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    Analisa Balok 400

    Uploaded by

    Rifki Saputra
    This document summarizes the design of a steel beam according to AISC 360 specifications. It includes: 1) Details of the UC 400x400x197 steel section being designed, including dimensions, properties, and classifications. 2) Calculations of the required flexural and shear strengths based on a major axis moment of 296.7 kNm. 3) Determination that the section is noncompact in flexure. 4) Calculation of nominal flexural and shear strengths considering factors like lateral-torsional buckling. 5) Comparison of required and design strengths, confirming the design is adequate for both flexure and shear.

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    © All Rights Reserved
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    Analisa Balok 400

    Uploaded by

    Rifki Saputra
    25 views3 pages
    This document summarizes the design of a steel beam according to AISC 360 specifications. It includes: 1) Details of the UC 400x400x197 steel section being designed, including dimensions, properties, and classifications. 2) Calculations of the required flexural and shear strengths based on a major axis moment of 296.7 kNm. 3) Determination that the section is noncompact in flexure. 4) Calculation of nominal flexural and shear strengths considering factors like lateral-torsional buckling. 5) Comparison of required and design strengths, confirming the design is adequate for both flexure and shear.

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    Original Title

    ANALISA BALOK 400

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    This document summarizes the design of a steel beam according to AISC 360 specifications. It includes: 1) Details of the UC 400x400x197 steel section being designed, including dimensions, properties, and classifications. 2) Calculations of the required flexural and shear strengths based on a major axis moment of 296.7 kNm. 3) Determination that the section is noncompact in flexure. 4) Calculation of nominal flexural and shear strengths considering factors like lateral-torsional buckling. 5) Comparison of required and design strengths, confirming the design is adequate for both flexure and shear.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    25 views3 pages

    Analisa Balok 400

    Uploaded by

    Rifki Saputra
    This document summarizes the design of a steel beam according to AISC 360 specifications. It includes: 1) Details of the UC 400x400x197 steel section being designed, including dimensions, properties, and classifications. 2) Calculations of the required flexural and shear strengths based on a major axis moment of 296.7 kNm. 3) Determination that the section is noncompact in flexure. 4) Calculation of nominal flexural and shear strengths considering factors like lateral-torsional buckling. 5) Comparison of required and design strengths, confirming the design is adequate for both flexure and shear.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    Project Job Ref.

    TUGAS AKHIR
    Section Sheet no./rev.

    JTS FT UNRAM 406*403 1


    Calc. by Date Chk'd by Date App'd by Date
    ILMANSYAH 10/10/2017

    STEEL MEMBER DESIGN (AISC 360)


    In accordance with AISC360 14th Edition published 2010 using the LRFD method
    Tedds calculation version 4.3.01

    Resistance factors
    Shear; v = 0.90
    Flexure; b = 0.90
    Tensile yielding; t,y = 0.90
    Tensile rupture; t,r = 0.75
    Compression; c = 0.90

    Design section 1
    Section details
    Section type; UC 400x400x197 (JIS)
    ASTM steel designation; A992
    Steel yield stress; Fy = 345 N/mm2
    Steel tensile stress; Fu = 448 N/mm2
    Modulus of elasticity; E = 199948 N/mm2

    UC 400x400x197 (JIS)
    21

    Section depth, d, 400 mm


    Section breadth, bf, 408 mm
    Weight of section, Weight, 1.776 kN/m
    Flange thickness, tf, 21 mm
    Web thickness, tw, 21 mm
    Area of section, A, 25070 mm2
    Radius of gyration about x-axis, rx, 168.157 mm
    Radius of gyration about y-axis, ry, 97.454 mm
    400

    Elastic section modulus about x-axis, Sx, 3544400 mm3


    Elastic section modulus about y-axis, Sy, 1167120 mm3
    Plastic section modulus about x-axis, Zx, 3992460 mm3
    21 Plastic section modulus about y-axis, Zy, 1793740 mm3
    Second moment of area about x-axis, Ix, 708881000 mm4
    Second moment of area about y-axis, Iy, 238093000 mm4
    21

