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    KKKH3353 - Structural Steel Design - Tension Members

    Uploaded by

    Zeyad Tareq Al Sarori
    This document discusses tension members in structural steel design. It defines tension members as members subjected to axial tensile forces and covers various types including members with and without holes. Key points covered include: - Calculating the design tensile resistance based on yielding of the gross section or fracture of the net section, whichever is lower. - Determining the net section area by deducting hole areas in staggered or non-staggered fastener arrangements. - Examples calculating tensile strength for a flat bar and angle member connected by bolts are provided to illustrate the design process.

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    KKKH3353 - Structural Steel Design - Tension Members

    Uploaded by

    Zeyad Tareq Al Sarori
    123 views21 pages
    This document discusses tension members in structural steel design. It defines tension members as members subjected to axial tensile forces and covers various types including members with and without holes. Key points covered include: - Calculating the design tensile resistance based on yielding of the gross section or fracture of the net section, whichever is lower. - Determining the net section area by deducting hole areas in staggered or non-staggered fastener arrangements. - Examples calculating tensile strength for a flat bar and angle member connected by bolts are provided to illustrate the design process.

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    KKKH3353 - Structural Steel Design - Tension members

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    This document discusses tension members in structural steel design. It defines tension members as members subjected to axial tensile forces and covers various types including members with and without holes. Key points covered include: - Calculating the design tensile resistance based on yielding of the gross section or fracture of the net section, whichever is lower. - Determining the net section area by deducting hole areas in staggered or non-staggered fastener arrangements. - Examples calculating tensile strength for a flat bar and angle member connected by bolts are provided to illustrate the design process.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    123 views21 pages

    KKKH3353 - Structural Steel Design - Tension Members

    Uploaded by

    Zeyad Tareq Al Sarori
    This document discusses tension members in structural steel design. It defines tension members as members subjected to axial tensile forces and covers various types including members with and without holes. Key points covered include: - Calculating the design tensile resistance based on yielding of the gross section or fracture of the net section, whichever is lower. - Determining the net section area by deducting hole areas in staggered or non-staggered fastener arrangements. - Examples calculating tensile strength for a flat bar and angle member connected by bolts are provided to illustrate the design process.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 21

    Faculty of Engineering and Built Environment

    Department of Civil Engineering

    KKKH3353
    STRUCTURAL STEEL DESIGN
    (Tension Member)
    Dr. Ahmed Wadood Al Zand
    PM Dr. Shahrizan Bin Baharom
    Example of Tension Members
    Plate and Angle in Tension with Bolts
    Typical Truss and Frame Connections
    Rupture of Net Section
    Block Shear Tear-Out
    Block Shear Tear-Out
    Design of Tension Members

     Axially loaded tension member (concentrically tensile force)


     Member without holes
     Member with holes

     Tension member with moments; Eccentrically tensile force-


    connected unsymmetrically or member itself unsymmetrical such
    as angles, tees and channels.
     Direct tensile failure and/or block tearing
    Axially Loaded
    Tension Members
    Eurocode 3 states that tensile resistance
    should be verified as follows:

    Nt,Ed ≤ Nt,Rd Tension check

    Nt,Ed is the tensile design effect


    Nt,Rd is the design tensile resistance
    Design Tensile Resistance, Nt,Rd

    Design tensile resistance Nt,Rd is limited either by:

    • Yielding of the gross cross-section (Npl,Rd)


    Or
    •Ultimate fracture failure (Nu,Rd) of the net cross-section (at holes for fasteners)

    whichever is the lesser.


    Yielding of Gross Cross-Section, Npl,Rd
    The Eurocode 3 design expression for yielding of the gross cross-section
    (plastic resistance) given as:

    Npl,Rd= A.fy / γM0

    This criterion is applied to prevent excessive deformation of the member.

