Slender Columns - Civil Engineering (CE) PDF Download

Instructional Objectives: At the end of this lesson, the student should be able to:

• define a slender column,
• give three reasons for its increasing importance and popularity,
• 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- Δ effect and moments due to minimum eccentricities in slender columns, Δ
• identify a column if sway or nonsway type,
• understand the additional moment method for the design of slender columns, 
•  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 l_ex/D and l_ey/b less than twelve are designated as short and otherwise, they are slender, where l_ex and l_ey are 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, 

(iii) innovative structural concepts – specially, the architect’s 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 Euler’s load, which is much less in comparison to that of short columns having equal area of cross-section. The buckling load is termed as Euler’s 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 elasticis”, 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 P_cr at which a member will fail by buckling is as follows: 

P_cr  =  π²EI /(kl)²

where E is the Young’s modulus I is the moment of inertia about the axis of bending, l 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 = Ar², where A is the area of cross-section of the column and r is the radius of gyration, the above equations can be written as,

P_cr  =  π²EA /(kl/r)²  (10.62)

Thus, P_cr of a particular column depends upon kl/r or slenderness ratio. It is worth mentioning that kl is termed as effective length l_e of the column. 

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,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 PIs (points of zero moment), the distance between the PIs 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. 

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 (PIs), if any, the distance between the PIs and the value of k. All these columns have the respective k values greater than one or at least one.

Figures 10.27.7 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, except that the lower ends of the portal frame are hinged. One of the two points of inflection (PIs) 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 Pcr of the hinge-hinge 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 PIs 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 P_cr of hinge-hinge column and effective lengths l_e = kl  of elastic and reinforced concrete columns with different boundary conditions and for a constant unsupported length l

Table 10.14: Critical loads in terms of P_cr of hinge-hinge column and effective lengths l_e = kl  of elastic and reinforced concrete columns with different boundary conditions and for a constant unsupported length l.

 2. * The critical load is slightly less than P_cr of hinge-hinge 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 Sl.No.6 approaches to hinge-hinge 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 hinge-hinge column P_cr and effective lengths le (equal to the distance between two points of inflection P_Is = kl) of elastic and reinforced concrete columns for a constant value of the unsupported length l.  

The stress-strain curve of concrete, as shown in Fig.1.2.1 of Lesson 2, reveals that the initial tangent modulus of concrete E_c is much higher than E_t (tangent modulus at higher stress level). Taking this into account in Eq.10.62, Fig.10.27.9 presents a plot of buckling load P_cr 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)_lim 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)_lim will fail by buckling, while columns having any value of kl/r less than (kl/r)lim 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 0.5l and l, 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. 

10.27.3  Slender Columns under Axial Load and Uniaxial Moment 

(A) Columns bent in single curvature 

 Figure 10.27.10a shows a column bent in single curvature under axial load P less than its critical load P_cr with constant moment Pe. The deflection profile marked by dotted line is due to the constant moment. However, there will be additional moment of P_y at a distance z from the origin (at the bottom of column) which will deflect the column further, as shown by the solid line. The constant moment P_e and additional moment P_y are shown in Fig.10.27.10b. Thus, the total moment becomes 

M  =  M_o + P_y  =  P(e + y)       (10.63) 

The maximum moment is P(e + Δ ) at the mid-height of the column. This, we can write 

M_max  =  M_o + P Δ  =  P(e + Δ )      (10.64) 

This is known as  P - Δ effect

Figure 10.27.11a shows another column whose bending is caused by a transverse load H. The bending moment at a distance z 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 P_cr, causes additional moment resulting in further deflection, marked by solid line in the figure. This additional deflection produces additional moment of P_y at a section z from the origin. The two bending moment diagrams are shown in Fig.10.27.11b. Here again, the total moment is 

M  =  M_o + P_y  =  H_z/2 + P_y  (10.65) 

The maximum moment at the mid-height of the column is  M_max  =  M_o,_max + P Δ  =  Hl/4 + P Δ  (10.66) 

The total moment in Eqs.10.63 and 10.65 consists of the moment M_o that acts in the presence of P and the additional moment caused by P (= P_y). The deflections  y can be computed from y_o, the deflections without the axial load from the expression

y  =  y_o[1/{1 – (P/P_cr)}]       (10.67) 

