Ch 1 structural analysis stiffness method

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Ch 1 structural analysis stiffness method

  1. 1. CHAPTER 1Structural Analysis—Stiffness Method1.1 IntroductionComputer programs for plastic analysis of framed structures havebeen in existence for some time. Some programs, such as those devel-oped earlier by, among others, Wang,1 Jennings and Majid,2 andDavies,3 and later by Chen and Sohal,4 perform plastic analysis forframes of considerable size. However, most of these computer pro-grams were written as specialist programs specifically for linear ornonlinear plastic analysis. Unlike linear elastic analysis computerprograms, which are commonly available commercially, computerprograms for plastic analysis are not as accessible. Indeed, very few,if any, are being used for daily routine design in engineering offices.This may be because of the perception by many engineers that theplastic design method is used only for certain types of usually simplestructures, such as beams and portal frames. This perception dis-courages commercial software developers from developing computerprograms for plastic analysis because of their limited applications. Contrary to the traditional thinking that plastic analysis is per-formed either by simple manual methods for simple structures or bysophisticated computer programs written for more general applica-tions, this book intends to introduce general plastic analysis methods,which take advantage of the availability of modern computationaltools, such as linear elastic analysis programs and spreadsheet applica-tions. These computational tools are in routine use in most engi-neering design offices nowadays. The powerful number-crunchingcapability of these tools enables plastic analysis and design to be per-formed for structures of virtually any size. The amount of computation required for structural analysis islargely dependent on the degree of statical indeterminacy of the
  2. 2. 2 Plastic Analysis and Design of Steel Structuresstructure. For determinate structures, use of equilibrium conditionsalone will enable the reactions and internal forces to be determined.For indeterminate structures, internal forces are calculated by consid-ering both equilibrium and compatibility conditions, through whichsome methods of structural analysis suitable for computer applica-tions have been developed. The use of these methods for analyzingindeterminate structures is usually not simple, and computers areoften used for carrying out these analyses. Most structures in practiceare statically indeterminate. Structural analysis, whether linear or nonlinear, is mostly basedon matrix formulations to handle the enormous amount of numericaldata and computations. Matrix formulations are suitable for computerimplementation and can be applied to two major methods of struc-tural analysis: the flexibility (or force) method and the stiffness (or dis-placement) method. The flexibility method is used to solve equilibrium and compat-ibility equations in which the reactions and member forces areformulated as unknown variables. In this method, the degree of stat-ical indeterminacy needs to be determined first and a number ofunknown forces are chosen and released so that the remaining struc-ture, called the primary structure, becomes determinate. The pri-mary structure under the externally applied loads is analyzed andits displacement is calculated. A unit value for each of the chosenreleased forces, called redundant forces, is then applied to the pri-mary structure (without the externally applied loads) so that, fromthe force-displacement relationship, displacements of the structureare calculated. The structure with each of the redundant forces iscalled the redundant structure. The compatibility conditions basedon the deformation between the primary structure and the redundantstructures are used to set up a matrix equation from which theredundant forces can be solved. The solution procedure for the force method requires selection ofthe redundant forces in the original indeterminate structure and thesubsequent establishment of the matrix equation from the compati-bility conditions. This procedure is not particularly suitable for com-puter programming and the force method is therefore usually usedonly for simple structures. In contrast, formulation of the matrix equations for the stiffnessmethod is done routinely and the solution procedure is systematic.Therefore, the stiffness method is adopted in most structural analysiscomputer programs. The stiffness method is particularly useful forstructures with a high degree of statical indeterminacy, althoughit can be used for both determinate and indeterminate structures.The stiffness method is used in the elastoplastic analysis describedin this book and the basis of this method is given in this chapter.
  3. 3. Structural Analysis—Stiffness Method 3In particular, the direct stiffness method, a variant of the general stiff-ness method, is described. For a brief history of the stiffness method,refer to the review by Samuelsson and Zienkiewicz.51.2 Degrees of Freedom and IndeterminacyPlastic analysis is used to obtain the behavior of a structure at collapse.As the structure approaches its collapse state when the loads are increas-ing, the structure becomes increasingly flexible in its stiffness. Itsflexibility at any stage of loading is related to the degree of statical inde-terminacy, which keeps decreasing as plastic hinges occur with theincreasing loads. This section aims to describe a method to distinguishbetween determinate and indeterminate structures by examining thedegrees of freedom of structural frames. The number of degrees of free-dom of a structure denotes the independent movements of the structuralmembers at the joints, including the supports. Hence, it is an indicationof the size of the structural problem. The degrees of freedom of a struc-ture are counted in relation to a reference coordinate system. External loads are applied to a structure causing movements atvarious locations. For frames, these locations are usually definedat the joints for calculation purposes. Thus, the maximum numberof independent displacements, including both rotational and transla-tional movements at the joints, is equal to the number of degrees offreedom of the structure. To identify the number of degrees of freedomof a structure, each independent displacement is assigned a number,called the freedom code, in ascending order in the global coordinatesystem of the structure. Figure 1.1 shows a frame with 7 degrees of freedom. Note that thepinned joint at C allows the two members BC and CD to rotate indepen-dently, thus giving rise to two freedoms in rotation at the joint. In structural analysis, the degree of statical indeterminacy isimportant, as its value may determine whether the structure 2 5 3 1 6 4 B C 7 A DFIGURE 1.1. Degrees of freedom of a frame.
