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Plastic Analysis and Design of Steel Structures This page intentionally left blank Plastic Analysis and Design of Steel Structures M. Bill Wong Department of Civil Engineering Monash University, Australia AMSTERDAM BOSTON HEIDELBERG LONDON NEW YORK OXFORD PARIS SAN DIEGO SAN FRANCISCO SINGAPORE SYDNEY TOKYO Butterworth-Heinemann is an imprint of Elsevier Butterworth-Heinemann is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA Linacre House, Jordan Hill, Oxford OX2 8DP, UK Copyright 2009, Elsevier Ltd. All rights reserved. No part of this publication may be reproduced, stored in a retri system, or transmitted in any or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Permissions may be sought directly from Elsevier’s Science num- ber of pinned joints p 4.5. Hence, fr 3 3 2 45 0 and the truss is a determinate structure. For the truss in Figure 1.5b, number of members n 2; number of pinned joints p 3. Hence, fr 3 2 2 3 0 and the truss is a determinate structure. Example 1.2 Determine the degree of statical indeterminacy for the frame with mixed pin and rigid joints shown in Figure 1.6. Solution. For this frame, a member is counted as one between two adjacent joints. Number of members 6; number of rigid or fixed joints 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 of pinned joints 3. Hence, fr 3 6 3 5 2 3 3 and the frame is an inde- terminate structure to the degree 3. 1.3 Statically Indeterminate StructuresDirect Stiffness The spring system shown in Figure 1.7 demonstrates the use of the stiffness in its simplest . The single degree of freedom structure consists of an object supported by a linear spring obeying Hooke’s law. For structural analysis, the weight, F, of the object and the spring constant or stiffness, K, are usually known. The purpose A B C E D F FIGURE 1.6. DeterminationofdegreeofstaticalindeterminacyinExample1.2. 6Plastic Analysis and Design of Steel Structures of the structural analysis is to find the vertical displacement, D, and the internal force in the spring, P. From Hooke’s law, F KD1.3 Equation 1.3 is in fact the equilibrium equation of the system. Hence, the displacement, D, of the object can be obtained by D FK1.4 The displacement, d, of the spring is obviously equal to D. That is, d D1.5 The internal force in the spring, P, can be found by P Kd1.6 In this simple example, the procedure for using the stiffness is demonstrated through Equations 1.3 to 1.6. For a struc- ture composed of a number of structural members with n degrees of freedom, the equilibrium of the structure can be described by a num- ber of equations analogous to Equation 1.3. These equations can be expressed in matrix as Ff gn1 K nnDfgn11.7 whereFf gn1is the load vector of size n 1 containing the external loads,K nnisthestructurestiffnessmatrixofsizen n corresponding to the spring constant K in a single degree system shown in Figure 1.7, andDfgn1is the displacement vector of size n 1 containing the unknown displacements at designated loca- tions, usually at the joints of the structure. K F D FIGURE 1.7. Load supported by linear spring. Structural AnalysisStiffness 7 The unknown displacement vector can be found by solving Equation 1.7 as Dfg K 1Ff g1.8 Details of the ation of Ff g, K , and Dfg are given in the following sections. 1.3.1 Local and Global Coordinate Systems A 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 the corresponding displacements are shown in Figure 1.8, where the axial forces are acting along the longitudinal axis of the member and the shear forces are acting perpendicular to its longitudinal axis. In Figure 1.8, Mi,j, yi,j bending moments and corresponding rotations at ends i, j, respectively; Ni,j, ui,jare axial forces and corresponding axial deations at ends i, j, respectively; and Qi,j, vi,jare shear forces and corresponding transverse displacements at ends i, j, respectively. The directions of the actions and movements shown in Figure 1.8 are positive when using the stiffness . As mentioned in Section 1.2, the freedom codes of a structure are assigned in its global coordinate system. An example of a member ing 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 is assigned a freedom code “zero,” otherwise a nonzero freedom code is assigned. The relationship for forces and displacements between local and global coordinate systems will be established in later sections. i j Mj, j Nj, uj Qj, vj Mi, iNi, ui Qi, vi FIGURE 1.8. Local coordinate system for member forces and displacements. 