Structural Analysis and Design of Welded Plate Girder and Span Calculation
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Added on  2023/06/08
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This article discusses the structural analysis and design of welded plate girder and span calculation. It covers the calculation of moments, shear forces, and stresses for different loads and provides guidelines for designing the girder. The article also includes solved assignments, essays, and dissertations on this topic available at Desklib.
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Solution Sway mechanism H(θh1) = Mpcθ+ Mpcθh1 h2 At BAt BAt C H =Mpcθ+Mpcθh1 h2 θh1 θh1 BCD θ θ A θh1 Beam mechanism VθL1= Mpbθ+ Mpb(θ+θL1 L2)+ MpbθL1 L2 At BAt CAt D VθL1= 2Mpbθ+ MpbθL1 L2+ MpcθL1 L2 V =2Mpbθ+MpbθL1 L2+MpcθL1 L2 θL1 θL1 θθL1 L2
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Thus the moment at E, from a free body diagram of ABC ∑MAbout C = 0 2VA+ 5HA– Mc= 0 Mc= 240KN.m Since there is a plastic hinged at C of value MpC= 240 KNm, we have the equilibrium H = 256 KN V = 300 KN Mpc= 120 KN Mpb= 240 KN L1 = 2, L2 = 3, h2 = 3, H1= 5 M ∑MAbout C = 0 2*300 + 5*256 – Mc0 Mc= 1880 KNM ∑MAbout C = 0 3HE– 120 = 0 HE= 40 KN Moment at B 256 *2.5 = 640 KNm Moment at D 1880 – 300 *3 = 980 KNm BMD 640 KNm120 KNm BCD
640 KNm A1880 KNm Part B Span of the welded girder = 15 m It carries a load of approximately 500 KN each The loads are at a distance of 4 m from each end of the girder The girder carries uniform distributed load of 30 KN/m, which include the self-weight of girder Designing the girder 500 KN30 kn/m500 KN ACEDB 5 m5 m5 m ∑MB= 0 RA* 15 = 500*10 + 500*5 + 20*15 * 15/2 = 650 KN RB= 500 + 500 + 30 * 15 – 650 = 800 KN Mc= 650 * 5 – 30 * 5 * 2.5 = 2875 KNm MB= 650 * 7.5 – 500 * 2.5 – 30 *7.5 * 7.5/2 = 2781.25 KNm The overall depth of girder = span/10 = 1500 mm Assume that the girder has 30 mm thick therefore the allowance Pb= 140 N/mm2 If we assume that the flange plate resist the bending moment then the moment will be resisted by lever arm of about 1460 mm Flange area = (2781.25 *106)/(1460 *140) = 1.36 * 104mm2
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If we assume the flange that was used is of dimensions 150 mm by 100 mm Assume the allowance shear stress is 100 N/mm2 Depth of web plate is 1300 mm and shear 650 KN at girder support Thickness of web plate = (650 *103)/(1300 *100) = 4.64 mm If a thickness of 20 mm was used Checking the bending stress IXX= (2 * 100 * 150 * 7002) + (20 *13003)/12 = 1.84 *1010mm4 fbc= (2781.25 *106*650)/ (1.84 * 1010) = 98.25 N/mm2(it is safe) load bearing stress are required at the support under the point loads The spacing should not exceed 1.5d = 11950 mm 180t = 180 *20 = 3600 Stiffeners under 500 KN Assuming the stiffeners each 100 mm by 30 mm Bearing stress = (500 * 103)/(2*135 * 30) = 61.72 (safe ) The area of centerline of web IXX= (30 * 3003)/12 = 6.75 * 107mm4 A = (300 * 20) + (100 * 30*2) = 12, 000 mm2 rXX=√6.75∗107 12000= 75 mm l/rXX= (0.7 *1300)/75 = 12.1 Pc= 149.95 N/mm2 Fc= (650 *103)/(12000) = 54.17 N/mm2 A smaller size stiffener was used End plate Assume the end plate of 20 mm thick was used
Checking the bearing Bearing stress = (650 * 103)/(450 *10) = 144.4 (safe ) Checking bthe section acting at the strut A = 450 * 10 + (200 *10) = 6500 mm2 IXX= (10 * 4503)/12 = 7.6 * 107mm4 rXX=√7.6∗107 6500= 108.1 mm l/rXX= (0.7 *1300)/108.1 = 8.4 Pc= 150.8 N/mm2 Fc= (650 *103)/(6500) = 100 N/mm2 Assume the weld stretch = 350 N/mm Approximately the welded length = (650 *103)/350 = 1875 mm Part B (ii) Minimum yield strength at nominal thickness 16 mm For steel S275 = 275 N/mm2 Tensile strength between 3 mm and 16 mm = 370 to 530 Mpa Dead load s.w = 20 KN point load w1d = 200 KN, w2d = 200 Kn imposed load udl = 40 KN/m point load, w1d = 300 KN, w2d = 300 KN 720 KN40 kn/m720 KN BEKGJ
