ) 0 ( / )(1 y/ ) a y = = (1 )a (1 / )( y/ ) 0 ( / )( y/ ) 0 u, = 0 u y, = 0 u, y = 1 y (1 )a u, yy 1 = a. Strain components follow from te strain-displacement relationsip u, 0 a yy = uy, y = / γ 1 y/ y u, y u + y, Stress components follow from te stress-strain relationsip.

4 ν ν E a a E yy ν 1 0 = / =. 1 ν 1 ν 0 0 (1 ν ) / 1 y/ y 1 ν y (1 ) Evaluation at te midpo ( y, ) = (, )/ gives ν a E yy = 1. 1 ν (1 ν ) / y

5 e kinematic assumptions of Bernoulli beam model are u = u( ) zw ( ), u y = 0 and uz = w ( ). e kinetic assumptions are = = 0. Derive te epressions of te normal force N( ), yy bending moment M( ), and sear force Q ( ) by using definitions N = d, M = zd and Q = zd, zz in wic te egrals are over te cross-section. Moments of te cross-section are = 1d, S = zd, and I = z d. Use te stress-strain and strain-displacement relationsips of a omogeneous, isotropic and linearly elastic material. Solution Virtual work epression, stress resultant epression etc. of an engineering model follow from te kinematic and kinetic assumptions and generic epression of linear elasticity teory. e kinematic assumptions of (te planar) Bernoulli beam model u = u( ) zw ( ), u y = 0 and uz = w ( ) mean tat te cross-sections move as rigid bodies in deformation. lso, cross-sections perpendicular to te ais remain perpendicular to te ais in deformation. e kinetic assumptions of te model are yy = zz = 0. Kinematic assumptions are used to simplify te generic strain-displacement relationsip u, u ( ) zw ( ) yy = uy, y = 0 u 0 zz z, z γ y u, y + uy, yz = uy, z + uz, y = 0+ 0 = 0 γ u u w ( ) w ( ) 0 + z z,, z = u ( ) zw ( ) (just one non-zero component!). Kinetic assumptions are used to simplify te strain-stress (or stress-strain) relationsip. 1 ν ν yy ν 1 ν = 0 = ν E E ν ν 1 0 ν zz 0 y 1 0 = yz G 0 z 1 = = E = E[ u ( ) zw ( )] z = z = 0. E

6 Constitutive equations of te beam model for te stress resultants follow from te definitions, N = [ ( ) ( )] ( ) ( ) d = E u zw d = Eu Sw, M = [ ( ) ( )] ( ) ( ) zd = E u zw zd = ESu EIw Q = zd = 0, (NO RIGH!) in wic = 1d, S = zd, and I = z d. ctually, te sear force Q of te Bernoulli beam model cannot be identically zero as equilibrium requires e.g. tat M Q = 0 in z plane bending. NOICE HIS. In an engineering model, te idea is to simplify te generic virtual work density epression δ y δγ y V = yy yy yz yz δ w δ δγ δ δγ zz zz z z consisting of work-conjugate pairs of strain and stress components. s te options, one may assume tat a stress component is zero or te corresponding strain component is zero (to be more precise: a quantity wose variation vanises). Stress-strain relationsip is not (directly) applicable to te stress components wose work conjugate strains vanis due to te kinematic assumptions of an engineering model. typical eample is te sear force Q of te Bernoulli beam model tat follows from te equilibrium equations.

7 e kinematic assumptions of imosenko (z-plane) beam model are u = u ( ) + zθ ( ), u y = 0 and uz = w ( ). e kinetic assumptions concerning stress are yy = zz = 0. Derive te epressions of te engineering strain and Caucy stress components of te model starting from te generic epressions 1 ν ν γy y 1 yy ν 1 ν 1 = yy E, yz = yz, ν ν 1 G zz zz γz z u, yy = uy, y u zz z, z and γ y u, y + uy, yz = uy, z + uz, y γ u + u z z,, z Solution Virtual work epression, stress resultant epression etc. of an engineering model follow from te kinematic and kinetic assumptions of te model.e kinematic assumptions of (te planar) imosenko beam model u = u ( ) + zθ ( ), u y = 0 and uz = w ( ) mean tat te cross-sections move as rigid bodies in deformation. e kinetic assumptions of te model are yy = zz = 0. Kinematic assumptions are used to simplify te generic strain-displacement relationsip u, u ( ) + zθ ( ) yy = uy, y = 0 u 0 zz z, z γ y u, y + uy, 0+ 0 yz = uy, z + uz, y = 0+ 0 γ u u w ( ) + θ ( ) + z z,, z = u ( ) + zθ ( ) γz = w ( ) + θ( ) (two non-zero components!). Kinetic assumptions are used to simplify te strain-stress (or stress-strain) relationsip. 1 ν ν yy ν 1 ν = 0 = ν E E ν ν 1 0 ν zz γy y 1 yz = yz G γz z γ z 1 = = E = Eu [ ( ) + zθ ( )] and E 1 = z z = Gw [ ( ) + θ( )]. G

