Karol Bzdawka Composite Column - Calculation Examples

Tampereen teknillinen yliopisto. Rakennustekniikan laitos. Rakennetekniikka. Tutkimusraportti 47 Tampere University of Technology. Department of Civil Engineering. Structural Engineering. Research Report 47 Karol Bzdawka Composite Column - Calculation Examples

Tampereen teknillinen yliopisto. Rakennustekniikan laitos. Rakennetekniikka. Tutkimusraportti 47 Tampere University of Technology. Department of Civil Engineering. Structural Engineering. Research Report 47 Karol Bzdawka Composite Column Calculation Examples Tampereen teknillinen yliopisto. Rakennustekniikan laitos Tampere 00

ISBN 978-95-5-3- (printed) ISBN 978-95-5-803-3 (PDF) ISSN 797-96

(54) PREFACE This study was done in Research Centre of Metal Structures in Hämeenlinna unit. The financial support of Hämeenlinnan kaupunki, Hämeen Ammattikorkeakoulu and Rautaruukki Oyj is gratefully acknowledged. The discussions with Mr. Arto Sivill, Finnmap Consulting Oy, and Mr. Aarne Seppänen, Rautaruukki Oyj, were very helpful when completing the study. This study is part of on-going research dealing with optimization of load bearing structures of metal based buildings. The previous report of Karol Bzdawka dealt with design of WQ-beams and cost estimations of different framing systems for office buildings. In this study the design of composite columns is presented based on EN standards. The design methods are implemented into MATLAB enabling in the future the applications of mathematical optimizations tools to be used to find better solutions for load bearing structures than now are available Hämeenlinna 0..00 Markku Heinisuo Professor of Metal Structures

(54) COMPOSITE COLUMN CALCULATION EXAMPLES TABLE OF CONTENTS Preface... Abstract... 3 Square column... 4. Ambient design... 4. Fire design (R60)... 8.3 Fire design at 0 C... 8 3 Circular column... 37 3. Ambient design... 37 4 Results... 48 5 Summary... 53 References... 54 Appendix A

3 (54) ABSTRACT Composite columns are often used in structures due to the ease and speed of erection, and high performance in fire situation. Concrete filled tubes are steel tubes that are in site filled with reinforced concrete. In normal situation column works as composite while in fire situation majority of the load is carried by the reinforced concrete core. There are numerous publications about this type of columns but all of them use very simplified methods. Especially regarding the calculation of the neutral axis of circular columns. Also the shear resistance of the column is often omitted. The reason for this document is to present the calculation of concrete-filled tubes in details. Also to investigate the influence of the bars arrangement in circular columns on their bending resistance. Calculations have been carried for ambient and fire temperatures. This report includes calculation examples and results in different design situations, for a number of various column cross-sections.

4 (54) SQUARE COLUMN. Ambient design Data for the calculation: col t 50mm 6mm width and height of the steel tube wall thickness of the steel tube n 4 number of reinforcing bars fi fi s s u s L eff 0mm 6mm 300mm 35mm 3600mm diameter of reinforcing bars diameter of stirrups stirrups spacing distance from the inner surface of the tube to the surface of the bar effective length of the column u.s fi.s col fi t col Figure. Symbols used for the basic design variables Material properties Steel Concrete Reinforcement f y 355MPa f ck 40MPa f sk 500MPa characteristic strength gamma a. gamma c.35 gamma s.5 material safety factor f ad f y gamma a f cd f ck gamma c f sd f sk gamma s design strengths f ad 3.7MPa f cd 9.6MPa f sd 434.8MPa E a 0000MPa E cm 35000MPa E s 0000MPa Young modulus

5 (54) Forces in the column in normal situation: N Sd 84.0kN M Sd 3.5kN m bigger bending moment (upper or lower end) M Sd 3.0kN m smaller bending moment (upper or lower end) (bending moments on the same side of the column have the same signs) V Sd M Sd M Sd L eff V Sd 67.9kN constant shear force Characteristics of the cross section Steel tube A a 4 ( col t) t A a 5856mm tube cross-section area col 4 ( col t) 4 I a col 3 ( col t) 3 W pa 4 4 I a 5.84 0 7 mm 4 tube second moment of area W pa 5.359 0 5 mm 3 plastic modulus Reinforcement n fi A s 4 A s 57mm cross-section area of the reinforcement I s col t u s fi A s I s 6.88 0 6 mm 4 second moment of area of the reinforcement Concrete core A cgross ( col t) A cgross 56644mm A c A cgross A s A c 55387mm gross cross-section area of the concrete nett cross-section area of the concrete ( col t) 4 I cgross I cgross.674 0 8 mm 4 gross second moment of area of the concrete I c I cgross I s I c.605 0 8 mm 4 nett second moment of area of the concrete Check if the reinforcement meets the requirement of the reinforcement level A s A cgross max 6.0 % >.% > min.5 % Requirement met

6 (54) Maximum design bending resistance for the cross-section is given by equation: W pc M max.rd W pa f ad W ps f sd f cd Where: W pa is the plastic section modulus of the steel tube W ps is the plastic section modulus of the reinforcing bars W pc is the plastic section modulus of the conctrete, in above equation only half is taken into account since concrete does not carry tension. W pa col 3 4 ( col t) 3 4 W pa 535.9cm 3 W ps A s col fi u s t W ps 93.0cm 3 (reinforcement area times the distance from the midline) ( col t) 3 W pc 4 W ps W pc 377.3cm 3 Thus: M max.rd W pa f ad W ps f sd W pc f cd M max.rd 6.9kN m For a column cross-section in pure bending the neutral axis is not the same as the axis of symmetry of the cross-section. New neutral axis has to be calculated from the equations of equilibrium. For details of used symbols see Figure. A.an h.n A.sn A.cn Figure. Symbols used in calculation of plastic bending resistance

7 (54) Design axial resistance of the concrete only is equal to: N pmrd A c f cd N pmrd 64kN Equilibrium equation: 0.5 N pmrd ( col t) h n f cd A sn f sd 4 t h n f ad A sn is the cross-section area of the reinforcement in range - h n to h n from the centerline of the column cross-section. In this case equal to zero because there are no bars lying on the centerline. A sn 0mm After rearranging the equilibrium equation, distance between the centerline and the neutral axis is given: h n N pmrd A sn f sd f cd col t ( ) f cd 4t f ad h n 55.453mm Difference between maximum plastic bending resistance M max.rd and plastic bending resistance around the neutral axis M pl.rd is equal to the bending resistance of the part of the cross section that is in distance h n from the centerline. W pcn M pl.rd M max.rd W pan f ad W psn f sd f cd Where: W pan t h n 4 W pan 36.9cm 3 W psn 0 cm 3 there are no bars in considered region ( col t) h n W pcn 4 W psn W pcn 73.9cm 3 M pl.rd M max.rd W pan f ad W psn f sd W pcn f cd M pl.rd 39.kN m Values of the resistance to axial force (EN 994-- [3], 6.7.3.) N plrd A a f ad A c f cd A s f sd N plrd 4077kN design resistance N plr A a f y A c f ck A s f sk N plr 493kN characteristic resistance

8 (54) Effective flexural and axial stiffnesses (EN 994--, 6.7.3.3) EI eff E a I a 0.6 E cm I c E s I s EI eff 95.4kN m N cr L eff EI eff N cr 4564.8kN Check if the long term effect need to be taken into account A a f ad N plrd 0.464 is the contribution ratio of the steel tube, has to be between 0. and 0.9 - only then column works as a composite N plr N cr 0.58 is the relative slenderness of the column, has to be smaller than (EN 994-- 6.7.3.3) vert 0.8 vert.49 According to [], if is smaller than vert, long term effects do not need to be taken into account. Calculation of the long term effect E c.eff E cm N G.Sd N Sd t E Where: t.50 is the creep coeff. acc. to EN 99-- [4], 3..4, calculated for relative humidity 80%, loading time 30 days, concrete C 40/50 N G.Sd is the permanent part of the load, in this case assumed 0.8 N Sd E c.eff 5.9GPa If the long term effect has to be taken into account, effective flexural stiffness is equal: EI eff E a I a 0.6 E c.eff I c E s I s EI eff 64.5kN m And the elastic critical normal force: N cr L eff EI eff N cr 9.5kN

9 (54) Design value of the effective flexural stiffness, used to determine the sectional forces. (EN 994-- 6.7.3.4) EI effii 0.9 E a I a 0.5 E cm I c E s I s EI effii 639.3kN m N cr.eff L eff EI effii N cr.eff 483.4kN Including the long term effects: EI effii.long 0.9 E a I a 0.5 E c.eff I c E s I s EI effii.long 454.4kN m N cr.eff.long L eff EI effii N cr.eff.long 483.4kN Stiffness to axial compression, used to determine the sectional forces.: EA E a A a E s A s E cm A c EA 343.MN Axial resistance (EN 994--, 6.7.3.5) N Rd N plrd Where: is the reduction factor for the relevant buckling mode, in terms of relative slenderness (EN 993-- [5], 6.3..) if 0. or N.Sd/N.plRd =< 0. than =. but =< 0.58 0.5 0. 0.76 Where: is an imperfection factor, equal to 0.49 (buckling curve C) min 0.796 N Rd N plrd N Rd 347.kN

