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. 3.92 0.53 11.83 DE 0.296 - 66.83 9.97 /r13.43 5.30 \p- 1.0 0.72 - 65.27 E ED 0.402 + 55.83 - 26.8 4.98 - 2.0 2.65 - 1.06 44.54 EE1 0.299 - 1.49 - 0.8 - 22.27 EE2 0.299 - 1.49 - 0.8 - 22.27
(b) Maximum hogging moment at D in beam and (d) maxlmum axla1 force in coluinn at D.
Figure 17.6 Fixed end ~llonlelltsare as follows .
MAB = -21.2 kNm = -MBA MBC = - 26.83 kN m = - MCB
Mcn = - 82.50 kN 111 = - MDc MDE = - 66.83 kN m = - ME
- - Joint Member D.F. F.E.M. Fitst C.0. Second C.O. Third Final Dist. [list. llist. I A AA1 0.284 6.02 - 0.25 0.57 6.34 AA? 0.284 6.02 -0.25 0.57 634 AB 0.432 - 21.20 9.I6 0.88 - 0.38 -2.00 0.86 - 12.86 ,,/\r I B BA 0.3 12 + 21.20 1.76 4.58 - 4.00 - 0.19 0.00 23.35 BB 1 0.206 1.16 - 2.64 0.00 - 1 .48 BB2 0.206 1.16 -2.64 0.00 1 -1.48 I BC: 0.276 - 26.83 1.55 8.24 -3.54 0.19 0.00 - 20.38 /\ C C:B 0.296 + 26.83 16.48 0.77 0.39 - 1.77 0.09 , 42.79 C:C: 1 0.2 19 12.19 0.29 0.07 12.55 CC2 0.2 19 1219 0.29 0.07 1255 CD 0.266 - 82.50 14.8 1 ,> - 2.08 0.35 1.46 0.08 -67.88 \r D DC: 0.296 + 82.50 -4.17/' 7.40 0.175 -0.17 1 83.66 1 DD1 0.2 19 - 3.43 1 7 -0.14 - 1.16 DD2 0.219 -3.43 2.4 1 -0.14 -1.16 DE. . 0.296 - 66.83 -4.64 ,-13.43 3.26 , 0.46 -0.19 -81.37
- 0.49 - 19.78 -0.49 1') 7s * The distribution factors for upper column and lower column is same, therefore Approximate Methods several steps in moment distribution are common to both at a joint. of Annlysis C D E CD DC DE ED Reaction due to udl 99.00 99.00 89.10 89.10 Reaction due to moment - 3.16 +3.16 +9.29 - 9.29 102.16 98.39 RD = 200.55 kN 1 The maximum moment is at support D
= 83.66 kN m and MI)E = - 81.37kN m) and maximum axial load in column is also at support, D RD = 200.55 kN. (c) Minimum moment at centre of BC
Figure 17.7 Fixed end moments are as follows :
Joint Member D.F. F.E.M. First C.O. Second C.O. Third Final I Dist. Dist. Dist. A AAi 0.284 15.00 1.15 0.87 17.02 AA2 0.284 AB 0.432 -52.80 22.81 ,7-4.05 1.75 ,,-3.06 1.32 -34.03 /I B BA 0.312 + 52.80 - 8.10 11.40 - 6.13~' 0.87 -0.78 50.06 ' BB 1 0.206 - 5.35 - 4.04 -0.51 -9.90 BBz 0.206 - 5.35 - 4.04 -0.50 - 0.90 BC 0.276 -26.83 -7.17, 8.24 -5.42, 1.62 -0.69 -30.25 C CB 0.296 +26.83 16.48 "- 3.58 3.25 -2.71 0.88 41.15 I 0.65 15.24 0.65 15.24 I i
. - .
".'.a[,, 1 ! ,. . ,\ri:~lyxcille strh~tirulct'ranlc. show11ill Figrlrc 17.8 Ior (ai rnaximun~span 1:!\vlielll ilk FG: (h)~li;~~ilu~~i\ colurnll n~c~lncnl a1 F: ;u~d(c) 111irri1nu111spi111 ~i>t;tiicn~in FG. ((7) Analyse lie fl(mr ABCI) of inlcrnldinfe thrn~eshown in Figure 17.0 by substitute frame method in lerlns of computing (a) maximum hogging moment at B, (b) maximum sagging moment in span CD. ;md !c)maximum axial lea$ in colunui at B. Tlic frames are spnccd a( 4.5 111 ccntrcs. Live lo;~dis 4 kN/in-. Thickness of Hoor slab is 150 mm. Floor finish is 1 kN/ln-. iwerall size of the heam is 230 x 350 mm ant1 siztl of column is 230 x 350 Inm. Thickness of wall may be assumed as 150 mln.