    408

    Analysis results
    Required flexural strength - Major axis; Mr,x = 296.7 kNm
    Required shear strength - Major axis; Vr,x = 302.85 kN
    Restraint spacing
    Major axis lateral restraint; Lx = 7000 mm
    Minor axis lateral restraint; Ly = 7000 mm
    Torsional restraint; Lz = 7000 mm
    Classification of sections for local buckling - Section B4
    Classification of flanges in flexure - Table B4.1b (case 10)
    Width to thickness ratio; bf / (2  tf) = 9.71
    Limiting ratio for compact section; pff = 0.38  [E / Fy] = 9.15
    Project Job Ref.
    TUGAS AKHIR
    Section Sheet no./rev.

    JTS FT UNRAM 406*403 2


    Calc. by Date Chk'd by Date App'd by Date
    ILMANSYAH 10/10/2017

    Limiting ratio for non-compact section; rff = 1.0  [E / Fy] = 24.08; Noncompact

    Classification of web in flexure - Table B4.1b (case 15)


    Width to thickness ratio; (d - 2  k) / tw = 14.95
    Limiting ratio for compact section; pwf = 3.76  [E / Fy] = 90.55
    Limiting ratio for non-compact section; rwf = 5.70  [E / Fy] = 137.27; Compact
    Section is noncompact in flexure
    Design of members for shear - Chapter G
    Required shear strength; Vr,x = 302.9 kN
    Web area; Aw = d  tw = 8400 mm2
    Web plate buckling coefficient; kv = 5
    (d - 2  k) / tw <= 2.24  (E / Fy)
    Web shear coefficient - eq G2-2; Cv = 1.000
    Nominal shear strength - eq G2-1; Vn,x = 0.6  Fy  Aw  Cv = 1737.5 kN
    Resistance factor; v = 1.00
    Design shear strength; Vc,x = v  Vn,x = 1737.5 kN
    Vr,x / Vc,x = 0.174
    PASS - Design shear strength exceeds required shear strength

    Design of members for flexure - Chapter F


    Required flexural strength; Mr,x = 296.7 kNm
    Plastic moment - eq F2-1; Mp,x = Fy  Zx = 1015.1 kips_ft

    Lateral-torsional buckling - Section F3.1


    Unbraced length; Lb = Ly_s1 = 7000 mm
    Limiting unbraced length for yielding - eq F2-5; Lp = 1.76  ry  (E / Fy) = 4131 mm
    Distance between flange centroids; ho = 384 mm
    c=1
    rts = 110 mm
    Limiting unbraced length for inelastic LTB - eq F2-6;Lr = 1.95  rts  E / (0.7  Fy)  ((J  c / (Sx  ho)) + ((J  c / (Sx  ho))2 +
    6.76  (0.7  Fy / E)2)) = 15311 mm
    LTB modification factor; Cb = 1.000
    Nominal flexural strength for lateral-torsional buckling - eq F2-2
    Mn,ltb,x = min(Cb  (Mp,x - (Mp,x - 0.7  Fy  Sx)  (Lb - Lp) / (Lr - Lp)), Mp,x) =
    1242.6 kNm
    Compression flange local buckling - Section F3.2
     = bf / (2  tf) = 9.714
    Nominal flexural strength for compression flange local buckling - eq F3-1;
    Mn,flb,x = Mp,x - (Mp,x - 0.7  Fy  Sx)  ( - pff) / (rff - pff) = 1356.7 kNm

    Design flexural strength - F1


    Nominal flexural strength; Mn,x = min(Mn,ltb,x, Mn,flb,x) = 1242.6 kNm
    Design flexural strength; Mc,x = b  Mn,x = 1118.4 kNm
    Mr,x / Mc,x = 0.265
    PASS - Design flexural strength exceeds required flexural strength
    Project Job Ref.
    TUGAS AKHIR
    Section Sheet no./rev.

    JTS FT UNRAM 406*403 3


    Calc. by Date Chk'd by Date App'd by Date
    ILMANSYAH 10/10/2017

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