8Plastic Analysis and Design of Steel Structures 1.4 Member Stiffness Matrix The structure stiffness matrix K is assembled on the basis of the equilibrium and compatibility conditions between the members. For a general frame, the equilibrium matrix equation of a member is Pf g Ke df g1.9 wherePf g is the member force vector, Ke is the member stiffness matrix, anddf g 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 Pf g Ni Qi Mi Nj Qj Mj 8 9 ; ; Ke K1100K1400 0K22K230K25K26 0K32K330K35K36 K4100K4400 0K52K530K55K56 0K62K630K65K66 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 ; df g ui vi yi uj vj yj 8 9 ; where the elements ofPf g anddf g are shown in Figure 1.8. 1.4.1 Derivation of Elements of Member Stiffness Matrix A member under axial forces Niand Njacting at its ends produces axial displacements uiand ujas shown in Figure 1.10. From the stress-strain relation, it can be shown that Ni EA L uiuj 1.10a Nj EA L ujui 1.10b j i 1 2 3 4 5 6 FIGURE 1.9. Freedom codes of a member in a global coordinate system. Structural AnalysisStiffness 9 where E is Young’s modulus, A is cross-sectional area, and L is length of the member. Hence, K11 K14 K41 K44 EA L . For a member with shear forces Qi, Qjand bending moments Mi, Mjacting at its ends as shown in Figure 1.11, the end displace- ments and rotations are related to the bending moments by the slope-deflection equations as Mi 2EI L 2yiyj 3 vjvi L 1.11a Mj 2EI L 2yjyi 3 vjvi L 1.11b Hence, K62 K65 6EI L2 , K63 2EI L , and K66 4EI L . By taking the moment about end j of the member in Figure 1.11, we obtain Qi MiMj L 2EI L2 3yi3yj 6 vjvi L 1.12a i j Ni ui uj Nj Original position Displaced position FIGURE 1.10. Member under axial forces. i vi vj Qj Original position Displaced position j Qi Mi Mj i j FIGURE 1.11. Member under shear forces and bending moments. 10Plastic Analysis and Design of Steel Structures Also, by taking the moment about end i of the member, we obtain Qj MiMj L Qi1.12b Hence, K22 K55 K25 K52 12EI L3 and K23 K26 K53 K66 6EI L2 . In summary, the resulting member stiffness matrix is symmetric about the diagonal Ke EA L 00 EA L 00 0 12EI L3 6EI L2 0 12EI L3 6EI L2 0 6EI L2 4EI L 0 6EI L2 2EI L EA L 00 EA L 00 0 12EI L3 6EI L2 0 12EI L3 6EI L2 0 6EI L2 2EI L 0 6EI L2 4EI L 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 1.13 1.5 Coordinates Transation In order to establish the equilibrium conditions between the member forces in the local coordinate system and the externally applied loads in the global coordinate system, the member forces are transed into the global coordinate system by force resolution. Figure 1.12 shows a member inclined at an angle a to the horizontal. 1.5.1 Load Transation The forces in the global coordinate system shown with superscript “g” in Figure 1.12 are related to those in the local coordinate system by Hg i Nicosa Qisina1.14a Vg i Nisina Qicosa1.14b Mg i Mi1.14c Structural AnalysisStiffness 11 Similarly, Hg j Njcosa Qjsina1.14d Vg j Njsina Qjcosa1.14e Mg j Mj1.14f In matrix , Equations 1.14a to 1.14f can be expressed as Fg e T Pf g1.15 where Fgefg is the member force vector in the global coordinate system and T is the transation matrix, both given as Fg e Hg i Vg i Mg i Hg j Vg j Mg j 8 9 ; and T cosasina0000 sinacosa0000 001000 000cosasina0 000sinacosa0 000001 2 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 5 1.5.2 Displacement Transation The displacements in the global coordinate system can be related to those in the local coordinate system by following the procedure simi- lar to the force transation. The displacements in both coordinate systems are shown in Figure 1.13. From Figure 1.13, ui ug i cosa vg i sina1.16a i Mi Ni Qi Mj Nj Qj Mg i Vg i Hg i j Mg j Vg j Hg j FIGURE 1.12. Forces in the local and global coordinate systems. 