Saouma V.E. Матричный структурный анализ и введение в конечные элементы (читают лекции примечания, Колорадо, 1999), , Saouma V.E. Matrix structural analysis and introduction to finite elements (lecture notes, Colorado, 1999)(613s).pdf:

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Saouma V.E. Матричный структурный анализ и введение в конечные элементы (читают лекции примечания, Колорадо, 1999),

Saouma V.E. Matrix structural analysis and introduction to finite elements (lecture notes, Colorado, 1999)(613s).pdf

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1. Two dimensional versus three dimensional; Should we model a single bay of a building, or the entire structure? 2. Frame or truss, can we neglect ﬂexural stiﬀness? 3. Rigid or semi-rigid connections (most imp ortant in steel structures) 4. Rigid supp orts or elastic foundations (are the foundations over solid rock, or over clay which may consolidate over time) 5. Include or not secondary memb ers (such as diagonal braces in a three dimensional analysis). 6. Include or not axial deformation (can we neglect the axial stiﬀness of a b eam in a building?) 7. Cross sectional prop erties (what is the moment of inertia of a reinforced concrete b eam?) 8. Neglect or not haunches (those are usually present in zones of high negative moments) 9. Linear or nonlinear analysis (linear analysis can not predict the p eak or failure load, and will underestimate the deformations). 10. Small or large deformations (In the analysis of a high rise building sub jected to wind load, the moments should b e ampliﬁed by the product of the axial load times the lateral deformation, P − ∆ eﬀects). 11. Time dep endent eﬀects (such as creep, which is extremely imp ortant in prestressed concrete, or cable stayed concrete bridges). 12. Partial collapse or local yielding (would the failure of a single element trigger the failure of the entire structure?). 13. Load static or dynamic (when should a dynamic analysis b e p erformed?). 14. Wind load (the lateral drift of a high rise building sub jected to wind load, is often the ma jor limitation to higher structures). 15. Thermal load (can induce large displacements, sp ecially when a thermal gradient is present.). 16. Secondary stresses (caused by welding. Present in most statically indeterminate structures). Victor Saouma Matrix Structural Analysis...

From the analysis, we ﬁrst obtain the nodal displacements, and then the element internal forces. Those internal forces vary according to the element typ e. For a two dimensional frame, those are the axial and shear forces, and moment at each node.
20 21...

                                            ...

The assignment of the element stiﬀness matrix term Kiej (note that e, i, and j are all known since we are looping on e from 1 to the numb er of elements, and then looping on the rows and S columns of the element stiﬀness matrix i, j ) into the global stiﬀness matrix Kkl is made through the LM vector (note that it is k and l which must b e determined).
15 Since the global stiﬀness matrix is also symmetric, we would need to only assemble one side of it, usually the upp er one. 16 17...

5–17 = −0.8, s = 0.6, E A = 15, 000 L −9 −9 9600 −10 −7200   4  −9600 5 7200
...

The contents of the input.m ﬁle which the user is to ﬁll out is given b elow:
%********************************************************************************************** % Scriptfile name: indat.m (EXAMPLE 2D-FRAME INPUT DATA) % % Main Program: casap.m % % This is the main data input file for the computer aided % structural analysis program CASAP. The user must supply % the required numeric values for the variables found in % this file (see user’s manual for instructions). % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %**********************************************************************************************...

_________________________________________________________________________ Load Case: 1 Nodal Loads: Node: 2 Fx = 1.875000e+001 Node: 2 Fy = -4.635000e+001 Elemental Loads: Element: 1 Point load = 0 at 0 from left Distributed load = 0 Element: 2 Point load = 0 at 0 from left Distributed load = 4.000000e-003 Displacements: (Node: 2 delta X) 9.949820e-001 (Node: 2 delta Y) -4.981310e+000 (Node: 2 rotate ) -5.342485e-004 Reactions: (Node: 1 (Node: 1 (Node: 1 (Node: 3 (Node: 3 (Node: 3...

Example 6-2: Beam Statics Matrix Considering the b eam shown in Fig. 6.2, we have 3 elements, each with 2 unknowns (v and m) plus two unknown reactions, for a total of 8 unknowns. To solve for those unknowns we have 2 equations of equilibrium at each of the 4 nodes. Note that in this problem we have selected as primary unknowns the shear and moment at the right end of each element. The left comp onents can b e recovered from equilibrium. From equilibrium we thus have:
                          ...

Example 6-7: Congruent Transformation of a Frame Assemble the stiﬀness matrix of the frame shown in Fig. 6.6 using the direct stiﬀness method, and the two congruent approaches. Solution:...

It can b e shown that the variation and derivation op erators are commutative
d dx (δ u)...

We still have to deﬁne δΠ. The ﬁrst variation of a functional expression is
δ δF = Π= ∂F ∂b δ u u a ∂F + ∂ u δu δF dx...

32 In one dimensional elements with initial strain (temp erature eﬀect, supp ort settlement, or other) such that: σx + ε0 (11.11) εx = x i E d ue to load nitial strain...

Diﬀerentiating the shap e functions from Eq. 10.35 we obtain:
   ∂  N1   ∂x  εx   0   εy =  γxy    {  x3 − x2  }   x2 y3  ∂
N1 ∂y...

10 1 −5 1 0 0  1    2  −20 3 20 0 1 0   0 0 1 6 5 35 2. Elimination of the ﬁrst column: (a) row 1=0.1(row 1) (b) row 2=(row2)+20(new row 1) (c) row 3=(row 3) -5(new row 1)
...

where λmax and λmin are the maximum and minimum eigenvalues of the coeﬃcient matrix.
95 In the decomp osition of a matrix, truncation errors may result in a loss of precision which has b een quantiﬁed by: s = p − log κ (3.34)...

Since scalar op erations are in general not applicable to vectors, we deﬁne A+B = B+A A×B = −B×A A = Ax i + Ay j + Az k A·B = |A||B| cos(A, B) = Ax Bx + Ay By + Az Bz A⊗B = grad A = ∇A = div A = ∇·A = = Laplacian ∇2 = ∇·∇ = j k Ax Ay Az Bx By Bz ∂A ∂A ∂A +j +k i ∂x ∂y ∂z i · ∂ ∂ ∂ +j +k (iAx + jAy + kAz ) ∂x ∂y ∂z ∂ Az ∂ Ax ∂ Ay + + ∂x ∂y ∂z 2A 2A ∂ ∂ ∂2A + + ∂ x2 ∂ y2 ∂z2
i...