Design steps of foot bridge , stringer and cross girder
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The document provides design steps for a foot bridge, including: 1. Design of the floor planks by checking for flexure stress, shear stress, and deflection to determine the required plank depth. 2. Design of the cross beams by checking flexure to determine the required beam section. 3. Design of the main girders by analyzing member forces in the truss under dead and live loads to select member sizes. 4. Design of the stringers by calculating loads and checking bending and shear stresses to select the stringer section. 5. Design of the cross girders by calculating loads and checking bending and shear stresses to select the cross girder section.
Design steps of foot bridge , stringer and cross girder
- 1. DESIGN STEPS OF FOOT BRIDGE By Gaurav Ghai Gauravghai18jan@gmail.com Basic Information Required:- 1. Type of Girders = Lattice Stype (Say) 2. Span of Girders = 16 m c/c of bearings 3. Cross Girder Spacing = 2 m 4. Walking Width = 2.50 m (Clear Width) 5. Pedestrian Load = 4000 N/M2 6. Flooring to be made of timber planks supported on cross girders Step 1 : Design of Flooring / Planks Dead Load of Planks 60 mm Thick = 8000 N/m3 Therefore = 60/1000 x 8000 = 480 N/m2 Live Load = 4000 N/m2 Total Load = Deal Load + Live Load Consider 1 meter wide strip of Plank A) Check for Flexure Stress:- Maximum BM = wl2/8 Equate Maximum BM = Moment of Resistance 1/6xfxbxd2 = Maximum BM Calculate Stress f in N/mm2 Permissible Bending Stress in Planks = 10 N/mm2 Therefore Calculated stress < Permissible Stress in Planks. If this check fails then increase the depth “d” of the Planks. B) Check for Shear Stress:- Maximum Shear Force = wl/2 N Shear Stress = Max. Shear Force /b/d N/mm2 Mean Shear Stress = 1.5 x Shear Stress N/mm2 Permissible Shear Stress = 0.8 N/mm2 Therefore Mean Shear Stress < Permissible Shear Stress Planks. If this Check Fails then increase the depth “d” of the Planks. C) Check For Deflection:- I i.e. Moment of Inertia of the 1m Wide Strip = bxd3/12 Maximum Deflection d = 5 wt4 / 384 EI Permissible Deflection = span/325 Therefore Maximum Deflection < Permissible Deflection. If this Check Fails then increase the depth “d” of the Planks.
- 2. Step 2 : Design of Cross Beams Clear Width of Footway=2.50 m C/C Distance between Girders = 2.50+0.15+0.15=2.80 m Load on One cross Beam = 2x2.5x(Dead Load+Live Load) Self wt. of cross beam = 300 N (Assumed) Total Load = Load on one cross beam + Self Wt. of cross beam A) Check for Flexure:- Maximum Bending Moment = Total Load x c/c distance between girders / 8 = Total Loadx2.8/8 N/m Permissible bending stress = 165 N/mm2 Therefore Calculated Maximum bending stress < Permissible Stress. Section modulus required = Maximum BM x 103 / Permissible Stress Check Section ISLC in Steel Table with section modulus greater than calculated above and provide that section for example ISLC 125 has 57100 mm3 of Section Modulus. Step 3 : Design of Main Girders Loads Dead Load transmitted from cross beam = = Total load of cross beam-2xclear walking width x Live Load / 2/distance between cross girder Weight of One Truss (assumed) = 400 N/m Total Load on one truss = Dead load as calculated above + Weight of one truss Live load on one truss = Live Load x Clear walking width / 2 Check for self weight of the truss i.e. 400N/m Assumed by Hudson Formula can be made: MEMBER L3L4 Height of ILD = a(l-a)/lh
- 3. Total Load on one truss = Live load on one truss+Total Dead load on one truss N/m Maximum Tension in member L3L4 = Area of the ILD x Load Intensity Allowing a Stress of 150 N/mm2 net area required for this member A = Maximum Tension in Member / Allowable Stress(150N/mm2) Weight of Two trusses including bracings = 0.785xA N/m Weight of one truss with bracings = Wt. of Two trusses/2 Weight of one truss assumed = 400 N/m < Weight of one truss with bracings calculated above , Hence Safe. Step 4 Analysis of the truss Dead Load on the Truss = Calculated Above Live Load on the truss = Calculated above Dead Load +Live Load Forces in the members of the truss: A) Top Chord Members (All are in compression) Say Member UoU1 Maximum Compression = (DeadLoad+Live Load)xarea of ILD Similarly Calculate for all top chord compression members B) Bottom Chord Members ( All are in Tension) Force in tension = force in compression in another member but the member is in tension. For example Compression in Member U1U2 = Tension in Member L2L3 C) Vertical Members Member U1L1 Area of + ILD Area of –ILD Dead Load force in member = Dead load on the truss x (Area of +ILD- Area of –ILD) Live Load compression = Live load on the truss x Area of +ILD Live Load Tension = Live load on the truss x Area of - ILD