0 (54) Utilization check: N Sd N Rd 0.567 <.0 Increse of the bending moment due to second order effects (EN 994--, 6.7.3.4) Second order effects may be allowed for by multiplying the bigger of the is geater or equal to.0. M Sd by a factor k, where k k N Sd N cr.eff Where: is an equivalent moment factor (EN 994--, Table 6.4) max 0.66 0.44 M Sd M Sd 0.44 0.440 k N Sd N cr.eff k 0.56 since k is smaller than.0 => k max( k ) k.0 Thus: M Sd k M Sd M Sd 3.5kN m Transverse shear Resistance of the steel tube (EN 993--, 6..6) col A v A s ( col) A v 68.3mm area in shear for RHS V plards A v f y 3 gamma a V plards 35.kN plastic shear resistance of the tube. Resistance of the concrete (EN 99--, 6.) d col t u s fi b w col t d b w 93.0mm 38.0mm effective depth of the cross-section smallest width of the cross section in the tensile area

(54) A sl A s A sn A sl 68.3mm area of the tensile reinforcement, properly anchored A sl b w d 0.04 ratio of the longitudinal reinforcement Resistance of the concrete without shear reinforcement. (6..) cp min N pmrd N plrd N Sd A c 0. f cd cp 5.9MPa cp is the compressive stress in concrete due to axial loading. Part of the axial loading that is coming into concrete is taken as the proportion of the concrete plastic resistance ( cross-section plastic resistance ( N plrd ) times axial load. N plrd ) and Design value of the shear resistance is given by: V Rdc C Rdc 00 min 0.0 f ck 3 k cp b w d With the minimum value of: V Rd.c.min v min k cp b w d Calculation of design shear resistance: C Rdc 0.8 gamma c k 0.5 C Rdc 0.33 recommended value recommended value V Rdc C Rdc 00 min 0.0 f ck MPa 3 k cp MPa b w d MPa V Rdc 64.kN Minimum value of the shear resistance: k min 00mm d 0.5 k.0 v min 3 0.035 k f ck MPa 0.5 v min 0.66 recommended value

(54) V Rd.c.min v min k cp MPa b w d MPa V Rd.c.min 69.6kN In the end design shear resistance is the bigger of the velues V Rdc and V Rd.c.min. V Rd.c max V Rdc V Rd.c.min V Rd.c 69.6kN Resistance of concrete cross-section with shear reinforcement (6..3) The shear resistance of a member with perpendicular shear reinforcement is equal to: V Rds A sw s z tan f sd but not greater than: f cd V Rdmax cw b w z tan cot Where: 45deg angle between concrete compression strut and column axis [degrees]. If not proven otherwise equal to 45 degrees. z 0.9 d z 73.7mm 0.6 f ck 50MPa 0.504 is the strength reduction factor for concrete cracked in shear cw is the coefficient taking account of the state of the stress in the compression chord, it depends on cp. cp min N pmrd N plrd N Sd A c cp 3.4MPa

3 (54) cw if cp 0 cw.5 cp if 0 f cp 0.5 f cd cd.5 if 0.5 f cd cp 0.5 f cd.5 cp if 0.5 f f cd cp cd The maximum effective cross-sectional area of the shear reinforcement, calculated for cot =, is given by: A swmax 0.5 cw f cd b w s f sd A swmax 53.7mm Cross sectional area of a single stirrup. A sw fi s 4 A sw 56.5mm Resistance of stirrups in tension: V Rds A sw s z tan f sd V Rds 4.kN Resistance of concrete chords in compression: f cd V Rdmax cw b w z tan cot V Rdmax 385.8kN Resistance of cross-section requiring shear reinforcement V Rd.reinf min V Rds V Rdmax V Rd.reinf 4.kN Resistance of the column to transverse shear. If number of longitudinal reinforcing bars is 0, resistance to shear is calculated from: V Rd V Rd.c V plards Otherwise it is equal to: V Rd max V Rd.c V Rd.reinf V plards V Rd 40.8kN

4 (54) Influence of shear on the resistance to bending has to be taken into account if the part of the shear force that is coming to the steel tube V a.sd is greater than half it's resistance V plards It is done by reduction of the design resistance of the steel tube, calculated using reduced yield strength in the part of the tube that carries shear - A v (EN 993--, 6..8). shear f yd Where: shear V a.sd V plards Where: V plards V a.sd is the plastic design resistance of the steel tbe only is the part of the shear force that is coming to the steel tube, assuming that the resistance of the concrete core to shear is fully utilized. V a.sd is calculated as follow: V a.sd max V Sd max V Rd.c V Rd.reinf 0 To simplify the calculation and avoid the reduction of the yield strength, limitation has been imposed that: V a.sd V plards 0.5 In the considered case: V a.sd 0.0N => V a.sd V plards 0.0 Utilization of the cross-section in shear. V Sd V Rd 0.6 < 0.5 For purposes of conditioning all the restraints to be <.0, above condition has been rewritten in another form: V Sd V Rd 0.33 <.0 (this is how the program checks utillities, thanks to this all the constraints are penalized with correct proportions) Shear does not affect the bending resistance.

5 (54) Resistance of column for bending (EN 994--, 6.7.3.3) M Rd M M pl.rd To determine the bending resistance of the cross-section, which is depending on the axial force, value of coefficient has to be callculated. A graphic method is used to determine. The method is explained in Figure 3. Symbols that are used are evaluated below. pm N pmrd N plrd pm 0.40 N Rd N plrd 0.796 d N Sd N plrd d 0.45 n M Sd M Sd 4 n 0.370 M max.rd M pl.rd.095

6 (54) N.Rd N.pl.Rd.0 =0.785.d=0.45.pm=0.40.n=0.365 0.5.pm=0.0 M.Rd M.pl.Rd mju.k mju mju.d.0 M.max.Rd M.pl.Rd =.095 Figure 3. Graphic method of calculating µ. 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0. 0.4 0.6 0.8..4 Figure 4. Graphic method using Matlab.

7 (54) For steel grades S35 - S355 the coefficient M is taken equal to 0.9 For steel grades S40 - S460 the coefficient M is taken equal to 0.8 In this case: 0.7436 Coefficient greater than.0 may be used only when bending moment M Ed in column depends directly on the action of the normal force N Ed. Otherwise the cross-sectional force that is causing the increase of the resistance should be reduced by 0% ( F 0.8 acc. to EN 994--, 7...(7) ) It has been checked, in this research, that reduction of the N Ed not always leads to the reduction of M Rd. Especially in fire situation. Thus the value has been modified so that only the amount that is above.0 is reduced by 0%. It has been done according to the folllowing expression. if.0 F if.0 0.7436 Thus: M Sd M pl.rd 0.739 < M 0.9 Results: EA 343.MN EI effii 6.4MN m N plrd 4077.4kN N pmrd 64.kN N Rd 347.kN M pl.rd 39.kN m M max.rd 6.9kN m M pl.rd 77.9kN m V Rd 40.8kN

8 (54). Fire design (R60) Material properties at elevated temperatures in fire situation. Temperatures for the steel tube, concrete core and the reinforcement of the concrete core were taken from: Betonitaytteisen terasliitto-pilarin suunnitteluohje []. The temperatures depend on: the fire class - in this case R60 the dimensions of the steel tube cross-section for reinforcement - on the distance from the inner surface of the tube to the axis of the rod of the longitudinal reinforcement. Denoted: u s + fi/ Also according to [], u s is equal to 35 mm if the outer tube dimension (for rectangular tubes - lesser of the dimensions) is not greater than 300 mm. Otherwise u s = 45 mm. In considered case: u s col t 35.0mm 50.0mm 6.0mm n 4 fi 0.0mm Effective lenght of the column in fire situation is equal to half the total column length: L eff. 0.5L eff Temperatures read from []: T a T c T s 9deg 475deg 488.5deg tube temperature, (Appendix 4) concrete temperature, (Appendix 3) steel temperature, ( Appendix 4, value interpolated for u s +fi/ = 45 mm ) Reduction factors for steel tube are taken from Table 6 []. Design yield strength in fire: k y. 0.0578 and f a. f y k y. => f a. 0.5MPa Young modulus in fire: k Ea. 0.06505 and E a. k Ea. E a => E a. 3.7GPa

9 (54) Reduction factors for concrete are taken from Table 7 []. Design compressive strength in fire: k c. 0.6375 and f c. f ck k c. => f c. 5.5MPa Young modulus in fire: cu. 0.0375 and E csec. k c. f ck cu. => E csec. 854.5MPa Reduction factors for reinforcement are taken from Table 8 []. Design compressive strength in fire: k s. 0.7005 and f s. f sk k s. => f s. 350.5MPa Young modulus in fire: k Es. 0.484 and E s. E s k Es. => E s. 87864.0MPa Forces in the column in fire situation: N Sd. 807.0kN M Sd..5kN m bigger bending moment (upper or lower end) M Sd. 3.0kN m smaller bending moment (upper or lower end) (bending moments on the same side of the column have the same signs) V Sd. M Sd. M Sd. L eff V Sd. 4.0kN constant shear force Maximum design bending resistance for the cross-section: Is calculated similarily to ambient situation but with material resistances in fire situation M max.rd. W pa f a. W ps f s. W pc f c. M max.rd. 85.4kN m Design axial resistance of the concrete only is equal to: N pmrd. A c f c. N pmrd. 4kN Calculation of the new neutral axis in fire is done using the equilibrium equation: 0.5 N pmrd. ( col t) h n. f c. A sn f s. 4 t h n. f a. Same as previoously A sn is equal to zero because there are no bars lying on the centerline. A sn 0mm

