Module 10 Compressio Members Versio 2 CE IIT, Kharagpur
Lesso 26 Short Compressio Members uder Axial Load with Biaxial Bedig Versio 2 CE IIT, Kharagpur
Istructioal Objectives: At the ed of this lesso, the studet should be able to: uderstad the behaviour of short colums uder axial load ad biaxial bedig, uderstad the cocept of iteractio surface, idetify the load cotour ad iteractio curves of P u -M u i a iteractio surface, metio the limitatio of direct applicatio of the iteractio surface i solvig the problems, explai the simplified method of desig ad aalysis of short colums uder axial load ad biaxial bedig, apply the IS code method i desigig ad aalysig the reiforced cocrete short colums uder axial load ad biaxial bedig. 10.26.1 Itroductio Beams ad girders trasfer their ed momets ito the corer colums of a buildig frame i two perpedicular plaes. Iterior colums may also have biaxial momets if the layout of the colums is irregular. Accordigly, such colums are desiged cosiderig axial load with biaxial bedig. This lesso presets a brief theoretical aalysis of these colums ad explais the difficulties to apply the theory for the desig. Thereafter, simplified method, as recommeded by IS 456, has bee explaied with the help of illustrative examples i this lesso. Versio 2 CE IIT, Kharagpur
10.26.2 Biaxial Bedig Figures 10.26.1a ad b preset colum sectio uder axial load ad uiaxial bedig about the pricipal axes x ad y, respectively. Figure 10.26.1c Versio 2 CE IIT, Kharagpur
presets the colum sectio uder axial load ad biaxial bedig. The eccetricities e x ad e y of Fig.10.26.1c are the same as those of Fig.10.26.1a (for e x ) ad Fig.10.26.1b (for e y ), respectively. Thus, the biaxial bedig case (case c) is the resultat of two uiaxial bedig cases a ad b. The resultat eccetricity e, therefore, ca be writte as (see Fig.10.26.1c): e = + (10.55) 2 2 1/ 2 ( e x ey ) Desigatig the momets of cases a, b ad c by M ux, M uy ad M u, respectively, we ca write: M = + u (10.56) 2 2 ( M ux M uy ) 1/ 2 ad the resultat M u is actig about a iclied axis, so that taθ = e x /e y = M uy /M ux (10.57) the agle of icliatio θ is measured from y axis. This iclied resultat axis shall also be the pricipal axis if the colum sectio icludig the reiforcig bars is axisymmetric. I such a situatio, the biaxial bedig ca be simplified to a uiaxial bedig with the eutral axis parallel to the resultat axis of bedig. The reiforced cocrete colum cross-sectios are, i geeral, oaxisymmetric with referece to the logitudial axis ad, therefore, the eutral axis is ot parallel to the resultat axis of bedig (θ is ot equal to λ i Fig.10.26.1c). Moreover, it is extremely laborious to fid the locatio of the eutral axis with successive trials. However, failure strai profile ad stress block ca be draw for a give locatio of the eutral axis. Figs.10.25.1d ad e preset the strai profile ad stress block, respectively, of the sectio show i Fig.10.25.1c. Versio 2 CE IIT, Kharagpur
10.26.3 Iteractio Surface Figure 10.26.2 ca be visualised as a three-dimesioal plot of P u -M ux - M uy, wherei two two-dimesioal plots of P u -M uy ad P u -M us are marked as case (a) ad case (b), respectively. These two plots are the iteractio curves for the colums of Figs.10.26.1a ad b, respectively. The evelope of several iteractio curves for differet axes will geerate the surface, kow as iteractio surface. The iteractio curve marked as case (c) i Fig.10.26.2, is for the colum uder biaxial bedig show i Fig.10.26.1c. The correspodig axis of bedig is makig a agle θ with the y axis ad satisfies Eq.10.57. It has bee explaied i Lesso 24 that a colum subjected to a pair of P ad M will be safe if their respective values are less tha P u ad M u, give by its iteractio curve. Extedig the same i the three-dimesioal figure of iteractio surface, it is also acceptable that a colum subjected to a set of P u, M uy ad M ux is safe if the set of values lies withi the surface. Sice P u is chagig i the directio of z, let us desigate the momets ad axial loads as metioed below: M uxz = desig flexural stregth with respect to major axis xx uder biaxial loadig, whe P u = P uz, Versio 2 CE IIT, Kharagpur