From Eq.10.64, we have M_max  =  M_o + P Δ  =  M_o + P Δ o[1/{1 – (P/P_cr)}]    (10.68) 

Equation 10.68 can be written as 

 (10.69) 
where  ψ depends on the type of loading and generally varies between ±0.20. Since P/P_cr is always less than one, we can ignore ψ (P/P_cr) term of Eq.10.69, to have 

Mmax  =  M_o/{1 – (P/P_cr)}       (10.70) 

where 1/{1 – (P/P_cr)} 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 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 

Figure 10.27.12a shows a column subjected to equal end moment of opposite signs. From the moment diagrams M_o and P_y (Figs.10.27.12b and c), it is clear that though M_o moments are maximum at the ends, the P_y 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_o is comparatively smaller than M_{o max} at the ends. Accordingly, the total maximum moment is moderately higher than M_{o max}. 
 From the above, it is evident that the moment M_o will be magnified most strongly if the section of M_{o max} coincides 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_o 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 M_o 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  

M_max  =  M_o Cm/{1 – (P/P_cr)}  (10.71) 

where Cm   =  0.6 + 0.4(M₁/M₂)  ≥  0.4      (10.72) 

The moment M₁ is smaller than M₂ and M₁/M₂ is positive if the moments produce single curvature and negative if they produce double curvature. It is further seen from Eq.10.72 that  C_m = 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_o are magnified when axial load P is applied. The deflection can be obtained from

 y  =  y_o [1/{1 – (P/4P_cr)}]       (10.73)

(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_o are at the ends of the column where the maximum values of positive and negative moments due to P also occur. Thus, the total moment shall be at the ends as the two effects are fully additive. 

Figure 10.27.14a shows a fixed portal frame, laterally braced so that no sidesway can occur. Figures 10.27.14b and c show the moments M_o 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 magnification of the moment either may not be true or shall be small. 

(D) Columns with different slenderness ratios 

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 Δ = 0, i.e., for a very short column. It should be noted that Δ 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 P_sc of the short column with moment M_sc = P_sc e. 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  caused by increasing value of P is significant. Finally, the long column fails at load P Δ
lc and moment M_lc = P_lc(e + Δ ). The loading path OB further reveals that the secondary moment P_lc Δ is comparable to the primary moment P_lce. Moreover, the failure load and the primary moment of the long column P_lc and P_lc e, respectively, are less than those of the short column (P_sc and P_sc e, respectively), though both the columns have the same cross-sectional 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. 

The loading path OC, on the other hand, is for a very long column when the lateral deflection Δ 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 P_cr, though this column has the same cross-sectional 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 M_o and P_y. 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 M_o and P_y. As a result, the moment magnification is either less or zero.  

4. Members of frames not braced laterally, the maxima of M_o and P_ymostly 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., le/l) from two parameters β₁ and β₂ which are obtained from the following expression: 

 (10.74)

where K_c and K_b are flexural stiffnesses of columns and beams, respectively. The quantities β₁ and β₂ 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 and bottom, respectively. Thus β₁ and β₂ for the frame shown in Fig.10.27.16 are as follows: 

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: 

   (10.77)

where ∑ P_u  =  sum of axial loads on all columns in the storey,
Δ_u  =  elastically computed first-order lateral deflection,
H_u  =  total lateral force acting within the storey, and 
h_z =  height of the storey.    

The column may be taken as no sway type if the value of Q is ≤ 0.4, otherwise, the column is considered as sway type.

10.27.6  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- Δ 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, cl.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 (P_y 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.,  P_u = P_ub. 

(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  =  M_o + P_y  =  M_o + P_ea  =  M_o + M_a  (10.78) 

where P is the factored design load P_u, M are the total factored design moments M_ux and M_uy about the major and minor axes, respectively; M_o are the primary factored moments M_oux and M_ouy about the major and minor axes, respectively; Ma are the additional moments M_ax and M_ay about the major and minor axes, respectively and ea are the additional eccentricities e_ax and e_ay along the minor and major axes, respectively. The quantities M_o and P of Eq.10.78 are known and hence, it is required to determine the respective values of ea, the additional eccentricities only

