V. SNiP SP16 2017 - I section with axial force and Bi-moment

Design a column subject to axial compressive force and biaxial moment per the SP 16.13330.2017 code.

Details

A 5 m tall, simply supported column has an HE650A section. The column is subject to a 80 kN axial load along with a uniformly distributed load of 30 kN/m in the local X axis and a uniformly distributed load of 2 kN/m in the local Y axis. The steel used has a modulus of elasticity of 206,000 MPa and a Ry = 235 MPa. γc = 1, γm = 1.05

Section Properties

D = 640 mm

B = 300 mm

tf = 26 mm

tw = 13.5 mm

A = 241.6 cm2

Ix = 175,200 cm4

Iy = 11,720 cm4

IT = 458 cm4

rx = 26.93 cm

ry = 6.97 cm

Validation

Ry = Ryn/ γm = 223.8 MPa

Rs = 0.58×Ry/ γm = 129.8 MPa

Shear force at support:

Qx = qx × L / 2 = 75 kN

Bending moment:

Mx = qx × L2 / 8 = 30 (5)2 / 8 = 93.75 kN·m

My = qy × L2 / 8 = 2 (5)2 / 8 = 6.25 kN·m

Design for Strength (Cl. 9.1.1)

$σ = N A n = 80 241.6 ( 10 ) − 1 = 3.311 < 0.1 R y = 22.38$

Ryn ≤ 440 N/mm2

τ = 0; i.e., < 0.5×Rs

So, as per Cl. 9.1.1, F.105 should not be checked. Rather F.106 needs to be checked.

$m e f = η × m$
where
 m = e×A / Wc = 5.172 (e = M/N = 93.75 / 80 = 1.172)
$λ x = k x l r x = 1.0 × 500 26.93 = 18.57$
$λ y = k y l r y = 1.0 × 500 6.97 = 71.74$
$λ _ x = λ x R y E = 18.58 223.8 206,000 = 0.612$
$λ _ y = λ y R y E = 71.74 223.8 206,000 = 2.364$

Therefore, $λ _ = min ⁡ ( λ x _ , λ y _ ) = 0.612$

Evaluate η, the coefficient of the shape of cross section vertical element when 5 < m ≤ 20, Af / Aw ≥ 1, and 0 ≤ λ ≤ 5:

 $η = 1.4 − 0.02 λ _ = ( 1.4 − 0.02 × 0.612 ) = 1.388$ (Table Д.2, Note 1)

So, $m e f = 1.388 × 5.172 = 7.18 < 20$ [As per F.(110)]

 $N A n ± M x y I x n ± M y x I y n ± B ω I ω n R y γ c ≤ 1$ (F.(106) )
where
 x = 150 mm y = 320 mm Bω = 0

So, the ratio is $3.311 + 93.75 ( 320 ) 175,200 × 10 − 2 + 6.25 ( 150 ) 11,720 × 10 − 2 223.8 × 1 = 0.127 < 1$

Design for Stability (Cl. 9.2.2)

From Table E.3, depending on conditional slenderness and reduced relative eccentricity:

ϕe = 0.203

 $N ϕ e × A × R y × γ c = 80 0.203 × 241.6 ( 10 ) − 1 × 223.8 × 1 = 0.073 < 1$ (F.(109) )

Design for Stability (Cl. 9.2.4)

Calculate the stability of eccentrically compressed elements of constant cross-section, out-of-plane bending moment in the plan of maximum stiffness (Ix > Iy), coinciding with the plane of symmetry:

 $N c × ϕ y × A × R y × γ c ≤ 1$ (F.111 )
where
 ϕy = $0.5(δ−δ2−39.48λ_2λ_2$ $λ_$ = conditional slenderness $= max ⁡ ( λ x _ , λ y _ ) = 2.365$ δ = $9.87 ( 1 − α + β λ _ ) + λ _ 2$, per Eq. 9 where α = 0.03 and β = 0.06 from Table 7.$= 9.87 [ 1 − 0.03 + 0.06 ( 2.365 ) ] + ( 2.365 ) 2 = 16.57$
$ϕ y = 0.5 16.57 − ( 16.57 ) 2 − 39.48 × ( 2.365 ) 2 ( 2.365 ) 2 = 0.826$
 $c max = c 5 ( 2 - 0.2 m x ) + c 10 ( 0.2 m x - 1 )$ (F.114)
where
α
=
0.65 + 0.05×mx = 0.909
β
=
1.0 for λy < 3.14 per Table 21
mx
=
relative eccentricity = m = 5.172
c5
=
$β 1 + ɑ × m x = 1 1 + 0.909 × 5.172 = 0.175 < 1.0$ per F.112
ψ
=
2.25 + 0.07ɑ = 2.507 where:
 $ɑ = 1.54 I t I y ( l e f h ) 2 = 1.54 458 11,720 ( 5,000 640 ) 2 = 3.673$ (Eq. G.4)
ϕ1
=
$ψ I y I x h l e f 2 E R y = 2.507 11,720 175,200 ( 640 5,000 ) 2 206,000 223.8 = 2.529$
ϕb
=
0.68 + 0.21×ϕ1 = 1.211 > 1.0, take ϕb = 1.0
c10
=
$1 1 + m x ( ϕ 1 / ϕ b ) = 1 1 + 5.172 × ( 0.826 / 1 ) = 0.190$ per F.113

Therefore, $c max = 0.175 ( 2 - 0.2 × 5.172 ) + 0.190 ( 0.2 × 5.172 - 1 ) = 0.176$