  4. 4. 4 Plastic Analysis and Design of Steel Structuresis globally unstable or stable. If the structure is stable, the degree ofstatical indeterminacy is, in general, proportional to the level of com-plexity for solving the structural problem. The method described here for determining the degree of staticalindeterminacy of a structure is based on that by Rangasami andMallick.6 Only plane frames will be dealt with here, although the methodcan be extended to three-dimensional frames.1.2.1 Degree of Statical Indeterminacy of FramesFor a free member in a plane frame, the number of possible displace-ments is three: horizontal, vertical, and rotational. If there are n mem-bers in the structure, the total number of possible displacements,denoted by m, before any displacement restraints are considered, is m ¼ 3n (1.1) For two members connected at a joint, some or all of the displa-cements at the joint are common to the two members and these com-mon displacements are considered restraints. In this method fordetermining the degree of statical indeterminacy, every joint is con-sidered as imposing r number of restraints if the number of commondisplacements between the members is r. The ground or foundationis considered as a noncounting member and has no freedom. Figure 1.2indicates the value of r for each type of joints or supports in a planeframe. For pinned joints with multiple members, the number of pinnedjoints, p, is counted according to Figure 1.3. For example, for a four-member pinned connection shown in Figure 1.3, a first joint iscounted by considering the connection of two members, a secondjoint by the third member, and so on. The total number of pinnedjoints for a four-member connection is therefore equal to three. In gen-eral, the number of pinned joints connecting n members is p ¼ n – 1.The same method applies to fixed joints. r=1 r=2 r=3 r=2 r=3 (a) Roller (b) Pin (c) Fixed (d) Pin (e) Rigid (≡ fixed)FIGURE 1.2. Restraints of joints.
  5. 5. Structural Analysis—Stiffness Method 5 No. of pins, p = 1 No. of pins, p = 2 No. of pins, p = 3FIGURE 1.3. Method for joint counting. No. of pins, p = 2.5FIGURE 1.4. Joint counting of a pin with roller support. For a connection at a roller support, as in the example shown inFigure 1.4, it can be calculated that p ¼ 2.5 pinned joints and that thetotal number of restraints is r ¼ 5. The degree of statical indeterminacy, fr, of a structure is deter-mined by X fr ¼ m À r (1.2) a. If fr ¼ 0, the frame is stable and statically determinate. b. If fr < 0, the frame is stable and statically indeterminate to the degree fr. c. If fr > 0, the frame is unstable. Note that this method does not examine external instability orpartial collapse of the structure.Example 1.1 Determine the degree of statical indeterminacy for thepin-jointed truss shown in Figure 1.5. (a) (b)FIGURE 1.5. Determination of degree of statical indeterminacy in Example 1.1.
  6. 6. 6 Plastic Analysis and Design of Steel StructuresSolution. For the truss in Figure 1.5a, number of members n ¼ 3; num-ber of pinned joints p ¼ 4.5. Hence, fr ¼ 3 Â 3 À 2 Â 4:5 ¼ 0 and the truss is a determinatestructure. For the truss in Figure 1.5b, number of members n ¼ 2;number of pinned joints p ¼ 3. Hence, fr ¼ 3 Â 2 À 2 Â 3 ¼ 0 and the truss is a determinatestructure.Example 1.2 Determine the degree of statical indeterminacy for theframe with mixed pin and rigid joints shown in Figure 1.6. C D B E A FFIGURE 1.6. Determination of degree of statical indeterminacy in Example 1.2.Solution. For this frame, a member is counted as one between twoadjacent joints. Number of members ¼ 6; number of rigid (or fixed)joints ¼ 5. Note that the joint between DE and EF is a rigid one,whereas the joint between BE and DEF is a pinned one. Number ofpinned joints ¼ 3. Hence, fr ¼ 3 Â 6 À 3 Â 5 À 2 Â 3 ¼ À3 and the frame is an inde-terminate structure to the degree 3.1.3 Statically Indeterminate Structures—DirectStiffness MethodThe spring system shown in Figure 1.7 demonstrates the use of thestiffness method in its simplest form. The single degree of freedomstructure consists of an object supported by a linear spring obeyingHooke’s law. For structural analysis, the weight, F, of the object andthe spring constant (or stiffness), K, are usually known. The purpose
  7. 7. Structural Analysis—Stiffness Method 7 K D FFIGURE 1.7. Load supported by linear spring.of the structural analysis is to find the vertical displacement, D, andthe internal force in the spring, P. From Hooke’s law, F ¼ KD (1.3) Equation (1.3) is in fact the equilibrium equation of the system.Hence, the displacement, D, of the object can be obtained by D ¼ F=K (1.4)The displacement, d, of the spring is obviously equal to D. That is, d¼D (1.5)The internal force in the spring, P, can be found by P ¼ Kd (1.6) In this simple example, the procedure for using the stiffnessmethod is demonstrated through Equations (1.3) to (1.6). For a struc-ture composed of a number of structural members with n degrees offreedom, the equilibrium of the structure can be described by a num-ber of equations analogous to Equation (1.3). These equations can beexpressed in matrix form as fFgnÂ1 ¼ ½K ŠnÂn fDgnÂ1 (1.7)where fFgnÂ1 is the load vector of size ðn  1Þ containing the externalloads, ½K ŠnÂn is the structure stiffness matrix of size ðn  nÞcorresponding to the spring constant K in a single degree systemshown in Figure 1.7, and fDgnÂ1 is the displacement vector of sizeðn  1Þ containing the unknown displacements at designated loca-tions, usually at the joints of the structure.