M uyz = desig flexural stregth with respect to mior axis yy uder biaxial loadig, whe P u = P uz, M ux1 = desig flexural stregth with respect to major axis xx uder uiaxial loadig, whe P u = P uz, ad M uy1 = desig flexural stregth with respect to mior axis yy uder uiaxial loadig, whe P u = P uz. The above otatios are also show i Fig.10.26.2. All the iteractio curves, metioed above, are i plaes perpedicular to xy plae. However, the iteractio surface has several curves parallel to xy plae, which are plaes of costat P u. These curves are kow as load cotour, oe such load cotour is show i Fig.10.26.2, whe P u = P uz. Needless to metio that the load is costat at all poits of a load cotour. These load cotour curves are also iteractio curves depictig the iteractio betwee the biaxial bedig capacities. 10.26.4 Limitatio of Iteractio Surface The mai difficulty i preparig a exact iteractio surface is that the eutral axis for the case (c) of Fig.10.26.1c will ot, i geeral, be perpedicular to the lie joiig the loadig poit P u ad the cetre of the colum (Fig.10.26.1c). This will require several trials with c ad λ, where c is the distace of the eutral axis ad λ agle made by the eutral axis with the x axis, as show i Fig.10.26.1c. Each trial will give a set of P u, M ux ad M uy. Oly for a particular case, the eutral axis will be perpedicular to the lie joiig the load poit P u to the cetre of the colum. This search makes the process laborious. Moreover, several trials with c ad λ, givig differet values of h (see Fig.10.26.1c), may result i a failure surface with wide deviatios, particularly as the value of P u will be icreasig. Accordigly, the desig of colums uder axial load with biaxial bedig is doe by makig approximatios of the iteractio surface. Differet coutries adopted differet approximate methods. Clause 39.6 of IS 456 recommeds oe method based o Bresler's formulatio, also kow as "Load Cotour Method", which is take up i the followig sectio. (For more iformatio, please refer to: "Desig Criteria for Reiforced Colums uder Axial Load ad Biaxial Bedig", by B. Bresler, J. ACI, Vol.32, No.5, 1960, pp.481-490). 10.26.5 IS Code Method for Desig of Colums uder Axial Load ad Biaxial Bedig Versio 2 CE IIT, Kharagpur
IS 456 recommeds the followig simplified method, based o Bresler's formulatio, for the desig of biaxially loaded colums. The relatioship betwee M uxz ad M uyz for a particular value of P u = P uz, expressed i o-dimesioal form is: ( / ) M 1 + ( / 1) ux M ux M uy M uy (10.58) 1 where M ux ad M uy = momets about x ad y axes due to desig loads, ad is related to P u /P uz, (Fig.10.26.3), where P uz = 0.45 f ck A c + 0.75 f y A sc = 0.45 A g + (0.75 f y - 0.45 f ck ) A sc (10.59) where A g = gross area of the sectio, ad A sc = total area of steel i the sectio M uxz, M uyz, M ux1 ad M uy1 are explaied i sec.10.26.3 earlier. It is worth metioig that the quatities M ux, M uy ad P u are due to exteral loadigs applied o the structure ad are available from the aalysis, whereas M ux1, M uy1 ad P uz are the capacities of the colum sectio to be cosidered for the desig. Equatio 10.58 defies the shape of the load cotour, as explaied earlier (Fig.10.26.2). That is why the method is also kow as "Load Cotour Method". The expoet of Eq.10.58 is a costat which defies the shape of the load Versio 2 CE IIT, Kharagpur
