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 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. 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 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. 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 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. 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 Â 1Þ containing the externalloads, ½K ŠnÂn is the structure stiffness matrix of size ðn Â nÞcorresponding to the spring constant K in a single degree systemshown in Figure 1.7, and fDgnÂ1 is the displacement vector of sizeðn Â 1Þ containing the unknown displacements at designated loca-tions, usually at the joints of the structure.
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. 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 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. 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 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. 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 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) Â 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 7Â gÃ 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. 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 Â g Ã 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 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:
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. 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 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. 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 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. 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 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 Â 104 0 À4:8 Â 104 À1:92 Â 104 0 À4:8 Â 104 0 6 8 Â 10 6 À8 Â 10 6 70 6 0 0 0 7 6 7 6 1:6 Â 105 4:8 Â 104 0 8 Â 104 7 0½ K e Š1 ¼ 6 g 6 7 4 71 6 1:92 Â 104 0 4:8 Â 10 7 6 7 4 Symmetric 8 Â 106 0 52 5 1:6 Â 10 3 2 1 2 3 4 5 6 3 2 Â 107 0 0 À2 Â 107 0 0 1 6 3 Â 105 3 Â 105 À3 Â 105 3 Â 105 7 2 6 0 7 6 7 g 6 4 Â 105 0 À3 Â 105 2 Â 105 7 3 ½K e Š2 ¼ 6 7 6 2 Â 107 0 0 74 6 7 4 Symmetric 3 Â 105 À3 Â 105 5 5 4 Â 105 6 2 4 5 6 7 8 9 3 1 Â 107 0 0 À1 Â 107 0 0 4 6 3:75 Â 104 7:5 Â 104 À3:75 Â 104 7:5 Â 104 7 5 6 0 7 6 7 g 6 2 Â 105 0 À7:5 Â 104 1 Â 105 7 6½Ke Š3 ¼ 6 7 6 1 Â 107 0 0 77 6 7 4 Symmetric 3:75 Â 104 À7:5 Â 10 458 2 Â 105 9 2 7 8 10 0 0 0 3 1:92 Â 104 0 À4:8 Â 104 À1:92 Â 104 0 4:8 Â 104 7 6 8 Â 106 À8 Â 106 7 8 6 0 0 0 7 6 7 g 6 1:6 Â 105 À4:8 Â 104 0 8 Â 104 7 10½ K e Š4 ¼6 47 0 6 1:92 Â 104 0 À4:8 Â 10 7 6 7 4 Symmetric 8 Â 106 0 5 0 5 1:6 Â 10 0 g By assembling from ½Ke Š of all members, the structure stiffnessmatrix is obtained:
26. 26 Plastic Analysis and Design of Steel StructuresThe load vector is given by 8 9 8 À3 9 0 1:354 Â 10 m 0 À9:236 Â 10À6 m 0 À6:770 Â 10À4 radian 0 1:354 Â 10À3 m = = À100 À1 À1:304 Â 10À3 m fF g ¼ and fDg ¼ ½KŠ fFg ¼ À4 0 À3:713 Â 10 radian 0 1:353 Â 10À3 m 0 À3:264 Â 10À6 m 0 6:733 Â 10À4 radian : ; : À4 ; radian 0 À4:059 Â 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 Â 10 Q 6 ÀC L ÀS 0 C S 0 7 j 6 7 À9:236 Â 10À6 6 L L L 7 : ; 7: ; Mj 6 12EI 12EI 6EI 12EI 12EI 7 À6:770 Â 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. 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 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
29. TABLE 1.1Fixed-end forces ÀbÁ wb þc MFi þ MFj w (load/length) Shear force at end i, QFi QFi ¼ 2 þ L L Shear force at end j, QFj QFj ¼ wb À QFi a b c j w h ii Bending moment at end i, MFi MFi ¼ 2 ðL À aÞ3 ðL þ 3aÞ À c3 ð4L À 3cÞ L 12L w h i Structural Analysis—Stiffness Method Bending moment at end j, MFj MFj ¼ À 2 ðL À cÞ3 ðL þ 3cÞ À a3 ð4L À 3aÞ 12L 2 P b 2a Shear force at end i, QFi QFi ¼ P 1þ L L a b Shear force at end j, QFj QFj ¼ P À QFii j L Bending moment at end i, MFi Pab2 MFj ¼ L2 Pa2 b Bending moment at end j, MFj MFj ¼ 2 L 29
30. 30 Plastic Analysis and Design of Steel Structures 0 2 80 kN 12 kN/m 3 0 0 1 A 1 B 45Њ 0 A B 4m 3m 2 5 C 4 C 3m (a) Frame with applied load (b) FreedomFIGURE 1.26. Example 1.4.at C. For both members AB and BC, E ¼ 2 Â 108 kN/m2, A ¼ 0.2 m2,I ¼ 0.001 m4.Solution. The structure has 5 degrees of freedom with a degree of stat-ical indeterminacy of 2. Freedom codes corresponding to the 5 degreesof freedom are shown in Figure 1.26b. The fixed-end force vector formember 1 is 8 9 0 24 = 16 f PF g ¼ 0 24 : ; À16The equivalent force vector is 8 9 0 0 À24 0 È g É À16 = 0 fPE g ¼ PE ¼ 0 1 À24 2 : ; 16 3which is added to the externally applied force to form 8 9 0 1 À104 2 = fF g ¼ 16 3 0 4 : ; 0 5 The orientations of the members are member 1: a ¼ 0 , member2: a ¼ –45 . The stiffness matrices of the members in the global coor-dinate system are
