Structural Design Of Hangar For Airport The structural design of any construction related to civil engineering will enhance the life of that structure. Hangar is a closed structure to hold aircraft or spacecraft in protec tive storage.Hangars are used for protection from weather, direct sunlight, maintenance, repair, manufacture, assembly and storage of aircraft on airfields, aircraft carriers and ships. This report explains about the manual analysis of design process carried out for h angar, and its importance in the construction for an airport. The checks and corrections are analyzed in STAAD PRO software. The results obtained in this detailed project report will prove to be economical from design point of view. '1.1 Introduction Structure in the civil point of view is very ver y important to be in proper design and detailing so that maximum benefit can be achieved by the people. An aircraft should be properly repaired and maintained after its certain use in transportationHence, the hangar is a place where the aircraft can be taken proper care,proper management and enhancement in the manufacturing parts can be made. If hangars are not provided then the aircraft company can face the problems due to its less maintenance and services. Even less maintenance and repairing may lead to large vital accidents in future; hence hangars should be designed keeping in mind the future point of view. Materials used in hangars mainly consist of wood, fabric and steel. Structure with span larger than 40 m can be regarded as long span structures and need to be carefully designed keeping a balance of all the aspects like its weight, deflections (sway) and foundation forces. There are many combinations of designing large spans, like conventional truss & RCC column combination, truss & steel columns, Pre-engineered building (PEB) etc. The design under discussion is a 50 5 0 meter clear span hangar for aircrafts. We have designed this hangar in 3D on STAAD software, for proper simulation of the load distribution uniformly in three co-ordinates system i.e. X, Y and Z. All the basic loads i.e. dead, de ad, live, wind, temperature, seismic etc. have been taken into consideration for designing of the frames. The structure has been designed under enclosed as well as open conditions for application of wind loads, because of the opening & closing of the large sized hangar doors. The basic philosophy of rigid frame design is b y adopting 'Fixed' or 'Pinned' column base conditions. A fixed column base is always alwa ys a sturdy frame and helps in controlling allowable deflection (side sway) in the frames. Steel designers always prefer fixed base to pinned base frames. On the contrary, for foundation designers, the design of foundations becomes a nightmare particularly in large span buildings. In fixed base design, the frame is rigid, but transfers heavy moments to the foundations. ' On weak soil, designing foundations becomes tedi ous task. Likewise, for pinned support, the frame does not transfer any moment to the foundation and only vertical & horizontal reactions affect the design of foundation. It looks simple bu t in case of large spans, controlling deflections of frame in pinned base condition is a challenging task. The simplest wayfor controlling this deflection is to increase the geometrical properties/sectional sizes of frame, but it is not advisable as it adds to the tonnage of the whole building, adding not only to the seismic forces but also adding to the cost subsequently. We need a solution wherein