Let us consider the columns of Figs.10.27.10 and 11 showing Δ as the maximum deflection at the mid-height section of the columns. The column of Fig.10.27.10, having a constant primary moment M_o, causes constant curvature φ , while the column of Fig.10.27.11, having a linearly varying primary moment with a maximum value of M_{o max} at the mid-height section of the column, has a linearly varying curvature with the maximum curvature of φ _max at the mid-height section the column. The two maximum curvatures can be expressed in terms of their respective maximum deflection Δ as follows:  

The constant curvature (Fig.10.27.10) (10.79)

The linearly varying curvature (Fig.10.27.11)   (10.80)

where le are the respective effective lengths kl of the columns. We, therefore, consider the maximum φ as the average value lying in between the two values of Eqs.10.79 and 80 as 

 (10.81)

Accordingly, the m axim um additio nal eccentricities  ea, which are equal to the maximum deflections Δ , can be written as 

 (10.82) 

Assuming the column undergoes a balanced failure when P_u = P_ub, the maximum curvature at the mid-height section of the column, shown in Figs.10.27.17a and b, can be expressed as given below, assuming

(i) the values of ε_c = 0.0035, ε_st = 0.002 and  d ′ / D = 0.1, and

(ii) the additional moment capacities are about eighty per cent of the total moment.

φ  =  eighty percent of {(0.0035 + 0.002)/0.9D} (see Fig.10.27.17c) 

or φ  =  1/200D  (10.83) 

Substituting the value of φ in Eq.10.82, e_a  =  D(l_e/D)²/2000        (10.84) 

Therefore, the additional moment M_a can be written as, 

M_a  =  P_y  =  P Δ  =  P_ea  =  (PD/2000) (le/D)²  (10.85) 

Thus, the additional moments M_ax and M_ay about the major and minor axes, respectively, are: M_ax  =  (P_uD/2000) (l_ex/D)²  (10.86) 

May  =  (P_ub/2000) (l_ey/b)²  (10.87) 

where P_u   =  axial load on the member,  

l_ex   =  effective length in respect of the major axis,

l_ey   =  effective length in respect of the minor axis,

D   =  depth of the cross-section 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 M_ax and M_ay 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 P_u > P_ub. 

(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 max of Eq.10.64), with which the additional moments of Eqs.10.86 and 87 are to be added, is to be taken as:

M_{o max}  =  0.4 M1 + 0.6 M₂  (10.88) 

and further  M_{o max} ≥  0.4 M₂  (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. 

To eliminate the possibility of total moment M_{u max} becoming less than M₂ for columns bent in double curvature (see Fig.10.27.12) with M₁ and M₂ having opposite signs, another condition has been imposed as M_{u max} ≥ M₂  (10.90) 
 The above recommendations are given in notes of cl.39.7.1 of IS 456.

(C)  Unbraced columns  

Unbraced frames undergo considerable deflection due to P-Δ effect. The additional moments determined from Eqs.10.86 and 87 are to be added with the maximum primary moment M_{o max} at the ends of the column. Accordingly, we have 
M_{o max}  =  M₂ + M_a  (10.91) 
The above recommendation is given in the notes of cl.39.7.1 of IS 456.

(D)  Columns undergoing compression failure (P_u > P_ub) 

 It has been mentioned in part A of this section that the expressions of additional moments given by Eqs.10.86 and 10.87 are for columns undergoing balanced failure (Fig.10.27.17). However, when the column causes compression failure, the e/D ratio is much less than that of balanced failure at relatively high axial loads. The entire section may be under compression causing much less curvatures. Accordingly, additional moments of Eqs.10.86 and 10.87 are to be modified by multiplying with the reduction factor k as given below:

(i)   For P_u > P_ubx:  k_ax  =  (P_uz – P_u)/(P_uz – P_ubx)     (10.92) 

(ii)  For P_u > P_uby:  k_ay  =  (P_uz – Pu)/(P_uz – P_uby)     (10.93) 

with a condition that kax and kay should be ≤ 1      (10.94) 

where P_u  =  axial load on compression member P_uz  is given in Eq.10.59 of Lesson 26 and is,

P_uz  =  0.45 f_ck A_c + 0.75 f_y A_st   …  (10.59) 

P_ubx, P_uby  =  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 (k_ax and k_ay) vary linearly from zero (when P_u = P_uz) to one (when P_u = P_ub). Since Eqs.10.92 and 10.93 are not applicable for P_u < P_ub, 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 mome nt metho d 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.