12Plastic Analysis and Design of Steel Structures vi ug i sina vg i cosa1.16b yi yg i 1.16c uj ug j cosa vg j sina1.16d vj ug j sina vg j cosa1.16e yj yg j 1.16f In matrix , Equations 1.16a to 1.16f can be expressed as df g T tDg e 1.17 whereDgefg is the member displacement vector in the global coordi- nate system corresponding to the directions in which the freedom codes are specified and is given as Dg e ug i vg i yg i ug j vg j yg j 8 9 ; and T tis the transpose of T . 1.6 Member Stiffness Matrix in Global Coordinate System From Equation 1.15, Fg e T Pf g T Ke df gfrom Equation 19 uj vj vg i ug i i j ui vi vg j ug j FIGURE 1.13. Displacements in the local and global coordinate systems. Structural AnalysisStiffness 13 T Ke T tDg e from Equation 117 Kg e Dg e 1.18 where Kge T Ke T t member stiffness matrix in the global coor- dinate system. An explicit expression for Kge is Kg e C2 EA L S2 12EI L3 SC EA L 12EI L3 0 1 A S6EI L2 C2 EA L S2 12EI L3 0 1 A SC EA L 12EI L3 0 1 A S6EI L2 S2 EA L C2 12EI L3 C6EI L2 SC EA L 12EI L3 0 1 A S2 EA L C2 12EI L3 0 1 A C6EI L2 4EI L S6EI L2 C6EI L2 2EI L C2 EA L S2 12EI L3 SC EA L 12EI L3 0 1 A S6EI L2 SymmetricS2 EA L C2 12EI L3 C6EI L2 4EI L 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5 1.19 where C cos a; S sin a. 1.7 Assembly of Structure Stiffness Matrix Consider part of a structure with four externally applied forces, F1, F2, F4, and F5, and two applied moments, M3 and M6, acting at the two joints p and q connecting three members A, B, and C as shown in Figure 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 of two conditions compatibility and equilibrium conditions at the joints. 1.7.1 Compatibility Condition At joint p, the global displacements are D1 horizontal, D2 vertical, and D3 rotational. Similarly, at joint q, the global displacements are D4 horizontal, D5 vertical, and D6 rotational. The compatibility condition is that the displacements D1, D2, and D3 at end p of mem- ber A are the same as those at end p of member B. Thus, ug jA ug i B D1, vg jA vg i B D2, and yg jA yg iB D3. The same condition applies to displacements D4, D5, and D6 at end q of both members B and C. 14Plastic Analysis and Design of Steel Structures The member stiffness matrix in the global coordinate system given in Equation 1.19 can be written as Kg e k11k12k13k14k15k16 k21k22k23k24k25k26 k31k32k33k34k35k36 k41k42k43k44k45k46 k51k52k53k54k55k56 k61k62k63k64k65k66 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 1.20 where k11 C2 EA L S2 12EI L3 , etc. For member A, from Equation 1.18, Hg j A k44AD1 k45AD2 k46AD31.21a Vg j A k54AD1 k55AD2 k56AD31.21b Mg j A k64AD1 k65AD2 k66AD31.21c Similarly, for member B, Hg i B k11BD1 k12BD2 k13BD3 k14BD4 k15BD5 k16BD6 1.21d Vg i B k21BD1 k22BD2 k23BD3 k24BD4 k25BD5 k26BD6 1.21e Mg i B k31BD1 k32BD2 k33BD3 k34BD4 k35BD5 k36BD6 1.21f 2 1 3 4 5 6 p q F1 F2 F4 F5 A BC M3 M6 FIGURE 1.14. Assembly of structure stiffness matrix [K]. Structural AnalysisStiffness 15 Hg j B k41BD1 k42BD2 k43BD3 k44BD4 k45BD5 k46BD6 1.21g Vg j B k51BD1 k52BD2 k53BD3 k54BD4 k55BD5 k56BD6 1.21h Mg j B k61BD1 k62BD2 k63BD3 k64BD4 k65BD5 k66BD6 1.21i Similarly, for member C, Hg i C k11CD1 k12CD2 k13CD3 1.21j Vg i C k21CD1 k22CD2 k23CD3 1.21k Mg i C k31CD1 k32CD2 k33CD3 1.21l 1.7.2 Equilibrium Condition Any of the externally applied forces or moments applied in a certain direction at a joint of a structure is equal to the sum of the member forces acting in the same direction for members connected at that joint in the global coordinate system. Therefore, at joint p, F1 Hg j A Hg i B 1.22a F2 Vg j A Vg i B 1.22b M3 Mg j A Mg i B 1.22c Also, at joint q, F4 Hg j B Hg i C 1.22d F5 Vg j B Vg i C 1.22e M6 Mg j B Mg i C 1.22f By writing Equations 1.22a to 1.22f in matrix 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 16Plastic Analysis and Design of Steel Structures F1 F2 M3 F4 F5 M6 8 9 ; k44A k11Bk45A k12Bk46A k13Bk14Bk15Bk16B k54A k21Bk55A k22Bk56A k23Bk24Bk25Bk26B k64A k31Bk65A k32Bk66A k33Bk34Bk35Bk36B k41Bk42Bk43Bk44B k11Ck45B k12Ck46B k13C k51Bk52Bk53Bk54B k21Ck55B k22Ck56B k23C k61Bk62Bk63Bk64B k31Ck65B k32Ck66B k33C 2 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 4 3 7 7 7 7 7 7 7 7 7 7 7 7 7
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