Biaxial Bending for Eurocode 2

Biaxial Bending for Eurocode 2 Guide

Brace Non-Slender Column DesignBiaxial Bending

Design of column between the ground floor and first floor. Assume the foundation is at the ground floor level. Hence, moments at the ground floor level are considered as zero amusing pin support conditions.

1. 400mm square column
2. Axial Load N_Ed = 3000kN
3. Moments at top M_y = 110kNm and M_z = 130kNm
4. Moments at Bottom are zero
5. f_ck = 30N/mm2
6. f_yk = 500N/mm2
7. Nominal Cover 25mm
8. Floor to Floor height 4000mm
9. Depth of the beam supported by the column 500mm

The following column design calculated is done for biaxial bending. This method is done with trial and error method as it has to be done until (M_Edz / M_Rdz)^a + (M_Edy / M_Rdy)^a  < 1. However, the 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 when a column design for biaxial bending. The old method is straight forward than the method considered following calculations because once we have found the biaxial bending moments, we can find the reinforcement area easily.

M_z  = 130 kNm

M_y  = 110 kNm

N_Ed = 3000 kN
Clear height               = 4000-500 = 3500mm                   

Effective length          = l_o       = factor * l

Factor                         = 0.85 (concise Eurocode 2, Table 5.1. This may more conservative).

l_o                                  = 0.85* 3500    = 2975mm                                 

Slenderness λ            = l_o/i

i   = radios of gyration       = h/√12

λ   = l_o/( h/√12 )       = 3.46*l_o/h    = 3.46*2975/400     = 25.73         

Limiting Slenderness λlim

λ_lim                               = 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-M_o1/M_o2

M_o1                              = 0 kNm

M_o2                              = 130kNm  where lMo2l ≥ lMo1l
C                                  = 1.7 – 0/130)  = 1.7

n                                  = N_Ed / (A_c*f_cd)

f_cd                                = f_ck / 1.5   = (30/1.5)*0.85   = 17

n                                  = 3000*1000 / (400*400*17)  = 0.94 

λ_lim                              = 20*0.7*1.1*1.7/√0.94    = 27 

λ_lim > 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.
M_Edy                             = Max{M_o2, M_oEd +M₂, M_o1 + 0.5M₂}

M_o2y                             = Max {M_top, M_bottom} + ei*N_Ed   = 110 + (2.975/400)*3000 ≥  Max(400/30, 20)*3000 = 110+22.3 ≥ 60  = 132.3kNm  > 60kNm

M_o1y                            = 0                                            

M_oEdy                           = 0.6*M_o2+ 0.4*M_o1 ≥ 0.4*M_o2 = 0.6*132.3 + 0.4* (0) ≥ 0.4*132.3  = 79.4≥ 52.9

M_2y                              = 0 , Column is not slender
M_Edy                             = Max{M_o2, M_oEd +M₂, M_o1 + 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.
M_Edz                             = Max{M_o2, M_oEd +M₂, M_o1 + 0.5M₂}M_o2z = 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                                            

M_oEdz                           = 0.6*M_o2+ 0.4*M_o1 ≥ 0.4*M_o2  = 0.6*152.3 + 0.4*0) ≥ 0.4*152.3  = 91.4≥ 60.9
M_2z                              = 0 , Column is not slender
M_Edz                             = Max{M_o2, M_oEd +M₂, M_o1 + 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,

M_Edz                             = 132.3kNm and M_Edy = 130kNm
Consider Critical Moment

M_Edz                             = 132.3kNm
M_Ed/ [b*(h₂)*f_ck]         = (132.3*10^6) / [400*(400^2)*30] = 0.07  = (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 referred 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.
A_s*f_yk / b*h*f_ck            = 0.3A_s   = 0.3*400*400*30 / 500 = 2880mm2

Provides six 25mm bars (As Provided 2940mm2)Six 25mm bars are provided for bending, but column have to be reinforced symmetrically. Therefor total no of base provided for the column are eight.

Check for Biaxial Bending

Further, a check is not required if

0.5 ≤ (λ_y/ λ_z) ≤  2.0 For rectangular column

and

0.2 ≥ (e_y/h_eq)/(e_z/b_eq) ≥ 5.0

Here λ_y and λ_z are slenderness ratios

λ_y is equal to λ_z as beam height is each direction is similar.

Therefore  λ_y/λ_z = 1 

Hence, λ_y/λ_z < 2 and > 0.5 OK

e_y/h_eq  =  (M_Edz / N_Ed) / h_eq

e_z/b_eq  =  (M_Edy / N_Ed) / b_eq     

(e_y/h_eq)/(e_z/b_eq) = M_Edz / M_Edy  

Here h=b=h_eq=b_eq, column is square

M_Edz  = 130 kNm

M_Edy  = 132.3 kNm  

(e_y/h_eq)/(e_z/b_eq)          = 130/132.3  = 0.98 > 0.2 and < 5

Therefore, biaxial check is required.
(M_Edz / M_Rdz)^a + (M_Edy / M_Rdy^)a ≤  1

M_Edz                             = 130kNm

M_Edy                             =132.3kNm

M_Rdz and M_Rdy are the moment resistance in respective direction, corresponding to an axial load N_Ed.

For symmetric reinforcement section

M_Rdz                            = M_Rdy

As Provided                 = 2940mm2
As*fyk / b*h*fck           = 2940*500/(400*400*30) = 0.31

N_Ed / (b*h*f_ck)             = 0.63

From the chart d₂/h =0.15

M_Ed/ [b*(h²)*_fck]         = 0.075

M_Ed                              = 0.098*400*400*400*30 = 144kNm
a                               = an exponent

a                               = 1.0 for N_Ed / N_Rd = 0.1

a                               = 1.5 for N_Ed / N_Rd = 0.7

N_Ed                              = 3000kN
N_Rd                              = Ac*f_cd + As*f_yd

N_Rd                              = 400*400*(0.85*30/1.5) + 3920* (500/1.15) = 4424kN                                      
N_Ed / N_Rd                     = 3000/4424 = 0.68

By interpolating, a   = 1.48
(M_Edz / M_Rdz)^a + (M_Edy / M_Rdy)^a =  (130 / 144)^1.48 + (132 / 144)^1.48  = 1.74 > 1                                                 

Hence, check for biaxial bending is not ok

Therefore, increase the number 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 opposite direction have not been considered).Therefore no of reinforcements, which are effective for moment resistance are eight (3920mm2) and twelve are effective for axial capacity (5880mm2). 

A_s*f_yk / b*h*f_ck            = 3920*500/(400*400*30) = 0.41

N_Ed / (b*h*f_ck)             = 0.63

From the chart d₂/h    =0.15

M_Ed/ [b*(h^2)*f_ck]      = 0.105

M_Ed                               = 0.105*400*400*400*30 = 202 kNm
a                               = an exponent

a                               = 1.0 for N_Ed / N_Rd = 0.1

a                               = 1.5 for N_Ed / N_Rd = 0.7

N_Ed                           = 3000kN
N_Rd                           = Ac*f_cd + As*f_ydN_Rd = 400*400*(0.85*30/1.5) + 3920* (500/1.15) = 4850kN
N_Ed / N_Rd                  = 3000/4850  = 0.62

By interpolating,

a                                = 1.43
(M_Edz / M_Rdz)^a + (M_Edy / M_Rdy)^a =  (130 / 202 )^1.43 + (132 / 202 )^1.43 = 1 (nearly equals to one)

Hence, provided reinforces are ok.

For more about Biaxial Bending for Eurocode 2 contact Structural Guide.