Design and Estimation of Dry Dock PDF [PDF]

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CHAPTER 1 INTRODUCTION 1.1



GENERAL



Dry dock is a dock into which the ship floats. The dock gates are closed behind it, the water is pumped out, and the ship rests on the docking blocks ready for its hull to be repaired or cleaned. There are various types of dry docks as follows,      



GRAVING DOCKS FLOATING DOCKS SHIPLIFTS SLIPWAYS TRANSFER SYSTEMS SMALL CRAFT LAUNCHING RAMPS



1.2



GRAVING DOCKS Graving docks are large, fixed basins built into ground at water’s edge, separated from the water by a dock gate. Its basic structure consists of a floor, sidewalls, head(front) wall and a dock gate. Alter may be incorporated into the side walls for structural stability.



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Fig 1.1



Fig 1.2



Fig 1.3



1.3



ADVANTAGES OF A BASIN DOCK  Long life expectancy of the basic structure.  Low maintenance costs. (Dock floor and walls can be built of granite or concrete which last a very long time with little maintenance)  There is no limit to the size of the basin dock.



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 There is no need to worry about ship/dock stability, pumping plans or longitudinal deflection of the dock while docking ships. (Ship stability and block loading must still be addressed, however)  The basin can be equipped with an intermediate gate that allows flooding of the aft half of the dock while the forward half remains dry. 1.4



DISADVANTAGES OF A BASIN DOCK  High initial construction cost.  The basin is a fixed structure, which cannot be moved. Makes it harder to re-sell thus harder to get financing.  Routing of men and material is difficult since floor is below grade.  Ventilation and lighting are not good because one has to work “in a hole”.  It is very difficult to enlarge a basin dock.  Transfer is not possible from a basin dock.  Usually slower to operate (Power is inversely proportional to size).



1.5



TYPES OF BASIN DOCKS There are 3 basic types of basin docks: 1) Full Hydrostatic Dock - A full hydrostatic dock uses its weight or an anchorage system to resist the full hydrostatic head at the maximum water table. 2) Fully Relieved Dock - A fully relieved dock uses a drainage system around the entire dock to drain away the water before it can build hydrostatic pressure on the walls and floor. 3) Partially Relieved Dock - A partially relieved dock uses a drainage system under the dock floor to eliminate the hydrostatic pressure on the floor only. The walls resist the full hydrostatic head. 1.6



ENTRANCE CLOSURES All basin docks must, of course, have an entrance closure that keeps water out of the dock once the ship is in and retracts out of the way for docking and undocking operations. The basic requirements of the entrance closure are:  Ease & speed of installation and removal  Water-tightness  Low maintenance  Feasibility of traffic movement across top  Cost 3



1.7



SLIDING OR ROLLING CAISSONS These are built up box sections with a sliding or rolling surface at the base. The gate slides or rolls into a notch built into the side of the dock. 1.8



ADVANTAGES Fast operating



1.9



DISADVANTAGES  Cleaning and maintenance of rollers or slide paths is difficult.  Operating mechanism is expensive  Major repairs require removal of gate  Recesses must be built into walls.



1.10 GRAVING DRY DOCK OPERATION Most basin docks flood entirely by gravity. A few docks have a super flooding feature, which allows pumping the water inside the dock to a greater elevation than the outside water although this greatly complicates the gate design. There are 3 basic methods of flooding basin docks,  Through culverts built into the walls and connected to floor openings spaced along the dock length.  Through culverts passing transversely under the dock floor near the entrance and with openings leading up to the floor.  Through pipes in the entrance closures (gates). Some common features that are usually incorporated into the basin dock flooding systems are:  Trash racks are placed over inlet openings to prevent the intake of solid matter. The racks should be removable for maintenance and replacement.  Vertical slots should be provided between the trash racks and the sluice gates to accommodate stop logs to shut off water for sluice gate maintenance.  Sluice gates (one for each intake tunnel) control the dock flooding. Basin docks usually have 2 separate dewatering systems.  The primary system, consisting of large high-capacity pumps, performs the main portion of the dock’s deballasting.



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 The secondary system, consisting of smaller pumps, collects the last few inches of water in the basin as well as rain water, flushing water and water from the under drain system. Sand sumps (settling basins) should be located in accessible areas of the water collector channels. These allow abrasive materials such as sand, grit, etc., to settle out of the water before reaching the pump impellers. In general, operation of a basin dock is easier than that of a floating dock. The operator does not have to be concerned with dock deflections, ability or differentially deballasting different ballast tanks under the vessel to provide proper lift as in a floating dock. Ship stability, block loadings and loading of floor slab must be considered, however. Because trim of the keel block line can not be easily adjusted care must be taken to properly trim the vessel to reasonable match the keel block trim or sue loads could develop. This could overload the blocks and affect the stability of the vessel as she lands. On some types of pressure-relieved docks, care must be taken not to dewater the basin too quickly, since the water table in the surrounding soils must be allowed to drop as the basin level drops. This can greatly increase the time required for a docking or undocking evolution. Prior to docking a vessel in a basin dock, the following minimum calculations should be performed:  Stability of vessel afloat  Stability of the vessel at landing on blocks  Stability of the vessel at hauling of side blocks (if applicable)  Block/slab loading calculations  Hurricane/Earthquake overturning calculation. 1.11 STAAD PRO ANALYSIS Our project involves analysis and design of dry dock using popular designing software STAAD Pro. In the initial phase of our project we have done calculations regarding loadings on dry dock. We have chosen STAAD Pro because of its various advantages. STAAD Pro features a state-of-the-art user interface, visualization tools, powerful analysis and design engines with advanced finite element and dynamic analysis capabilities. From model generation, analysis and design to 5



visualization and result verification, STAAD Pro is the professional’s choice of steel , concrete, timber, aluminums, and cold –formed steel design of low and high-rise buildings, culverts, petrochemical plants, tunnels, bridges, piles and much more. It is used to generate the model, which can than be analyzed using the STAAD engine. After analysis and design is completed, the GUI can also be used to view the results.



1.12 OBJECTIVES  To design a dry dock for Chennai harbor using Indian Standard codal provisions.  To draw and draft the layout using AutoCad software package.  To analyze the same using StaadPro software, to serve all types of ships.



1.13 NEED FOR STUDY  There is a British era slipway only present in our Chennai port for maintenance of small ships weighting up to 1000tonnes. So a dry dock is to be constructed in order for maintenance of ships up to 10000tonnes.  This will increase the standard of our Chennai port.  It provides various repair and maintenance processes for all types of ships under the weight and dimensions limits.



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CHAPTER 2 LITERATURE REVIEW 2.1



BUREAU OF YARDS AND DOCKS, FOURTEENTH NAVAL DISTRICT(1940)



Design and construction as main contractor of dry dock, length 130m, width 27m, depth 10.80m DESIGN 1. Organization a. Bureau: The constructions of the dry dock were standard construction, in accordance with plans supplied by the Bureau. b. Station: Designs were not developed by the station. c. Architect and Engineer: Assistance in design was given, as consultants, by F.R. Harris, Inc., of New York, consulting engineers. 2. Criteria (General): Although the Robbins Dry Dock, at Erie Basin, New York,has been designed and constructed using (in part) tremie concrete floor slab (circa 1927), relatively few criteria existed, at the time that the work being discussed was begun, for a structure of the magnitude of Dry Dock No. 2. The conventional (circa 1910), braced, sheet-pile cofferdam method, employed in the construction of Dry Dock No. 1, had failed, and thus necessitated reconstruction, applying the costly laborious floating-caisson design which had consumed some six years. A repetition of the experience was to be avoided. Certain data as the "Robbins" and similar structures were available - and were availed of. Briefly, the philosophy of the design assumed that, when the deck was unwatered, the combined weights of (1) the floor slab and sidewalls; (2) a small part of the frictional wedge of the backfill on the sidewalls; and (3) U.S. NAVAL BASE, PEARL HARBOR, DRY DOCK No. 2 HAER HI-66 some 30% of the theoretical uplift value of the H-section steel piles; would resist the hydrostatic upward pressure. The dock floor was designed as a beam to transmit 7



