Compression Members Version 2 CE IIT, Kharagpur SlenderColu Version 2 CE IIT, Kharagpur Instructional Objectives: At the end of this lesson, the student should be able to: 0 define a slender column, o give three reasons for its increasing importance and popularity, o explain the behaviour of slender columns loaded concentrically, - explain the behaviour of braced and unbraced single column or a part of rigid frame, bent in single or double curvatures, - roles and importance of additional moments due to P- A effect and moments due to minimum eccentricities in slender columns, o identify a column if sway or nonsway type, - understand the additional moment method for the design of slender columns, o apply the equations or use the appropriate tables or charts of SP-16 for the complete design of slender columns as recommended by IS 456. 11.27.1 Introduction Slender and short are the two types of columns classified on the basis of slenderness ratios as mentioned in sec.10.21.5 of Lesson 21. Columns having both Iex/Dand Ie,/b less than twelve are designated as short and otherwise, they are slender, where Iexand Ieyare the effective lengths with respect to major and minor axes, respectively; and D and b are the depth and width of rectangular columns, respectively. Short columns are frequently used in concrete structures, the design of such columns has been explained in Lessons 22 to 26, loaded concentrically or eccentrically about one or both axes. However, slender columns are also becoming increasingly important and popular because of the following reasons: (i) the development of high strength materials (concrete and steel), (ii) improved methods of dimensioning and designing with rational and reliable design procedures, Version 2 CE IIT, Kharagpur (iii) innovative structural concepts specially, the architects expectations for creative structures. Accordingly, this lesson explains first, the behaviour of slender elastic columns loaded concentrically. Thereafter, reinforced concrete slender columns loaded concentrically or eccentrically about one or both axes are taken up. The design of slender columns has been explained and illustrated with numerical examples for easy understanding. 10.27.2 Concentrically Loaded Columns It has been explained in Lessons 22 to 26 that short columns fail by reaching the respective stresses indicating their maximum carrying capacities. On the other hand, the slender or long columns may fail at a much lower value of the load when sudden lateral displacement of the member takes place between the ends. Thus, short columns undergo material failure, while long columns may fail by buckling (geometric failure) at a critical load or Eu|ers load, which is much less in comparison to that of short columns having equal area of cross-section. The buckling load is termed as Eulers load as Euler in 1744 first obtained the value of critical load for various support conditions. For more information, please refer to Additamentum, De Curvis e|asticis, in the Methodus inveiendi Lineas Curvas maximi minimive proprietate gaudentes Lausanne and Geneva, 1744. An English translation of this work is given in Isis No.58, Vol.20, p.1, November 1933. The general expression of the critical load PC,at which a member will fail by buckling is as follows: PC,= n2EI/(kI)2 where E is the Youngs modulus I is the moment of inertia about the axis of bending, I is the unsupported length of the column and k is the coefficient whose value depends on the degree of restraints at the supports. Expressing moment of inertia I = A12,where A is the area of crosssectionof the column and r is the radius of gyration, the above equations can be written as, PC,= n2EA/(kl/r)2 (10.62) Thus, PC,of a particular column depends upon kl/r or slenderness ratio. It is worth mentioning that kl is termed as effective length /6 of the column. Version 2 CE IIT, Kharagpur 9; igpgg L %Pi;&'@r g ; pig sf Pi E Fiéiivfilatiltfir Fig.1037.1:Qeiumn - hirmad Fig w.2r.2:ceriummu 33;mmgmgiprkm1) at bothands,, (ta; Q5) F {:F1.3r * P *74PM I .3 Evie] 3* IE2 RE.i :F,|Zt§i'%!l. ofiensmztidn F {Pg *3P "5 :Fiig.*'E..2?.t3:: Cieiumnsupported on arnzasm/teams iki} Figures 10.27.1 and 2 show two elastic slender columns having hinge supports at both ends and fixed supports against rotation at both ends, Version 2 CE IIT, Kharagpur respectively. Figure 10.27.3 presents a column of real structure whose end supports are not either hinged or fixed. It has supports partially restrained against rotation by the top and bottom beams. Each of the three figures shows the respective buckled shape, points of inflection Pls (points of zero moment), the distance between the Pls and the value of k. All the three columns, having supports at both ends, have the k values less than one or at most one. By providing supports at both ends, one end of the column is prevented from undergoing lateral movement or sidesway with respect to the other end. P e_25e,_,, Rt. P; I. F£}§E§£ of ittffm:-tim Column »»fixed at an free at other and an w 3} Version 2 CE IIT, Kharagpur .1. pg aa s I S m. s Paiérai anecmn Fig.-m.2?:5: ratati-mnai§y fi:-tad at birth and, (k. m 1} Version 2 CE IIT, Kharagpur ha E "an er..-~. 3*. . isto£2 an9,,g. Cltjm ;;lIl§i1lEél37l'§gil@$tl"~%.EET@~ilIf at bath lens it *2 1 However, cantilever columns are entirely free at one end, as shown in Fig.10.27.4. Figure 10.27.5 shows another type of column, rotationally fixed at both ends but one end can move laterally with respect to the other. Like that of Fig.10.27.3, a real column, not hinged, fixed or entirely free but restrained by top and bottom beams, where sideway can also take place. Each of these three figures, like those of Figs.10.27.1 to 3, presents the respective buckled shape, points of inflection (Pls), if any, the distance between the Pls and the value of k. All these columns have the respective kvalues greater than one or at least one. Version 2 CE IIT, Kharagpur V EPntnt oflvnflseciien Fig. '1:.2?'.?: Erased portal frame {Ea:1 1.2?..