Karol Bzdawka Composite Column - Calculation Examples

Transcription

1 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

2 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

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4 (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 Markku Heinisuo Professor of Metal Structures

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

6 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.

7 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

8 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 mm 4 tube second moment of area W pa 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 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 mm 4 gross second moment of area of the concrete I c I cgross I s I c 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

9 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

10 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 mm 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 [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

11 8 (54) Effective flexural and axial stiffnesses (EN 994--, ) 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 kN Check if the long term effect need to be taken into account A a f ad N plrd is the contribution ratio of the steel tube, has to be between 0. and 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 ) 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

12 9 (54) Design value of the effective flexural stiffness, used to determine the sectional forces. (EN ) 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--, ) N Rd N plrd Where: is the reduction factor for the relevant buckling mode, in terms of relative slenderness (EN [5], 6.3..) if 0. or N.Sd/N.plRd =< 0. than =. but =< Where: is an imperfection factor, equal to 0.49 (buckling curve C) min N Rd N plrd N Rd 347.kN

13 0 (54) Utilization check: N Sd N Rd <.0 Increse of the bending moment due to second order effects (EN 994--, ) 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 M Sd M Sd 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

14 (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 k f ck MPa 0.5 v min 0.66 recommended value

15 (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 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

16 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

17 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.

18 5 (54) Resistance of column for bending (EN 994--, ) 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 d N Sd N plrd d 0.45 n M Sd M Sd 4 n M max.rd M pl.rd.095

19 6 (54) N.Rd N.pl.Rd.0 =0.785.d=0.45.pm=0.40.n= 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 Figure 4. Graphic method using Matlab.

20 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: 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 Thus: M Sd M pl.rd < M 0.9 Results: EA 343.MN EI effii 6.4MN m N plrd kN 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

21 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 and f a. f y k y. => f a. 0.5MPa Young modulus in fire: k Ea and E a. k Ea. E a => E a. 3.7GPa

22 9 (54) Reduction factors for concrete are taken from Table 7 []. Design compressive strength in fire: k c and f c. f ck k c. => f c. 5.5MPa Young modulus in fire: cu and E csec. k c. f ck cu. => E csec MPa Reduction factors for reinforcement are taken from Table 8 []. Design compressive strength in fire: k s and f s. f sk k s. => f s MPa Young modulus in fire: k Es and E s. E s k Es. => E s MPa Forces in the column in fire situation: N Sd kN 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

23 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 mm 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 cm 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--, ) 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--, ) 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.

24 (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 kN 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 Where: is an imperfection factor, equal to 0.49 (buckling curve C) min N Rd. N plrd. N Rd. 57.kN Utilization check: N Sd. N Rd <.0

25 (54) Increse of the bending moment due to second order effects. (EN 994--, ) 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 M Sd. M Sd k N Sd. N cr.eff. k 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

26 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: f c. v min k MPa v min 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 cw. depends on cp.. cp. min N pmrd. N plrd. N Sd. A c cp. 0.4MPa

27 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 kN 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

28 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 < 0.5 Shear does not affect the bending resistance. Resistance of column for bending (EN 994--, ) 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 N Rd. N plrd d. N Sd. N plrd. d n. M Sd. M Sd. 4 n. 0.4 M max.rd. M pl.rd..80

29 6 (54) Figure 5. Calculation of µ in fire situation.

30 7 (54) The bending resistance in fire situation is checked as follows:.456 if.0 F if Thus: M Sd. M pl.rd. 0.8 < M 0.9 Results: EA 93.MN EI effii. 476.kN m N plrd kN 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

31 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 and f a. 0 f y k y. 0 => f a MPa k Ea and E a. 0 k Ea. 0 E a => E a GPa Concrete charcteristics in fire situation at 0 deg.: k c and f c. 0 f ck k c. 0 => f c MPa cu and E csec. 0 k c. 0 f ck => cu. 0 Reinforcement material charcteristics in fire situation at 0 deg.: E csec GPa k s and f s. 0 f sk k s. 0 => f s MPa k Es and E s. 0 E s k Es. 0 => E s GPa 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 kN M Sd. 0.5kN m bigger bending moment (upper or lower end) M Sd kN 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

