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    Analysis and Design of Tension Members (LRFD)

    Uploaded by

    Amira Hachem
    1) The document discusses the analysis and design of tension members subjected to pure axial force. Examples given include trusses, braces, cables, and tie rods. 2) Design is based on either yielding of the gross section or rupture of the net section. The net section is calculated based on bolt hole dimensions and locations. 3) Examples are provided to demonstrate the design calculations using Load and Resistance Factor Design (LRFD) and Allowable Stress Design (ASD) methods. Calculations include determining the nominal strength based on yielding or rupture, and checking that design strengths are sufficient based on the applied loads.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Analysis and Design of Tension Members (LRFD)

    Uploaded by

    Amira Hachem
    26 views17 pages
    1) The document discusses the analysis and design of tension members subjected to pure axial force. Examples given include trusses, braces, cables, and tie rods. 2) Design is based on either yielding of the gross section or rupture of the net section. The net section is calculated based on bolt hole dimensions and locations. 3) Examples are provided to demonstrate the design calculations using Load and Resistance Factor Design (LRFD) and Allowable Stress Design (ASD) methods. Calculations include determining the nominal strength based on yielding or rupture, and checking that design strengths are sufficient based on the applied loads.

    Original Description:

    Steel Design Tension Members

    Original Title

    Tension 1

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    © © All Rights Reserved

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    1) The document discusses the analysis and design of tension members subjected to pure axial force. Examples given include trusses, braces, cables, and tie rods. 2) Design is based on either yielding of the gross section or rupture of the net section. The net section is calculated based on bolt hole dimensions and locations. 3) Examples are provided to demonstrate the design calculations using Load and Resistance Factor Design (LRFD) and Allowable Stress Design (ASD) methods. Calculations include determining the nominal strength based on yielding or rupture, and checking that design strengths are sufficient based on the applied loads.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    26 views17 pages

    Analysis and Design of Tension Members (LRFD)

    Uploaded by

    Amira Hachem
    1) The document discusses the analysis and design of tension members subjected to pure axial force. Examples given include trusses, braces, cables, and tie rods. 2) Design is based on either yielding of the gross section or rupture of the net section. The net section is calculated based on bolt hole dimensions and locations. 3) Examples are provided to demonstrate the design calculations using Load and Resistance Factor Design (LRFD) and Allowable Stress Design (ASD) methods. Calculations include determining the nominal strength based on yielding or rupture, and checking that design strengths are sufficient based on the applied loads.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 17

    Analysis and Design of Tension Members

    (LRFD)

    Member is subjected to pure axial force (Pu) passing


    through c.g. and perpendicular to cross section.

    - No Buckling

    - Simple Design

    - Care for design and detailing of connections, Load

    should be concentric.
    Examples:
    a- Trusses

    b- Braces

    c- Cables / Hangers

    d- Tie Rods

    Sections:

    Rods /Angles/Channels/Cables/ IPE / Built-up


    Staggered Holes:

    At sec 1:
    Anet = bxt – 1xdhxt = t(b-dh) … (Full force Pu in plate)

    At sec 2:

    Anet = bxt – 2dht + x t …. (Full Pu in Plate)


    1) Anet = 1.2(32 – 3x1.8) = 31.92 cm2 (Full Pu)

    2) Anet =1.2x32-3x1.8x1.2+1.2( )

    = 35.12 cm2 (4/5 Pu)

    3) Anet =1.2x32-4x1.8x1.2 +2( )

    = 36.16 cm2 (Pu)

    4) Anet=1.2x32-5x1.8x1.2 +4( )

    = 40.4 cm2 (Pu)

    5) Anet =1.2x32 -4x1.8x1.2 +2( )

    = 36.16 cm2 (4/5 Pu)


    Failure Modes:

    1) Yielding: elongation at gross section (a-a)


     excessive deformations at connections.
    Pn ≥ Ag Fy
    Where Ag = b x t

    2) Rupture @ least-net area section


     Strain- hardening is reached quickly @ net area section.
    Pn ≥ Aeff Fu
    Where Aeff = effective net area ≤ Anet

    = U Anet where U≤1

    - Anet = Ag - Aholes + xt For Staggered Bolts

    Then, for sec (b-b) or (c-c): Anet= b x t – n(dh)(t) = t (b-n dh)

    Where n = # of bolts/ section

    S = Pitch distance between 2 consecutive bolts.

    g = Gage distance between 2 consecutive bolts.

     Note: - At section (b-b)  Full (Pn) in plate.

    - At section (c-c)  2(Pn/4) = Pn/2 remains in plate.

