 # V. EC3 French NA - Column with Axial Load

Calculate the axial and bending capacities and interaction ratios of a column using the French NA to EC3.

## Details

The 5 m tall column as a fixed base and is free to translate and rotate at the top. The column is subject to a 25 kN compressive load and moments of 5 kN·m about the X axis and 10 kN·m about the Z axis. The section is an HD320X127, grade S275 steel.

## Validation

Bending Capacity

Moment capacity:

The critical moment is given by:

$M c r = C 1 π 2 E I ( k L ) 2 { ( k k w ) 2 I w I + ( k L ) 2 G I t π 2 E I + ( C 2 Z g ) 2 − C 2 Z g }$
where
 C1 = 2.578 C2 = 1.554 $π 2 E I y k L 2$ = 7,477,200 $k k w 2 I w I y$ = 22,394 $k L 2 G I t π 2 E I y$ = 23,740 C2Zg = 1.554×160 = 248.6

Therefore, Mcr = 1,540.6 kN·m

From the French NA, λLT, 0 = 0.2 + 0.1×b / h = 0.2 + 0.1× 300 / 327 = 0.292, β = 1.0

So, $λ L T = w y f y M c r = 2,149 × 10 3 × 275 1,540.6 × 10 6 = 0.619$

H/b = 320/300 = 1.067 < 2. So, from Table 6.5 of Eurocode 3, $αLT=0.4−0.2bhλLT2=0.4−0.2300327(0.619)2=0.323$.

From Cl. 6.3.2.3 of Eurocode3:

ФLT = 0.5[1+αLTLT- λLT, 0) + β× λLT2] = 0.5[1 + 0.323×(0.619 - 0.292) +0.75 × 0.6192] = 0.746

So, $χLT=ФLT+ФLT2-βλLT2-1=0.861$

So, kc = 0.9 and thus modification factor, f = 1 - 0.5(1 - 0.9)×[1 -2(0.619 - 0.8)2] = 0.953

$χLT,mod⁡=χLT⁡/f=0.903<1/λLT2=2.603$

Compression Capacity

Critical compressive values:

$λ cy = A × f y N cry = 0.770$
$λ cz = A × f y N crz = 0.422$

Determine χy

h/b = 1.067 < 1.2 and tf = 20.5 mm < 100 mm

So, for y-y axis, use curve "c". Hence, α = 0.49 [Table 6.2, EC3]

So, ф = 0.5[1+α (λ - 0.2) +λ2)] = 0.936

$χ y = 1 ф + ф 2 - λ 2 0.681$ (governs)

Determine χz

h/b = 1.067 < 1.2 and tf = 20.5 mm < 100 mm

So, for z-z axis, use curve "b". Hence, α = 0.34 [Table 6.2, EC3]

So, ф = 0.5[1+α (λ - 0.2) +λ2)] = 0.627

$χ z = 1 ф + ф 2 - λ 2 0.917$ (does not govern)

Compression capacity:

χy×NRk = 0.681×7,477 = 3,020.2 kN

Compression ratio: 25 kN / 3,020.2 kN = 0.008

Critical Axial Loads for Flexure and Flexural Torsional Buckling

From NCCI document SN001a-EN-FU:

$N c r , T = 1 i 0 2 G I t + π 2 E I w l T 2$
where
 $i o 2$ = $i y 2 + i z 2 + y o 2 + z o 2$; here y0 = z0 = 0iy = 75.68 mm and iz = 138.23 mm = 24,835 mm2
$N c r , T F = i 0 2 2 ( i y 2 + i z 2 ) N c r , y + N c r , T - N c r , y + N c r , T 2 - 4 N c r , y N c r , T i y 2 + i z 2 i 0 2$

Interaction Check

From Annex A, Table A.1 of EC3, we get the auxiliary terms:

$μ z = 1 - N E d N c r z 1 - χ z N E d N c r z = 1.000$
$μ y = 1 - N E d N c r y 1 - χ y N E d N c r y = 0.999$

wz = wplz / welz = 1.116

wy = wply / wely = 1.525 > 1.5

$C m z , 0 = 0.79 + 0.21 × 1 + 0.36 ( 1 − 0.33 ) 25 24 , 943 = 1.000 > 1 − N E d N c r , z$
$C m y , 0 = 0.79 + 0.21 × 1 + 0.36 ( 1 − 0.33 ) 25 4 , 477 = 1.001 > 1 − N E d N c r , y$