&8i: l;§;tt'bm£',i:r§d portal!;ff'1'.t"l:fE{: 3*2'; Figures 10.27.? and 8 present two reinforced concrete portal frames, a typical reinforced concrete rigid frame. Columns of Fig.10.27.7 are prevented from sidesway and those of Fig.10.27.8 are not prevented from sidesway, respectively, when subjected to concentric loadings. The buckled configuration of the frame, prevented from sidesway (Fig.10.27.7) is similar to that of Fig.10.27.3, Version 2 CE IIT, Kharagpur except that the lower ends of the portal frame are hinged. One of the two points of inflection (Pls) is at the lower end of the column, while the other PI is slightly below the upper end of the column, depending on the degree of restraint. The value of k for such a frame is thus less than 1. The critical load is, therefore, slightly more than PC, of the hingehinge column of Fig.10.27.1. The buckled configuration of the other portal frame of Fig.10.27.8, where sidesway is not prevented, is similar to the column of Fig.10.27.4 when it is made upside down, except that the upper end is not fixed but partially restrained by the supporting beam. In this case, the value of k exceeds 2, depending on the degree of restraint. One of the two Pls is at the bottom of the column. The critical load of the column of Fig.10.27.8 is much less than that of the column of Fig.10.27.1. Table 10.14: Critical loads in terms of Per of hingehinge column and effective lengths I9 = kl of elastic and reinforced concrete columns with different boundary conditions and for a constant unsupported length I Support conditions Critical load Effective length PC, /9 = kl No. A Elastic sinle Fig. No. columns sideswa both ends no sideswa Partially restrained against rotation by top and bottom Between PC, I > kl > I/2 10.27 3 cross- beams, no sideswa Fixed at one end and . entirely free at other end sidesway not 2 I, one PI is on imaginary extension revented 5. Rotationally fixed at both ends sidesway not I, one PI is on imaginary . revented extension Partially restrained against rotation at both ends sidesway not revented B Reinforced 10.27.5 concrete columns sideswa Between zero and slightly less than PC, * l< kl < a 10.27.6 2. * The critical load is slightly less than P0, of hingehinge column (Sl.No.1), when cross-beams are very rigid compared to columns, i.e., the case under Sl.No.6 approaches the case under Sl.No.1. The critical load is zero when cross-beams are very much flexible compared to columns, i.e., the case under S|.No.6 approaches to hingehinge column of Sl.No.1, allowing sidesway. In that case, it becomes unstable and hence, carries zero load. Table 10.14 presents the critical load in terms of that of hingehinge column Po, and effective lengths /6 (equal to the distance between two points of inflection Pls = kl) of elastic and reinforced concrete columns for a constant value of the unsupported length I. The stress-strain curve of concrete, as shown in Fig.1.2.1 of Lesson 2, reveals that the initial tangent modulus of concrete EC is much higher than E; (tangent modulus at higher stress level). Taking this into account in Eq.10.62, Fig.10.27.9 presents a plot of buckling load PC,versus kl/r. It is evident from the plot that the critical load is reducing with increasing slenderness ratio. For very short columns, the limiting factored concentric load estimated from Eq.10.39 of Lesson 24 will be found to be less than the critical load, determined from Eq.10.62. The column, therefore, will fail by direct crushing and not by buckling. We can also find out the limiting value of kl/r when the crushing load and the buckling load are the same. The (kl/r),,-,,,is shown in Fig.10.27.9. The limiting value of kl/r also indicates that a column having kl/r more than (kl/r),,-,,, will fail by Version 2 CE IIT, Kharagpur buckling, while columns having any value of kl/r less than (kl/r),,-,,,will fail by crushing of concrete. The following are the observations of the discussions about the concentrically loaded columns: 1. As the slenderness ratio kl/r increases, the strength of concentrically loaded column decreases. 2. The effective length of columns either in single members or parts of rigid frames is between 0.5! and I, if the columns are prevented from sidesway by bracing or otherwise. The actual value depends on the degree of end restraints. 3. The effective length of columns either in single members or parts of rigid frames is always greater than one, if the columns are not prevented from sidesway. The actual value depends on the degree of end restraints. 4. The critical load of braced frame against sidesway is always significantly larger than that of the unbraced frame. tweed t.E?.§{a}: C5f=ef%3a»z::tis.::rle Fig. 1«2?..§@§fb}:f Mum-e:nts. Fiig.*%f.E?'.1: Enfumnbesrlt in singlemwature, Q} Version 2 CE IIT, Kharagpur 10.27.3 Slender Columns under Axial Load and Uniaxial Moment (A) Columns bent in single curvature Figure 10.27.10a showsa column bent in single curvature under axial load P less than its critical load PC,with constant moment Pe. The deflection profile marked by dotted line is due to the constant moment. However, there will be additional moment of Py at a distance 2 from the origin (at the bottom of column) which will deflect the column further, as shown by the solid line. The constant moment Pe and additional moment Py are shown in Fig.10.27.10b. Thus, the total moment becomes M = Mo+Py (10.63) = P(e+y) The maximum moment is P(e + A) at the midheight of the column. This, we can write Mme, = Mo+PA (10.64) This is known as = P(e+ A) P - A effect. ga p:f§i E-IE2 1=l1i.2?,.E:1: tax}:Easeeieetiansl ?%g:.. lI3.2?*.*ttf{b}:Moments Cxeirumnlament. in aeirrgltesr:umat:u:re, Version 2 CE IIT, Kharagpur Figure 10.27.11a shows another column whose bending is caused by a transverse load H. The bending moment at a distance 2 from the origin (bottom of the column) is Hz/2 causing deflection of the column marked by dotted line in the figure. The axial load P, less than its critical load P0,, causes additional moment resulting in further deflection, marked by solid line in the figure. This additional deflection produces additional moment of Py at a section 2 from the origin. The two bending moment diagrams are shown in Fig.10.27.11b. Here again, the total moment is M = Mo+Py = Hz/2+Py (10.65) The maximum moment at the midheight of the column is Mmax= MomaX+PA = (10.66) The total moment in Eqs.10.63 and 10.65 consists of the moment Mo that acts in the presence of P and the additional moment caused by P (= Py). The deflections y can be computed from ya, the deflections without the axial load from the expression Y = yo[1/{1 - (P/Pcr)}] (10.67) From Eq.10.64, we have Mmax= (10.68) = Equation 10.68 can be written as MW 2 My1+ 1//(P/P6,) 1'(P/Pcr) (10.69) where V depends on the type of loading and generally varies between i 0.20. Since P/PC,is always less than one, we can ignore 1,u(P/Pcr) term of Eq.10.69, to have Mme, = M,/{1 (10.70) (P/Pc,)} where 1/{1 (P/Pc,)} is the moment magnification factor. In both the cases above (Figs.10.27.10 and 11), a direct addition of the maximum moment caused by Version 2 CE IIT, Kharagpur transverse load or otherwise, to the maximum moment caused by P gives the total maximum moment as that is the most unfavourable situation. However, this is not the case for situation taken up in the following. (B) Columns bent in double curvature if 5.5 5: i2 M l M H:=,K1:&\ gr3:: gjmv i fl? E 2, A ax a *3 ,3) , ,.. ""MWl§§ Eli:-znder t'>Er§Umn (bi)l!.