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 l_ex and l_ey = 7.0 and 6.0 m, respectively; unsupported length l = 8 m; factored load P_u = 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 

l_ex/D  =  7000/450  =  15.56 > 12,

l_ey/b  =  6000/350  =  17.14  >  12.

Hence, the column is slender with respect to both the axes.

Step 2: Minimum eccentricities and moments due to minimum eccentricities (Eq.10.3 of Lesson21) 

e_{x min} =  l/500 + D/30  =  8000/500 + 450/30  =  31.0  >  20 mm

e_{y min} =  l/500 + b/30  =  8000/500 + 350/30  =  27.67  >  20 mm 

M_ox (Min. ecc.)  =  Pu(e_{x min})  =  (1700) (31) (10-3)  =  52.7 kNm

M_oy (Min. ecc.)  =  Pu(e_{y min})  =  (1700) (27.67) (10-3)  =  47.04 kNm

Step 3:  Additional eccentricities and additional moments  

Method 1:  Using Eq. 10.84

e_ax  =  D(l_ex/D)2/2000  =  (450) (7000/450)2/2000  =  54.44 mm

e_ay  =  b(l_ex/b)2/2000  =  (350) (6000/350)2/2000  =  51.43 mm

M_ax  =  P_u(e_ax)  =  (1700) (54.44) (10-3)  =  92.548 kNm

May  =  P_u(e_ay)  =  (1700) (51.43) (10-3)  =  87.43 kNm

Method 2:  Table I of SP-16

For l_ex/D  =  15.56, Table I of SP-16 gives:

e_ax/D  =  0.1214, which gives  e_ax = (0.1214) (450)  =  54.63 mm

For l_ey/D  =  17.14, Table I of SP-16 gives:

e_ay/b  =  0.14738, which gives  e_ay = (0.14738) (350)  =  51.583 mm  

It is seen that values obtained from Table I of SP-16 are comparable with those obtained by Eq. 10.84 in Method 1.

Step 4:  Primary moments and primary eccentricities (Eqs.10.88 and 89) 

M_ox  =  0.6M2 – 0.4M₁  =  0.6(70) – 0.4(30)  =  30 kNm, which should be  ≥ 0.4 M₂ (= 28 kNm).   Hence, o.k.

M_oy  =  0.6M2 – 0.4M₁  =  0.6(60) – 0.4(30)  =  24 kNm, which should be  ≥ 0.4 M₂ (= 24 kNm).   Hence, o.k.

Primary eccentricities: 

e_x  =  M_ox/P_u  =  (30/1700) (103)  =  17.65 mm

e_y  =  M_oy/P_u  =  (24/1700) (103)  =  14.12 mm 

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,  M_ox = 52.7 kNm and M_oy = 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

M_ux  =  M_ox + (Modification factor) (M_ax)  =  52.7 + (0.55) (92.548)          

=  52.7 + 50.9  =  103.6 kNm

M_uy  =  M_oy + (Modification factor) (M_ay)  =  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 P_u= 1700 kN and M_u = 1.15  . So, we have  M _u = (1.15){(103.6)² + (95.13)2}^1/2 = 161.75 kNm. With these values of P_u (= 1700 kN) and M_u (= 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_u/f_ck bD  =  1700(103)/(30)(350)(450)  =  0.3598

M_u/f_ck bD²  =  161.75(106)/(30)(350)(450)(450)  =  0.076

We have to interpolate the values of  p/f_ck for  d′/D = 0.134 obtained from Charts 44 (for d′/D = 0.1) and 45 ( d′/D = 0.15). The values of p/f_ck 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  = 0.134 is 1.704 per cent, which gives A
d′/D s_c = (1.704)(350)(450)/100 = 2683.8 mm². We use 4-25 + 4-20 (1963 + 1256 = 3219 mm²), to have  p provided = 2.044 per cent giving p/f_ck = 0.068.

Step 8:  Calculation of balanced loads P_b

The values of P_bx and P_by are determined using Table 60 of SP-16. For this purpose, two parameters k₁ and k₂ are to be determined first from the table. We have p/f_ck = 0.068, d ′ / D = 0.134 and d ′ / b = 0.173. From Table 60, k₁ = 0.19952 and k₂ = 0.243 (interpolated for d ′ / D = 0.134) for P_bx. So, we have: P_bx/f_ckbD = k₁ + k₂ (p/f_ck) = 0.19952 + 0.243(0.068) = 0.216044, which gives P_bx = 0.216044(30)(350)(450)(10^-3) = 1020.81 kN.