  8. 8. 8 Plastic Analysis and Design of Steel Structures The unknown displacement vector can be found by solvingEquation (1.7) as fDg ¼ ½K ŠÀ1 fFg (1.8)Details of the formation of fFg, ½KŠ, and fDg are given in the followingsections.1.3.1 Local and Global Coordinate SystemsA framed structure consists of discrete members connected at joints,which may be pinned or rigid. In a local coordinate system for a mem-ber connecting two joints i and j, the member forces and thecorresponding displacements are shown in Figure 1.8, where the axialforces are acting along the longitudinal axis of the member and theshear forces are acting perpendicular to its longitudinal axis. In Figure 1.8, Mi,j, yi,j ¼ bending moments and correspondingrotations at ends i, j, respectively; Ni,j, ui,j are axial forces andcorresponding axial deformations at ends i, j, respectively; and Qi,j,vi,j are shear forces and corresponding transverse displacements atends i, j, respectively. The directions of the actions and movementsshown in Figure 1.8 are positive when using the stiffness method. As mentioned in Section 1.2, the freedom codes of a structure areassigned in its global coordinate system. An example of a memberforming part of the structure with a set of freedom codes (1, 2, 3, 4,5, 6) at its ends is shown in Figure 1.9. At either end of the member,the direction in which the member is restrained from movement isassigned a freedom code “zero,” otherwise a nonzero freedom code isassigned. The relationship for forces and displacements between localand global coordinate systems will be established in later sections. Qj, vj Mj, qj Nj, uj Qi, vi j Mi, qi Ni, ui iFIGURE 1.8. Local coordinate system for member forces and displacements.
  9. 9. Structural Analysis—Stiffness Method 9 5 6 j 4 2 3 i 1FIGURE 1.9. Freedom codes of a member in a global coordinate system.1.4 Member Stiffness MatrixThe structure stiffness matrix ½K Š is assembled on the basis of theequilibrium and compatibility conditions between the members. Fora general frame, the equilibrium matrix equation of a member is f Pg ¼ ½ K e Š f dg (1.9)where fPg is the member force vector, ½Ke Š is the member stiffnessmatrix, and fdg is the member displacement vector, all in the mem-ber’s local coordinate system. The elements of the matrices in Equa-tion (1.9) are given as 8 9 2 3 8 9 > Ni > > > K11 0 0 K14 0 0 > ui > > > >Q > > > 6 0 >v > > > > i> > > 6 K22 K23 0 K25 K26 7 7 > i> > > < = 6 0 < = Mi 6 K32 K33 0 K35 K36 7 7; fdg ¼ yif Pg ¼ ; ½Ke Š ¼ 6 > Nj > > > 6 K41 0 0 K44 0 0 7 7 > uj > > > > Qj > > > 4 0 > > > vj > > > > > K52 K53 0 K55 K56 5 > > > > : ; : ; Mj 0 K62 K63 0 K65 K66 yjwhere the elements of fPg and fdg are shown in Figure 1.8.1.4.1 Derivation of Elements of Member Stiffness MatrixA member under axial forces Ni and Nj acting at its ends producesaxial displacements ui and uj as shown in Figure 1.10. From thestress-strain relation, it can be shown that EA À Á Ni ¼ ui Àuj (1.10a) L EA À Á Nj ¼ uj Àui (1.10b) L
  10. 10. 10 Plastic Analysis and Design of Steel Structures uj Nj Original position j Displaced position i Ni uiFIGURE 1.10. Member under axial forces.where E is Young’s modulus, A is cross-sectional area, and L is length EAof the member. Hence, K11 ¼ ÀK14 ¼ ÀK41 ¼ K44 ¼ . L For a member with shear forces Qi, Qj and bending momentsMi, Mj acting at its ends as shown in Figure 1.11, the end displace-ments and rotations are related to the bending moments by theslope-deflection equations as À Á! 2EI 3 vj Àvi Mi ¼ 2yi þyj À (1.11a) L L À Á! 2EI 3 vj Àvi Mj ¼ 2yj þyi À (1.11b) L L 6EI 2EI 4EI Hence, K62 ¼ ÀK65 ¼ 2 , K63 ¼ , and K66 ¼ . L L L By taking the moment about end j of the member in Figure 1.11,we obtain À Á! Mi þMj 2EI 6 vj Àvi Qi ¼ ¼ 2 3yi þ3yj À (1.12a) L L L Displaced position qj vj j qi Qj Mj vi i Original position Qi MiFIGURE 1.11. Member under shear forces and bending moments.