Figure 17.9
17.3 HORIZONTAL OR LATERAL LOADING
A building frame usually carries horizontal or lateral load as shown in Figure 17.10 (a) due to wind or earthguake and deforms due to such loads as shown in Figure 17.10 (b). From this exaggerated shape of deformed frame, we can assume that point of contraflexure lies approximately at centres of column and centres of beam.
Figure 17.10 (a) :Building Frnme Subjected to Rpre 17.10 (b) :Deflected Shape of Frame Lated Loading There are two methods, viz. (a) portal method, and (b) cantilever method, which are ApprUNIIWl+:LVlrluUU> generally employed for the approximate analysis of frames subjected to horizontal loading. of Analysis 17.3.1 Portal Method The following assumptions are made to simplify the given framed structure to a statically determinate one : (a) Point of centraflexure is located at mid height of each column, (b) Point of contraflexure is located at centre of each beam, and (c) The horizontal shear is divided among all the columns on the basis that each interior column takes twice the shear carried by the exterior column. In the third assumption, exterior column corresponds to single portal leg and interior column corresponds to two portal legs. Hence, it is reasonable to assume interior column to carry twice the shear of exterior column. Steps for Analysis (a) The shear in columns of a particular storey is found out by equating to storey shear, i.e lateral load on the frame above that section. (b) Moments at the end of a column are determined by multiplying the column shear with half the length of the column. (c) The moments at the ends of beams is computed by considering the equilibrium of each joint separately. The column moments obtained from Step (b) is known and the beam moment which is unknown can be determined. (d) Since the moment in be'un is known from Step (c) the reactions and hence, shear in beam is determined. (e) Axial force in column is the algebraic sum of reactions from the beams. (f) Finally, carry out the check of = 0 and xFy= 0 at the supports. Limitations (a) This method is useful for low rise building. (b) This method is unsuitable for building frames having different number of bays at different tloors. Example 17.2 Using portal method, analyse the frame shown in Figure 17.11 for shear force, bending moment and axial force in all members. Area of each column is same, i.e. unity.
Figure 17.11 Solution Step (1) : Shear in Each Storey Let P be the shear in each exterior column of one storey, then 2 P shall be the shear in each interior column of that storey. Miscellaneous (a) For the first storey, total horizontal shear resisted by the column is equal to Topics total horizontal forces acting above XI XI as shown in Figure 17.12.
PI + 2P1 + 2P1+ PI = 120 + 80 r 200 kN P1 = 33.33 kN and 2P1 = 66.67 kN (b) Similarly, for second storey, total horizontal shear resisted by the column is equal to total horizontal forces acting above X2 X2 as shown in Figure 17.12. P2 + 2P2 + 2P2 + P2 = 80 kN PZ = 13.33 kN and 2 P2 = 26.67 kN
Figure 17.12 : Horizontd Shear for each Storey
Step (2) : Moments at the Ends of Columns As there is a point of contratlexure at the centre of column, the moments can be detennined by considering the free body diagram of column members. Refer Figure 17.12 and consider column AE. Due to column shear, joint moment at E, MEA, = P x ($1 = 83.33 (clrkwise). Coilsidering equilibrium at the cross section, equal and opposite to joint moment. column end moment will be
It is represented in Figure 17.13. Similarly, for all columns, the joint moments and column end moments can be calculated as shown in Figure 17.1 3 (a) md (h!
