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# C I V I L E N G I N E E R I N G Structural Analysis Worsak Kanok Nukulchaii

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### C I V I L E N G I N E E R I N G Structural Analysis Worsak Kanok Nukulchaii

1. 1. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiSTRUCTURAL ANALYSISWorsak Kanok-NukulchaiAsian Institute of Technology, ThailandKeywords: Structural System, Structural Analysis, Discrete Modeling, Matrix Analysisof Structures, Linear Elastic Analysis.Contents1. Structural system2. Structural modeling.3. Linearity of the structural system. S4. Definition of kinematics LS5. Definitions of statics6. Balance of linear momentum7. Material constitution R O8. Reduction of 3D constitutive equations for 2D plane problems. TE9. Deduction of Euler-Bernoulli Beams from Solid. -E10. Methods of structural analysis11. Discrete modeling of structures12. Matrix force method AP13. Matrix displacement method O14. Trends and perspectivesGlossary H CBibliographyBiographical Sketch C ESTo cite this chapter ESummary N PLThis chapter presents an overview of the modern method of structural analysis based onUdiscrete modeling methods. Discrete structural modeling is suited for digitalcomputation and has lead to the generalization of the formulation procedure. The two Mprincipal methods, namely the matrix force method and the matrix displacement method,are convenient for analysis of frames made up mainly of one-dimensional members. Of SAthe two methods, the matrix displacement method is more popular, due to its naturalextension to the more generalized finite element method. Using displacements as theprimary variables, the stiffness matrix of a discrete structural model can be formedglobally as in the case of the matrix displacement method, or locally by considering thestiffness contributions of individual elements. This latter procedure is generally knownas the direct stiffness method. The direct stiffness procedure allows the assembly ofstiffness contributions from a finite number of elements that are used to model anycomplex structure. This is also the procedure used in a more generalized method knownas the Finite Element Method©Encyclopedia of Life Support Systems (EOLSS)
2. 2. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai1. Structural SystemStructural analysis is a process to analyze a structural system in order to predict theresponses of the real structure under the excitation of expected loading and externalenvironment during the service life of the structure. The purpose of a structural analysisis to ensure the adequacy of the design from the view point of safety and serviceabilityof the structure. The process of structural analysis in relation to other processes isdepicted in Figure 1. S LS R O TE -E AP O H C C ES E N PL Figure 1. Role of structural analysis in the design process of a structure.UA structural system normally consists of three essential components as illustrated in MFigure 2: (a) the structural model; (b) the prescribed excitations; and (c) the structuralresponses as the result of the analysis process. In all cases, a structure must be idealized SAby a mathematical model so that its behaviors can be determined by solving a set ofmathematical equations. Figure 2. Definition of a structural system©Encyclopedia of Life Support Systems (EOLSS)
3. 3. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiA structural system can be one-dimensional, two-dimensional or three-dimensionaldepending on the space dimension of the loadings and the types of structural responsesthat are of interest to the designer. Although any real-world structure is strictly three-dimensional, for the purpose of simplification and focus, one can recognize a specificpattern of loading under which the key structural responses will remain in just one ortwo-dimensional space. Some examples of 2D and 3D structural systems are shown inFigure 3. S LS R O TE -E AP O H C C ES E N PLU M SA Figure 3. Some examples of one, two and three-dimensional structural models.2. Structural ModelingA structural (mathematical) model can be defined as an assembly of structural members(elements) interconnected at the boundaries (surfaces, lines, joints). Thus, a structuralmodel consists of three basic components namely, (a) structural members, (b) joints(nodes, connecting edges or surfaces) and (c) boundary conditions.©Encyclopedia of Life Support Systems (EOLSS)
4. 4. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai(a) Structural members: Structural members can be one-dimensional (1D) members(beams, bars, cables etc.), 2D members (planes, membranes, plates, shells etc.) or in themost general case 3D solids.(b) Joints: For one-dimensional members, a joint can be rigid joint, deformable joint orpinned joint, as shown in Figure 4. In rigid joints, both static and kinematics variablesare continuous across the joint. For pinned joints, continuity will be lost on rotation aswell as bending moment. In between, the deformable joint, represented by a rotationalspring, will carry over only a part the rotation from one member to its neighbor offsetby the joint deformation under the effect of the bending moment. S LS R O TE -E AP O H C C ES E N PLU M SA Figure 4. Typical joints between two 1D members. (c) Boundary conditions: To serve its purposeful functions, structures are normallyprevented from moving freely in space at certain points called supports. As shown inFigure 5, supports can be fully or partially restrained. In addition, fully restrainedcomponents of the support may be subjected to prescribed displacements such asground settlements.©Encyclopedia of Life Support Systems (EOLSS)