Determine λ0: C1 = 1, C2 = 0, Mcr = 1,606 kN·m

$λ 0 = 2 , 149 × 10 3 × 275 1 , 606 × 10 6 = 0.607$
$λ 0 , min ⁡ = 0.2 C 1 ( 1 − N E d N c r , y ) × ( 1 − N E d N c r TF ) 4 = 0.321$
$a L T = 1 ⁢ − I t / I z = 0.993$
$ε z = M z , E d N E d × A w e l z = 3.350$
$C m z = C m z , 0 + ( 1 − C m z , 0 ) ε z a L T 1 + ε z a L T = 1.000$

Cmy = Cmy,0 = 1.001

$C m L T = C m z 2 a L T ( 1 − N E d N c r , y ) × ( 1 − N E d N c r T ) = 0.995 < 1$

So, CmLT = 1.0, λmax = max.(λzy) = λy = 0.770

$b L T = 0.5 a L T λ 0 2 M y E d χ L T × M p l y . R d × M z E d M p l z . R d = 0.5 × 0.993 × 0.607 2 5 0.909 × 258.3 × 10 591 ≈ 0$
$η p l = N E d N R k / Γ M 1 = 25 4 , 436 = 0.006$
$C z z = 1 + ( w z − 1 ) [ ( 2 − 1.6 w z C m z 2 λ max ⁡ − 1.6 w z C m z 2 λ max ⁡ 2 ) η p l − b L T ] = 1 + ( 1.116 − 1 ) [ ( 2 − 1.6 1.116 × 1 2 × 0.770 − 1.6 1.116 × 1 2 × 0.770 2 ) 0.006 − 0 ] = 1.000$
$c L T = 10 a L T λ 0 2 5 + λ y 4 × M z , E d C m z χ L T M p l z , R d = 0.013$
$C z y = 1 + ( w y − 1 ) [ ( 2 − 14 C m y 2 λ max ⁡ 2 w y 5 ) η p l − c L T ] = 0.996$
$C z y > 0.6 w y w z × w e l y w p l y = 0.456$
$d L T = 2 a L T λ 0 0.1 + λ y 4 × M z , E d C m z × χ L T × M p l z , R d × M y , E d C m y × M p l y , R d = 2 × 0.993 0.607 0.1 + 0.770 4 10 1 × 0.903 × 591 5 1.000 × 258.3 = 0.001$
$C y z = 1 + ( w z − 1 ) [ ( 2 − 14 C m z 2 λ max ⁡ 2 w z 5 ) η p l − d L T ] = 0.998$
$e L T = 1.7 a L T λ 0 0.1 + λ y 4 × M z , E d C m z × χ L T × M p l z , R d = 1.7 × 0.993 0.607 0.1 + 0.770 4 10 1 × 0.903 × 591 = 0.043$
$C y y = 1 + ( w y − 1 ) [ ( 2 − 1.6 w y C m y 2 λ max ⁡ − 1.6 w y C m y 2 λ max ⁡ 2 ) η p l − e L T ] = 1 + ( 1.5 − 1 ) [ ( 2 − 1.6 1.5 × 1.001 2 × 0.770 − 1.6 1.5 × 1.001 2 × 0.770 2 ) 0.006 − 0.043 ] = 0.980$

Wely/Wply = 0.656 < Cyy. So, Cyy = 0.980

Interaction factors:

$K z z = C m z C m L T μ z 1 − N E d N c r z 1 C z z = 1.001$
$K z y = C m y μ z 1 − N E d N c r y 1 C z y 0.6 w y w z = 0.701$
$K y z = C m z C m L T μ y 1 − N E d N c r z 1 C y z 0.6 w z w y = 0.518$
$K y y = C m y μ y 1 − N E d N c r y 1 C y y = 1.023$

Check for Clause 6.3.3-661:

$N E d χ Z N R k γ M 1 + K Z Z M z , E d χ L T M z , R k γ M 1 + K z y M y , E d M y , R k γ M 1 = 25 0.9174 × 4,436 + 1.001 10 0.903 × 591 + 0.701 5 258.3 = 0.038$

Check for Clause 6.3.3-662:

$N E d χ y N R k γ M 1 + K y z M z , E d χ L T M z , R k γ M 1 + K y y M y , E d M y , R k γ M 1 = 25 0.6809 × 4,436 + 0.518 10 0.903 × 591 + 1.023 5 258.3 = 0.038$

## Results

Table 1.
Moment capacity, Mckd (kN·m) 591.0 591.0 none
Critical moment, Mcr (kN·m) 1,540.6 1,536.7 negligible
Bending capacity, MB (kN·m) 533.7 534.4 none
Critical load for torsional buckling, Ncr,T (kN) 13,889 13,898.0 negligible
Critical load for torsional-flexural buckling, Ncr,TF 13,889 13,898.0 negligible
Compression interaction, Cl. 6.3.1.1 0.008 0.008 none
Bending and compression interaction, Cl 6.3.3-661 0.038 0.038 none
Bending and compression interaction, Cl. 6.3.3-662 0.038 0.037 3% Rounding difference in small numbers.

