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
I_z= 447 cm⁴
I_y = 234 cm⁴
Z_z = 91 cm³
Z_y = 68.2 cm³
J = 486 cm⁴
C_w = 0 cm⁶
$r_{z} = \sqrt{\frac{I_{z}}{A}} = \sqrt{\frac{447}{23.4}} = 4.371 cm$
$r_{y} = \sqrt{\frac{I_{y}}{A}} = \sqrt{\frac{234}{23.4}} = 3.162 cm$

Partial safety factors:

Γ_M0 = 1.0

Γ_M1 = 1.0

Γ_M2 = 1.25

Validation

Section Classification

ɑ = 1

C = D - 2t = 120 - 2 × 6.3 = 107.4 mm

ε = \sqrt{\frac{235}{f_{y}}} = \sqrt{\frac{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

\frac{C}{t} = \frac{107.4}{6.3} = 17.05 < \frac{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.

N_{pl,Rd} = A_{g} \times f_{y} / γ_{M0} = 2,340 \times 275 / 1.0 \times {(10)}^{-3} = 643.5 kN

N_{u,Rd} = 0.9 A_{net} \times f_{u} / γ_{M2} = 0.9 \times 2,340 \times 295 / 1.25 \times {(10)}^{-3} = 497.0 kN

The tensile capacity is the minimum of N_pl,Rd and N_u,Rd, thus: N_t,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.

N_{c,Rd} = A_{g} \times f_{y} / γ_{M0} = 2,340 \times 275 / 1.0 \times {(10)}^{-3} = 643.5 kN

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

N_{b,Rd} = χ \times A \times f_{y} / γ_{M1}

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

Along the Z axis:

ƛ_{z} = \sqrt{A \times f_{y} / N_{cr}} = L_{cr} / i_{z} \times λ_{1}

where

L_cr	=	5,000 mm
i_z	=	43.71 mm
λ₁	=	93.9×ε = 93.9 × 0.924 = 86.8

ƛ_z = 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} = \frac{1}{Ф + \sqrt{Ф^{2} - {ƛ_{z}}^{2}}} = 0.4605

N_{b,Rd} = 0.4605 \times 2,340 \times 275 / 1.0 = 296.3 kN

Along the Y axis:

ƛ_{y} = \sqrt{A \times f_{y} / N_{cr}} = L_{cr} / i_{y} \times λ_{1}

where

i_y

31.62 mm

ƛ_z = 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} = \frac{1}{Ф + \sqrt{Ф^{2} - {ƛ_{y}}^{2}}} = 0.2644

N_{b,Rd} = 0.2644 \times 2,340 \times 275 / 1.0 = 170.2 kN

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

Ratio per Eq. 6.9: M_Ed / M_c,Rd = 25.0 / 170.2 = 0.147

Bending Capacity

Along Z axis:

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

Check for bending capacity per Cl. 6.2.5:

For a Class 1 section:

M_{c,Rd} = W_{ply} \times f_{y} / γ_{M0} = 91 \times 275 / 1.0 \times {(10)}^{-3} = 25.03 kN·m

Ratio per Eq. 6.12: M_Ed / M_c,Rd = 10.0 / 25.03 = 0.400

Along Y axis:

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

Check for bending capacity per Cl. 6.2.5:

For a Class 1 section:

M_{c,Rd} = W_{ply} \times f_{y} / γ_{M0} = 68.2 \times 275 / 1.0 \times {(10)}^{-3} = 18.76 kN·m

Ratio per Eq. 6.12: M_Ed / M_c,Rd = 5.0 / 18.76 = 0.267

Shear Capacity

Along Z direction:

A_{v} = \frac{A \times B}{D + B} = \frac{2,340 \times 80}{120 + 80} = 936 {mm}^{2}

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

V_{c,Rd} = V_{pl,Rd} = A_{v} \times (\frac{f_{y}}{\sqrt{3}}) / γ_{M0} = 936 \times (\frac{275}{\sqrt{3}}) / 1.0 = 148.6 kN

Along Y direction:

A_{v} = \frac{A \times D}{D + B} = \frac{2,340 \times 120}{120 + 80} = 1,404 {mm}^{2}

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

V_{c,Rd} = V_{pl,Rd} = A_{v} \times (\frac{f_{y}}{\sqrt{3}}) / γ_{M0} = 1,404 \times (\frac{275}{\sqrt{3}}) / 1.0 = 222.9 kN