  11. 11. Structural Analysis—Stiffness Method 11Also, by taking the moment about end i of the member, we obtain Mi þMj Qj ¼ À ¼ ÀQi (1.12b) LHence, 12EI 6EIK22 ¼ K55 ¼ ÀK25 ¼ ÀK52 ¼ and K23 ¼ K26 ¼ ÀK53 ¼ ÀK66 ¼ 2 . L3 L In summary, the resulting member stiffness matrix is symmetricabout the diagonal: 2 3 EA EA 6 L 0 0 À 0 0 7 6 L 7 6 6EI 7 6 12EI 6EI 12EI 7 6 0 0 À 3 7 6 L3 L2 L L2 7 6 7 6 6EI 4EI 6EI 2EI 7 6 0 0 À 2 7 6 L2 L 7 6 L L 7 ½Ke Š ¼ 6 EA 6 EA 7 7 (1.13) 6À 0 0 0 0 7 6 L L 7 6 7 6 7 6 12EI 6EI 12EI 6EI 7 6 0 À 3 À 2 0 À 2 7 6 L L L3 L 7 6 7 6 6EI 2EI 6EI 4EI 7 4 0 0 À 2 5 L2 L L L1.5 Coordinates TransformationIn order to establish the equilibrium conditions between the memberforces in the local coordinate system and the externally applied loadsin the global coordinate system, the member forces are transformedinto the global coordinate system by force resolution. Figure 1.12shows a member inclined at an angle a to the horizontal.1.5.1 Load TransformationThe forces in the global coordinate system shown with superscript “g”in Figure 1.12 are related to those in the local coordinate system by g Hi ¼ Ni cos a À Qi sin a (1.14a) Vig ¼ Ni sin a þ Qi cos a (1.14b) g Mi ¼ Mi (1.14c)
  12. 12. 12 Plastic Analysis and Design of Steel Structures Qj Mj Nj Qi g j Vj Vg i Mi Ni g Mj α Mg i i Hj g Hg iFIGURE 1.12. Forces in the local and global coordinate systems. Similarly, g Hj ¼ Nj cos a À Qj sin a (1.14d) Vjg ¼ Nj sin a þ Qj cos a (1.14e) g Mj ¼ Mj (1.14f) In matrix form, Equations (1.14a) to (1.14f) can be expressed as È gÉ Fe ¼ ½T ŠfPg (1.15) gwherefFe g is the member force vector in the global coordinate systemand ½T Š is the transformation matrix, both given as 8 g9 2 3 Hi cos a Àsin a 0 0 0 0 g V i 6 sin a cos a 0 0 0 07 g 6 7 M 6 7 È gÉ i = 6 0 0 1 0 0 07 Fe ¼ 6 and ½T Š ¼ 6 7: 0 cos a Àsin a 0 7 g Hj 6 0 0 7 g V j 6 4 0 0 sin a cos a 0 5 7 0 g : ; Mj 0 0 0 0 0 11.5.2 Displacement TransformationThe displacements in the global coordinate system can be related tothose in the local coordinate system by following the procedure simi-lar to the force transformation. The displacements in both coordinatesystems are shown in Figure 1.13. From Figure 1.13, ui ¼ ug cos a þ vi sin a i g (1.16a)
  13. 13. Structural Analysis—Stiffness Method 13 vj uj vi θj vg i θi vg j ui ug i α ug jFIGURE 1.13. Displacements in the local and global coordinate systems. vi ¼ Àug sin a þ vi cos a i g (1.16b) yi ¼ yg i (1.16c) uj ¼ ug cos a þ vj sin a j g (1.16d) vj ¼ Àug sin a þ vj cos a j g (1.16e) yj ¼ yg j (1.16f) In matrix form, Equations (1.16a) to (1.16f) can be expressed as È É f dg ¼ ½ T Š t D g e (1.17)where fDg g is the member displacement vector in the global coordi- enate system corresponding to the directions in which the freedomcodes are specified and is given as 8 g9 ui vg i g È g É yi = De ¼ g uj g v j g : ; yjand ½T Št is the transpose of ½T Š.1.6 Member Stiffness Matrix in Global Coordinate SystemFrom Equation (1.15), È gÉ Fe ¼ ½T ŠfPg ¼ ½T Š½Ke Šfdg from Equation ð1:9Þ
  14. 14. 14 Plastic Analysis and Design of Steel Structures È É ¼ ½T Š½Ke Š½T Št Dg e from Equation ð1:17Þ Â g ÃÈ É ¼ K e Dge (1.18)  gwhere Ke Š ¼ ½T Š½Ke Š½T Št ¼ member stiffness matrix in the global coor-dinate system. g An explicit expression for ½Ke Š is 2 0 1 0 1 0 1 3 EA 12EI EA 12EI A 6EI EA 12EI EA 12EI A 6EI 6 C2 6 þ S2 3 SC@ À 3 ÀS 2 À@C2 þ S2 3 A ÀSC@ À 3 ÀS 2 7 6 L L L L L L L L L L 7 7 6 0 1 0 1 7 6 7 6 EA 12EI 6EI EA 12EI A EA 12EI 6EI 7 6 S2 þ C2 3 C ÀSC@ À 3 À@S2 þ C2 3 A C 2 7 6 L2 L 7 6 L L L L L L 7 6 7 6 4EI 6EI 6EI 2EI 7 6 ÀC 7 gà 66 S 7 L L2 L2 L 7 Ke ¼ 6 6 0 1 7 7 6 6EI 7 6 EA 12EI EA 12EI A 7 6 C2 þ S2 3 SC@ À 3 S 2 7 6 L L L L L 7 6 7 6 7 6 EA 12EI 6EI 7 6 Symmetric S2 þ C2 3 ÀC 2 7 6 L L L 7 6 7 6 7 6 4EI 7 4 5 L (1.19)where C = cos a; S = sin a.1.7 Assembly of Structure Stiffness MatrixConsider part of a structure with four externally applied forces, F1, F2,F4, and F5, and two applied moments, M3 and M6, acting at thetwo joints p and q connecting three members A, B, and C as shown inFigure 1.14. The freedom codes at joint p are {1, 2, 3} and at joint q are{4, 5, 6}. The structure stiffness matrix [K] is assembled on the basis oftwo conditions: compatibility and equilibrium conditions at the joints.1.7.1 Compatibility ConditionAt joint p, the global displacements are D1 (horizontal), D2 (vertical),and D3 (rotational). Similarly, at joint q, the global displacements areD4 (horizontal), D5 (vertical), and D6 (rotational). The compatibilitycondition is that the displacements (D1, D2, and D3) at end p of mem-ber A À Á the same as those at end p of member B. Thus, are À gÁðug ÞA ¼ ug B ¼ D1, ðvj ÞA ¼ vi B ¼ D2, and ðyg ÞA ¼ ðyg ÞB ¼ D3. The j i g j isame condition applies to displacements (D4, D5, and D6) at end qof both members B and C.