5. 5. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Figure 5. Boundary conditions of supports for 1D members. S3. Linearity of the Structural System LSAssumptions are usually observed in order for the structural system to be treated aslinear: R O(a) The displacement of the structure is so insignificant that under the applied loads, the TE deformed configuration can be approximated by the un-deformed configuration in -E satisfying the equilibrium equations.(b) The structural deformation is so small that the relationship between strain and AP displacement remains linear.(c) For small deformation, the stress-strain relationship of all structural members falls O in the range of Hooke’s law, i.e., it is linear elastic, isotropic and homogeneous. H CAs a result of (a), (b) and (c), the overall structural system becomes a linear problem; C ESconsequently the principle of superposition holds.4. Definition of Kinematics E N PLa) MotionsUAs shown in Figure 6, the motion of any particle in a body is a time parameter family of Mits configurations, given mathematically by SA ^x = x ( P, t ) (1)~ ~ ˆwhere x is a time function of a particle, P, in the body. During a motion, if relativepositions of all particles remain the same as the original configuration, the motion iscalled a “rigid-body motion”.b) DisplacementDisplacement of a particle is defined as a vector from its reference position to the newposition due to the motion. If P0 is taken as the reference position of P at t = t0, then thedisplacement of P at time t = t1 is©Encyclopedia of Life Support Systems (EOLSS)
6. 6. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai ^ ^u ( P, t1 ) = x( P, t1 ) − x( P, t0 ) (2)~ ~ ~c) DeformationThe quantitative measurement of deformation of a body can be presented in manyforms:(1) Displacement gradient matrix ∇ u ~ ⎡ u1,1 u1,2 u1,3 ⎤ ⎢ ⎥ S∇u = ⎡ui , j ⎤ = ⎢u2,1 u2,2 ⎣ ⎦ u2,3 ⎥ (3a) ⎢ ⎥ LS ⎢ u3,1 u3,2 ⎣ u3,3 ⎥ ⎦ RNote that this matrix is not symmetric. O TE(2) Deformation gradient matrix F -EF = I + ∇u or Fij = δ ij + ui , j AP (3b) Owhere I is the identity matrix and δ ij is Kronecker delta. Like the displacement H Cgradient matrix, the deformation gradient matrix is not symmetric. C ES E N PLU M SA Figure 6. Motion of a body and position vectors of a particle.(3) Infinitesimal strain tensor (for small strains) 1ε ij = (ui , j + u j ,i ) (4) 2©Encyclopedia of Life Support Systems (EOLSS)
7. 7. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai ⎡ 1 1 ⎤ ⎢ u1,1 (u + u2,1 ) (u1,3 + u3,1 ) ⎥⎡ ε11 ε12 ε13 ⎤ ⎢ 2 1,2 2 ⎥⎢ ⎥ ε 23 ⎥ = ⎢ (u + u3,2 ) ⎥ 1⎢ε 21 ε 22 ⎢ u2,2 2 2,3 ⎥ (5)⎢ε 31 ε 32⎣ ε 33 ⎥ ⎢ ⎦ ⎥ ⎢( SYM ) u3,3 ⎥ ⎢ ⎣ ⎥ ⎦The strain tensor is symmetric. Its symmetric shear strain components in the oppositeoff-diagonal positions can be combined to yield the engineering shear strain γ ij asdepicted in Figure 7 as,γ ij = ε ij + ε ji = ui, j + u j ,i S (6) LS R O TE -E AP Figure 7. Engineering shear strain γ ij OSince ε is symmetric, there is no need to work with all the 9 components, strain tensor H C ~is often rewritten in a vector form as C ESε = [ε11 ε 22 ε 33 γ 12 γ 23 γ 31 ]T (7) E~ N PL(4) Rotational Strain tensor, illustrated in Figure 8, is defined asU 1ωij = (u j ,i − ui , j ) M (8a) 2 SAor in matrix form: ⎡ 1 1 ⎤ ⎢ 0 (u − u ) (u3,1 − u1,3 ) ⎥ 2 2,1 1,2 2 ⎢ ⎥ω = ⎢ (u1,2 − u2,1 ) (u3,2 − u2,3 ) ⎥ 1 1 0 (8b)~ ⎢2 2 ⎥ ⎢1 1 ⎥ ⎢ (u1,3 − u3,1 ) (u − u ) 0 ⎥ ⎢2 ⎣ 2 2,3 3,2 ⎥ ⎦©Encyclopedia of Life Support Systems (EOLSS)
8. 8. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S LS R O TE -E Figure 8. Illustration of a rotational strain ω 12 APSome notes on the properties of ε and ω O ~ ~ Ha) One can easily show that C C ESui, j = ε ij − ωij (9) Eand that a displacement gradient is decomposable into 2 parts ε and − ω . N ~ ~ PLb) If ui, j = u j ,i then ω = 0 (pure deformation)U ~ Mc) If ui , j = −u j ,i then ε = 0 (pure rotation) ~ SA5. Definitions of StaticsStress is defined as internal force per unit deformed area, distributed continuouslywithin the domain of a continuum that is subjected to external applied forces. When thedeformation is small, stress can be approximated by internal force per un-deformed areabased on its original configuration. This type of stress is represented by the Cauchystress tensor in the form of ⎡σ 11 σ 12 σ 13 ⎤ ⎡ ⎤ ⎢σ = ⎣σ ij ⎦ = ⎢σ 21 σ 22 σ 23 ⎥ ⎥ (10) ⎢σ 31 σ 32 σ 33 ⎥ ⎣ ⎦©Encyclopedia of Life Support Systems (EOLSS)
9. 9. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiAs illustrated in Figure 9, the two subscript indices i and j corresponding to xi and x jrefer to the normal vector of the surface and the direction associated with the actinginternal force. S LS R O TE Figure 9. Definition of Cauchy Stress Tensor. -E6. Balance of Linear Momentum APBased on the 2nd law of Newton, the rate of linear momentum change equals the total Oforce applied in the same direction. Thus, at any deformed state Bt of a deformable Bbody, the rate of total linear momentum change on the left-handed side equals to body H Cforce, b , over the whole body Bt and traction force τ on the boundary ∂Bt respectively. B C ESd ∫ ρ u dv = ∫ ρt b dv + ∫ τ dadt B t ~ (11) E ~ ~ t B t ∂B t N PLwhere ρt is the unit mass of the body at the Bt state and u is the velocity of a particle inU Bthe body. In componential form, using the relationship between the Cauchy stress tensor Mand the traction as illustrated in Figure 10, one can present Eq.(11) in tensor notation as SAd ∫ ρ u dv = ∫ ρt b j dv + ∫ σ ij ni dadt B t j (12) t B t ∂B twhere ρt is mass density, Bt is a deformed body at time t, ∂ Bt is its boundary and n is Bits unit vector. Eq.(11) is the global balance of linear momentum. It applies to a body atthe state of Bt (deformed configuration corresponding to a time station t). Integrals over BBt and ∂ Bt are not explicit as the deformed configuration itself depends on u (t ) . This is B ~therefore a nonlinear problem.©Encyclopedia of Life Support Systems (EOLSS)
