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    Plate Girder Design

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    kalpanaadhi
    This document summarizes the design of a plate girder railway bridge using limit state design principles. Key points include: - The plate girder was modeled in STAAD and designed to support a total factored load of 1.5(2918.5 + 961.2) kN. - The optimum depth of the web was calculated as 2000 mm with a thickness of 60 mm. - Bending and shear capacities were calculated using plastic and slender section assumptions. The calculated bending capacity of 9339.07 kN was greater than the design moment of 9093.75 kN. - Shear capacity was calculated using both simple post-critical and tension field methods, with

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    Attribution Non-Commercial (BY-NC)
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    Plate Girder Design

    Uploaded by

    kalpanaadhi
    639 views8 pages
    This document summarizes the design of a plate girder railway bridge using limit state design principles. Key points include: - The plate girder was modeled in STAAD and designed to support a total factored load of 1.5(2918.5 + 961.2) kN. - The optimum depth of the web was calculated as 2000 mm with a thickness of 60 mm. - Bending and shear capacities were calculated using plastic and slender section assumptions. The calculated bending capacity of 9339.07 kN was greater than the design moment of 9093.75 kN. - Shear capacity was calculated using both simple post-critical and tension field methods, with

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    Plate Girder Design

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    This document summarizes the design of a plate girder railway bridge using limit state design principles. Key points include: - The plate girder was modeled in STAAD and designed to support a total factored load of 1.5(2918.5 + 961.2) kN. - The optimum depth of the web was calculated as 2000 mm with a thickness of 60 mm. - Bending and shear capacities were calculated using plastic and slender section assumptions. The calculated bending capacity of 9339.07 kN was greater than the design moment of 9093.75 kN. - Shear capacity was calculated using both simple post-critical and tension field methods, with

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    639 views8 pages

    Plate Girder Design

    Uploaded by

    kalpanaadhi
    This document summarizes the design of a plate girder railway bridge using limit state design principles. Key points include: - The plate girder was modeled in STAAD and designed to support a total factored load of 1.5(2918.5 + 961.2) kN. - The optimum depth of the web was calculated as 2000 mm with a thickness of 60 mm. - Bending and shear capacities were calculated using plastic and slender section assumptions. The calculated bending capacity of 9339.07 kN was greater than the design moment of 9093.75 kN. - Shear capacity was calculated using both simple post-critical and tension field methods, with

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    IIT MADRAS

    LIMIT STATE DESIGN OF PLATE GIRDER RAILWAY BRIDGE

    CE3320: DESIGN OF STEEL STRUCTURAL SYSTEMS

    SUBMITTED BY: SIVAPRASAD K V (CE08B038) SREENATH PALERI (CE08B071)

    STAAD

    MODEL

    Plate section.

    girder

    Plate girders with Lateral bracings

    Design of Elements

    Design of Plate Girder

    References

    Total dead load = 22.5 kN/ m Total Load = 2918.5

    Live load for bending, (for span of 15 m) = 2356 kN Impact load = 961.2

    Total factored load = 1.5 (2918.5 + 961.2 ) Maximum bending moment in our plate girder. = X w l /w 8 Bending moment = 9093.75- 5820/(2 X 25 X x2/ 2) Assume thickness of the (web = 12 mm) = 9093.76 58.2 x2 = 5819.55 kN

    = 9093.75 kNm

    Therefore optimum depth d = (MK/f y.t.w )

    So optimum depth to carry the moment = (9094X 106)/(250 X 12) Adopt the depth of the web as 2000 mm Net area of tension flange:= 1741.1 mm

    Cl. 11.2

    Assume the flange width as h/0.3 ie, 0.3 times the web depth = 600 mm Z p = M d X m0 /f y = 3.676 X 107 mm3 So required thickness = 50 mm

    M/(F au X Lever arm) = (9094 X 106 )/(150 X 2000) = 30313.33 mm2

    Z p of the section considers that flanges only take the bending moment = 6.15 X 107 mm3 > Z p req

    Calculation of Moment Capacity


    Effective length = 0.7 X 25 = 17.5 m f = 1 b = (600 -12) / 2 = 294 mm b/t f = 294 / 50 = 5.98 < 9.4, plastic section d/t w = 2000/12 = 166 > 15.7 , slender The section is classified as slender. The flange is plastic and the web is slender. Therefore the web cannot be loaded to plastic moment Iz = 2 X [
    6000503 12

    8.2.2 IS 800 2007

    section

    capacity. Hence it is disregarded in bending strength calculations. + 500 50 1025^2]

    I y = 2 X 50 X 6003 / 12 Z p = 2 X (600 X 50 X 1025 ) = 6.15 X 107 mm3 Torsion constant I t = b i f i 3/ 3 = 5.1152 X 107 mm4 I w = (1 - f ) f I y h y 2 f = 0.5 h y = (d - t f ) = 2000 = 1.8 X 1015 mm6 I w = (1 0.5 ) X 0.5 X 1.8 X 109 X 20002 E = 2 X 105 N/mm2 G = 0.769 X 105 MPa
    4

