# V.SNiP SP16 - I Section with Axial Load

Verify the design capaciy of an I section used in a column per the SP16.13330.2011 code.

## Details

The column is 7.5 m tall with a pinned base and a roller (free to move vertically and rotate) at the top. The column has a 3,500 kN axial load. The section used is an HD320X127. Steel grade is S235.
• E = 205,000 MPa
• Ry = 235 MPa (yield strength)
• γc1 = 1, γc2 = 1

## Validation

Check for Axial Force

 $NAn×Ry×γc≤1$ [Cl. 7.1.1, Eqn. 5]
 $Nϕ×A×Ry×γc≤1$ [Cl. 7.1.3, Eqn. 7]
where
 A = cross sectional area = 161 cm2

Slenderness ratios:

Ky = Kx = 0.75

$λx=KxLrx=0.75×7.50.138=40.76$
$λy=KyLry=0.75×7.50.0757=74.3$
$λ¯x=λxRyE=40.76235205,000=1.38$
$λ¯y=λyRyE=74.3235205,000=2.516$
 $ϕ=0.5∂-∂2+39.48×λ¯2λ¯2$ [Eqn. 8]
 $∂=9.87×(1-α+β×λ¯)+λ¯2$ [Eqn. 9]
where
 α = 0.04 (From Table 7) β = 0.09 (From Table 7) $λ¯$ = 2.516 (maximum slenderness ratio)
$∂=9.87×[1-0.04+0.09×(2.516)]+(2.516)2=18.04$
 $Nϕ×A×Ry×γc=3,5000.7385×0.0161×235(103)×1.0=1.25>1$ [Cl. 7.1.3, Eqn. 7]

Check Web Slenderness

$λ¯w=hefTwRyE=23811.5235205,000=0.700$
 $λ¯uw=1.2+0.35λ¯=1.2+0.35(2.516)=2.08$ [Table 9; Eqn. 24]

Check Flange Slenderness

$λ¯f=befTfRyE=12720.5235205,000=0.210$
 $λ¯uf=0.36+0.1λ¯=0.36+0.1(2.516)=0.612$ [Table 10; Eqn. 37]

## Results

Table 1. Comparison of results
Cl. 7.1.1 0.925 0.92 negligible
Cl. 7.1.3 1.25 1.25 none
$λ¯w$ 0.700 0.699 negligible
$λ¯uw$ 2.08 2.08 none
$λ¯f$ 0.210 0.209 negligible
$λ¯uf$ 0.612 0.611 negligible

## STAAD.Pro Input File

The file C:\Users\Public\Public Documents\STAAD.Pro CONNECT Edition\Samples\ Verification Models\09 Steel Design\Russia\SNiP SP16\SNiP SP16 - I Section with Axial Load.std is typically installed with the program.

STAAD PLANE
START JOB INFORMATION
ENGINEER DATE 18-Mar-19
END JOB INFORMATION
INPUT WIDTH 79
UNIT METER KN
JOINT COORDINATES
1 0 0 0; 2 0 7.5 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 HD320X127
CONSTANTS
MATERIAL STEEL ALL
SUPPORTS
1 PINNED
2 FIXED BUT FY MZ
2 FY -3500
PERFORM ANALYSIS
PARAMETER 1
CODE RUSSIAN
KY 0.75 ALL
KZ 0.75 ALL
SGR 1 ALL
TRACK 2 ALL
CHECK CODE ALL
FINISH


                       STAAD.PRO CODE CHECKING - (SP 16.13330.2011)   V2.0
********************************************
ALL UNITS ARE - KN METRE
========================================================================
MEMBER     CROSS          RESULT/   CRITICAL COND/    RATIO/    LOADING/
SECTION NO.      N             Mx            My      LOCATION
========================================================================
*     1  I      HD320X127     FAIL      SP cl.7.1.3      1.25         1
3.500E+03 C    0.000E+00    0.000E+00   0.000E+00
1  I      HD320X127     PASS      SP cl.7.1.1      0.92         1
3.500E+03 C    0.000E+00    0.000E+00   0.000E+00
MATERIAL DATA
Steel                         = S235        EN10025-2
Modulus of elasticity         = 206.E+06 kPa
Design Strength (Ry)          = 235.E+03 kPa
SECTION PROPERTIES (units - m, m^2, m^3, m^4)
Member Length                 = 7.50E+00
Gross Area                    = 1.61E-02
Net Area                      = 1.61E-02
x-axis      y-axis
Moment of inertia (I)         :   308.E-06    924.E-07
Section modulus (W)           :   193.E-05    616.E-06
First moment of area (S)      :   107.E-05    470.E-06
Radius of gyration (i)        :   138.E-03    757.E-04
Effective Length              :   5.63E+00    5.63E+00
Slenderness                   :   407.E-01    743.E-01
DESIGN DATA (units -kN,m) SP16.13330.2011
Axial force                   :   350.0E+01
x-axis      y-axis
Moments                       :   0.000E+00    0.000E+00
Shear force                   :   0.000E+00    0.000E+00
Bi-moment                     :   0.000E+00 Value of Bi-moment not being entered!!!
CRITICAL CONDITIONS FOR EACH CLAUSE CHECK
F.(7)  N/(FI*A*Ry*GammaC)= 350.0E+01/( 7.40E-01* 1.613E-02* 235.0E+03* 1.00E+00)= 1.25E+00>1
7.3 LAMBDA_w&amp;lt;=LAMBDA_uw= 6.99E-01&amp;lt;= 2.08E+00 OK
LAMBDA_f&amp;lt;=LAMBDA_uf= 2.09E-01&amp;lt;= 6.11E-01 OK
F.(5)  N/(An*Ry*GammaC)= 350.0E+01/( 1.61E-02* 235.0E+03* 1.00E+00)= 9.23E-01=&lt;1