10. 10. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Figure 10 Relationship between Cauchy stress tensor and traction. SFor the case of small displacements and infinitesimal strains, Bt ≈ B0 and δ Bt ≈ δ B0 . LSThen, all integrals can be carried over Bo and ∂ B0 , Eq.(11), which governs the body as B Ra whole, can be localized to obtain equations of motion which govern any particle Owithin the body. This is achieved through the use of the Gauss theorem which TEtransforms surface integral to a volume integral, i.e., -E ∫ σ ij ni da = ∫ σ ij ,i dv AP (13)∂ Bo Bo OHence, Eq.(12) is reduced to H C ∫ ρo u j dv = ∫ ρo b j dv + ∫ σ ij ,i dv C (14) ESBo Bo Bo EConsequently, the tensor form of local balance of linear momentum, also known as the N PLequation of motion isUρo u j = ρo b j + σ ij ,i (15a) MThe left-handed side term is inertia force and the two right-handed side terms are SArespectively the body force and the internal force all in xj direction.7. Material ConstitutionSo far, the balance of linear momentum is applicable to any solid material. In thepresent scope, only linear, elastic, isotropic and homogeneous material is considered. Inmost practical structures under service loads, the material deformation is limited toinfinitesimal small strains. Under this situation, its stress-strain relationship can beassumed to vary linearly, as illustrated in Figure 11. The standard Hooke’s three-dimensional stress-strain relationship can be expressed as©Encyclopedia of Life Support Systems (EOLSS)
11. 11. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S LSFigure 11. Stress-strain relationship in most structural materials can be assumed linear if strain is limited to a small magnitude. R O⎧σ 11 ⎫ ⎡λ + 2μ λ λ 0 ⎤ ⎧ ε11 ⎫ TE 0 0⎪σ ⎪ ⎢ ⎥⎪ ⎪ -E⎪ 22 ⎪ ⎢ λ λ + 2μ λ 0 0 0 ⎥ ⎪ε 22 ⎪⎪σ 33 ⎪ ⎢ λ⎪ ⎪ λ λ + 2μ 0 0 0 ⎥ ⎪ε 33 ⎪ ⎪ ⎪ ⎬=⎢⎨⎪σ 12 ⎪ ⎢ 0 0 0 2μ 0 AP ⎥ ⎨ ⎬ (15c) 0 ⎥ ⎪ε12 ⎪ O⎪σ 23 ⎪ ⎢ 0 0 0 0 2μ 0 ⎥ ⎪ε 23 ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎩σ 31 ⎭ ⎣ 0 2 μ ⎦ ⎩ε 31 ⎪ H⎪ ⎪ ⎪ ⎭ C 0 0 0 0 C ESor Eσ ij = λδ ij ε kk + 2με ij (15d) N PLUin which the two Lame’s constants νE M Eλ= and μ = . (1 + ν )(1 − 2ν ) 2(1 + ν ) SA8. Reduction of 3D Constitutive Equations for 2D Plane ProblemsThere are two special cases of 3D solid that can be reduced to 2D plane problems, asfollows:a) Plane strain problemWhen the longitudinal dimension of a prismatic solid is relatively long, strains in thelongitudinal direction can be negligible as shown in Figure 12.©Encyclopedia of Life Support Systems (EOLSS)
12. 12. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S Figure 12. Plane strain reduction from 3D prismatic solid LSFrom the 3D constitutive equations in Eq.(15c), by prescribing ε 33 = ε13 = ε 23 = 0 , theremaining constitutive equation can be expressed as R⎧σ 11 ⎫ ⎡λ + 2μ λ ⎤ ⎧ ε11 ⎫ O TE 0⎪ ⎪ ⎢ ⎥ ⎪ε ⎪ -E⎨σ 22 ⎬ = ⎢ λ λ + 2μ 0 ⎥ ⎨ 22 ⎬ (16a)⎪ ⎪ 2μ ⎥ ⎩ε12 ⎪ ⎪⎩σ 12 ⎭ ⎢ 0 ⎣ 0 ⎦ ⎭ AP Oor H Cσ ij = λδ ij ε kk + 2με ij (i , j = 1, 2) (16b) C ESNote that while σ 22 + σ 33 is not zero due to the Poisson’s ratio effect and can be Eobtained as N PLσ 33 = λ (ε11 + ε 22 ) (16c)U(b) Plane stress problem MWhen the transverse dimension of a solid is relatively small as illustrated in Figure 13, SAsay in x3 direction, the transverse stress components, namely, σ 33 , σ 13 and σ 23 can beassumed negligible. Consequently the only stress components need to be considered inplane stress problems are σ 11 , σ 22 , σ 12 all acting in the plane.Referring to the third equation of Eq.(15), in views of the zero value of σ 33 , one can λobtain ε 33 = − (ε11 + ε 22 ) . Then substituting ε 33 in terms of ε11 and ε 22 in the λ + 2μfirst and second equation of Eq.(15) leads to:©Encyclopedia of Life Support Systems (EOLSS)
13. 13. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai⎧σ 11 ⎫ ⎡λ + 2 μ λ 0 ⎤ ⎧ ε11 ⎫⎪ ⎪ ⎢ ⎥⎪ ⎪⎨σ 22 ⎬ = ⎢ λ λ + 2μ 0 ⎥ ⎨ε 22 ⎬ (17a)⎪ ⎪ ⎢ ⎥⎪ ⎭ 2 μ ⎦ ⎩ε12 ⎪⎩σ 12 ⎭ ⎣ 0 0orσ ij = λδ ij ε kk + 2με ij (i,j = 1,2) (17b)where 2λμ νEλ= = (18) S λ + 2μ 1 − ν 2 LSNote that the form of the constitutive equation for the plane stress problems is exactly thesame as that for the plane strain problems, except that λ is used instead of λ . R O TETable 1 summarizes the three sets of equations of which the number is exactly the same as -Ethe number of unknowns in both 2D and 3D cases. To solve for the unknowns, one cancombine the set of equations to obtain a single governing equation in terms of thedisplacement unknowns, u j , as follows: AP Oρo u j = ρo b j + (λ + μ )uk ,kj + μ u j ,kk (19) H CThis governing equation is a second-order partial differential equation with respect to C ESspace and time. E N PLU M SA Figure 13. Non-zero stress components in 2D plane stress problems©Encyclopedia of Life Support Systems (EOLSS)
14. 14. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai9. Deduction of Euler-Bernoulli Beams from SolidThin beam is one of the most common 1D components of a structural system, definedby its bending capability about the beam axis. Thin beam is governed by Euler-Bernoulli beam theory, based on the following assumptions:Assumption 1. Transverse dimensions of a beam are small in comparison with thelongitudinal dimension; hence, normal stresses in the two transverse directions areassumed to be negligible as illustrated in Figure 14.Assumption 2. Shear distortion in a thin beam is negligible. Therefore, its plane sectionswhich are normal to the beam axis remain plane and normal to the deformed beam axisas illustrated in Figure 15. S LS Equations 2D 3D 1 Local Balance of Linear Momentum 2 3 ρo u j = ρo b j + σ ij ,i ( j = 1, n ) R 2 Constitutive Law O 3 6 σ ij = λ δ ij ε kk + 2με ij (i, j = 1, n) TE -E 3 Strain-Displacement Relations 3 6 1 ε ij = (ui , j + u j ,i )(i, j = 1, n) AP 2 O Total number of equations 8 15 Unknowns H C 1 Displacements ui (i = 1, n) 2 3 Strains ε ij (i, j = 1, n) C 2 3 6 ES 3 Stresses σ ij (i, j = 1, n) 3 6 E Total number of unknowns 8 15 N PL Table 1. Summary Of Equations And Unknowns For 2D/3D SolidsU M SA Figure 14. All transverse stress components in thin beam are negligible.©Encyclopedia of Life Support Systems (EOLSS)
15. 15. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Figure 15. Plane section remains plane in thin beam S(a) Effect of Assumption 1 LSFrom the 3D constitutive equations R⎧σ 11 ⎫ ⎡λ + 2μ λ λ 0 O 0 0 ⎤ ⎧ ε11 ⎫ TE⎪σ ⎪ ⎢ ⎪ ⎪⎪ 22 ⎪ ⎢ λ λ + 2μ λ 0 ⎥ ⎪ε 22 ⎪ -E 0 0 ⎥⎪σ 33 ⎪ ⎢ λ⎪ ⎪ λ λ + 2μ 0 0 0 ⎥ ⎪ε 33 ⎪ ⎪ ⎪⎨ ⎬=⎢ ⎥⎨ ⎬ AP (20)⎪σ 12 ⎪ ⎢ 0 0 0 2μ 0 0 ⎥ ⎪ε12 ⎪⎪σ 23 ⎪ ⎢ 0 2μ 0 ⎥ ⎪ε 23 ⎪ O 0 0 0⎪ ⎪ ⎢ ⎥⎪ ⎪⎪σ 31 ⎪ ⎣ 0⎩ ⎭ 0 0 0 0 2 μ ⎦ ⎪ε 31 ⎪ ⎩ ⎭ H C CEliminating (σ 22 + σ 33 ) between the first equation of Eq.(20), i.e., ESσ 11 = (λ + 2μ )ε11 + λ (ε 22 + ε 33 ) E (21) N PLand the summation of the second and the third equations, i.e.,Uσ 22 + σ 33 = 2λε11 + 2(λ + μ )(ε 22 + ε 33 ) = 0 (22) M SAleads to (3λ + 2μ )σ 11 = με11 (23) (λ + μ ) νE ESubstituting λ = and μ = in Eq.(23) yields (1 − 2ν )(1 + ν ) 2(1 + ν )σ 11 = Eε11 (24)©Encyclopedia of Life Support Systems (EOLSS)