the sway of the frame can be controlled and the section sizes are also not increased. ' Definitions Span : The centre to centre distance between two supports of roof truss is called span. Spacing: Horizontal distance between two consecutive trusses is known as spacing Principal rafter: Member of roof truss from end support to ridge point is known as principal rafter. Main tie: Bottom chord member of the roof truss. Sag tie: Vertical member joining ridge point and midpoint o f main tie. Ridge line: Line joining ridge point of one truss to ridge point of another truss. Eaves line: Line joining eave point of one truss to eave point of another truss. Purlin: Member supported on panel points of two consecutive roof trusses. Web members: Vertical or inclined members joining top chord and bottom chord members. Panel point: Points on principal rafter where vertical or inclined me mbers connected to principal rafter are called panel points. Panel: The portion of roof truss between two consecutive panel point on principal rafter. Rise: Vertical distance from main tie to the ridge point. Pitch: Ratio of rise and span. Wind bracing: Two roof truss are connected by cross members to stabilize it against action of wind. ' Objectives Identify the current and long-term projected airplane fleet, including the number of airplanes that will use the facility, annual use of airplanes, and maintenance-hour requirements. Identify the shops and support functions that will be required to support the hangar functions. Assess company standards for interior and exterior finish quality, security, facility maintenance, access for people with disabilities, and corporate image. Assess airplane layout requirements. As with hangar height, the internal dimen sions of the hangar will be determined by the types and number of airplanes housed and the work performed. In addition, possible insurance or regulatory authority clearance requirements as well as maintenance policy will affect the internal dimensions of the hangar. Areas to consider include required horizontal clearances around airplanes including stands and floor- or roof-supported maintenance docks, the ability to move airplanes while others are in the
hangar, tail-in versus nose-in airplane-parking configurations, building setback requirements, the proximity to adjacent buildings, the use of tail doors, and the installation of floor airplane powersupply stations.. Corrosion control and cleaning can be carried out whenever required. Heavy maintenance, overhauling and modification can be done. ' CHAPTER-3: PLAN OF WORK 1. Determining Dimensions of hangar. 2. Determination of Load on purlin and its design. 3. Determinationof Load on Truss and deciding the Configuration of Truss. 4. Structurally analyzing the truss. 5. Structure design of truss. 6. Designing Column for truss. 7. Design of foundation. ' Chapter-4: 4.1 Materials Used 4.1.1 AutoCAD AutoCAD is the software developed by the Autodesk Company US. AutoCAD is the software by which we design the drawings and figures of the truss members and its configurations. It is a very easy and accurate software to draw the engineering structures. This software is useful in preparing all the reinforced concrete detailed drawings and at the time of construction, these drawings are most useful for the construction engineer. All the dimensions and scale of the drawing can be easily changed by this software. Layout can also be prepared with the help of this software. In this project, detailed drawing of the trusses and its loading is clearly shown by this software. 