this upward pressure, or thrust, to the under sides of the walls. As will be elsewhere noted in the text of this report, the construction methods stipulated in the plans and specifications were supplemented, and to some extent modified, by the disclosures reveled by experimental fieldwork. Several weeks before the Japanese attack of December 7, 1941, (and less than twenty months after its construction was begun), Dry Dock No. 2 had been brought to a stage of completion such that it could be and was - used to repair Navy craft affected by the "blitz." Criteria developed from this dock's design and construction (and from those of Dry Dock No. 4, Philadelphia Navy Yard, constructed concurrently) were of inestimable value in facilitating (and thus expediting) "rush" completion of eight of the world's largest dry docks, all built by the Navy under war-time pressure; one of them, the recently completed Dry Dock No. 5, at Pearl Harbor. DETAILS OF SITE The dry docks' (Nos. 2 and 3) location is well suited to the function of docking deep-drift ships, dry Dock No. 2 is on the northerly water frontage of the Pearl Harbor Navy Yard, adjacent to the site of previously-constructed Dry Dock No.1; repair and transportation facilities, power and water, were readily accessible, and had been extensively developed for use by Dry Dock No. 1. Core-boring tests had been made during 1938 and 1939. They showed an overlay of adobe over (successively) volcanic tuff; volcanic sand (loose, strong, hard);limestone, coral-reef formation (hard, coarse, and fine, silty); below the elevation of the floor slab, compact clay (brown and gray); and, still lower, loose, fragmentary limestone formations, extending indefinitely. Tests were run, too, to determine the extent of the abrasive and corrosive effects of coral and salt water on (structural) metal. With the test results known, it was decided that the site was suitable for the projects construction. Designs were developed and the work begun.



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2.2



SOLETANCHE BACHY (2001) -CONCARNEAU DRY DOCK



Design and construction as main contractor of dry dock, length 130m, width 27m, depth 10.80m DESIGN The dry dock is 130m long, 27m wide and 10.80m deep, controlled on the seaward side by trolley-mounted sliding gate. The remote end has a spiral access ramp for more efficient operational use by the commercial companies operating there. A pump room is provided to control wash water and gate leakage. Three pumps can discharge up to 4000 m3 per hour to dewater the dock in four hours when a ship is being docked. There are all the usual fittings conventionally found in harbors works such as bollards, capstans and winches. WORKS One of the challenges facing the consortium was how to deal with the mud covering the lagoon bed to depths of up to 7 meters, considering that the finished dock was to be surrounded by earth platforms for normal harbor operations, with a specified bearing capacity of at least 3 tones per square meter. Excavation of the mud would have been difficult and disposal even more problematical, and it was decided to consolidate it in situ by preloading. Apart from the excavation for the dock itself, therefore, all the mud has been left in place. An interceptor channel was dug to divert the river around the lagoon, then the lagoon was emptied to expose the mud. A geo-textile was laid over the whole area and covered with the same thickness of free-draining gravel. Strip drains were sunk from this platform down to bedrock in a 1-metre square array. The subsequent weight of the fill gradually expelled the water from the mud through sumps collecting the water in the free-draining layer. Settlement of approximately one meter was observed before construction work proper could commence. The dock sidewalls were built as diaphragm walls, tied back at the top with passive anchors to sheet piling and fixed at the bottom by the concrete floor of the dock. 9



The floor is a drained raft to prevent the build-up of uplift. Works on the dock entrance proceeded behind a watertight cofferdam built in the port: pump room, floor under the gate, gate recess (rock excavation with concrete and nail support). The contract required a turnkey graving facility, and ancillary works included a perimeter road around the dock, drinking water, electricity and gas supply, fire-fighting system, two-storey control building and all fittings for ship docking (keel blocks, winched cradles, etc.). One of the last operations was assembly of the dock gate, by assembling four caissons to form a single unit 28m long, 11m high and 4m thick, weighing 170 tonnes. The gate was launched by a nearby boat hoist, towed to station and sunk onto its trolleys, standing ready on their rail tracks.



Fig 2.1



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2.3



MURRAYLANDS DRY DOCK(2009)



INTRODUCTION Dynamic Project Delivery (DPD) was engaged by the Murray lands Regional Development Board (MRDB) on behalf of the Murray lands Dry Dock Working Party to undertake an evaluation to determine the viability of building a dry dock facility in the Mid Murray Council region. DPD were also asked to define the optimum site for construction. The four key stages for the evaluation of the potential to establish a Murray lands Dry Dock facility identified were: 1. To identify an ideal site for the construction of a dry dock 2. To obtain costs, timelines and parameters for the construction of the dry dock 3. To identify potential funding or investors for the construction of the dry dock 4. To recommend the ownership and management structure for the dry dock. The Murray lands Dry Dock Working Party determined that firm concept designs must be obtained and endorsed prior to lodging a Pre-Lodgment Agreement / Development Application, and prior to funding being sought. The development of conceptual designs will therefore form an interim stage between this report and the pre-lodgment process. DRY DOCK DESIGN The Murray lands Dry Dock Working Party determined that firm concept designs must be obtained and endorsed prior to lodging a Pre-Lodgment Agreement / Development Application, and prior to funding being sought. The development of conceptual designs will therefore form an interim stage between this report and the pre-lodgment process. EXAMPLES OF EXISTING DRY DOCKS WERE INVESTIGATED AND STUDIED �Randell Dry Dock , Mannum, was installed in 1873 by William 11



Randell. The dry dock was actually built at Milang, by A.H. Landseer, and towed across Lake Alexandrina by the steamer Nildesperandum. It was during the boomdays that the dock and wharf were used to their capacity due to a huge trading enterprise built by J.G. Arnold. The dry dock now has a heritage listing. � South Brisbane Dry Dock was designed by William D Nesbit, chief engineer for Harbours & Rivers, in 1875. It was constructed between 1876 and 1881 by J & A Overend. The busy Brisbane port required a substantial facility for the maintenance, repair and refitting of commercial ships and Harbours & Rivers dredges, barges and other vessels. The dock was originally 320 feet (97.54 metres) long, but was extended to 420 feet (12.81 metres). The width at the top is 24.08 metres and 16.15 metres at the bottom. The overall depth is 9.75 metres with 5.79 metres at the entrance sill. The caisson (dock gate) was manufactured by the notable firm of RR Smellie & Co. of Brisbane. It is probably the largest locally made wrought iron composition in Queensland. The dry dock site is incorporated in the Queensland Maritime Museum which includes many moveable heritage items, such as the HMAS Diamantina which resides in the dock. � Sutherland Dry Dock, Sydney, NSW, was constructed as a dry dock between 1882 and 1890 under the supervision of the engineer Louis Samuel to supplement the capacity of the smaller Fitzroy dock. Its gate or caisson was originally operated by a steam-driven engine, but later changed to an electric motor in 1915. The dock has been modified several times since then – in 1913 to accommodate the battle cruiser HMAS Australia and in 1927 for the docking of the cruisers HMAS Australia and Canberra. � Entec – Wallsend, Tyneside UK - The proposed dry dock replaces the existing slipways, which are inclined and fall into the River Tyne. These are of reinforced concrete construction, founded over significant areas on bearing piles of steel, concrete and timber.



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Chapter 3 METHODOLOGY



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1CLRD .AmEe PDDiPs leDsOg as tRn ni dagTa vnrtn iaed eowma wf ien nta ghl soy ds s



i t



i



i Fig 3.1 FLOW CHART OF DRY DOCK DESIGN AND ANALYSIS



3.1



LIMIT STATE METHOD The design process of structural planning and design requires not only imagination and conceptual thinking but also sound knowledge of science of structural engineering besides the knowledge of practical aspects, such as recent design codes, bye laws, backed up by ample experience, intuition and 14



judgement. The purpose of standards is to ensure and enhance the safety, keeping careful balance between economy and safety. This design process includes the design of dry dock components Manually. The components of dry dock designed in this process are as follows, 1. Staircase 2. Slab 3. Retaining wall 4. Steel section 5. pile The analysis of the bending moment and deflection is done by the STAAD Pro software. 3.1.1 STAIRCASE DESIGN This design is based limit state method. Tread and rise is taken from the book ‘design and consruction of dry docks’ by ‘b.k mazurkiewicz’. Then using the indiam standard codes IS 456:2000 to calculate its other dimensions and to check whether the design is safe manually. 3.1.2 SLAB DESIGN This design is based limit state method. Depth is taken from the book ‘design and consruction of dry docks’ by ‘b.k mazurkiewicz’. Based on Ly / Lx Value all the slabs are designed in two way method.Then using the indiam standard codes IS 456:2000 to calculate its other dimensions and to check whether the design is safe manually. A two way slab having aspect ratio Ly / Lx < 2 is generally economical compared to one way slab because steel along the spans acts as main steel and transfers the load to all its four supports. The two way action is advantageous essentially for large spans and for live loads greater than 3kN/m^2. for short spans and light loads, steel required for two way slab does not differ appreciably as compared to steel for one way slab because of the requirement of minimum steel.



