Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 61 views

    Factored Reaction (W) Factored Tensile Force On The Plate W/2

    Uploaded by

    Sudhanshu Shekhar Singla
    1. The nominal shear capacity of a bolt is calculated as 450 kN. 2. The bearing capacity of a bolt is calculated as 112 kN. 3. Approximately 10 bolts are required based on the calculations.

    Copyright:

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

    Factored Reaction (W) Factored Tensile Force On The Plate W/2

    Uploaded by

    Sudhanshu Shekhar Singla
    61 views9 pages
    1. The nominal shear capacity of a bolt is calculated as 450 kN. 2. The bearing capacity of a bolt is calculated as 112 kN. 3. Approximately 10 bolts are required based on the calculations.

    Original Description:

    bolt

    Original Title

    Bolt

    Copyright

    © © All Rights Reserved

    Available Formats

    XLSX, PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    1. The nominal shear capacity of a bolt is calculated as 450 kN. 2. The bearing capacity of a bolt is calculated as 112 kN. 3. Approximately 10 bolts are required based on the calculations.

    Copyright:

    © All Rights Reserved
    Download as XLSX, PDF, TXT or read online from Scribd
    Download as xlsx, pdf, or txt
    61 views9 pages

    Factored Reaction (W) Factored Tensile Force On The Plate W/2

    Uploaded by

    Sudhanshu Shekhar Singla
    1. The nominal shear capacity of a bolt is calculated as 450 kN. 2. The bearing capacity of a bolt is calculated as 112 kN. 3. Approximately 10 bolts are required based on the calculations.

    Copyright:

    © All Rights Reserved
    Download as XLSX, PDF, TXT or read online from Scribd
    Download as xlsx, pdf, or txt
    You are on page 1of 9

    FACTORED REACTION(W) = 850 KN

    FACTORED TENSILE FORCE ON THE PLATE W/2 = 425 KN

    1 NOMINAL SHEAR CAPACITY OF BOLT :

    Vdsb = fub ( nn x Anb+ ns x Asb) x


    sqrt3 x ϒmb

    specification Grade /classifications yield stress fu , ULTIMATE


    fy , tensile stress
    N/mm 2
    N/mm2
    MIN. NOMINAL MIN. NOMINAL
    IS 1367 3.6 330 180 190 300
    (part 3) 4.6 400 240 240 400
    4.8 420 320 340 400
    5.6 500 300 300 500
    5.8 520 400 420 500
    6.8 600 480 480 600
    8.8 (D<16mm) 800 660
    9.8 900 720
    10.9 1040 940
    12.9 1220 1100

    NOMINAL DIAMETER OF BOLT


    BOLT GRADE
    ULTIMATE STRENGTH OF BOLT
    YIELDING STRENGTHOF BOLT
    ULTIMATE STRENGTH OF PLATE
    YIELDING STRENGTHOF PLATE
    NO. OF SHEAR PLANE WITH THREAD INTERCEPTING SHEAR PLANE
    NO. OF SHEAR PLANE WITHOUT THREAD INTERCEPTING SHEAR PLANE
    NOMNAL PLANE SHANK AREA OF BOLT = (πd2)/4
    NET SHEAR AREA OF BOLT HREAD = 0.78 x(πd2)/4
    NUM. OF SHEAR PLANE
    Vdsb = 45.272432 KN

    2 BEARING CAPACITY OF BOLT :

    Vdpb = 2.5 x Kb x d x t
    ϒmb

    DIAMETER d = 20
    GROSS AREA do = 22
    SUM OF CONNECTED PLATES THICKNESS t = 11.6
    EDGE DISTANCE e = 40
    PITCH DISTANCE p = 80

    i) e/3do = 0.606061
    ii) p/3do - 0.25 = 0.962121
    iii) fub/fu = 0.97561
    iv) = 1
    kb is minimum of above BOLT VALUE =
    kb = 0.60606061 NO OF BOLT =
    SAY =
    Vdpb = 112.4848 KN

    3 TENSILE STRENGTH :

    Tdb= Tnb/ϒmb
    Tnb= 0.9*fu*An < fyb*Asb*(ϒmb/ϒm0)
    An = (b-ndo+∑p si/4gi)t
    2

    NET EFFECTIVE AREA An =


    Tdb =
    =

    6 APPROX. NO. OF BOLT n = SQRT(6M/n1pVsd)


    ECCENTRIICITY e = 500 mm
    MOMENT OF CONNECTION DUE TO FACTORED LOAD M = 212500000 N
    NO. OF VERTICAL ROWS OF BOLTS (ASSUMED) n1 = 4
    PITCH OF VERTICAL BOLTS p = 80
    DESIGN STRENGTH OF BOLT Vsd = 45272.432
    = 9.3813036
    SAY = 10 (PER ROW)
    7 ECCENTRIC BOLTED CONNETIONS

    i) RESISTANCE AGAINST TRANSLATION = P/n

    FACTORED LOAD = 425 KN


    NO. OF BOLTS ON ONE BRACKET = 40
    P/n = 10.625 KN
    ii) WITH RESPECT TO CENTROID

    HORIZONTAL DISTANCE OF PARTICULAR BOLT TO CENTROID = 90


    VERTICAL DISTANCE OF PARTICULAR BOLT TO CENTRIOD = 360

    ∑x2+∑y2 = 4 mm2

    K= P*e
    ∑x +∑y2
    2

    = 53.125 KN/mm
    SAY = 53.5 KN/mm
    RESISTING FORCE AGAINST ROTATION
    Sa = K*ra
    = 53.5 * ra

    TOTAL VERTICAL COMPONET ON BOLT V= P + K*ra sinϴ


    n
    = 15.44 KN

    TOTAL HORIZONTAL COMPONET ON BOLT H = Sa * cosϴ


    = 19.26 KN

    RESULTANT = 24.68483745 N

    CHECK FOR STRENGTH :