16. 16. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai −λNote that (ε 22 + ε 33 ) = ε = −2νε11 due to the Poisson’s ratio effect. λ + μ 11(b) Stress resultantsN11 = ∫ σ 11dA (Normal stress resultant) (25) AN12 = ∫ σ 12 dA (Transverse shear resultant) (26) AM = ∫ σ 11 x2 dA (Bending moment) (27) S A LS(c) Effect of Assumption 2. R O TE -E AP O H C C ES Figure 16. Displacement field can be represented by the displacements and rotation of E the corresponding plane section based on thin beam. assumptions. N PLFrom Figure 16, the longitudinal displacement field at any point in the beam body canUbe expressed in terms of the longitudinal displacement at the corresponding point on thebeam axis and the effect of the rotation of the plane section, i.e., Mu1 ( x1 , x2 ) = U1 ( x1 ) − x2θ (28) SAIn views of the assumption that plane section remains plane and normal to beam axis,one can show that the rotation of the plane section equal the slope of the beam axis, i.e., dU 2 dU 2θ= . Substituting θ with leads to dx1 dx1 dU 2u1 ( x1 , x2 ) = U1 ( x1 ) − x2 (x ) (29) dx1 1©Encyclopedia of Life Support Systems (EOLSS)
17. 17. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiIf the normal strains in the transverse direction are negligible, the transversedisplacement field in the beam body is equal to the corresponding transversedisplacement of at the beam axis, i.e.,u2 ( x1 , x2 ) = U 2 ( x1 ) (30)(d) Strains resultantsMeasurements of deformation in a thin beam are made on a small segment of the beam inthe form of strain resultants. The main strain resultant is the normal strain of a smallsegment lim ΔX1 , i.e., ΔX1 →0 S du1ε11 ( x1 x2 ) ≡ (31) LS dx1 RSubstituting Eq.(29) in Eq.(31) leads to O TE d 2U 2 -E dU1ε11 ( x1 x2 ) ≡ ( x1 ) − x2 2 ( x1 ) (32) dx1 dx1The second thin beam assumption of zero shear can be confirmed from AP O H ∂ u1 ∂ u2 ⎧ dU 2 ⎫ dU 2 C ∂ ⎪ ⎪γ 12 ( x1 x2 ) = + = ⎨U1 − x2 ⎬+ =0 (33) ∂ x2 ∂ x1 ∂x2 ⎪ ⎩ dx1 ⎪ dx1 ⎭ C ES(e) Constitutive equations E N PLAt the material level, the relationship between the only pair of the stress and strain tensors isgiven in Eq.(24), i.e., σ 11 = Eε11 . Applying ∫ ( )dA on both sides of Eq.(24) in views ofU A MEq.(32) leads to SA ⎛ ⎞ dU ( x ) d 2U 2 dU ( x )N11 = ∫ Eε11 ( x1 , x2 )dA = E ⎜ ∫ dA ⎟ 1 1 + ∫ x2 dA. = EA 1 1 (34) ⎜ ⎟ 2 A ⎝ A ⎠ dx1 A dx1 dx1 h/2Note that ∫ x2 dA ≡ 0 i.e. ∫ bdx2 = 0 when the origin of x2 axis is located at the A −h / 2centroid of the plane section. Applying 2 x d Aon both sides of Eq.(24) in views of Eq.(32) leads to©Encyclopedia of Life Support Systems (EOLSS)
18. 18. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai dU1 d 2U 2 d 2U 2M = ∫ σ 11 x2 dA = E ( ∫ x2 dA) − E ( ∫ x2 dA) 2 2 = EI 2 (35) A A dx1 A dx1 dx1where the sectional moment of inertia, I3 = ∫ x2 dA . 2(f) Beam equation of motionThe equations of motion, Eq.(15), are applicable to any particle within a continuum.These equations will be used to derive a set of governing equations for thin beammembers. S(i) Based on Eq.(15), by applying the cross-sectional integration over both sides of theequation of motion in x1 direction, the longitudinal equation of motion can be derived LSas follows: R∫ ρ0u1dA = ∫ ρ0b1dA + ∫ (σ 11,1 + σ 21,2 + σ 31,3 )dA O (36) TEA A A -EIn view of Eq.(29) and the definition of stress resultants in Equations 25 and 26, Eq.(36) canbe further expanded as follows: AP O dU 2 ∂σ ∂σ∫ ρ0 [U1 − x2 ]dA = F1 + ∫ 11 dA + ∫ 21 dA ∂ x1 σ x2 (37) H dx1 CA A A C ES dN11 dN 21ρ0 AU1 = F1 + + (38) dx1 dx2 E N PLAs N 21 is not a function of x2 , the last term on the right-handed side can be omitted,Uthus, M dN11ρ0 AU1 = F1 + (39) dX1 SA(ii) Based on Eq.(15), by applying ∫ () x2 dA over both sides of the equation, the bending Aequation of motion can be derived as follows:∫ ( ρ0u1 ) x2 dA = ∫ ( ρ0b1 ) x2 dA + ∫ (σ 11,1 + σ 21,2 ) x2 dA (40)A A ASubstituting Eq.(29) and considering zero value of the first term on the right-handedside yield©Encyclopedia of Life Support Systems (EOLSS)
19. 19. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai ⎡ dU 2 ⎤ ∂σ 11 ∂σ∫ ρ0 ⎢U1 − x2 ⎣ ⎥ x2 dA = 0 + ∫ dx1 ⎦ ∂ x1 x2 dA + ∫ 21 x2 dx2 b ∂ x2 (41)A A AOmitting the zero terms and integrating by parts the last term of Eq.(41) lead to dU 2 d d− ρ0 ∫ x2 dA 2 = ∫ σ11 x2 dA + ∫ dx2 (σ 21 x2 )dx2b − ∫ σ 21dA (42) A ∂ x1 dx1 A A AUsing the definitions in Eqs. (26) and (27) and in view of the vanishing second term on theright-handed side, the bending equation of motion can be obtained as S dU 2 dM− ρ0 I 3 = − N12 LS (43) dx1 dx1 R ∫ ()dA(iii) Based on Eq.(15), by applying O in x2 direction over both sides of the TE A -Eequation, the transverse equation of motion can be derived as follows: ∂σ 12 ∂σ 22 ∂σ 12 AP∫ ρo u2 dA = ∫ ρ0b2 dA + ∫ ( ∂ x1 + ∂ x2 + ∂ x3 ) dA (44) OA A A H CRemoving the zero terms and defining the vertical distributed unit load q = ∫ ρ0 b2 dA C ES Alead to E d∫ ρ0U 2 dA = q + dx1 ∫ σ12 dA N (45) PLAUFinally, upon substituting the definition of the sectional shear term from Eq.(26), the Mtransverse equation of motion for thin beam can be obtained as SA dN12ρ0 AU 2 = q + (46) dx1Table 2 summarizes the three sets of equations that govern the behaviors of thin beamproblems. The number of equations is seven which is exactly the same as the number ofunknowns. To solve for the unknowns, one can combine the three set of equations to obtaintwo sets of governing equations in terms of the displacement unknowns as follows: d 2U1EA 2 + F1 = ρ 0 AU1 (47) dx1©Encyclopedia of Life Support Systems (EOLSS)
