V. EC3 German NA - Column with Axial Load

Calculate the axial and bending capacities and interaction ratios of a column using the German 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:

M_{c k d} = \frac{W_{p l y} f_{y}}{γ_{M 0}} = \frac{2,149 {(10)}^{3} 275}{1.0 {(10)}^{6}} = 591.0 kN·m

The critical moment is given by:

M_{c r} = C_{1} \frac{π^{2} E I}{{(k L)}^{2}} {\sqrt{{(\frac{k}{k_{w}})}^{2} \frac{I_{w}}{I} + \frac{{(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
$\frac{π^{2} E I_{y}}{{(k L)}^{2}}$	=	7,477,200
${(\frac{k}{k_{w}})}^{2} \frac{I_{w}}{I_{y}}$	=	22,394
$\frac{{(k L)}^{2} G I_{t}}{π^{2} E I_{y}}$	=	23,740
C₂Z_g	=	1.554×160 = 248.6

Therefore, M_cr = 1,540.6 kN·m

From the German NA, λ_{LT, 0} = 0.4, β = 0.75

So, $λ_{L T} = \sqrt{\frac{w_{y} f_{y}}{M_{c r}}} = \sqrt{\frac{2,149 \times 10^{3} \times 275}{1,540.6 \times 10^{6}}} = 0.619$

H/b = 320/300 = 1.067 < 2. So, from Table 6.5 of Eurocode 3, α_LT = 0.34.

From Cl. 6.3.2.3 of Eurocode3:

Ф_LT = 0.5[1+α_LT (λ_LT- λ_LT,
0) + β× λ_LT²] = 0.5[1 + 0.34×(0.619 - 0.4) +0.75 × 0.619²] = 0.681

So, $χ_{L T} = {(Ф_{L T} + \sqrt{{Ф_{L T}}^{2} - β {λ_{L T}}^{2}})}^{- 1} = 0.908$

$k c = \frac{1}{\sqrt{C_{1}}} = \frac{1}{\sqrt{2.578}} = 0.623$ and thus modification factor, f = 1 - 0.5(1 - 0.623)×[1 -2(0.619 - 0.8)²] = 0.824

χ_{L T, \mod} = χ_{L T} / f = 0.824 = 1.102 > 1

So, $χ_{L T, \mod} = 1.0$

M_{B} = \frac{χ_{L T} w_{y} f_{y}}{γ_{M 1}} = \frac{1.0 \times 2,149 {(10)}^{3} \times 275}{1.1} = 537.3 kN·m

Compression Capacity

Critical compressive values:

N_{cry} = \sqrt{\frac{π^{2} E I_{y}}{{(k l)}^{2}}} = 7,477 kN

N_{crz} = \sqrt{\frac{π^{2} E I_{z}}{{(k l)}^{2}}} = 24,943 kN

N_c,Rd = A_gf_y / γ_M0 = 16,130 × 275 / 1.0 (10)^-3 = 4,436 kN for a Class 1 section.

Check the flexural buckling resistance per Cl. 6.3.1.1.

N_b,Rd = χ×A×f_y / γ_M1 for a Class 1 section.

Buckling curves for I section are found in Table 6.2. Imperfection factor, ɑ, values are found in Table 6.1.

Along Z

h/b = 1.067 < 2

t_f = 20.5 < 100

So, for the z-z axis, use curve "b". Hence, ɑ = 0.34 (EC3 Table 6.2).

λ_{cz} = \sqrt{\frac{A \times f_{y}}{N_{crz}}} = \frac{L_{cr}}{i_{z} \times λ_{1}}

where

L_cr	=	5,000 mm
i_z	=	138.2 mm
λ₁	=	$93.9 ε = 93.9 \sqrt{\frac{235}{f_{y}}} = 93.9 \sqrt{\frac{235}{275}} = 86.80$

Thus, $λ_{cz} = \frac{5,000}{138.2 \times 86.80} = 0.417$

ф = 0.5[1+α (λ_z - 0.2) +λ²)] = 0.624

χ_{z} = \frac{1}{ф + \sqrt{ф^{2}} - λ^{2}} = 0.919

Compression capacity about Z:

N_b,Rd = χ_z×A×f_y / γ_M1 = 0.919×16,130×275 / 1.1 = 3,707.5 kN

Along Y

For the y-y axis, use curve "c". Hence, ɑ = 0.49 (EC3 Table 6.2).