 Similarly, for P_by:   d ′ / b = 0.173, p/f_ck = 0.068. From Table 60 of SP-16, k₁= 0.19048 and k₂ = 0.1225 (interpolated for d ′ / b = 0.173). This gives P_by/f_ckbD = 0.19048 + 0.1225(0.068) = 0.19881, which gives P_by = (0.19881)(30)(350)(450)(10^-3) = 939.38 kN.  

Since, the values of P_bx and P_by are less than P_u, the modification factors are to be used. 

Step 9:  Determination of P_uz

Method 1:  From Eq.10.59 of Lesson 26

P_uz  =  0.45 f_ck A_g + (0.75 fy – 0.45 f_ck) A_sc        

=  0.45(30)(350)(450) + {0.75(415) – 0.45(30)}(3219)  =  3084.71 kN

Method 2: Using Chart 63 of SP-16  We get P_uz/A_g  = 19.4 N/mm² from Chart 63 of SP-16 using p = 2.044 per cent. Therefore, P_uz = (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

k_ax  =  (P_uz – P_u)/(P_uz – P_ubx)  …  (10.92)  

or k_ax  =  (3084.71 – 1700)/(3084.71 – 1020.81)  =  0.671 and

k_ay  =  (P_uz – P_u)/(P_uz – P_uby)  …  (10.93)  

or k_ay  =  (3084.71 – 1700)/(3084.71 – 939.39)  =  0.645

 The values of the two modification factors are different from the assumed value of 0.55 in Step 5. However, the moments are changed and the section is checked for safety. 

Method 2:  From Chart 65 of SP-16  From Chart 65 of SP-16, for the two parameters, P_bx/P_uz = 1020.81/3084.71 = 0.331 and P_u/P_uz = 1700/3084.71 = 0.551, we get k_ax = 0.66. Similarly, for the two parameters, P_by/P_uz = 939.38/3084.71 = 0.3045 and P_u/P_uz =  0.551, we have kay = 0.65. Values of k_ax and k_ay are comparable with those of Method 1.

Step 11:  Total moments incorporating modification factors

M_ux  =  M_ox (from Step 4) + (k_ax) M_ax (from Step 3)          

=  52.7 + 0.671(92.548)  =  114.8 kNm

M_uy  =  M_oy (from Step 4) + k_ay (M_ay) (from Step 3)          

=  47.04 + (0.645)(87.43)  =  103.43 kNm.

Step 12:  Uniaxial moment capacities  

The two uniaxial moment capacities M_ux1 and M_uy1 are determined as stated:

(i) For M_ux1, by interpolating the values obtained from Charts 44 and 45, knowing the values of P_u/f_ckbD = 0.3598 (see Step 7), p/f_ck = 0.068 (see Step 7), d ′ / D = 0.134 (see Step 7),

(ii) for M_uy1, by interpolating the values obtained from Charts 45 and 46, knowing the same values of P_u/f_ckbD and p/fck as those of (i) and d ′ / D = 0.173 (see Step 7). The results are given below:

(i)   M_ux1/f_ckbD²  =  0.0882 (interpolated between 0.095 and 0.085)

(ii)  M_uy1/f_ckbb²  =  0.0827 (interpolated between 0.085 and 0.08)

So, we have, M_ux1 = 187.54 kNm and M_uy1 = 136.76 kNm.

Step 13:  Value of  αn Method 1:  From Eq.10.60 of Lesson 26  

We have P_u/P_uz = 1700/3084.71 = 0.5511. From Eq.10.60 of Lesson 26, we have  α_n = 0.67 + 1.67 (P_u/P_uz) = 1.59.

Method 2: Interpolating the values between (Pu/Puz) = 0.2 and 0.6  

The interpolated value of  α_n = 1.0 + (0.5511 – 0.2)/0.6 = 1.5852. Both the values are comparable. We use  α_n = 1.5852. 

Step 14:  Checking of column for safety 

Method 1: From Eq.10.58 of Lesson 26  We have in Lesson 26:  

 Here, putting the values of M_ux, M_ux1, M_uy, M_uy1 and α_n , we get: (114.8/187.54)1.5452 + (103.43/136.76)1.5852 = 0.4593 + 0.6422 = 1.1015. Hence, the section or the reinforcement has to be revised. 

