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RAM | STAAD | ADINA Wiki Russian SP 16.13330 bending check verification
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    Russian SP 16.13330 bending check verification

       
      Applies To 
       
      Product(s): STAAD.Pro
      Version(s): All
      Environment:  N/A
      Area:  Steel Design
      Subarea:  Verifications
      Original Author: Anisurya Ghosh, BSW-Structural Design, Bentley Kolkata
       
    •  Introduction:

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

    • Validation Problem:

     

    A 5 m long beam formed from a HD320X127 I section (Steel grade: S235), is checked for an applied UDL of -100 KN/m throughout the span.

     

    • Input File:

    STAAD PLANE
    INPUT WIDTH 79
    UNIT METER KN
    JOINT COORDINATES
    1 0 0 0; 2 5 0 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 FX MZ
    LOAD 1 LOADTYPE None TITLE LOAD CASE 1
    MEMBER LOAD
    1 UNI GY -100
    PERFORM ANALYSIS
    PRINT FORCE ENVELOPE NSECTION 2 ALL
    PARAMETER 1
    CODE RUSSIAN
    BEAM 1 ALL
    *CMN 5 ALL
    SGR 1 ALL
    GAMC2 1.1 ALL
    GAMC1 1.1 ALL
    TB 1 ALL
    DFF 200 ALL
    *BMT 100 ALL
    TRACK 2 ALL
    CHECK CODE ALL
    FINISH

    • Manual Calculation:

    Section Properties:

     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

    Maximum Bending Moment = 312.5 KN-m

    Material Properties:

    S235

    E = 206000 MPa, fy = 235 MPa, γc1 = 1.1, γc2 = 1.1

    Check for Flexure:

    We need to satisfy eqn. 41 of SNiP SP16.13330.2011:

    M = 312.5, Wn,min = Wx = 1926.5 cm3, Ry = 235 MPa, γc2 = 1.1

    Then, M/( Wn,min*Ry* γc) = 0.6275 < 1 (OK)

    STAAD.Pro value is 0.628 [Matching]

     

    Check for Shear:

    We need to satisfy eqn. 42 of SNiP SP16.13330.2011:

    Q = q.l/2 = 250 KN, S = 1070 cm3, I = 30820 cm4, Rs = 0.58*235 MPa, γc2 = 1.1, tw = 11.5 mm

    Then, QS/(I*tw*Rs*γc) = 0.5034 < 1.0 (OK)

    Check for Lateral Torsional Buckling:

    We need to satisfy eqn. 42 of SNiP SP16.13330.2011

     

    Check for Combined Shear & Flexure:

    We need to satisfy eqn. 44 of SNiP SP16.13330.2011


    Check for Deflection:


    (5*100*54)/(384*206000*1000*30820*10-8) = 0.0128 m

    STAAD.Pro value is 0.0128 m [Matching]

    Maximum member deflection = l/200 = 0.025 m

    STAAD.Pro value is 0.0128 m [Matching]

    Ratio = 0.0128/0.025 = 0.512

    STAAD.Pro value is 0.51 [Matching]

     

    • Summary Table:

    Item STAAD.Pro Value Hand Calculated Value % Deviation
    Flexure Ratio 0.628 0.6275 0.0796
    LTB Ratio 0.63 0.6275 0.3968
    τxy 0 0 0.0000
    Ratio for combined shear & Flexure 0.546 0.546 0.0000
    λub 0.865 0.865 0.0000
    λb 0.5629 0.5629 0.0000
    Maximum Deflection 0.0128 0.0128 0.0000
    Deflection Ratio 0.51 0.512 0.3922
      • SNiP SP 16.13330.2011
      • SNiP_I section
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      • History
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      • Anisurya Ghosh Created by Bentley Colleague Anisurya Ghosh
      • When: Fri, Sep 19 2014 5:06 AM
      • Modestas Last revision by Bentley Colleague Modestas
      • When: Thu, Apr 27 2023 7:35 AM
      • Revisions: 11
      • Comments: 0
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