32 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 kN 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 mm 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 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 cm 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 kN m

33 30 (54) Values of the resistance to axial force (EN 994--, ) N plrd. 0 A a f a. 0 A c f c. 0 A s f s. 0 N plrd kN design resistance N plr. 0 A a f a. 0 A c f c. 0 A s f s. 0 N plr kN characteristic resistance Effective flexural and axial stiffnesses (EN 994--, ) EI eff. 0 E a. 0 I a 0.8 E csec. 0 I c E s. 0 I s EI eff kN m N cr. 0 L eff. 0 EI eff. 0 N cr 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 E a. 0 I a 0.5 E csec. 0 I c E s. 0 I s EI effii kN m N cr.eff. 0 L eff. 0 EI effii. 0 N cr.eff kN 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 MN Axial resistance N Rd. 0 0 N plrd but =< 0 N plr. 0 N cr

34 3 (54) Where: is an imperfection factor, equal to 0.49 (buckling curve C) 0 min N Rd. 0 0 N plrd. 0 N Rd kN Utilization check: N Sd. 0 N Rd <.0 Increse of the bending moment due to second order effects (EN 994--, ) 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 M Sd. 0 M Sd k 0 N Sd. 0 N cr.eff. 0 k k 0 max k 0 k Thus: M Sd. 0 k 0 M Sd. 0 M Sd. 0.5kN m

35 3 (54) Transverse shear Resistance of the steel tube (EN 993--, 6..6) V plards. 0 A v f a. 0 3 V plards kN 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 MPa 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 kN Minimum value of the shear resistance: f c. 0 v min k MPa v min recommended value V Rd.c.min. 0 v min. 0 k cp. 0 MPa b w d MPa V Rd.c.min kN Design shear resistance. V Rd.c. 0 max V Rdc. 0 V Rd.c.min. 0 V Rd.c kN

36 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: f c. 0 50MPa cw. 0 depends on cp. 0. cp. 0 min N pmrd. 0 N plrd. 0 N Sd. 0 A c cp MPa cw. 0 if cp. 0 0 cw. 0.5 cp. 0 if 0 f cp f c. 0 c. 0.5 if 0.5 f c. 0 cp 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 cw f c. 0 b w s f s. 0 A swmax mm Resistance of stirrups in tension: A sw 56.5mm V Rds. 0 A sw s z f s. 0 tan V Rds kN

37 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 kN 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 kN 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 In the considered case: V a.sd N => V a.sd. 0 V plards Utilization of the cross-section in shear. V Sd. 0 V Rd < 0.5 Shear does not affect the bending resistance.

38 35 (54) Resistance of column for bending (EN 994--, ) 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 N Rd. 0 N plrd d. 0 N Sd. 0 N plrd. 0 d n. 0 0 M Sd. 0 M Sd. 0 4 n M max.rd. 0 M pl.rd Figure 6. Calculation of µ in fire situation.

39 36 (54) The bending resistance in fire situation is checked as follows: Thus: if F if 0.0 M Sd. 0 0 M pl.rd < M 0.9 Results: EA MN EI effii kN m N plrd kN N pmrd kN N Rd kN M pl.rd kN m M max.rd kN m 0 M pl.rd kN m V Rd kN

40 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

41 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 mm 4 tube second moment of area W pa 6 ( col) 3 ( col t) 3 W pa 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 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 cm 3 plastic modulus of the concrete cross-section

42 39 (54) ( col t) 4 I cgross 64 I cgross mm 4 gross second moment of area of the concrete I I c 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 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).

43 40 (54) h n 55.4mm Thus: A sn 808mm W psn 47mm 3 W pan 3784mm 3 W pcn 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--, ) 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--, ) 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 kN 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 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 ) vert 0.8 vert.3 According to [], if is smaller than vert, long term effects do not need to be taken into account.

44 4 (54) Design value of the effectibve flexural stiffness, used to determine the sectional forces. (EN ) 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--, ) 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 N Rd N plrd N Rd kN Utilization check: N Sd N Rd <.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.