Figure 17.13 (hi Elid Mon~crlttsat Storey - 2 63 33 TMEA
Figure 17.13 Step (3) : Moments at the End of Beam Approximate Methods of Analysis Considering the equilibrium at each joint due to joint moments as shown in Figure 17.13 (a) and (b), the joint moments for beams can be computed as given under : Joint I CM = 0 --+ MIE+ MIJ = 0
MIJ = - MIE = - 26.67 kN In (anticlockwise) Left end moment in beam IJ (taking equal and opposite), MIJ = + 26.67 kN m (clockwise) Since the point of contraflexure lies at the centre of IJ, right end moment in beam 1J MJ1 = .- MIJ = 26.67 kN m (clockwise) which is explained in Step (4). Therefore, joint moment (considering equal and opposite) MJ1 = - 26.67 kN m (anticlockwise) .Joint J CM = 0 + MJr+Mj~+MjK= 0;
Thus, MJK = - (MJI + MJF) = - [ - 26.67 (anticlockwise) + 53.33 (clockwise)] = - 26.67 kN m (anticlockwise) From free body diagram of beam JK, end moments MKJ = MJK = 26.67 kN m (clockwise) Similarly, moments in members at joints K and L can be worked out and final result is shown in Figure 17.13. Joint E CM=O+ MEA+MEF+MEI=O;
Thus, MEF = - (MEA+ MEJ = - [83.33 (clockwise) + 26.67 (clockwise)] = - 110 k.Nm (anticlockwise) From free body diagram of beam EF, ME^ = 110 k.Nm (clockwise) = MFE Joint F
Thus, MFG = - (MFE + MFB + MFJ)
= - [ - 110 (anticlockwise) + 166.67 (clockwise) + 53.33 (clockwise)] = - 1 10 kN m (anticlockwise) From free body diagram of beam EF, MFG = 110 kN m (clockwise) = MGF Similarly, moments in members at joint G andH can be worked out and is shown in Figure 17.13. Step (4) : Shears in the Beams Consider a beam element 1-2 as shown in Figure 17.14 (a) which is subjected to end moments M1 and M2. As there is no lateral load considering F, = 0. S1 + S2 = 0 4 S2 = - S1 therefore, S2 will be acting downwards. Miscellaneous As there is a hinge in the centre of the beam and moment at hinge is zero. Topics M1 = S1 (Ll2) (clock wise) is given moment. M2 = S2 (Ll2) which is also clock wise. As S1 and S2 are equal and opposite, M1 = M2. For beam IJ, the end moments are Mlj 'and MJIboth clockwise as shown in Figure 17.13. At the end of a beam SIJwill be upward and SJIwill be downword. Taking the equilibrium near the joint, SIJ at the joint I will be downward while SIJ near joint J is upward as shown in Figure 17.14 (b).
Figure 17.14 (a) Figure 17.14 (b) Step (5) :Axial Forces in Columns For determination of axial forces in column the equilibrium of joints are considered at first joint I CFy= 0 AIE + SIJ = 0 4AIE = - SIJ = 13.33 kN (upward) The equal and opposite axial force AIE= 13.33 kN (downward) is directed towards the hinge in the Figure 17.15. The presence of hinge will not effect the axal force AEI, when equilibrium of column member is considered. Therefore, axial force at member end = 13.33 kN (upward) and axial force at joint E = 13.33 kN (downward). The axial forces in other columns can be computed which is shown in Figure 17.15.
Figure 17.15 : Axial Force in Columns and Shear in Beams (Portal Method) From Figure 17.13, the end moments for free bodies of columns and beams are Approximate Methods considered (ignoring the end moments for joints) and Figure 17.16 is redrawn. of Analysis
Figure 17.16 :Bending Moments at Ends of Columns and Beams (Portal Method) The bending moment'diagram (depicted on tension side) can be drawn and is shown in Figure 17.17.
EYgure 17.17 :Bending Moment Diagram of fnune (depicted on temion side) SAQ 2
Xnaly;~the fra~riesshown i11 Figure 17.18 (;I) ;uld Figu1.e 17. I ti (bj by pi)rt:il !l!i:ll~ ){ i Miscellaneous 17.3.2 Cantilever Method Topics This method is based on the assumption that the frame is acting like a cantilever beam with the column cross sectional areas as the fibers in a beam. The assumptions made in the analysis are as follows , (a) Point of contraflexure (i.e., zero bending moment) is located at the mid height of the column in each storey. (b) Point of contreflexure is located at the mid point of each beam, and (c) The axial force in each column is proportional to its distance from the centre of gravity of the areas of column group at that level. The third assumption includes the effects