     Limit State for Yielding in gross-section:

    Pn=Fy Ag... (AISC –EqnD2_1)

    Φt Pn = 0.9 Fy Ag... (Design tensile Strength by


    LRFD)

     Limit State for Rupture in net-section:

    Pn= Fu Aeff … (AISC – D2_2)

    Φt Pn = 0.75 Fu Aeff
    Ex 1

    LRFD:
     For yielding of G. Section:
    Ag = 5 x ½ = 2.5 in2
    ksi in²
    And Pn (nominal strength) = Ag Fy = 36 x 2.5 = 90 kips

    Then Φt Pn (design strength) = 0.9 x 90 = 81 kips

     For Rupture of N. Section:

    dh = db + ’’ = + = ’’
    in²
    Aeff = U (Anet) = 1 x 1.75 = 1.75 in2
    ksi in²
    And Pn = Fu Aeff = 58 x 1.75 = 101.5 Kips

    Then Φt Pn = 0.75 x 101.5 = 76.1 Kips governs

    ASD:

    = = 53.9 k …. (Yielding) corresponds to 0.6 FY

    And = = 50.8 k … (Fracture) corresponds to 0.5 Fu


    Ex 2

    Given:
    A-36 Steel …. L 3 ½ x 3 ½ x 3/8 …… 7/8’’ bolts
    Then dh = 7/8 + 1/8 = 1’’

    LRFD
     For yielding of G. Section:
    Ag = 2.5 in2 (from manual tables)

    Hence Φt Pn = 0.9 (36 x 2.5) = 0.9 x 90 = 81 kips


     For Rupture of N. Section:
    An = 2.5 – 1.0 (3/8’’ x 1’’) = 2.125 in2

    Where Φt Pn = 0.75 Fu Aeff


    In this example, assume u = 0.85 i.e Aeff = 0.85 An = 1.806 in2

    Hence Φt Pn = 0.75 x 58 x 1.806 = 78.5 kips governs

    IF PD = 35 kips and PL = 15 kips


    Combination 1: Pu = 1.4 PD = 1.4 x 35 = 49 kips

    Combination 2: Pu = 1.2 PD + 1.6 PL = 1.2x35 +1.6x15 = 66 kips

    (Note that when PL < 8 PD then comb 2 controls)

    Then Pu = 66 kips < Φt Pn = 78.5 kips OK

    ASD
    Pa = PD + PL = 35 + 15 = 50 kips

     Yield: ft = = = 20 ksi < Ft = 0.6 Fy = 21.6 ksi

     Fracture: ft = = = 27.7 ksi < Ft = 0.5 Fu = 29 ksi

    Hence the member is satisfactory


    Ex 3

    Given

    A 36 steel - 2L 5x3x5/16 - 4 bolts Φ1/2’’ (Assume Ae= 0.75 An)

    LRFD (For one angle):


    in²
     Yield: Pn=Fy Ag = 36 x 2.41 =86.76 kips

     Φt Pn = 78.1 kips

     Rupture: (An)1L =2.41 – 2 x [ x ( + ) ] = 2.0193 in2 then

    Ae=0.75x2.0193 = 1.514 in2

     Φt Pn =0.75 x 1.52 x 58 = 65.86 kips

    Therefore for 2Ls  Φt Pn= 2 x 66.15 = 131.72 kips

    Check: if Pu < Φt Pn = 131.72 kips

    ASD:

     Yield: = = 51.95 kips

     Fracture: = = 44.1 kips

    Then for 2Ls = 2x44.1 = 88.2 kips Check if Pa ≤ = 88.2 kips


    LRFD:

     Yield: Φt Pn= 0.9 x 36 x 10 = 324 kips governs

     Fracture: Φt Pn=0.75(58 x Ae)

    Where Ae= u x An (in this example assume u=1)

     find least An??


    in²
    - sec ABCD = 10 -1(0.4 x 7/8) – 1(0.65 x 7/8) = 9.08 in2 (Full Pn)
    - sec EFBGCD

    = 10 – 2(0.4 x 7/8 + 0.65 x 7/8) +( x 0.4) + 2( )( )

    = 8.8 in2 (Full Pn)


    - sec EFBCD:

    10 -2(0.65x 7/8) -1(0.4 x7/8) + ( )( ) =8.77 in2 governs

    Hence Φt Pn = 0.75(58x8.77) = 381.5 kips


    In Transition zone, “Shear Lag Phenomenon”:
    Use Aeffb= U Anet in Bolted Connections

    Aeff = U Agross in Welded Connections

    Where U = reduction factor accounts for no-uniform stress distribution

     U= 1 -

    where : distance from centroid of connected area to plane of


    connection.

    And L: distance between 1st and last bolt in the line = length of line
    with max number of bolts (if 2 lines)

     U  Ae and if L  U  Ae

    Note: if ‘< , connect long leg & make short leg unconnected.

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