32. 32 Plastic Analysis and Design of Steel Structures 90.8 155.5 42.8 B B A A 26.3 111.7 C 111.7 C 26.3 (a) Shear force diagram (b) Bending momentFIGURE 1.27. Results of Example 1.4.1.12 Treatment of PinsExample 1.3 demonstrates the analysis of a frame with a pin at joint D.The way to treat the pin using the stiffness method for structural anal-ysis is to allow the members attached to the pinned joint to rotateindependently, thus leading to the creation of different freedom codesfor rotations of individual members. When carrying out elastoplasticanalysis (Chapter 4) for structures using the stiffness method, theplastic hinges, behaving in a way similar to a pin, are formed in stagesas the loads increase. In assigning different freedom codes to representthe creation of plastic hinges in an elastoplastic analysis, the numberof degrees of freedom increases by one every time a plastic hinge isformed. For a structure with a high degree of statical indeterminacy,the increase in the number of freedom codes from the beginningof the elastoplastic analysis to its collapse due to instability inducedby the formation of plastic hinges may be large. Elastoplastic analysisusing this method for simulating pin behavior, hereafter called theextra freedom method, therefore requires increasing both the numberof equilibrium equations to be solved and the size of the structurestiffness matrix ½K Š, thus increasing the storage requirements for thecomputer and decreasing the efficiency of the solution procedure. Inorder to maintain the size of ½KŠ and maximize computational effi-ciency in an elastoplastic analysis, the behavior of a pin at the endsof the member can be simulated implicitly by modifying the memberstiffness matrix ½Ke Š. This latter method for pin behavior simulatedimplicitly in the member stiffness matrix is called the condensationmethod, which is described next.1.12.1 Condensation MethodThe rotational freedom for any member can be expressed explicitlyoutside the domain of the stiffness matrix. In doing so, the rotational
33. Structural Analysis—Stiffness Method 33freedom is regarded as a variable dependent on other displacementquantities and can be eliminated from the member stiffness matrix.The process of elimination is called condensation and hence the nameof this method. In using the condensation method, while the stiffness matrix ofthe member needs to be modified according to its end connection con-dition, the internal loads associated with that member also need to bemodified. There are three cases that need to be considered for a mem-ber. They are (i) pin at end j, (ii) pin at end i, and (iii) pins at both ends.Case i: Pin at end jConsider part of a structure shown in Figure 1.28. The freedom codesfor member 2 with a pin at end j are f1; 2; 3; 4; 5; X g where the rota-tional freedom X is treated as a dependent variable outside the struc-ture equilibrium equation, leaving the member with only 5 freedomcodes pertaining to the structure stiffness matrix ½K Š. Note that therotational freedom code ‘6’ belongs to member 3. From Equation (1.28) for member 2 with internal loads, 2 3 EA EA 6 L 0 0 À 0 0 7 6 L 7 6 7 6 12EI 6EI 12EI 6EI 7 6 0 0 À 78 9 66 L3 L2 L3 L 2 78 7 u 9 8 0 9 Ni 6 7 Q 6 i 6 0 6EI 4EI 6EI 2EI 7 i 6 0 À 7 vi QFi Mi = 6 L2 L L2 L 7 y = M 7 = i Fi ¼6 7 þ Nj 6 EA 6À EA 7 u 0 Qj 6 L 0 0 0 0 7 j 6 L 7 vj QFj : ; 7: ; : ; MjX 6 12EI 6EI 12EI 7 yjX 6EI 7 MFj 6 0 À À À 2 7 6 0 6 L3 L2 L3 L 7 6 7 6 6EI 2EI 6EI 4EI 7 4 0 0 À 5 L2 L L2 L (1.29) 2 3 5 1 i 1 4 2 3 X j 6FIGURE 1.28. Member with a pin at end j.
34. 34 Plastic Analysis and Design of Steel Structureswhere the rotation at end j is yjX corresponding to a rotational freedomcode ‘X’. Expanding the last equation in Equation (1.29) and givenMjX ¼ 0 for a pin, yjX can be derived as 3 1 3 MFj yjX ¼ À vi À y þ i vj À (1.30) 2L 2 2L 4EI=L By substituting Equation (1.30) into the other equations of Equa- Â Ãtion (1.29), a modified 5 Â 5 member stiffness matrix, Kej , and amodified fixed-end force vector, fPFj g, for a member with pin at end jare obtained: 8 9 8 9 Ni ui Qi = Â Ã vi = È É Mi ¼ Kej yi þ PFj (1.31) Nj uj : ; : ; Qj vjwhere 2 3 EA EA 6 L 0 0 À 0 7 6 L 7 6 7 6 3EI 3EI 3EI 7 6 0 0 À 3 7 6 L3 L2 L 7 6 7 6 3EI 7 Â Ã 6 3EI 3EI 7 Kej ¼ 6 0 0 À 2 7 (1.32) 6 L2 L L 7 6 7 6 EA EA 7 6À 0 7 6 0 0 7 6 L L 7 6 7 6 3EI 3EI 3EI 7 4 0 À À 0 5 L3 L2 L3 8 9 0 8 9 Q À 3MFj 0 Fi 0 2L Q Fi = È É MFj = 0 PFj ¼ MFi ¼ MFi À (1.33) 2 0 0 0 :Q ; Fj QFj þ 3MFj : 2L ; Equation (1.33) represents the support reactions equal to those ofa propped cantilever beam. Explicit expressions for the coefficients infPFj g are given in Table 1.2 in Section 1.12.1.4.