45 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

46 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 k f ck MPa 0.5 v min 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

47 44 (54) 0.6 f ck 50MPa 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

48 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 Utilization of the cross-section in shear. V Sd V Rd 0.0 < 0.5

49 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--, ) 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 N Rd N plrd.000 d N Sd N plrd d n M Sd M Sd 4 n M max.rd M pl.rd.6

50 47 (54) Figure 0. Graphic method for calculating. Thus: M Sd M pl.rd 0.38 < M 0.9 Results: EA 544.6MN EI effii 33.MN m N plrd N pmrd kN 660.kN N Rd M pl.rd kN 386.4kN m M max.rd M pl.rd 43.kN m 40.0kN m V Rd 08.kN

51 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

52 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 0T x 50 x 5 4T x 50 x 5 4T x 80 x 5 0T x 80 x 5 4T x 80 x 5 4T x 80 x 5 4T x 80 x 5 4T x 00 x 5 0T x 00 x 5 4T x 00 x 5 4T x 00 x 5 4T x 00 x 6 0T x 00 x 6 4T x 00 x 6 4T x 00 x 6 4T x 0 x 6 0T x 0 x 6 4T x 0 x 6 4T x 0 x 6 4T x 50 x 6 0T x 50 x 6 4T x 50 x 6 4T x 50 x 6 4T x 50 x 6 8T x 50 x 6 8T x 300 x 8 0T x 300 x 8 4T x 300 x 8 4T x 300 x 8 4T x 300 x 8 8T x 300 x 8 8T x 350 x 0 0T x 350 x 0 4T x 350 x 0 4T x 350 x 0 4T x 350 x 0 8T x 350 x 0 8T x 400 x 0 0T x 400 x 0 4T x 400 x 0 4T x 400 x 0 4T

53 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 0T ,3 x 5 4T ,3 x 5 4T ,3 x 5 4T , x 5 0T , x 5 4T , x 5 4T , x 5 4T , x 5 6T x 5 0T x 5 4T x 5 4T x 5 4T x 5 6T x 6 0T x 6 4T x 6 4T x 6 4T x 6 6T ,9 x 6 0T ,9 x 6 4T ,9 x 6 4T ,9 x 6 6T ,9 x 6 4T ,9 x 6 6T ,9 x 6 8T ,6 x 6 0T ,6 x 6 4T ,6 x 6 4T ,6 x 6 6T ,6 x 6 4T ,6 x 6 6T ,6 x 6 8T ,4 x 8 0T ,4 x 8 6T ,4 x 8 T ,4 x 8 6T ,4 x 8 6T ,4 x 8 8T x 8 0T x 8 8T x 8 8T x 8 8T x 0 0T x 0 8T

54 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 0T ,3 x 5 4T ,3 x 5 4T ,3 x 5 4T , x 5 0T , x 5 4T , x 5 4T , x 5 4T , x 5 6T x 5 0T x 5 4T x 5 4T x 5 4T x 5 6T x 6 0T x 6 4T x 6 4T x 6 4T x 6 6T ,9 x 6 0T ,9 x 6 4T ,9 x 6 4T ,9 x 6 6T ,9 x 6 4T ,9 x 6 6T ,9 x 6 8T ,6 x 6 0T ,6 x 6 4T ,6 x 6 4T ,6 x 6 6T ,6 x 6 4T ,6 x 6 6T ,6 x 6 8T ,4 x 8 0T ,4 x 8 6T ,4 x 8 T ,4 x 8 6T ,4 x 8 6T ,4 x 8 8T x 8 0T x 8 8T x 8 8T x 8 8T x 0 0T x 0 8T

55 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 0T x 0 x 5 4T x 50 x 6 0T x 50 x 6 4T x 50 x 6 4T x 00 x 8 0T x 00 x 8 4T x 00 x 8 4T x 00 x 8 4T 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 0T x 00 x 5 4T x 50 x 6 0T x 50 x 6 4T x 50 x 6 4T x 300 x 8 0T x 300 x 8 4T x 300 x 8 4T x 300 x 8 4T

56 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 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.

57 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 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 Eurocode 3: Design of steel structures Part -: General rules and rules for buildings

58 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