0 (54) After rearranging the equilibrium equation, distance between the centerline and the neutral axis is given: h n. N pmrd. A sn f s. f c. col t ( ) f c. 4t f a. h n. 07.67mm Plastic bending resistance of the cross section is equal: W pcn. M pl.rd. M max.rd. W pan. f a. W psn. f s. f c. Where: W pan. t h n. 4 W pan. 39.0cm 3 W psn. 0 cm 3 there are no bars in considered region ( col t) h n. W pcn. 4 W psn. W pcn. 756.9cm 3 M pl.rd. M max.rd. W pan. f a. W psn. f s. W pcn. f c. M pl.rd. 47.4kN m Values of the resistance to axial force (EN 994--, 6.7.3.) N plrd. A a f a. A c f c. A s f s. N plrd. 973kN design resistance N plr. A a f a. A c f c. A s f s. N plr. 973kN characteristic resistance Effective flexural and axial stiffnesses (EN 994--, 6.7.3.3) EI eff. 0.9 E a. I a 0.8 E csec. I c 0.9E s. I s EI eff. 645.kN m N cr. L eff. EI eff. N cr. 50.6kN Long term effects in fire situation are not taken into account.

(54) Design value of the effective flexural stiffness, used to determine the sectional forces. EI effii. 0.9 E a. I a 0.5 E csec. I c E s. I s EI effii. 476.kN m N cr.eff. L eff. EI effii. N cr.eff. 4496.5kN Long term effects in fire situation are not taken into account. Stiffness to axial compression, used to determine the sectional forces.: EA E a. A a E s. A s E csec. A c EA 93.MN Axial resistance N Rd. N plrd. Where: but =< N plr. N cr. 0.67 0.5 0. 0.80 Where: is an imperfection factor, equal to 0.49 (buckling curve C) min 0.769 N Rd. N plrd. N Rd. 57.kN Utilization check: N Sd. N Rd. 0.53 <.0

(54) Increse of the bending moment due to second order effects. (EN 994--, 6.7.3.4) Second order effects may be allowed for by multiplying the bigger of the M Sd. by a factor k, where k is greater or equal to.0. k N Sd. N cr.eff. Where: max 0.66 0.44 M Sd. M Sd. 0.44 0.545 k N Sd. N cr.eff. k 0.664 k max k k.0 Thus: M Sd. k M Sd. M Sd..5kN m Transverse shear Resistance of the steel tube (EN 993--, 6..6) V plards. A v f a. 3 V plards..3kn Resistance of the concrete (EN 99--, 6.) Resistance of the concrete without shear reinforcement. (6..) cp. min N pmrd. N plrd. N Sd. A c 0. f c. cp. 5.MPa Design value of the shear resistance is given by: V Rdc. C Rdc 00 min 0.0 f c. 3 k cp. b w d With the minimum value of: V Rd.c.min. v min. k cp. b w d

3 (54) Calculation of design shear resistance: C Rdc 0.8 V Rdc. C Rdc 00 min 0.0 f c. MPa 3 k cp. MPa b w d MPa V Rdc. 6.kN Minimum value of the shear resistance: 3 0.5 f c. v min. 0.035 k MPa v min. 0.500 recommended value cp. V Rd.c.min. v min. k MPa b w d MPa V Rd.c.min. 58.kN Design shear resistance. V Rd.c. max V Rdc. V Rd.c.min. V Rd.c. 6.kN Resistance of concrete cross-section with shear reinforcement (6..3) The shear resistance of a member with perpendicular shear reinforcement is equal to: V Rds. A sw s z f s. tan but not greater than: f c. V Rdmax. cw. b w z. tan cot Where:. 0.6 f c. 50MPa. 0.539 cw. depends on cp.. cp. min N pmrd. N plrd. N Sd. A c cp. 0.4MPa

4 (54) cw. if cp. 0 cw..5 cp. if 0 f cp. 0.5 f c. c..5 if 0.5 f c. cp. 0.5 f c..5 cp. if 0.5 f f c. cp. c. The maximum effective cross-sectional area of the shear reinforcement, calculated for cot =, is given by: A swmax. 0.5 cw.. f c. b w s f s. A swmax. 749.mm Resistance of stirrups in tension: A sw 56.5mm V Rds. A sw s z f s. tan V Rds..5kN Resistance of concrete chords in compression: f c. V Rdmax. cw. b w z. tan cot V Rdmax. 355.0kN Resistance of cross-section requiring shear reinforcement V Rd.reinf. min V Rds. V Rdmax. V Rd.reinf..5kN Resistance of the column to transverse shear. If number of longitudinal reinforcing bars is 0, resistance to shear is calculated from: V Rd. V Rd.c. V plards. Otherwise it is equal to: V Rd. max V Rd.c. V Rd.reinf. V plards. V Rd. 84.5kN To simplify the calculation and avoid the reduction of the yield strength, limitation has been imposed that: V a.sd. max V Sd. max V Rd.c. V Rd.reinf. 0

5 (54) V a.sd. V plards. 0.5 In the considered case: V a.sd. 0.0N => V a.sd. V plards. 0.0 Utilization of the cross-section in shear. V Sd. V Rd. 0.048 < 0.5 Shear does not affect the bending resistance. Resistance of column for bending (EN 994--, 6.7.3.3) M Rd. M M pl.rd. To determine the bending resistance of the cross-section, which is depending on the axial force, value of coefficient has to be callculated. A graphic method is used to determine. (see Figure 5.) Symbols that are used are evaluated below. pm. N pmrd. N plrd. pm. 0.76 N Rd. N plrd. 0.769 d. N Sd. N plrd. d. 0.409 n. M Sd. M Sd. 4 n. 0.4 M max.rd. M pl.rd..80

6 (54) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0. 0.4 0.6 0.8..4.6.8 Figure 5. Calculation of µ in fire situation.

7 (54) The bending resistance in fire situation is checked as follows:.456 if.0 F if.0.3405 Thus: M Sd. M pl.rd. 0.8 < M 0.9 Results: EA 93.MN EI effii. 476.kN m N plrd. 973.0kN N pmrd. 4.4kN N Rd. 57.kN M pl.rd. 47.4kN m M max.rd. 85.4kN m M pl.rd. 63.5kN m V Rd. 84.5kN

8 (54).3 Fire design at 0 C Material properties at room temperatures in fire situation. Temperatures of steel tube, and concrete core is 0 degrees Celsius. Material properties are taken at temperature 0 degrees. Steel tube material charcteristics in fire situation at 0 deg.: k y. 0.000 and f a. 0 f y k y. 0 => f a. 0 355.0MPa k Ea. 0.000 and E a. 0 k Ea. 0 E a => E a. 0 0.0GPa Concrete charcteristics in fire situation at 0 deg.: k c. 0.000 and f c. 0 f ck k c. 0 => f c. 0 40.0MPa cu. 0 0.0050 and E csec. 0 k c. 0 f ck => cu. 0 Reinforcement material charcteristics in fire situation at 0 deg.: E csec. 0 6.0GPa k s. 0.000 and f s. 0 f sk k s. 0 => f s. 0 500.0MPa k Es. 0.000 and E s. 0 E s k Es. 0 => E s. 0 0.0GPa Effective lenght of the column in fire situation is equal to the column length: L eff. 0 L eff Forces in the column in fire situation: N Sd. 0 807.0kN M Sd. 0.5kN m bigger bending moment (upper or lower end) M Sd. 0 8.0kN m smaller bending moment (upper or lower end) (bending moments on the same side of the column have the same signs) V Sd. 0 M Sd. 0 M Sd. 0 L eff V Sd. 0 8.kN constant shear force

9 (54) Maximum design bending resistance for the cross-section: Is calculated similarily to ambient situation but with material resistances. M max.rd. 0 W pa f a. 0 W ps f s. 0 W pc f c. 0 M max.rd. 0 30.3kN m Design axial resistance of the concrete only is equal to: N pmrd. 0 A c f c. 0 N pmrd. 0 5kN Calculation of the new neutral axis is done using the equilibrium equation: 0.5 N pmrd. 0 ( col t) h n. 0 f c. 0 A sn f s. 0 4 t h n. 0 f a. 0 Same as previoously A sn is equal to zero because there are no bars lying on the centerline. A sn 0mm After rearranging the equilibrium equation, distance between the centerline and the neutral axis is given: h n. 0 N pmrd. 0 A sn f s. 0 f c. 0 col t ( ) f c. 0 4t f a. 0 h n. 0 6.405mm Plastic bending resistance of the cross section is equal: W pcn. 0 M pl.rd. 0 M max.rd. 0 W pan. 0 f a. 0 W psn. 0 f s. 0 f c. 0 Where: W pan. 0 t h n. 0 4 W pan. 0 45.cm 3 W psn. 0 0 cm 3 there are no bars in considered region ( col t) h n. 0 W pcn. 0 4 W psn. 0 W pcn. 0 897.4cm 3 M pl.rd. 0 M max.rd. 0 W pan. 0 f a. 0 W psn. 0 f s. 0 W pcn. 0 f c. 0 M pl.rd. 0 68.3kN m

