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RAM | STAAD | ADINA Wiki SNiP_Axial Check_I section Design
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    SNiP_Axial Check_I section Design

       
      Applies To 
       
      Product(s): STAAD.Pro
      Version(s): All
      Environment:  N/A
      Area:  Wiki
      Subarea:  Axial Checks as per SNiP SP 16.13330.2011
      Original Author: Anisurya Ghosh
    BSW-Structural Design, Bentley Kolkata
       

     

     

     

     

     

     

     

     

    • Introduction:

    The following is a validation problem for a detailed calculation of the axial check for the SNiP SP 16 I section design.

    • Validation Problem:

    A 7.5 m long column formed from a HD320X127 I section (Steel grade: S235), is checked for an applied concentrated load of 3500 KN at the tip.

     

    • Input File:

    STAAD PLANE
    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
    LOAD 1 LOADTYPE None TITLE LOAD CASE 1
    JOINT LOAD
    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

    • Manual Calculation:

    Section Properties:

    HD320X127

    D= 320 mm, B= 300 mm, Tf= 20.5 mm, Tw= 11.5 mm, Ix= 30820 cm4, Iy= 9239 cm4, Wx= 1926.5 cm3, Sx= 1070 cm3, J= 225.1 cm4, H= 2.069*10^12 mm6

    rx = 13.8 cm, ry = 7.57 cm, An = 161 cm2

    Material Properties:

    S235

    E = 205000 MPa, fy = 235 MPa, γc1 = 1, γc2 = 1

    Check for Axial Force:

    • As per clause 7.1.1, we need to satisfy eqn. 5 of SP-16.13330.2011:

    N/(An*Ry*γc) < = 1

    So, 3500/(161*10-4*235*1000*1.0) = 0.9251 < 1 [OK]

    STAAD.Pro value is 0.92 [Matching]

    KY = KZ = 0.75

    λx = KL/r = 7.5*0.75/0.138 = 40.76

    λy = KL/r = 7.5*0.75/0.0757 = 74.3

    λ̅x = λx*√(Ry/E) = 1.38

    λ̅y = λy*√(Ry/E) = 2.516

     λ̅ = 2.516

     

    • As per clause 7.1.3, we need to satisfy eqn. 5 of SP-16.13330.2011:

    N/(ф*A*Ry*γc) < = 1 [eqn. 7]

    From eqn. 8, we get:

    Ф = 0.5*[∂-√(∂2-39.48*λ̅2)]/(λ̅)2

    From eqn. 9, ∂ = 9.87*(1-α+β*λ̅) + λ̅2

    From table 7, we get α = 0.04, β = 0.09

    So, ∂ = 9.87*(1-0.04+0.09*2.516) + 2.5162 = 18.040

    Now, ф = 0.5*[18.04-√(18.042-39.48*2.5162)]/2.5162 = 0.7385 < 7.6/λ̅2, i.e. 1.2006 [OK]

    N/(ф*A*Ry*γc) = 3500/(0.7385*0.0161*235000*1.0) = 1.2526 > 1.0

    STAAD.Pro value is 1.25 [Matching]

     

    • As per clause 7.3.2: λ̅w < λ̅uw

    hef = D-4*tf = 320-4*20.5 mm = 238 mm

    λ̅w = (hef/tw)*√(Ry/E) = 0.693

    STAAD.Pro value is 0.699 [Matching]

    From table 9, < λ̅uw = 1.2+0.35*λ̅ = 2.0806 (eqn. 24)

    STAAD.Pro value is 2.08 [Matching]

    Hence, λ̅w < λ̅uw [OK]

    • As per clause 7.3.8: λ̅f < λ̅uf

    bef = 0.5*(B-2*tw) = 0.5*(300-4*11.5) = 127 mm

    λ̅f = (bef/tf)*√(Ry/E) = 0.209

    STAAD.Pro value is 0.209 [Matching]

    From table 10, < λ̅uf = 0.36+0.1*λ̅ = 0.6116 (eqn. 37)

    STAAD.Pro value is 0.611 [Matching]

    Hence, λ̅f < λ̅uf [OK]

    • Summary Table:
    Item STAAD.Pro Value Hand Calculated Value % Deviation
    Ratio-Cl 7.1.1 0.92 0.9251 0.5543
    Ratio-Cl 7.1.3 1.25 1.2526 0.2080
    λ̅uw  2.08 2.0806 0.0288
    λ̅uf 0.611 0.6116 0.0982

     

      • SNiP_Axial Check_I section Design
      • SNiP SP 16.13330.2011
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      • Anisurya Ghosh Created by Bentley Colleague Anisurya Ghosh
      • When: Mon, Sep 22 2014 5:10 AM
      • Steve Crabtree Last revision by Steve Crabtree
      • When: Fri, Jun 24 2016 11:44 AM
      • Revisions: 6
      • Comments: 0
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