3.1.3 CANTILEVER RETAINING WALL DESIGN Retaining walls are structures designed to restrain soil to unnatural slopes. They are used to bound soils between two different elevations often in areas of terrain possessing undesirable slopes or in areas where the landscape needs to be shaped severely and engineered for more specific purposes like hillside farming or 15



roadway overpasses. This design is based limit state method. Depth is taken from the book ‘design and consruction of dry docks’ by ‘b.k mazurkiewicz’. Based on L y / Lx Value all the slabs are designed in two way method.Then using the indiam standard codes IS 456:2000 to calculate its other dimensions and to check whether the design is safe manually. 3.1.4 GATE DESIGN It is designed under the conditions of limit state method. Its dimensions are assumed considering heavy loads. The design is processed by considering a I section of steel. Its properties are taken from steel tables and the calculations are made manually to calculate the load of the gate. 3.1.5 PILE DESIGN It is designed considering the load to act on the foundation. Based on the load the depth is decided and pile is designed using the Indian standard codes in IS 6403:1981.then the check of the pile design is done manually. 3.2



CADD DRAWING Its is an software used for drawing the different views of the structures and reinforcement details. It helps the presentation easy and also in correction of dimensions to make the design safe.



Chapter 4 DESIGN AND ANALYSIS 4.1



PILE DESIGN 16



4.1.1 REQUIREMENTS FOR PILE DESIGN    



Types of soil layers Thickness of various layers in soil Standard penetration values (N) Skin friction of soil



4.1.2 SOIL PROPERTIES TABLE 4.1 SOIL PROPERTIES



DESCRIPTION OF SOIL Water



THICKNESS OF SOIL



SPT VALUE(N)



7



0



Silty clay (layer 1)



7



1



Clay sand



1



6



Silty clay (layer 2)



1



19



Silty sand



6



100



cemented sand(layer 1)



10



100



cemented sand(layer 2)



1



100



Hard rock



-



-



4.1.3 DESIGN OF PILE GIVEN DATA Diameter of pile (D) =1m 17



Length of pile (L) = 7m Fcu = 35N/mm2 Cross sectional area of pile (Ac) = (π*D2)/ 4 = 0.785m2 Perimeter of pile = π*D = 3.14m Structural capacity = 0.25* Fcu * Ac = 0.25*35*0.785KN = 6872.23KN END BEARING CAPACITY SPT (N) =100 qb =14*N*(L/D)(KN/m2) L = 7m D =1m qb qb



=14*100*(7/1) KN/m2 = 9800 KN/m3



End bearing capacity = qb * Ac = 9800*0.785 = 7693KN Factor of safety



= 2.5



Allowable end bearing capacity = End bearing capacity / Factor of safety =3077.2KN SKIN FRICTION FOR PILE (NEGATIVE SKIN FRICTION) QS = 9*N*π*D*L Skin friction =5145.92KN Factor of safety =2.5 Factored skin friction= 2058.368KN Total axial load allowed = end bearing + skin friction = 3077.2+2058.368KN Load bearing of pile = 5135.56KN 18



SKIN FRICTION FOR THE PILE (TOTAL SKIN FRICTION) Total skin friction = 24937.95KN Factor of safety = 2.5 Allowable skin friction = 9975.18KN Weight of pile = area of pile * unit weight of concrete = 0.785*25 = 19.63KN



4.2



RETAINING WALL DESIGN



4.2.1 REQUIREMENTS FOR RETAINING WALL DESIGN      



Bulk density for each layer Surcharge pressure Hydraulic pressure Dry soil density pressure Soil pressure due each layer Wave pressure 19



4.2.2 BULK DENSITY TABLE 4.2 BULK DENSITY



Soil texture



Critical bulk density range (g/cc)



clay, silt loam



1.4-1.55



silty clay, silty clay loam, silt



1.4-1.45



clay loam



1.45-1.55



Loam



1.45-1.6



sandy clay



1.55-1.65



sandy clay loam



1.55-1.75



sandy loam



1.55-1.75



sandy loam



>1.75



Based on Harris 1990 and Morris and Lowery 1988



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4.2.3 SOIL PRESSURE TABLE 4.3.1 SOIL PRESSURE Soil layer



Depth(z )



Bulk density(γunsat )



Cohesion(c )



Angle of friction(φ )



Rankine’s coefficient(ka )



Void ratio(e )



Dry density(γd )



Water



7



1



0



30



0.33



0.3



0.77



Silty clay(layer 1) Clayey sand



7



1.45



0



30



0.33



0.3



1.12



1



1.65



0



30



0.33



0.3



1.27



Silty clay(later 2) Silty sand



1



1.45



0



30



0.33



0.3



1.12



6



1.75



0



30



0.33



0.3



1.35



Concrete(laye r 1) Concrete(laye r 2)



10



2.4



0



30



0.33



0.3



1.85



1



2.4



0



30



0.33



0.3



1.85



TABLE 4.3.2 SOIL PRESSURE Soil layer



γsat



γsat -10



K0



Q



Surcharge pressure (q* K0)



Hydraulic head pressure(γw*z)



Total pressure



Cummalativ e pressure



72.5 70



Dry soil density pressure(γsat -10)( q* K0) -39.03 -32.59



Water Silty clay(layer 1) Clayey sand Silty clay(later 2) Silty sand Concrete(laye r 1) Concrete(laye r 2)



2.31 3.35



-7.69 -6.65



0.7 0.7



20 20



14 14



47.47 51.41



47.47 98.88



3.81 3.35



-6.19 -6.65



0.7 0.7



20 20



14 14



10 10



-4.33 -4.66



19.67 19.34



118.55 137.89



4.04 5.54



-5.96 -4.46



0.7 0.7



20 20



14 14



60 100



-25.03 -31.22



48.97 82.78



186.86 269.64



5.54



-4.46



0.7



20



14



10



-3.12



20.88



290.52



21



4.2.4 WAVE PRESSURE



Fig 4.1



GIVEN DATA High sighted wave (ds)= 2.5m wave length L @ D = 2.5m is 48.7m D = 11m h= 11m R = 5m r =5m Hb =3m A1 = 0.6 + 0.5 { [(4*π*D)/ L ] * (1/ [sinh(4*π*D)/48.7])} =0.6 + 0.5 { [(4*π*11) / 48.7 ] * (1/ [sinh(4*π*11)/48.7])} A1 = 0.82 m2 A2 = {(h-ds) / 3h } * {Hb / ds}2 = {(11-6) / 3*11 } * {3 / 6}2 A2 =0.038 m2 A3 = {1-(ds/D)} * {1- (1/cosh[2*π*D / L])} = {1-(6/11)} * {1- (1/cos11[2*π*11 / 48.7])} A3 =0.282 m2 22



P1 = (A1+A2)*W*Hb =(0.82+0.038)*10*3 P1 =25.74 KN/m2 P2 = A3*P1 =0.282*25.74 P2 =7.26KN/m2 P3 ={1-(r /R)} – P1 ={1-(5/5)} – 25.74 P3 =5.148 KN/m2 F = {0.5*(p1+p2)*ds } + { 0.5 (p1+p3)* (ds+hc)} = {0.5*(25.74+7.26)*6 } + { 0.5 (25.74+5.148)* (6+2)} =34.65 + 123.552 F =158.202KN



M =F* hc =158.202*2 M =316.404KN.m



4.2.5 DESIGN OF RETAINING WALL GIVEN DATE Assume the following, 23



Safe bearing capacity(p) = 2000KN/m3 Height of embankment above ground level = 10m Density of soil (w) =18 KN/m3 Angle of repose = 300 Friction between soil and concrete (µ) = 0.5 Use M35 grade concerte and Fe500 steel bars SOLUTION Step 1 Dimensions of retaining wall Minimum depth of foundation = ((p/w)*((1-sin�) / (1+sin�))0.5) = ((2000/18)*(1/3)0.5) = 12.35m Overall depth (H) = 10+12.35 m = 22.35m => 23m Thickness of base slab = (H/12) = (23/12) = 1.92m => 2m Thickness of stem at base = 2m Height of stem(h) = H - 2 = 23- 2 = 21m Width of base slab (b) = 0.5H to 0.6H = 13m Width of heel slab = (((2/3)*13) – 2) = 8.67 – 2 = 6.67 m Width of toe slab = 4.35m Step 2 Design of stem Height of stem (h) = 21m 24