4.1.2 Staad Pro STAAD or (STAAD.Pro) is a structural analysis and design computer program o riginally developed by Research Engineers International inYorba Linda, CA.The commercial version STAAD.Pro is one of the most widely used structural analysis and design software. It supports several steel, concrete and timber design codes. It is 3D design software which shows the detailing and is an advanced software compared to AutoCAD. In this the an alysis and checks of the project is carried out. ' 4.2 Methodology used The roof truss member is analyzed by the method of joints.After analysis of this truss we come to conclusion whether the member is in compression or in tension. The member being in tension is designed by the following steps. Determining the gross section area by the factored axial tension on the member and yield stress ??mo=T/fy ??mo Selection of trial section from the steel table. The roof truss comprises of welded joints no bolting is ca rried out, computation of three strength
of member is carried out. Tdg=design strength due to yielding of gross section Tdn= design strength due to rupture of critical section Tdb= design strength due to block shear The min value of the above strength is taken and the section selected should be great the minimum value The design strength is less than the T the trial section is changed, and higher section is selected The slenderness ratio (KL/r) check it the value should be less than effective slenderness ratio. The member being in compression is designed by following steps: Based on length and loading in compression, section classification is carried out in various terms such as ??= (250/fy)0.5 =1 b/tf = 9.4< ?? for plastic state d/tw = >42?? for slender Determining the effective length for the member kl = 0.65L (0.65 for welded joints) ' Buckling curve classification is carried out. h/bf = >1.2 bucking class curve 'C' The slenderness ratio (KL/r) is obtained and fcd is calculated. The design compressive strength is calculated ( Pd= Ae x fcd) If the value of Pd is less than the loading on the member, the selection of member is changed. After determining the compression and tension member, the roo f truss is resting on the sole plate and base plate, the thickness of this plate is determined as per bending stresses. The jointing of the roof truss to the sole plate and the base plate is done by the rag bolt and is connected to leg of the angle section. The whole loading is transferred from the column to the foundation. The ISHB section in column and strip footing is selected due to high uplift wind force. The gable columns are provided at the intermediate joints of the roof truss. As the wind load is comparatively high as compared to dead load and live load, bracing is provided in roof truss. The bracing such as principal rafter bracing, wall bracing, tie runner, eave girder, side runner are provided. ' 4.2.1. Determination of loads on truss A). Dead Load (D.L.) Weight of roofing material Assume the G.I. sheets @ 61 N/m2 = 61 ?? 203.96 = 12441.56 N Weight of purlin Weight acting on plan area=282.528 N/m2
= 200 ?? 282.528 = 56505.6 N Self-weight of roof truss = 10 (50/3+ 5)N/m2 = 216.66 N/m2 Self-weight of roof truss = 216.66 ?? 200 = 43332 N Weight of wind bracing Assume the weight as 12 N/m2 =12 ?? 200 = 2400 N Total Dead Load = 12441.56 + 56505.6 + 43332 + 2400 = 114.67 KN On one side of roof truss, there are 10 Full panel points (P.P.) Therefore, D.L per Panel point=98173.56/10 = 9817.356 N D.L. on I.P.P (full P.P) =11.46 KN D.L.on E.P.P = 5.733 KN B). Live Load (L.L.) L.L on purlin = 750 ' 20 (??-10) = 750 - 20 (11.309-10) = 723.82 > 400 N/m^2 L.L. on roof truss =2/3??723.82 = 482.546 N/m^2 Total L.L. = 482.546 ?? 200 = 96509.2 N L.L. on I.P.P =96509.2/10 = 9.65KN L.L. on E.P.P =96509.2/(2 ??10000) = 4.825 KN I.P.P Total D.L= 19.467 KN E.P.P Total (D.L. + L.L.) = 9.7335 KN Weight of purlin = 18 KN Roof truss = 43.33 KN Weight of wind bracing = 26 KN