35. Structural Analysis—Stiffness Method 35 The member stiffness matrix in the global coordinate systemÂcan Ãbe derived as before using a modified transformation matrix, Tj ,which is given as 2 3 cos a Àsin a 0 0 0 6 7 6 sin a cos a 0 0 0 7 6 7 Â Ã 6 7 6 0 Tj ¼ 6 0 1 0 0 7 (1.34) 7 6 7 6 0 0 0 cos a Àsin a 7 4 5 0 0 0 sin a cos a Accordingly, for a member with a pin at end j, the member stiff-ness matrix in the global coordinate system is h i Â ÃÂ ÃÂ Ã g t Kej ¼ Tj Kej Tj 2 0 1 0 1 0 1 3 6 C2 EA þ S2 3EI SC@ EA 3EI A À 3 ÀS 3EI À@C2 EA 3EI þ S2 3 A ÀSC@ EA 3EI A À 3 7 6 L L3 L L L2 L L L L 7 6 7 6 7 6 7 6 7 6 7 6 0 1 0 17 6 7 6 EA 3EI 3EI EA 3EI A EA 3EI 7 6 2 þ C2 3 ÀSC@ À 3 À @S2 þ C2 3 A 7 6 S C 2 7 6 L L L L L L L 7 6 7 6 7 6 7 6 7 6 7 6 7 6 3EI 3EI 3EI 7 ¼6 S ÀC 2 7 6 L L2 L 7 6 7 6 7 6 7 6 7 6 7 6 0 1 7 6 7 6 EA 3EI EA 3EI A 7 6 6 C2 þ S2 3 SC@ À 3 7 7 6 L L L L 7 6 7 6 7 6 7 6 7 6 7 6 EA 3EI 7 4 Symmetric S2 þ C2 3 5 L L (1.35) The modified fixed-end force vector in the global coordinate sys-tem,fPg g, can be derived in a way similar to Equation (1.27). Ej There are two ways to calculate the member forces. The firstway is to use Equation (1.24), for which the end rotation at end j ofthe member in fDg g is replaced by yjX calculated from Equation e(1.30). The second way is to use a form similar to Equation (1.24): Â ÃÂ Ãt È É fPg ¼ Kej Tj Dg e (1.36a)where, through Equations (1.32) and (1.34),
36. 36 Plastic Analysis and Design of Steel Structures 2 3 EA EA EA EA 6 C L S 0 ÀC ÀS 6 L L L 77 6 7 6 3EI 7 6 3EI 3EI 3EI 3EI 7 6 ÀS 3 C S ÀC 3 7 6 L L3 L2 L3 L 7 6 7 6 7 Â ÃÂ Ãt 6 6 3EI 3EI 3EI 3EI 3EI 7 Kej Tj ¼ 6 ÀS L2 C S ÀC 2 7 6 L2 L L2 L 7 7 (1.36b) 6 7 6 7 6 EA EA EA EA 7 6 ÀC ÀS 0 C S 7 6 L L L L 7 6 7 6 7 6 3EI 3EI 3EI 3EI 3EI 7 4 S ÀC À ÀS C 3 5 L3 L3 L2 L3 L The 5 Â 1 member displacement vector fDg g in Equation (1.36a) eis extracted from fDg according to the 5 freedom codes f1; 2; 3; 4; 5gshown in Figure 1.28 for the member.Case ii: Pin at end iThis case is shown in Figure 1.29 where member 2 has a pin at end iwith an independent rotational freedom code Y. The freedom codesfor member 2 with a pin at end i are f1; 2; Y; 4; 5; 6g. Note that the free-dom code 3 belongs to member 1. By writing 8 9 8 9 Ni ui Qi vi = = MiY yiY fP g ¼ ; f dg ¼ Nj uj Qj vj : ; : ; Mj yjand given MjY ¼ 0 for a pin at end i, yiY can be derived as 3 3 1 MFi yiY ¼ À vi þ vj À yj À (1.37) 2L 2L 2 4EI=L 2 3 1 Y 5 i 6 1 2 4 3 jFIGURE 1.29. Member with a pin at end i.