Maximum working moment in stem (M) = (Cp *w*h3) / 6 Where Cp = (1-sin�) / (1+sin�) = (1/3) M = ((1/3)*18*213) / 6 = 9261 KN.m Factored bending moment (Mu) = 1.5*M = 1.5*9261 KN.m = 13891.5 KN.m Limiting thickness of stem at base: Mu = 0.138*fck*b*d2 13891.5 *106 = 0.138*35*1000*d2 d = 1695.9mm adopt effective depth of stem(=2m) and at top(=1m) Mu / (b*d2) = 13891.5*106 / (1000*20002) = 3.47 Step 3 Main reinforcement FROM SP 16, (From table 4) Pt = 0.951 Ast = (Pt*b*d) / 100 = (0.951*1000*2000) / 100 = 19020 mm2 Provide 10 bars of 50mm diameter Astpro = 10*(π/4)*502 mm2 = 19634.9 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*502) / 19634.9 mm = 100 mm Provide 10 bars of 50mm diameter at 100mm centre to centre spacing Distribution reinforcement 25



Ast = 0.12%*b*d = (0.12 / 100)*1000*2000 mm2 = 2400 mm2 Provide 4 bars of 30mm diameter Astpro = 4* (π / 4)*302 mm2 = 2827.43 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*302) / 2827.43 mm = 250 mm Provide 4 bars of 30mm diameter at 250mm centre to centre spacing Step 4 Stability calculations Heel projection = ((2/3)*13) - 2 = 6.67 m



Load calculations Load W1=8.67*2*25 W2 = (4.23*2*25) + 459 W3 = 6.67*21*18 W4 = 5*4.33*25



Magnitude of loads (KN) 440 675.5



Distance from ‘a’ (m) (8.67 / 2) = 4.33 8.67 + (4.33/2) = 10.8 (6.67 /2) = 3.35 10.84



2521.25 541.25 ξW = 4178



Earth pressure P0 = K0*γ*z = 0.8*18*21 = 302.4 KN / m2 Z = ξM / ξW = 5.67 m 26



Moment (KN.m) 1905.2 7319.04 8408.36 5864.44 ξM = 23496.72



Eccentricity (e) = ( z – (b/2)) = 5.67 – (13/2) = -0.83 (b/6) = 13/6 = 2.17 Therefore , e < (b/6) Maximum and minimum pressure at base Pmin = (ξW / b)* ( 1 + (6e / b)) = (4178/13)* ( 1+(6(-0.83) / 13) = 198.26 KN/m2 Pmax = (ξW / b)* ( 1 - (6e / b)) = (4178/13)* ( 1- (6(-0.83) / 13) = 444.49KN/m



Fig 4.2



(318.15 / 6.23 ) = (x / 2) x = 100.52 KN/m (246.33 / 13) = (x / 6.67) x = 126.33 KN/m2 Step 5 Design of heel slab



27



Load W3 = 6.67*21*18 Self weight = 6.67*2*25



Magnitude of load(KN) 2521.261 333.5



Distance from ‘a’(m) 6.67 / 2 = 3.33 6.67 / 2 = 3.33



Moment (KN.m) 8408.4 1112.22 ξM = 9520.62



Deduction Uplift pressure = 198*6.67 (ghi) = 0.5*6.67*126.33



1322.39



6.67 / 2 = 3.33



4410.17



421.3



6.67 / 2 = 3.33



1405.03 ξM = 5815.2



Maximum bending moment in heel slab (M)= 9520.62 - 5815.2 KN.m = 3705.42 KN.m Ultimate moment (Mu) =1.5*M = 1.5*3705.42 KN.m = 5558.13 KN.m Mu / (b*d2) = 5558.13 *106 / (1000*20002) = 1.39 Step 6 Main reinforcement FROM SP 16, (From table 4) Pt = 0.34 Ast = (Pt*b*d) / 100 = (0.34*1000*2000) / 100 = 6800 mm2 Provide 4 bars of 50mm diameter Astpro = 4*(π/4)*502 mm2 = 7853.9 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*502) / 7853.9 mm = 250 mm Provide 4 bars of 50mm diameter at 250 mm centre to centre spacing 28



Distribution reinforcement Ast = 0.12%*b*d = (0.12 / 100)*1000*2000 mm2 = 2400 mm2 Provide 4 bars of 30mm diameter Astpro = 4* (π / 4)*302 mm2 = 2827.43 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*302) / 2827.43 mm = 250 mm Provide 4 bars of 30mm diameter at 250mm centre to centre spacing Step 7 Check for safety against sliding Total horizontal earth pressure (P) = (Ka*w*H2) / 2 = ((1/3)*18*233) / 2 = 1587 KN Maximum possible friction force (W) = 0.5*4178 KN = 2039KN Hence factor of safety against sliding =(W/ P) = 2039 / 1587 = 1.31 < 1.5 Hence a shear key has to be deigned. Step 8 Design of shear key Passive force (Pp ) = Kp*P Kp = (1+sin�) / (1-sin�) =3 P = 1323 KN/m Pp = 3*1323 KN/m2 29



= 3969 KN/m2 If ‘a’ is depth of shear key = 2 m. Total passive force (Pp) = Pp*a = 3969*2 = 7938 KN Factor of safety against sliding = (W+ Pp) / P = (2039 + 7938)/ 1587 = 6.3 > 1.5 Step 9 Check for shear stress at junction of stem and base slab Net working shear force (V) = (1.5*P) – W = (1.5*1587) – 2089 = 291.5 KN Factored shear force (Vu) = 1.5* 291.5 KN = 437.25 KN Nominal shear stress (τv) = Vu / (b*d) = (437.25*103 ) / (1000*200) = 0.218 N/mm2 Pt = (100*Astpro) / (b*d) = (100*19634.6) / (1000*2000) = 0.98 From IS 456:2000 , (TABLE 19, PAGE NO:73) Using Pt = 0.98, we get, τC = 0.84 N/mm2 τC > τv Hence it is safe.



30



4.3 SLAB DESIGN 4.3.1 REQUIREMENTS FOR SLAB DESIGN:    



Hydraulic conductivity Hydraulic gradient Seepage flow Uplift pressure



4.3.2 HYDRAULIC CONDUCTIVITY (By Marsily 1986) TABLE 4.4



MEDIUM



K (m/s)



Coarse gravel



10-1 - 10-2



Sand and gravel



10-1 - 10-5



Fine sand, silts, loess



10-5 - 10-9



Clay, shale, glacial till



10-9 - 10-13



4.3.3 HYDRAULIC GRADIENT Water level (h1) = 12m Water table level(h2) = 7m Length of dry dock(L) = 160m Hydraulic gradient(i)



= (h1-h2)/l 31



= (12-7) / 160 = 0.03



4.3.4 SEEPAGE FLOW BY DARCY’S LAW, Void ratio (e) = 0.3 Porosity (n) = 0.23 Hydraulic conductivity (k) = (1/109) (m/s) Hydraulic gradient (i) = 0.03 Seepage flow = K*i*A = (1/109) *0.03 * 160 * 50 = 0.25cm/s 4.3.5 UPLIFT PRESSURE



32



Fig 4.3



4.3.6 DESIGN OF SLAB 4.3.6.1



DESIGN OF SLAB WITH STAIRCASE LOAD



GIVEN DATA Depth (D) = 5m (Depth is taken from the book ‘DESIGN AND CONSRUCTION OF DRY DOCKS’ by ‘B.K MAZURKIEWICZ ) Assume the following Ly =10m Lx =10m M35 grade concrete with Fe500 steel bars is used Ly / Lx = 10 / 10 => 1 < 2(FROM IS 456:2000, PAGE NO:90) Therefore design two way slab SOLUTION 33



Step 1 Effective span (lx) = (Lx + cover) = (10 + 0.15) m = 10.15 m Effective span (lx) = 10.15 m Step 2 Load calculations i.



Dead load due to self weight of concrete = D*1*unit weight of concrete = 5*1*25 KN/m = 125 KN/m



ii.



Dead load due to weight of staircase = 76.53 KN/m Total dead load = 201.53 KN/m



iii.



Live load due to water = 10*(area excluding staircase area) = 10*(100 – 11.22) KN/m = 887.8 KN/m Live load = 887.8 KN/m



iv.