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Europe\EC3 French NA-Column with Axial Load.std is typically installed with the program.

The following design parameters are used:

• The French NA is specified using NA 4
• Fixed support: CMM 2.0
• Fixed base and free at other end: CMN 0.7
• The value of C2 1.554 is specified
• Program calculated kc per Table 6.6 of EN 1993-1-1:2005: KC 0
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 12-June-20
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 5 0;
MEMBER INCIDENCES
1 1 2;
MEMBER PROPERTY EUROPEAN
1 TABLE ST HD320X127
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+08
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-05
DAMP 0.03
END DEFINE MATERIAL
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 FIXED
2 FY -25 MX 5 MZ 10
PERFORM ANALYSIS
PARAMETER 1
CODE EN 1993-1-1:2005
NA 4
CMM 2 ALL
C2 1.554 ALL
FU 295000 ALL
PY 275000 ALL
KC 0 ALL
CMN 0.7 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH


                         STAAD.PRO CODE CHECKING - NF EN 1993-1-1:2005
********************************************
NATIONAL ANNEX - NF EN 1993-1-1/NA
PROGRAM CODE REVISION V1.14 BS_EC3_2005/1
STAAD SPACE                                              -- PAGE NO.    3
ALL UNITS ARE - KN   METE (UNLESS OTHERWISE Noted)
FX            MY             MZ       LOCATION
=======================================================================
1 ST   HD320X127   (EUROPEAN SECTIONS)
PASS     EC-6.3.3-661       0.038         1
25.00 C          5.00         -10.00        0.00
=======================================================================
MATERIAL DATA
Modulus of elasticity    =  205 kN/mm2
Design Strength  (py)    =  275  N/mm2
SECTION PROPERTIES (units - cm)
Member Length =    500.00
Gross Area =  161.30          Net Area =  161.30
z-axis          y-axis
Moment of inertia        :    30820.004        9239.001
Plastic modulus          :     2149.000         939.100
Elastic modulus          :     1926.250         615.933
Shear Area               :       81.998          51.728
Radius of gyration       :       13.823           7.568
Effective Length         :      500.000         500.000
DESIGN DATA (units - kN,m)   EUROCODE NO.3 /2005
Section Class            :   CLASS 1
GM0 :  1.00          GM1 :  1.00          GM2 :  1.25
z-axis          y-axis
Slenderness ratio (KL/r) :         36.2           66.1
Compression Capacity     :       4078.2         3045.5
Tension Capacity         :       3426.0         3426.0
Moment Capacity          :        591.0          258.3
Reduced Moment Capacity  :        591.0          258.3
Shear Capacity           :       1301.9          821.3
BUCKLING CALCULATIONS (units - kN,m)
Lateral Torsional Buckling Moment       MB =  534.4
co-efficients C1 & K : C1 =2.570 K =1.0, Effective Length= 5.000
Lateral Torsional Buckling Curve :
Elastic Critical Moment for LTB,               Mcr   =  1536.7
Compression buckling curves:     z-z:  Curve b   y-y:  Curve c
Critical Load For Torsional Buckling,          NcrT  = 13898.0
Critical Load For Torsional-Flexural Buckling, NcrTF = 13898.0
STAAD SPACE                                              -- PAGE NO.    4
CRITICAL LOADS FOR EACH CLAUSE CHECK (units- kN,m):
CLAUSE        RATIO  LOAD     FX       VY      VZ      MZ      MY
EC-6.3.1.1     0.008     1    25.0      0.0     0.0   -10.0     5.0
EC-6.2.9.1     0.020     1    25.0      0.0     0.0   -10.0     5.0
EC-6.3.3-661   0.038     1    25.0      0.0     0.0   -10.0     5.0
EC-6.3.3-662   0.037     1    25.0      0.0     0.0   -10.0     5.0
EC-6.3.2 LTB   0.019     1    25.0      0.0     0.0   -10.0     5.0
Torsion has not been considered in the design.
_________________________
************** END OF TABULATED RESULT OF DESIGN **************