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Unit 17 Approximate Methods of Analysis

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Unit 17 Approximate Methods of Analysis UNIT 17 APPROXIMATE METHODS OF ANALYSIS Structure 17.1 Introduction Objectives 17.2 Substitute Frame Method 17.3 Horizontal or Lateral Loading 17.3.1 Portal Method 17.3.2 Cantilever Method 17.4 Mixed Approximate Methods Based on Existing Exact and Approximate Methods 17.5 Summary 17.6 Keywords 17.7 Answers to SAQs INTRODUCTION In statically determinate structures, the analysis of a structure is made without the knowledge of the cross-sectional dimensions of the members, whereas cross section and members' elastic properties are prerequisite for the analysis of indeterminate structures. Therefore, it becomes necessary to perform approximate analysis for deciding preliminary selection of member sizes. Approximate methods are also useful for quick checking the results of exact analysis. It is also a powerful tool for carrying out spontaneous scrutiny of design, which involves large amount of analysis. In approximateanalysis, the statically indeterminate structure is simprified to a statically determinate structure by making suitable assumptions based on the experience of the analyst. Then, the analysis is carried out by using principles of statics. The validity of the results is based upon assumptions made in the analysis. In this unit, substitute frame method is described in Section 17.2. The approximate methods such as portal and cantileyer method are explained in Section 17.3.1 and 17.3.2, respectively. Objectives After studying this unit, you should be able to conceptualise and appreciate the use of approximate methods of analysis, find out internal forces, i.e. axial force, shear force and bending moment in any member of a building frame by method of substitute frame method due to vertical loading, compute internal forces in beams and columns of a plane frame subjected to lateral loading, by using portal and cantilever methods, and find out shear and moment due to transverse loading by generalising and approximating the facts based on existing exact methods. Role of Approximate Methods in Design Procedure The complete analysis of a structure usually requires a knowledge of sizes of all its members which are determined by design decisions. These decisions must be based on the knowledge of internal forces in the structure that are result of an analysis. Therefore, to begin with the structural engineer must make initial guess which is called "preliminary" design. The preliminary design is often based on approximate analysis and is strongly influenced by past experience and judgement of an engineer. Having determined the initial set of member sizes, a more detailed analysis may be made to determine the forces and displacements. These may then lead to redesign and subsequent reanalysis. This cycle continues till stresses and deflection criteria specified in the code is satisfied at a minimum cost. Therefore, in this cyclic process, approximate analysis plays an important role in the preliminary design. Miscellaneous Topics 17.2 SUBSTITUTE FRAME METHOD The building frame is a three dimensional space structure having breadth, height and length i.e. x, y and z coordinates. The manual analysis of space structure is tedious and time consuming. Therefore, approximation is made and the space frame is divided into several plane frames inx and z directions. Then the analysis of these plane frames is carried out. Even an analysis of inultistoreyed plane frame is laborious and time-consuming. Therefore, further simplified assumptioilsis made and analysis of roof or floor beam is made by considering this beam alongwith colunuls of upper and lower storeys. Columns are considered as fixed at far ends. Such a simplified beam-column arrangement is called a substitute frame. 14 15 16 Figure 17.2 (a) : Substitute Frame at Roof Level I I GROUND FLOOR Figure 17.1 :Typical Plane Frame Figure 17.2 (b) :Substitute Frame at First Floor Level Normally, a building frame is subjected to vertical as well as horizontal loads. The vertical loads consist of dead load and live load. The dead load comprises of self weight of beams, slabs, columns. wall, finishes, water proofing course etc. The horizontal loads consist of wind forces and earthquake forces. In order to evaluate ultimate load or factored load, the dead load and live load are multiplied by a factor which is known as partial safety~factorqf load or simply a load,factor. This factor is 1.50. In order to evaluate minimum possible dead load on the span which is self weight, sometimes tliis dead load is multiplied by a factor 0.90 for