Floor finished = 0.6 K N/m Total load (W) = 1089.93 KN/m Ultimate load (Wu) = 1.5*total load =1.5*1089.93 KN/m = 1634.89 KN/m Ultimate load (Wu) = 1634.89 KN/m



Step 3 Moment and shear calculations From IS 456:2000 , (TABLE 26 , PAGE NO:91) Assume the condition => ONE LONG EDGE DISCONTINOUS And using Ly / Lx =1.0, we get ,  Negative moment at continuous span (αx = 0.037)  Positive moment at mid-span (αx = 0.028) 34



Bending moment From IS 456:2000 , (CLAUSE D-1.1 , PAGE NO:91) Mux (+ve) = αx*Wu*(lx)2 = 0.028*1634.89*(10.15)2 = 4716.06 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1634.89*(10.15)2 = 6231.94 KN. Long span direction Mux (+ve) = αx*Wu*(lx)2 = 0.028*1634.89*(10.15)2 = 4716.06 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1634.89*(10.15)2 = 6231.94 KN.m Shear force Vux = 0.5*Wu*lx = 0.5*1634.89*10.15 = 8297.07 KN. Step 4 Reinforcement details FROM IS 456:2000 , (G-1.1.b, PAE NO:96) a)



M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 4716.06*106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 4716.06*106 = 2175000Ast – 6.21Ast2 Ast = 2181.89 mm2 Provide 8 bars of 20mm diameter Astpro = 8*(π / 4)*202 mm2 = 2513.27 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*202) / 2513.27 mm = 125 mm. 35



Provide 8 bars of 20mm diameter at 125mm centre to centre spacing b ) M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 6231.94*106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 6231.94*106 = 2175000Ast – 6.21Ast2 Ast = 2889.09 mm Provide 10 bars of 20mm diameter Astpro = 10*(π / 4)*202 mm2 = 3141.5 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*202) / 3141.5 mm = 100 mm. Provide 10 bars of 20mm diameter at 100mm centre to centre spacing Step 5 a) Check for depth: Mmax = 0.138*fck*b*d2 6231.94*106 = 0.138*35*1000* d2 d = 1.135 m < (depth (D) = 5 m) hence it is safe. b) Check for shear: FROM IS 456:2000 (CLAUSE 31.6.2.1 , PAGE NO:57) τv = Vu / (b*d) = (8297.07 *103) / (1000*5000) = 1.7 Pt = (100*Astpro) / (b*d) = (100*3141.5) / (1000*5000) = 0.1 From IS 456:2000 , (TABLE 19, PAGE NO:73) Using Pt = 0, we get τC = 0.29 τC < τv 36



From IS 456:2000 , (TABLE 20, PAGE NO:73) τC max= 3.7 τC max >τv hence it is safe 4.3.6.2 DESIGN OF SLAB WITH CEMENT MORTAR LOAD GIVEN DATA Depth (D) = 4m (depth is taken from the book ‘DESIGN AND CONSRUCTION OF DRY DOCKS’ by ‘B.K MAZURKIEWICZ ) Assume the following Ly =10m Lx =10m M35 grade concrete with Fe500 steel bars is used Ly / Lx = 10 / 10 => 1 < 2(FROM IS 456:2000, PAGE NO:90) Therefore design two way slab SOLUTION Step 1 Effective span (lx) = (Lx + cover) = (10 + 0.15) m = 10.15 m Effective span (lx) = 10.15 m Step 2 Load calculations i.



Dead load due to self weight of concrete = D*1*unit weight of concrete = 4*1*25 KN/m = 100 KN/m



ii.



Dead load due to weight of cement mortar = area of cement mortar* unit Weight of concrete = 0.5*10*1*25 KN/m = 125 KN/m Total dead load = 225 KN/m



iii.



Live load due to water = 10*area of slab = 10*100 KN/m 37



= 1000 KN/m Live load = 1000 KN/m iv.



Floor finished = 0.6 KN/m Total load (W) = 1225.6 KN/m Ultimate load (Wu) = 1.5*total load =1.5*1225.6 KN/m = 1838.4 KN/m Ultimate load (Wu) = 1838.4 KN/m



Step 3 Moment and shear calculations From IS 456:200 , (TABLE 26 , PAGE NO:91) Assume the condition => ONE LONG EDGE DISCONTINUOUS And using Ly / Lx =1.0, we get ,  Negative moment at continuous span (αx = 0.028)  Positive moment at mid-span (αx = 0.037) Bending moment From IS 456:2000 , (CLAUSE D-1.1 , PAGE NO:91) Mux (+ve) = αx*Wu*(lx)2 = 0.028*1838.4*(10.15)2 = 5303.1 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1838.4*(10.15)2 = 7007.67 KN.m Long span direction Mux (+ve) = αx*Wu*(lx)2 = 0.028*1838.4*(10.15)2 = 5303.1 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1838.4*(10.15)2 = 7007.67 KN.m 38



Shear force Vux = 0.5*Wu*lx = 0.5*1838.4*10.15 = 9329.88 KN. Step 4 Reinforcement details FROM IS 456:2000 , (G-1.1.b, PAE NO:96) a)



M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 5303.1*106 = 087*500*Ast*4000 (1 – ((500*Ast) / (1000*4000*35))) 5303.1*106 = 1740000Ast – 6.21Ast2 Ast = 3081.65 mm2 Provide 10 bars of 22mm diameter Astpro = 10*(π / 4)*222 mm2 = 3801.32 mm2



Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*222) / 3801.32 mm = 100 mm. Provide 10 bars of 22mm diameter at 100mm centre to centre spacing b)



M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 7007.67*106 = 087*500*Ast*4000 (1 – ((500*Ast) /(1000*4000*35))) 7007.67*106 = 1740000Ast – 6.21Ast2 Ast = 4087 mm2 Provide 10 bars of 24mm diameter Astpro = 10*(π / 4)*242 mm = 4532.89 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*242) / 4532.89 mm = 100 mm. Provide 10 bars of 24mm diameter at 100mm centre to centre spacing



Step 5 Check for depth Mmax = 0.138*fck*b*d2 39



7007.67*106 = 0.138*35*1000* d2 d = 1.204 m < (depth (D) = 4 m) hence it is safe. Check for shear FROM IS 456:2000 (CLAUSE 31.6.2.1 , PAGE NO:57) τv = Vu / (b*d) = (9329.88 *103) / (1000*4000) = 2.3 Pt = (100*Astpro) / (b*d) = (100*4532.89) / (1000*4000) = 0.11 From IS 456:2000 , (TABLE 19, PAGE NO:73) Using Pt = 0.11, we get < τC = 0.29 τC < τv From IS 456:2000 , (TABLE 20, PAGE NO:73) τC max= 3.7 τC max >τv hence it is safe. 4.3.6.3 DESIGN OF SLAB WITH SHIP AND KEEL BLOCK LOAD GIVEN DATA Depth (D) = 5m Weight of Keel block = 200KN/m (depth and weight of keel block is taken from the book ‘DESIGN AND CONSRUCTION OF DRY DOCKS’ by ‘B.K MAZURKIEWICZ ) Assume the following Weight of ship = 10000 tonnes Ly =10m Lx =10m 40



M35 grade concrete with Fe500 steel bars is used Ly / Lx = 10 / 10 => 1 < 2(FROM IS 456:2000, PAGE NO:90) Therefore design two way slab SOLUTION Step 1 Effective span (lx) = (Lx + cover) = (10 + 0.15) m = 10.15 m Effective span (lx) = 10.15 m Step 2 Load calculations 1. Dead load due to self weight of concrete = D*1*unit weight of concrete = 5*1*25 KN/m = 125 KN/m 2. Dead load due to weight of keel block = 200 KN/m Total dead load = 325 KN/m 3. Live load due to water = 10*(area of slab) = 10*(100) KN/m = 1000 KN/m 4. Live load due to ship = 99640 / 140 = 711.7 KN/m Live load = 1711.7 KN/m 5. Floor finished = 0.6 KN/m Total load (W) = 2037.3 KN/m Ultimate load (Wu) = 1.5*total load =1.5*2037.3KN/m = 3055.95 KN/m Ultimate load (Wu) = 3055.95 KN/m Step 3 Moment and shear calculations From IS 456:200 , (TABLE 26 , PAGE NO:91) 41