cotour ad depeds o the value of P u. For low value of the axial load, the load cotour is approximated as a straight lie ad, i that case, = 1. O the other had, for high values of axial load, the load cotour is approximated as a quadrat of a circle, whe = 2. For itermediate load values, the value of lies betwee 1 ad 2. Chart 64 of SP-16 presets the load cotour ad Fig.10.26.3 presets the relatioship betwee ad P u /P uz. The mathematical relatioship betwee ad P u /P uz is as follows: = 1.0, whe P u /P uz 0.2 = 0.67 + 1.67 P u /P uz, whe 0.2 < (P u /P uz ) < 0.8 = 2.0, whe (P u /P uz ) 0.8 (10.60) 10.26.6 Solutio of Problems usig IS Code Method The IS code method, as discussed i sec.10.26.5, ca be employed to solve both the desig ad aalysis types of problems. The oly differece betwee the desig ad aalysis type of problems is that a trial sectio has to be assumed icludig the percetage of logitudial reiforcemet i the desig problems. However, these data are available i the aalysis type of problems. Therefore, a guide lie is give i this sectio for assumig the percetage of logitudial reiforcemet for the desig problem. Further, for both types of problems, the eccetricities of loads are to be verified if they are more tha the correspodig miimum eccetricities, as stipulated i cl.25.4 of IS 456. Thereafter, the relevat steps are give for the solutio of the two types of problems. (a) Selectio of trial sectio for the desig type of problems As metioed i sec.10.24.2(i) of Lesso 24, the prelimiary dimesios are already assumed durig the aalysis of structure (mostly statically idetermiate). Thus, the percetage of logitudial steel is the oe parameter to be assumed from the give P u, M ux, M uy, f ck ad f y. Pillai ad Meo (Ref. No. 4) suggested a simple way of cosiderig a momet of approximately 15 per cet i excess (lower percetage up to 5 per cet if P u /P uz is relatively high) of the resultat momet M = + u 2 2 (1.15) ( M ux M uy ) 1/ 2 (10.61) as the uiaxial momet for the trial sectio with respect to the major pricipal axis xx, if M ux Muy; otherwise, it should be with respect to the mior pricipal axis. Versio 2 CE IIT, Kharagpur
The reiforcemet should be assumed to be distributed equally o four sides of the sectio. (b) Checkig the eccetricities e x ad e y for the miimum eccetricities Clause 25.4 of IS 256 stipulates the amouts of the miimum eccetricities ad are give i Eq.10.3 of sec.10.21.11 of Lesso 21. However, they are give below as a ready referece. e xmi e ymi greater of (l/500 + b/30) or 20 mm greater of (l/500 + D/30) or 20 mm. (10.3) where l, b ad D are the usupported legth, least lateral dimesio ad larger lateral dimesio, respectively. The clause further stipulates that for the biaxial bedig, it is sufficiet to esure that the eccetricity exceedig the miimum value about oe axis at a time. (c) Steps for the solutio of problems The followig are the steps for the solutio of both aalysis ad desig types of problems while employig the method recommeded by IS 456. (i) Verificatio of eccetricities It is to be doe determiig e x = M ux /P u ad e y = M uy /P u from the give data of P u, M ux ad M uy ; ad e xmi ad e ymi from Eq.10.3 from the assumed b ad D ad give l. (ii) Assumig a trial sectio icludig logitudial reiforcemet This step is eeded oly for the desig type of problem, which is to be doe as explaied i (a) above. (iii) Determiatio of M ux1 ad M uy1 Use of desig charts should be made for this. M ux1 ad M uy1, correspodig to the give P u, should be sigificatly greater tha M ux ad M uy, respectively. Redesig of the sectio should be doe if the above are ot satisfied for the desig type of problem oly. (iv) Determiatio of P uz ad The values of P uz ad ca be determied from Eqs.10.59 ad 10.60, respectively. Alteratively, P uz ca be obtaied from Chart 63 of SP-16. Versio 2 CE IIT, Kharagpur
