 # V. EC3 - Tube Section with Axial Load

Verify the adequacy of a cantilever tube section subject to axial force and bending per EN 1993-1-1:2005 (no national annex used).

## Details

The member is a 5 m long cantilever. The member is subject to a 25 kN axial compressive load along with moments of 10 kN·m about the major axis and 5 kN·m about the minor axis at the free end. The steel is grade S275. The section is a European TUB120806.

Section Properties

• A = 23.4 cm2
• Depth, D = 120 mm
• Width, B = 80 mm
• t = 6.3 mm
• Iz= 447 cm4
• Iy = 234 cm4
• Zz = 91 cm3
• Zy = 68.2 cm3
• J = 486 cm4
• Cw = 0 cm6

Partial safety factors:

ΓM0 = 1.0

ΓM1 = 1.0

ΓM2 = 1.25

## Validation

Section Classification

ɑ = 1

C = D - 2t = 120 - 2 × 6.3 = 107.4 mm

$ε = 235 f y = 235 275 = 0.924$

As per Table 5.2:

C / t = 107.4 / 6.3 = 17.05 < 50×ε2 = 50 (0.924)2 = 42.7

$C t = 107.4 6.3 = 17.05 < 396 ε 13 ɑ - 1 = 30.36$

Hence, this is a Class 1 section.

Slenderness Ratio

The slenderness ratio = kL / r = 1 × 5,000 / 31.62 = 158.1

Axial Tension

Determine axial tension capacity per Cl. 6.2.3.

The tensile capacity is the minimum of Npl,Rd and Nu,Rd, thus: Nt,Rd = 497.0 kN.

No axial tension in section, so by observation no need to check ratio.

Axial Compression

Determine axial compression capacity per Cl. 6.2.4 for a Class 1 section.

Next, check the flexural buckling resistance per Cl. 6.3.1.3:

$N b,Rd = χ × A × f y / γ M1$

From Table 6.1: the imperfection factor, ɑ = 0.21.

Along the Z axis:
$ƛ z = A × f y / N cr = L cr / i z × λ 1$
where
 Lcr = 5,000 mm iz = 43.71 mm λ1 = 93.9×ε = 93.9 × 0.924 = 86.8

ƛz = 1.318

$Ф = 0.5 [ 1 + ɑ ( ƛ z - 0.2 ) + ƛ z 2 ] = 0.5 [ 1 + 0.21 ( 1.318 - 0.2 ) + ( 1.318 ) 2 ] = 1.486$
$χ z = 1 Ф + Ф 2 - ƛ z 2 = 0.4605$
Along the Y axis:
$ƛ y = A × f y / N cr = L cr / i y × λ 1$
where
 iy = 31.62 mm

ƛz = 1.822

$Ф = 0.5 [ 1 + ɑ ( ƛ y - 0.2 ) + ƛ y 2 ] = 0.5 [ 1 + 0.21 ( 1.822 - 0.2 ) + ( 1.822 ) 2 ] = 2.330$
$χ y = 1 Ф + Ф 2 - ƛ y 2 = 0.2644$

The compression capacity is the minimum of Nc,Rd and Nb,Rd, thus: Nc,Rd = 170.2 kN.

Ratio per Eq. 6.9: MEd / Mc,Rd = 25.0 / 170.2 = 0.147

Bending Capacity

Along Z axis:

Maximum bending moment in the section: MEd,z = 10.0 kN·m.

Check for bending capacity per Cl. 6.2.5:

For a Class 1 section:

Ratio per Eq. 6.12: MEd / Mc,Rd = 10.0 / 25.03 = 0.400

Along Y axis:

Maximum bending moment in the section: MEd,y = 5.0 kN·m.

Check for bending capacity per Cl. 6.2.5:

For a Class 1 section:

Ratio per Eq. 6.12: MEd / Mc,Rd = 5.0 / 18.76 = 0.267

Shear Capacity

Along Z direction:

Check for shear capacity per Cl. 6.2.6 for plastic design (Class 1):

Along Y direction:

Check for shear capacity per Cl. 6.2.6 for plastic design (Class 1):

No shear in section, so by observation no need to check ratio.

Lateral Torsional Buckling

$M c r = C 1 π 2 E I ( k L ) 2 [ ( k k w ) 2 I w I z + ( k L ) 2 G I t π 2 E I z + ( C 2 Z g ) 2 − C 2 Z g ]$
where
 C1 = 1.0 C2 = 1.0 $π 2 E I y k L 2$ = 189,400 $k k w 2 I w I y$ = 0 $k L 2 G I T π 2 E I y$ = 2,023,000 C2Zg = 1.0×40 = 40

Therefore,

As per Cl. 6.3.2.1, tube sections are not susceptible to lateral-torsional buckling, so χLT = 1.0.