Assume the condition => ONE LONG EDGE DISCONTINUOUS And using Ly / Lx =1.0, we get ,  Negative moment at continuous span (αx = 0.028)  Positive moment at mid-span (αx = 0.037) Bending moment From IS 456:2000 , (CLAUSE D-1.1 , PAGE NO:91) Mux (+ve) = αx*Wu*(lx)2 = 0.028*3055.95 *(10.15)2 = 8815.28 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*3055.95 *(10.15)2 = 11648.69 KN.m Long span direction Mux (+ve) = αx*Wu*(lx)2 = 0.028*3055.95 *(10.15)2 = 8815.28 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*3055.95 *(10.15)2 = 11648.69 KN.m Shear force Vux = 0.5*Wu*lx = 0.5*3055.95 *10.15 = 15508.94 KN. Step 4 Reinforcement details FROM IS 456:2000 , (G-1.1.b, PAE NO:96) a)



M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 8815.28 *106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 8815.28 *106 = 2175000Ast – 6.21Ast2 Ast = 4100.99 mm2 Provide 10 bars of 24mm diameter Astpro = 10*(π / 4)*242 mm2 = 4523.19 mm2 42



Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*202) / 2513.27 mm = 100 mm. Provide 10 bars of 24mm diameter at 100mm centre to centre spacing b)



M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 11648.69 *106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 11648.69 *106 = 2175000Ast – 6.21Ast2 Ast = 5440.22 mm2 Provide 14 bars of 24mm diameter Astpro = 14*(π / 4)*242 mm2 = 6333.45 mm2



Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*242) / 6333.45 mm = 100 mm. Provide 14 bars of 24mm diameter at 100mm centre to centre spacing Step 5 Check for depth Mmax = 0.138*fck*b*d2 11648.69 *106 = 0.138*35*1000* d2 d = 1.552 m < (depth (D) = 5 m) hence it is safe. Check for shear FROM IS 456:2000 (CLAUSE 31.6.2.1 , PAGE NO:57) τv = Vu / (b*d) = (15508.94 *103) / (1000*5000) = 3.1 Pt = (100*Astpro) / (b*d) = (100*6333.45) / (1000*5000) = 0.12 43



From IS 456:2000 , (TABLE 19, PAGE NO:73 Using Pt = 0.12 τC = 0.29 τC < τv From IS 456:2000 , (TABLE 20, PAGE NO:73) τC max= 3.7 τC max >τv hence it is safe. 4.3.6.4 DESIGN OF SLAB WITH GATE LOAD GIVEN DATA Depth (D) = 5m (depth is taken from the book ‘DESIGN AND CONSRUCTION OF DRY DOCKS’ by ‘B.K MAZURKIEWICZ ) Assume the following Weight of gate = 5115.4 KN Ly =10m Lx =10m M35 grade concrete with Fe500 steel bars is used Ly / Lx = 10 / 10 => 1 < 2 Therefore design two way slab SOLUTION Step 1 Effective span (lx) = (Lx + cover) = (10 + 0.15) m = 10.15 m Effective span (lx) = 10.15 m Step 2 Load calculations a. Dead load due to self weight of concrete = D*1*unit weight of concrete = 5*1*25 KN/m 44



= 125 KN/m b. Dead load due to weight of gate = 5115.4 / 30 = 170.5 KN/m Total dead load = 295.51 KN/m c. Live load due to water = 10*(area of slab) = 10*(100) KN/m = 1000 KN/m Live load = 1000 KN/m d. Floor finished = 0.6 KN/m Total load (W) = 1296.11 KN/m Ultimate load (Wu) = 1.5*total load =1.5*1296.11 KN/m = 1944.17 KN/m Ultimate load (Wu) = 1944.17 KN/m Step 3 Moment and shear calculations: From IS 456:2000 , (TABLE 26 , PAGE NO:91) Assume the condition => ONE LONG EDGE DISCONTINUOUS And using Ly / Lx =1.0, we get ,  Negative moment at continuous span (αx = 0.028)  Positive moment at mid-span (αx = 0.037)  Bending moment From IS 456:2000 , (CLAUSE D-1.1 , PAGE NO:91) Mux (+ve) = αx*Wu*(lx)2 = 0.028*1944.17 *(10.15)2 = 5608.2 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1944.17 *(10.15)2 = 7410.8 KN.m Long span direction 45



Mux (+ve) = αx*Wu*(lx)2 = 0.028*1944.17 *(10.15)2 = 5608.2 KN.m Mux (-ve) = αx*Wu*(lx)2 = 0.037*1944.17 *(10.15)2 = 7410.8 KN.m Shear force Vux = 0.5*Wu*lx = 0.5*1944.17 *10.15 = 9866.66 KN. Step 4 Reinforcement details FROM IS 456:2000 , (G-1.1.b, PAE NO:96) a) M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 5608.2 *106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 5608.2 *106 = 2175000Ast – 6.21Ast2 Ast = 2597.66 mm2 Provide 8 bars of 22mm diameter Astpro = 8*(π / 4)*222 mm2 = 3041.06 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*222) / 3041.06 mm = 125 mm. Provide 8 bars of 22mm diameter at 125mm centre to centre spacing b) M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck))) 7410.8 *106 = 087*500*Ast*5000 (1 – ((500*Ast) / (1000*5000*35))) 7410.8 *106 = 2175000Ast – 6.21Ast2 Ast = 3440.7 mm2 Provide 10 bars of 22mm diameter Astpro = 10*(π / 4)*222 mm2 = 3801.3 mm2 Spacing = (1000*ast) / Astpro 46



= (1000 * (π / 4)*222) / 3801.3 mm = 100 mm. Provide 10 bars of 22mm diameter at 100mm centre to centre spacing Step 5 Check for depth Mmax = 0.138*fck*b*d2 7410.8*106 = 0.138*35*1000* d2 d = 1.238 m < (depth (D) = 5 m) hence it is safe. Check for shear FROM IS 456:2000 (CLAUSE 31.6.2.1 , PAGE NO:57) τv = Vu / (b*d) = (9866.66 *103) / (1000*5000) = 1.97 From IS 456:2000 , (TABLE 19, PAGE NO:73) Pt = (100*Astpro) / (b*d) = (100*3801.3) / (1000*5000) = 0.11 From IS 456:2000 , (TABLE 19, PAGE NO:73) Using Pt = 0.11 τC = 0.29 τC < τv From IS 456:2000, (TABLE 20, PAGE NO:73) τC max= 3.7 τC max >τv hence it is safe.



4.4



DESIGN OF GATE



Step 1 47



DETERMINATION OF FACTORED LOAD Service load = 160 KN / m Load factor =3 Factored load (w) = service load * load factor = 160 * 3 Factored load (w) = 480 KN / m Step 2 BENDING MOMENT M = (W*L2) / 8 =(480*103*102) / 8 M = 6*106 N.m SHEAR FORCE V =(W*L) / 2 =(480*103*10) / 2 V =2.4 * 106 N Step 3 PLASTIC SECTION MODULUS ZP = MγM / FY = (6*106 * 1.1 ) / 500 ZP = 13.2*103mm3



Step 4 Consider a section ISMB 450 A=92.27cm2 D= 450mm bf=150mm tf=174mm tw=9.4mm rz=18.15cm Zez=1350.7 cm3 Zpz = 1533.36cm3 Zpz / Zez = 1.15 (shape factor) 48



Step 5 To check B / tf = 125 /17.4 = 7.18 < 10.5€ From table 2, (IS 800 : 2007) It is considered to be class 2 . Hence the section is compact. Step 6 Check for shear (i) Vd = {fy / (√3*γm)} * h*tw ={500 / (√3*1.1)} * 10*103*9.4 =24.6*106 N Vd > V Hence safe



(ii) check for high / low shear case = 0.6Vd =0.6*24.6*106 =14.76*106 V 12 steps. Number of steps = 12 steps. Step 2 a) Effective span (l) = ((number of steps) * (tread)) + width of landing beams = ((12)*(0.27)) + 0.5 m = 3.74m. Effective span(l) = 3.74m. b) Thickness (t) = span / 20 = 3.74 / 20 =0.18m => 0.2m Thickness (t) = 0.2m. c) Effective depth (d) = D - cover = 0.2-0.02 m = 0.18m Effective depth (d) = 0.18m. Step 3 Load calculations i.



Dead load on slab (slope) (ws ) = D*1*unit weight of concrete = 0.2*1*25 KN/m = 5 KN/m. 52



ii.



Dead load of slab (horizontal) (w) = (ws(R2+T2)0.5) / T = (5 (0.172+0.272) 0.5) / 0.27 KN/m = 5.9 KN/m



iii.



Dead load of one step = (1/2)*R*T*25 = 0.5*0.17*0.27*25 KN = 0.57 KN Dead load per metre length (w1 ) = dead load of one step / T = 0.57 / 0.27 KN/m = 2.12 KN/m



iv.