λ_{cz} = \sqrt{\frac{A \times f_{y}}{N_{crz}}} = \frac{L_{cr}}{i_{z} \times λ_{1}}

where

L_cr	=	5,000 mm
i_y	=	75.68 mm
λ₁	=	$93.9 ε = 93.9 \sqrt{\frac{235}{f_{y}}} = 93.9 \sqrt{\frac{235}{275}} = 86.80$

Thus, $λ_{cy} = \frac{5,000}{75.68 \times 86.80} = 0.761$

ф = 0.5[1+α (λ_z - 0.2) +λ²)] = 0.827

χ_{y} = \frac{1}{ф + \sqrt{ф^{2}} - λ^{2}} = 0.687

Compression capacity about Z:

N_b,Rd = χ_z×A×f_y / γ_M1 = 0.687×16,130×275 / 1.1 = 2,768.6 kN

Thus, the compression capacity, N_b,Rd = 2,768.6 kN

Compression ratio: 25 kN / 2,768.6 kN = 0.009

Critical Axial Loads for Flexure and Flexural Torsional Buckling

From NCCI document SN001a-EN-FU:

N_{c r, T} = \frac{1}{i_{0}^{2}} (G I_{t} + \frac{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 = 0

i_y = 75.68 mm and i_z = 138.23 mm

= 24,835 mm²

N_{c r, T} = \frac{1}{24,835} [\frac{20,500 \times 225.1 {(10)}^{4}}{2.6} + \frac{π^{2} (205,000) 2.069 {(10)}^{12}}{{5,000}^{2}}] = 13,889 kN

N_{c r, T F} = \frac{i_{0}^{2}}{2 (i_{y}^{2} + i_{z}^{2})} [N_{c r, y} + N_{c r, T} - \sqrt{{(N_{c r, y} + N_{c r, T})}^{2} - 4 N_{c r, y} N_{c r, T} \frac{i_{y}^{2} + i_{z}^{2}}{i_{0}^{2}}}]

= \frac{1}{2} [24,943 + 13,889 - \sqrt{{(24,943 + 13,889)}^{2} - 4 (24,943) (13,889) \frac{1}{1}}] = 13,889 kN

Interaction Check

From Annex B, Table B.3 of EN 1993:1-1:2005,

W_z = W_plz / W_elz = 2,149 / 1,926 = 1.116

W_y = W_ply / W_ely = 939 / 615.9 = 1.525 > 1.5; W_y = 1.5

ψ = 1.0

So, C_mz = 0.6 + 0.4ψ = 1.0

C_mLT = C_mz = 1.0

C_my = 0.6 + 0.4ψ = 1.0

Interaction factors:

K_{z z} = C_{m z} [1 + (λ_{z} - 0.2) \frac{N_{E d}}{χ_{z} \times N_{R k} \times γ_{M 1}}] = 1 [1 + (0.422 - 0.2) \frac{25}{0.917 \times 4, 436}] = 1.001

K_{y y} = C_{m y} [1 + ({2 λ}_{y} - 0.6) \frac{N_{E d}}{χ_{y} \times N_{R k} \times γ_{M 1}}] = 1 [1 + (2 \times 0.770 - 0.6) \frac{25}{0.681 \times 4, 436}] = 1.008

K_{y z 1} = 0.6 K_{z z} = 0.601

K_{y z 2} = 1 + \frac{0.1 \times λ_{y}}{C_{m L T} - 0.25} \frac{N_{E d}}{χ_{y} \times N_{R k} \times γ_{M 1}} = 1 - \frac{0.1 \times 0.770}{(1 - 0.25)} \frac{25}{0.681 \times 4, 436} = 0.999

So, K_yz = maximum(K_yz1, K_yz2) = 0.999

K_zy = 0.6×K_yy = 0.6(1.008) = 0.605

Check for Clause 6.3.3-661:

