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    13 - Torsion in Concrete Beams

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    hmuranjan
    Here are the key steps to design the torsion reinforcement for girder A2-A3 based on the information provided: 1. Calculate the factored torque Tu = wuLn2/24 = 1000*30*302/24 = 375,000 in-lbs 2. Check the concrete strength in torsion: - Acp = bwh + k1h2 = 30*24 + 4*(24)2 = 17,280 in2 - Pcp = 2h + 2(bw) = 2*24 + 2*(30*24) = 1,728 in - Tc = 3√f'cAcp/4Pcp = 3√4000*17,280

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    13 - Torsion in Concrete Beams

    Uploaded by

    hmuranjan
    79 views9 pages
    Here are the key steps to design the torsion reinforcement for girder A2-A3 based on the information provided: 1. Calculate the factored torque Tu = wuLn2/24 = 1000*30*302/24 = 375,000 in-lbs 2. Check the concrete strength in torsion: - Acp = bwh + k1h2 = 30*24 + 4*(24)2 = 17,280 in2 - Pcp = 2h + 2(bw) = 2*24 + 2*(30*24) = 1,728 in - Tc = 3√f'cAcp/4Pcp = 3√4000*17,280

    Original Description:

    calculate torsion in beams

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    Here are the key steps to design the torsion reinforcement for girder A2-A3 based on the information provided: 1. Calculate the factored torque Tu = wuLn2/24 = 1000*30*302/24 = 375,000 in-lbs 2. Check the concrete strength in torsion: - Acp = bwh + k1h2 = 30*24 + 4*(24)2 = 17,280 in2 - Pcp = 2h + 2(bw) = 2*24 + 2*(30*24) = 1,728 in - Tc = 3√f'cAcp/4Pcp = 3√4000*17,280

    Copyright:

    © All Rights Reserved
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    79 views9 pages

    13 - Torsion in Concrete Beams

    Uploaded by

    hmuranjan
    Here are the key steps to design the torsion reinforcement for girder A2-A3 based on the information provided: 1. Calculate the factored torque Tu = wuLn2/24 = 1000*30*302/24 = 375,000 in-lbs 2. Check the concrete strength in torsion: - Acp = bwh + k1h2 = 30*24 + 4*(24)2 = 17,280 in2 - Pcp = 2h + 2(bw) = 2*24 + 2*(30*24) = 1,728 in - Tc = 3√f'cAcp/4Pcp = 3√4000*17,280

    Copyright:

    © All Rights Reserved
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    of 9

    Torsion in Girders

    The beams framing into


    girder A2-A3 transfer a
    A2 B2 moment of wuln2/24 into the
    girder. This moment acts
    about the longitudinal axis
    Mu = Mu = wuln2/10 Mu = wuln2/11 of the girder as torsion.
    wuln2/24

    B3
    A torque will also be
    A3
    induced in girder B2-B3
    due to the difference
    between the end moments
    in the beams framing into
    the girder.
    Girder A2-A3
    A2

    A3

    A2 A3

    Torsion Diagram
    Strength of Concrete in Torsion
    Acp2
    Tu  Tc  3
    4 f 'c
    Pcp
    where,
    Acp = smaller of bwh + k1h2f or bwh + k2hf(h - hf)
    Pcp = smaller of 2h + 2(bw+ k1hf) or 2(h + bw) + 2k2(h - hf)

    k1 k2
    Slab one side of web
    4 1
    Slab both sides of web
    8 2

    As for shear, ACI 318 allows flexural members be designed for


    the torque at a distance ‘d’ from the face of a deeper support.
    Redistribution of Torque
    p. 44 notes-underlined paragraph
    When redistribution of forces and
    moments can occur in a statically
    indeterminate structure, the maximum
    torque for which a member must be
    designed is 4 times the Tu for which the
    torque could have been ignored. When
    redistribution cannot occur, the full
    factored torque must be used for design.

    In other words, the design Tmax = 4Tc if other


    members are available for redistribution of forces.
    Torsion Reinforcement
    • Stresses induced by torque are resisted with
    closed stirrups and longitudinal reinforcement
    along the sides of the beam web.

    • The distribution of torque along a beam is


    usually the same as the shear distribution
    resulting in more closely spaced stirrups.
    Design of Stirrup Reinforcement
    4 ( Av  2 At ) f y
    st  s   12"
    1.28 Aoh At f y Ph
    Tu 3bw f 'c 8

    1
    s  1
    sv  s1t
    Aoh = area enclosed by the centerline of the closed
    transverse torsional reinforcement

    At = cross sectional area of one leg of the closed ties


    used as torsional reinforcement

    st = spacing required for torsional reinforcment only

    sv = spacing required for shear reinforcement only

    S = stirrup spacing
    Design of Longitudinal
    Reinforcement
    At Ph  5 f 'c Acp At  At 25bw
    Al    Ph  where, 
    st  fy st  st fy

    Al = total cross sectional area of the


    additional longitudinal reinforcement
    required to resist torsion

    Ph = perimeter of Aoh

    max. bar spacing = 12”


    minimum bar diameter = st/24
    Cross Section Check
    To prevent compression failure due
    to combined shear and torsion:

       
    Vu 2
    bwd
    Tu Ph
    2
    1.7 Aoh
    2
     7.5 f 'c
    Design of Torsion Reinforcement
    for Previous Example p. 21 notes

    Design the torsion reinforcement for girder


    A2-A3.

    30 ft 30 ft 30 ft 30 ft

    24 ft

    24 ft

    24 ft

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