30 (54) Values of the resistance to axial force (EN 994--, 6.7.3.) N plrd. 0 A a f a. 0 A c f c. 0 A s f s. 0 N plrd. 0 493kN design resistance N plr. 0 A a f a. 0 A c f c. 0 A s f s. 0 N plr. 0 493kN characteristic resistance Effective flexural and axial stiffnesses (EN 994--, 6.7.3.3) EI eff. 0 E a. 0 I a 0.8 E csec. 0 I c E s. 0 I s EI eff. 0 6989.3kN m N cr. 0 L eff. 0 EI eff. 0 N cr. 0 938.kN Long term effects in fire situation are not taken into account. Design value of the effective flexural stiffness, used to determine the sectional forces. EI effii. 0 0.9 E a. 0 I a 0.5 E csec. 0 I c E s. 0 I s EI effii. 0 465.0kN m N cr.eff. 0 L eff. 0 EI effii. 0 N cr.eff. 0 0787.3kN Long term effects in fire situation are not taken into account. Stiffness to axial compression, used to determine the sectional forces.: EA 0 E a. 0 A a E s. 0 A s E csec. 0 A c EA 0 379.9MN Axial resistance N Rd. 0 0 N plrd. 0 0 0 0 0 but =< 0 N plr. 0 N cr. 0 0 0.67 0 0.5 0 0. 0 0 0.79

3 (54) Where: is an imperfection factor, equal to 0.49 (buckling curve C) 0 min 0 0 0 0 0.775 N Rd. 0 0 N plrd. 0 N Rd. 0 386.9kN Utilization check: N Sd. 0 N Rd. 0 0.473 <.0 Increse of the bending moment due to second order effects (EN 994--, 6.7.3.4) Second order effects may be allowed for by multiplying the bigger of the M Sd. by a factor k, where k is greater or equal to.0. 0 k 0 N Sd. 0 N cr.eff. 0 Where: 0 max 0.66 0.44 M Sd. 0 M Sd. 0 0.44 0 0.496 0 k 0 N Sd. 0 N cr.eff. 0 k 0 0.596 k 0 max k 0 k 0.000 Thus: M Sd. 0 k 0 M Sd. 0 M Sd. 0.5kN m

3 (54) Transverse shear Resistance of the steel tube (EN 993--, 6..6) V plards. 0 A v f a. 0 3 V plards. 0 386.3kN Resistance of the concrete (EN 99--, 6.) Resistance of the concrete without shear reinforcement. (6..) cp. 0 min N pmrd. 0 N plrd. 0 N Sd. 0 A c 0. f c. 0 cp. 0 8.0MPa Design value of the shear resistance is given by: V Rdc. 0 C Rdc 00 min 0.0 f c. 0 3 k cp. 0 b w d With the minimum value of: V Rd.c.min. 0 v min. 0 k cp. 0 b w d Calculation of design shear resistance: C Rdc 0.8 V Rdc. 0 C Rdc 00 min 0.0 f c. 0 MPa 3 k cp. 0 MPa b w d MPa V Rdc. 0 86.5kN Minimum value of the shear resistance: 3 0.5 f c. 0 v min. 0 0.035 k MPa v min. 0 0.66 recommended value V Rd.c.min. 0 v min. 0 k cp. 0 MPa b w d MPa V Rd.c.min. 0 83.9kN Design shear resistance. V Rd.c. 0 max V Rdc. 0 V Rd.c.min. 0 V Rd.c. 0 86.5kN

33 (54) Resistance of concrete cross-section with shear reinforcement (6..3) The shear resistance of a member with perpendicular shear reinforcement is equal to: V Rds. 0 A sw s z f s. 0 tan but not greater than: f c. 0 V Rdmax. 0 cw. 0 b w z. 0 tan cot Where:. 0 0.6 f c. 0 50MPa. 0 0.504 cw. 0 depends on cp. 0. cp. 0 min N pmrd. 0 N plrd. 0 N Sd. 0 A c cp. 0 4.7MPa cw. 0 if cp. 0 0 cw. 0.5 cp. 0 if 0 f cp. 0 0.5 f c. 0 c. 0.5 if 0.5 f c. 0 cp. 0 0.5 f c. 0.5 cp. 0 if 0.5 f f c. 0 cp. 0 c. 0 The maximum effective cross-sectional area of the shear reinforcement, calculated for cot =, is given by: A swmax. 0 0.5 cw. 0. 0 f c. 0 b w s f s. 0 A swmax. 0 799.3mm Resistance of stirrups in tension: A sw 56.5mm V Rds. 0 A sw s z f s. 0 tan V Rds. 0 6.4kN

34 (54) Resistance of concrete chords in compression: f c. 0 V Rdmax. 0 cw. 0 b w z. 0 tan cot Resistance of cross-section requiring shear reinforcement V Rd.reinf. 0 min V Rds. 0 V Rdmax. 0 V Rdmax. 0 V Rd.reinf. 0 50.9kN 6.4kN Resistance of the column to transverse shear. If number of longitudinal reinforcing bars is 0, resistance to shear is calculated from: V Rd. 0 V Rd.c. 0 V plards. 0 Otherwise it is equal to: V Rd. 0 max V Rd.c. 0 V Rd.reinf. 0 V plards. 0 V Rd. 0 47.8kN To simplify the calculation and avoid the reduction of the yield strength, limitation has been imposed that: V a.sd. 0 max V Sd. 0 max V Rd.c. 0 V Rd.reinf. 0 0 V a.sd. 0 V plards. 0 0.5 In the considered case: V a.sd. 0 0.0N => V a.sd. 0 V plards. 0 0.0 Utilization of the cross-section in shear. V Sd. 0 V Rd. 0 0.07 < 0.5 Shear does not affect the bending resistance.

35 (54) Resistance of column for bending (EN 994--, 6.7.3.3) M Rd. 0 M 0 M pl.rd. 0 To determine the bending resistance of the cross-section, which is depending on the axial force, value of coefficient has to be callculated. A graphic method is used to determine. (see Figure 6). Symbols that are used are evaluated below. pm. 0 N pmrd. 0 N plrd. 0 pm. 0 0.450 0 N Rd. 0 N plrd. 0 0 0.775 d. 0 N Sd. 0 N plrd. 0 d. 0 0.367 n. 0 0 M Sd. 0 M Sd. 0 4 n. 0 0.66 M max.rd. 0 M pl.rd. 0.7 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0. 0.4 0.6 0.8..4 Figure 6. Calculation of µ in fire situation.

36 (54) The bending resistance in fire situation is checked as follows: Thus: 0 0.9657 0 0 if 0.0 0 0.9657 0 F if 0.0 M Sd. 0 0 M pl.rd. 0 0.083 < M 0.9 Results: EA 0 379.9MN EI effii. 0 465.0kN m N plrd. 0 49.7kN N pmrd. 0 5.5kN N Rd. 0 386.9kN M pl.rd. 0 68.3kN m M max.rd. 0 30.3kN m 0 M pl.rd. 0 59.kN m V Rd. 0 47.8kN

37 (54) 3 CIRCULAR COLUMN 3. Ambient design Data for the calculation: col t 355.6mm 6mm width and height of the steel tube wall thickness of the steel tube n 6 number of reinforcing bars fi fi s s u s L eff 5mm 8mm 375mm 45mm 000mm diameter of reinforcing bars diameter of stirrups stirrups spacing distance from the inner surface of the tube to the surface of the bar effective length of the column u.s fi.s t fi col Figure 7. Symbols used for the basic design variables. Material properties are the same is in previous example Steel Concrete Reinforcement f y 355MPa f ck 40MPa f sk 500MPa characteristic strength gamma a. gamma c.35 gamma s.5 material safety factor f ad 3.7MPa f cd 9.6MPa f sd 434.8MPa design strengths E a 0000MPa E cm 35000MPa E s 0000MPa Young modulus

38 (54) Forces in the column in normal situation: N Sd 54.0kN M Sd 3.5kN m bigger bending moment (upper or lower end) M Sd 3.0kN m smaller bending moment (upper or lower end) V Sd.3kN constant shear force Characteristics of the cross section Steel tube A a 4 ( col) ( col t) A a 6590mm tube cross-section area I a 64 ( col) 4 ( col t) 4 I a.007 0 8 mm 4 tube second moment of area W pa 6 ( col) 3 ( col t) 3 W pa 7.334 0 5 mm 3 plastic modulus Reinforcement A bar fi 4 A bar 49mm cross-section area of a single bar A s n A bar A s 945mm cross-section area of the reinforcement dist col t u s fi dist 4.3mm distance between the axis of the outermost bar to the midline of the column W ps A bar dist 4 sin 6 W ps 4.4cm 3 plastic modulus of the reinforcement cross-section I s A bar dist 4 sin 6 I s.94 0 7 mm 4 second moment of area of the reinforcement Concrete core A cgross 4 ( col t) A cgross 975mm A c A cgross A s A c 89780mm gross cross-section area of the concrete nett cross-section area of the concrete W pc 6 ( col t) 3 W ps W pc 6536.5cm 3 plastic modulus of the concrete cross-section