37. Structural Analysis—Stiffness Method 37 The corresponding matrices for this case can be derived in a waysimilar to Case i mentioned earlier. The results are 8 9 8 9 Ni ui Qi = vi = Nj ¼ ½Kei Š uj þ ðPFi Þ (1.38) Q j v j : ; : ; Mj yjwhere 2 3 EA EA 6 L 0 À 0 0 7 6 L 7 6 7 6 3EI 3EI 3EI 7 6 0 0 À 7 6 L3 L3 L2 7 6 7 6 7 6 EA 7 6 EA 7 ½Kei Š ¼ 6 À L 6 0 L 0 0 7 7 (1.39) 6 7 6 7 6 3EI 3EI 3EI 7 6 0 À 0 À 2 7 6 L3 L3 L 7 6 7 6 7 6 3EI 3EI 3EI 7 4 0 0 À 5 L2 L2 L 8 9 0 8 9 3MFi Q À 0 Fi 2L 00 Q Fi = 0 = fPFi g ¼ 0 ¼ (1.40) 00 3MFi Q QFj þ Fj 2L 00 : ; MFj MFi MFj À : 2 ; Equation (1.40) represents the support reactions equal to those ofa propped cantilever beam. Explicit expressions for the coefficients infPFi g are given in Table 1.2 in Section 1.12.1.4. 2 3 cos a Àsin a 0 0 0 6 sin a cos a 0 0 07 6 7 ½Ti Š ¼ 6 0 6 0 cos a Àsin a 077 (1.41) 4 0 0 sin a cos a 05 0 0 0 0 1
38. 38 Plastic Analysis and Design of Steel Structures Accordingly, for the member with pin at end i, the member stiff-ness matrix in the global coordinate system is 2 0 1 0 1 0 1 3 6 C2 EA þ S2 3EI SC@ EA 3EIA À 3 À@C2 EA 3EI þ S2 3 A ÀSC @EA À 3EIA 3EI 7 ÀS 2 7 6 L3 L3 6 L L L L L L L 7 6 0 1 0 1 7 6 7 6 3EI 7 6 EA 3EI @EA À 3EIA @S2 EA þ C2 3EIA 7 6 S2 þ C2 3 ÀSC À C 2 7 6 L L L L3 L L3 L 7 6 0 1 7Â gÃ 6 6 7 7 Kei ¼ 6 EA 3EI EA 3EIA 3EI 7 6 6 C2 þ S2 3 SC@ À 3 S 2 7 6 L L L L L 7 7 6 7 6 EA 3EI 3EI 7 6 S2 þ C2 3 ÀC 2 7 6 L L L 7 6 7 6 7 6 3EI 7 4 Symmetric 5 L (1.42)and 2 3 EA EA EA EA 6 C L S ÀC ÀS 0 7 6 L L L 7 6 7 6 3EI 3EI 3EI 3EI 3EI 7 6 ÀS 3 C S ÀC 7 6 L L3 L3 L3 L2 7 6 7 6 7 6 EA EA EA EA 7 ½Kei Š½Ti Št ¼ 6 ÀC L 6 ÀS L C L S L 0 7 7 (1.43) 6 7 6 3EI 3EI 3EI 3EI 3EI 7 6 S ÀC ÀS À 2 7 6 C 7 6 L3 L3 L3 L3 L 7 6 7 6 3EI 7 4 ÀS 3EI C 3EI S 3EI ÀC 3EI 5 L2 L2 L2 L2 L The 5 Â 1 member displacement vector fDg g is extracted from efDg according to the 5 freedom codes f1; 2; 4; 5; 6g for the member.Case iii: Pins at both ends i and jThis case is shown in Figure 1.30 where member 2 has a pin at bothends i and j. The freedom codes for member 2 are f1; 2; Y; 4; 5; X g.Note that the freedom codes 3 and 6 belong to members 1 and 3,respectively. 2 3 1 Y 5 1 i 3 X 4 2 j 6FIGURE 1.30. Member with a pin at end i.