stability criteria. Therefore, Wn,in = D.L. or 0.9 D.L, and W ,,,,, = 1.5 (D.L+ L;.L) The effect of a loaded span on the farther spans is much smaller. The n~oment,shearand reaction in any element is mainly due to loads on the spans very close to it. Therefore it is . recommended to put live load on alternate spans and adjacent spans in order to cause severe effect at a desired location or section. L Figure 17.3 (a) : Maximum Hogging Moment at D Figure 17.3 (b) :Max. Sagging Moment at Centre of BC Approximate Methods of Analysis Figure 17.3 (c) :Maximum Sagging Moment Figure 17.3 (d) : Maximum Column Force in a Column at the Centl-e of CD at D, i.e. Maximum Shear in Beam CD and DE Figure 17.3 (e): Arrangement of Loads for Maximum Bending Moment in a Cdumn at B Table 17.1 shows the arrangement of live load (LL) on spans in addition to dead load (DL) 011 all spans depending upon critical condition. Table 17.1 - - - - - I 1 SNo. / Critical Condition 1 Liveiad(LL) on spans Reference 1 1 I I Maximum hogging moment at D 1 DE and CD I Figure 17.3 (a) I 1 2. 1 Maximum ragging moment at centre of B 1 BC: and DE 1 Figure 17.3 (h) ~ I--- -- - - Minimum sagging moment at centre of C:D AB, CD. and EF Figure 17.3 (c) Maximum axial force in a c01umli at D, i.e. C:D and DE ' Figure 17.3 (dl H-, maximum shear in beam C:D and DE , Maximum moment in columi~at B Longer span on one side Figure 17.7 (e) 1 of columli The restraining effect of any member forming a joint depends also upon Llle restraining condition existing at the other end. The other end may have following three conditions : (a) Freely supported or hinged. (b) Partially restrained. br (c) Rigidly fixed. I11 n~oslof the framed structures the far end is considered as rigidly fixed because of monolithic construction of a joint. I11 a substitute frame, ullbalanced moment at a joint IS distributed in columns and beams depending upon their ratio of stiffnesses. Steps for the Analysis (a) Select a substitdte frame, by taking-floor beam with columns of lower and upper storeys fixed at far ends. (b) , Cross sectional dimensions of beams and columns may be chosen such that moment of inertia of beam is 1.5 to 2 times that of a column and find distribution factors at a joint considering stiffnesses of beams and columns. (e). Calculate the dead load and live load on beam. Live load should be placed in such a way that it causes worst effect at the section considered i.e alternate and adjacent loading should be adopted. -.. Miscellaneous (d) Find the initial fixed end moments and analyse this frame by moment Topics distribution method. (e) Finally draw shear and moment diagram indicating values at critical section. Limitations (a) Height of all columns should be same in a particular storey. (b) Sway of substitute frame is ignored even during unsymmetrical loading. Example 17.1 Analyse the substitute frame shown in Figure 17.4 for (a) Maximum sagging moment at centre of span BC, (b) Maximum hogging moment at D, (c) Minimuni possible nioment at centre of BC,and (d) Maximum axial force in column at D. Assume frames are spaced at 3.5 ni centres. Other data is as follows : Thiclcness of floor slab = 120 mm Live load = 2 wm2 Floor finish = 1 kN/m2 Size of beam (overall) = 230 x 450 mm Size of column = 230 x 375 mm SEC. 22 Solution Calculation of Loads (a) Live load : ql = 3 w/m2 hadon beam = 3 x 3.5 m = 10.5 kN/m for all spans. (b) Dead load : (considering density of RCC as 25 w/m3) (1) D.L. of beam = 25 x 0.23 x 0.33 = 1.8975 kN/m (2) Floor fish = 1 kN/m2 (3) Slab load = 0.12 x 25 = 3 kN/n12 (4) Floor finish + Slab load = 4 kN/m2 (5) Load on beam = 4 x 3.5 m = 14 kN/m Total dead load (1) + (5) = 15.8975 kN/m Calcularion of Disrriburion Factor Approximate Methods of Analysis 1 I 1 Relative stiffness. I Total Relative 1 Distribution 1 Member Stiffness (x k) factor = k/Ck Fucrored Loads w,,, = 1.5 (wd+ wl) = 1.5 (15.8975 + 10.5) - (a) Maxiirium sagging moment at centre of BC Figure 17.5 Fixed end moments are as follows : MA^ = - ( 15.90x42) = - 21.20 kN m = - MBA 12 MBC = - 66.825 kN 111 = - MCB McD = - 33.125kNm = - MDc MDE = - 66.825 kN m = - M,cD Joint Member . D.F. F.E.M. Fitst C.O. Second C.O. Third Final Dist. Dist. Dist. A AA1 0.284 - 2.02 -0.017 3.98 AAz 0.284 - 2.02 - 0.017 3.98 AB 0.432 -21.20 9.16 - 3.07 ,f 0.06 - 0.036 - 7.96 B BA 0.312 + 21.20 14.24 0.12 \- 1.53 0.97 39.58 BB1 0.206 0.08 0.64 10.1 2 B-332 0.206 0.08 0.64 10.12 BC 0.276 - 66.83 12.59 C CB 0.296 + 66.83 -9.97 CC 1 0.2P9 - 7.38 - 0.53 - 10.27 CC2 0.2 19 -7.38 - 0.53 - 10.27 C: D 0.266 - 33.13 - 8.96\ I) DC: 0.266 + 33.13 8.96 / -4.48 4.76 /'- 1.43 0.65 41.59 DD1 0.219 3.92 0.53 11.83 DDz 0.219.
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