20. 20. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai d 4U 2 dθEI 3 4 − q = ρ0 I 3 − ρ0 AU 2 (48) dx1 dx1For static problems, the time-dependent terms can be omitted, and the equations ofmotions reduce to the equilibrium equations of thin beams as follows: d 2U1EA + F1 = 0 (49) dx1 d 4U 2EI 3 4 −q =0 (50) S dx1 LS10. Methods of Structural Analysis RTo design safe structures, structural engineers must fully understand the structural Obehaviors of these structures. In the long past, structural engineers gained the TEknowledge into the structural behaviors by carrying out experimentations using a -Ephysical model of the real structure in the laboratory. Based on the test results, thebehaviors of the prototype structure can be understood and generalized. However, the APphysical modeling has its limitations as it is expensive and time-consuming. Thus,mathematical modeling has been a viable alternative. O H Equations Number C 1 Local Balance of Linear Momentum 3 C dN11 ES + F1 = ρ0 AU1 dx1 E N PL dM − N12 = ρ0 I 3θU dx1 M dN12 + q = ρ0 AU 2 SA dx1 2 Constitutive Law 2 N11 = EAε11 M = − EI 3φ 3 Strain-Displacement Relations 2©Encyclopedia of Life Support Systems (EOLSS)
21. 21. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai dU1 ε11 = dx1 d 2U 2 φ = 2 dx1 Total number of equations 7 Unknowns (Figure 17) Number 1 Displacements U ,U 2 1 2 2 Strain resultants ε11 , φ 2 3 Stress resultants M , N11 , N12 3 Total number of unknowns 7 S Table 2. Summary of Equations and Unknowns for Thin Beams LS R O TE -E AP O H C Figure 17. Definition of unknown variables in a thin beam segment C ESAs shown in the diagram of Figure 18, mathematical modeling is an idealization of aspecific structural behavior that is expressed by mathematical equations. For simple Estructures, mathematical model is a convenient tool for predicting a specific behavior of Nthe structures. For more complex structural system, a structural model may consist of an PLassembly of structural members. Analytical solution of its mathematical model requiresUsolving a large-scale boundary value problem. Thus, the solution of large-scalestructural system based on mathematical modeling is rather impractical, if not Mimpossible at all. SAAfter the advent of computer technology, discrete modeling has become an industrialstandard in structural analysis. Like mathematical modeling, the formulation of discretestructural model is based on a mathematical foundation, to enable its large-scalesolution with the help of computational tools. Basically, each structural member in thestructural model must be discretized so that its key properties can be represented atselected discrete variables associated with the member. The assembly of membercontributions leads to a sizable set of algebraic equations associated with thesevariables, which can be systematically solved using computers.©Encyclopedia of Life Support Systems (EOLSS)
22. 22. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S LS Figure 18. Three modeling techniques and their relationships. R11. Discrete Modeling of Structures O TE(a) Definition of discrete structures -EIn discrete structural model, finite number of control points (joints) will be designated APand corresponding discrete variables are assigned to these joints, for which the solutionwill be obtained. Interior (member) variables are treated as dependent variables, which Owould be determined once the joint variables are known. The difference in the classical Hcontinuous model and the discrete model is demonstrated in Figure 19. C C ES E N PLU M SA Figure 19. Continuous model versus discrete model.Therefore a discrete structure is a network of joints or nodes interlinked by 1D, 2D or3D deformable members or any of their combinations. Joints are usually established fora member at points of geometric discontinuities and applied loads.(b) Definition of discrete quantitiesIn the continuous system, one deals with a particle within the domain of the continuum.Interesting quantities in a continuum can be categorized as statics and kinematics.©Encyclopedia of Life Support Systems (EOLSS)
23. 23. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiStatics consist of external force and internal force in the form of stress tensors, whilekinematics consist of displacements and deformation in the form of strain tensors.In the discrete system, an element is associated with a set of selected nodes, onlythrough which the element interacts with other elements. The external work doneassociated with each of these nodes is the product of ‘a vector of force, R ’ and itscorresponding ‘vector of displacement, r ’. The internal work done of an element isrepresented by the product of ‘a vector of action, S ’ and the corresponding ‘vector ofdeformation, v ’.(c) Discrete representation of internal work done: SIf a structural member is free of interior member loads, the distribution of stresses willfollow a specific pattern. Consequently, a set of discrete actions ( S ) and deformations LS( v ) can be defined to represent the total internal work done, i.e., R ∫σ T ε dV = S T v O (51) TEB0 -EGenerally, for a 2D beam member, it can be shown that three quantities are needed for APthe pair of action and deformation for their product to represent the overall internalwork done as O H ⎧θ 1 ⎫ ⎧ v1 ⎫ C ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 2⎪∫ σ T ε dV = M 1M 2 N ⎨θ 2 ⎬ = S1S 2 S 3 C ⎨v ⎬ ES (52)B0 ⎪Δ⎪ ⎪ 3⎪ ⎪ ⎪ ⎩ ⎭ ⎪v ⎪ ⎩ ⎭ E N PLwhere the three action terms M 1M 2 N are the bending moments at the two ends ofUthe beam and its axial force respectively, while the corresponding three deformation Mterms θ 1 θ 2 Δ are the two chord rotations and the axial elongation respectively, asshown in Figure 20. SA©Encyclopedia of Life Support Systems (EOLSS)
24. 24. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S Figure 20 Definition of actions and deformations for 2D bending members LSIt could be shown that the relationship between and , i.e., , can be obtained as R ⎡ L 1 L 1 O ⎤⎧v 1⎫ ⎢ 3EI + GA′L − + 6 EI GA′L 0 ⎥ 1 ⎧ ⎫ TE⎪ ⎪ ⎢ L ⎥ ⎪S ⎪ -E⎪ 2⎪ ⎢ ⎪ ⎪ 0 ⎥ ⎨S 2 ⎬ 1 L 1⎨v ⎬ = ⎢ − + + ⎥ (53)⎪ 3 ⎪ ⎢ 6 EI GA′L 3EI GA′L ⎥⎪ 3⎪ AP⎪v ⎪⎩ ⎭ ⎢ L ⎥ ⎪S ⎪ ⎩ ⎭ 0 0 ⎢ ⎣ ⎥ AE ⎦ O H CThe corresponding beam stiffness k can be obtained as k = f −1 C ES12. Matrix Force Method EMatrix Force Method employs the virtual force principle to form a system of flexibility N PLequations. In Figure 21, a diagram shows the relationship between statics R and S , andUthat between the kinematics v and r , as well as the constitutive equation linking S andv . Note that if one defines S = BR , one can prove by using virtual principle that Mr = BT v . SAIf the size of S is exactly the same as the number of equilibrium equations, this is thecase of statically determinate structures. The solution of S can be obtained explicitlyfrom the equilibrium equations alone, i.e., S = BR . Subsequently, other part of thesolution can be determined one after another.If the size of S is greater than the number of equations, this is the case of staticallyindeterminate structures. The solution of S cannot be determined directly from theequilibrium equations alone. The three sets of equations, namely the equilibrium, thematerial constitution and the compatibility, must be combined to obtain a system offlexibility equations, r = F R to obtain r . Subsequently other parts of the solution, v©Encyclopedia of Life Support Systems (EOLSS)