8 e sear stress (and stress resultant Q ) of te imosenko beam model follows from a constitutive equation.

9 Kul Finite element metod I; Formulae collection GENERL 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: 1 ν ν 1 yy ν 1 ν = yy E ν ν 1 zz zz γy y 1 yz = yz G γz z E G = (1 + ν ) or 1 ν ν ν E yy ν 1 ν ν = yy [ E] yy (1 ν)(1 ν) + ν ν 1 ν zz zz zz y γy yz = G yz γ z z 1 ν 0 E [ E] = ν ν 0 0 (1 ν ) / 1 ν ν 0 E [ E] = ν 1 ν 0 (1 + ν)(1 ν) 0 0 (1 ν ) / Strain-displacement: u, yy = uy, y u zz z, z γ y u, y + uy, yz = uy, z + uz, y γ u + u z z,, z ELEMEN CONRIBUION (constant load) Bar (aial): F E 1 1 u f = F u 1 E 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

10 FX ux1 FX Po loads: F Y = uy1 F Y F Z u Z1 F Z PRINCIPLE OF VIRUL WORK MX θ X1 MX M Y = θy1 M Y M M θ Z1 Z1 Z e δw = δw + δw δw = δw = 0 δa δw = δwdω e E Ω Bar: δw = δu Eu,, δw = δuf Bar (torsion): δ w = δφ GI φ δ w = δφm, rr, Beam (z): Beam (y): δw = δ w EI w δw = δwf z, yy, δw = δ v, EIzzv, δw = δvf y Beam (Bernoulli): S, z Sy u, δu δ w = δ v E S I I v δφ GI φ w, S w δ y Iyz Iyy,, z zz zy,, rr, δu 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 [ ] v, y δu, y + δv, u, y + v, δ w δ u f = δ v f y Plane-strain (y): δu, u, δw = δv, y te [ ] v, y δu, y + δv, u, y + v, δ w δ u f = δ v f y Kircoff-plate (y):

11 δ w, w 3, t δw = δw, yy [ E] w, yy 1 δ w, y w, y Reissner-Mindlin plate (y): δw = δwf z δθ, θ 3, t δ w, y δφ w, y φ δ w = δφ, y [ E] φ, y tg 1 δ w, + δθ w, + θ δφ, δθ, y φ, θ, y δw = δwf z Body (yz): δ δγ y y yy yy yz yz δ w = δ δγ δ δγ zz zz z z δu f δ y δw = v f δ w f z or, u,, y + δ,, y +, δu δu 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 PPROXIMIONS (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 VIRUL WORK EXPRESSIONS δ uxi FXi δθ Xi MXi Rigid body (force): δw = δuyi FYi + δθyi MYi δu Zi F Zi δθ Zi M Zi

12 Bar (aial): δw 1 E 1 1 u1 δu = δu 1 1 u δw δu1 f 1 = δ u 1 Bar (torsion): δw δθ1 GIrr 1 1 θ 1 δθ 1 1 θ = δw δθ1 m 1 = ) δθ 1 Beam (z): δw z uz1 y1 EI yy θ y1 3 z z y θ y δu δθ = δu u δθ δw δuz1 6 δθ y1 z f = δuz 1 6 δθ y Beam (y): δw y uy1 z1 EIzz θz1 3 y uy z θz δu δθ = δ u δθ δw δuy1 6 δθz1 fy = ) δ uy 1 6 δθ z CONSRINS Frictionless contact: n u = 0 Jo: ub = u Rigid body (link): ub = u + θ ρb. θb = θ

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Kul Finite element method I, Exercise 08/2016 Kul-49.3300 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