39. Structural Analysis—Stiffness Method 39 In this case, substitute MiY ¼ MjX ¼ 0 into Equation (1.29), weobtain vj À vi MFj À 2MFi yiY ¼ þ (1.44a) L 6EI=L vj À vi MFi À 2MFj yjX ¼ þ (1.44b) L 6EI=Land 8 9 8 9 Ni ui Q Â = i Ã vi = È É ¼ Keij þ PFij (1.45) Nj uj : ; : ; Qj vjwhere 2 3 1 0 À1 0 EA 6 0 0 0 0 7 6 7 f Keij g ¼ (1.46) L 4 À1 0 1 0 5 0 0 0 0 8 9 8 9 À 0 Á 0 MFi þ MFj 000 QFi À È É Q = L = Fi PFij ¼ ¼ (1.47) 0 À 0 Á 000 : ; Q þ MFi þ MFj QFj Fj : L ; Equation (1.47) represents the support reactions equal to those ofa simply supported beam. Explicit expressions for the coefficients inÈ É PFij are given in Table 1.2 in Section 1.12.1.4. Â Ã It is noted that Keij is in fact the stiffness matrix of a trussmember. The transformation matrix for the member in this case is 2 3 C ÀS 0 0 Â Ã 6S C 0 0 7 Tij ¼ 6 40 7 (1.48) 0 C ÀS 5 0 0 S C The corresponding stiffness matrix in the global coordinate sys-tem for a member with pins at both ends is 2 2 3 C CS ÀC2 ÀCS h i EA 6 S2 ÀCS ÀS2 7 g 6 7 Keij ¼ 6 7 (1.49) L 4 C 2 CS 5 Symmetric S2
40. 40 Plastic Analysis and Design of Steel Structuresand 2 3 C S ÀC ÀS h i Ãt EA 6 0 g Â 6 0 0 0 77 Keij Tij ¼ (1.50) L 4 ÀC ÀS C S 5 0 0 0 0Modified Fixed-End Force VectorThe explicit expressions for the coefficients of the modified fixed-endforce vectors given in Equations (1.33), (1.40), and (1.47) are summar-ized in Table 1.2.TABLE 1.2Modified fixed-end forces for members with pins À Á 0 0 wb b þ c M w (load/length) QFi ¼ 2 þ Fi j L L 0 0 i QFj ¼ wb À QFi a b c 0 w h i MFi ¼ 2 ðb þ 2cÞb 2L2 À c2 À ðb þ cÞ2 L 8L 00 00 QFi ¼ P À QFj P 00 Pa2 i QFj ¼ ðb þ 2LÞ j 2L3 À 2 Á a b 00 Pb L À b2 MFi ¼ L 2L2 Pb 3b b3 Moment ¼ 2À þ 3 2 L L under load w (load/length) 000 wb bi QFi ¼ þc L 2 a b c 000 wb b QFj ¼ þa L L 2 2 000 x À a2 Q Mmax ¼w at x ¼ a þ Fi 2 w P 000 Pb QFi ¼i j L 000 Pa a b QFj ¼ L L Pab Moment ¼ L under load
41. Structural Analysis—Stiffness Method 41Procedure for Using Condensation Method 1. For any joints with pins, determine whether the connecting members have (a) no pin, (b) pin at end i, (c) pin at end j, or (d) pins at both ends. 2. Use the appropriate stiffness matrix for the cases just given for all members. 3. Assign freedom codes to each joint. 4. Assemble the structure stiffness matrix ½K Š. 5. After solving the structure equilibrium equation, calculate the angle of rotation yjX or yiY for each pin using Equations (1.30), (1.37), or (1.44). Calculate the member forces accordingly.1.12.2 Methods to Model PinThere are a number of ways to model a joint with a pin using the for-mulations given in the previous section. Consider a pinned joint con-necting two members 1 and 2. There are four ways of formulation foruse in the stiffness method of analysis as shown in Figure 1.31.Figure 1.31a is based on the extra freedom method where both mem-bers 1 and 2 have independent rotations D3 and D4 using the full6 Â 6 member stiffness matrix. Figure 1.31b is based on the condensa-tion method for member 1 using the formulation for pin at end j asgiven in Section 1.12.1.1, whereas member 2 retains use of the full6 Â 6 member stiffness matrix. Figure 1.31c is also based on the con-densation method for member 2 using the formulation for pin at end i 2 2 1 X 1 3 4 3 1 1 2 2 (a) (b) 2 2 X 1 3 1 Y Y 1 1 2 2 (c) (d)FIGURE 1.31. Modeling pin at a joint.
42. 42 Plastic Analysis and Design of Steel Structuresas given in Section 1.12.1.2, whereas member 1 retains use of the full6 Â 6 member stiffness matrix. Figure 1.31d is based on the condensa-tion method using the formulation for pin at end j for member 1 andpin at end i for member 2.Example 1.5 Determine the displacements and forces in the beamABC with a pin at B shown in Figure 1.32. Ignore the effect of axialforce. E ¼ 2000 kN/m2, I ¼ 0.015 m4. 