25. 25. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchaiand S can be determined. The formulation procedure for the matrix force methodapplicable to statically indeterminate structures will be presented next. S LS R O Figure 21. Notation of quantities and their relationships in matrix force method TE -EFor the purpose of demonstration, a simple indeterminate structure with 3 degrees ofstatically indeterminacy under joint loads R , temperature change ΔT and prescribed ˆ ˆ APdisplacements x and q will be used in the formulation of the flexibility equations based Oon the matrix force method. H CThe following step by step formulation procedure is used: C ES(a) Define an associated determinate base structure. EA statically determinate base structure will be selected after releasing kinematics Ncorresponding to the selected redundants X as shown on the right-handed side of PLFigure 22.U(b) Decomposition of the structure into 3 base structures. MIn Figure 23, the indeterminate structure is decomposed into 3 cases of the base SAstructure under (a) applied joint loads, (b) the redundant forces and (c) thermal changes.The sum of the kinematics corresponding to the redundants must be compatible to theirreal values of the total structure before their release. Thus, there are three compatibilityequations to be solved for the three unknown redundants X . The total value of eachquantity is the sum of its contributions from the three cases, as listed in Table 3.©Encyclopedia of Life Support Systems (EOLSS)
26. 26. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S Figure 22. A prototype indeterminate structure with 3 degrees of static indeterminacy LS and its base structure after 3 redundants are removed. R O TE -E AP O H C Figure 23. The superposition of three cases of base structure to satisfy the real C ES kinematics corresponding to the redundants. E N PL Statics KinematicsU Actions S = S R + S X ( ST = 0) Deformations v = vR + v X + vT r = rR + rX + rT M Forces R Displacements Redundants X Kinematics of X x = xR + x X + xT ˆ SA Reactions Q = QR + QX (QT = 0) Settlements ˆ q Table 3. List of statics and kinematics(c) Determination of the transformation matrices ( BR , BX , DR and DX ) andthermal strain ( VT )As the base structure is statically determinate, actions and reactions can be obtainedeasily using the equilibrium equation alone. As a result, the transformation matrices©Encyclopedia of Life Support Systems (EOLSS)
27. 27. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiBR , BX , DR and DX can be obtained, as well as the deformation VT due to the freeexpansion effect of the temperature change, as illustrated in Figure 24. S LS Figure 24. Analysis of 3 systems of statically determinate structures R(d) Determining the unknown redundants X using virtual force method: O TEApplying virtual forces δ X corresponding to the redundants on the base structure as the -Evirtual force system, Figure 25, one can obtain the equality of external and internalvirtual works as: AP O δ WE = δ WI H δ X T x + δ QT q = δ S T v C ˆ X ˆ X ( ) ( ) ( vR + vX + vT ) Cδ X T x + δ X T DT q = δ X T B T ES ˆ X ˆ X ˆ X ˆ ( x + DT q = BT f S R + f S X + vT ) (54) E X N x + DT q = ⎡ BT f BR ⎤ R + ⎡ BT f BX ⎤ X + BT vT PL ˆ X ˆ ⎣ X ⎦ ⎣ X ⎦ XU x + DT q = FXR R + FXX X + BT vT ˆ X ˆ X M x + xq = xR + x X + xT ˆ ˆ SA©Encyclopedia of Life Support Systems (EOLSS)
28. 28. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Figure 25. Application of virtual force to obtain kinematics corresponding to the redundants.The last equation assures that the prescribed kinematics x + xq at the supports are ˆ ˆsatisfied by the summation of the three cases. If there is no support settlement, both ˆ ˆ x and xq are zero. Physically the size of X must be adjusted to ensure this condition;thus, this equation is solved for the value of X , i.e.,⎣ ⎦ {⎡ FXX ⎤ { X } = x + DT q − FXR R − BT vT → X ˆ X ˆ X } (55)Note the two subscripts of the flexibility matrix FXR = BT f BR . The first refers to the S Xresulting displacement vector x while the second refers to the applying force vector R . LSSimilarly, the two subscripts in the flexibility matrix FXX = BT f BX refer to the X Rdisplacement x and the force vector X respectively. O TE(e) Obtain other parts of the solution. -ESimilar to Eq.(54), one obtains by virtual force principle the following: AP Or + DR q = FRR R + FRX X + BR vT T ˆ T (56) H Cwhere FRR = BR f BR and FRX = BR f BX . T T C ESOnce the unknown redundants X are solved, the remaining quantities can be obtained Eas shown in Table 4. Finally, the step by step procedure of the Matrix Force Method is Nsummarized in Table 5. PLU Statics Kinematics M Actions S = BR R + BX X Deformations v = f BR R + f BX X + vT r = FRR R + FRX X + BR vT − DR q SA T T R ˆ Forces Displacements Redundants X Kinematics x = FXR R + FXX X + BT vT − DT q ˆ X X ˆ of X Reactions Q = DR R + DX X Settlements ˆ q Table 4. Final results of statics and kinematics after X is solved. 1 ˆ Define a base structure and identify X and x 2 ˆ Define f , R, r , S , v and Q (if there is settlement q )©Encyclopedia of Life Support Systems (EOLSS)