    DEPTH OF PLATE h=
    NEUTRAL AXIS h/7 =
    TOTAL TENSION EXERTED BY BOLTS ∑y =
    ∑y2 =

    K = P * e
    ∑y2 + (2h/21) * ∑y

    = 106250
    (CLAUSE.10.3.3 IS 800:2007)
    PAGE 75
    REDUCTION FACTOR

    Properties
    Elongation
    percentage (min)

    24
    22
    14
    20
    10
    80
    12
    10
    9
    8

    = UNITS
    d = 20 mm
    = 4.6
    fub = 400 N/mm2
    fyb = 240 N/mm2
    fu = 410 N/mm2
    fy = 250 N/mm2
    nn = 1
    ns = 0
    Asb = 314.1593 mm2
    Anb = 245.0442 mm2
    = 1

    (CLAUSE.10.3.4 IS 800:2007)
    PAGE 75

    mm
    mm
    mm
    mm
    mm

    45.27243 KN
    9.387611
    9.5

    (CLAUSE.10.3.4 IS 800:2007)
    PAGE 76
    (CLAUSE.6.3.1 IS 800:2007)
    PAGE 32

    245.0442
    224.64 < 85679.8
    SATISFIED
    DESIGN STRENGTH OF YIELDING OF GROSS SECTION (PLATE)

    Tdg = (Ag*fy/ϒm0)

    WIDTH OF PLATE b = 200 mm


    THICKNESS OF PLATE t = 8 mm
    AREA OF PLATE Ag = 1600 mm2
    Yield Stress Of Material fy = 250 N/mm2
    Partial Safety Factor ϒmo = 1.1

    Tdg = 363636.4 KN

    RUPTURE OF CRITICAL SECTION

    Tdn = (0.9*fu*An)/ϒml

    Yield Stress Of Material fy = 250 N/mm2


    NET EFFECTIVE AREA An = (b-ndo+∑p2si/4gi)t

    Threaded Rods :
    Tdn = (0.9*fu*An)/ϒml
    Net Root Area At Threaded Section An

    Single Angles :

    Tdn = (0.9*fu*An)/ϒml + (ß*Ago*fy)/ϒm0

    ß = 1.4 - 0.076* w*fy*bs <= 0.9*fu*ϒm0 =>0.7


    t*fu*Lc fy*ϒml

    Outstand leg Width w =


    Shear Leg Width bs =
    Length Of End Connection,that is Distance
    between the outer most bolts in the end
    joint measured along the load direction
    or the length of weld along the load direction
    Lc =

    Tdn = alpha*(0.9*fu*An)/ϒml

    No. Of bots alpha


    1-2 = 0.6
    3= 0.7
    >=4 = 0.8
    alpha = 0.8
    Net Area of Cross Section An =
    Net Area of Conneted Leg Anc =
    Gross Area Of Outstanding Leg Ago =
    Thickness Of Leg t =
    Tdn =

    Channel or I-section connected through Flanges

    An = Gross Area - Ar. Of Bolt Holes - Half Area Of Web

    Block Shear Faliure

    Tdb = smaller of

    = Avg*fy + 0.9*Atn*fu (Against Shear Yield And tensile fracture)


    sq(3)*ϒm0 ϒml

    = 0.9*Avn*fu + Avg*fy
    sq(3)* ϒml ϒm0

    Minimum Gross and Net Area in Shear Along bolt


    line Parallel to external force respectively Avg,Avn =

    Minimum Gross And Net Area in Tension From


    The bolt Hole to the toe of the angle end bolt
    line perpendicular to the line of force resp. Atg,Atn =

    ultimate and yield stress of material resp. fu.fy =

    NOTE:-
    block shear faliure can be prevented by increasing distance between bolts

    Design of Tension Members :


    1.)Gross sectional Area Required Ag = T * ϒm0
    fy

    Factored load T = kN
    fy =
    ϒm0 = 1.1

    Effective Net Area


    Anc = k1*k2*k3*k4*An

    Net Area Of Section Determined - ar. Of holes An =


    Ductility Factor k1 =
    Hole Forming Factor k2 =
    drilled holes 1
    puched 0.85
    Geometry Factor k3 =
    Shear leg Factor k4 =

    Anc = k4*An

    slenderness ratio

    Least radius of gyration r = sqrt(I/A) = t/sqrt(12)

    Slenderness ratio l/r =

    You might also like