  15. 15. Structural Analysis—Stiffness Method 15 5 2 6 q 4 3 B C F4 1 M6 p F1 F5 A M3 F2FIGURE 1.14. Assembly of structure stiffness matrix [K]. The member stiffness matrix in the global coordinate systemgiven in Equation (1.19) can be written as 2 3 k11 k12 k13 k14 k15 k16 6 k21 k22 k23 k24 k25 k26 7 6 7  g à 6 k31 k32 k33 k34 k35 k36 7 Ke ¼ 6 6 k41 7 (1.20) 6 k42 k43 k44 k45 k46 7 7 4 k51 k52 k53 k54 k55 k56 5 k61 k62 k63 k64 k65 k66 EA 12EIwhere k11 ¼ C2 þ S2 3 , etc. L L For member A, from Equation (1.18), g Hj ¼ ::::: þ ::::: þ ::::: þ ðk44 ÞA D1 þ ðk45 ÞA D2 þ ðk46 ÞA D3 (1.21a) A Vjg ¼ ::::: þ ::::: þ ::::: þ ðk54 ÞA D1 þ ðk55 ÞA D2 þ ðk56 ÞA D3 (1.21b) A g Mj ¼ ::::: þ ::::: þ ::::: þ ðk64 ÞA D1 þ ðk65 ÞA D2 þ ðk66 ÞA D3 (1.21c) A Similarly, for member B,À gÁ Hi B ¼ ðk11 ÞB D1 þ ðk12 ÞB D2 þ ðk13 ÞB D3 þ ðk14 ÞB D4 þ ðk15 ÞB D5 þ ðk16 ÞB D6 (1.21d)À gÁ Vi B ¼ ðk21 ÞB D1 þ ðk22 ÞB D2 þ ðk23 ÞB D3 þ ðk24 ÞB D4 þ ðk25 ÞB D5 þ ðk26 ÞB D6 (1.21e)À g Á Mi B ¼ ðk31 ÞB D1 þ ðk32 ÞB D2 þ ðk33 ÞB D3 þ ðk34 ÞB D4 þ ðk35 ÞB D5 þ ðk36 ÞB D6 (1.21f)
  16. 16. 16 Plastic Analysis and Design of Steel Structures g Hj ¼ ðk41 ÞB D1 þ ðk42 ÞB D2 þ ðk43 ÞB D3 þ ðk44 ÞB D4 þ ðk45 ÞB D5 þ ðk46 ÞB D6 B (1.21g) Vjg ¼ ðk51 ÞB D1 þ ðk52 ÞB D2 þ ðk53 ÞB D3 þ ðk54 ÞB D4 þ ðk55 ÞB D5 þ ðk56 ÞB D6 B (1.21h) g Mj ¼ ðk61 ÞB D1 þ ðk62 ÞB D2 þ ðk63 ÞB D3 þ ðk64 ÞB D4 þ ðk65 ÞB D5 þ ðk66 ÞB D6 B (1.21i) Similarly, for member C, À gÁ Hi C ¼ ðk11 ÞC D1 þ ðk12 ÞC D2 þ ðk13 ÞC D3 þ ::::: þ ::::: þ ::::: (1.21j) À gÁ Vi C ¼ ðk21 ÞC D1 þ ðk22 ÞC D2 þ ðk23 ÞC D3 þ ::::: þ ::::: þ ::::: (1.21k) À g Á Mi C ¼ ðk31 ÞC D1 þ ðk32 ÞC D2 þ ðk33 ÞC D3 þ ::::: þ ::::: þ ::::: (1.21l)1.7.2 Equilibrium ConditionAny of the externally applied forces or moments applied in a certaindirection at a joint of a structure is equal to the sum of the memberforces acting in the same direction for members connected at thatjoint in the global coordinate system. Therefore, at joint p, À gÁ g F1 ¼ Hj þ Hi B (1.22a) A À Á F2 ¼ Vjg þ Vig B (1.22b) A À gÁ g M3 ¼ Mj þ Mi B (1.22c) A Also, at joint q, À gÁ g F4 ¼ Hj þ Hi C (1.22d) B À Á F5 ¼ Vjg þ Vig C (1.22e) B À gÁ g M6 ¼ Mj þ Mi C (1.22f) BBy writing Equations (1.22a) to (1.22f) in matrix form using Equations(1.21a) to (1.21l) and applying this operation to the whole structure,the following equilibrium equation of the whole structure is obtained:
  17. 17. 8 9 8 9 2 3 6 7 F1 6 ðk44 ÞA þ ðk11 ÞB ðk45 ÞA þ ðk12 ÞB ðk46 ÞA þ ðk13 ÞB ðk14 ÞB ðk15 ÞB ðk16 ÞB 7 D1 6 7 6 7 F2 6 ðk54 ÞA þ ðk21 ÞB ðk55 ÞA þ ðk22 ÞB ðk56 ÞA þ ðk23 ÞB ðk24 ÞB ðk25 ÞB ðk26 ÞB 7 D2 6 7 6 7 M3 Structural Analysis—Stiffness Method = 6 ðk64 ÞA þ ðk31 ÞB ðk65 ÞA þ ðk32 ÞB ðk66 ÞA þ ðk33 ÞB ðk34 ÞB ðk35 ÞB ðk36 ÞB 7 D3 = 6 7 ¼6 7 F4 6 6 ðk41 ÞB ðk42 ÞB ðk43 ÞB ðk44 ÞB þ ðk11 ÞC ðk45 ÞB þ ðk12 ÞC ðk46 ÞB þ ðk13 ÞC 7 D4 6 7 7 F5 6 ðk51 ÞB ðk52 ÞB ðk53 ÞB ðk54 ÞB þ ðk21 ÞC ðk55 ÞB þ ðk22 ÞC ðk56 ÞB þ ðk23 ÞC 7 D5 7 6 6 7 M6 6 4 ðk61 ÞB ðk62 ÞB ðk63 ÞB ðk64 ÞB þ ðk31 ÞC ðk65 ÞB þ ðk32 ÞC ðk66 ÞB þ ðk33 ÞC 5 7 D4 : ; : ; (1.23) 17
  18. 18. 18 Plastic Analysis and Design of Steel Structureswhere the “l” stands for matrix coefficients contributed from theother parts of the structure. In simple form, Equation (1.23) can bewritten as fF g ¼ ½ K Š fD gwhich is identical to Equation (1.7). Equation (1.23) shows how thestructure equilibrium equation is set up in terms of the load vectorfFg, structure stiffness matrix ½K Š, and the displacement vector fDg. Close examination of Equation (1.23) reveals that the stiffnesscoefficients of the three members A, B, and C are assembled into ½K Šin a way according to the freedom codes assigned to the members.Take member A as an example. By writing the freedom codes in theorder of ends i and j around the member stiffness matrix in the globalcoordinate system shown in Figure 1.15, the coefficient, for example,k54 , is assembled into the position [2, 1] of ½K Š. Similarly, the coeffi-cient k45 is assembled into the position [1, 2] of ½K Š. The coefficientsin all member stiffness matrices in the global coordinate system canbe assembled into ½K Š in this way. Since the resulting matrix is sym-metric, only half of the coefficients need to be assembled. A schematic diagram showing the assembly procedure for thestiffness coefficients of the three members A, B, and C into ½K Š is gshown in Figure 1.16. Note that since ½Ke Š is symmetric, ½K Š is alsosymmetric. Any coefficients in a row or column corresponding to zerofreedom code will be ignored.1.8 Load VectorThe load vector fFg of a structure is formed by assembling the individ-ual forces into the load vector in positions corresponding to the direc-tions of the freedom codes. For the example in Figure 1.14, the loadfactor is given as that shown in Figure 1.17. 