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Flange Splice for Plate Girder - Welding and Bolt Connection

Learn about the proper method for joining flanges in plate girders through complete penetration butt welds and bolt connections. Ensure the strength and stability of the splice plates and bolts used.

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Flange Splice for Plate Girder - Welding and Bolt Connection

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  1. Chapter 11 Flange splice

  2. Introduction In shop, flanges shall be joined by complete penetration but welds, where possible as the web. These welds shall develop the full strength of the plates. The flange plate can be obtained in length up to 15 m and thus in plate girder over 15 m span we have to arrange field splice at which the web plate and the flange plates are cut. Flange joints should not be preferably located at points of maximum stresses. Where splice plates are used, their area shall be not less than 5 % in excess of the flange plate. There shall be enough bolts on each side of the splice to develop the maximum strength of the flange plate.

  3. One splice plate covers the joint of the flange or splice plates on both sides of the flange plate may be used. For example if we have flange plate 32020 mm we can use one splice plate 320(20 – 22) mm use bolts M22, holes =24 mm. Clause 6.2.2

  4. -   Net area of flange plate = (32 - 42.4)2 = 44.80 cm2 -         Number of bolts (for one side), (single shear bolts), M22, HSB GRADE (8.8) = m  m = qb = 0.25 Fub = 0.25x8 = 2 t/cm2 Rs. sh = qbxAs x n = 2x3.03x1= 6.06 t (n = No. of shear planes) y = 3.6 t/cm2 (St 52)  m= Use 20 bolts (4x5) (6.4.1)-P(100) P(101)

  5. Or we can use One splice plate 32012 mm from the outside and two splice plate 12012 mm from the inside of the flange plate Net area of splice plates     = (321.2 + 2121.2) - (221.22.4) = 67.20 – 11.52 = 55.68 cm2. Number of bolts (double shear or bearing bolts, M22(8.8) = n  n =

  6. qb = 0.25 Fub = 0.25 x 8 = 2 t/cm2 Rd. sh = qbx Asx n = 2 x 3.03 x 2 = 12.12 t (n = No. of shear planes) Rb = Fbx d x min t Fb = x Fu  depends on end distance (S), (Table 6.2) (6.4.2)-P(101)

  7. End distance in direction of force     1.2 1.0 0.8 0.6 e 2.5d 2 d 1.5d 3d  Code page (102) Edge distance = 2.5  = 1.0  Fb = 1.0 x Fu = 5.2 t/ cm2 Rb = 5.2 x 2.2 x 2.0 = 22.88 t Back

  8. y = 3.6 t/cm2 (St 52)  n = Use 8 bolts (24)

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