So, the ratio is $80 0.176 × 0.826 × 242 ( 10 ) − 1 × 223.9 × 1 = 0.102 < 1$

Design for Stability (Cl. 9.2.9)

 $N ϕ exy × A × R y × γ c ≤ 1$ (F.111 )
where
 ϕey = 0.333 (per Table Д.3) ϕexy = $ϕ ey ( 0.6 c 3 + 0.4 c 4 ) = 0.333 ( 0.6 0.176 3 + 0.4 0.176 4 ) = 0.198$
So, the ratio is $80 0.198 × 242 ( 10 ) − 1 × 223.9 × 1 = 0.075 < 1$

Results

Ratio per Cl. 9.1.1 0.127 0.127 none
Ratio per Cl. 9.2.2 0.073 0.0729 negligible
Ratio per Cl. 9.2.4 0.102 0.099 negligible
Ratio per Cl. 9.2.9 0.075 0.074 negligible
mef 7.18 7.16 negligible
mx 5.172 5.18 negligible
C 0.176 0.179 negligible
ϕey 0.333 0.33 negligible
ϕexy 0.198 0.199 negligible

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Russia\SNiP SP16 2017 - I section with axial force and Bi-moment.std is typically installed with the program.

STAAD SPACE
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 5 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+08
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-05
DAMP 0.03
TYPE STEEL
STRENGTH FY 253200 FU 407800 RY 1.5 RT 1.2
END DEFINE MATERIAL
MEMBER PROPERTY EUROPEAN
1 TABLE ST HE650A
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FY MX MZ
2 FY -80
1 UNI GX 30
1 UNI GZ 2
PERFORM ANALYSIS
PARAMETER 1
CODE RUSSIAN
TB 2 ALL
GAMM 2 ALL
ENSGR 1 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH


                       STAAD.PRO CODE CHECKING - (SP 16.13330.2017)   V1.0
********************************************
ALL UNITS ARE - KN METRE
========================================================================
SECTION NO.      N             Mx            My      LOCATION
========================================================================
1  I      HE650A        PASS      SP cl.9.1.1      0.13         1
8.000E+01 C    9.375E+01    6.250E+00   2.500E+00
1  I      HE650A        PASS      SP cl.9.2.2      0.07         1
8.000E+01 C    9.375E+01    6.250E+00   2.500E+00
1  I      HE650A        PASS      SP cl.9.2.4      0.10         1
8.000E+01 C    9.375E+01    6.250E+00   2.500E+00
1  I      HE650A        PASS      SP cl.9.2.9      0.07         1
8.000E+01 C    9.375E+01    6.250E+00   2.500E+00
MATERIAL DATA
Steel                         = S235       EN10025-2
Modulus of elasticity         = 206.E+06 kPa
Design Strength (Ry)          = 224.E+03 kPa
SECTION PROPERTIES (units - m, m^2, m^3, m^4)
Member Length                 = 5.00E+00
Gross Area                    = 2.42E-02
Net Area                      = 2.42E-02
x-axis      y-axis
Moment of inertia (I)         :   175.E-05    117.E-06
Section modulus (W)           :   548.E-05    781.E-06
First moment of area (S)      :   307.E-05    603.E-06
Radius of gyration (i)        :   269.E-03    696.E-04
Effective Length              :   5.00E+00    5.00E+00
Slenderness                   :   186.E-01    718.E-01
DESIGN DATA (units -kN,m) SP16.13330.2017
Axial force                   :   800.0E-01
x-axis      y-axis
Moments                       :   937.5E-01    625.0E-02
Shear force                   :   0.000E+00    0.000E+00
Bi-moment                     :   0.000E+00 Value of Bi-moment not being entered!!!
Stress-strain state checked as:   Class    2
CRITICAL CONDITIONS FOR EACH CLAUSE CHECK
F.(106) (N/A+Mx*y/Ix+My*x/Iy+B*w/Iw)/(Ry*GammaC)=
( 800.0E-01/ 2.4E-02+ 937.5E-01* 3.20E-01/ 1.75E-03+ 625.0E-02* 1.50E-01/
1.17E-04+ 0.000E+00* 2.50E-01/ 1.10E-05)/( 223.8E+03* 1.00E+00)
= 1.27E-01=&lt;1
cl.9.2.2  m_ef=eta*mx= 1.38E+00* 5.18E+00= 7.16E+00
F.(109)  N/(FIe*A*Ry*GammaC)= 800.0E-01/( 2.03E-01* 2.42E-02* 223.8E+03* 1.00E+00)
= 7.29E-02=&lt;1
F.(114)  c=c5(2-0.2*mx)+c10*(0.2*mx-1.0)=
1.82E-01*(2-0.2* 5.18E+00)+ 1.08E-01*(0.2* 5.18E+00-1.0)= 1.79E-01
c_max= 3.20E-01
F.(111)  N/(c*FIy*A*Ry*GammaC)= 8.00E+01/(1.79E-01*8.26E-01* 2.42E-02* 223.8E+03* 100.E-02)
= 998.E-04=&lt;1
F.(117)  FIexy=FIey*(0.6*c**(1/3)+0.4*c**(1/4))=
3.33E-01*(0.6*1.79E-01**(1/3)+0.4*1.79E-01**(1/4))= 1.99E-01
F.(116) N/(FIexy*A*Ry*GammaC)= 800.0E-01/( 1.99E-01* 2.42E-02* 223.8E+03*1.00E+00)
=7.41E-02=&lt;1