1 2 3 k11 k12 k13 k14 k15 k16 k21 k22 k23 k24 k25 k26 k31 k32 k33 k34 k35 k36 [Keg]A = k41 k42 k43 k44 k45 k46 1 k51 k52 k 53 k54 k55 k56 2 k61 k62 k63 k64 k65 k66 3FIGURE 1.15. Assembly of stiffness coefficients into the structure stiffnessmatrix.
  19. 19. Structural Analysis—Stiffness Method 19 [K g ] of member A e [K g ] of member B e 1 2 3 4 5 6 1 [K ] of structure 2 3 4 5 6 [K g ] of member C eFIGURE 1.16. Assembly of structure stiffness matrix. Freedom codes • F1 1 F2 2 F3 3 {F } = F4 4 F5 5 F6 6 •FIGURE 1.17. Assembly of load vector.1.9 Methods of SolutionThe displacements of the structure can be found by solving Equation(1.23). Because of the huge size of the matrix equation usually encoun-tered in practice, Equation (1.23) is solved routinely by numericalmethods such as the Gaussian elimination method and the iterativeGauss–Seidel method. It should be noted that in using these
  20. 20. 20 Plastic Analysis and Design of Steel Structuresnumerical methods, the procedure is analogous to inverting the struc-ture stiffness matrix, which is subsequently multiplied by the loadvector as in Equation (1.8): fDg ¼ ½K ŠÀ1 fFg (1.8) The numerical procedure fails only if an inverted ½K Š cannot befound. This situation occurs when the determinant of ½K Š is zero,implying an unstable structure. Unstable structures with a degree ofstatically indeterminacy, fr, greater than zero (see Section 1.2) willhave a zero determinant of ½K Š. In numerical manipulation by compu-ters, an exact zero is sometimes difficult to obtain. In such cases, agood indication of an unstable structure is to examine the displace-ment vector fDg, which would include some exceptionally largevalues.1.10 Calculation of Member ForcesMember forces are calculated according to Equation (1.9). Hence, f Pg ¼ ½ K e Š f dg (1.24) ¼ ½Ke Š ½T Št fDg g ewhere fDg g is extracted from fDg for each member according to its efreedom codes and 2 3 EA EA EA EA C S 0 ÀC ÀS 0 6 L L L L 7 6 7 6 7 6 6EI 7 6 ÀS 12EI C 12EI 6EI 12EI S 3 ÀC 3 12EI 7 6 L3 L3 L2 L L L2 7 6 7 6 7 6 2EI 7 6 ÀS 6EI 6EI 4EI 6EI ÀC 2 6EI 7 6 C 2 S 2 7 6 L2 L L L L L 7 t 6 7 ½Ke Š½T Š ¼ 6 7 6 EA EA EA EA 7 6 ÀC ÀS 0 C S 0 7 6 L L L L 7 6 7 6 7 6 12EI 12EI 6EI 12EI 12EI 6EI 7 6 S ÀC 3 À 2 ÀS 3 C 3 À 2 7 6 L3 L L L L L 7 6 7 6 7 6 4EI 7 4 6EI 6EI 2EI 6EI 6EI 5 ÀS 2 C 2 S 2 ÀC 2 L L L L L L For the example in Figure 1.14,
  21. 21. Structural Analysis—Stiffness Method 21 2 3 EA EA EA EA 6 C L S 0 ÀC ÀS 0 7 6 L L L 7 6 7 6 7 6 12EI 12EI 6EI 12EI 12EI 6EI 7 8 9 6 ÀS L3 6 C L3 L2 S L3 ÀC L3 7 L 2 7 8 D1 9 Ni 6 6 7 7 Q 6 i 6 7D 2 6 ÀS 6EI 6EI 4EI 6EI 6EI 2EI 7 C S ÀC 7 Mi 6 = 6 L2 L2 L L2 L2 L 7 D 7 3 =f Pg ¼ ¼ 6 7 Nj 6 6 7 7 D 6 ÀC EA ÀS EA 0 C EA S EA 0 7 4 6 L L L L 7 Qj 6 7 D5 6 7 : 6 ; Mj 6 S 12EI 12EI 6EI 12EI 12EI 6EI 7 D6 ; 7: 6 ÀC À ÀS C À 2 7 6 L3 L3 L2 L3 L3 L 7 6 7 6 7 6 4EI 7 4 ÀS 6EI C 6EI 2EI S 6EI ÀC 6EI 5 L2 L2 L L2 L2 L In summary, the procedure for using the stiffness method to calculatethe displacements of the structure and the member forces is as follows. 1. Assign freedom codes to each joint indicating the displace- ment freedom at the ends of the members connected at that joint. Assign a freedom code of “zero” to any restrained displacement. 2. Assign an arrow to each member so that ends i and j are defined. Also, the angle of orientation a for the member is defined in Figure 1.18 as: j α iFIGURE 1.18. Definition of angle of orientation for member. 3. Assemble the structure stiffness matrix ½K Š from each of the member stiffness matrices. 4. Form the load vector fFg of the structure.
