Design a deck type plate girder railway bridge for single tract

Design a deck type plate girder railway bridge for single tract B.G. main line loading for the following data:

1.      Effective span                          :           24 m.

2.      Spacing of plate girder             :           1.9 m c/c

3.      Weight of stock rails                :           440 N/m (90 lb/yard rails)

4.      Weight of guard rails                :           260 N/m

5.      Weight of fastenings etc           :           280 N/m of track

6.      Timber sleepers                       :           250 mm x 150 mm x 2.8 m @0.4 m c/c

7.      Density of timber                     :           7.4 KN/m³

Take permissible stresses as per Railway Steel Bridge Code.

Solution:

            Step 1 Computation of loads, B.M. and S.F. in Plate girder

            Weight of main rails per metre run of track                = 0.44 x 2        = 0.88 KN/m

            Weight of guard rails per metre run of track              = 0.26 x 2        = 0.52 KN/m

            Weight of fastening per metre run of track @ 0.28 KN/m                 = 0.25 KN/m

            Weight of sleeper per metre run of track =             = 1.94 KN/m

                                                                                                Total Load     = 3.62 KN/m

            Self weight of girder for B.G. main line loading is w = 0.78 L         = 18.72 KN/m

                        Total dead load per meter run of track           = 3.62 + 18.72 = 22.34 KN/m

                                                            Total Dead Load       = 22.34 x 24    = 536.2 KN

                                                            Dead load per girder   = 0.5 x 536.2   = 268.1 KN

            EUDL for B.M., for BG (revised) for loaded length L = span = 24 m is 2034 KN

                        CDA = 0.15 + (8/(6+L)) = 0.15 + (8/(6+24)) = 0.417

            Impact load (I.L.) for B.M., = 2034 x 0.417 = 848.2 KN

            Hence total load per track = 536.2 + 2034 + 848.2 = 3418.4

            Hence load per girder, for B.M. = 0.5 x 3418.4 = 1709.2 KN

                        M_max = (WL/8) = (1709.2 x 24) / 8 = 5127.6 KN/m

            This occurs at the mid span of each girder. B.M. at distance x from the mid span is

                        M_x = 5127.6 – (1709.2/24) x X²/2 = 5127.6 – 35.61X²

            Fro end sections, L = 24 m

            Hence EUDL for shear = 2231 KN

                                    CDA = 0.15 + (8/(6+24)) = 0.417

                        Impact load,                = 2231 x 0.417            = 930.3 KN

            Total (L.L + I.L)                     = 2231 + 930.3 = 3161.3 KN per track

            Hence, (L.L + I.L)                  = 0.5 x 3161.3 ≈ 1580.6 KN

            S.F. due to (L.L + I.L)                        = 0.5 x 1580.6≈           = 790.3KN

            Dead load per girder, W_d        = 268.1 KN

            Dead load shear per girder      = 0.5 x 268.1 = 134.1 KN

            Total S.F. at each end of a girder       = 790.3 + 134.1 = 924.4 KN

            For any section x from one end, L = l – x

            Let W_e be the total L.L. + I.L. one length L for each girder F_x = R_A = (W_e (l-x))/2l

            For section at 2 m from end, L = 24 – 2 = 22 m

            EUDL for shear = 2102 KN

            CDA = 0.436

            (L.L + I.L) = 2102 (1 + 0.436) = 3018.4 KN

            For both girders, (L.L + I.L) for each girder = W_e = 0.5 x 3018.4 = 1509.2 KN

            F₂ = (1509.2 (24 – 2)) / (2 x 24) = 691.7 KN

            Also, S.F. due to dead load = ((W_d)/2)-((W_d)/l)x = (268.1/2)-(268.1/24)x =134.05-11.17x

            At x = 2m, F = 134.05 – 111.17 x 2 = 111.7 KN

                        Total shear = 691.7 + 111.7 = 803.4 KN

Step 2. Design of central section of plate girder:

            M_max = 5127.6 KN-m.

            Economical depth d = 1.1 √(M / t_w σ_b) = 1.1. √(5127.6 / 10 x 138) = 2120 mm

            D = 1.1. √(5127.6E6 / 12 x 138) = 1936 mm

            Hence adopt d = 2000 mm and t_w = 12 mm.

