Biaxial Design

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Example 02
Brace Non Slender Column Design
Biaxial Bending
Desing of column between ground floor and first floor. Aassume foundation is at ground floor level. Hence, moments at ground floor level are zero.

• 400mm square column
• Axial Load NEd = 3000kN
• Moments at top My = 110kNm and Mz = 130kNm
• Moments at Bottom are zero
• fck 30N/mm2
• fyk 500N/mm2
• Nominal Cover 25mm
• Floor to Floor height 4000mm
• Depth of the beam supported by the column 500mm

Mz                        = 130 kNm
My                       = 110 kNm
NEd                      = 3000 kN

Clear height        = 4000-500
                           = 3500mm
Effective length    = lo
                            = factor * l
Factor                  = 0.85 (concise Eurocode 2, Table 5.1. This may more conservative).
lo                         = 0.85* 3500
                            = 2975mm

Slenderness λ   = lo/i

i                       = radios of gyration

                        = h/√12

λ                      = lo/( h/√12 )

                        = 3.46*lo/h

                        = 3.46*2975/400

                        = 25.73

Limiting Slenderness λlim

λlim                     = 20ABC/√n

A                      = 0.7 if effective creep factor is unknown

B                      = 1.1 if mechanical reinforcement ratio is unknown          

C                      = 1.7 - rm              

                        = 1.7-Mo1/Mo2

Mo1                  = 0 kNm

Mo2                  = 130kNm  where lMo2l ≥ lMo1l

C                      = 1.7 - 0/130)

                         = 1.7

n                       = NEd / (Ac*fcd)

fcd                    = fck / 1.5

                         = (30/1.5)*0.85
                         = 17
n                       = 3000*1000 / (400*400*17)

                         = 0.94

 λlim                         = 20*0.7*1.1*1.7/√0.94

                         = 27

 λlim > λ hence, column is not slender.

Calculation of design moments

Column has moments in both directions.

First find the critical moment.
Consider My,

Assume there is no moment at ground floor level as foundations are at that level.

MEdy                 = Max{Mo2, MoEd +M2, Mo1 + 0.5M2}

Mo2y                 = Max {Mtop, Mbottom} + ei*NEd

                        = 110 + (2.975/400)*3000 ≥  Max(400/30, 20)*3000

                        = 110+22.3 ≥ 60

                        = 132.3kNm  >  60kNm

Mo1y                = 0

                                            

MoEdy                   = 0.6*Mo2+ 0.4*Mo1 ≥ 0.4*Mo2

                       = 0.6*132.3 + 0.4*(0) ≥ 0.4*132.3

                       = 79.4≥ 52.9

M2y                      = 0 , Column is not slender

MEdy                 = Max{Mo2, MoEd +M2, Mo1 + 0.5M2}

                        = Max{132.3, 79.4+0, 0 + 0.5*0} 

                             = 132.3 kNm

Consider Mz,

Assume there is no moment at ground floor level as foundations are at that level.

MEdz                 = Max{Mo2, MoEd +M2, Mo1 + 0.5M2}

Mo2z                 = Max {Mtop, Mbottom} + ei*NEd

                        = 130 + (2.975/400)*3000 ≥  Max(400/30, 20)*3000

                        = 130+22.3 ≥ 60

                        = 152.3 kNm  >  60 kNm

Mo1z                  = 0

                                            

MoEdz                  = 0.6*Mo2+ 0.4*Mo1 ≥ 0.4*Mo2

                       = 0.6*152.3 + 0.4*(0) ≥ 0.4*152.3

                       = 91.4≥ 60.9

M2z                      = 0 , Column is not slender

MEdz                 = Max{Mo2, MoEd +M2, Mo1 + 0.5M2}

                        = Max{152.3, 91.4+0, 0 + 0.5*0} 

                             = 152.3 kNm

Imperfection needs to be considered only in one direction, which has the most unfavorable effect.

Therefore,

MEdz = 132.3kNm and MEdy = 130kNm

Consider Critical Moment MEdz = 132.3kNm

MEd / [b*(h^2)*fck]  = (132.3*10^6) / [400*(400^2)*30]                       

                                 = 0.07

NEd / (b*h*fck)        = (3000*10^6) / (400*400*30

                                 = 0.63

Assume 25mm diameter bars as main reinforcement and 10mm bars as shear links

d2                  = 25+10+25/2

                     = 47.5mm

d2/h                = 47.5 / 400

                      = 0.12

Note: d2/h = 0.15 chart is reffed to find the reinforcement area, but it is more conservative. Interpolation between charts d2/h = 0.10 and d2/h = 0.15 can be used to find more accurate answer.