Wind load basic wind speed Vb= 39 m/s Design wind speed Vz = Vb K1 K2 K3 K1 = 1.06 (from table 1 IS 875 Pg 3) K2= 1.10 (for class B type and having terrain category 1) K3 = 1 Vz = 39 ?? 1.10 ?? 1 ?? 1.06 =45.474 m/s ' Design wind pressure PZ = 0.6 V_z^2 = 0.6?? 45.4742 = 1247 N/m^2 TABLE 1. Wind angle Pressure coefficient Cpe+Cpi A xPd FORCE (KN) Cpe Cpi WL LW (KN) WL LW WL LW 0 -1.12 -0.4 -0.7 -1.82 -1.1 190.869 -347.38 -209.95 0.7 -0.42 +0.3 190.869 -80.164 57.26 90 -0.79 -0.6 -0.7 -1.49 -1.3 190.869 -284.39 -248.12 0.7 -0.09 +0.1 190.869 -17.17 19.08 WL on IPP =- 485.69/ 10 = -48.57 KN WL on EPP = -48.57/2 = -24.29 KN ' 4.2.2.Design of purlin. >> Self weight of roofing material = 61 ?? 2.5498 ?? 8 = 1244.302 N >>Self weight of purlin = 288 ?? 8 =2304 N >>Total Dead load = 1244.302 + 2304 = 3548 N >>Component of D.L. perpendicular to the Principal rafter >>Cos 11.309 = DL/3548= 3479.14 N Live load = 750 ' 20 (??-10) = 750 ' 20 (11.309-10) = 723.82 > 400 N/m^2 L.L. = 723.82 ?? 25 ?? 8 = 14476.4 N
Cos 11.309 = LL/4476.4 L.L = 14195.3243 N Wind Load (W.L.) W.L. = -1247 ?? (2.549 ?? 8) = -24952 N Load Combination D.L = 3.5 KN L.L. = 14.195 KN W.L. = - 24.95 KN D.L.+ L.L = 3.5 + 14.195 = 17.695 KN D.L.+ W.L = 3.5 ' 24.95 = - 21.45 KN D.L.+ L.L + W.L = 3.5 + 14.95 - 24.95 = - 7.255 KN U.D.L = w=21.47/8= 2.68 KN/m M =(2.68 ??8^2)/10= 17.15 KN.m ' Required Zez = M_z/F_y =(17.15 '??10'^6)/250 = 68600 mm^3 = 68.6 cm^3 Minimum depth of purlin = D = L/ 45 = 8000/45 = 177.77 mm Minimum width of purlin = B = L/60 = 8000/60 = 33.33 mm Izz required Permissible deflection = L/ 180 = 8000/180 = 44.44 m Actual deflection =(5 ??WL^4)/(384 ??EI) 44.44 =(5 ??268 '??8000'^4)/(384 ??2 '??10'^(5 )??I_z ) Iz = 16.08 ?? 106 mm^4 Now select section ISWB ' 200 Depth = 200 > 177.77mm Width = 140 > 133.33mm Iz = 26.26 ?? 106> 16.08 ?? 106 Actual deflection = (5 ??WL^4)/(384 ??EI) =(5 ??268 '??8000'^4)/(384 ??2 '??10'^(5 )??26.26 '??10'^6 ) = 27.21 mm < 44.44 mm ' C=-VE T=+VE 4.2.3. Analysis of truss with dead load, live load and wind load (with load combinations) DL LL WL 1.5(DL+LL) 1.5(DL+WL) 1.2(DL+LL+WL) max A,A1 0 0 4.85 0 7.275 5.82 7.275 T A,B -5.73 -4.825 24.84 -15.8325 28.665 17.142 28.665 T
B,A1 -210.75 178.188 897.5 -583.407 1030.125 610.2744 1030.125 T B,B1 180.72 152.796 -752.46 500.274 -857.61 -502.7328 857.61 C A1,A2 184.291 155.821 781.93 -510.168 896.4585 530.1816 896.4585 T A1,B1 60.831 51.468 -251.46 168.4485 285.9435 -166.9932 285.9435 C B1,A2 -97.378 -82.39 402.54 -269.652 457.743 267.3264 457.743 T B1,B2 256.76 217.132 1066.79 710.838 -1215.045 -711.4776 1215.045 C A2,A3 -261.84 221.431 1111.68 724.9065 1274.76 754.0908 1274.76 T A2,B2 34.183 28.951 -139.12 94.701 -157.4055 -91.1832 157.4055 C B2,A3 -48.342 -40.942 196.75 -133.926 222.612 128.9592 222.612 T B2,B3 290.943 246.082 -1206 805.537 -1372.585 -802.77 1372.586 C A3,A4 296.704 250.954 1221.24 -821.487 1386.804 808.2984 1386.804 T A3,B3 15.89 13.51 -31.172 44.1 -22.923 -2.1264 44.1 T B3,A4 -20.684 -17.586 40.577 -57.405 29.8395 2.7684 57.405 C B3,B4 304.184 257.34 -1232 842.286 -1391.724 -804.5712 1391.724 C A4,A5 310.203 262.435 1257.2 -858.957 1420.495 821.4744 1420.49 T A4,B4 1.782 1.608 23.52 5.085 