of columns having different cross sectional areas. The stress intensities can be obtained by method analogous to that used for determining the distribution of normal stress intensities on a transverse section of a cantilever beam. Steps for the Analysis (a) Determine centre of gravity of column groups from areas of columns (column area may be assumed unity). (b) Consider the equilibrium of the particular storey at the section of point of contraflexure in columns. Other column axial forces are expressed in terms of one unknown column axial force. as they are assumed proportional to the distance from centre of gravity. Taking moment of external loads and axial forces in columns about any point at the section, the unknown forces in the columns are evaluated. (c) The beam shear is calculated by using condition, CF, = 0 at each joint separately. The column axial force is known from Step (b). (d) The moment at the end of the beam is beam shear multiplied by half the length of beam. Since the point nf contraflexure is assunied at the centre of the beam. (e) The column moments are determined by consideriilg condition M = 0 at each joint separately. The moment at the ends of the beams is known from Step (d). (f) Finally, carry out check of CF, = 0 and CF, = 0 at tlle supports. Limitations (a) This method is useful for tall narrow buildings. (b) The method is not applicable to fraines having different cross sectional areas of tlle same columi at different floors. Example 17.3 Using the cantilever inetlx~d,analyse the frame shown in Figure 17.1 1 for shear force. bending inoinent and axial force in all members. Area of each column IS same, i.e. unity. Solution Step (1) : Centre of Gravity of Column CSroup Let cross sectional area of each colurrul be 'a'. By taking ~nonientsof areas of columns about column AI, we get the centroid of colunln group from column A1 is at a distance of x which is computed as below :
(ax 0) +(o x4) + ((1 x 10) + ((7 x 15) X = (4 x 0)
Step (2) : Axial Forces in Column As we have assumed that the frame is acting as a cantilever, therefore, axial forces in columns will be in proportion to the distance from CG of column group. Let PAEbe the axial force in column AE, the axial force in the other column shall be as shown in Figure 17.19. Approximate Methods of Analysis
As the point of contraflexure lies at the centre of column, the moment is zero at such locations. Taking moments about XI (point of contraflexure in column AE),
(80 x 6.5) + (120 x 2.5) + (P~~x4.0)- (PCGX 10) - (PDHX 15) = O 820 + (0.448 PA^ x 4.0) - (0.379 PAE x 10.0) - (1.069 PAE x 15) = 0 Giving, PA^ = +45.47 kN, PcG = - 17.23 kN PBF = +20.37 kN, PDH = -48.61 kN
. Rpre 17.20 I Similarty, for top storey (Referring to Figure 17.20) I - - PFj = 0.448 PEI , I PGK = - 0.379 PE! ,and PHL = - 1.069 PEI Taking moment about X2 point of contratlexure in column EI, (80 x 2.0) + (PFJx 4.0) - (PGKx 10.0) - (PHLx 15.0) = 0
160 + (0.448 P,yI x 4.0) - f0.379 PEI x 10.0) - (1.069 PE1 X 15) = 0 Giving, Pm = 8.87 kN, PG,y = - 3.36 ldl PFJ = 3.97 kN, PHL = - 9.48 kN Miscellaneous Topics
zby= 0; 315 + PIE = 0 sIJ = - PIE = - 8.87 kN (downward) = 8.87 kN (upward) Shear at the end I in the free body diagram of beam IJ is SIJ = - 8.87 kN (upward) = 8.87 kN (downward) Beam IJ c Fy = 0; sJI + SIJ = 0 SJI = - SIJ = - 8.87 kN (downward) = 8.87 kN (upward) Joint J F, = 0; SJI + PJF + SJK = 0 SJK = - (PJF + SJI) = - r8.87 (downward) + 3.97 (downward)] = 12.84 kN (upward) Therefore, in heam JK, at the end J, SJK = - 12.84 kN (upward) = 12.84 kN (upward) Beam JK Fy = 0; SKJ + SJK = 0 SKJ = - SJK = - 12.84 kN (downward) = 12.84 kN (upward) Similarly, shear in the other beams can be determined and are shown in Figure 17.21.
Figure 17.21 :Axial Forces in Cola~lnsmd Shear in Beama (Cndilever Method) Step (4) : Moments at the Ends of Beams Approximate Methods of Analysis As discussed in portal method, refering Figure 17.14 (a), we have MI =Six[:] and M2=s2x[:) Both are equal and in same direction,
= 17.74 kN m (clockwise)
= 38.52 kN m (clockwise) Moments in all other beams are determined in the similar manner and are shown in Figure 17.22.