As*fyk / b*h*fck      = 0.3

                           As    = 0.3*400*400*30 / 500

                             = 2880mm2

Provides six 25mm bars (As Provided 2940mm2)

Six 25mm bars are provided for bending, but colulmn have to be reinforced symmetrically. Therefor total no of base provided for the column are eight.

Check for Biaxial Bending
Further check is not required if 
0.5 ≤ ( λy/ λz) ≤  2.0 For rectangular column
and
0.2 ≥ (ey/heq)/(ez/beq) ≥ 5.0
Here λy and λz are slenderness ratios

λy is equal to λz as beam height is each directions are similar.
therefore  λy/λz = 1
Hence, λy/λz < 2 and > 0.5 OK

ey/heq  =  (MEdz / NEd) / heq
ez/beq  =  (MEdy / NEd) / beq
    

(ey/heq)/(ez/beq) = MEdz / MEdy  Here h=b=heq=beq, column is square

MEdz  = 130 kNm
MEdy  = 132.3 kNm 
(ey/heq)/(ez/beq) = 130/132.3
                                = 0.98 > 0.2 and < 5
Therefore, biaxial check is required.

(MEdz / MRdz)^a + (MEdy / MRdy)^a ≤  1

MEdz            = 130kNm
MEdy           =132.3kNm

MRdz and MRdy are the moment resistance in respective direction, corresponding to an axial load NEd.

For symetric reinforcement section

MRdz            = MRdy

As Provided  = 2940mm2

As*fyk / b*h*fck        = 2940*500/(400*400*30)
                             = 0.31

NEd / (b*h*fck)          = 0.63
From the chart d2/h =0.15

MEd / [b*(h^2)*fck]   = 0.075

MEd                           = 0.098*400*400*400*30
                             = 144kNm

a                               = an exponent
a                               = 1.0 for NEd / NRd = 0.1
a                               = 1.5 for NEd / NRd = 0.7

NEd                                    = 3000kN

NRd                                    = Ac*fcd + As*fyd
NRd                                    = 400*400*(0.85*30/1.5) + 3920*(500/1.15)
                                     = 4424kN

NEd / NRd                        = 3000/4424
                                 = 0.68
By interpolating
a                               = 1.48

(MEdz / MRdz)^a + (MEdy / MRdy)^a =  (130 / 144)^1.48 + (132 / 144)^1.48

                                                  = 1.74 > 1
Hence, Check for biaxial bending is not ok
Therefore, increase the noumner of bars to 12 (In this design, the reinforcement along the two sides have only taken to account for this design and effect of the other reinforcement bars at in opposit direction have not been considered).
Therefore no of reinforcements, which are effctive for moment resistance are eight (3920mm2) and twelve are effctive for axial capasity (5880mm2). 

As*fyk / b*h*fck        = 3920*500/(400*400*30)
                             = 0.41

NEd / (b*h*fck)          = 0.63
From the chart d2/h =0.15

MEd / [b*(h^2)*fck]   = 0.105

MEd                          = 0.105*400*400*400*30
                             = 202 kNm

a                               = an exponent
a                               = 1.0 for NEd / NRd = 0.1
a                               = 1.5 for NEd / NRd = 0.7

NEd                                    = 3000kN

NRd                                     = Ac*fcd + As*fyd
NRd                                    = 400*400*(0.85*30/1.5) + 3920*(500/1.15)
                                  = 4850kN

NEd / NRd                            = 3000/4850
                                 = 0.62
By interpolating
a                                = 1.43

(MEdz / MRdz)^a + (MEdy / MRdy)^a =  (130 / 202 )^1.43 + (132 / 202 )^1.43

                                                  = 1 (nearly equals to one)
Hence, Provided reinforcemts are ok.

Note :

This column is design for biaxial bending. This mathod is like trial and error method. Becouse, this has to be done untill (MEdz / MRdz)^a + (MEdy / MRdy)^a  < 1, but Manual for the design of concrete building structures to Eurocode 2 and the book, Reinforced concrete design to Eurocode 2 written by Bill Mosley, John Bungey and Ray Hulse have suggested the old method, which is in BS 8110.

Old method is more straight forward than this method, because once we have find the biaxial moment then we can find the reinforcement area easily.