MIT2_094S11_lec20

MIT2_094S11_lec20 - 2.094 — Finite Element Analysis of...

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Unformatted text preview: 2.094 — Finite Element Analysis of Solids and Fluids Fall ‘08 Lecture 20 - Beams, plates, and shells Prof. K.J. Bathe MIT OpenCourseWare Timoshenko beam theory The fiber moves up and rotates and its length does not change. Principle of virtual displacement � EI L � � � �T β β � dx + (Ak )G 0 0 L (Linear Analysis) � dw −β dx �T � � �L dw − β dx = wT pdx dx 0 (20.1) Two-node element: Three-node element: For a q -node element, �T � ˆ u = w1 · · · wq θ1 · · · θq ˆ w = Hw u ˆ β = Hβ u � � Hw = h1 · · · hq 0 · · · 0 � � Hβ = 0 · · · 0 h1 · · · hq dx J= dr (20.2) (20.3) (20.4) (20.5) (20.6) (20.7) 85 MIT 2.094 20. Beams, plates, and shells dw ˆ = J −1 Hw,r u dx � �� � (20.8) dβ ˆ = J −1 Hβ,r u dx � �� � (20.9) Bw Bβ Hence we obtain � �1 � T EI Bβ Bβ det(J )dr + (Ak )G −1 1 � T ˆ (Bw − Hβ ) (Bw − Hβ ) det(J )dr u −1 � 1 = T Hw p det(J )dr (20.10) −1 ˆ Ku = R (20.11) K is a result of the term inside the bracket in (20.10) and R is a result of the right hand side. For the 2-node element, w1 = θ1 = 0 (20.12) w2 , θ2 = ? (20.13) γ= 1+r w2 − θ2 L 2 (20.14) We cannot make γ equal to zero for every r (page 404, textbook). Because of this, we need to use about 200 elements to get an error of 10%. (Not good!) Recall almost or fully incompressible analysis: Principle of virtual displacements: � � �T � � � C � dV + �v (κ�v )dV = R V (20.15) V u/p formulation � � T �� C � �� dV − �v pdV = R V V � �p � p + �v dV = 0 κ V (20.16) (20.17) But now we needed to select wisely the interpolations of u and p. We needed to satisfy the inf-sup condition � qh � · vh dVol ≥β>0 (20.18) nf sup Vol �i��� ���� �qh ��vh � qh ∈Qh vh ∈Vh 86 MIT 2.094 20. Beams, plates, and shells 4/1 element: We can show mathematically that this element does not satisfy inf-sup condition. But, we can also show it by giving an example of this element which violates the inf-sup condition. v1 = �, above � v2 = 0 ⇒ � · vh for both elements is positive and the same. Now, if I choose pressures as qh �vh dVol = 0, hence (20.18) is not satisfied! (20.19) Vol 9/3 element satisfies inf-sup 9/4-c satisfies inf-sup Getting back to beams �L � � EI β βdx + (AkG) 0 � 0 L � � dw − β γ AS dx = R dx (20.20) L � � γ AS γ − γ AS dx = 0 (20.21) dw − β, dx (20.22) 0 where γ= from displacement interpolation 87 MIT 2.094 20. Beams, plates, and shells γ AS = Assumed shear strain interpolation (20.23) 2-node element, constant shear assumption. From (20.21), � L � 0 � �L dw − β ��dx = γ AS γ AS ��dx γ AS dx 0 � +1 � ⇒− −1 ⇒ γ AS = 1+r θ2 2 � · Reading: Sec. 4.5.7 (20.24) L dr + w2 = γ AS · L 2 (20.25) w2 − L θ2 2 L (20.26) γ AS (shear strain) is equal to the displacement-based shear strain at the middle of the beam. Use γ AS in (20.20) to obtain a powerful element. For “our problem”, γ AS = 0 � ⇒ E I 0 L θ2 2 (20.27) � � β β � dx = M β � =L x (20.28) hence L w2 = �� � � 2 1 · L θ2 = M ⇒ EI L ⇒ θ2 = ML , EI w2 = (20.29) M L2 2EI (20.30) (exact solutions) 88 MIT 2.094 20. Beams, plates, and shells Plates Reading: Fig. 5.25, p. 421 ⎧ ⎪ w = w(x, y ) is the transverse displacement of the mid-surface ⎨ v = −zβy (x, y ) ⎪ ⎩ u = −zβx (x, y ) (20.31) For any particle in the plate with coordinates (x, y, z ), the expressions in (20.31) hold! We use w= q � hi wi (20.32) i=1 βx = − βy = + q � i=1 q � i hi θ y (20.33) i hi θ x (20.34) i=1 where q equals the number of nodes. Then the element locks in the same way as the displacement-based beam element. 89 MIT OpenCourseWare http://ocw.mit.edu 2.094 Finite Element Analysis of Solids and Fluids II Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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