Ratio per Eq. 6.12: MEd / Mb,Rd = 10.0 / 24.92 = 0.400

Check for Interaction

From Table B.3 in Annex B of EC3:

Ψ = 1.0

Cmy = Cmz = CmLT = 0.6 + 0.4×Ψ = 1.0

NRk = A × fy = 643.5 kN

From Table B.1 of Annex B, the interaction factors:

$K zz = C mz 1 + ( λ z - 0.2 ) N Ed χ z N Rk γ M1 = 1.0 1 + ( 1.318 - 0.2 ) 25 0.4605 × 643.5 1 = 1.094 > 1 + 0.8 N Ed χ z N Rk γ M1 = 1.067$

Kzz = 1.067

$K yy = C my 1 + ( λ y - 0.2 ) N Ed χ y N Rk γ M1 = 1.0 1 + ( 1.822 - 0.2 ) 25 0.2644 × 643.5 1 = 1.238 > 1 + 0.8 N Ed χ y N Rk γ M1 = 1.118$

Kyy = 1.118

Kyz = 0.6×Kzz = 0.641

Kzy = 0.6×Kyy = 0.671

For Kyz, consider Table B.2 as well:

$K yz = 1 - 0.1 λ y C mLT - 0.25 N Ed χ y N Rk γ M1 = 0.964 < 1 - 0.1 C mLT - 0.25 N Ed χ y N Rk γ M1 = 0.980$

So, Kyz = 0.980

Check for Clause 6.3.3-661:

$N Ed χ z N Rk γ M1 + K zz M z,Ed χ LT M z,Rk γ M1 + K zy M y,Ed M y,Rk γ M1 = 25 0.4605 × 643.5 + 1.067 10 1.0 × 25.03 + 0.671 5 18.76 = 0.690$

Check for Clause 6.3.3-662:

$N Ed χ y N Rk γ M1 + K yz M z,Ed χ LT M z,Rk γ M1 + K yy M y,Ed M y,Rk γ M1 = 25 0.2644 × 643.5 + 0.980 10 1.0 × 25.03 + 1.118 5 18.76 = 0.837$

## Results

Table 1. Comparison of results
Section Class Class 1 Class 1 none
Slenderness Ratio 158.1 158.1 none
Tension Capacity (kN) 497.0 497 none
Compression Capacity (Major) (kN) 296.3 296.3 none
Compression Capacity (Minor) (kN) 170.2 170.2 none
Moment Capacity (Major) (kN·m) 25.03 25 negligible
Moment Capacity (Minor) (kN·m) 18.76 18.8 negligible
Shear Area (Major) (cm2) 9.36 9.36 none
Shear Area (Minor) (cm2) 14.04 14.04 none
Shear Capacity (Major) (kN) 148.6 148.6 none
Shear Capacity (Minor) (kN) 222.9 222.9 none
Mcr (kN·m) 261.8 258.5 negligible
MB (kN·m) 25.03 25 negligible
Ratio per Cl. 6.3.1.1 0.147 0.147 none
Ratio per Cl. 6.2.9.1 0.400 0.4 none
Ratio per Cl. 6.3.3-661 0.690 0.69 none
Ratio per Cl. 6.3.3-662 0.837 0.837 none

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

The following design parameters are used:
• Fixed end supports: CMM 2
• Cantilever member: CMN 0.7
• Use Cl. 6.3.2.2 to determine χLT: MTH 1
• The values of C1 1.0 and C2 1.0 are specified.
• The values of Fy and Fu are specified directly using PY 275000 and FU 295000.
STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 05-May-21
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 TUB120806
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
CMM 2 ALL
C1 1 ALL
C2 1 ALL
FU 295000 ALL
PY 275000 ALL
MTH 1 ALL
CMN 0.7 ALL
TRACK 2 ALL
CHECK CODE ALL
PRINT MEMBER PROPERTIES
FINISH


                         STAAD.PRO CODE CHECKING - EN 1993-1-1:2005
********************************************
NATIONAL ANNEX - NOT USED
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   TUB120806   (EUROPEAN SECTIONS)
PASS     EC-6.3.3-662       0.837         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 =   23.40          Net Area =   23.40
z-axis          y-axis
Moment of inertia        :      447.000         234.000
Plastic modulus          :       91.000          68.200
Elastic modulus          :       74.500          58.500
Shear Area               :        9.360          14.040
Radius of gyration       :        4.371           3.162
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) :        114.4          158.1
Compression Capacity     :        296.3          170.2
Tension Capacity         :        497.0          497.0
Moment Capacity          :         25.0           18.8
Reduced Moment Capacity  :         25.0           18.8
Shear Capacity           :        148.6          222.9
BUCKLING CALCULATIONS (units - kN,m)
Lateral Torsional Buckling Moment       MB =   25.0
co-efficients C1 & K : C1 =1.000 K =1.0, Effective Length= 5.000
Lateral Torsional Buckling Curve : CURVE d
Elastic Critical Moment for LTB,               Mcr   =   258.5
Compression buckling curves:     z-z:  Curve a   y-y:  Curve a
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     0.147     1    25.0      0.0     0.0   -10.0     5.0
EC-6.2.9.1     0.400     1    25.0      0.0     0.0   -10.0     5.0
EC-6.3.3-661   0.690     1    25.0      0.0     0.0   -10.0     5.0
EC-6.3.3-662   0.837     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 **************