37.953 32.292 37.953 T B4,A5 -2.19 -1.976 -28.9 -6.249 -46.635 -39.6792 46.635 C B4,B5 305.455 258.488 1215.21 845.914 1364.632 -781.5204 1364.633 C ' A5,A6 -311.5 263.606 1250 -862.659 1407.75 809.8728 1407.75 T A5,B5 -9.94 -8.271 70 -27.3165 90.09 62.1468 90.09 T B5,A6 11.72 9.752 -82.542 32.208 -106.233 -73.284 106.233 C B5,B6 300 253.32 1171.47 829.98 -1307.205 -741.78 1307.205 C A6,A7 305.164 258.335 1214.84 845.2485 1364.514 781.6092 1364.514 T A6,B6 -20.155 -16.88 110.77 -55.5525 135.9225 88.482 135.9225 T B6,A7 23.056 243.942 -126.72 400.497 -155.496 168.3336 400.497 C B6,B7 288.8 19.31 -1110.1 462.165 -1231.95 -962.388 1231.95 C A7,A8 -294 248.771 1161.52 814.1565 1301.28 742.4988 1301.28 T A7,B7 -29.42 -24.654 147.98 -81.111 177.84 112.6872 177.84 T B7,A8 32.9 231.615 -165.44 396.7725 -198.81 118.89 396.7725 C B7,B8 297.86 27.564 1039.13 488.136 -1111.905 -856.4472 1111.905 C A8,A9 -279 239.199 1095 777.2985 1224 692.1612 1224 T A8,B8 -37.95 -31.838 182.73 -104.682 217.17 135.5304 217.17 T B8,A9 41.68 34.973 -200.72 114.9795 -238.56 -148.8804 238.56 C B8,B9 280.612 217.143 -956.72 746.6325 -1014.162 -550.758 1014.162 C
A9,P -262.98 -221.44 1019.75 -726.63 1135.155 642.396 1135.155 T B9,B9 -41.87 -38.593 215.66 120.6945 260.685 162.2364 260.685 T B9,P 42.29 41.81 -233.33 126.15 -286.56 -179.076 286.56 C B9,O 263.34 201.062 -866.86 696.603 -905.28 -482.9496 905.28 C ' 4.2.4.Design of truss members and connections FORCE area CM2 TYPE OF SECTION Provided member l r area (k*l)/r fcd Compression load VERTICAL 28.28 1.24 25-25-2 60-60-4.8 1000 22.2 1001 29.27 220.438 220.65 168.44 7.41 50-50-4.5 100-100-4 1500 38.9 1495 25.06 222.96 333.32 94.7 4.16 38-38-3.2 60-60-4.8 2000 22.2 1001 58.55 196.45 196.645 44.1 1.94 25-25-2.6 60-60-4.8 2500 22.2 1001 73.19 177.203 177.38 37.953 1.66 25-25-2 60-60-4.8 3000 22.2 1001 87.83 153.314 153.46 90.07 3.96 40-40-3.2 60-60-4.8 3500 22.2 1001 102.477 127.78 127.90 135.922 5.98 50-50-3.6 60-60-4.8 4000 22.2 1001 117.11 105.036 105.14 177.84 7.82 60-60-4 60-60-4.8 4500 22.2 1001 131.75 86.46 86.54 217.17 9.55 60-60-4.8 100-100-4 5000 38.9 1495 83.54 160.63 240.14 260.685 11.47 80-80-4 100-100-4 5500 38.9 1495 91.90 145.76 217.91 DIAGONAL 1030.1 45.32 180-180-8 220-220-8 2898.5 86 6619 21.90 224.85 1488.28 457.74 20.14 100-100-6 150-150-4 3201.6 59.3 2295 35.09 216.43 496.70 222.61 9.79 60-60-4.8 100-100-4 3535.5 38.9 1495 59.075 195.94 292.93 29.89 1.31 25-25-2 100-100-4 3905.1 38.9 1495 65.252 188.175 281.32 43.63 1.91 25-25-2.6 100-100-4 4301.2 38.9 1495 71.87 179.195 267.89 32.20 1.41 25-25-2 100-100-4 4717 38.9 1495 78.81 179.19 267.88 400.49 17.62 100-100-5 150-150-4 5147.8 59.3 2295 56.42 198.57 455.71 396.77 17.45 100-100-5 150-150-4 5590.2 59.3 2295 61.27 193.349 443.73 115 5.06 60-60-2.6 150-150-4 6041.5 59.3 2295 66.22 186.92 428.98 126.5 5.566 60-60-2.6 150-150-4 6500 59.3 2295 71.24 180.14 413.42 60-60-4.8 14 Number of members 100-100-4 15 150-150-4 10 220-220-8 2 Tie 220-220-10 2 Rafter ' 4.2.5. Design of column # Calculation OfLoad Compressive force in column Roofing material load = 20.392 KN Tie runner load = 10 KN