(v) Checkig the adequacy of the sectio This is doe either usig Eq.10.58 or usig Chart 64 of SP-16. 10.26.7 Illustrative Example Problem 1: Desig the reiforcemet to be provided i the short colum of Fig.10.26.4 is subjected to P u = 2000 kn, M ux = 130 knm (about the major pricipal axis) ad M uy = 120 knm (about the mior pricipal axis). The usupported legth of the colum is 3.2 m, width b = 400 mm ad depth D = 500 mm. Use M 25 ad Fe 415 for the desig. Solutio 1: Step 1: Verificatio of the eccetricities Give: l = 3200 mm, b = 400 mm ad D = 500 mm, we have from Eq.10.3 of sec.10.26.6b, the miimum eccetricities are: e xmi = greater of (3200/500 + 400/30) ad 20 mm = 19.73 mm or 20 mm = 20 mm Versio 2 CE IIT, Kharagpur
e ymi = greater of (3200/500 + 500/30) ad 20 mm = 23.07 mm or 20 mm = 23.07 mm Agai from P u = 2000 kn, M ux = 130 knm ad M uy = 120 knm, we have e x = M ux /P u = 130(10 6 )/2000(10 3 ) = 65 mm ad e y = M uy /P u = 120(10 6 )/2000(10 3 ) = 60 mm. Both e x ad e y are greater tha e xmi ad e ymi, respectively. Step 2: Assumig a trial sectio icludig the reiforcemet We have b = 400 mm ad D = 500 mm. For the reiforcemet, 2 2 1/ 2 M u = 1.15 ( M ux + M uy ), from Eq.10.61 becomes 203.456 knm. Accordigly, we get P u /f ck bd = 2000(10 3 )/(25)(400)(500) = 0.4 M u /f ck bd 2 = 203.456(10 6 )/(25)(400)(500)(500) = 0.0814 Assumig d' = 60 mm, we have d'/d = 0.12. From Charts 44 ad 45, the value of p/f ck is iterpolated as 0.06. Thus, p = 0.06(25) = 1.5 per cet, givig A sc = 3000 mm 2. Provide 12-20 mm diameter bars of area 3769 mm 2, actual p provided = 1.8845 per cet. So, p/f ck = 0.07538. Step 3: Determiatio of M ux1 ad M uy1 We have P u /f ck bd = 0.4 ad p/f ck = 0.07538 i step 2. Now, we get M ux1 /f ck bd 2 from chart correspodig to d' = 58 mm (Fig.10.26.4) i.e., d'/d = 0.116. We iterpolate the values of Charts 44 ad 45, ad get M ux1 /f ck bd 2 = 0.09044. So, M ux1 = 0.0944(25)(400)(500(500)(10-6 ) = 226.1 knm. For M ux1, d'/b = 58/400 = 0.145. I a similar maer, we get M uy1 = 0.0858(25)(400)(400)(500)(10-6 ) = 171.6 knm. As M ux1 ad M uy1 are sigificatly greater tha M ux ad M uy, respectively, redesig of the sectio is ot eeded. Step 4: Determiatio of P uz ad From Eq.10.59, we have P uz = 0.45(25)(400)(500) + {0.75(415) - 0.45(25)}(3769) = 3380.7 kn. Versio 2 CE IIT, Kharagpur
Alteratively, Chart 63 may be used to fid P uz as explaied. From the upper sectio of Chart 63, a horizotal lie AB is draw at p = 1.8845, to meet the Fe 415 lie B (Fig.10.26.5). A vertical lie BC is draw from B to meet M 25 lie at C. Fially, a horizotal lie CD is draw from C to meet P uz /A g at 17. This gives P uz = 17(400)(500) = 3400 kn. The differece betwee the two values, 19.3 kn is hardly 0.57 per cet, which is due to the error i readig the value from the chart. However, ay oe of the two may be employed. Now, the value of is obtaied from Eq.10.60 for P u /P uz = 2000/3380.7 = 0.5916, i.e., 0.2 < P u /P uz < 0.8, which gives, = 0.67 + 1.67 (P u /P uz ) = 1.658. Alteratively, may be obtaied from Fig.10.26.3, draw to scale. Step 5: Checkig the adequacy of the sectio Usig the values of M ux, M ux1, M uy, M uy1 ad i Eq.10.58, we have (130/226.1) 1.658 + (120/171.6) 1.658 = 0.9521 < 1.0. Hece, the desig is safe. Versio 2 CE IIT, Kharagpur