    = 6.305 X 1010 mm4 = 1.8 X 10 9 mm4

    Z e = I z / (2100/2) = I z / 1050 = 6 X 107 mm3

    = 200 X 123 / 3 + 2 X 600 X 503/ 3

    2 105 1.8 107 2 2 10 1.810 [0.769 105 5.1152 107 + ] M cr = 2 (17.5 103 )2 (17.5 103 )^2

    15

    LT = ( p Z p f y / M cr ) = 1.07 > 0.4

    = 1.3423 X 1010 kNm

    8.2.2.1 IS 800 2007

    = (1 X 6.15 X 107 X 250) / 13423

    Lateral buckling governs. LT = 0.49

    LT = 1 / [ Lt +( Lt 2 - LT 2)]0.5 1 Lt = 0.5[1 + LT ( LT 0.2 ) + LT 2 Lt = 1.2886 LT = 0.56 < 1

    f bd = LT f y / m0 M d = b Z p f bd

    Design bending strength = 782.7.105 kNm < 9093.75 kNm = 6.15 X 107 X 127.27

    = 127.27 MPa

    Hence increase the flange thickness to 60mm Z p = 2 X 600 X 60 X 1025 M d ~ 9339.07 kN > 9093 kN Hence take 60mm as the final thickness of the web.
    R

    = 7.38 X 107 mm3

    Calculation Of Shear Capacity


    Resistance to shear buckling:d/t w > 67 for web without stiffeners d/t w > 67 ( K v /5.35) for web without stiffeners. Assume the stiffeners are provided at 2m intervals. c/d = 2000/2000 = 1 so K v = 5.35 + 4/ (c/d)2 = 7.35
    5

    8.4.2.1 IS 800 2007

    67 ( K v /5.35) = 67 ( 7.35/5.35) = 78

    d/t w = 2000 / 12 = 166 > 78 hence shear buckling strength governs the design.

    a.) Simple post Critical Method


    Shear strength (normal) V n = m0 = A v X b a) b = f yw / 3 ; for w = 0.8 c) b = f yw / 3 ; w > 1.2 b is the shear strength corresponding to web buckling b) b = [1 0.8 ( w 0.8)] f yw / 3 ; 0.8 < w 1.2 w = (f y /(3 cr,w )) cr = (K v 2 E) / (12 (1 - 2)(d/t w )2) For the assumed section = 47.83 Mpa cr = (7.35 X 2 X 2 X 105)/ 12 (1 - 0.32 )(166.6)2 = 1.73 > 1.12 8.4.2.2 (a) IS 800 2007

    w = 250/(3 X 4783) b = 250 / 3 w 2 = 48.28 Mpa V cr = A v X b = t wd X b = 250 / 3 X 1.732

    V d = 1171.2/ 1.1

    = 2000 X 12 X b = 1171.2 kN = 1064.7 kN

    b.) Tension Field Method


    The tension field method, based on the post shear buckling strength, 8.4.2.2 (b) may be used for webs with intermediate transverse stiffeners, in IS 800 2007 addition to the transverse stiffeners at supports, provided the panel under tension fielda action, or the end posts peovide anchorage for the tension fields and if c/d 1.0, where c, d are the spacing of the

    transverse stiffeners and depth of the web respectievley. V n = V tf Where, Where,

    In the tension field method, the nominal shear resistance, V n , is given by V tf = [A v b + 0.9 w tf t w f v sin] V p b = buckling strength f v = yield strength of the tension field obtained from = 1.5 b sin2 = 62.71 = [ f yw 2 - 3 b 2 + 2]0.5 = 1.5 X 48.28 X sin600 = tan-1(d/c)/1.5 = 48.26 MPa

    = inclination of the tension field = tan-1(1)/1.5 f v = 181.1 MPa = 300

    w tf = the width of the tension field, given by: f yw = yield stress of the web d = depth of the web = dcos (c s c s t )sin

    c = spacing of stiffeners in the web b = shear stress corresponding to buckling of the web. tension flange respectievley, as obtained from: s = (2/sin) [M fr / f yw t w ]0.5 c
    7

    s c , s t = anchorage lengths of tension field along the compression and

    where

    M fr = 0.25 b f t f 2 f yf [ 1 {N f / (b f t f f yf / mo )}2] N f is the axial force in the flange. Assume an anchorage of 40% = (1171.2 + 782.35) = 1953.55 kN Hence w tf is found to be 800 mm.

    V tf = [A v b + 0.9 X 800 X 12 X 181.1 X sin300] So the design shear strength = V tf / 1.1 = 1775.95 kN It can be seen that the strength has increased by 66% from that of a post buckling shear strength design.

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