*1,., «diagram {iii @345 diagram rid}Mg,4% Pi,r diagram {e}».MEe'rn.ativ'e Py diagram elf-lg. if3..2?..t;2:: Slerndesr column under axial load and abend.ing, bent in double curvsttire Figure 10.27.12a shows a column subjected to equal end moment of opposite signs. From the moment diagrams Mo and Py (Figs.10.27.12b and c), it is clear that though Mo moments are maximum at the ends, the Py moments are maximum at some distance from the ends. The total moment can be either as shown in d or in e of Fig.10.27.12. In case of Fig.10.27.12d, the maximum moment remains at the ends and in Fig.10.27.12e, the maximum moment is at some distance from the ends, where M, is comparatively smaller than M, maxat the ends. Accordingly, the total maximum moment is moderately higher than M, max- From the above, it is evident that the moment M, will be magnified most strongly if the section of Momaxcoincides with the section of maximum value of y, as in the case of column bent in single curvature of Figs.10.27.10 and 11. Similarly, if the two moments are unequal but of same sign as in Fig.10.27.10, the moment M, will be magnified but not so much as in Fig.10.27.10. On the other hand, if the unequal end moments are of opposite signs and cause bending in double curvature, there will be little or no magnification of Mo moment. This dependence of moment magnification on the relative magnitudes of the two moments can be expressed by modifying the earlier Eq.10.70 as Version 2 CE IIT, Kharagpur Mm = M0 C,,/{1(P/Pc,)} (10.71) where cm = 0.6+0.4(M1/M2) 2 0.4 (10.72) The moment M1 is smaller than M2 and M1/M2is positive if the moments produce single curvature and negative if they produce double curvature. It is further seen from Eq.10.72 that Cm= 1, when M1 = M2 and in that case, Eq.10.71 becomes the same as Eq.10.70. For the column of Fig.10.27.12a, the deflections caused by M, are magnified when axial load P is applied. The deflection can be obtained from Y = yo[1/{1-(P/4Pcr)}] (10.73) P i fa} aortas? frame {ab}lvfwuments {:13trim-an-is :1 cf}fwfn-manta fmm. H fr;c:sm$3 -from H 1%HF RI. anFfznft of i§§E1:;tis:Js"s Fig. 'lI.;.~'.?.t3: Ff:-:e:z:lip=r3rtaif"rarr1=es -la'1tara;Ely urnbracacl (C) Portal frame laterally unbraced and braced Here, the sidesway can occur only for the entire frame simultaneously. A fixed portal frame, shown in Fig.10.27.13a, is under horizontal load H and compression force P. The moments due to H and P and the total moment diagrams are shown in Fig.10.27.13b, c and d, respectively. The deformations of the frame due to H are shown in Fig.10.27.13a by dotted curves, while the solid curves are the magnified deformations. It is observed that the maximum values of positive and negative M, are at the ends of the column where the maximum Version 2 CE IIT, Kharagpur values of positive and negative moments due to P also occur. Thus, the total momentshall be at the ends as the two effectsare fullyadditive. r§ajlPr::i:ta§frs:rg1:e {gag mnment gagMomentcm is:n Fig. "E .2?.l.4: Fixed portalfranrae »»latetvaily braced Figure 10.27.14a shows a fixed portal frame, laterally braced so that no sideswaycan occur. Figures 10.27.14b and c show the momentsMo and due to P_ It is seen that the maximum values of the two different moments do not occur at the same location.As a result,the magnificationof the momenteither may not be true or shall be small. (D) Columns with different slenderness ratios Version 2 CE IIT, Kharagpur SB asLmg crmlumnimatexrial re t 2: twig ecaéimmn §:bt;%EY§§El%r§§.§ failure} Fig. *:l.2;i?.l§:: Eha'§f%-'l..%lF mt:$ilerr»erl3:EL§fT|:§"§ Figure 10.27.15 shows the interaction diagram of P and M at the midheight section of the column shown in Fig.10.27.10. Three loading paths OA, OB and OC are also shown in the figure for three columns having the same crosssectional area and the eccentricity of loads but with different slenderness ratios. The three columns are loaded with increasing P and M (at constant eccentricity) up to failure. The loading path OA is linear indicating A = 0, i.e., for a very short column. It should be noted that A should be theoretically zero only when either the effective length or the eccentricity is zero. In a practical short column, however, some lateral deflection shall be there, which, in turn will cause additional moment not more than five per cent of the primary moment and may be neglected. The loading path OA terminates at point A of the interaction diagram, which shows the failure load Pscof the short column with moment M30= Psce. The short column fails by crushing of concrete at the mid-height section. This type of failure is designated as material failure, either a tension failure or a compression failure depending on the location of the point A on the interaction curve. The load path OB is for a long column, where the deflection A caused by increasing value of P is significant. Finally, the long column fails at load Pic and moment M,c = P,c(e+ A). The loading path OB further reveals that the secondary moment P,cA is comparable to the primary moment Pice. Moreover, the failure load and the primary moment of the long column Picand Pice, respectively, are less than those of the short column (PS0and PS0e, respectively), though both the columns have the same crosssectional areas and eccentricities but different slenderness ratios. Here also, the mid-height section of the column undergoes material failure, either a compression failure or a tension failure, depending on the location of the point B on the interaction diagram. Version 2 CE IIT, Kharagpur The loading path OC, on the other hand, is for a very long column when the lateral deflection A is so high that the slope of the path dP/dM at C is zero. The column is so slender that the failure is due to buckling (instability) at a comparatively much low value of the load P0,,though this column has the same crosssectional area and the eccentricity of load as of the other two columns. Such instability failure occurs for very slender columns, specially when they are not braced. The following points are summarised from the discussion made in sec.10.27.3. 1. Additional deflections and moments are caused by the axial compression force P in columns. The additional moments increase with the increase of kl/r, when other parameters are equal. 