\frac{N_{E d}}{χ_{Z} \frac{N_{R k}}{{γ M}_{1}}} + K_{Z Z} \frac{M_{z, E d}}{χ_{L T} \frac{M_{z, R k}}{{γ M}_{1}}} + K_{z y} \frac{M_{y, E d}}{\frac{M_{y, R k}}{{γ M}_{1}}} = \frac{25 \times 1.1}{0.9174 \times 4,436} + 1.001 \frac{10 \times 1.1}{1.0 \times 591} + 0.605 \frac{5 \times 1.1}{258.3} = 0.038

Check for Clause 6.3.3-662:

\frac{N_{E d}}{χ_{y} \frac{N_{R k}}{{γ M}_{1}}} + K_{y z} \frac{M_{z, E d}}{χ_{L T} \frac{M_{z, R k}}{{γ M}_{1}}} + K_{y y} \frac{M_{y, E d}}{\frac{M_{y, R k}}{{γ M}_{1}}} = \frac{25 \times 1.1}{0.681 \times 4,436} + 0.999 \frac{10 \times 1.1}{1.0 \times 591} + 1.008 \frac{5 \times 1.1}{258.3} = 0.049

Results

Table 1.
Result Type	Reference	STAAD.Pro	Difference
Moment capacity, M_ckd (kN·m)	591.0	591.0	none
Critical moment, M_cr (kN·m)	1,540.6	1,541.5	negligible
Bending capacity, M_B (kN·m)	537.3	537.2	negligible
Critical load for torsional buckling, N_cr,T (kN)	13,889	13,898.0	negligible
Critical load for torsional-flexural buckling, N_cr,TF	13,889	13,898.0	negligible
Compression interaction, Cl. 6.3.1.1	0.009	0.009	none
Bending and compression interaction, Cl 6.3.3-661	0.038	0.038	none
Bending and compression interaction, Cl. 6.3.3-662	0.049	0.049	none

STAAD.Pro Input

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

The following design parameters are used:

The German NA is specified using NA 10
Fixed support: CMM 2.0
Fixed base and free at other end: CMN 0.7
Program calculated kc per Table 6.6 of EN 1993-1-1:2005: KC 0

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 05-Aug-2021
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
LOAD 1 LOADTYPE None  TITLE LOAD CASE 1
JOINT LOAD
2 FY -25 MX 5 MZ 10
PERFORM ANALYSIS
PRINT MEMBER PROPERTIES ALL
PARAMETER 1
CODE EN 1993-1-1:2005
NA 10
CMM 2 ALL
FU 295000 ALL
PY 275000 ALL
KC 0 ALL
CMN 0.7 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH

STAAD.Pro Output

                         STAAD.PRO CODE CHECKING - DIN EN 1993-1-1:2010-12
                         ********************************************
                         NATIONAL ANNEX - DIN EN 1993-1-1/NA:2010-12
 PROGRAM CODE REVISION V1.14 BS_EC3_2005/1
      STAAD SPACE                                              -- PAGE NO.    4
 ALL UNITS ARE - KN   METE (UNLESS OTHERWISE Noted)
 MEMBER     TABLE       RESULT/   CRITICAL COND/     RATIO/     LOADING/
                          FX            MY             MZ       LOCATION
 =======================================================================
       1 ST   HD320X127   (EUROPEAN SECTIONS)
                           PASS     EC-6.3.3-662       0.049         1
                       25.00 C          5.00         -10.00        0.00
 =======================================================================
   MATERIAL DATA                
      Grade of steel           =  USER          
      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     
      Squash Load              :   4435.75
      Axial force/Squash load  :     0.006
      GM0 :  1.00          GM1 :  1.10          GM2 :  1.25
                                      z-axis          y-axis
      Slenderness ratio (KL/r) :         36.2           66.1
      Compression Capacity     :       3707.4         2768.6
      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 =  537.2
      co-efficients C1 & K : C1 =2.578 K =1.0, Effective Length= 5.000
      Lateral Torsional Buckling Curve : Curve b
      Elastic Critical Moment for LTB,               Mcr   =  1541.5
      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.    5
   CRITICAL LOADS FOR EACH CLAUSE CHECK (units- kN,m):
    CLAUSE        RATIO  LOAD     FX       VY      VZ      MZ      MY   
   EC-6.3.1.1     0.009     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   0.049     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 **************