39 (54) ( col t) 4 I cgross 64 I cgross 6.84 0 8 mm 4 gross second moment of area of the concrete I I c 6.650 0 8 mm 4 nett second moment of area c I cgross I s of the concrete Check if the reinforcement meets the requirement of the reinforcement level A s A cgross max 6.0 % > 3.% > min.5 % Requirement met Maximum design bending resistance for the cross-section is given by equation: M max.rd W pa f ad W ps f sd W pc f cd M max.rd 43.kN m Calculation of the location of the neutral axis. A.an h.n A.sn A.cn Figure 8. Symbols used in calculation of plastic bending resistance Design axial resistance of the concrete only is equal to: N pmrd A c f cd N pmrd 660kN Equilibrium equation: 0.5 N pmrd 0.5 A cn f cd A sn f sd A an f ad Due to the shape of the column cross-section, the calculation of h n is very difficult. An iterational method has been used to compute this. Concrete core has been divided into 00.000 layers, h n has been moving away from the centerline one layer at a time and equilibrium equation has been checked at each step. The first value of h n for which the right side of the equation got bigger than the left is the neutral axis (Appendix).

40 (54) h n 55.4mm Thus: A sn 808mm W psn 47mm 3 W pan 3784mm 3 W pcn 985770 mm 3 With this data M pl.rd can be now calculated. M pl.rd M max.rd W pan f ad W psn f sd W pcn f cd M pl.rd 386.4kN m Values of the resistance to axial force (EN 994--, 6.7.3.) N plrd A a f ad A c f cd A s f sd N plrd 6067kN design resistance N plr A a f y A c f ck A s f sk N plr 7403kN characteristic resistance Effective flexural and axial stiffnesses (EN 994--, 6.7.3.3) EI eff E a I a 0.6 E cm I c E s I s EI eff 395.5kN m N cr L eff EI eff N cr 96605.0kN Check if the long term effect need to be taken into account A a f ad N plrd 0.35 is the contribution ratio of the steel tube, has to be between 0. and 0.9 - only then column works as a composite N plr N cr 0.77 is the relative slenderness of the column, has to be smaller than (EN 994-- 6.7.3.3) vert 0.8 vert.3 According to [], if is smaller than vert, long term effects do not need to be taken into account.

4 (54) Design value of the effectibve flexural stiffness, used to determine the sectional forces. (EN 994-- 6.7.3.4) EI effii 0.9 E a I a 0.5 E cm I c E s I s EI effii 334.6kN m N cr.eff L eff EI effii N cr.eff 8776.kN Stiffness to axial compression, used to determine the sectional forces.: EA E a A a E s A s E cm A c EA 544.6MN Axial resistance (EN 994--, 6.7.3.5) N Rd N plrd Where: is the reduction factor for the relevant buckling mode, in terms of relative slenderness (EN 993--, 6.3..) if 0. or N.Sd/N.plRd =< 0. than =. N Sd N plrd 0.089 N Rd N plrd N Rd 6067.4kN Utilization check: N Sd N Rd 0.089 <.0 Transverse shear Resistance of the steel tube (EN 993--, 6..6) A v A s A v 875.0mm area in shear for RHS V plards A v f y 3 gamma a V plards 048.kN plastic shear resistance of the tube.

4 (54) Resistance of the concrete (EN 99--, 6.) As a simplification the heigth and width od the column cross-section core are taken as 0.8 of the actual core diameter []. Core shear resistance is calculated as for rectangular section. (Figure 9.) d col-t Figure 9. Effective width and depth of the cross section. d 0.8 col t ( ) u s fi d 7.4mm effective depth of the cross-section b w 0.8 ( col t) b w 74.9mm effective width of the cross section A sl A s A sn A sl 876.6mm area of the tensile reinforcement, properly anchored A sl b w d 0.03 ratio of the longitudinal reinforcement Resistance of the concrete without shear reinforcement. (6..) cp min N pmrd N plrd N Sd A c 0. f cd cp.6mpa Design value of the shear resistance is given by: V Rdc C Rdc 00 min 0.0 f ck 3 k cp b w d With the minimum value of: V Rd.c.min v min k cp b w d

43 (54) Calculation of design shear resistance: C Rdc 0.8 gamma c k 0.5 C Rdc 0.33 recommended value recommended value V Rdc C Rdc 00 min 0.0 f ck MPa 3 k cp MPa b w d MPa V Rdc 58.kN Minimum value of the shear resistance: k min 00mm d 0.5 k.0 v min 3 0.035 k f ck MPa 0.5 v min 0.607 recommended value V Rd.c.min v min k cp MPa b w d MPa V Rd.c.min 60.0kN In the end design shear resistance is the bigger of the velues V Rdc and V Rd.c.min. V Rd.c max V Rdc V Rd.c.min V Rd.c 60.0kN Resistance of concrete cross-section with shear reinforcement (6..3) The shear resistance of a member with perpendicular shear reinforcement is equal to: V Rds A sw s z tan f sd but not greater than: f cd V Rdmax cw b w z tan cot Where: z 45deg 0.9 d angle between concrete compression strut and column axis [degrees]. If not proven otherwise equal to 45 degrees. z 95.6mm

44 (54) 0.6 f ck 50MPa 0.504 is the strength reduction factor for concrete cracked in shear cw is the coefficient taking account of the state of the stress in the compression chord, it depends on cp. cp min N pmrd N plrd N Sd A c cp.6mpa cw if cp 0 cw.09 cp if 0 f cp 0.5 f cd cd.5 if 0.5 f cd cp 0.5 f cd.5 cp if 0.5 f f cd cp cd The maximum effective cross-sectional area of the shear reinforcement, calculated for cot =, is given by: A swmax 0.5 cw f cd b w s f sd A swmax 98.4mm Cross sectional area of a single stirrup. A sw fi s 4 A sw 00.5mm Resistance of stirrups in tension: V Rds A sw s z tan f sd V Rds.8kN Resistance of concrete chords in compression: f cd V Rdmax cw b w z tan cot V Rdmax 437.4kN Resistance of cross-section requiring shear reinforcement V Rd.reinf min V Rds V Rdmax V Rd.reinf.8kN

45 (54) Resistance of the column to transverse shear. If number of longitudinal reinforcing bars is 0, resistance to shear is calculated from: V Rd V Rd.c V plards Otherwise it is equal to: V Rd max V Rd.c V Rd.reinf V plards V Rd 08.kN Influence of shear on the resistance to bending has to be taken into account if the part of the shear force that is coming to the steel tube V a.sd is greater than half it's resistance V plards It is done by reduction of the design resistance of the steel tube, calculated using reduced yield strength in the part of the tube that carries shear - A v (EN 993--, 6..8). shear f yd Where: shear V a.sd V plards Where: V plards V a.sd is the plastic design resistance of the steel tube only is the part of the shear force that is coming to the steel tube, assuming that the resistance of the concrete core to shear is fully utilized. V a.sd is calculated as follow: To avoid the reduction of the yield strength, limitation has been imposed: V a.sd V plards 0.5 V a.sd max V Sd max V Rd.c V Rd.reinf 0 In the considered case: V a.sd 6.3kN => V a.sd V plards 0.059 Utilization of the cross-section in shear. V Sd V Rd 0.0 < 0.5

46 (54) Rewritten in another form: V Sd V Rd 0. <.0 Shear does not affect the bending resistance. Resistance of column for bending (EN 994--, 6.7.3.3) M Rd M M pl.rd To determine the bending resistance of the cross-section, which is depending on the axial force, value of coefficient has to be calculated. A graphic method is used to determine (Figure.) Symbols that are used are evaluated below. pm N pmrd N plrd pm 0.438 N Rd N plrd.000 d N Sd N plrd d 0.089 n M Sd M Sd 4 n 0.465 M max.rd M pl.rd.6

47 (54) 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. 0 0 0. 0.4 0.6 0.8..4 Figure 0. Graphic method for calculating. Thus: 0.8.0377 M Sd M pl.rd 0.38 < M 0.9 Results: EA 544.6MN EI effii 33.MN m N plrd N pmrd 6067.4kN 660.kN N Rd M pl.rd 6067.4kN 386.4kN m M max.rd M pl.rd 43.kN m 40.0kN m V Rd 08.kN

48 (54) 4 RESULTS Following some resistances of different column cross-section are presented. The results are for three design situations: At ambient temperature In fire situation (R60), at elevated temperature In fire situation, at 0 C