9 m7 m9 m Total weight = 1.2 D.L + 1.6 L.L Load w1d = 1.2*200 + 1.6*300 = 720 KN Load w2d = 1.2 * 200 + 1.6* 300 = 720 KN ∑MJ= 0 RB* 25 = 720*16 + 720*9 + 40*25 * 25/2 = 1220 KN RJ= 720 + 720 + 40 * 25 – 1220 = 1220 KN ME= 1220 * 9 – 40 * 9 * 4.5 = 9360 KNm MK= 1220 * 12.5 – 720 * 3.5 – 40 *12.5 * 12.5/2 = 9605 KNm Girder section The overall depth of girder = span/10 = 25000/10 = 2500 mm Take the cover to be 40 mm thick therefore the allowance stress bending Pb= 275N/mm2 If we assume that the flange plate resist the bending moment then the moment will be resisted by lever arm of about 2460 mm Flange area = (9605 *106)/(2460 *275) = 1.42 * 104mm2 If we assume the flange that was used is of dimensions 300 mm by 50 mm Depth of web plate is 2400 mm and shears 1220 KN at girder support Thickness of web plate = (1220 *103)/(2400 *100) = 5.08mm If a thickness of 10 mm was used Checking the bending stress IXX= (2 * 50 * 300 * 12252) + (10 *24003)/12 = 1.156 *1010mm4 fbc= (9605 *106*1200)/ (1.156 * 1010) = 997 N/mm2(it is safe) for the web the ratio d/t = 2400/10 = 240 the for the intermediate stiffener must be provided
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Load bearing stress is required at the support under the point loads The spacing should not exceed 1.5d = 3600 mm 180t = 180 *10 = 1800 From the table of allowable shear stress assume 100 N/mm2 Stiffeners under 720 KN Try two stiffeners each 200 mm by 20 mm Bearing stress = (720 * 103)/(2*175 * 20) = 103 N/mm2(safe ) The area of centerline of web IXX= (20 * 3303)/12 = 6.0 * 107mm4 A = (500 * 10) + (200 * 20*2) = 13, 000 mm2 rXX=√6.0∗107 13000= 67.94 mm l/rXX= (0.7 *2400)/67.94 = 24.7 Pc= 142.4 N/mm2 fc= (720 *103)/(13000) = 55.38 N/mm2 it is advisable to use smaller size stiffener take weld strength approximately to 450 N/mm length = (720*103)/450 = 1600 mm End plate Assume the end plate of 20 mm thick was used Checking the bearing
Bearing stress = (1220 * 103)/(500 *20) = 22 N/mm2 Maximum stiffener = 11t = 11 * 20 = 220 mm (safe) Checking the section acting at the strut A = 500 * 20 + (1900 *10) = 11.9 * 103mm2 IXX= (20 * 5003)/12 = 20/83 * 107mm4 rXX=√20.83∗107 11900= 132.3 mm l/rXX= (0.7 *2400)/132.3 = 12.7 Pc= 147.2 N/mm2 Fc= (1220*103)/ (11900) = 102.5 N/mm2 Try the weld strength = 500 N/mm Approximately the welded length = (1220 *103)/500 = 2440mm Intermediate stiffeners Use stiffeners of dimensions of each 100 mm by 10 mm Maximum value = 10t = 10 *20 = 200 Moment of inertia = I = (10 *2103)/12 = 7.73 * 105mm4 Distance between stiffeners = 1800 mm Required thickness of web = 1300/ 180 = 7.22 mm For shear strength, t = 5.08 I = (1.5 *12003*7.353)/(104*18002) = 3.2 *106mm Web to flange web Horizontal shear per weld = (1220 *103*300 *50 *625)/(1.156 *1010*2) = 495 N/mm ( hence ok)
Part B(iii) Design consideration 1.0Considering the types plate girder Plate girder construction involves use of welded steel plate which together will form an I section 2.0stresses and loads application of loads to the plate girder are through stanchions, floor slab and floor beams, which will be carried by the girder
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the flange of the plate will carry the bending moments while the web will have a resistance to the shear force. To avoid plate from failing and working effectively the rules from BS449 will be used to govern the with or thickness ratios of the plates to ensure no buckling takes place, the rules also will govern the positions of both the intermediate and stiffeners. 3.0plate girder depth the depth should always be about 1/10 of the span for average loading. The lightly loaded girders the depth will be between 1/15 to 1/20 The flange depth is always about 1/3 of the depth Web buckling are prevented by provision of stiffeners 4.0plate girders deflections Clause 15 of BS449 will be used determining the deflection requirement 5.0permissible stresses if the plate thickness will exceed 40 mm a lower stress must be used the permissible bending stress to be used are provided in BS 449 6.0bending stresses Clause 17, 27 and 32 of BS 449 sets section area for the girders Properties for plate girders that includes moment of inertia, area, modulus of section and radii of gyration are calculated from the first principle. Maximum outstand for flange plate are always provided in BS 449, like for example Compression flange = 16t Tension flange = 20t t is the thickness 7.0shear stress of the web will be determined by the formula fq=shear depthofwebplate∗thicknessoftheweb Allowable shear stress will depenf on the value of d/t and stiffeners spacing If the ratio of d/t exceed 85 vertical stiffeners will be required at a distance which will not exceed 1.5d The thickness should be more than 1/180