Purlin = 24 KN Eaves = 5 KN Truss = 77.20 KN Girts = 17.5 KN Side wall = 125.568 KN Self-weight = 50.57 KN Total compressive force = 382.23 KN Tensile force in column Vertical uplift force = 0.8 ?? 0.75 ?? 1247 ?? 15 ?? 25 = 280.575KN Moment at base Net horizontal force at tie level due to wind on roof = sin 11.303 ?? 25.49 ?? 8 ?? 0.75 ?? 1247 = 36 KN 0.65 0.75 0.4Pd 0.7Pd Pd= 0.75 ?? 1247 ?? 8 = 7.482 KN/m ' Deflection of column AB =(T '??15'^3)/3EI+ (2.25 '??15'^3)/8EI Deflection of column CD =(5.25 '??15'^3)/8EI+ (36 '??15'^3)/3EI- (T '??15'^3)/3EI Equating deflection of AB and CD (T '??15'^3)/3EI+ (2.25 '??15'^3)/8EI= (5.25 '??15'^3)/8EI+ (36 '??15 '^3)/3EI- (T '??15'^3)/3EI (2T '??15'^3)/3EI- (36 '??15'^3)/3EI= (5.25 '??15'^3)/8EI- (2.25 '?? 15'^3)/8EI ('15'^(3 )??(2T-36))/3EI= (3 '??15'^3)/8EI 2T-36 = 9 2T= 45 T = 22.5 KN Moment at base of AB = 2.25 ?? 15 ?? 7.5 + 22.5 ?? 15 = 590.625KN.m Moment at base of CD = 5.25 ?? 15?? 7.5 + 36.?? 15 ' 22.5 ?? 15 = 793.125KN.m
considering the increase in allowable stress for wind load combination forces, the reduced forces are : Tensile force = 0.85 ?? 280.575 = 238.488 KN Moment = 0.75 ?? 793.125 = 600 KN.m ' Design of section Compressive force = 382.25 KN Tensile force = 238.488 KN Base moment = 600 KN.m For built up section ISHB 450 with 40 mm thick and 400 mm long plate on both sides Rxx = 230.6 mm ,Ryy = 102.2 mm A= 437.89 cm2 = 43789 mm2 W= 343.7 Kg/m KL/R=(0.8 ??1500)/23.06= 52.038 Fcd = 183 Mpa Allowable compressive force = 183 ?? 43789 = 8013.387 KN Check for deflection ISHB 450 Ixx = 40349.9 + 213.33 ?? 2 = 40776.566 cm^4 Deflection at top = (535.138 '??15'^4 '??100'^3)/( 8 ??2.047 '??10'^6 ??40776.566)+ (1376.146 '??15'^3 '??100'^3)/(3 ??2.047 '??10'^6 ??40776.586)= 59.12 > 10 cm'''' Not ok Check for compressive stress due to bending = 6116207.9/(780.94 ??8787) = 0.891 < 1 ok Check for simultaneous action of bending and axial tension and moment =15290.5/(437.89 ??1500)+ 6116207.9/(1650 ??8787) = 0.445 < 1 ok ' For deflection providing two extra plate of 40 x 400 mm at both side of column New M1 = 40776.56 + 41666.66 ?? 2 = 83443.226 KN.m = (535.168 '??15'^4 '??100'^3)/(8 ??2.047 '??10'^6 ??83448.226) = 19.269 ' 11.4108 =7.85
)^0.5 = 73.49 mm Fb= 0.45 ??f_ck ??y ??b ' n = 0.45 ?? 40 ?? 73.49 ?? (700 ' 382.7 ?? 103) = 543.27 KN Hence provide 3 bolts of 32 mm diameter. Nu=5.5 '??40 ?? '400'^1.5 = 784.244 > 543.27 KN So 400 mm anchor length Maximum BM B.M = 73.49 ?? 700 ?? 0.45 ?? 40 ?? ((370 ' 78.5/2))/'10'^6 = 306.26 KN.m Moment capacity of base plate = 1.2 ?? 250/1.1 ?? t^2 ?? 700/6 = 31818.18 t^2 t = '((306.26 ?? 106)/31818.18)= 100 mm thickness ' Truss shoe angle design WL reaction = -150KN DL reaction = 382.23KN Therefore wind load tension per bolt ft ft= (382.23-150)/4 = 58.0575 KN Wind load shear per bolt = 36/4 = 9KN Try 4- M30 4.6 grade black bolt Fs=??/4 ?? d^(2 )?? 0.78 ?? 400= 220.540KN Ft =??/(4 ) ?? d^(2 )?? 0.75 ??400= 212.058KN (9/220.540)^2+ (58.0575/212.058)^2= 0.0766<1 Try 2 ' ISA (200 ?? 200 ?? 18) X = 2 ?? tan'''60'^?? ' ??14.82+2 = 53.34 cm Thickness t = 2.5 cm Moment in angle = 58.0575 ??14.82 = 860.412 KN.m Bending stress = (860.412 ?? 6)/(53.34 ?? 1.8 ?? 1.8) = 298.72 < 320 N/mm^2 therefore safe. Cap plate design:Moment = 58.0575 ?? 15.385 = 893.215 KN.m X = 2 ?? tan60?? ?? 15.385+2 = 55.3 cm Size of plate = (630 ?? 500 ?? 20)mm Bending stress = (893.215 ?? 6)/(55.3 ?? 2 ?? 2) = 242.28 N/mm^2< 320 N/mm^2 therefore safe. ' 4.2.7. Design Of Tie Runner. Portion of wind load from gable end along the ridge will be transferred as axial load to tie runner provided along the length of building at tie level. L/W = 48/50 = 0.96< 4
