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    Bar Bending Schedule For Beam

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    truth finder
    This document provides information about the bar bending schedule for a beam, including: - The beam has a clear span of 3.5m and is reinforced with 2 bars of 12mm diameter on the top and bottom, with 8mm diameter stirrups spaced at 150mm. - It calculates the cutting lengths of the top and bottom bars as 8.86m each, requiring 1 bar of 12mm diameter for each. - It determines 25 pieces of stirrups are needed, each with a cutting length of 1.44m, for a total stirrup length of 46m requiring 2 bars of 8mm diameter.

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    Bar Bending Schedule For Beam

    Uploaded by

    truth finder
    30 views15 pages
    This document provides information about the bar bending schedule for a beam, including: - The beam has a clear span of 3.5m and is reinforced with 2 bars of 12mm diameter on the top and bottom, with 8mm diameter stirrups spaced at 150mm. - It calculates the cutting lengths of the top and bottom bars as 8.86m each, requiring 1 bar of 12mm diameter for each. - It determines 25 pieces of stirrups are needed, each with a cutting length of 1.44m, for a total stirrup length of 46m requiring 2 bars of 8mm diameter.

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    This document provides information about the bar bending schedule for a beam, including: - The beam has a clear span of 3.5m and is reinforced with 2 bars of 12mm diameter on the top and bottom, with 8mm diameter stirrups spaced at 150mm. - It calculates the cutting lengths of the top and bottom bars as 8.86m each, requiring 1 bar of 12mm diameter for each. - It determines 25 pieces of stirrups are needed, each with a cutting length of 1.44m, for a total stirrup length of 46m requiring 2 bars of 8mm diameter.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    30 views15 pages

    Bar Bending Schedule For Beam

    Uploaded by

    truth finder
    This document provides information about the bar bending schedule for a beam, including: - The beam has a clear span of 3.5m and is reinforced with 2 bars of 12mm diameter on the top and bottom, with 8mm diameter stirrups spaced at 150mm. - It calculates the cutting lengths of the top and bottom bars as 8.86m each, requiring 1 bar of 12mm diameter for each. - It determines 25 pieces of stirrups are needed, each with a cutting length of 1.44m, for a total stirrup length of 46m requiring 2 bars of 8mm diameter.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 15

    Bar Bending Schedule for Beam

    EXAMPLE -1

    • At the bottom of the figure, the beam has a clear span of 3.5
    m, which has 2 numbers and 12 mm dia bars on top and
    bottom with 8 mm dia stirrups on the 150 mm and clear
    cover with 25 mm clear cover on both ends and sides of the
    beam
    GIVEN DATA

    Clear Span of Beam = 3500 mm


    Development Length Ld = 41d (assumption)
    Clear Cover on any ends = 25 m
    Bottom – 2 numbers of 12⌀ (DIA)
    Top – 2 numbers of 12⌀ (DIA)
    Stirrups spacing – 8⌀ (DIA) @ 150mm clear cover
    CUTTING LENGTH OF TOP BAR

    Length of 1 Rod Cutting length of top bar = l + (Ld x 2) – (2 x c.c)


    Clear Cover on 2 ends
    l = Clear Span of Beam
    Ld = Development length (Anchorage) Ld on 2 sides (41d)
    d = dia of bar
    c.c = clear cover (2 ends)
     
    Length of 1 Rod Cutting length of top bar = 3.5 + (41d x 2) – (2 x
    25) = 3.5 + (41(12) x 2) – (2 x 25)
     
    = 3500+(2 x 41d) – 2 x 25 = 3500 + (2 x 41 x 12) – 50
     
    Length of 1 Rod Cutting length of top bar = 4434
    Length of one Rod Cutting length of top bar is 4.43. we have total bar two
    bars

    = 2 x 4.43 = 8.86m

    Total length of top bar = 8.86m bar  – 1bar  of 12 mm needed


     
    FIND CUTTING LENGTH OF BOTTOM BAR

    Length of 1 Rod Cutting length of bottom bar = l + (Ld x 2) – (2 x c.c) Clear


    Cover on 2 ends
     
    Follow above procedure
     
    Total length of bottom bar = 8.86m bar – 1bar of 12 mm needed
    CUTTING THE LENGTH OF STIRRUPS

    The cross-sectional area of the beam is 300 mm x 400 mm

    Take it has,

    300mm is side A
    400mm is side B

    Minimum clear cover to the reinforcement: 15 mm to the bars in slabs, 25 mm to bars
    in beams and columns. In large columns, say 450 mm in thickness, the cover should be
     40 mm.
    SIDE A SIDE B
    A =300 – 2 Side clear cover B =400 –  2 x Top & Bottom cover
       
    A = 300  – 2 x clear cover B = 400 – 2 x clear cover
       
    A = 300  – 2 x 40 B = 400 – 2 x 40
       
    A = 300  – 80 B = 400 – 80
       
    A = 220 mm B = 320 mm
    NO OF STIRRUPS REQUIRED

    As mentioned Stirrups spacing – 8⌀ (DIA) @ 150mm clear cover

    Formula = L/150 +1

    L/150 + 1 = 3500/150 + 1 = 23.33 + 1 = 24.33 = 25nos


     
    CUTTING LENGTH OF ONE STIRRUP
    Formula:
     
    Cutting Length of Stirrups = Perimeter of Shape + Total hook length – Total Bend Length
     
    = 2(a+b) + 2(9d) – 3(2d) – 2 (3d)

    = 2(300+400) + 2(9x8) – 3(2x8) – 2(3x8)

    = 2(700) + 2(72) – 3(16) – 2(24)

    = 1400 + 144 – 48 – 48

    = 1448mm say as 1.44m


     
    We have a total of 25 nos of stirrups, which are going to use,
     
    Total length

    = 25 x 1.84

    CUTTING THE LENGTH OF RECTANGULAR STIRRUPS = 46 m


    RESULT:

    Diameter of Bar Numbers Cutting Length Total Length No of bar

    Top Bar 12mm 2 4.3m 8.6 m 1

    Bottom Bar 12mm 2 4.3m 9.3 m 1

    Stirrups 8mm 25 1.44m 46m 2


    EXAMPLE – 2

    • At the bottom of the figure, the beam has a clear span of 5.5
    m, which has 3 numbers and 12 mm dia bars on top and the
    bottom consists of two layers of the bottom layer (2
    numbers of 16mm dia) and one layer of the top (2 numbers
    of 20mm dia). with 8 mm dia stirrups on 3 Zones, where
    Zone A, C has stirrups of 150 mm spacing & Zone B, has
    stirrups of 200 mm spacing. and clear cover with 25 mm
    clear cover on both ends and sides of the beam

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