2. Laterally braced compression members and bent in single curvature have the same or nearby locations of the maxima of both Mo and Py. Thus, being fully additive, they have large moment magnification. 3. Laterally braced compression members and bent in double curvature have different locations of the maxima of both Mo and Py. As a result, the moment magnification is either less or zero. 4. Members of frames not braced laterally, the maxima of Mo and Py mostly occur at the ends of column and cause the maximum total moment at the ends of columns only. Additional moments and additional deflections increase with the increase of kl/r. 10.27.4 Effective Length of Columns Annex E of IS 456 presents two figures (Figs.26 and 27) and a table (Table 26) to estimate the effective length of columns in frame structures based on a research paper, Effective length of column in multistoreyed building by R.H. Wood in The Structural Engineer Journal, No.7, Vol.52, July 1974. Figure 26 is for columns in a frame with no sway, while Fig.27 is for columns in a frame with sway. These two figures give the values of k (i.e., I9/I) from two parameters ,6,and/2,which are obtained from the following expression: ,3 = ZK,/ZK, +219, (10.74) where Kc and Kb are flexural stiffnesses of columns and beams, respectively. The quantities ,3, and,6, at the top and bottom joints A and B, respectively, are determined by summing up the K values of members framing into a joint at top Version 2 CE IIT, Kharagpur and bottom, respectively. Thus ,3, and,3, for the frame shown in Fig.10.27.16 are as follows: l = (Kc + Kct)/(Kc+ Kct + Kbt + Kb2) (10.75) g = (Kc + Kcb)/(Kc+ Kcb+ Kb3+ Kb4) (10.76) However, assuming idealised conditions, the effective length in a given plane may be assessed from Table 28 in Annex E of IS 456, for normal use. 10.27.5 Determination of Sway or No Sway Column Clause E-2 of IS 456 recommends the stability index Q to determine if a column is a no sway or sway type. The stability index Q is expressed as: Q = Z P,,Au/Huh, (10.77) whereZ Pu = sumof axialloadsonallcolumnsinthestorey, All Hu = elastically computed firstorder lateral deflection, = total lateral force acting within the storey, and Version 2 CE IIT, Kharagpur h, = height of the storey. The column may be taken as no sway type if the value of Q is S 0.4, otherwise, the column is considered as sway type. 1027.6 Design of Slender Columns The design of slender columns, in principle, is to be done following the same procedure as those of short columns. However, it is essential to estimate the total moment i.e., primary and secondary moments considering P-A effects. These secondary moments and axial forces can be determined by second-order rigorous structural analysis particularly for unbraced frames. Further, the problem becomes more involved and laborious as the principle of superposition is not applicable in second-order analysis. However, c|.39.7 of IS 456 recommends an alternative simplified method of determining additional moments to avoid the laborious and involved secondorder analysis. The basic principle of additional moment method for estimating the secondary moments is explained in the next section. 10.27.7 Additional Moment Method In this method, slender columns should be designed for biaxial eccentricities which include secondary moments (Py of Eq.10.63 and 10.65) about major and minor axes. We first consider braced columns which are bent symmetrically in single curvature and cause balanced failure i.e., Pu = Pub. (A) Braced columns bent symmetrically in single curvature and undergoing balanced failure For braced columns bent symmetrically in single curvature, we have from Eqs.10.63 and 10.65, M = Mo+Py = Mo+Pea = M,,+M, (10.78) where P is the factored design load Pu, M are the total factored design moments Mux and Muy about the major and minor axes, respectively; M, are the primary factored moments Mouxand Mouyabout the major and minor axes, respectively; Ma are the additional moments Max and May about the major and minor axes, respectively and ea are the additional eccentricities eaxand ea, along the minor and major axes, respectively. The quantities Mo and P of Eq.10.78 are known and hence, it is required to determine the respective values of ea, the additional eccentricities only. Version 2 CE IIT, Kharagpur Let us consider the columns of Figs.10.27.10 and 11 showing A as the maximum deflection at the midheight section of the columns. The column of Fig.10.27.10, having a constant primary moment Mo, causes constant curvature (is,while the column of Fig.10.27.11, having a linearly varying primary moment with a maximum value of Mo maxat the midheight section of the column, has a linearly varying curvature with the maximum curvature of ¢maxat the midheight section the column. The two maximum curvatures can be expressed in terms of their respective maximum deflection A as follows: max= 8A/lf The constant curvature (Fig.10.27.10) (15 (10.79) The linearlyvarying curvature(Fig.10.27.11)¢m = 12A/If (10.80) where /6 are the respective effective lengths kl of the columns. We, therefore, consider the maximum ¢ as the average value lying in between the two values of Eqs.10.79 and 80 as ¢m : 10A/if (10.81) Accordingly, the maximum additional eccentricities ea, which are equal to the maximum deflections A, can be written as ea = A = ¢z§/10 (10.82) Version 2 CE IIT, Kharagpur ta}; Lctngitudinet section gut Cmsss sentient %wmg gs;St-rratnpmfite ti ,.'31? Ei*5,,+#1, EH {Br5:? Fig. 10.2?.*l?: Maettmaumcurvature at midnheigzht seatingwhe, E iP,,,, Assuming the column undergoes a balanced failure when Pu = Pub,the maximum curvature at the midheight section of the column, shown in Figs.10.27.17a and b, can be expressed as given below, assuming (i) the values of ac = 0.0035, 5,, = 0.002 and d/D = 0.1, and (ii) the additional moment capacities are about eighty per cent of the total moment. ¢ = eighty per cent of {(0.0035 + 0.002)/0.9D} (see Fig.10.27.17c) or ¢ = 1/200D (10.83) Substituting the value of ¢ in Eq.10.82, ea = D(Ie/D)2/2000 (10.84) Therefore, the additional moment Ma can be written as, Version 2 CE IIT, Kharagpur Ma = Py = PA = Pea = (PD/2000)(Ia/D)2 Thus, the additional moments Max and May about the major and minor axes, respectively, are: Max= (PUD/2000) (lax/D)2 (10.86) Ma,= (Pub/2000) (Ia,/b)2 (10.87) where Pu = axial load on the member, IGX effective length in respect of the major axis, lay = effective length in respect of the minor axis, D = depth