5 kN 1 0 0 0 0 0 0 0 A B C A C 4m 2m 2 B 3 1 2 (a) Beam with pin at B (b) Freedom codes−Extra Freedom MethodFIGURE 1.32. Example 1.5.Solution(i) Extra Freedom Method When the axial force effect is ignored, a zero freedom code isassigned to the axial deformation of the members. Thus, the structurehas a total of 3 degrees of freedom shown in Figure 1.32b. For all matrices, only the coefficients corresponding to nonzerofreedom codes will be shown. For member 1, 2 0 0 0 0 1 2 3 :: :: :: :: :: :: 0 6 :: :: :: :: :: 7 0 6 7 6 :: :: :: :: 7 0 ½Ke Š1 ¼ 6 g 6 7 6 :: :: :: 7 0 7 4 Symmetric 5:625 À11:25 5 1 30 2 For member 2, 20 1 3 0 0 03 :: :: :: :: :: :: 0 6 45 45 :: :: :: 7 1 6 7 6 60 :: :: :: 7 3 ½Ke Š2 ¼ 6 g 6 7 6 :: :: :: 7 0 7 4 Symmetric :: :: 5 0 :: 0
43. Structural Analysis—Stiffness Method 43Hence, 2 3 50:625 À11:25 45 ½K Š ¼ 4 À11:25 30 0 5 45 0 60 8 9 À5 = fF g ¼ 0 : ; 0 By solving the structure equilibrium equation fFg ¼ ½K ŠfDg forfDg, we obtain 8 9 8 9 D1 = À0:395 = fDg ¼ D2 ¼ À0:148 : ; : ; D3 0:296 Figure 1.33 shows the deflection and rotations of the members. A 0.395 m C 0.148 B 0.296FIGURE 1.33. Deflection and rotations. The member forces for member 1 are 2 38 9 8 9 :: :: :: :: 0 0 0 0 6 :: :: :: :: À5:625 11:25 7 0:556 6 7 0 6 :: :: :: :: À11:25 7 = = 15 7 0 2:223 f Pg1 ¼ 6 6 :: :: :: :: 7 ¼ 6 0 0 7 0 0 4 :: :: :: :: 5:625 À11:25 5 À0:395 À0:556 : ; : ; :: :: :: :: À11:25 30 À0:148 0 The member forces for member 2 are 2 38 9 8 9 :: 0 0 :: :: :: 0 0 6 :: 45 6 45 :: :: :: 7 À0:395 À4:444 7 6 :: 45 7 0:296 = = 60 :: :: :: 7 0 f Pg2 ¼ 6 6 :: ¼ 6 0 0 :: :: :: 7 7 0 0 4:444 4 :: À45 À45 :: :: :: 5 0 : ; : ; :: 45 30 :: :: :: 0 À8:891 The member forces for the structure are shown in Figure 1.34. 2.223 kNm 8.891 kNm 0.556 kN 4.444 kN 0.556 kN 4.444 kNFIGURE 1.34. Member forces.
44. 44 Plastic Analysis and Design of Steel Structures(ii) Method of Condensation In using this method, the stiffness matrix of member 1 iscondensed so that rotation at end j, denoted as X in Figure 1.35,becomes a dependent variable. The freedom codes of the structureare also shown in Figure 1.35. 1 0 0 0 0 0 0 0 A B C 1 X 2 2FIGURE 1.35. Freedom codes: method of condensation. The stiffness matrix of member 1 is given by Equation (1.32) as 20 0 0 0 1 3 :: :: :: :: :: 0 h i 6 :: :: :: :: 7 0 Â Ã 6 7 Kej 1 ¼ Kej ¼ 6 g 6 :: :: :: 7 0 7 1 4 Symmetric :: 5 0 1:4063 1 For member 2, 20 1 2 0 0 03 :: :: :: :: :: :: 0 6 45 45 :: :: :: 7 1 6 7 6 60 :: :: :: 7 2 ½Ke Š2 ¼ 6 g 6 7 6 :: :: :: 7 0 7 4 Symmetric :: :: 5 0 :: 0 Hence, the structure stiffness matrix, of size 2 Â 2, can be assem-bled as ! 46:406 45 ½K Š ¼ 45 60The load vector is À5 fF g ¼ 0By solving fFg ¼ ½K ŠfDg for fDg, we obtain D1 À0:395 fDg ¼ ¼ D2 0:296
45. Structural Analysis—Stiffness Method 45The rotation yjX for member 1 can be obtained from Equation (1.30) as 3 1 3 3 yjX ¼ À v i À yi þ vj ¼ ðÀ0:395Þ ¼ À0:148 2L 2 2L 2Â4The member forces for member 2 can be calculated using Equation(1.36a) Â ÃÂ Ãt fPg1 ¼ Kej Tj fDg g e 2 38 9 8 9 :: :: :: :: 0 0 0 6 :: :: :: :: À1:406 7 0 0:556 = = 6 7 ¼ 6 :: :: :: :: À5:625 7 6 7 0 ¼ 2:223 4 :: :: :: :: 5 0 0 0 : ; : ; : :: :: :: 1:406 À0:395 À0:556which are the same as those calculated before.1.13 Temperature EffectsMost materials expand when subject to temperature rise. For a steelmember in a structure, the expansion due to temperature rise isrestrained by the other members connected to it. The restraint imposedon the heated member generates internal member forces exerted on thestructure. For uniform temperature rise in a member, the internal mem-ber forces are axial and compressive, and their effects can be treated inthe same way as for internal loads described in Section 1.11.1.13.1 Uniform TemperatureThe fixed-end force vector fPF g for a steel member shown inFigure 1.36 subject to a temperature rise of ðT À To Þ, where T isthe current temperature and To is the ambient temperature of themember, is given by 8 9 NFi 0 = 0 f PF g ¼ (1.51) NFj 0 : ; 0 NEi NFi NFj NEj i jFIGURE 1.36. Fixed-end forces for member subject to temperature rise.