Method 2: Chart 64 of SP-16  

The point having the values of (M_ux/M_ux1) = 114.8/187.54 = 0.612 and (M_uy/M_uy1) = 103.43/136.76 = 0.756 gives the value of P_u/P_z more than 0.7. The value of P_u/P_uz here is 0.5511 (see Step 13). So, the section needs revision.  We revise from Step 7 by providing 8-25 mm diameter bars (= 3927 mm², p = 2.493 per cent and p/f_ck = 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 P_b  

Table 60 of SP-16 gives k₁ = 0.19952, and k₂ = 0.243. We have p/f_ck = 0.0831 now. So, P_bx  =  {0.19952 + (0.243)(0.0831)} (30)(350)(450)(10^-3)  =  1038.145 kN. Similarly, k₁ = 0.19048, k2 = 0.1225 and p/f_ck = 0.0831 give P_by  =  {0.19048 + (0.1225)(0.0831)} (30)(350)(450)(10^-3)  =  948.12 kN.  

The values of P_bx and P_by are less than P_u (= 1700 kN).

So, modification factors are to be incorporated.

Step R9:  Determination of Puz (Eq. 10.59 of Lesson 26) 

P_uz = 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) 

k_ax  =  (3295.514 – 1700)/(3295.514 – 1038.145)  =  0.707

k_ay  =  (3295.514 – 1700)/(3295.514 – 948.12)  =  0.68

Step R11:  Total moments incorporating modification factors 

M_ux  =  52.70 + 0.707(92.548)  =  118.13 kNm

M_uy  =  47.04 + 0.68(87.43)  =  106.49 kNm 

Step R12:  Uniaxial moment capacities  

Using Charts 44 and 45 for M_ux1 and Charts 45 and 46 for M_uy1, we get (i) the coefficient 0.1032 (interpolating 0.11 and 0.10) and (ii) the coefficient 0.0954 (interpolating 0.1 and 0.09) for M_ux1 and M_uy1, respectively.  

M_ux1  =  (0.1032)(30)(350)(450)(450)(10^-6)  =  219.429 kNm

M_uy1  =  (0.0954)(30)(450)(350)(350)(10^-6)  =  157.77 kNm

Step R13:  Value of  α n (Eq.10.60 of Lesson 26) 

P_u/P_uz  =  1700/3295.514  =  0.5158 which gives  

α_n  =  1 + (0.5158 – 0.2)/0.6  =  1.5263

Step R14:  Checking of column for safety (Eq.10.58 of Lesson 26)  

(118.13/219.424)^1.5263 + (106.49/157.77)^1.5263  =  0.3886 + 0.5488  = 0.9374 < 1.0 

 Hence, the revised reinforcement is safe. The section is shown in Fig.10.27.18. 

Q.7: Check the column of Fig.10.27.19, if subjected to an axial factored load of P_u = 1500 kN only when the unsupported length of the column = l = 8.0 m,  l_ex = l_ey = 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 L_ex/D  =  6000/400  =  15  > 12 L_ey/b  =  6000/300  =  20  > 12 The column is slender about both the axes.

Step 2: Minimum eccentricities and moments due to minimum eccentricities (Eq.10.3 of Lesson 21)

e_{x min}  =  l/500 + D/30  =  8000/500 + 400/30  =  29.33 mm > 20 mm

e_{y min}  = 8000/500 + 300/30  =  26 mm > 20 mm

M_x due to min. e_cc.  =  P_u (e_{x min})  =  1500(29.33)  =  43.995 kNm

M_y due to min. e_cc.  =  P_u (e_{y min})  =  1500(26.0)   =  39.0 kNm

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) 

e_ax  =  D(l_ex/D)²/2000  =  400(6000/400)²/2000  =  45 mm  >  e_{x min} (= 29.23 mm) 
e_ay  =  b(l_ey/b)²/2000  =  300(6000/300)²/2000  =  60 mm  >  e_{y min} (= 26 mm) 

Step 5:  Calculation of balance loads Pbx and Pby

Given A_sc =  3927 mm² (8 bars of 25 mm diameter give  p = 3.2725 per cent. So,  

p/f_ck = 0.1636. Using 8 mm diameter lateral tie, d′ = 40 + 8 + 12.5 = 

60.5 mm giving d ′ /D = 60.5/400 = 0.15125 ≅ 0.15 and d ′ /b = 60.5/300 = 0.2017 ≅ 0.20. 