of the crosssection at right angles to the major axis, and b = width of the member. Clause 39.7.1 of IS 456 recommends the expressions of Eqs.10.86 and 87 for estimating the additional moments Max and May for the design. These two expressions of the additional moments are derived considering the columns to be braced and bent symmetrically undergoing balanced failure. Therefore, proper modifications are necessary for different situations like braced columns with unequal end moments with the same or different signs, unbraced columns and columns causing compression failure i.e., when Pu > Pub. (B) Braced columns subjected to unequal primary moments at the two ends For braced columns without any transverse loads occurring in the height, the primary maximum moment (Mo maxof Eq.10.64), with which the additional moments of Eqs.10.86 and 87 are to be added, is to be taken as: Momax = M1 + M2 (10.88) (10.89) where M2 is the larger end moment and M1 is the smaller end moment, assumed to be negative, if the column is bent in double curvature. Version 2 CE IIT, Kharagpur To eliminatethe possibilityof total moment Mu maxbecomingless than M2 for columnsbent in double curvature(see Fig.10.27.12) with M1 and M2 having oppositesigns,anotherconditionhas been imposedas Mu max 2 M2 (10.90) The above recommendationsare given in notesof c|.39.7.1 of IS 456. (C) Unbraced columns Unbracedframes undergoconsiderabledeflectiondue to P-A effect. The additionalmomentsdeterminedfrom Eqs.10.86 and 87 are to be added with the maximum primary moment Mo maxat the ends of the column. Accordingly,we have Momax = M2 '5' Ma (10.91) The above recommendationis given in the notes of c|.39.7.1 of IS 456. (D) Columns undergoing compression failure (Pu > Pub) It has been mentioned in part A of this section that the expressionsof additionalmoments given by Eqs.10.86 and 10.87 are for columns undergoing balanced failure (Fig.10.27.17). However, when the columncauses compression failure, the e/D ratio is much less than that of balanced failure at relativelyhigh axial loads. The entire section may be under compressioncausing much less curvatures.Accordingly,additionalmoments of Eqs.10.86 and 10.87 are to be modifiedby multiplyingwith the reductionfactork as given below: For Pu > Pubx: kax = (Puz Pu)/(Puz Pubx) (10.92) Pu > (10.93) kay = (Puz Pu)!(Puz with a conditionthat kaxand keyshouldbe 3 1 (10.94) where Pu = axial load on compressionmember Pu, is given in Eq.10.59 of Lesson26 and is, Puz = fckAc+ ... Version 2 CE IIT, Kharagpur P,,,,X,Puby = axial loads with respect to major and minor axes, respectively, corresponding to the condition of maximum compressive strain of 0.0035 in concrete and tensile strain of 0.002 in outermost layer of tension steel. It is seen from Eqs.10.92 and 10.93 that the values of k (kaxand key)vary linearly from zero (when Pu = Puz)to one (when Pu = Pub).Since Eqs.10.92 and 10.93 are not applicable for Pu < Pub, another condition has been imposed as given in Eq.10.94. The above recommendations are given in cl.39.7.1.1 of IS 456. The following discussion is very important for the design of slender columns. Additional moment method is one of the methods of designing slender columns as discussed in A to D of this section. This method is recommended in cl.39.7 of IS 456 also. The basic concept here is to enhance the primary moments by adding the respective additional moments estimated in a simple way avoiding laborious and involved calculations of second-order structural analysis. However, these primary moments under eccentric loadings should not be less than the moments corresponding to the respective minimum eccentricity, as stipulated in the code. Hence, the primary moments in such cases are to be replaced by the minimum eccentricity moments. Moreover, all slender columns, including those under axial concentric loadings, are also to be designed for biaxial bending, where the primary moments are zero. In such cases, the total moment consisting of the additional moment multiplied with the modification factor, if any, in each direction should be equal to or greater than the respective moments under minimum eccentricity conditions. As mentioned earlier, the minimum eccentricity consideration is given in cl.25.4 of IS 456. 10.27.8 illustrative Example The following illustrative example is taken up to explain the design of slender columns. The example has been solved in step by step using (i) the equations of Lessons 21 to 27 and (ii) employing design charts and tables of SP16, to compare the results. Version 2 CE IIT, Kharagpur 5%f:3fTf5T?}fi."*;'fi?3FE9i~3l.%5 .3 g. l..E?.."EE: ?z:::1bl&1=i1"; 1 Problem 1: Determine the reinforcement required for a braced column against sidesway with the following data: size of the column = 350 x 450 mm (Fig.10.27.18); concrete and steel grades = M 30 and Fe 415, respectively; effective lengths Iexand I6, = 7.0 and 6.0 m, respectively; unsupported length I = 8 m; factored load Pu = 1700 kN; factored moments in the direction of larger dimension = 70 kNm at top and 30 kNm at bottom; factored moments in the direction of shorter dimension = 60 kNm at top and 30 kNm at bottom. The column is bent in double curvature. Reinforcement will be distributed equally on four sides. Solution 1: Step 1: Checking of slenderness ratios Iex/D = 7000/450 = 15.56 > 12, Iey/b = 6000/350 = 17.14 > 12. Hence, the column is slender with respect to both the axes. Step 2: Minimum eccentricities and moments eccentricities (Eq.10.3 of Lesson21) ex,,,,-,,= I/500 + D/30 = 8000/500+ due to minimum 450/30 = 31.0 > 20 mm eymin= I/500 + b/30 = 8000/500 +350/30 = 27.67 > 20 mm Version 2 CE IIT, Kharagpur Max(Min.ecc.) = Pu(6xmin) = (1700)(31)(103) = 52.7kNm Mgy (Min.ecc.)