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 R_y = 235 MPa. γ_c = 1, γ_m = 1.05

Section Properties

D = 640 mm

B = 300 mm

t_f = 26 mm

t_w = 13.5 mm

A = 241.6 cm²

I_x = 175,200 cm⁴

I_y = 11,720 cm⁴

I_T = 458 cm⁴

r_x = 26.93 cm

r_y = 6.97 cm

Validation

R_y = R_yn/ γ_m = 223.8 MPa

R_s = 0.58×R_y/ γ_m = 129.8 MPa

Shear force at support:

Q_x = q_x × L / 2 = 75 kN

Bending moment:

M_x = q_x × L² / 8 = 30 (5)² / 8 = 93.75 kN·m

M_y = q_y × L² / 8 = 2 (5)² / 8 = 6.25 kN·m

Design for Strength (Cl. 9.1.1)

σ = \frac{N}{A_{n}} = \frac{80}{241.6 {(10)}^{- 1}} = 3.311 < 0.1 R_{y} = 22.38

R_yn ≤ 440 N/mm²

τ = 0; i.e., < 0.5×R_s

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

m_{e f} = η \times m

where

e×A / W_c = 5.172 (e = M/N = 93.75 / 80 = 1.172)

λ_{x} = \frac{k_{x} l}{r_{x}} = \frac{1.0 \times 500}{26.93} = 18.57

λ_{y} = \frac{k_{y} l}{r_{y}} = \frac{1.0 \times 500}{6.97} = 71.74

{\overline{λ}}_{x} = λ_{x} \sqrt{\frac{R_{y}}{E}} = 18.58 \sqrt{\frac{223.8}{206,000}} = 0.612

{\overline{λ}}_{y} = λ_{y} \sqrt{\frac{R_{y}}{E}} = 71.74 \sqrt{\frac{223.8}{206,000}} = 2.364

Therefore, $\overline{λ} = \min (\overline{λ_{x}}, \overline{λ_{y}}) = 0.612$

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

η = 1.4 - 0.02 \overline{λ} = (1.4 - 0.02 \times 0.612) = 1.388

(Table Д.2, Note 1)

So, $m_{e f} = 1.388 \times 5.172 = 7.18 < 20$ [As per F.(110)]

\frac{\frac{N}{A_{n}} \pm \frac{M_{x} y}{I_{x n}} \pm \frac{M_{y} x}{I_{y n}} \pm \frac{B_{ω}}{I_{ω n}}}{R_{y} γ_{c}} \leq 1

(F.(106) )

where

x	=	150 mm
y	=	320 mm
B_ω	=	0

So, the ratio is $\frac{3.311 + \frac{93.75 (320)}{175,200 \times 10^{- 2}} + \frac{6.25 (150)}{11,720 \times 10^{- 2}}}{223.8 \times 1} = 0.127 < 1$

Design for Stability (Cl. 9.2.2)

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

ϕ_e = 0.203

\frac{N}{ϕ_{e} \times A {\times R}_{y} \times γ_{c}} = \frac{80}{0.203 \times 241.6 {(10)}^{- 1} \times 223.8 \times 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 (I_x > I_y), coinciding with the plane of symmetry:

\frac{N}{c \times ϕ_{y} \times A {\times R}_{y} \times γ_{c}} \leq 1

(F.111 )

where

ϕ_y	=	$\frac{0.5 (δ - \sqrt{δ^{2} - 39.48 {\overline{λ}}^{2}}}{{\overline{λ}}^{2}}$
$\overline{λ}$	=	conditional slenderness $= \max (\overline{λ_{x}}, \overline{λ_{y}}) = 2.365$
δ	=	$9.87 (1 - α + β \overline{λ}) + {\overline{λ}}^{2}$ , 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 \frac{16.57 - \sqrt{{(16.57)}^{2} - 39.48 \times {(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

0.65 + 0.05×m_x = 0.909

1.0 for λ_y < 3.14 per Table 21

m_x

relative eccentricity = m = 5.172

c₅

\frac{β}{1 + ɑ \times m_{x}} = \frac{1}{1 + 0.909 \times 5.172} = 0.175 < 1.0

per F.112

2.25 + 0.07ɑ = 2.507 where:

ɑ = 1.54 \frac{I_{t}}{I_{y}} {(\frac{l_{e f}}{h})}^{2} = 1.54 \frac{458}{11,720} {(\frac{5,000}{640})}^{2} = 3.673

(Eq. G.4)

ϕ₁

ψ \frac{I_{y}}{I_{x}} {(\frac{h}{l_{e f}})}^{2} \frac{E}{R_{y}} = 2.507 \frac{11,720}{175,200} {(\frac{640}{5,000})}^{2} \frac{206,000}{223.8} = 2.529

ϕ_b

0.68 + 0.21×ϕ₁ = 1.211 > 1.0, take ϕ_b = 1.0

c₁₀

\frac{1}{1 + m_{x} (ϕ_{1} / ϕ_{b})} = \frac{1}{1 + 5.172 \times (0.826 / 1)} = 0.190

per F.113

Therefore, $c_{\max} = 0.175 (2 - 0.2 \times 5.172) + 0.190 (0.2 \times 5.172 - 1) = 0.176$

So, the ratio is $\frac{80}{0.176 \times 0.826 \times 242 {(10)}^{- 1} \times 223.9 \times 1} = 0.102 < 1$

Design for Stability (Cl. 9.2.9)

\frac{N}{ϕ_{exy} \times A \times R_{y} \times γ_{c}} \leq 1

(F.111 )

where

ϕ_ey	=	0.333 (per Table Д.3)
ϕ_exy	=	$ϕ_{ey} (0.6 \sqrt[3]{c} + 0.4 \sqrt[4]{c}) = 0.333 (0.6 \sqrt[3]{0.176} + 0.4 \sqrt[4]{0.176}) = 0.198$

So, the ratio is

\frac{80}{0.198 \times 242 {(10)}^{- 1} \times 223.9 \times 1} = 0.075 < 1

Results

Result Type	Reference	STAAD.Pro	Difference
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
m_ef	7.18	7.16	negligible
m_x	5.172	5.18	negligible
C	0.176	0.179	negligible
ϕ_ey	0.333	0.33	negligible
ϕ_exy	0.198	0.199	negligible

STAAD.Pro Input

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
LOAD 1 LOADTYPE None  TITLE LOAD CASE 1
JOINT LOAD
2 FY -80
MEMBER LOAD
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 Output

                       STAAD.PRO CODE CHECKING - (SP 16.13330.2017)   V1.0
                       ********************************************
   ALL UNITS ARE - KN METRE
   ========================================================================
   MEMBER     CROSS          RESULT/   CRITICAL COND/    RATIO/    LOADING/
              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