Alteratively, Chart 64 may be used to determie the poit (M ux /M ux1 ), (M uy /M uy1 ) is withi the curve of P u /P uz = 0.5916 or ot. Here, M ux /M ux1 = 0.5749 ad M uy /M uy1 = 0.6993. It may be see that the poit is withi the curve of P u /P uz = 0.5916 of Chart 64 of SP-16. Step 6: Desig of trasverse reiforcemet As per cl.26.5.3.2c of IS 456, the diameter of lateral tie should be > (20/4) mm diameter. Provide 8 mm diameter bars followig the arragemet show i Fig.10.26.4. The spacig of lateral tie is the least of : (a) 400 mm = least lateral dimesio of colum, mm), (b) 320 mm = sixtee times the diameter of logitudial reiforcemet (20 (c) 300 mm Accordigly, provide 8 mm lateral tie alterately @ 250 c/c (Fig.10.26.4). 10.26.8 Practice Questios ad Problems with Aswers Q.1: Explai the behaviour of a short colum uder biaxial bedig as the resultat of two uiaxial bedig. A.1: See sec. 10.26.2 Q.2: Draw oe iteractio surface for a short colum uder biaxial bedig ad show typical iteractio curves ad load cotour curve. Explai the safety of a colum with referece to the iteractio surface whe the colum is uder biaxial bedig. A.2: Q.3: A.3: Q.4: A.4: See sec.10.26.3 ad Fig.10.26.2. Discuss the limitatio of the iteractio curve. See sec.10.26.4. Illustrate the IS code method of desig of colums uder biaxial bedig. See sec.10.26.5. Versio 2 CE IIT, Kharagpur
Q.5: Aalyse the safety of the short colum of usupported legth 3.2 m, b = 450 mm, D = 500 mm, as show i Fig.10.26.6, havig 12-16 mm diameter bars as logitudial reiforcemet ad 8 mm diameter bars as lateral tie @ 250 mm c/c, whe subjected to P u = 1600 kn, M ux = 120 knm ad M uy = 100 knm. Use M 25 ad Fe 415. A.5: Step 1: Verificatio of the eccetricities From the give data: l = 3200 mm, b = 450 mm ad D = 500 mm, e xmi = 3200/500 + 450/30 = 21.4 > 20 mm, so, 21.4 mm e ymi = 3200/500 + 5000/30 = 23.06 > 20 mm, so, 23.06 mm e x = M ux /P u = 120(10 3 )/1600 = 75 mm e y = M uy /P u = 100(10 3 )/1600 = 62.5 mm So, the eccetricities e x ad e y are >> e xmi ad e ymi. Step 2: Determiatio of M ux1 ad M uy1 Versio 2 CE IIT, Kharagpur
Give data are: b = 450 mm, D = 500 mm, f ck = 25 N/mm 2, f y = 415 N/mm 2, P u = 1600 kn, M ux = 120 knm, M uy = 100 knm ad A sc = 2412 mm 2 (12-16 mm diameter bars). We have p = (100)(2412)/(450)(500) = 1.072 per cet, ad d'/d = 56/500 = 0.112, d'/b = 56/450 = 0.124, P u /f ck bd = 1600/(25)(450)(500) = 0.2844 ad p/f ck = 1.072/25 = 0.043. We get M ux1 /f ck bd 2 from Charts 44 ad 45 as 0.09 ad 0.08, respectively. Liear iterpolatio gives M ux1 /f ck bd 2 for d'/d = 0.112 as 0.0876. Thus, M ux1 = (0.0876)(25)(450)(500)(500) = 246.376 knm Similarly, iterpolatio of values (0.09 ad 0.08) from Charts 44 ad 45, we get M uy1 /f ck db 2 = 0.085 for d'/b = 0.124. Thus M uy1 = (0.085)(25)(500)(450)(450) = 215.156 knm Step 3: Determiatio of P uz ad From Eq.10.59, P uz = 0.45(25)(450)(500) + {0.75(415) - 0.45(25)}(2412) = 3254.85 kn. This gives P u /P uz = 1600/3254.85 = 0.491574. From Eq.10.60, 1.4909. = 0.67 + 1.67(P u /P uz ) = 0.67 + 1.67(0.491574) = Step 4: Checkig the adequacy of the sectio From Eq.10.58, we have: (120/246.376) 1.4909 + (100/215.156) 1.4909 = 0.6612 < 1. Hece, the sectio is safe to carry P u = 1600 kn, M ux = 120 knm ad M uy = 100 knm. 10.26.9 Refereces 1. Reiforced Cocrete Limit State Desig, 6 th Editio, by Ashok K. Jai, Nem Chad & Bros, Roorkee, 2002. 2. Limit State Desig of Reiforced Cocrete, 2 d Editio, by P.C.Varghese, Pretice-Hall of Idia Pvt. Ltd., New Delhi, 2002. 3. Advaced Reiforced Cocrete Desig, by P.C.Varghese, Pretice-Hall of Idia Pvt. Ltd., New Delhi, 2001. 4. Reiforced Cocrete Desig, 2 d Editio, by S.Uikrisha Pillai ad Devdas Meo, Tata McGraw-Hill Publishig Compay Limited, New Delhi, 2003. Versio 2 CE IIT, Kharagpur