29. 29. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai 3 Form BR , BX , DR , DX and vT (if temperature changes are given) 4 Form FRR , FRX , FXR and FXX 5 Solve for X from ⎣ ⎦ { ⎡ FXX ⎤ { X } = x + DT q − FXR R − BT vT ˆ X ˆ X } 6 Obtain the rest from Table 4. Table 5. Step by step Procedure for Matrix Force Method13. Matrix Displacement MethodMatrix Displacement Method employs the virtual displacement principle to form asystem of stiffness equations. Figure 26 shows a diagram of the relationship between Skinematics v and r and the statics R and S , as well as the constitutive equationlinking S and v . Note that if one defines v = a r , one can prove by using virtual LSprinciple that R = aT S . R OIn a structure in which the deformation v can be obtained directly from known TEdisplacement vector r , this structure is referred as a kinematically determinate structure. -EIn usual cases, displacements in a structure are unknowns and therefore most practicalstructures are kinematically indeterminate. AP OIn Matrix Displacement Method, it is necessary to identify the degree of kinematicindeterminacy ( o KI ), also known as the number of degrees of freedom (dof). In frame H Cstructures, this refers to the number of independent discrete displacements, i.e., alldisplacements that vary varied independently. In a 2D frame structure, each rigid joint C EShas 3 degrees of freedom, while in 3D frame structure, each rigid joint has 6 degrees offreedom. If there are internal constraints, some joint degrees of freedom could become Einter-dependent. Thus, the size of independent degrees of freedom will reduce. The need Nto eliminate dependency among joint degrees of freedom is to avoid solving a singular PLequation system.UFigure 26 illustrates the deformation modes associated with each degree of freedom Mindependent of the others. When constraints are prescribed in the system, the number ofdegrees of freedom will be reduced. Figure 27 illustrates the same structure except that SAthree constraints are introduced, i.e., the three members are not allowed to elongate orshorten after assuming that their axial deformation is negligible. Under these constraints,deformation modes in (a) (b) (d) and (e) in Figure 26 are not feasible, while one newdeformation mode arises as shown in Figure 27. Theoretically, the degree of kinematicindeterminacy can be calculated from the number of joint degrees of freedom less thenumber of internal constraints. In contrary to the force method which derives kinematicsfrom statics, the displacement method derives statics from kinematics. The relationshipsamong the displacement r , the deformation v , the action S and the nodal force R aredepicted in Figure 28.©Encyclopedia of Life Support Systems (EOLSS)
30. 30. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiFor the purpose of demonstration, a simple kinematically indeterminate structure underjoint loads R , temperature change ΔT and prescribed displacements x and q as shown ˆ ˆin Figure 29 will be used in the formulation of the stiffness equations based on thematrix displacement method. The following step by step formulation procedure is used: ˆ(a) Define R, r , S , v, X , x, Q and q (Figure 29).(b) Decompose the total structure to separate the effects of r , x and ΔT ˆAs shown in Figure 30, the independent effects of r , x and ΔT are considered ˆseparately and their contributions into each quantity of statics and kinematics will besuperposed as listed in Table 6. S(c) Derive the matrices ar , ax , cr , cx , k , vT , qT and ST LS R O TE -E AP O H C C ES E N PLU M Figure 26. Degree of Kinematic Indeterminacy equals joint degrees of freedom if there is no constraint in the system. SA©Encyclopedia of Life Support Systems (EOLSS)
31. 31. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai S Figure 27. Degree of Kinematic Indeterminacy equals joint degrees of freedom less LS number of constraints.Based on the geometric change, one can obtain the deformed elastic curve as a result of R Oa unit displacement of r by which ar and cr can be derived. Similarly for the effect of TE ˆsettlement x , one can obtain a x and cx . For the effect of temperature change, r and -E x have to be clamped; thus ST can be obtained while deformation vT is normally zero.However, if constraint exists by means of allowing for qT , vT can be non-zero. Figure AP30 presents the symbolic results of all these quantities. The member stiffness matrix Ok relating v to S can be obtained from f −1 . H C C ES E N PLU M SAFigure 28. Notation of quantities and their relationships in matrix displacement method(d) Determination of r by virtual displacement method.Using the virtual displacement system on the right-handed side of Figure 31, applyingvirtual displacement δ r corresponding to the external force R , one can obtain theequality of external and internal virtual works as:©Encyclopedia of Life Support Systems (EOLSS)
32. 32. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai δ WE = δ WI δ r T R + δ qr Q = δ vr S T T ( ) (δ r T R + δ r T cr Q = δ r T ar T T ) ( Sr + S x + ST ) R + cr Q = ar ( k vr + k vx + ST ) T T (57) R + cr Q = ⎡ ar k ar ⎤ r + ⎡ ar k ax ⎤ x + ar ST T T T ˆ T ⎣ ⎦ ⎣ ⎦ R + cr Q = K rr r + K xx x + ar ST T ˆ T R + Rq = Rr + Rx + RT S LS R O TE -E AP O Figure 29 Quantities needed in the formulation of stiffness equations by Matrix Displacement Method. H C C ES E N PLU M SA Figure 30. The superposition of effects due to three cases of kinematic excitation©Encyclopedia of Life Support Systems (EOLSS)
33. 33. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Figure 31. Application of virtual displacement δ r to obtain external force R .The last equation above equates external forces to internal forces as individual effects of Sthe three cases. Note that Kij = ai k a j is a stiffness matrix. The first subscript in the LSstiffness matrices Kij refer to the internal force result while the second subscript refers ˆto the displacement cause. Thus, the term K rx x will result in Rx which corresponds to R Othe contribution to internal force vector R as a result of applying the settlement vector TE ˆ x. -EOnce the displacement vector r is obtained from the system of stiffness equations, i.e., { }⎡ K rr ⎤ {r} = R + cr Q − K xx x − ar ST⎣ ⎦ T ˆ T AP (58) O Hthe remaining quantities can be obtained as shown in Table 6. Finally, the step by step Cprocedure of the Matrix Displacement Method is summarized in Table 7. C ES14. Trends and Perspectives EDiscrete approach of structural modeling is suited for digital computation and has lead N PLto the generalization of the formulation procedure. The two principal methods, namelyUthe matrix force method and the matrix displacement method, are convenient foranalysis of frames made up mainly of one-dimensional members. Of the two methods, Mthe matrix displacement method is more popular, due to its natural extension to themore generalized finite element method. SA Kinematics Statics v = ar r + ax x + vT S = k ar r + k ax x + ST Deformation Action Displacement r Force R = K rr r + K xx x + ar ST − cr Q ˆ T T Settlement ˆ x Reaction X = K xr r + K xx x + aT ST − cT Q x x Dependent dof q = qr + qx + qT Dependent force Q©Encyclopedia of Life Support Systems (EOLSS)