            Area of web A_w = t_w x d = 12 x 2000 = 24000 mm²

            Area of flange A_f = (M/σ_bt x d) = (5127.6E6/138 x 2000) = 18578 mm²

            Net are of flange plates A_p = (2/3)A_f – (1/8)A_w = (2/3)x18578 – (1/8)x24000 = 9386 mm²

            Keep gross area of plates = 1.2 x 9386 = 11263 mm²

            Provide two plates 500 x 12 mm each having A_p = 2x500x12 = 12000 mm²

            Net area of flange angles A_a` = A_f /3 = 18578 / 3 = 6193 mm²

            Keep gross area of flange angles A_a = 1.25 x 6193 = 7740 mm²

            Gross area of each angle = 7740 / 2 = 3870 mm²

            Choose 2 ISA 200 x 150 x 12 mm @ 13.8 kg/m, each having A = 4056 mm²

Step 3. Design of riveted jointing connecting flange angles with the web:

            (L.L. + I.L.) per girder, per mm run = (0.5 x 224.6) / 1200 (1+1) = 0.187 KN/mm

            Dead load of track, per girder = 0.5 x 3.62 = 1.81E-3 KN/mm

            Total load w on compression flange = 0.187 + 0.00181 = 0.189 KN/mm

(a)   For compression flange:

P = 59.86 / (((924.4/2000) + (14112/14112+4000))² + 0.189²) = 147.2 mm

(b)   For tension flange:

P = ((59.86 x 2000)/924.4) x ((7080 + 3000)/7080) = 184.4 mm

                        Max. permissible pitch = 16 x 12 = 192 mm.

                        Hence provide uniform pitch of 180 mm throughout the length.

Step 4. Design of riveted joint connecting flange plates to flange angles:

(a)   Connection of flange plate to flange angle in compression flange, at ends

V = 924.4 KN and one plate is available in top flange.

                        A_f = 8112 + 1200/2 = 14112 mm²

                        Bearing = 21.5 x 12 x 232E-3 = 59.86 KN, R = 36.3 KN

                                    P = ((36.3 x 2000)/924.4) x ((14112 + 4000)/6000) = 237 mm

(b)   Connection of flange plate to flange angles in tension flange at 3.66 m ends:

A_f = 7080 + 10968 / 2 = 12564 mm²

                        (1/8)A_w = 3000 mm². V = 703 KN.

                        P = ((36.3 x 2000)/703) x ((12564 + 3000)/10968/2)) = 293 mm.

Step 6. Design of end stiffener:

            End reaction = 924.4 KN

            Permissible bearing stress = 185 N/mm²

            Bearing area required = (((3/4) x 924.4E3)/185) = 3748 mm²

            4t_a (150 – 10) = 3748

            Which gives t_a = 6.7 mm

            However provide 150 x 75 x 8 mm angles having A = 1742 mm²

            Actual length of stiffener = 2000 – 2 x 12 = 1976 mm

            Effective length of stiffener = (3/4) x 1976 = 1482 mm

            l/r = 1482/69.3 = 21.4

            Hence Load carrying capacity = 132.65 x 10568E-3 = 1402 KN.

Design of riveted connection for the end stiffener:

            Strength of 20mm dia. rivets in double shear = 2 x (Π/4) x 21.5² x 100E-3 = 72.6 KN

            Strength of rivets in bearing on web = 21.5 x 12 x 233E-3 = 59.68 KN

            No. of rivets required = 924.4/59.86 = 16

            Hence provide 8 rivets, in each row, thus making a total of 16 rivets.

Step 7. Design of intermediate stiffeners:

            V = 924.4 KN

            Shear stress, f_s = (924.4E3 / 2000 x 12) = 38.5 N/mm²

            Hence for d = 1700 mm, t_w = 12 mm and f_s = 38.5 N/mm²

            While for overall depth of 2438 mm, depth of 2048 mm.

            Try 2 ISA 125 x 75 x 8,

            Outstand face of web = 125 mm

            f_s = (682.6E3 / 2000 x 12) = 28.4 N/mm²

            Hence Max. spacing = 1900 mm.

            Maximum permissible spacing = 1830 mm as per railway bridge code.        

The Design procedure is completed.

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I am Thomas Britto here to share my experiences in the civil engineering field to all my readers.Today many students are struggling to buy books at high prices. So I decided to start a blog and share my experience and knowledge with all my readers.

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