This set of Prestressed Concrete Structures Multiple Choice Questions & Answers (MCQs) focuses on “Prestressing of Tanks”.
1. Prestressed concrete tanks have been widely used for the storage of:
Explanation: Prestressed concrete tanks have been widely used for storage of fluids, such as water, oil, gas, sewage, granular materials like cement, process liquids and chemicals, slurries and more recently cryogens water storage tanks of large capacity are invariably made of prestressed concrete recent applications include special forms of prestressed concrete tanks, which are triaxially prestressed and serve as containment vessels and biological shields for nuclear reactors.
2. Prestressed concrete although it is water tight, it is not:
a) Gas tight
b) Liquid tight
c) Vapour tight
d) Material tight
Explanation: Tanks have been built for storing liquid oxygen at 230 degrees with capacities up to one million liters and prestressed concrete, although water tight, it is not gas tight were vapours under pressure are to be stored and in such cases, a thin membrane linear of steel provides rigidity and increases the steel tensile capacity of the pretressed concrete.
3. The metal linear concept in prestressed tanks has proved to be success in case of:
a) Air tanks
b) Water tanks
c) Fluid tanks
d) Vapour tanks
Explanation: The metal linear concept has proved so successful that it is being increasingly used in America, even for large water tanks and in the case of sanitary structures like sludge digestion tanks, spherical shapes are preferred and for practical reasons, the tank is made up of a top and bottom conical shell connected by a circular cylindrical intermediate portion.
4. In the case of large tanks, the base slabs is subdivided by:
Explanation: In the case of large tanks, the bars slab is subdivided by joints which are sealed by water stops and the floor slabs are cast in panels and according to the British standard the maximum length of side of such panels should not exceed 7.5m for reinforcement slabs and 6m for nominal slabs and they may be formed out of 50 to 80mm thick gunite reinforced with 0.5 percent of steel distributed in each of the principal directions.
5. The nominal reinforcement provided for floor slabs stipulated by Indian standard code is not less than:
Explanation: The Indian standard code stipulates the floor slabs of tanks resting on the ground should be provided with a nominal reinforcement of not less than 0.15 percent and the floor slabs should be cast in panels of area not more than 4.5m2 with contraction or expansion joints and these slabs are to be cast over a layer of concrete not less than 75mm thick with a sliding layer of bitumen paper provided to prevent the bond between the screed and the floor slab.
6. In the fixed base joint the junction is between the tank wall and:
Explanation: In fixed base joint the junction is between the tank wall and footing is the most vulnerable location as far as leakage is concerned and hence in the case of tanks storing penetrating liquids, it is necessary to form the wall and footing in monolithic construction and this type of connection is generally well suited for shallow tanks with diameters up to 30m, where the fixing moment developed at the wall base does not result in excessively high stresses and congestion of reinforcement.
7. When a sliding joint is made what is interposed at the junction of wall and base:
Explanation: A sliding joint is made by interposing rubber or neoprene pads at the junction of the wall and the base and the preload engineering company has developed this type of sliding base in which a vertical water stop is inserted between two rubber strips and in the present state of art, single neoprene pads have also used and the main function of these pads is to allow for free horizontal movement of the wall relative to the base by shear deformation of the rubber joint, which does not exceed a critical value of 30 degrees.
8. The most common method of wire wrapping for circular tanks consists of:
a) VBR machine
b) Slump cone
c) Cassagrande apparatus
d) Traction machine
Explanation: The most common method of wire wrapping circular tanks consists of a traction machine, and it is suspended from a trolley which runs along the top of the tank walls and the high tensile wire is drawn through a die while it is wound on the tank to achieve the designed tension in the wire and as a precaution the wires are anchored by clips, the wall at regular intervals to ensure that in the event of wire fracture, the winding does not get detached.
9. Calculate minimum wall thickness given a cylindrical prestressed water tank of internal diameter 30m over a depth of 7.5m and the permissible compressive stress at transfer is 13n/mm2 and the maximum compressive stress under working pressure is 1n/mm2 and the loss ratio is 0.75?
Explanation: D = 30m, H = 7.5m, Nd = 720n/mm, ɳ= 0.75, fct = 13n/mm2, pressure is 1n/mm2
T = Nd/ɳfct-fmin.w = 720/ (0.75X 13) – (1) = 82.3mm.
10. Calculate circumferential prestress of a cylindrical prestressed concrete water tank given that the thickness is 12mm, loss ratio is 0.75, the maximum stress under working pressure is 1n/mm2(Nd value is 720)?
Explanation: Nd = 720, fmin.w = 1, ɳ= 0.75, t = 120mm
Fc = Nd/ ɳt+ fmin.w/ ɳ = 720/0.75 x 120+1/0.75 = 9.4n/mm2.
11. Calculate vertical prestressing force if wires of 5mm diameter with an initial stress of 1000n/mm2 are available for circumferential winding and Freyssinet cables made up of 12 wires of 8mm diameter stressed to 1200n/mm2 are to be used for vertical prestressing?
Explanation: 5mm diameter wires stress is 1000n/mm2, 12 wires of 8mm diameter are stressed to 1200n/mm2, fc= (12x1000x200)/(1000) = 2400kn.
12. Calculate circumferential prestress if loss ratio 0.75, thickness is 120mm, working pressure is 1n/mm2 and Nd is given as 840n/mm?
Explanation: Given Nd = 840, fmin.w = 1, ɳ= 0.75, t = 120mm,
Fc = Nd/ ɳt+ fmin.w/ ɳ = 840/0.75×120+ 1/0.75 = 10.75n/mm2.
13. Calculate the spacing of 5mm wires having a loss ratio of 0.075, compressive stress is 10.75n/mm2, 5mm diameter wires stress is 1000n/mm2, 12 wires of 8mm diameter are stressed to 1200n/mm2(Nd = 840n/mm2)?
Explanation: ɳ= 0.075, t = 120mm, internal diameter is 30×103, Nd = 840
S =2×840/0.075x1000x20/10.75x30x103x120) = 11.6mm.
14. Calculate the maximum vertical moment due to prestress if given self weight moment is 16.5kn/m, thickness is 0.115m and loss ratio is 0.0075?
Explanation: Mw = 16.5kn/m, t = 0.115, ɳ= 0.075
Mt = Mw x Wt / ɳ 16500(0.11/0.075) = 25.4×106nmm/m.
15. Find vertical prestressing force if characteristic strength is 8.2, wires are stressed at 1000n/mm2, diameter is 150mm?
Explanation: fc = 8.2, stress = 1000, diameter = 150mm
F = (8.2 x 1000x 150)/1000 = 1230kn.
Sanfoundry Global Education & Learning Series – Prestressed Concrete Structures.
To practice all areas of Prestressed Concrete Structures, here is complete set of 1000+ Multiple Choice Questions and Answers.