49 (54) S355 C 40/50 A500 HW Table. Resistances of square columns. COLUMN AMBIENT FIRE - R60 FIRE at 0 C steel tube reinf. N.pl.Rd M.pl.Rd N.pm.Rd N.fi.pl.Rd M.fi.pl.Rd N.fi.pm.Rd N.0.pl.Rd M.0.pl.Rd N.0.pm.Rd 50 x 50 x 5 0T00 57 57 58 33 4 65 84 64 784 50 x 50 x 5 4T 700 63 567 450 8 58 0 70 766 50 x 50 x 5 4T6 84 67 557 550 54 83 75 75 80 x 80 x 5 0T00 986 84 856 634 6 564 399 94 56 80 x 80 x 5 4T 69 93 843 780 3 555 607 04 38 80 x 80 x 5 4T6 3 99 83 893 8 548 768 4 80 x 80 x 5 4T0 495 06 89 039 4 539 977 9 06 80 x 80 x 5 4T5 78 6 798 399 37 55 330 3 077 00 x 00 x 5 0T00 38 06 070 848 7 769 89 8 444 00 x 00 x 5 4T6 654 4 046 3 75 398 39 4 00 x 00 x 5 4T0 837 33 03 6 30 74 3407 49 394 00 x 00 x 5 4T5 34 46 0 634 48 77 373 64 365 00 x 00 x 6 0T00 550 4 047 848 9 753 3067 38 44 00 x 00 x 6 4T6 876 4 03 08 3 736 3437 58 38 00 x 00 x 6 4T0 3059 50 00 343 35 76 3645 69 363 00 x 00 x 6 4T5 3345 63 989 6 47 7 3970 83 335 0 x 0 x 6 0T00 939 5 8 30 05 3554 69 73 0 x 0 x 6 4T6 365 73 58 39 8 006 394 93 698 0 x 0 x 6 4T0 3449 84 45 630 43 996 43 06 680 0 x 0 x 6 4T5 3735 00 4 9 58 979 4457 4 65 50 x 50 x 6 0T00 3568 99 678 565 4 444 4345 66 50 x 50 x 6 4T6 3894 5 655 76 3 44 475 5 34 50 x 50 x 6 4T0 4077 39 64 973 47 4 493 68 5 50 x 50 x 6 4T5 4364 60 60 03 64 394 548 9 87 50 x 50 x 6 8T0 4586 7 604 38 84 380 550 308 65 50 x 50 x 6 8T5 559 307 56 84 7 344 65 350 09 300 x 300 x 8 0T00 5405 377 390 544 8 347 6543 4 36 300 x 300 x 8 4T0 595 49 353 933 70 3 7 48 376 300 x 300 x 8 4T5 60 457 33 396 05 90 7447 5 348 300 x 300 x 8 4T3 6709 503 94 3776 50 53 803 566 3098 300 x 300 x 8 8T5 6996 58 73 4048 87 33 8350 587 3069 300 x 300 x 8 8T3 80 595 99 5007 70 60 9503 678 969 350 x 350 x 0 0T00 766 636 37 374 5 3398 984 709 4356 350 x 350 x 0 4T0 85 695 389 459 7 3358 976 777 4306 350 x 350 x 0 4T5 84 76 368 4600 55 3336 0087 83 477 350 x 350 x 0 4T3 899 779 33 507 0 397 0664 874 47 350 x 350 x 0 8T5 907 797 30 5487 68 375 0990 900 499 350 x 350 x 0 8T3 03 888 3036 6700 393 397 44 008 4099 400 x 400 x 0 0T00 933 845 479 5070 69 4707 34 943 5776 400 x 400 x 0 4T0 98 98 44 565 49 4666 89 07 576 400 x 400 x 0 4T5 009 957 40 5954 96 4643 7 07 5697 400 x 400 x 0 4T3 066 03 483 6560 77 4603 794 48 5647

50 (54) S355 C 40/50 A500 HW Table. Resistances of circular columns, with some of the bars lying on the centerline COLUMN AMBIENT FIRE - R60 FIRE at 0 C steel tube reinf. N.pl.Rd M.pl.Rd N.pm.Rd N.fi.pl.Rd M.fi.pl.Rd N.fi.pm.Rd N.0.pl.Rd M.0.pl.Rd N.0.pm.Rd 68,3 x 5 0T00 4 49 583 38 4 35 698 55 787 68,3 x 5 4T 594 55 570 539 0 37 906 6 769 68,3 x 5 4T6 737 58 559 66 4 3 068 66 755 68,3 x 5 4T0 90 6 546 8 6 304 76 70 737 9, x 5 0T00 03 86 07 85 8 773 567 97 374 9, x 5 4T6 49 03 994 30 755 937 6 34 9, x 5 4T0 6 0 980 44 39 744 346 6 33 9, x 5 4T5 898 9 959 75 49 78 347 38 95 9, x 5 6T5 396 9 930 0 60 706 39 49 56 73 x 5 0T00 968 38 60 538 3 434 3667 55 73 73 x 5 4T6 394 6 586 884 45 43 4037 83 4 73 x 5 4T0 3477 74 57 079 6 40 446 98 3 73 x 5 4T5 3764 9 55 48 84 38 457 8 094 73 x 5 6T5 46 06 5 873 00 356 50 35 055 73 x 6 0T00 309 6 585 556 5 49 397 8 40 73 x 6 4T6 3535 85 56 907 48 407 497 08 08 73 x 6 4T0 379 96 548 05 63 395 4505 090 73 x 6 4T5 4005 57 448 85 376 4830 4 06 73 x 6 6T5 4403 7 498 894 0 350 58 59 0 33,9 x 6 0T00 498 33 64 46 69 583 6 3056 33,9 x 6 4T0 4707 79 7 983 87 3 576 35 3006 33,9 x 6 4T5 4993 30 06 33 0 6087 343 978 33,9 x 6 6T5 539 30 77 3770 4 8 6538 365 938 33,9 x 6 4T3 550 337 69 3930 67 74 6663 386 98 33,9 x 6 6T3 653 367 4683 0 6 7403 4 863 33,9 x 6 8T3 6804 438 073 5435 84 078 843 503 799 355,6 x 6 0T00 4874 84 747 30 8 837 6048 38 3709 355,6 x 6 4T0 5383 334 70 360 03 799 666 378 3659 355,6 x 6 4T5 5670 360 689 3933 39 777 695 409 3630 355,6 x 6 6T5 6067 380 660 4394 63 747 7403 43 359 355,6 x 6 4T3 677 400 65 45 94 739 758 458 3580 355,6 x 6 6T3 689 43 604 577 3 690 868 495 356 355,6 x 6 8T3 748 507 557 603 38 64 9008 58 345 406,4 x 8 0T00 6778 485 3547 403 53 3807 8343 543 4788 406,4 x 8 6T0 754 559 349 4986 5 3747 90 63 473 406,4 x 8 T0 8306 647 3435 5868 79 3687 0077 733 4637 406,4 x 8 6T5 797 597 3460 548 6 373 9698 676 4670 406,4 x 8 6T3 8733 66 3404 6363 30 3653 056 753 4595 406,4 x 8 8T3 9385 753 3356 76 408 360 30 858 453 457 x 8 0T00 868 63 456 5379 69 504 06 697 60 457 x 8 8T0 986 755 445 6553 56 4958 7 853 6009 457 x 8 8T5 9759 86 4409 73 346 49 9 935 5953 457 x 8 8T3 0774 945 4335 8384 47 488 3075 076 585 508 x 0 0T00 059 947 554 697 4 6359 3036 060 748 508 x 0 8T0 609 096 5467 8089 336 674 49 35 738