34. 34. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-Nukulchai Table 6. Final results of statics and kinematics after r is solved. 1 o Identify KI or degrees of freedom of the problem and select appropriate vector of r 2 ˆ Define k R, r , S , v , X , x , Q and q 3 Form ar , ax , cr , cx and vT , ST (if temperature changes are given) 4 Form K rr , K rx , K xr and K xx 5 Solve for r from ⎣ ⎦ { ⎡ K rr ⎤ {r} = R + cr Q − K xx x − ar ST T ˆ T } 6 Obtain the rest from Table 7. S Table 8 Step by step Procedure for Matrix Displacement Method LSUsing displacements as the primary variables, the stiffness matrix of a discrete Rstructural model can be formed globally as in the case of the matrix displacement Omethod, or locally by considering the stiffness contributions of individual elements. TEThis latter procedure is generally known as the direct stiffness method. The direct -Estiffness procedure allows the assembly of stiffness contributions from a finite numberof elements that are used to model any complex structure. The basic procedure for APdeveloping the stiffness of an element is to assume appropriate shape functions thatrepresent the displacement field over the element. Based on these assumed displacement Ofields, the element stiffness can be formulated in terms of nodal degrees of freedom. H CThe direct stiffness process was in fact the main reason how the powerful general- C ESpurposed finite element programs can be developed for any discrete structural model.Today, many commercial software packages are widely available for linear elasticanalysis based on the stiffness method. These commercial software packages may help Ereduce the work load of structural engineers, but will not decrease their accountability N PLon the reliability the analysis results. Thus, it is important that structural engineersUunderstand the fundamentals of the analysis, whether the analysis is performed by theengineers themselves or with the help of a computer program. MGlossary SABase structure: The choice of a statically determinate structure reduced from the original statically indeterminate structure after removing the redundant kinematics.Boundary The imposition of either force or displacement along a boundarycondition: of the structureCauchy stress: A definition of stress associated with the deformed configuration.Compatibility: The term used to describe the conformity between the strains and the displacement field of a deformed structure.Constitutive The equations used to describe the relationship between stressesequations: and strains of a material.Deformation The gradient of deformation in the neighborhood of a particle.gradient:©Encyclopedia of Life Support Systems (EOLSS)
35. 35. CIVIL ENGINEERING - Structural Analysis - Worsak Kanok-NukulchaiDeformation: A change in the relative positions of particles of a body.Degrees of Displacement variables associated with nodes of a discretefreedom: structure.Direct stiffness A method to directly assemble the contributions of individualmethod: members or elements of a structure to form the stiffness matrix of a structure.Discrete model of A representation of the structure to be analyzed by a computerstructures: method by which the primary variables are determined numerically.Discrete A method to model structures by a finite number of elementsmodeling: (members) that are associated with a set of discrete variables to be determined numerically.Displacement: The disposition measurement of a particle in a body or a structure. SElastic: Property of a material that rebounds to its original position from LS deformation upon the release of the applied force.Engineering shear A measurement of shear distortion of a body commonly used instrain: engineering. REquilibrium: all directions. O A state in which the internal forces balance all external forces in TE -EEuler-Bernoulli A beam theory by which the plane section of a beam remainsbeams: plane and normal to the neutral axis.Excitation: Any disturbance to the structure including external forces, AP prescribed displacement, accelerations, etc. OFinite element A method to model structures by a finite number of typicalmethod: elements associated with a set of discrete variables to be H C determined numerically.Infinitesimal A linearized strain after neglecting the nonlinear high-order terms C ESstrain: due to the smallness of the strain.Internal work Work done by the product of stress and strain. Edone: NIsotropic: A description of material property that is invariant to all PL directions.UJoints: A point where two or more one-dimensional members meet, about which a set of degrees of freedom are specified.. MKinematics: Quantities associated with geometry, the position changes or the SA deformation of geometry. This term is used in opposition to the term “statics”.Linear elastic A structural analysis that assumes linear stress-strain relationshipanalysis: for all elastic structural members.Mathematical A mathematical representation of the structure.model:Matrix analysis of A term used to describe structural analysis of discrete structuralstructures: system using matrix operations.Matrix An analysis of discrete structural model of which displacementsdisplacement are primary variables.method:Matrix force An analysis of discrete structural model of which force parameters©Encyclopedia of Life Support Systems (EOLSS)