Design of Plate Girder and Analysis of Structural Mechanisms
Verified
Added on 2023/06/08
|13
|2217
|251
AI Summary
This article discusses the design considerations for plate girders, including depth and flange design. It also covers the analysis of structural mechanisms, including sway and beam mechanisms, and provides solved assignments, essays, and dissertations on the topic. Course codes, names, and universities are not mentioned.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
Solution Q1a) Sway mechanism H(θh1) = Mpcθ+ Mpcθh1 h2 At BAt BAt C H =Mpcθ+Mpcθh1 h2 θh1 θh1 BCD θ θ A θh1 Q1b) 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
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
θθL1 L2 iii)Equilibrium DMpc EHE ∑Mabout D = 0 3HE– Mpc= 0 HE=Mpc 3=120 3=40KN ∑Fx= 0 ∑Mabout E = 0 -2V + 2H + 3H = 0 5 * 80 = 2V V = 200 KN 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
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
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 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
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
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 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)
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Part B(iii) Plate girder design considerations 1.0plate girder depth Recommended depth is determined by multiplying 1/10 * the span for average loading.
The depth should be between1 15¿1 20if one is dealing with lightly loaded girders In determination of the flange depth we multiply the overall depth by 1/3 Stiffeners are used to prevent Web buckling 2.0plate girders deflections The deflection of the plate girder is determined by referring to Clause 15 of BS449 in order to determine the deflection requirement 3.0permissible stresses According to BS449 it is recommended that whenever the plate thickness will exceed 40 mm a lower stress must be used The allowable permissible bending stress to be used are provided in BS 449 4.0Considering the types plate girder Plate girder construction involves use of welded steel plate which together will form an I section 5.0Bending stresses Bending stresses will be determined by referring to Clause 17, 27 and 32 of BS 449 sets section area for the girders In consideration all the characteristics of the plate girders that includes moment of inertia, area, modulus of section and radii of gyration are calculated from the first principle. Refer to BS 449 in determining maximum outstand for flange plate, Where, Compression flange = 16t Tension flange = 20t t is the thickness 6.0Stresses and loads Loads subjected to the plate girder are from stanchions, floor slab and floor beams, in which it is carried by the girder Whereas, 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. 7.0Shear stress of the web Shear stress of the web is determined by the formula fq=shear depthofwebplate∗thicknessoftheweb Allowable shear stress will depend on the value of d/t and stiffeners spacing If the ratio of d/t more than 85 vertical stiffeners will be required at a distance which will not exceed 1.5d The thickness should be more than1 180