Rgure 17.12 : LII~Alu~t~ents .and Joint Mo~ne~~ts(Cantilever Method)
Step (5) : Moments ;I[ I 11 r I Figure 17.23 :Bending Moment Dingram (Ckntilever Method) SAQ 3 Analyve the frames shown in ~i~uri17.18 (a) and (b)by Canlilever Method. - - 17.4 MIXED APPROXIMATEMETHODS BASED ON EXISTING EXACT AND APPROXIMATE METHODS Portal frames subjected to vertical loading can also be analysed based upon approximations derived from certain facts in existing exact methods. The study of bending moment diagrams of single span beam with different end conditions and different loading reveals that, * beam having both ends fixed and carrying uniformly distributed load has point of contraflexure or point of zero bending moment at 0.21 L from fixed ends, * a beam having both ends hinged, carrying uniformly distributed load has zero moment points only at the supports, * a beam having both'ends fixed and carrying point load at centre, pointy of contraflexure is at 0.25 L from supports, and * a beam having one end fixed and other hinged and carrying point load at centre, the contraflexure point is at 0.27 L from fixed support. Approximate Methotls It is concluded therefore, that the point of zero moment lies at a distance varying from zero of Analysis at support to 0.27 L from support for different ends conditions and different loadings. Hence, it can be generalised and assumed for approximate analysis that a point of contraflexure~sat 0.1 L from the ends and the analysis of indeterminate frame is made by converting the frame into determinate one, after introducing points of contraflexure as shown in Figure 17.24. Figure 17.24 (a) :Assumed Point of Contratlexure Figure 17.24 (b) :Determinate Structure SAQ 4 Analysl: the frame shown in Figure 17.25 by the inixed method. Figure 17.25 SUMMARY Analysis of indeterminate structures using approximate methods is quicker method of scrutinizing the structures. This is also useful for preliminary design. Substitute frame method is used for analysis of roof or floor beams along with columns of upper and lower storeys. It is usdd for vertical loading. Portal and cantilever methods of approximate analysis are used to compute internal forces in multistoreyed multibay frames subjected to horizontal forces. Low rise frames may be analysed by using portal method which is based on following assumptions : 1 (a) Shear force in exterior column = - [ Shear Force in interior column] 2 (b) Points of contraflexure are located at centre of beams & columns. Tall and narrow frames are analysed by using cantilever method, which is based on following assumptions : (a) Axial force in columns are proportional to distance from the C.G. of the cross sectional areas of the columns. (b) Point of contraflexure is at centre of columns and beams. Mixed method based on summary of exact results of beams subjected to vertical loading assumes that there are points of centraflexure at a distance of 0.10 of span length from supports. In case of portal frames moment at a joint in column is found out by equilibrium condition and moment at the base of column is simply a carry over. 17.6 KEY WORDS Substitute Frame : Roof or floor beam with column of upper and lower storeys fixed at far ends. Plane Frame : It is a frame having member only in one plane, say xy or yz. Space frame comprises of several such plane frames in both planes xy and yz. Portal Method : Each bay of portal is tested separately and hence, the shear in exterior column is half of the shear in interior column, therefore it is named as mrtal method. Cantilever Method : Frame is tested as cantilever fixed at base and force in any column (fibre of frame) is proportional to the distance of centre of gravity and hence, this is known as cantilever method. Conventional : Any method of analysis which is established by using the Method conventional theory of force displacement relation such as consistent deformation method, three moment theorem, moment distribution method, slope deflection method and strain energy method etc. Point of : It is the point where bending moment changes sign, i.e. bending Contraflexure moment at this point is zero. 17.7 ANSWERS TO SAQs SAQ 2 (a) MEI = 10.0 kN m (ACW) MEF = 43.33 kN m (CW) Ma= 33.33 kN m (ACW) MGF = 43.34 kN m (CW) MGK = 20 kN m (ACW) MG~= 43.33 kN m (CW) MCx = 66.67 kN m (ACW) (b) Shear in column for top storey = 40 kN Shear in column for middle storey = 100 kN Shear in column for lower storey = 160 kN SAQ 3 (a) ME/ = 6.10 kN m (ACW) M,qF = 26.49 kN m (CW) ME* = 20.39 kN m (ACW) MGF= 51.38 kN m (CW) MG~= 23.88 kN m (ACW) Approximate Methocls of Analysis MGH = 52.20 kN m (CW) MGC = 79.70 kN m (ACW) (b) For top storey, centre of gravity of two columns EG and FH is to be evaluated and for bottom two storeyes the centre of gravity of three column should be evaluated I SAQ 4 Assume point of cantratlexure at 1 m from points E, F, C, and D. Analyse the detenninate beam, between points of cantraflexureand then, analyse cantilever of 1 m length. I McE = 450 kN m (CW) MpD = 900 kN m (ACW) MCA = 450 kN m (CW) FURTHER READING Reddy, C. S., Basic Structural Analysis, Tata McGraw Hill Publishing Company Ltd., 1996. Naris, C. H. and Wilbur, J. B., Elementary Structural Analysis, McGraw Hill Book Company Inc., New Y ork, International Student Edition. Rao Prakash, D. S., Structural Analysis -A United Approach, University Press (India) Ltd., Hyderabad. Junarkar, S. B., Mechanics ofStmture, Vol 11, Charotar Publishing House, Anand, 1989. Kinny, J. S., Indeterminate Structural Analysis, Oxford and IBH Publishing Company Pvt. Ltd., New Delhi.