V. SNiP SP16 2017 - Interaction check of a column

Design a column per the SP 16.13330.2017 code.

Details

A 6.78m tall, simply supported column has a European HE600B section. The column is subject to a 310 kN axial load along with a uniformly distributed load of 50 kN/m in the local Y axis. The steel used has a modulus of elasticity of 206,000 MPa and a R_y = 239 MPa. γ_c = 1, γ_m = 1.05

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 = q_x × L / 2 = 171.8 kN

M_x = q_x × L²/ 8 = 295.0 kN·m

Bending moment:

M_x = q_x × L² / 8 = 171.8 (6.78)² / 8 = 295.0 kN·m

Design for Strength (Cl. 9.1.1)

σ = \frac{N}{A_{n}} = \frac{310}{270 {(10)}^{- 1}} = 11.48 < 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 = 4.51 (e = M/N = 0.9516)

λ_{x} = \frac{k_{x} l}{r_{x}} = \frac{1.0 \times 6.87}{0.2517} = 27.3

λ_{y} = \frac{k_{y} l}{r_{y}} = \frac{1.0 \times 6.87}{0.0708} = 97.0

{\bar{λ}}_{x} = λ_{x} \sqrt{\frac{R_{y}}{E}} = 27.3 \sqrt{\frac{223.8}{206, 000}} = 0.900

{\bar{λ}}_{y} = λ_{y} \sqrt{\frac{R_{y}}{E}} = 97.0 \sqrt{\frac{223.8}{206, 000}} = 3.198

Therefore, $\bar{λ} = \min (\bar{λ_{x}}, \bar{λ_{y}}) = 0.900$

η = (1.90 - 0.1 m) - 0.02 (6 - m) \bar{λ} = (1.90 - 0.1 \times 4.51) - 0.02 (6 - 4.51) 0.900 = 1.422

So, $m_{e f} = 1.422 \times 4.51 = 6.41 < 20$ [As per F.(110)]

A_{f} / A_{w} = 90 / 90 = 1

\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	=	300 mm
B_ω	=	0

So, the ratio is $\frac{11.48 + \frac{295.0 (300)}{171, 000 \times 10^{- 2}}}{223.8 \times 1} = 0.28 < 1$

Design for Stability (Cl. 9.2.2)

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

ϕ_e = 0.2144

\frac{N}{ϕ_{e} \times A {\times R}_{y} \times γ_{c}} = \frac{310}{0.2144 \times 270 {(10)}^{- 1} \times 223.8 \times 1} = 0.24 < 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	=	$\frac{0.5 (δ - \sqrt{δ^{2} - 39.48 {\bar{λ}}^{2}}}{{\bar{λ}}^{2}}$
$\bar{λ}$	=	conditional slenderness $= \max (\bar{λ_{x}}, \bar{λ_{y}}) = 3.198$
δ	=	$9.87 (1 - α + β \bar{λ}) + {\bar{λ}}^{2}$ , per Eq. 9

From Table 7, α = 0.03 and β = 0.06.

δ = 9.87 [1 - 0.03 + 0.06 (3.198)] + {(3.198)}^{2} = 21.70

ϕ_{y} = 0.5 \frac{21.70 - \sqrt{{(21.70)}^{2} - 39.48 \times {(3.198)}^{2}}}{{(3.198)}^{2}} = 0.660

c = \frac{β}{1 + α {\times m}_{x}}

(Cl. 9.2.5)

where

α	=	0.65 + 0.05×m_x = 0.875
$\bar{λ_{y}}$	=	3.14
δ_c	=	$9.87 [1 - 0.03 + 0.06 (3.14)] + {(3.14)}^{2} = 21.29$
ϕ_c	=	$0.5 \frac{21.29 - \sqrt{{(21.29)}^{2} - 39.48 \times {(3.14)}^{2}}}{{(3.14)}^{2}} = 0.674$
β	=	$\sqrt{\frac{ϕ_{c}}{ϕ_{y}}} = \sqrt{\frac{0.674}{0.661}} = 1.009$

Therefore, $c = \frac{1.009}{1 + 0.875 \times 4.51} = 0.204 < 1$

So, the ratio is $\frac{310}{0.204 \times 0.66 \times 270 {(10)}^{- 1} \times 223.9 \times 1} = 0.38 < 1$

Calculate C_max Per Annex E.1

c_{\max} = \frac{2}{1 + δ β + \sqrt{{(1 + δ β)}^{2} + \frac{16}{μ} {(α - \frac{e_{x}}{h})}^{2}}}