Dead load due to finishes (w2 ) = 0.6 KN/m Total dead load = (5+5.9+0.57+2.12+0.6) KN/m = 13.63 KN/m. Dead load = 13.63 KN/m



v.



Live load = unit weight of water * area of staircase = 10 KN/m3 * 3.74 m2 = 37.4 KN/m. Live load = 37.4 KN/m



Total load (W) = dead load + live load = (13.63+37.4) KN/m = 51.03 KN/m Ultimate load (Wu ) = 1.5 * total load = (1.5 * 51.03 ) KN/m = 76.54 KN/m. Total ultimate load (Wu) = 76.54 KN/m. Step 4 a) Bending moment (M) = (Wu * l2) / 8 = (76.54 * 3.742) / 8 KN.m = 133.83 KN.m Bending moment (M) = 133.83 KN.m 53



b) Shear force (V) = (Wu* l) / 2 = (76.54 * 3.74) / 2 KN = 143.1 KN Shear force (V) =143.1 KN. Step 5 a) Main reinforcement details FROM IS 456:2000 , (G-1.1.b, PAE NO:96) M = 0.87*fy*Ast*d (1 – ((fy*Ast) / (b*d*fck)) 133.83*106 = 087*500*Ast*180 (1 – ((500*Ast) / (1000*180*35))) 133.83*106 = 78300Ast – 6.21 Ast2 Ast = 2038.7 mm2 Provide 16 bars of 16mm diameter Astpro = 16*(π / 4)*162 mm2 = 3216.99 mm2 Spacing = (1000*ast) / Astpro = (1000 * (π / 4)*162) / 3216.99 mm = 62.5 mm => 100mm. Provide 16 bars of 16mm diameter at 100mm centre to centre spacing. b) Distribution reinforcement details Ast = (0.3%) * b* d = (0.3 / 100)*1000*180 = 540 mm2 Provide 8 bars of 10mm diameter Astpro = 8*( π / 4)*102 = 628.31 mm2 Spacing = (1000*ast) / Astpro = (1000*( π / 4)*102 ) / 628.31 = 125 mm. Provide 8 bars of 10mm diameter at 125mm centre to centre spacing. Step 6 c) Check for depth 54



M = 0.138*fck*b*d2 133.82*106 = 0.13*35*1000* d2 d = 0.171m < (Effective depth (d) = 0.18m) hence it is safe d) Check for shear: FROM IS 456:2000 (CLAUSE 31.6.2.1 , PAGE NO:57) τv = V / (b*d) = (143*103) / (1000*180) = 0.794 From IS 456:2000 , (TABLE 19, PAGE NO:73) Pt = (100*Astpro) / (b*d) = (100*3216.99) / (1000*180) = 1.78 From IS 456:2000 , (TABLE 19, PAGE NO:73) Using Pt = 1.78 and from interpolation of 0.82 and 0.86 τC = 0.84 τC > τv Hence it is safe.



CADD DRAWINGS



55



Fig 4.6 PLAN



56



Fig 4.7 CROSS SECTION



57



Fig 4.8 RETAINING WALL 58



Fig 4.9 SLAB WITH STAIRCASE LOAD



59



Fig 4.10 SLAB WITH CEMENT MORTAR LOAD 60



Fig 4.11 SLAB WITH SHIP AND KEEL BLOCK LOAD 61



Fig 4.12 SLAB WITH GATE LOAD 62



Fig 4.13 STAIRCASE 63



STAADPRO ANALYSIS



Fig 4.14 SIDE VIEW



Fig 4.15 FRONT VIEW 64



Fig 4.16 BOTTOM VIEW



Fig 4.17 WHOLE STRUCTURE 65



CHAPTER 5 ESTIMATION OF DRYDOCK CONCRETE ESTIMATION RATIO OF THE CONCRETE (1 : 2 : 3) Cement = 15.2 / 1+2+3 = 2.53 per Cu.m Sand = 2.53 x 2 = 5.067 per Cu.m Ballast = 2.53 x 3 = 7.59 per Cu.m RATES ASSUMED Cement = 7650.00 per Cu.m Sand = 700.00 per Cu.m Ballast = 650.00 per Cu.m ESTIMATION OF SLAB DATAS Lenght of Slab = 10 m Breadth of Slab = 10 m Depth of Slab = 5 m ESTIMATION Volume of Slab



Cement Sand Ballast



= lxbxh = 10 x 10 x 5 = 500 Cu.m



= 2.53 x 500 = 1265 Cu.m = 5.067 x 500 = 2533.5 Cu.m = 7.59 x 500 = 3795 Cu.m



TOTAL COST FOR ONE SLAB Cement = Rs. 96,77,250 Sand = Rs. 17,73,450 Ballast = Rs. 24,66,750 TOTAL COST FOR 80 SLABS Cement = Rs. 77,41,80,000 Sand = Rs. 14,18,76,000 Ballast = Rs. 19,73,40,000 66



Total cost for slabs =Rs 111,33,96,000 ESTIMATION OF PILE DATAS No. Of Piles = 80 Height of Pile = 7m Height of Cylinder = 5 m Height of Cone = 2m ESTIMATION FOR CYLINDER Volume of Cylinder



=



π r 2h



= π x 0.5 x 2 x 5 = 3.925 Cu.m Cement Sand Ballast



= 2.53 x 3.925 = 9.93 Cu.m = 5.067 x 3.925 = 19.88 Cu.m = 7.59 x 3.925 = 29.79 Cu.m



COST ESTIMATE Cement = Rs. 75,965 Sand = Rs. 13,916 Ballast = Rs. 19,364 FOR CONE Volume of Cone



Cement Sand Ballast



= 1/3 π r2 h = 1/3 x π x 0.5 x 2 x 2 = 0.523 Cu.m



= 2.53 x 0.523 = 1.323 Cu.m = 5.067 x 0.523 = 2.65 Cu.m = 7.59 x 0.523 = 3.969 Cu.m



COST ESTIMATE Cement = Rs. 10,120 Sand = Rs. 1,855 Ballast = Rs. 2,580 TOTAL COST OF THE PILE (CYLINDER + CONE) 67



Cement Sand Ballast



= Rs. 86,085 = Rs. 15,771 = Rs. 21,943



TOTAL COST FOR 80 PILES Cement = Rs. 68,86,916 Sand = Rs. 12,61,680 Ballast = Rs. 17,55,468 Total cost for piles =Rs 99,04,064 ESTIMATION OF RETAINING WALL THE WALL IS TAKEN AS DIFFERENT SHAPES 1 Triangle 2 Rectangle in Stem 3 Rectangle in Base Slab 4 Square in Shear Key 1.TRIANGLE Volume of Triangle = ½ x l x b = ½ x 1 x 21 = 10.5 Cu.m Cement Sand Ballast



= 2.53 x 10.5 = 26.565 Cu.m = 5.067 x 10.5 = 53.20 Cu.m = 7.59 x 10.5 = 79.69 Cu.m



COST ESTIMATE Cement = Rs. 2,03,222 Sand = Rs. 37,240 Ballast = Rs. 51,798 2.RECTANGLE IN STEM Volume of Rectangle = lxbxh = 1 x 1 x 21 = 21 Cu.m Cement Sand



= 2.53 x 21 = 53.13 Cu.m = 5.067 x 21 = 106.407 Cu.m 68



Ballast



= 7.59 x 21



= 159.39 Cu.m



COST ESTIMATE Cement = Rs. 4,06,444 Sand = Rs. 74,485 Ballast = Rs. 1,03,603 3.RECTANGLE IN BASE SLAB Volume of Rectangle = lxbxh = 13 x 1 x 2 = 26 Cu.m Cement Sand Ballast



= 2.53 x 26 = 65.78 Cu.m = 5.067 x 26 = 131.742 Cu.m = 7.59 x 26 = 197.3 Cum



COST ESTIMATE Cement = Rs. 5,03,217 Sand = Rs. 92,220 Ballast = Rs. 1,28,271



4.SQUARE IN SHEAR KEY Volume of square = l x b x h = 2x2x1 = 4 Cu.m Cement Sand Ballast



= 2.53 x 4 = 10.12 Cu.m = 5.067 x 4 = 20.26 Cu.m = 7.59 x 4 = 30.36 Cu.m



COST ESTIMATE Cement = Rs. 77,418 Sand = Rs. 14,182 Ballast = Rs. 19,734 TOTAL COST OF RETAINING WALL PER CU.M (STEM + BASE SLAB + SHEAR KEY) Cement = Rs. 11,90,301 69