46. 46 Plastic Analysis and Design of Steel Structureswhere NFi ¼ ÀNFj ¼ ET AaðT À To Þ (1.52) ET ¼ modulus of elasticity at temperature T, A ¼ cross-sectional area, a ¼ coefficient of linear expansion.As before, the equivalent force vector is fPE g ¼ ÀfPF g. In Equation (1.52), ET is often treated as a constant for low tem-perature rise. However, under extreme loading conditions, suchas steel in a fire, the value of ET deteriorates significantly over arange of temperatures. The deterioration rate of steel at elevatedtemperature is often expressed as a ratio of ET =Eo . This ratio hasmany forms according to the design codes adopted by differentcountries. In Australia and America, the ratio of ET =Eo is usuallyexpressed as ET T ¼ 1:0 þ 2 3 for 0 C T 600 C Eo T 5 2000 ln4 1100 0 1 (1.53a) T A 690@1 À 1000 ¼ for 600 C T 1000 C T À 53:5 In Europe, the ratio of ET =Eo , given in tabulated form in the Euro-code, can be approximated as ET T ¼ 1 À eÀ9:7265Â0:9947 (1.53b) Eo Although the coefficient of linear expansion a also varies withtemperature for steel, its variation is insignificantly small. There-fore, a constant value is usually adopted. The overall effect of risingtemperature and deteriorating stiffness for a steel member is that thefixed-end compressive force increases initially up to a peak at about500 C, beyond which the compressive force starts to decrease. Thevariation of the fixed-end compressive force, expressed as a dimen-sionless ratio relative to its value at 100 C using a varyingmodulus of elasticity according to Equation (1.53a), is shown inFigure 1.37. For comparison purpose, the variation of the fixed-endcompressive force using a constant value of ET is also shown inFigure 1.37.
47. Structural Analysis—Stiffness Method 47 Axial force /(Axial force at 100؇C) 12.0 10.0 8.0 Constant modulus of elasticity 6.0 4.0 Varying modulus of elasticity 2.0 0.0 0 200 400 600 800 1000 T (؇C)FIGURE 1.37. Variation of fixed-end compressive force with temperature.1.13.2 Temperature GradientFor a member subject to a linearly varying temperature across its crosssection with Tt ¼ temperature at the top of the cross section and Tb ¼temperature at the bottom of the cross section, the fixed-end forcevector is given by 8 9 NFi 0 = MFi f PF g ¼ (1.54) NFj 0 : ; MFjwhere Z NFi ¼ ÀNFj ¼ sdA (1.55a) A s ¼ ET a ð T À T o Þ (1.55b) In Equation (1.55a), the integration is carried out for the wholecross section of area A. The stress s at a point in the cross section cor-responds to a temperature T at that point. In practice, integration isapproximated by dividing the cross section into a number of horizon-tal strips, each of which is assumed to have a uniform temperature.
48. 48 Plastic Analysis and Design of Steel Structures Consider a member of length L with a linearly varying tempera-ture in its cross section subject to an axial force N. If the crosssection of the member is divided into n strips and the force instrip i with cross-sectional area Ai and modulus of elasticity Ei is Ni,then, for compatibility with a common axial deformation u forall strips, N1 L Ni L Nn L u¼ ¼ ::: ¼ ¼ ::: ¼ (1.56) E1 A 1 Ei A i E n AnFor equilibrium, N ¼ N1 þ ::: þ Ni þ ::: þ Nn (1.57) Substituting Equation (1.56) into Equation (1.57), we obtain P n Ei Ai 1 N¼ u (1.58) L By comparing Equation (1.58) with Equation (1.9), it can be seenthat for a member with a linearly varying temperature across its crosssection, P n Ei Ai 1 K11 ¼ ÀK14 ¼ (1.59) L Equation (1.59) can be rewritten as P n Eo mi Ai 1 K11 ¼ ÀK14 ¼ (1.60) Lin which Ei mi ¼ (1.61) Eo The value of mi can be obtained from Equations (1.53a) or (1.53b).The use of Equation (1.60) is based on the transformed sectionmethod, whereby the width of each strip in the cross section isadjusted by multiplying the original width by mi and the total areais calculated according to the transformed section. The stiffness coefficients for bending involving EI can also beobtained using the transformed section method. The curvature ofthe member as a result of bowing due to the temperature gradientacross the depth of the cross section is given as ðTt À Tb Þ k¼a (1.62) d
49. Structural Analysis—Stiffness Method 49 MEi MFi MFj MEj Tt d i j TbFIGURE 1.38. Fixed-end moments under temperature gradient.Hence, the fixed-end moments at the ends of the member, as shownin Figure 1.38, are ðTt À Tb Þ MFi ¼ ÀMFj ¼ ÀEa I a (1.63) dwhere d is depth of cross section. Similar to the calculation of the axial stiffness coefficient inEquation (1.60), EaI in Equation (1.63) is calculated numerically bydividing the cross section into a number of horizontal strips, each ofwhich is assumed to have a uniform temperature. The width of eachstrip in the cross section is adjusted by multiplying the original widthby mi so that X n X n Ea I ¼ Ei I i ¼ Eo mi Ii (1.64) 1 1where Ii is calculated about the centroid of the transformed section.Example 1.6 Determine the axial stiffness EA and bending stiffness EIfor the I section shown in Figure 1.39. The section is subject to a line-arly varying temperature of 240 C at the top and 600 C at the bottom.Use the European curve [Equation (1.53b)] for the deterioration rate of tf tw d BFIGURE 1.39. Example 1.6.