= P,,(ey,,,,-,,) = (1700)(27.67)(103)= 47.04kNm Step 3: Additional eccentricities and additional moments Method 1: Using Eq. 10.84 eax= D(lex/D)2/2000 = (450)(7000/450)3/2000 = 54.44mm ea, = b(Iex/b)2/2000 = (350)(6000/350)3/2000 = 51.43mm 5 II Pu(eax)= (1700)(54.44)(103) = 92.548kNm May= Pu(eay) = (1700)(51.43)(103)= 37.43kNm Method 2: Table I of SP-16 For Iex/D = 15.56, Table I of SP-16 gives: eax/D = 0.1214, whichgives eax= (0.1214) (450) = 54.63 mm For Ie,/D = 17.14, Table I of SP-16 gives: ea,/b = 0.14738, whichgives eay= (0.14738) (350) = 51.583 mm It is seen that values obtainedfrom Table I of SP-16 are comparablewith those obtainedby Eq. 10.84 in Method 1. Step 4: Primary moments and primary eccentricities (Eqs.10.88 and 89) Max = 0.6M20.4M1 = 0.6(70)0.4(30) 0.4 M2 (= 23 kNm). Hence, o.k. May = 0.6M2 0.4M1 = 0.6(60) 0.4 M2 (= 24 kNm). Hence, o.k. = 30 kNm, whichshouldbe 2 0.4(30) = 24 kNm, whichshouldbe 2 Primaryeccentricities: ex = Max/Pu= (30/1700)(103)= 17.65mm ey = Mo,/Pu= (24/1700) (103)= 14.12mm Version 2 CE IIT, Kharagpur Since, both primary eccentricities are less than the respective minimum eccentricities (see Step 2), the primary moments are revised to those of Step 2. So, M0,,= 52.7 kNm and Mo, = 47.04 kNm. Step 5: Modification factors To determine the actual modification factors, the percentage of longitudinal reinforcement should be known. So, either the percentage of longitudinal reinforcement may be assumed or the modification factor may be assumed which should be verified subsequently. So, we assume the modification factors of 0.55 in both directions. Step 6: Total factored moments = 52.7 + 50.9 = 103.6 kNm Mu, = M0,,+ (Modification factor) (May) = 47.04 + (0.55) (87.43) = 47.04 + 48.09 = 95.13 kNm Step 7: Trial section (Eq.10.61 of Lesson 26) The trial section is determined from the design of uniaxial bending with Pu = 1700kNandM, = 1.15(M; +Mfy)2.So,we haveM, = (1.15){(103.6)2 + (95.13)2}"2 = 161.75kNm.Withthesevaluesof P, (= 1700kN) andM, (= 161.75 kNm), we use chart of SP-16 for the d/D = 0.134. We assume the diameters of longitudinal bar as 25 mm, diameter of lateral tie = 8 mm and cover = 40 mm, to get d = 40 + 8 + 12.5 = 60.5 mm. Accordingly, d/D = 60.5/450 = 0.134 and d/b = 60.5/350 = 0.173. We have: P,/fckbD = 1700(103)/(30)(350)(450) = 0.3598 M,/fckbD2 = 161.75(106)/(30)(350)(450)(450) = 0.076 We have to interpolate the values of p/fck for d/D = 0.134 obtained from Charts 44 (for d/D = 0.1) and 45 (d/D = 0.15). The values of p/fck are 0.05 and 0.06 from Charts 44 and 45, respectively. The corresponding values of p are 1.5 and 1.8 per cent, respectively. The interpolated value of p for d'/ D = 0.134 is 1.704per cent, whichgivesAse= (1.704)(350)(450)/100 = 2683.8mm2.We use4-25+ 4-20(1963+ 1256= 3219mm'), to have p provided= 2.044percent giving p/fck= 0.068. Step 8: Calculation of balanced loads Pb Version 2 CE IIT, Kharagpur The values of PM and Pby are determined using Table 60 of SP-16. For this purpose, two parameters k1 and kg are to be determined first from the table. We have p/fck = 0.068, d/D = 0.134 and d/b = 0.173. From Table 60, k1 = 0.19952 and kg = 0.243 (interpolated for d'/ D = 0.134) for Pbx. So, we have: Pbx/fckbD= k1 + kg (p/fog)= 0.19952 + 0.243(0.068) = 0.216044, which gives PM = 0.216044(30)(350)(450)(10'3) = 1020.81kN. Similarly,for Pby: d/b = 0.173,p/fck= 0.068. From Table 60 of SP-16,k1 = 0.19048 and kg = 0.1225 (interpolated for d/b = 0.173). This gives Pb,/fckbD= 0.19048 + 0.1225(0.068) = 0.19881, which gives Pby = (0.19881)(30)(350)(450)(10'3) = 939.38kN. Since, the values of Pm and Pbyare less than Pu, the modification factors are to be used. Step 9: Determination of Pu, Method 1: From Eq.10.59 of Lesson 26 Puz = 0.45 fckAg + (0.75 fy 0.45 fck)Asc = 0.45(30)(350)(450) +{0.75(415)0.45(30)}(3219) = 3084.71 kN Method 2: Using Chart 63 of SP-16 WegetPug/Ag = 19.4N/mm2 fromChart63 of SP-16usingp = 2.044per cent. Therefore,Pu, = (19.4)(350)(450)(10'3) = 3055.5 kN, which is in good agreement with that of Method 1. Step 10: Determination of modification factors Method 1: From Eqs.10.92 and 10.93 kax = (PuzPu)/(PuzPubx) or kax = (3084.71 -1700)/(3084.71 (10-92) 1020.81) = 0.671 and key = (Pug Pu)/(Pug Puby) or (10.93) kay = (3084.71 -1700)/(3084.71 -939.39) The values of the two modification factors = 0.645 are different from the assumed value of 0.55 in Step 5. However, the moments are changed and the section is checked for safety. Version 2 CE IIT, Kharagpur Method 2: From Chart 65 of SP-16 From Chart 65 of SP-16, for the two parameters, Pbx/Pu, = 1020.81/3084.71 = 0.331 and P,/Pu, = 1700/3084.71 = 0.551, we get kax= 0.66. Similarly, for the two parameters, Pb,/Pu = 939.38/3084.71 = 0.3045 and P,/Pu = 0.551, we have kay = 0.65. Values of kax and key are comparable with those of Method 1. Step 11: Total moments incorporating modification factors Mux = M0,,(from Step 4) + (kax)Max(from Step 3) = 52.7 + 0.671(92.548) = 114.8 kNm Muy = Moy(from Step 4) + kay(May)(from Step 3) = 47.04+ (0.645)(87.43) = 103.43 kNm. Step 12: Uniaxial moment capacities The two uniaxial moment capacities Muxt and Muy1are determined as stated: (i) For Mm, by interpolating the values obtained from Charts 44 and 45, knowing the values of P,/fckbD = 0.3598 (see Step 7), p/fck = 0.068 (see Step 7), d'/ D = 0.134 (see Step 7), (ii) for Muy1,by interpolatingthe values obtainedfrom Charts 45 and 46, knowing the same values of P,/fckbD and p/fck as those of (i) and d'/ D = 0.173 (see Step 7). The results are given below: (i) Mm/r,kbD2= 0.0882(interpolated between0.095and0.085) (ii) Muyi/fckbbz = 0.0827(interpolated between 0.085and0.08) So, we have, Maxi = 187.54 kNm and Muy1= 136.76 kNm. Step 13: Value of an Method 1: From Eq.10.60 of Lesson 26 We have P,/Pu, = 1700/3084.71 = 0.5511. From Eq.10.60 of Lesson 26, we have a,, = 0.67 + 1.67 (P,/Puz)= 1.59. Method 2: Interpolating the values between (Pg/Puz)= 0.2 and 0.6 The interpolated value of an = 1.0 + (0.5511 0.2)/0.6 = 1.5852. Both the values are comparable. We use an = 1.5852. Version 2 CE IIT, Kharagpur Step 14: Checking of column for safety Method 1: From Eq.10.58 of Lesson 26 We have in Lesson 26: (Mm/Mm)"+(Muy/Muy1)"" S1 (10.58) Here, Putting the values of Mux, Mm, Muy, MW1 and or, we get: (114.8/187.54)-5452 + (103.43/136.76)-5852 = 0.4593+ 0.6422= 1.1015.Hence, the section Method or the reinforcement 2: Chart has to be revised. 64 of SP-16 The point having the values of (Mux/Mm) = 114.8/187.54 = 0.612 and (Mu,/M,,y1)= 103.43/136.76 = 0.756 gives the value of P,/P, more than 0.7. The value of P,/Pu, here is 0.5511 (see Step 13). So, the section needs revision. We revisefromStep7 by providing8-25mmdiameterbars(= 3927mm2, p = 2.493 per cent and p/fck = 0.0831) as the longitudinal reinforcement keeping the values of b and D unchanged. The revised section is checked furnishing the repeated calculations from Step 8 onwards. The letter R is used before the number of step to indicate this step as revised one. Step R8: Calculation of balanced loads Pb Table 60 of SP-16 gives k1 = 0.19952, and kg = 0.243. We have p/fck 0.0831now. So, Pbx = {0.19952+ (0.243)(0.0831)} (30)(350)(450)(10'3) 1038.145 kN. Similarly, k1 = 0.19048, K2= 0.1225 and p/fck = 0.0831 give Pby {0.19048+ (0.1225)(0.0831)} (30)(350)(450)(10'3) = 948.12kN. The values of PM and Pbyare less than Pu (= 1700 kN). So, modification factors are to be incorporated. Step R9: Determination of Pu, (Eq. 10.59 of Lesson 26) Puz= 0.45(30)(350)(450) + {0.75(415) 0.45(30)}(3927) = 3295.514 kN. Step R10: Determination of modification factors (Eqs.10.92 and 10.93) kax = (3295.514 1700)/(3295.514 1038.145) = 0.707 Kay = (3295.514 1700)/(3295.514 948.12) = 0.68 Step R11: Total moments incorporating modification factors Version 2 CE IIT, Kharagpur Mux 52.70 + 0.707(92.548) = 118.13 kNm Muy = 47.04 + 0.68(87.43) = 106.49 kNm Step R12: Uniaxial moment capacities Using Charts 44 and 45 for Mm and Charts 45 and 46 for MW1,we get (i) the coefficient0.1032 (interpolating0.11 and 0.10) and (ii) the coefficient0.0954 (interpolating0.1 and 0.09) for Mm and MW1,respectively. Mm = (0.1032)(30)(350)(450)(450)(10'6) = 219.429kNm Muyt= (0.0954)(30)(450)(350)(350)(10'6) = 157.77kNm Step R13: Value of an (Eq.10.60 of Lesson 26) Pu/PUZ- 1700/3295.514 = 0.5158 whichgives or71 = 1+(0.51580.2)/0.6 = 1.5263 Step R14: Checking of column for safety (Eq.10.58 of Lesson 26) (118.13/219.424)-5263 + (106.49/157.77)"5263 o.9374< 0.3886 + 0.5488 = 1.0 Hence, the revised reinforcement is safe. The section is shown in Fig.10.27.18. 10.27.9 ractice Questions and Problems with Answers Q.1: Define a slender column. Give three reasons for its increasingimportance and popularity. A.1: See sec. 10.27.1. Q.2: Explainthe behaviourof a slendercolumnsubjectedto concentricloading. ExplainEu|ers load. A.2: See sec.10.27.3. Q.3: Choose the correct answer. (A) As the slenderness ratio increases, the strength of concentrically loaded column: (i) increases (ii) decreases Version 2 CE IIT, Kharagpur (B) For braced columns, the effective length is between (i) land 2/ (ii) 0.5! and 2/ (iii) 0.5! and I (C) The critical load of a braced frame is (i) always larger than that of an unbraced column (ii) always smaller than that of an unbraced column (iii) sometimes larger and sometimes smaller than that of an unbraced column A.3: A. (ii), B. (iii), C. (i) Q.4: Explain the behaviour of slender columns under axial load and uniaxial bending, bent in single curvature. A.4: Part (A) of sec. 10.27.3. Q.5: Explain the behaviour of slender columns under axial load and uniaxial bending, bent in double curvature. A.5: Part (B) of sec. 10.27.3. Q.6: Explain the behaviour of columns in portal frame both braced and unbraced. A.6: Part (C) of sec. 10.27.3. E53 Etvfriiflfré? 3% 1| is; §=>va;2?.fE Eli: Version 2 CE IIT, Kharagpur Q.7: Check the column of Fig.10.27.19, if subjected to an axial factored load of Pu = 1500 kN only when the unsupported length of the column = I= 8.0 m, Iex= Iey= 6.0 m, D = 400 mm, b = 300 mm, using concrete of M 20 and steel grade in Fe 415. A.7: Solution: Step 1: Slenderness ratios Le,/D = 6000/400 = 15 >12 Ley/b = 6000/300 = 20 >12 The column Step is slender about both the axes. 2: Minimum eccentricities and moments eccentricities (Eq.10.3 of Lesson 21) due to minimum ex ,,,,-,,= V500 + D/30 = 8000/500 + 400/30 = 29.33 mm > 20 mm e,,,,,,-,,= 8000/500 + 300/30 = 26 mm > 20 mm Step 3: Primary moments Since the column is concentrically loaded, the primary moments are zero. Therefore, the additional moments must be greater than the respective moments due to minimum eccentricity. Step 4: Additional eccentricities and moments (Eq.10.84) eax= D(IeX/D)2/2000 = 400(6000/400)2/2000 = 45 mm > em,-,,(= 29.23 mm) ea, = b(Ie,/b)2/2000 = 300(6000/300)2/2000 = 60 mm > e,,,,,-n (= 26 mm) Step 5: Calculation of balance loads Pb, and Pb, GivenAsc= 3927 mm2(8 bars of 25 mm diametergive p = 3.2725per cent. So, p/fck = 0.1636. Using 8 mm diameter lateral tie, d = 40 + 8 + 12.5 = Version 2 CE IIT, Kharagpur 60.5 mm giving d/D = 60.5/400 = 0.15125 5 0.15 and d/b = 60.5/300 = 0.2017 5 0.20. From Table 60 of SP-16, we get k1 = 0.196 and kg = 0.061. Thus, we have: PM = {0.196+ (0.061)(0.1636)}(20)(300)(400)(1 03) = 494.35kN Similarly, for Pby:k1 = 0.184 and kg = -0.011, we get Pby= {0.184 (0.011)(0.1636)}(20)(300)(400)(10'3) = 437.281kN Since, PM and Pbyare less than Pu (= 1500 kN), modification factors are to be incorporated. Step 6: Determination of Puz(Eq.10.59 of Lesson 26) Puz= 0.45(20)(300)(400) +{0.75(415)-0.45(20)}(3927)(103) = 2266.94 kN Step 7: Determination of modification factors Kax= (2266.94 -1500)/(2266.94 494.35) = 0.433 and key= (2266.94 437.281) = 0.419 1500)/(2266.94 Step 8: Additional moments and total moments Max 15oo(o.433)(45) = 29.2275 kNm May = 1500(0.419)(60) = 37.71 kNm Since, primary moments are zero as the column is concentrically loaded, the total moment shall consist of the additional moments. But, as both the additional moments are less than the respective moment due to minimum eccentricity, the revised additional moments are: Max = 43.995 kNm and May = 39.0 kNm, which are the total moments also. Thus, we have: Mux= 43.995 kNm, Muy= 39.0 kNm and Pu = 1500 kN. Step 9: Uniaxial moment capacities Version 2 CE IIT, Kharagpur We have, P,/fck bD = {1500/(20)(300)(400)}(1000) = 0.625, p/fc = 0.1636 and d/D = 0.15 for Mm; and d/b = 0.2 for Mm. The coefficientsare 0.11 (from Chart 45) and 0.1 (from Chart 46) for Mm and Mu , respectively. So, we get, Mux,= 0.11(20)(300)(400)(400)(10'6) = 225.28kNm,and MW,= 0.1(20)(300)(300)(400)(10'6) = 72.0kNm Step 10: Value of an (Eq.10.60 of Lesson 26) Here, P,/Pu, = 1500/2266.94 = 0.6617. So, we get an = 1.0 + (0.4617/0.6) = 1.7695 Step 11: Checking the column for safety (Eq.10.58 of Lesson 26) (Mm/Mm) +(Muy/Muy1)" S1 Here, (43.995/225.23)-7695 + (39.0/72.0)-7695 = 0.0556 + 0.3379 = 0.3935 < 1 Hence, the column is safe to carry Pu = 1500 kN. 11.27.10 References 1. ReinforcedConcreteLimit State Design,6" Edition,by Ashok K. Jain, Nem Chand & Bros, Roorkee, 2002. 2. LimitStateDesignof ReinforcedConcrete,2 Edition,by P.C.Varghese, PrenticeHa|| of India Pvt. Ltd., New Delhi, 2002. 3. Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of India Pvt. Ltd., New Delhi, 2001. 4. ReinforcedConcreteDesign,2 Edition,by S.UnnikrishnaPillai and Devdas Menon, Tata McGrawHi|| Publishing Company Limited, New Delhi, 2003. 5. Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam, Oxford & |.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004. 6. ReinforcedConcreteDesign, 13 RevisedEdition,by S.N.Sinha,Tata McGrawHi|| Publishing Company. New Delhi, 1990. 7. ReinforcedConcrete,6" Edition,by S.K.Ma||ickand A.P.Gupta,Oxford & IBH Publishing Co. Pvt. Ltd. New Delhi, 1996. 8. Behaviour, Analysis & Design of Reinforced Concrete Structural Elements, by |.C.Sya| and R.K.Ummat, A.H.Whee|er & Co. Ltd., Allahabad, 1989. Version 2 CE IIT, Kharagpur 9. ReinforcedConcreteStructures,3 Edition,by |.C.Sya|and A.K.Goe|, A.H.Whee|er & Co. Ltd., Allahabad, 1992. 