From Table 60 of SP-16, we get k₁ = 0.196 and k₂ = 0.061. Thus, we have: 

P_bx  =  {0.196 + (0.061)(0.1636)}(20)(300)(400)(10^-3) = 494.35 kN  

Similarly, for P_{b y}: k₁  = 0.184 and k₂ = -0.011, we get

P_by  =  {0.184 - (0.011)(0.1636)}(20)(300)(400)(10^-3) = 437.281 kN  

Since, P_bx an d P_by are less than P_u (= 1500 kN), modification factors are to be incorporated.

Step 6:  Determination of Puz (Eq.10.59 of Lesson 26) 

P_uz  =  0.45(20)(300)(400) + {0.75(415) – 0.45(20)}(3927)(10^-3)  =  2266.94 kN 

Step 7:  Determination of modification factors

k_ax = (2266.94 – 1500)/(2266.94 – 494.35)  =  0.433  and

k_ay = (2266.94 – 1500)/(2266.94 – 437.281)  =  0.419

Step 8:  Additional moments and total moments  

M_ax  =  1500(0.433)(45)  =  29.2275 kNm M_ay  =  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: M_ax = 43.995 kNm and M_ay = 39.0 kNm, which are the total moments also.  

Thus, we have:

M_ux = 43.995 kNm, M_uy = 39.0 kNm and P_u = 1500 kN.

Step 9:  Uniaxial moment capacities  

We have, P_u/f_ck  bD = {1500/(20)(300)(400)}(1000) = 0.625, p/f_ck = 0.1636 and d ′ /D = 0.15 for

M_ux1; and d ′ /b = 0.2 for M_uy1.

The coefficients are 0.11 (from Chart 45) and 0.1 (from Chart 46) for M_ux1 and M_uy1, respectively.

So, we get,  M_ux1  =  0.11(20)(300)(400)(400)(10^-6)  =  225.28 kNm, and

M_uy1  =  0.1(20)(300)(300)(400)(10^-6)  =  72.0 kNm

Step 10:  Value of  α n (Eq.10.60 of Lesson 26)  

Here, P_u/P_uz   =  1500/2266.94  =  0.6617. So, we get 

α_n = 1.0 + (0.4617/0.6)  =  1.7695

Step 11:  Checking the column for safety (Eq.10.58 of Lesson 26) 

Here, (43.9 95/225.28)1.7695 + (39.0/72.0)1.7695  =  0.0556 + 0.3379  =  0.3935  <  1 
 Hence, the column is safe to carry  P_u = 1500 kN. 

11.27.11  Test 27 with Solutions 

Maximum Marks  =  50,    
 Maximum Time  =  30 minutes
 Answer all questions.

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: P_u = 2000 kN, concrete grade = M 20, steel grade = Fe 415, unsupported length l = 8.0 m, l_ex = 7.0 m, l_ey = 6.0 m, A_sc = 6381 mm2 (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 in the direction of shorter dimension. 
′
A.TQ.1: Solution  The following are the common steps for all three cases.

Step 1:  Slenderness ratios

l_ex/D = 7000/500 = 14 > 12 and  l_ey/b = 6000/400 = 15 > 12 The column is slender about both axes.

Step 2: Minimum eccentricities and moments due to minimum eccentricities (Eq.10.3 of Lesson 21)

e_{x min}  =  l/500 + D/30  =  8000/500 + 500/30  =  32.67 mm > 20 mm, and  

e_{y min}  =  l/500 + b/30  =  8000/500 + 400/30  =  29.34 mm > 20 mm

M_x (min. e_cc.)  =  2000(32.67)(10-3)  =  65.34 kNm, and

M_y (min. e_cc.)  =  2000(29.34)(10^-3)   =  58.68 kNm

Step 3:  Additional eccentricities and moments due to additional eccentricities (Eq.10.84)

e_ax  =  D(l_ex/D)²/2000  =  500(7000/500)²/2000  =  49 mm  >  e_{x min} (= 32.67 mm) 

e_ay  =  b(l_ey/b)²/2000  =  400(6000/400)²/2000  =  45 mm  >  e_{y min} (= 29.34 mm) 

M_ax  =  P_u(e_ax)  =  (2000)(49)(10^-3)  =  98 kNm, and M_ay  =  P_u(e_ay)  =  (2000)(45)(10^-3)  =  90 kNm

Step 4:  Calculation of balanced loads  

Using d ′ /D  = 0.121 and p/f_ck = 3.1905/20 = 0.159525 in Table 60 of SP16, we have k₁ = 0.20238 and k₂ = 0.2755 (by linear interpolation).