  22. 22. 22 Plastic Analysis and Design of Steel Structures 5. Calculate the displacement vector fDg by solving for fDg ¼ ½K ŠÀ1 fFg. 6. Extract the local displacement vector fDg g from fDg and cal- e culate the member force vector fPg using fPg ¼ ½Ke Š½T Št fDg g. e1.10.1 Sign Convention for Member Force DiagramsPositive member forces and displacements obtained from the stiffnessmethod of analysis are shown in Figure 1.19. To plot the forces in con-ventional axial force, shear force, and bending moment diagrams, it isnecessary to translate them into a system commonly adopted forplotting. The sign convention for such a system is given as follows.Axial ForceFor a member under compression, the axial force at end i is positive(from analysis) and at end j is negative (from analysis), as shown inFigure 1.20.Shear ForceA shear force plotted positive in diagram is acting upward (positivefrom analysis) at end i and downward (negative from analysis) atend j as shown in Figure 1.21. Positive shear force is usually plottedin the space above the member.FIGURE 1.19. Direction of positive forces and displacements using stiffnessmethod. Compressive j iFIGURE 1.20. Member under compression.
  23. 23. Structural Analysis—Stiffness Method 23 j iFIGURE 1.21. Positive shear forces.Bending MomentA member under sagging moment is positive in diagram (clockwiseand negative from analysis) at end i and positive (anticlockwise and pos-itive from analysis) at end j as shown in Figure 1.22. Positive bendingmoment is usually plotted in the space beneath the member. In doingso, a bending moment is plotted on the tension face of the member. j iFIGURE 1.22. Sagging moment of a member.Example 1.3 Determine the member forces and plot the shear forceand bending moment diagrams for the structure shown inFigure 1.23a. The structure with a pin at D is subject to a vertical forceof 100 kN being applied at C. For all members, E ¼ 2 Â 108 kN/m2,A ¼ 0.2 m2, and I ¼ 0.001 m4.Solution. The freedom codes for the whole structure are shown inFigure 1.23b. There are four members separated by joints B, C, andD with the member numbers shown. The arrows are assigned to 100 2 5 8 kN 6 3 1 4 7 9 B D B D C 2 C 3 10 2m 4m 5m 1 0 4 0 0 0 A E 0 0 A E (a) Frame with applied load (b) Freedom codesFIGURE 1.23. Example 1.3.
  24. 24. 24 Plastic Analysis and Design of Steel Structuresindicate end i (tail of arrow) and end j (head of arrow). Thus, the orien-tations of the members are Member 1: a ¼ 90 Member 2: a ¼ 0 Member 3: a ¼ 0 Member 4: a ¼ 270 or –90 g The ½Ke Š for the members with the assigned freedom codes forthe coefficients is 0 0 0 1 2 3 2 3 1:92  104 0 À4:8  104 À1:92  104 0 À4:8  104 0 6 8  10 6 À8  10 6 70 6 0 0 0 7 6 7 6 1:6  105 4:8  104 0 8  104 7 0½ K e Š1 ¼ 6 g 6 7 4 71 6 1:92  104 0 4:8  10 7 6 7 4 Symmetric 8  106 0 52 5 1:6  10 3 2 1 2 3 4 5 6 3 2  107 0 0 À2  107 0 0 1 6 3  105 3  105 À3  105 3  105 7 2 6 0 7 6 7 g 6 4  105 0 À3  105 2  105 7 3 ½K e Š2 ¼ 6 7 6 2  107 0 0 74 6 7 4 Symmetric 3  105 À3  105 5 5 4  105 6 2 4 5 6 7 8 9 3 1  107 0 0 À1  107 0 0 4 6 3:75  104 7:5  104 À3:75  104 7:5  104 7 5 6 0 7 6 7 g 6 2  105 0 À7:5  104 1  105 7 6½Ke Š3 ¼ 6 7 6 1  107 0 0 77 6 7 4 Symmetric 3:75  104 À7:5  10 458 2  105 9 2 7 8 10 0 0 0 3 1:92  104 0 À4:8  104 À1:92  104 0 4:8  104 7 6 8  106 À8  106 7 8 6 0 0 0 7 6 7 g 6 1:6  105 À4:8  104 0 8  104 7 10½ K e Š4 ¼6 47 0 6 1:92  104 0 À4:8  10 7 6 7 4 Symmetric 8  106 0 5 0 5 1:6  10 0 g By assembling from ½Ke Š of all members, the structure stiffnessmatrix is obtained:
  25. 25. 2 3 2:0019  107 0 4:8  104 À2  107 0 0 0 0 0 0 6 8:3  106 3  105 0 À3  105 3  105 0 0 0 0 7 6 7 6 5:6  105 0 À3  105 2  105 0 0 0 0 7 6 7 6 3  107 À1  107 7 Structural Analysis—Stiffness Method 6 0 0 0 0 0 7 6 3:375  105 À2:25  105 0 À3:75  104 7:5  104 0 7½K Š ¼ 6 6 7 7 6 6  105 0 À7:5  104 1  105 0 7 6 1:0019  107 0 0 4:8  104 7 6 7 6 Symmetric 8:0375  106 À7:5  104 0 7 6 7 4 2  105 0 5 5 1:6  10 25