50. 50 Plastic Analysis and Design of Steel Structuresthe modulus of elasticity. Eo at ambient temperature ¼ 210,000 MPa.A ¼ 7135 mm2, I ¼ 158202611 mm4, B ¼ 172.1 mm, d ¼ 358.6 mm,tw ¼ 8 mm, tf ¼ 13 mm.Solution. The section is divided into 24 strips, 4 in each of the flangesand 16 in the web. The temperature at each strip is taken as the tem-perature at its centroid. The area of each strip is transformed by multi-plying its width by ET/Eo.The total area of the transformed section ¼ 4549.9 mm2.The centroid of the transformed section from the bottom edge ¼ 239.0 mm.Total EA for the section ¼ 210000 Â 4549.9 ¼ 9.555 Â 108 N.The second moment of area of the transformed section ¼ 8.422 Â 107 mm4.Total EI for the section ¼ 210000 Â 8.422 Â 107 ¼ 1.769 Â 1013 N mm2.Problems1.1. Determine the degree of indeterminacy for the beam shown in Figure P1.1. FIGURE P1.1. Problem 1.1.1.2. Determine the degree of indeterminacy for the beam shown in Figure P1.2. FIGURE P1.2. Problem 1.2.1.3. Determine the degree of indeterminacy for the continuous beam shown in Figure P1.3.
51. Structural Analysis—Stiffness Method 51 FIGURE P1.3. Problem 1.3.1.4. Determine the degree of indeterminacy for the frame shown in Figure P1.4. FIGURE P1.4. Problem 1.4.1.5. The structure ABC shown in Figure P1.5 is subject to a clockwise moment of 5 kNm applied at B. Determine the angles of rotation at A and B using 1. Extra freedom method 2. Condensation method Ignore axial force effect. EI ¼ 30 kNm2. 5 kNm A B 3m 3m C FIGURE P1.5. Problem 1.5.1.6. The structure shown in Figure P1.6 is fixed at A and C and pinned at B and subject to an inclined force of 300 kN. Determine the forces in the structure and plot the bending moment and shear force diagrams. E ¼ 2 Â 108 kN/m2, I ¼ 1.5 Â 10À5 m4, A ¼ 0.002 m2.
52. 52 Plastic Analysis and Design of Steel Structures 300 kN C 45Њ 2m 45Њ B 4m A FIGURE P1.6. Problem 1.6.1.7. The frame ABC shown in Figure P1.7 is pinned at A and fixed to a roller at C. A bending moment of 100 kNm is applied at B. Plot the bending moment and shear force diagrams for the frame. Ignore the effect of axial force in the members. E ¼ 210000 kN/m2, I ¼ 0.001 m4. 100 kNm B C 5m 4m A FIGURE P1.7. Problem 1.7.1.8. A beam ABC shown in Figure P1.8 is pinned at A and fixed at C. A vertical force of 5 kN is applied at B. Determine the displace- ments of the structure and plot the bending moment and shear force diagrams. Ignore axial force effect. E ¼ 2000 kN/m2, I ¼ 0.015 m4. 5 kN A B C 4m 2m FIGURE P1.8. Example 1.8
53. Structural Analysis—Stiffness Method 531.9. Use the stiffness method to calculate the member forces in the structure shown in Figure P1.9. E ¼ 2 Â 105 N/mm2, A ¼ 6000 mm2, I ¼ 2 Â 107 mm4. 5m 10 kN/m C 30Њ A B 5m FIGURE P1.9. Problem 1.9.1.10. Plot the shear force and bending moment diagrams for the con- tinuous beam shown in Figure P1.10. Ignore axial force effect. E ¼ 3 Â 105 N/mm2, I ¼ 2 Â 107 mm4. 60 kN 10 kN/m A B 5m 5m 10m FIGURE P1.10. Problem 1.10.1.11. Determine the axial stiffness EA and bending stiffness EI for the I section shown in Figure 1.39. The section is subject to a line- arly varying temperature of 150 C at the top and 400 C at the bottom. Use the European curve [Equation (1.53b)] for the deterioration rate of the modulus of elasticity. Eo at ambient temperature ¼ 210000 MPa. A ¼ 7135 mm2, I ¼ 158202611 mm4. B ¼ 172.1 mm, d = 358.6 mm, tw ¼ 8 mm, tf ¼ 13 mm.Bibliography1. Wang, C. K. (1963). General computer program for limit analysis. Proceed- ings ASCE, 89(ST6).2. Jennings, A., and Majid, K. (1965). An elastic-plastic analysis by computer for framed structures loaded up to collapse. The Structural Engineer, 43(12).3. Davies, J. M. (1967). Collapse and shakedown loads of plane frames. Proc. ASCE, J. St. Div., ST3, pp. 35–50.
54. 54 Plastic Analysis and Design of Steel Structures4. Chen, W. F., and Sohal, I. (1995). Plastic design and second-order analysis of steel frames. New York: Springer-Verlag.5. Samuelsson, A., and Zienkiewicz, O. C. (2006). History of the stiffness method. Int. J. Num. Methods in Eng., 67, pp. 149–157.6. Rangasami, K. S., and Mallick, S. K. (1966). Degrees of freedom of plane and space frames. The Structural Engineer, 44(3), pp. 109–111.