And b/W = 15/50 = 0.3<0.5 From table 4 of IS 875(PART 3) 1987, external wind pressure = + 0.7 with internal pressure + 0.5, maximum pressure =1.2 Factored wind load on intermediate runner = 1.5 ??1.274 ?? 1.2 ?? 8 ?? (15/2 +6/2) =192.63KN Min r required =((8000 ?? 0.85))/250 = 27.2 mm Try ( 150 X 150 X 4) A =2295mm^2 Rmin =59.3mm Design strength due to rupture of critical section Tdg= A_g f_y/??_mo Tdg= ((2295 ??250 ))/1100 = 521.59 > 192 KN Ok Try ( 100x 100 x 4) A = 1495mm2 Tdg=A_g f_y/??_mo Tdg= ((1495 ?? 250 ))/1100 = 339.77 > 192.63 KN OK ' 4.2.8. Design of Eave Girder. Design wind pressure= 1.274 KN/m^2 Maximum force coefficient= 1.65 Eave girder bracing are connected between gable column which are assumed to be spaced at 8m center to center and truss bottom joints. Wind load on internal joints =(1.274 ??1.2 ??8 ??( 15/2 + 6/2 ) ?? 1.5 = 1.5 KN Reaction =(9 ?? 192.63 + 2 ?? 96.315) = 963.15 KN Using The Method Of Joint The maximum force in the bracing at the end of the eave girder Fbr= ((963.15 - 96.315))/cos'32.005 = ?? 1022.20 KN Members of the girder are subjected to reversal of stresses and hence they have been checked for compression as well as tension. Length of bracing = '(5^2+ 8^2 ) =9.43m Required rmin= 9434/350 = 26.95 mm Try section 150 x 150 x 4 with following properties A= 2295 mm2 Rmin =59.3mm Section classification ?? = (250/f_y )^0.5 =1 ' Using section 220 x 220 x 8
Fcd =170 Mpa Axial capacity in compression = (170?? 6619)/1000 = 1125.23 > 1022.22 KN Axial tensile capacity = (6619 ?? 250)/1100 = 1504.32 > 1022.20 KN Design strength due to rupture of critical section. Tdn= (0.6 ?? 6619 ?? 410)/(1.25 ?? 1000) = 1302.62 > 1022.20 KN ' 4.2.9. Design of rafter bracing member. Design wind pressure = 1.247 KN/m^2 Max force coefficient = -1.65 Factor wind load = 1.5 ?? 1.24 ?? 1.65 ?? 5 ??4?? sec11.307 = 62.8KN Length of bracing ='(5^2+8^2 ) =9.43m TRY (150x150 x 4) ' A'_g=2295 mm^2 Rmin = 59.3mm L/r = (9430 ?? 0.65) /59.3 = 103.36< 400 Tensile Capacity Tdg= A_(g ?? ) f_y/??_mo Tdg= ((2295 ?? 250 ))/1100 = 521.59 > 242.12 KN Ok Rupture Of Critical Section Tdn= '' ??A'_(g ) '?? f'_y/??_mo = ((2295 ?? 250))/1100 = 312.95> 242.12 KN Ok ' 4.2.10. Wall bracing Force from tie level bracing = 13.13 + 63.55 = 16.68 KN Wind drag from side walls = 0.025 ?? 1.247 ?? 3.94 ?? 8 ?? 6 = 5.89 KN Total compression on eaves beam = 82.575 KN Maximum tension in bracing = (82.57 ?? '(8^2+ '3.94'^2 ))/8 = 92.04 KN Length of bracing = 10.96m kl/r = ((0.8 ?? 15000))/59.3 = 202.36 Fcd=36.3 Compression capacity =36.3 ?? 3436
= 124.726 KN> 92 KN So required section= 150 x 150 x 5 ' 4.2.11 Gable columns. Height= 15m Wind force per column= 0.7 ?? 1247 ?? 5 ?? (5/2) = 10911.25 N/m Mmax = (10911.25 ?? 152)/8 = 308.878 KN.m Critical Stress Calculation CS = (1.2 ??10.1 '??10'^6 ??91813 ??55)/(3060.1 '??1500'^2 ) ?? '(1+(0.162 ??64.14 '??1500'^2)/(91813 '??55'^2 )) = 9237.80kg/cm^2 = 906.149 > 410 Mpa Ok Section required is ISMB 600 ' 4.2.12. Design of pedestal Self wt of pedestal= 0.94 ?? 0.8 ?? 3 ?? 2500 =55.32 KN Net downward load = 31.85KN Moment due to shear force at base of pedestal = 975 KN.m Total moment at base=1650 KN.m Therefore design compression= 47.775KN Design moment =2475KN.m Fck=40 N/mm^2 MU = '(0.0959/(F_(ck )?? B ?? d_2 )) PU ='(0.000165/(F_(ck )?? B ?? d_2 )) (d')/D= 0.05 P_u/F_ck = 0.03 Provide min reinforcement =2% Therefore area of longitudinal steel= 2% ?? 940 ?? 800 = 15040 mm^2 Provide 12 bars of 40 mm bars Diameter greater i) 5mm ii) ??(dia of main bar) =1/4(40) =10 mm Least lateral dimension =800 mm 16times dia of main bar = (16??40)=640mm 48 times dia of tie =480 mm Provide 10mm lateral tie at 300 mm c/c as per code ' 4.2.13. Design of footing Moment=1650 KN.m
