Prestressed Concrete Structures MCQ Set 1
1. In post tensioning, the concrete units are cast by:
Answer: a [Reason:] In post tensioning, the concrete units are first cast by incorporating ducts or grooves to house the tendons and when the concrete attains sufficient strength, the high tensile wires are tensioned by means of jack bearing on the end face of the member and anchored by wedges or nuts.
2. After the tensioning operation, the space between the tendons and the ducts are:
Answer: b [Reason:] The focus are transmitted to the concrete by means of the end anchorages and when the cable is curved through the radial pressure between the cable and the ducts and the space between the tendons and the duct is generally grouted after the tensioning operation.
3. The Freyssinet system of post tensioning anchorages was developed in:
Answer: c [Reason:] Freyssinet system of post tensioning anchorages which was developed in 1939, gave impetus to the various new systems devised over the years and at present, according to abeles there are over 64 patent post prominent systems used worldwide.
4. The Freyssinet system of prestressing involves the arrangement of wires of number:
Answer: c [Reason:] Freyssinet system of prestressing involves the arrangement of high tensile steel wires of about 12 in number with diameter of 5mm to 8mm forming a group into a cable, cylinder is provided as an anchorage and it has a conical hole at its centre provided with heavy hoop reinforcement these cylinders are is provided as an anchorage and it has a conical hole at its centre provided with heavy hoop reinforcement.
5. In Freyssinet system which element behaves as a guide?
a) Spiral spring
b) Hallow spring
c) Trapezoidal spring
d) Cable spring
Answer: a [Reason:] In Freyssinet system of prestressing spiral spring is provided inside the cable for sufficient clearance between the wires and it even provides a cement grouted channel and the spiral spring behaves as a guide in transferring the reaction to concrete.
6. One of disadvantage of Freyssinet system is:
a) Stresses in the wires are not similar
b) Rapid attainment of stretching force
c) Safeguarding of wires is economical
d) Projection of plug left in concrete
Answer: a [Reason:] The stresses in the wires are not similar, even though all the wires are been stretched together, Freyssinet system involves the application of heavy jacks which are economical force applied for stretching is about 250kn to 500kn which is maximum and is still inadequate.
7. In Gifford-Udall system, the number of methods for prestressing members are:
Answer: c [Reason:] Gifford-Udall system gave three methods for prestressing members by small wedging grips, anchoring of wires by wedging, application of tendons of H.T type with a diameter of about 28mm and this method of prestressing involves the anchoring of wires by wedges, due to which the wires directly fit into the tapered recesses present over the bearing plate.
8. In Gifford-Udall system, which method involves prestressing by H.T type tendon?
a) Method 1
b) Method 2
c) Method 3
d) Method 4
Answer: c [Reason:] Gifford-Udall system consists of 3 methods in that third method of prestressing involves the application of a tendon of H.T type with a diameter of about 28mm and it is quite stronger than 6mm diameter wires.
9. In Lee-McCall system, the prestressing tendons are in the form of:
a) Steel bars
b) Tension bars
c) Compression bars
Answer: a [Reason:] In Lee-McCall system, the prestressing tendons are in the form of high tensile alloy steel bars and the bar size ranges from 12-40mm diameter and after tensioning, the bars are anchored by special threaded nuts which bear against a distribution plate.
10. The drawback of Lee-McCall system is it cannot use:
a) Straight tendons
b) Elliptical tendons
c) Curved tendons
d) Trapezoidal tendons
Answer: c [Reason:] In this system, the forces are transmitted by the bearing at the end blocks while the system eliminates the loss of stress due to anchorage slip by screwing a nut and washer against end blocks and the drawback of this system is that it cannot use curved tendons.
11. The method of tensioning Magnel-Balton system involves:
a) Hydraulic jack and wires
b) Hydraulic jack and bars
c) Multi strand hydraulic jack
d) Jack inserted at centre of beam
Answer: a [Reason:] The method of tensioning Manel-Balton system involves hydraulic jack tensioning two wires at a time, the arrangement of Magnel-Balton system constitutes of a rectangular cable section provided with layers of wires of diameter about 5mm to 8mm such that each layer is supplied with four wires and the arrangement of wires in the geometric pattern is to be considered by supplying spacers/grills at uniform intervals throughout the length of the cable, an adequate resistance against friction for the wires in the of sandwich plates.