This gives  P_bx  =  {0.20238 + 0.2755(0.159525)}(20)(400)(500)(10^-3) = 983.32 kN  

Similarly, d ′ /b =  0.15125 and p/f_ck = 0.159525 in Table 60 of SP-16 gives k₁ = 0.1957 and k₂ = 0.198625 (by linear interpolation).

So, we get Pby  =  {0.1957 + 0.198625(0.159525)}(20)(400)(500)(10^-3) = 909.54 kN  

Both P_bx and P_by are smaller than P_u (= 2000 kN).

Hence, modification factors are to be incorporated.

Step 5:  Calculation of Puz (Eq.10.59 of Lesson 26)

P_uz  = 0.45 f_ck A_g + (0.75 f_y – 0.45 f_ck)

A_sc  =  0.45(20)(400)(500) + {0.75(415) – 0.45(20)}(6381)  =  3728.66 kN

Step 6:  Modification factors and revised additional moments (Eqs.10.92 and 10.93) 

k_ax = (3728.66 - 2000)/(3728.66 – 983.32)  =  0.6297,  and

k_ay = (3728.66 - 2000)/(3728.66 – 909.54)  =  0.6132 The revised additional moments are:

M_ax  =  98(0.6297)  =  61.71 kNm, and

M_ay  =  90(0.6132)  =  55.19  kNm Now, the different cases are explained.

Case (i):  Braced column in single curvature 

Primary moments = 0.4 M1 + 0.6 M₂, but should be equal to or greater than 0.4 M₂ and moment due to minimum eccentricities. So, we get,

M_ox  =  largest of 58 kNm, 28 kNm and 65.34 kNm = 65.34 kNm

M_oy  =  largest of 48 kNm, 24 kNm and 58.68 kNm = 58.68 kNm  Additional moments are

M_ax = 61.71 kNm and M_ay = 55.19 kNm (incorporating the respective modification factors).  

Total moments  =  M_ux  =  M_ox + M_ax  =  65.34 + 61.71  =  127.05 kNm > 65.34 kNm (moment due to minimum eccentricity), and

M_uy  =  M_oy + M_ay  =  58.68 + 55.19  =  113.87 kNm > 58.68 kNm (moment due to minimum eccentricity).

Case (ii):  Braced column in double curvature

Primary moments = - 0.4 M₁ + 0.6 M_2, but should be equal to or greater than 0.4M₂ and the moment due to minimum eccentricity. So, we get,

M_ox  =  largest of 26 kNm, 28 kNm and 65.34 kNm = 65.34 kNm Moy  =  largest of 24 kNm, 24 kNm and 58.68 kNm = 58.68 kNm  

Additional moments are M _ax =  61.71 kNm and M_ay = 55.19 kNm  

Final moments  =  M_ux  =  M_ox + M_ax  =  65.34 + 61.71  =  127.05 kNm > 65.34 kNm (moment due to minimum eccentricity), and

M_uy =  58.68 + 55.19  =  113.87 kNm > 58.68 kNm (moment due to minimum eccentricity).

Case (iii):  Unbraced column

Primary moments = M₂ and should be greater than or equal to moment due to minimum eccentricity.  M_ox  =  70 kNm > 65.34 kNm (moment due to minimum eccentricity), and M_oy = 60 kNm > 58.68 kNm (moment due to minimum eccentricity).  

Additional mome nts are M _ax =  61.71 kNm and M_ay = 55.19 kNm   

Final moments  =  M_ux  =  M_ox + M_ax  =  70.0 + 61.71  =  131.71 kNm > 65.34 kNm (moment due to minimum eccentricity), and 

M_uy =  M_oy + M_ax  =  60.0 + 55.19  =  115.19 kNm > 58.68 kNm (moment due to minimum eccentricity). 

 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- Δ 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.