Safe bearing capacity of soil= 16000 kg/m^2 Unit weight of soil =1500 kg/m^2 Try a footing (3 ?? 1 ?? 0.7)m per meter consider for strip footing Weight of soil above footing W3 = ((3 ?? 8)-(0.94 ?? 0.8)) ?? 2 ??1500 = 684.188KN W2= (3 ?? 8) ??0.7 ??2500 = 42000 kg =412.02 KN Total vertical load = 684.188+ 412.02+ 31.85 = 1128.058 KN Eccentricity of resultant vertical force =1128.05/1650 =0.68 > 0.5 Therefore, base pressure distribution is triangular with part of the footing lifting up :- Width of footing in contact with soil = ( 3/2 ' 0.68) ?? 8 =2.46 m Maximum pressure = ((1128.05 ?? 2))/(1.5 ??8) = 188.08 KN/m^2<196.20 KN/m^2 Pressure at c = ((188.08) (2.46-0.68))/2.46 = 136 .09 KN/ m^2 BM due at B due to back fill in concrete = 1500 ??3+2500 ?? 0.7 = 6250 Kg/m^2 Maximum BM at section g = (136.09 ?? (0.682/2)+(188.08 ' 136.09)?? (0.682/3) )??1.5 = 59.216 KN.m ' Maximum factored BM at section B = (1.5 ??625 ??0.682/2) = 3181.5 kg.m = 31.27 KN.m Effective depth =65 cm Area of steel = ((0.12 ?? 100 ?? 65))/100 = 78 mm^2 per m width Use 12 nos of 415 grade at 100 mm c/c. ' Chapter-5 5.1 RESULT Design of connection: 6 mm weld throughout. Design of column: ISHB 450 with 4cm thick and 40cm long plate on both sides.4cm thick and 40cm long plate on both sides on inner edges. Design of tie runner: Box Section:100 x 100 x4 mm Design of eave girder:220 x 220 x 8 mm Design of principal rafter bracing: Box Section 150 x 150 x 5 mm
Design of purlin: ISWB 200 Design of girts:200 x 100 x 5 mm Design of wall bracing:150 x 150 x 5 mm Supporting member of cap plate 2ISA 200 x 200 x 8 mm Cap Plate Design: 650 x 500 x 20 mm Base plate 900 x 700 x 100 mm Gable column ISMB 600 Anchor bolt 6 numbers of 32 diameter of 4.6 grade. Cap plate bolts 4 numbers of 30 mm diameter 4.6 grade. Tie rafter Strut1 60 x 60 x 4 mm Strut2 100 x 100 x 4 mm Strut 3 150 x 150 x 4 mm Tie 220 x 220 x 8 mm Principal rafter 220 x 220 x 10 mm ' Outcomes ' 5.3Conclusion The procedure of structural design of hanga r in airport according to this project would p rove to be economical and even the safety can be maintained in the structure. We must use built- up section for the column for 50m span because another section may fail in deflection. We can save the material and increase the stiffness of structure with the help of welded connection. Tubular section can serve all aspects and is economical among all sections for short span and lower duty structures. Sole plate and base plate are used for large span trusses. Anchor length of rag bolt increases with increase in span and decreases with increase in grade of concrete. Backfilling of soil in footing trenches will increase the footing capacit y against uplift wind forces. RCC strip footing can be used for large span trusses to resist the uplift load of wind. This strip footing acts as a combined footing and connects all the columns including gable and main columns. Higher dead load and live load can be resisted with the help of higher depth of corrugation galvanized iron sheets As a result; the spacing b etween the purlins can be increased. Door of hangar comprises of same tubular sections as that of wall b racing. ' Chapter-6 References IS 800-2007 General Construction in Steel. SP: 38-1987 Typified design for structures with steel roof trusses. SP: 6(1)-1964 Handbook for Structural Engineers Steel Section. SP:16-1980 Design aids for reinforced concrete. IS 4923 Specification for tubular box section.
IS 9595 Specification for welding. IS 456:2000 Plain and Reinforced Concrete Code for Practice. N Subramanian, Steel Structures, Oxford Publications. Design of Steel Structures, R.P. Rethaliya,AtulPrakashan. Auto-Cad Staad'Pro Tata-Structure Steel Table Sour ce: Essay
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