    A is the gross cross-sectional area


    fy is the yield tensile strength of plate/member
    γM0 is the partial factor which equal to 1.0
    Ultimate resistance of net section, Nu,Rd
    And for the ultimate resistance of the net cross-section (defined in clause
    6.2.2.2), the Eurocode 3 design expression is:

    Nu,Rd= 0.9.Anet.fu / γM2

    Anet is the net cross-sectional area to account (excluding the bolts’ holes)
    fu is the ultimate tensile strength of plate/member
    γM2 is the partial factor which equal to 1.25
    Non-Staggered fasteners
    For a non-staggered arrangement of fasteners, the total area to be deducted
    should be taken as the sum of the sectional areas of the holes on any line (A-A)
    perpendicular to the member axis that passes through the centreline of the holes.

    p2
    b

    Anet = A – n.d0.t
    p1 p1
    A is the gross cross-sectional area
    Anet is the net cross-sectional area to account (excluding the bolt holes)
    n is the number of bolt holes (bolts in the line of cutting)
    d0 is the diameter of bolt holes
    t is the material thickness
    Staggered fasteners
    For a staggered arrangement of fasteners, the total area to be deducted should
    be taken as the greater of the maximum sum of the sectional areas of the holes on
    any line (A-A & B-B) perpendicular to the member axis

    Flat member (plate)


    Anet = t.(b – n.do+ Ʃ(0.25.p12/p2)) p2
    b
    Any member (Angle, C-section…etc)
    Anet = A -t.(n.do+ Ʃ(0.25.p12/p2))
    p1 p1
    p1 is the staggered pitch of two consecutive holes
    p2 is the spacing of the centres of the same two holes measured perpendicular
    to the member axis
    Simple Tension Member

    Angle connected by a single row of bolts


    Single angles in tension connected by a single row of bolts through one leg,
    may be treated as concentrically loaded, but with an effective net section, to
    give the design ultimate tensile resistance as below.

    β2 and β3 are reduction factors dependent upon the bolt spacing (pitch) p1.
    Anet is the net area of the angle.
    Angles connected by a single row of bolts

    0.7

    0.4

    p1 ≤2.5do p1≥5.0do
    2.5do>p1 <5.0do
    Example 1
    A flat bar 200 mm wide x 25 mm thick is to be used as a tie (tension member).
    Erection conditions require that the bar be constructed together with a lap slice
    using six M20 bolts. Calculate the tensile strength of the bar assuming grade
    S275 steel (fy = 275 Mpa, fu= 410 MPa)
    .
    Solution
    Npl,Rd= A.fy / γM0 =(25x200)x275/1.0 = 1375000 N = 1375 kN

    n =2, d= 20 mm, d0 = 22mm, t= 25 mm

    Anet = A – n.d0.t = 25x200 – 2x22x25 = 3900 mm2

    Nu,Rd= 0.9.Anet.fu / γM2 = 0.9x3900x410/1.25 = 1151280 N = 1151.28 kN

    Then Nt,Rd is the smallest value of Nu,Rd & Npl,Rd =1151.28 kN

    The applied factored tension force T (Nt,Ed ) should be less or equal to Nt,Rd
    Example 2
    Bolted connection Try a 150×75×10 unequal angle, bolted (with a line of four 22
    mm HSFG bolts, at 125 mm centres) through the longer leg. Calculate the tensile
    strength of the bar assuming grade S275 steel (fy = 275 Mpa, fu= 410 MPa)
    Solution
    A =2170 mm2, d= 22 mm, d0 = 24mm, p1= 125 mm, t= 10 mm

    Npl,Rd= A.fy / γM0 =(2170)x275/1.0 = 596750 N = 596.7 kN

    5xdo =5x24 = 120 mm


    p1= 125 mm > 5xdo , then β3=0.7 (table)

    Anet = A – n.d0.t = 2170 – 1x24x10 = 1930 mm2

    Nu,Rd= β3.Anet.fu / γM2 = 0.7x1930x410/1.25 = 443128 N = 443.1 kN

    Then Nt,Rd is the smallest value of Nu,Rd & Npl,Rd =443.1 kN

    The applied factored tension force T (Nt,Ed ) should be less or equal to Nt,Rd (443.1 kN)

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