5. Limit State Desig of Reiforced Cocrete Structures, by P.Dayaratam, Oxford & I.B.H. Publishig Compay Pvt. Ltd., New Delhi, 2004. 6. Reiforced Cocrete Desig, 1 st Revised Editio, by S.N.Siha, Tata McGraw-Hill Publishig Compay. New Delhi, 1990. 7. Reiforced Cocrete, 6 th Editio, by S.K.Mallick ad A.P.Gupta, Oxford & IBH Publishig Co. Pvt. Ltd. New Delhi, 1996. 8. Behaviour, Aalysis & Desig of Reiforced Cocrete Structural Elemets, by I.C.Syal ad R.K.Ummat, A.H.Wheeler & Co. Ltd., Allahabad, 1989. 9. Reiforced Cocrete Structures, 3 rd Editio, by I.C.Syal ad A.K.Goel, A.H.Wheeler & Co. Ltd., Allahabad, 1992. 10. Textbook of R.C.C, by G.S.Birdie ad J.S.Birdie, Wiley Easter Limited, New Delhi, 1993. 11. Desig of Cocrete Structures, 13 th Editio, by Arthur H. Nilso, David Darwi ad Charles W. Dola, Tata McGraw-Hill Publishig Compay Limited, New Delhi, 2004. 12. Cocrete Techology, by A.M.Neville ad J.J.Brooks, ELBS with Logma, 1994. 13. Properties of Cocrete, 4 th Editio, 1 st Idia reprit, by A.M.Neville, Logma, 2000. 14. Reiforced Cocrete Desiger s Hadbook, 10 th Editio, by C.E.Reyolds ad J.C.Steedma, E & FN SPON, Lodo, 1997. 15. Idia Stadard Plai ad Reiforced Cocrete Code of Practice (4 th Revisio), IS 456: 2000, BIS, New Delhi. 16. Desig Aids for Reiforced Cocrete to IS: 456 1978, BIS, New Delhi. 10.26.10 Test 26 with Solutios Maximum Marks = 50, Maximum Time = 30 miutes Aswer all questios. Versio 2 CE IIT, Kharagpur
TQ.1: Aalyse the safety of the short square colum of usupported legth = 3.5 m, b = D = 500 mm, as show i Fig.10.26.7, with 12-16 mm diameter bars as logitudial reiforcemet ad 8 mm diameter bars as lateral tie @ 250 mm c/c, whe subjected to P u = 1800 kn, M ux = 160 knm ad M uy = 150 knm. A.TQ.1: Step 1: Verificatio of the eccetricities From the give data: l = 3500 mm, b = D = 500 mm, we have mm e mi i both directios (square colum) = (3500/500) + (500/30) = 23.67 e x = 160(10 3 )/1800 = 88.88 mm ad e y = 150(10 3 )/1800 = 83.34 mm Therefore, e x ad e y >> e mi. Versio 2 CE IIT, Kharagpur
Step 2: Determiatio of M ux1 ad M uy1 We have the give data: b = D = 500 mm, f ck = 25 N/mm 2, f y = 415 N/mm 2, P u = 1800 kn, M ux = 160 knm, M uy = 150 knm ad A sc = 2412 mm 2 (12-16 mm diameter bars). The percetage of logitudial reiforcemet p = 241200/(500)(500) = 0.9648 per cet, ad d'/d = 56/500 = 0.112 ad p/f ck = 0.9648/25 = 0.03859. Liear iterpolatio of values of M ux1 /f ck bd 2 from Charts 44 ad 45 for d'/d = 0.112 is obtaied as 0.08. Thus, M ux1 = (0.08)(25)(500)(500)(500) = 250 knm M uy1 = M ux1 = 250 knm (square colum) Step 3: Determiatio of P uz ad From Eq.10.59, P uz = 0.45(25)(500)(500) + {0.75(415) - 0.45(25)}(2415) = 3536.1 kn. P u /P uz = 1800/3536.1 = 0.509. From Eq.10.60, = 0.67 + 1.67(0.509) = 1.52. Step 4: Checkig the adequacy of the sectio From Eq.10.58, we have: (160/250) 1.52 + (150/250) 1.52 = 0.967 < 1. knm. Hece, the sectio ca carry P u = 1800 kn, M ux = 160 knm ad M uy = 150 10.26.11 Summary of this Lesso This lesso explais the behaviour of short colums uder axial load ad biaxial bedig with the help of iteractio surface, visualised as a three-dimesioal plot of P u -M ux -M uy. The iteractio surface has a set of iteractio curves of P u -M u ad aother set of iteractio curves of M uxz -M uyz at costat P uz, also kow as load cotour. The desig ad aalysis of short colums are also explaied with the help of derived equatios ad desig charts of SP-16. Numerical examples i the illustrative example, practice problems ad test will help i uderstadig the applicatio of the theory i solvig the aalysis ad desig types of problems of short colums uder axial load ad biaxial bedig. Versio 2 CE IIT, Kharagpur