10.Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited, New Delhi, 1993. 11.Designof ConcreteStructures,13" Edition,by Arthur H. Nilson,David Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company Limited, New Delhi, 2004. 12.Concrete Technology, by A.M.Neville and J.J.Brooks, ELBS with Longman, 1994. 13.Propertiesof Concrete,4 Edition, 13 Indian reprint, by A.M.Neville, Longman, 2000. 14.Reinforced ConcreteDesignersHandbook,10"Edition,by C.E.Reyno|ds and J.C.Steedman, E & FN SPON, London, 1997. 15.lndianStandardPlain and ReinforcedConcrete Code of Practice(4 Revision), IS 456: 2000, BIS, New Delhi. 16.Design Aids for Reinforced Concrete to IS: 456 1978, BIS, New ReinforcedConcreteLimit State Design,5" Edition,by Ashok K. Jain, Nem Chand & Bros, Roorkee, 1999. 11.27.11 Test 27 with Maximum Marks = 50, Answer all questions. I Solutions Maximum Time = 30 minutes TQ.1: Determine the primary, additional and total moments of the column shown in Fig.10.27.20 for the three different cases: (i) When the column is braced against sidesway and is bent in single curvature. (ii) When the column is braced against sidesway and is bent in double curvature. (iii) When the column is unbraced. Use the following data: Pu = 2000 kN, concrete grade = M 20, steel grade = Fe 415, unsupported length/= 8.0 m, /ex= 7.0 m, I. = 6.0 m, Ase= 6381mm2 (12-25 mm diameter bars), lateral tie = 8 mm diameter @ 250 mm c/c, d = 60.5 mm, D = 500 mm and b = 400 mm. The factored moments are: 70 kNm at top and 40 kNm at bottom in the direction of larger dimension and 60 kNm at top and 30 kNm at bottom A.TQ.1: Solution in the direction of shorter dimension. The following are the common steps for all three cases. Step 1: Slenderness ratios Iex/D= 7000/500 = 14 > 12 and Ie,/b = 6000/400 = 15 > 12 The column Step is slender about both axes. 2: Minimum eccentricities and moments eccentricities (Eq.10.3 of Lesson 21) ex ,,,,-,, due to minimum I/500 + D/30 = 8000/500 + 500/30 = 32.67 mm > 20 mm, and eymin = I/500 + b/30 = 8000/500 + 400/30 = 29.34 mm > 20 mm MX(min.ecc.) = 2000(32.67)(10'3) = 65.34kNm,and My(min.ecc.)= 2000(29.34)(103) = 58.68kNm Step 3: Additional eccentricities eccentricities (Eq.10.84) and moments due to additional eax= D(IeX/D)2/2000 = 500(7000/500)2/2000 = 49 mm > exmm (= 32.67 mm) Version 2 CE IIT, Kharagpur ea, = b(Ie,/b)2/2000 = 400(6000/400)2/2000 = 45 mm > 9ymin(= 29.34 mm) Max P,,(eaX) = (2ooo)(49)(1o'3) = 93 kNm,and Pu(eay) = (2ooo)(45)(1o'3) = 90kNm May Step 4: Calculation of balanced loads Using d/D = 0.121 and p/fck = 3.1905/20 = 0.159525 in Table 60 of SP16, we have k1 = 0.20238 and kg = 0.2755 (by linear interpolation). This gives Pbx= {o.2o233+ o.2755(o.159525)}(2o)(4oo)(5oo)(10'3) = 933.32kN Similarly, d/b = 0.15125 and p/fc = 0.159525 in Table 60 of SP-16 gives k1 = 0.1957 and kg = 0.198625 (by linear interpolation). So, we get Pby= {0.1957+ 0.198625(0.159525)}(20)(400)(500)(10'3) = 909.54kN Both Pm and Pby are smaller than Pu (= 2000 kN). Hence, modification factors are to be incorporated. Step 5: Calculation of Puz(Eq.10.59 of Lesson 26) Puz = fck + fck)Asc = 0.45(20)(400)(500) + {0.75(415) 0.45(20)}(6381) Step 6: = 3728.66 kN Modification factors and revised additional moments (Eqs.10.92 and 10.93) Kax= (3728.66 2000)/(3728.66 983.32) = 0.6297, and kay= (3728.66 2000)/(3728.66 909.54) = 0.6132 The revised additional moments are: Max = 98(0.6297) = 61.71 kNm, and Ma, = 90(0.6132) = 55.19 kNm Now, the different cases are explained. Case (i): Braced column in single curvature Version 2 CE IIT, Kharagpur Primary moments = 0.4 M. + 0.6 Mg, but should be equal to or greater than 0.4 Mg and moment due to minimum eccentricities. So, we get, Max largest of 58 kNm, 28 kNm and 65.34 kNm = 65.34 kNm May = largest of 48 kNm, 24 kNm and 58.68 kNm = 58.68 kNm Additional moments are Max = 61.71 kNm and May = 55.19 kNm (incorporating the respective modification factors). Total moments = Max = Max+ Max = 65.34 + 61.71 = 127.05 kNm > 65.34 kNm (moment due to minimum eccentricity). and due to minimum eccentricity). Case (ii): Braced column in double curvature Primary moments = 0.4 M. + 0.6 Mg, but should be equal to or greater than 0.4M2 and the moment due to minimum eccentricity. So, we get, Max largest of 26 kNm, 28 kNm and 65.34 kNm = 65.34 kNm May = largest of 24 kNm, 24 kNm and 58.68 kNm = 58.68 kNm Additional moments are Max= 61.71 kNm and May= 55.19 kNm 65.34 kNm (moment due to minimum eccentricity). and Muy = 58.68 + 55.19 = minimum eccentricity). Case (iii): 113.87 kNm > 58.68 kNm (moment due to Unbraced column Primary moments = M2 and should be greater than or equal to moment due to minimum eccentricity. Max = 70 kNm > 65.34 kNm (moment due to minimum eccentricity). and May= 60 kNm > 58.68 kNm (moment due to minimum eccentricity). Additional moments are Max= 61 .71 kNm and May= 55.19 kNm 65.34 kNm (moment due to minimum eccentricity). and Version 2 CE IIT, Kharagpur due to minimum eccentricity). 10.27.12 Summary of this Lesson This lesson mentions the reasons of increasing importance and popularity of slender columns and explains the behaviour of slender columns loaded concentrically or eccentrically. The role of minimum eccentricity that cannot be avoided in any practical column is explained for slender columns. The moments due to minimum eccentricities in both directions should be taken into account for a slender column loaded concentrically as it should be designed under biaxial bending. On the other hand, the given primary moments are also to be checked so that they are equal to or greater than the respective moments due to minimum eccentricity for all slender columns. Both braced and unbraced columns, bent in single or double curvatures, are explained. The importance of modification factors of the additional moments due to P-A effect is explained. Effective lengths and important parameter to determine the slenderness ratios are illustrated for different types of support conditions either in single column or when the column is a part of rigid frames. Additional moment method, a simple method for the design of slender columns, is explained, which is recommended in IS 456. Numerical problems in illustrative example, practice problem and test questions will help in understanding and applying the method for the design of slender columns, as stipulated in IS 456. Direct computations from the given equations as well as use of design charts and tables of SP-16 are illustrated for the design. Version 2 CE IIT, Kharagpur