Sand Ballast



= Rs. 2,18,126 = Rs. 3,03,407



TOTAL COST FOR THE LENGTH OF 160 M (LEFT) Cement = Rs. 19,04,48,280 Sand = Rs. 3,49,00,208 Ballast = Rs. 4,85,45,120 TOTAL COST FOR THE LENGTH OF 160 M (RIGHT) Cement = Rs. 19,04,48,280 Sand = Rs. 3,49,00,208 Ballast = Rs. 4,85,45,120 TOTAL COST FOR THE LENGTH OF 50 M (REAR) Cement = Rs. 5,95,15,087 Sand = Rs. 1,09,06,315 Ballast = Rs. 1,51,70,350 TOTAL COST FOR THE WHOLE RETAINING WALL Cement = Rs. 44,04,11,647 Sand = Rs. 8,07,06,731 Ballast = Rs. 11,22,61,790 Total cost for retaining wall = Rs 63,33,80,168 TOTAL COST OF CONCRETE Cement =Rs 122,14,78,563 Sand =Rs 22,38,44,411 Ballast =Rs 31,13,57,258 TOTAL COST OF CONCRETE FOR DRY DOCK= RS 175,66,80,232 STEEL ESTIMATION SLAB WITH STAIRCASE LOAD Main reinforcement 20mm dia @ 2.47kg/m , straight bars @ 125mm c/c No of bars = {(10-0.3) / 0.1} + 1 = 98 bars 70



Length = 10-0.3+(18*0.02) = 10.06m Bent up bars @100mm c/c = {(10-0.3) / 0.1}= 97 bars Length = 10.06+0.03 = 10.09m Total length = (98*10.06)+(97*10.09) =1964.61m Weight = 1978.65*2.47 = 4852.59kg



Distribution reinforcement 20mm dia @ 2.47kg/m , 125mmc/c No of bars =(10/0.125) + 1 = 81 bars Length =10-0.3+(18*0.02) = 10.06m Total length = 10.06*81 = 814.86m Weight = 814.86*2.47kg= 2012.7 kg Total weight = 6865.29kg Cost 1 slab =Rs 60*6865.29 = Rs 4,11,917.4 Cost for 9 slabs =Rs 9*4,11,917.4 =Rs 37,07,256.6 SLAB WITH CEMENT MOTAR LOAD Main reinforcement 24mm dia @ 3.33kg/m , straight bars @ 100mm c/c No of bars = {(10-0.3) / 0.1} + 1 = 98 bars Length = 10-0.3+(18*0.024) = 10.132m Bent up bars @100mm c/c 71



= {(10-0.3) / 0.1}= 97 bars Length = 10.132+0.03 = 10.162m Total length = (98*10.132)+(97*10.162) =1978.65m Weight = 1978.65*3.55= 7025 kg Distribution reinforcement 20mm dia @ 2.99kg/m , 100mmc/c No of bars =(10/0.1) + 1 = 101 bars Length =10-0.3+(18*0.022) = 10.096m Total length = 10.096*101 = 1019.696m Weight = 1019.697*2.99 = 3048.89 kg Total weight = 10073.89kg Cost 1 slab =Rs 60*10073.89 = Rs 6,04,433.4 Cost for 26 slabs =Rs 26*6,04,433.4=Rs 157,15,368.4 SLAB WITH SHIP AND KEEL BLOCK LOAD Main reinforcement 24mm dia @ 3.35kg/m , straight bars @ 100mm c/c No of bars = {(10-0.3) / 0.1} + 1 = 98 bars Length = 10-0.3+(18*0.024) = 10.132m Bent up bars @100mm c/c = {(10-0.3) / 0.1}= 97 bars Length = 10.132+0.03 = 10.162m 72



Total length = (98*10.132)+(97*10.162) =1978.65m Weight = 1978.65*3.55 = 7025 kg Distribution reinforcement 24mm dia @ 3.55kg/m , 100mmc/c No of bars =(10/0.1) + 1 = 101 bars Length =10-0.3+(18*0.024) = 10.132m Total length = 10.132*101=1023.33m Weight = 1023.33*3.55= 3632.82 kg Total weight = 10657.83 kg Cost 1 slab =Rs 60*10073.89= Rs 6,39,469.8 Cost for 42 slabs =Rs 42*6,39,469.8 =Rs 268,57,731.6 SLAB WITH GATE LOAD Main reinforcement 22mm dia @ 2.99kg/m , straight bars @ 100mm c/c No of bars = {(10-0.3) / 0.1} + 1 = 98 bars Length = 10-0.3+(18*0.022) = 10.096m Bent up bars @100mm c/c = {(10-0.3) / 0.1}= 97 bars Length = 10.132+0.03 = 10.126m Total length = (98*10.132)+(97*10.162)=1971.63m Weight = 1971.63*2.99= 5895.514 kg 73



Distribution reinforcement 22mm dia @ 2.99kg/m , 125mmc/c No of bars =(10/0.125) + 1 = 81 bars Length =10-0.3+(18*0.022) = 10.096 Total length = 10.096*81=817.776m Weight = 817.77*2.99= 2445.15 kg Total weight = 8340.324 kg Cost 1 slab =Rs 60*8340.324= Rs 5,00,419.44 Cost for 3 slabs =Rs 3*5,00,419.44 =Rs 15,01,258.32 Total cost for reinforcement of slabs=Rs 4,77,81,514.92 RETAINING WALL STEM Main reinforcement 50mm dia @ 15.432 kg/m @ 100mm c/c No of bars = {(370-0.10) / 0.1} + 1 = 3700 bars Length = 23-top cover-bottom cover+2 hooks = 23-0.05-0.05+(18*0.05) = 23.8m For both sides multiply by 2 =23.8*2+4m( shear key)=51.6m Total lenght = 51.6*3700 = 190920m Weight = 190920*15.432= 2946277.44kg Distribution reinforcement 30mm dia @5.56kg/m @250 c/c 74



No of bars = {(25-0.05-0.05) / 0.250 } +1 = 101 Length = 370 – 2 covers + 6 hooks = 370-0.1+(6*9*0.03)=371.52m Total length = 371.52*101 = 37523.52m Weight = 37523.52*5.56 = 208630.77kg BASE SLAB Main reinforcement 50mm dia @ 15.432 kg/m @ 250mm c/c (TOP AND BOTTOM) No of bars = [{(370-0.10) / 0.25} + 1] * 2(TOP AND BOTTOM) = 2962 bars Length = 13-0.05-0.05+(18*0.05)= 13.8m Total length = 13.8*2962 = 40875.6 m Weight = 40875.6 *15.432= 630792.26kg Distribution reinforcement 30mm dia @5.56kg/m @250 c/c No of bars = [{(13-0.05-0.05) / 0.250 } +1]*2 (TOP AND BOTTOM)= 106 bars Length = 370-0.1+(18*0.03) =370.44m Total length = 370.44*106 = 39266.64m Weight = 39266.64m*5.56=218322.518kg Total weight =4004023.348kg Total cost for reinforcement of retaining wall= Rs 24,02,41,400.9 TOTAL COST FOR REINFORCEMENT OF DRY DOCK = Rs28,80,22,915 75



TOTAL COST OF DRY DOCK = RS204,47,03,148



CHAPTER 6



CONCLUSION Dry dock components slab, retaining wall , staircase , and gate as being designed and analysed by INIDAN STANDARD CODAL , ‘B.K Mazurkiewicz’ (dec 1981)- ‘Design and consruction of dry docks’ AND STADD PRO.



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Finally the project reveals that structure of dry dock is adopted in Indian climatic conditions for economical way of ship and boat under the limits of dry dock designed for repairing and maintaining works.



Refernces



1. Soletanche bachy (2001) - ‘CONCARNEAU DRY DOCK’ , Concarneau – France, journal on design of dry dock.



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2. Francis wentworth-shields, john mowlem company& edmund nuttall sons & company (july 1933) – ‘KING GEORGE V GRAVING DOCK’Southampton's Western , journal on on design of dry dock. 3. ‘B.K Mazurkiewicz’(dec 1981)- ‘Design and consruction of dry docks’ - Book on dry docks. 4. Krishnaraju. N (2003, third edition) “Design of Reinforced Concrete Structures” 5. Dock master training manual 6. INDIAN STANDARD CODE BOOKS: IS 456:2000 IS 456:2007 IS 456:1978 IS 456 (PART 2):1989 IS 6403:1981



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