(F.(E.1) )

where

δ	=	4×ρ / μ
ρ	=	$\frac{I_{x} + I_{y}}{A h^{2}} + α^{2}$
μ	=	$8 ω + \frac{0.156 {\times I}_{t} {λ_{y}}^{2}}{A h^{2}}$
ω	=	$\frac{I_{ω}}{I_{y} {h_{c}}^{2}} = \frac{{h_{c}}^{2} \times b^{3} \times t_{f} / 24}{I_{y} {h_{c}}^{2}} = \frac{b^{3} \times t_{f}}{24 I_{y}} = \frac{{(30 cm)}^{3} \times 3 cm}{24 \times 13, 530 {cm}^{4}} = 0.249$
I_t	=	$\frac{k}{3} Σ b_{i} {t_{i}}^{3} = \frac{1.29}{3} [2 (30 cm) {(3 cm)}^{3} + (60 cm - 2 \times 3 cm) {(1.55 cm)}^{3}] = 783.1 {cm}^{4}$ (k = 1.29)
e_x	=	M / N = 295 / 310 = 0.952
B	=	$1 + 2 (β / ρ) (e_{x} / h_{c}) = 1$

As per Table E.6, α = 0, β = 0

μ = 8 (0.249) + \frac{0.156 \times (783.1) {(97.0)}^{2}}{270 \times {(60)}^{2}} = 3.17

ρ = \frac{171, 000 + 13, 530}{270 \times 60^{2}} + 0 = 0.190

δ = \frac{4 \times 0.19}{3.17} = 0.24

c_{\max} = \frac{2}{1 + 0.24 (0) + \sqrt{{(1 + 0.24 \times 0)}^{2} + \frac{16}{3.17} {(0 - \frac{0.952}{0.6})}^{2}}} = 0.43

Results

Result Type	Reference	STAAD.Pro	Difference
Ratio per Cl. 9.1.1	0.28	0.28	none
Ratio per Cl. 9.2.2	0.24	0.24	none
Ratio per Cl. 9.2.4	0.38	0.38	none
m_ef	6.41	6.41
m_x	4.51	4.51	none
C	0.204	0.204	none
C_max	0.43	0.40	negligible
Φ_y	0.66	0.66	none

STAAD.Pro Input

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Russia\SNiP SP16 2017 - Interaction check of a column.std is typically installed with the program.

STAAD SPACE
START JOB INFORMATION
ENGINEER DATE 02-Sep-20
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 6.87 0;
MEMBER INCIDENCES
1 1 2;
DEFINE MATERIAL START
ISOTROPIC STEEL
E 2.05e+008
POISSON 0.3
DENSITY 76.8195
ALPHA 1.2e-005
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 HE600B
****************************************
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 -310
MEMBER LOAD
1 UNI GX 50
********************************
PERFORM ANALYSIS
*********************************
PARAMETER 1
CODE RUSSIAN
TB 1 ALL
ENSGR 1 ALL
GAMM 2 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      HE600B        PASS      SP cl.9.1.1      0.28         1
                          3.100E+02 C    2.950E+02    0.000E+00   3.435E+00
      1  I      HE600B        PASS      SP cl.9.2.2      0.24         1
                          3.100E+02 C    2.950E+02    0.000E+00   3.435E+00
      1  I      HE600B        PASS      SP cl.9.2.4      0.38         1
                          3.100E+02 C    2.950E+02    0.000E+00   3.435E+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                 = 6.87E+00
      Gross Area                    = 2.70E-02
      Net Area                      = 2.70E-02
                                         x-axis      y-axis
      Moment of inertia (I)         :   171.E-05    135.E-06
      Section modulus (W)           :   570.E-05    902.E-06
      First moment of area (S)      :   321.E-05    696.E-06
      Radius of gyration (i)        :   252.E-03    708.E-04
      Effective Length              :   6.87E+00    6.87E+00
      Slenderness                   :   273.E-01    970.E-01
   DESIGN DATA (units -kN,m) SP16.13330.2017
      Axial force                   :   310.0E+00
                                         x-axis      y-axis
      Moments                       :   295.0E+00    0.000E+00
      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    1
   CRITICAL CONDITIONS FOR EACH CLAUSE CHECK
      F.(106) (N/A+Mx*y/Ix+My*x/Iy+B*w/Iw)/(Ry*GammaC)=
              ( 310.0E+00/ 2.7E-02+ 295.0E+00* 3.00E-01/ 1.71E-03+ 0.000E+00* 1.50E-01/
               1.35E-04+ 0.000E+00* 2.49E-01/ 1.10E-05)/( 223.8E+03* 1.00E+00)
               = 2.83E-01=&lt;1
      cl.9.2.2  m_ef=eta*mx= 1.42E+00* 4.51E+00= 6.41E+00
      F.(109)  N/(FIe*A*Ry*GammaC)= 310.0E+00/( 2.14E-01* 2.70E-02* 223.8E+03* 1.00E+00)
               = 2.39E-01=&lt;1
      F.(112)  c=beta/(1+alfa*mx)= 1.01E+00/(1+8.75E-01* 4.51E+00)= 2.04E-01
                    c_max= 4.00E-01
      F.(111)  N/(c*FIy*A*Ry*GammaC)
               = 0.31E+03/( 0.20E+00* 0.66E+00* 0.27E-01* 223.8E+03* 1.00E+00)
               = 3.80E-01=&lt;1