  26. 26. 26 Plastic Analysis and Design of Steel StructuresThe load vector is given by 8 9 8 À3 9 0 1:354  10 m 0 À9:236  10À6 m 0 À6:770  10À4 radian 0 1:354  10À3 m = = À100 À1 À1:304  10À3 m fF g ¼ and fDg ¼ ½KŠ fFg ¼ À4 0 À3:713  10 radian 0 1:353  10À3 m 0 À3:264  10À6 m 0 6:733  10À4 radian : ; : À4 ; radian 0 À4:059  10The member forces can be calculated using fPg ¼ ½Ke Š½T Št fDg g. e For member 1, where C ¼ cos 90 ¼ 0 and S ¼ sin 90 ¼ 1, 2 3 EA EA EA EA C S 0 ÀC ÀS 0 6 L L L L 7 6 7 6 7 6 12EI 12EI 6EI 12EI 12EI 6EI 7 6 ÀS C S ÀC 7 8 9 66 L3 L3 L2 L3 L3 2 78 L 7 9 Ni 6 6 7 0 Qi 6 7 6 ÀS 6EI C 6EI 4EI S 6EI ÀC 6EI 2EI 7 7 0 M 6 = L2 L2 L L2 L2 L 7 = i 6 7 0f Pg1 ¼ ¼6 7 À3 Nj 6 6 EA EA EA EA 7 1:354  10 Q 6 ÀC L ÀS 0 C S 0 7 j 6 7 À9:236  10À6 6 L L L 7 : ; 7: ; Mj 6 12EI 12EI 6EI 12EI 12EI 7 À6:770  10À4 6EI 7 6 S ÀC À ÀS C À 2 7 6 L3 L3 L2 L3 L3 L 7 6 6 7 6 4EI 7 4 6EI 6EI 2EI 6EI 6EI 5 ÀS 2 C S ÀC L L2 L L2 L2 L 8 9 73:9 kN À6:5 kN 10:8 kNm = ¼ À73:9 kN 6:5 kN : ; À43:3 kNm Similarly, 8 9 8 9 8 9 6:5 kN 6:5 kN 26:1 kN 73:9 kN À26:1 kN 6:5 kN 43:3 = kNm À104:5 = kNm 0 kNm =fPg2 ¼ ; fPg3 ¼ ; and fPg4 ¼ À6:5 kN À6:5 kN À26:1 kN À73:9 kN 26:1 kN À6:5 kN : ; : ; : ; 104:5 kNm 0 kNm 32:5 kNm
  27. 27. Structural Analysis—Stiffness Method 27 Shear force diagram 73. 6.5 6.5 26. Bending moment diagram 43.3 104. 10.8 32.5FIGURE 1.24. Shear force and bending moment diagrams.The shear force and bending moment diagrams are shown inFigure 1.24.1.11 Treatment of Internal LoadsSo far, the discussion has concerned externally applied loads actingonly at joints of the structure. However, in many instances, externallyapplied loads are also applied at locations other than the joints, suchas on part or whole of a member. Loads being applied in this mannerare termed internal loads. Internal loads may include distributedloads, point loads, and loads due to temperature effects. In such cases,the loads are calculated by treating the member as fixed-end, andfixed-end forces, including axial forces, shear forces, and bendingmoments, are calculated at its ends. The fictitiously fixed ends ofthe member are then removed and the effects of the fixed-end forces,now being treated as applied loads at the joints, are assessed usingthe stiffness method of analysis. In Figure 1.25, fixed-end forces due to the point load and the uni-formly distributed load are collected in a fixed-end force vector fPF gfor the member as
  28. 28. 28 Plastic Analysis and Design of Steel Structures QEi QFi 100 kN QFj QEj MFi MEj 20 kN/m i j MFj MEiFIGURE 1.25. Fixed-end forces. 8 9 0 QFi = MFi f PF g ¼ (1.25) 0 QFj : ; MFjThe signs of the forces in fPF g should follow those shown in Figure1.19. In equilibrium, fixed-end forces generate a set of equivalentforces, equal in magnitude but opposite in sense and shown as QEi ,MEi , QEj , MEj , being applied at the joints pertaining to both ends iand j of the member. The equivalent force vector is expressed as 8 9 0 ÀQ Fi = ÀMFi fP E g ¼ (1.26) 0 ÀQFj : ; ÀMFj If necessary, fPE g is transformed into the global coordinate sys-tem in a similar way given in Equation (1.15) to form È gÉ PE ¼ ½T ŠfPE g (1.27)which is added to the load vector fFg of the structure in accordancewith the freedom codes at the joints. Final member forces are calcu-lated as the sum of the forces obtained from the global structural anal-ysis and fixed-end forces fPF g. That is, fPg ¼ ½Ke Šfdg þ fPF g (1.28) Fixed-end forces for two common loading cases are shown inTable 1.1.Example 1.4 Determine the forces in the members and plot the bend-ing moment and shear force diagrams for the frame shown inFigure 1.26a. The structure is fixed at A and pinned on a roller support

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