5 (54) S355 C 40/50 A500 HW Table 3. Resistances of circular columns, without any bars lying on the centerline COLUMN AMBIENT FIRE - R60 FIRE at 0 C steel tube reinf. N.pl.Rd M.pl.Rd N.pm.Rd N.fi.pl.Rd M.fi.pl.Rd N.fi.pm.Rd N.0.pl.Rd M.0.pl.Rd N.0.pm.Rd 68,3 x 5 0T00 4 49 583 38 4 35 698 55 787 68,3 x 5 4T 594 54 570 539 8 37 906 60 769 68,3 x 5 4T6 737 57 559 66 3 068 64 755 68,3 x 5 4T0 90 6 546 8 5 304 76 69 737 9, x 5 0T00 03 86 07 85 8 773 567 97 374 9, x 5 4T6 49 0 994 755 937 34 9, x 5 4T0 6 08 980 44 30 744 346 33 9, x 5 4T5 898 9 959 75 45 78 347 34 95 9, x 5 6T5 396 37 930 0 69 706 39 56 56 73 x 5 0T00 968 38 60 538 3 434 3667 55 73 73 x 5 4T6 394 60 586 884 3 43 4037 77 4 73 x 5 4T0 3477 70 57 079 4 40 446 90 3 73 x 5 4T5 3764 86 55 48 6 38 457 08 094 73 x 5 6T5 46 5 873 06 356 50 4 055 73 x 6 0T00 309 6 585 556 5 49 397 8 40 73 x 6 4T6 3535 83 56 907 33 407 497 05 08 73 x 6 4T0 379 95 548 05 43 395 4505 7 090 73 x 6 4T5 4005 57 448 65 376 4830 35 06 73 x 6 6T5 4403 34 498 894 07 350 58 66 0 33,9 x 6 0T00 498 33 64 46 69 583 6 3056 33,9 x 6 4T0 4707 76 7 983 56 3 576 308 3006 33,9 x 6 4T5 4993 98 06 33 8 6087 33 978 33,9 x 6 6T5 539 38 77 3770 49 8 6538 37 938 33,9 x 6 4T3 550 334 69 3930 9 74 6663 373 98 33,9 x 6 6T3 653 383 4683 6 6 7403 436 863 33,9 x 6 8T3 6804 43 073 5435 75 078 843 494 799 355,6 x 6 0T00 4874 84 747 30 8 837 6048 38 3709 355,6 x 6 4T0 5383 38 70 360 78 799 666 364 3659 355,6 x 6 4T5 5670 350 689 3933 89 777 695 389 3630 355,6 x 6 6T5 6067 386 660 4394 73 747 7403 439 359 355,6 x 6 4T3 677 388 65 45 38 739 758 433 3580 355,6 x 6 6T3 689 446 604 577 43 690 868 508 356 355,6 x 6 8T3 748 499 557 603 308 64 9008 57 345 406,4 x 8 0T00 6778 485 3547 403 53 3807 8343 543 4788 406,4 x 8 6T0 754 563 349 4986 77 3747 90 636 473 406,4 x 8 T0 8306 777 3435 5868 467 3687 0077 883 4637 406,4 x 8 6T5 797 605 3460 548 30 373 9698 684 4670 406,4 x 8 6T3 8733 676 3404 6363 37 3653 056 768 4595 406,4 x 8 8T3 9385 74 3356 76 398 360 30 844 453 457 x 8 0T00 868 63 456 5379 69 504 06 697 60 457 x 8 8T0 986 75 445 6553 6 4958 7 85 6009 457 x 8 8T5 9759 87 4409 73 353 49 9 99 5953 457 x 8 8T3 0774 930 4335 8384 49 488 3075 060 585 508 x 0 0T00 059 947 554 697 4 6359 3036 060 748 508 x 0 8T0 609 09 5467 8089 34 674 49 3 738

5 (54) S355 C 40/50 A500 HW Table 4. Resistances of rectangular columns about major axis. COLUMN AMBIENT FIRE - R60 FIRE at 0 C steel tube reinf. N.pl.Rd M.pl.Rd N.pm.Rd N.fi.pl.Rd M.fi.pl.Rd N.fi.pm.Rd N.0.pl.Rd M.0.pl.Rd N.0.pm.Rd 00 x 0 x 5 0T00 60 77 69 33 6 5 937 86 836 00 x 0 x 5 4T0 747 84 60 37 9 47 08 94 83 50 x 50 x 6 0T00 476 45 973 735 640 967 6 34 50 x 50 x 6 4T 659 60 960 844 0 63 375 79 96 50 x 50 x 6 4T6 80 7 949 930 7 65 3337 9 8 300 x 00 x 8 0T00 4048 9 548 449 3 85 4839 35 090 300 x 00 x 8 4T6 4373 35 54 673 47 66 509 365 058 300 x 00 x 8 4T0 4557 343 5 800 60 55 547 385 040 300 x 00 x 8 4T5 4843 370 490 47 94 37 5743 47 0 S355 C 40/50 A500 HW Table 5. Resistances of rectangular columns about minor axis. COLUMN AMBIENT FIRE - R60 FIRE at 0 C steel tube reinf. N.pl.Rd M.pl.Rd N.pm.Rd N.fi.pl.Rd M.fi.pl.Rd N.fi.pm.Rd N.0.pl.Rd M.0.pl.Rd N.0.pm.Rd 0 x 00 x 5 0T00 60 5 69 33 3 5 937 57 836 0 x 00 x 5 4T0 747 53 60 37 4 47 08 60 83 50 x 50 x 6 0T00 476 97 973 735 7 640 967 08 34 50 x 50 x 6 4T 659 0 960 844 0 63 375 4 96 50 x 50 x 6 4T6 80 06 949 930 65 3337 8 8 00 x 300 x 8 0T00 4048 548 449 5 85 4839 36 090 00 x 300 x 8 4T6 4373 8 54 673 7 66 509 55 058 00 x 300 x 8 4T0 4557 37 5 800 34 55 547 65 040 00 x 300 x 8 4T5 4843 48 490 47 49 37 5743 79 0

53 (54) 5 SUMMARY Calculation of the resistances of different concrete-filled tubes has been presented. For square column the resistance calculation is presented in 3 situations: At ambient temperature At elevated temperature in fire situation (fire class R60) At 0 C in fire situation For circular column only at ambient temperature The calculation of h n for round column takes into account reinforcing bars lying in the area between -h n and +h n from the centerline. It is also possible that only part of the bar is in this range. Matlab program has been written to iteratively find the location of the neutral axis. Concrete core of the column had been divided into 00 000 layers, the axis had been moved one layer at a time until the equilibrium of the internal forces was obtained. Including the reinforcement A an in the calculation lead to observation that in some cases the bending resistance of a column varies up to 57% (at elevated temperature), depending on the arrangement of the bars. The difference is maximum 7% in normal design situation. Also calculation of shear resistance of columns has been presented. Shear resistance have to be calculated for columns that are rigidly connected with the beams, and thus carry the shear loads. Assumption has been made that shear stresses in the steel tube, due to shear, are less than half the yield strength and do not affect the bending resistance. For the circular column the shear area of the concrete has been taken as square with the size of 0.8 times the core diameter. In the end the resistances of different cross-sections of columns, in 3 design situations, have been presented.

54 (54) REFERENCES: [] Teräsrakenneyhdistys, Betonitäytteisen teräsliittopilarin suunnitteluohje, 004 [] Suunnittelun sovellutusohjeet ja Betoninormien RakMK B4 suunnitteluosa, RakMK B ja B, by 6, Suomen Betoniyhdistys r.y., Jyväskylä 984. pp. 75, Figure S 34. [3] EN 994-- Eurocode 4: Design of composite steel and concrete structures Part -: General rules and rules for buildings [4] EN 99-- Eurocode : Design of concrete structures Part -: General rules and rules for buildings [5] EN 993-- Eurocode 3: Design of steel structures Part -: General rules and rules for buildings

A (/) APPENDIX Part of the MATLAB script that calculates h n, W psn and A sn needed for the calculation of plastic bending resistance of a circular cross-section column. Presented example regards column with 6 reinforcing bars, with none of them lying on the centerline. r = (col-*t)/; dist = r-us-fi/; i = 00000; layer = r/i; area = 0; x = 0; j = 0; y = 0; Asn = 0*Abar; while (0.5*Ac-(area-0.5*Asn))*fcd-4*(r+t/)*asin(y/(r+t/))*t*(fyd)-Asn*(fsd) >0 j = j+; y = (j-0.5)*layer; x = (r^-y^)^0.5; area = area+*layer*x; hn = y; if hn <= (sin(pi/6)*dist-fi/) Wpsn = 0; Asn = 0; elseif hn >= (sin(pi/6)*dist-fi/) && hn < (sin(pi/6)*dist+fi/) rbar = fi/; a = abs(hn-(sin(pi/6)*dist)); if hn >= (sin(pi/6)*dist)); alfa = pi+*asin(a/rbar); elseif hn < (sin(pi/6)*dist)); alfa = pi-*asin(a/rbar); end Asingle = rbar^/*(alfa-sin(alfa)); % Asingle is the area of the section of the disc xc = 4/3*rbar*(sin(alfa/))^3/(alfa-sin(alfa)); % xc is distance from the center of A to the axis of the bar Wpsn = 4* Asingle * (sin(pi/6)*dist-xc); Asn = 4* Asingle; elseif hn >= (sin(pi/6)*dist+fi/) Wpsn = 4* Abar*sin(pi/6)*dist; Asn = 4* Abar; end end