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

Lateral Torsional Buckling

M_{c r} = C_{1} \frac{π^{2} E I}{{(k L)}^{2}} [\sqrt{{(\frac{k}{k_{w}})}^{2} \frac{I_{w}}{I_{z}} + \frac{{(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
$\frac{π^{2} E I_{y}}{{(k L)}^{2}}$	=	189,400
${(\frac{k}{k_{w}})}^{2} \frac{I_{w}}{I_{y}}$	=	0
$\frac{{(k L)}^{2} G I_{T}}{π^{2} E I_{y}}$	=	2,023,000
C₂Z_g	=	1.0×40 = 40

Therefore, $M_{cr} = 1.0 \times 189,400 [\sqrt{0 + 2,023,000 + 40^{2}} - 40] = 261.8 kN·m$

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

M_{b,Rd} = χ_{LT} W_{y} \times f_{y} / γ_{M1} = 1.0 \times 91 \times 275 / 1.0 \times {(10)}^{-3} = 25.03 kN·m

Ratio per Eq. 6.12: M_Ed / M_b,Rd = 10.0 / 24.92 = 0.400

Check for Interaction

From Table B.3 in Annex B of EC3:

Ψ = 1.0

C_my = C_mz = C_mLT = 0.6 + 0.4×Ψ = 1.0

N_Rk = A × f_y = 643.5 kN

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

K_{zz} = C_{mz} \frac{1 + (λ_{z} - 0.2) N_{Ed}}{χ_{z} N_{Rk} γ_{M1}} = 1.0 \frac{1 + (1.318 - 0.2) 25}{0.4605 \times 643.5 1} = 1.094 > 1 + \frac{0.8 N_{Ed}}{χ_{z} N_{Rk} γ_{M1}} = 1.067

K_zz = 1.067

K_{yy} = C_{my} \frac{1 + (λ_{y} - 0.2) N_{Ed}}{χ_{y} N_{Rk} γ_{M1}} = 1.0 \frac{1 + (1.822 - 0.2) 25}{0.2644 \times 643.5 1} = 1.238 > 1 + \frac{0.8 N_{Ed}}{χ_{y} N_{Rk} γ_{M1}} = 1.118

K_yy = 1.118

K_yz = 0.6×K_zz = 0.641

K_zy = 0.6×K_yy = 0.671

For K_yz, consider Table B.2 as well:

K_{yz} = 1 - \frac{0.1 λ_{y}}{C_{mLT} - 0.25} \frac{N_{Ed}}{χ_{y} N_{Rk} γ_{M1}} = 0.964 < 1 - \frac{0.1}{C_{mLT} - 0.25} \frac{N_{Ed}}{χ_{y} N_{Rk} γ_{M1}} = 0.980

So, K_yz = 0.980

Check for Clause 6.3.3-661:

\frac{N_{Ed}}{χ_{z} \frac{N_{Rk}}{γ_{M1}}} + K_{zz} \frac{M_{z,Ed}}{χ_{LT} \frac{M_{z,Rk}}{γ_{M1}}} + K_{zy} \frac{M_{y,Ed}}{\frac{M_{y,Rk}}{γ_{M1}}} = \frac{25}{0.4605 \times 643.5} + 1.067 \frac{10}{1.0 \times 25.03} + 0.671 \frac{5}{18.76} = 0.690

Check for Clause 6.3.3-662:

\frac{N_{Ed}}{χ_{y} \frac{N_{Rk}}{γ_{M1}}} + K_{yz} \frac{M_{z,Ed}}{χ_{LT} \frac{M_{z,Rk}}{γ_{M1}}} + K_{yy} \frac{M_{y,Ed}}{\frac{M_{y,Rk}}{γ_{M1}}} = \frac{25}{0.2644 \times 643.5} + 0.980 \frac{10}{1.0 \times 25.03} + 1.118 \frac{5}{18.76} = 0.837

Results

Table 1. Comparison of results
Result Type	Reference	STAAD.Pro	Difference
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
M_cr (kN·m)	261.8	258.5	negligible
M_B (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

STAAD.Pro Input

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 χ_LT: MTH 1
The values of C1 1.0 and C2 1.0 are specified.
The values of F_y and F_u 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
LOAD 1 LOADTYPE Dead  TITLE LOAD CASE 1
JOINT LOAD
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 Output

                         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)
 MEMBER     TABLE       RESULT/   CRITICAL COND/     RATIO/     LOADING/
                          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                
      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 =   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     
      Squash Load              :    643.50
      Axial force/Squash load  :     0.039
      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 **************