- 4. Extreme Value i)Dead Load force in member +Live Load compression ii)Dead Load force in member – Live Load Tension Similarly it is done for all the vertical members. D) Diagonal Members Member U1L2 Force in the member = Force in U1L1 cosec (Theta) Note: Calculate both for compression and Tension Forces as occurring in Vertical force calculations. E) Design of Top Chord Member Maximum force in Top Chord Member Length of Top Chord Panel = 2 m From Steel Table Try 2L 60x60x8 mm Area = 1792 mm2 r=18 mm Slenderness ration = l/r Corresponding to l/r Safe Compressive stress = 85 N/mm2 Safe Load = 85 x Area N
- 5. DESIGN STEPS OF STRINGERS Gaurav Ghai Gauravghai18jan@gmail.com Stringer Spacing = 2m c/c Cross Girder Spacing = 4 m c/c Stringer Support = Super Imposed Dead Load (SIDL)+Live Load +Impact Load SIDL (Super Imposed Dead Load): Weight of Stock Rails = 0.60 kN/m Weight of meter guard rails = 0.40 kN/m Size of Timber Sleepers used = 2.80 m x 250 mm x 250 mm @450 mm c/c spacing Unit weight of Timbers = 7.50 kN/m3 Step 1 Loads i) Dead Load Weight of stock rails per track per meter, (2x0.60) = 1.20 kN/m Weight of Guard Rails per track per meter, (2x0.40)=0.80 kN/m Weight of Fastenings (assumed) = 0.20 kN/m Weight of sleepers = (100/45)x(2.8x250x250x7.50/1000/1000) = 2.90 kN/m Self weight of stringers per track per meter (assumed) = 0.200 kN/m The dead load per track per meter = 7.12 kN/m Span of stringers = 4 m Total dead load per track for two stringers = 7.12x4 = 28.48 kN. ii) Dead Load, Live Load and Impact Load Consider the span is fully loaded. Impact factor = 1.00 Impact Load = 1.00 x L.L For Bending = (½)x BM due to Dead+Live+Impact Load For Shear = (½)x SF due to Dead+Live+Impact Load Step 2 Bending moment: BM occurs at the center BM=ML/8 = Mx4/8 kN-m Step 3 Shear Force: At the End SF= Total SF calculated above /2 kN
- 6. Step 4 Section Modulus: Z = M/Sigma cbc where Sgma cbc is permissible stress in bending = 165 N/mm2 From steel table select ISLB section whose Zxx is greater than Z calculated above and note down its following things: Zxx , h and tw Step 5 Shear Stress: Tv= Shear Force at the end/ height of section selected above / thickness of web tw and it should always be less than 100N/mm2 otherwise revise the section. Hence Safe. Provide the detail of the section provided. The connection of stringer with cross girder is deigned for end reaction.
- 7. DESIGN STEPS OF CROSS GIRDER Gaurav Ghai Gaurav Ghai STEP 1 LOADS: Effective span of cross girder = 4m. The dead load is transferred as the point load at 1m from the ends of the cross girder. Self weight of cross girder = 3kN/m (assumed) , it acts as a uniformly distributed load. The Cross girder is subjected to maximum live load when both the adjacent spans are loaded. EUDLL for 8m span = 1056 kN Impact factor=0.909 Impact Load = 1056x0.909 =960 kN Live Load + Impact Load = 2016 kN Half of the Live Load and Impact load is carried by the cross-girder as two point loads at 1m from each end. Point load due to super imposed dead load (1/2 x Dead load per track for two stringers as calculated in design of stringers) Point load due to live load and impact load ½ x (1/2x 2016) Total point load = Sum of the two calculated above kN Draw the loading diagram with udl and point load at 1m from the edges. The Reaction at the ends= Total point load+ udl load x span length kN Step 2 Bending Moment: At center M = Reaction at each endxspan/2-point loadx1-udl loadxspan/2x1 kN-m Allowable Stress in Bending = 165 N/mm2 Step 3 Section Modulus Z = M/Sigma cbc where Sigma cbc = 165 N/mm2 From steel table select ISWB section having Zxx greater than Z calculated above. Step 4 Shear Stress
- 8. Tv= Shear Force at the end/ height of section selected above / thickness of web tw and it should always be less than 100N/mm2 otherwise revise the section. Hence Safe. Provide the detail of the section provided. The connection of stringer with cross girder is deigned for end reaction.