(3) Talonrakennustekniikan tutkimusraportit v. 998 00 47 Bzdawka, K., Composite column calculation examples. 46 Bzdawka, K., Optimisation of a steel frame building. TUT 009. 04 p. + 38 45 Leivo, V., Ohje uimahallien ja kylpylöiden lattioiden liukkauden ehkäisemiseen. TTY 009. xx s. + x liites. 44 Leivo, V., Uimahallien laattalattioiden liukkaus. TTY 009. xx s. + x liites. 43 Vinha, J., Viitanen, H., Lähdesmäki, K., Peuhkuri, R., Ojanen, T., Salminen, K., Paajanen, L., Strander, T. & Iitti, H. Rakennusmateriaalien ja rakenteiden homehtumisriskin laskennallinen arviointi. TTY 009. xx s. + liites. 4 4 Rauhala, J., Kylliäinen, M., Eristerapatun betoniseinän ilmaäänen eristävyys. TTY 009. 9 s. + 83 liites. 4 4 Aho, H., Korpi, M. (toim.) Ilmanpitävien rakenteiden ja liitosten toteutus asuinrakennuksissa. TTY 009. 00 s. 4 40 Vinha, J., Korpi, M., Kalamees, T., Jokisalo, J., Eskola, L., Palonen, J., Kurnitski, J., Aho, H., Salminen, M., Salminen, K., Keto, M. Asuinrakennusten ilmanpitävyys, sisäilmasto ja energiatalous. TTY 009. 39 Leivo, V, Rantala, J., Maanvastaisten rakenteiden mikrobiologinen toimivuus. TTY 006. 57 s. + 55 liites. 34 38 Heinisuo, M., Aalto, A., Stiffening of Steel Skeletons Using Diaphragms. TUT 006. 3 p. 7 app. 37 Kylliäinen, M., Talonrakentamisen akustiikka. 36 Varjonen, S., Mattila, J., Lahdensivu, J. & Pentti, M., Conservation and Maintenance of Concrete Facades Technical Possibilities and Restrictions. TUT 006. 9 p. 35 Heinisuo, M., Ylihärsilä, H., All metal structures at elevated temperatures. TUT 006. 54 p. 37 app. 34 Aho, H., Inha, T., Pentti, M., Paloturvallinen rakentaminen EPS-eristeillä. TTY 006. 06 s. + 38 liites. 33 Haukijärvi, M., Varjonen, S., Pentti, M., Julkisivukorjausten turvallisuus. TTY 006. 5 s. + liites. 3 Heinisuo, M., Kukkonen, J., Design of Cold-Formed Members Following New EN 993--3. TUT 3 Vinha, J., Korpi, M., Kalamees, T., Eskola, L., Palonen, J., Kurnitski, J., Valovirta, I., Mikkilä, A., Jokisalo, J., Puurunkoisten pientalojen kosteus- ja lämpötilaolosuhteet, ilmanvaihto ja ilmatiiviys. TTY 005. 0 s. + 0 liites. 30 Vinha, J., Käkelä, P., Kalamees, T., Valovirta, I. Puurunkoisten ulkoseinärakenteiden lämpö- ja 9 Vinha, J., Valovirta, I., Korpi, M., Mikkilä, A., Käkelä, P. Rakennusmateriaalien rakennusfysikaaliset ominaisuudet lämpötilan ja suhteellisen kosteuden funktiona. TTY 005. 0 s. + liites. 8 Leivo, V., Rantala, J., Lattialämmitetyn alapohjarakenteen rakennusfysikaalinen toiminta. TTY 7 Lahdensivu, J., Luonnonkiviverhottujen massiivitiiliseinien vaurioituminen ja korjausperiaatteet. TTY 003. 56 s. + 9 liites. 6 Leivo, V., Hirsirakennuksen yläpohjan tiiviys vaikutus lämpöenergiankulutukseen. TTY 003. 63 s. 5 Kylliäinen, M., Uncertainty of impact sound insulation measurements in field. TUT 003. 63 p. + 50 4 Myllylä, P., Lod, T. (toim.), Pitkäikäinen puurakenteinen halli, toimiva kosteustekniikka ja Postiosoite Käyntiosoite Puhelin (03) 35 www.tut.fi PL 600 Korkeakoulunkatu 5 Faksi (03) 35 8 FIN-330 Tampere 3370 Tampere

(3) 3 Mattila, J., Pentti, M., Suojaustoimien tehokkuus suomalaisissa betonijulkisivuissa ja parvekkeissa. Leivo, V., Rantala, J., Moisture Behavior of Slab-on-Ground Structures. TUT 003. 00 p. + Leivo, V., Rantala, J., Maanvastaiset alapohjarakenteet kosteustekninen mitoittaminen ja korjaaminen. TTKK 00. 33 s. + liites. 0 Leivo, V., Rantala, J., Maanvastaisten alapohjarakenteiden kosteustekninen toimivuus. TTKK 003. 9 Lindberg, R., Wahlman, J., Suonketo, J., Paukku, E., Kosteusvirtatutkimus. TTKK 00. 9 s. + 3 8 Hietala, J., Kelluvan betonilattian kaareutuminen, osa II. 7 Vinha, J., Käkelä, P., Kalamees, T., Comparison of the Moisture Behaviour of Timber-Framed Wall Structures in a One-Family House. lkaistaan lähiaikoina) 6 Vinha, J., Käkelä, P., Kalamees, T., Puurunkoisten seinärakenteiden kosteusteknisen toiminnan 5 Junttila, T., Venäjän rakennusalan säädöstö ja viranomaishallinto, osa 4 3 Junttila, T., Lod, T., Aro, J., Rakennusinvestointihankkeen toteuttaminen Moskovassa. TTKK 00. Junttila, T., (toim.), Venäjän rakentamisen oppikirja.osa B: Talonrakennustekniikka. TTKK 00. Junttila, T., (toim.) Venäjän rakentamisen oppikirja. Osa A: Liiketoimintaympäristö ja rakennushankkeen johtaminen. TTKK 00. 73 s. + liites 0 356. + 33 09 Junttila,T., (toim.) Kiinteistöjohtaminen Suomessa ja Venäjällä. Edellytykset kiinteistöalan 08 Hietala, J., Kelluvan betonilattian kaareutuminen. TTKK 00. 80 s. + 7 liites. 07 Binamu, A., Lindberg, R., The Impact of Air Tightness of The Building Envelope on The Efficiency of Ventilation Systems with Heat Recovery. 06 Leivo, V., Rantala, J., Maanvaraisten alapohjarakenteiden kosteuskäyttäytyminen. TTKK 000. 4 05 Junttila, T. (toim.), Venäjän federaation kaavoitus- ja rakennuslaki. 04 Niemelä, T., Vinha, J., Lindberg, R., Carbon Dioxide Permeability of Cellulose-Insulated Wall Structures. TUT 000. 46 p. + 9 app. 03 Vinha, J., Käkelä, P., Water Vapour Transmission in Wall Structures Due to Diffusion and Convection. 0 Suonketo, J., Pessi, A-M., Pentti, M., 0 Pessi, A-M., Suonketo, J., Pentti, M., Raunio-Lehtimäki, A. Betonielementtijulkisivujen 00 Pentti, M., Haukijärvi, M., Betonijulkisivujen saumausten suunnittelu ja laadunvarmistus. TTKK 99 Torikka, K., Hyypöläinen, T., Mattila, J., Lindberg, R., Kosteusvauriokorjausten laadunvarmistus. 98 Mattila, J., Peuhkurinen, T., Lähiökerrostalon lisärakentamishankkeen tekninen esiselvitysmenettely. Korjaus- ja LVIS-tekninen osuus. TTKK 999. 48 s. 97 Kylliäinen, M., Keronen, A., Lisärakentamisen rakennetekniset mahdollisuudet lähiöiden Postiosoite Käyntiosoite Puhelin (03) 35 www.tut.fi PL 600 Korkeakoulunkatu 5 Faksi (03) 35 8 FIN-330 Tampere 3370 Tampere

3(3) 96 Vinha, J., Käkelä, P., Vesihöyryn siirtyminen seinärakenteissa diffuusion ja konvektion vaikutuksesta. 95 Leivo, V. (toim.), Opas kosteusongelmiin rakennustekninen, mikrobiologinen ja lääketieteellinen 94 Pentti, M., Hyypöläinen, T., Ulkoseinärakenteiden kosteustekninen suunnittelu. TTKK 999. 50 s. 93 Lepo, K., Laatujärjestelmän kelpoisuus. TTKK 998. 0 s. + 50 liites. 9 Berg, Malinen, P., Leivo, V., Internal Monitoring of The Technology Programme for Improving Product Development Efficiency in Manufacturing Industries Rapid Programme. TUT 998. 8 s. + 93 liites. 9 Berg, P., Salminen, K., Leivo, V., Nopeat tuotantojärjestelmät teknologiaohjelman painoalueet vuosille 998-000 sekä ohjelman arviointi- ja ohjaussuunnitelma. TTKK 998. 55 s. + 37 liites. 90 Lindberg, R., Keränen, H., Teikari, M., Ulkoseinärakenteen vaikutus rakennuksen energiankulutukseen. TTKK 998. 34 s. + 6 liites. 89 Pentti, M., Huttunen, I., Vepsäläinen, K., Olenius, K., Betonijulkisivujen ja parvekkeiden korjaus. Tutkimusraportin hinta:, ellei toisin ole mainittu. Oikeus hinnanmuutoksiin pidätetään. Hintoihin lisätään alv 8 %. Myynti: Juvenes-Yhtiöt Oy, Tampereen teknillisen yliopiston kirjakauppa, Rakennustalo Korkeakoulunkatu 5, 3370 Tampere Puh.(03) 35 35, faksi (03) 35 9, TTY.kirjakauppa@juvenes.fi tai Tampereen teknillinen yliopisto,terttu Mäkipää puh. (03) 35 4804, terttu.makipaa@tut.fi Postiosoite Käyntiosoite Puhelin (03) 35 www.tut.fi PL 600 Korkeakoulunkatu 5 Faksi (03) 35 8 FIN-330 Tampere 3370 Tampere

Tampereen teknillinen yliopisto PL 57 330 Tampere Tampere University of Technology P.O.B. 57 FIN-330 Tampere, Finland