12. Which system of prestressing is widely used in India?
Answer: d [Reason:] The Freyssinet anchorage system is widely used in Europe and India consists of a cylinder with a conical interior through which the high tensile wires pass and against the walls of which the wires are wedged by a conical plug lined longitudinally with grooves to house the wires.
Prestressed Concrete Structures MCQ Set 2
1. The prestressed concrete bridge decks generally comprise
a) Precast pretensioned
b) Precast post tensioned
c) Partially pretensioned
d) Partially post tensioned
Answer: a [Reason:] Pretensioned prestressed concrete bridge decks generally comprise pretensioned units used in conjunction with cast in situ concrete, resulting in composite bridge decks which are ideally suited for small and medium spans in the range of 20 to 30m and in general pretensioned girders are provided with straight tendons and to use of seven wire strands has been found to be advantageous in comparison with plain or indented wires.
2. The precast prestressed I and inverted T beams have been standardized by:
a) Pulverization association
b) Cement and concrete association
c) Brick association
d) Steel association
Answer: b [Reason:] The precast prestressed I and inverted T beams have been standardized by the cement and concrete association for use in the construction of bridge decks of spans varying from 7 to 36m and standard I and T units are widely employed in highway bridge beams in USA.
3. The post tensioning is ideally suited for prestressing of:
a) Short span girders
b) Long span girders
c) Effective span
d) Limited span
Answers: b [Reason:] Post tensioning is ideally suited for prestressing long span girders at the site of construction, without the need for costly factory type installations like pretensioning beds and segmental construction is ideally suited for post tensioning work and in this method, a number of segments can be combined by prestressing resulting in an integrated structure.
4. In India, a large number of long span bridges have been constructed using:
a) Simply supported beam
Answer: d [Reason:] In India a large number of long span bridges have been constructed and some of the notable examples being the barak bridge at silchar built in 1960 with a main span of 130m and the lunha bridge in assam with a span of 130m between the bearings.
5. Long span continuous prestressed concrete bridges are built of which type of box girders:
a) Single celled
b) 3 celled
c) Multi celled
d) 4 celled
Answer: c [Reason:] Long span continuous prestressed concrete bridges are built of which type of box girders multi celled box girders segments of variable depth using the post tensioning system, and typical cross sections of post tensioned prestressed concrete bridge decks and salient features of the cantilever construction method using cast insitu segments and precast concrete elements.
6. Calculate Permissible stresses compressive stresses in concrete at transfer and working loads, as recommended given fct = 15n/mm2, fst = 0.8, fcw is 12n/mm2?
Answer: a [Reason:] Permissible stresses compressive stresses in concrete at transfer and working loads, as recommended fct = 15n/mm2 < 0.4fci = (0.45×35) = 15.75n/mm2, fst ratio, ɳ = 0.80, fcw = 12n/mm2 < 0.33fck = (0.33×40) = 13.2n/mm2, fu = ftw = 0.
7. Calculate bending moment of dead load given total load = 14n/mm2, Dead load bending moment dead weight of slab = 12kn/m2, dead weight of w.c = 1.76kn/m2, span length is 10.4m?
Answer: c [Reason:] Dead load bending moment dead weight of slab = 12kn/m2, dead weight of w.c = 1.76kn/m2, total load = 14n/mm2,
Dead load bending moment (mg) = (14×10.42)/8 = 190knm.
8. Calculate effective length of Road on a highway given the overall thickness is 500mm and thickness of wearing coarse is 80mm ( x is 3.6m)?
Answer: a [Reason:] Overall thickness = 500mm and thickness of wearing coarse = 80mm, x is 3.6m
Effective length of road = (3.6+2(0.5+0.08)) = 4.76m.
9. Calculate the mid support section given that Ap = 7980mm2, fck = 60n/mm2, b = 800mm, fp = 1862n/mm2, bw = 200mm, Mu(required) = 16865knm, Dt = 400mm, d = 1750mm, failure by yielding of steel?
Answer: b [Reason:] Ap = 7980mm2, fck = 60n/mm2, b = 800mm, fp = 1862n/mm2, bw = 200mm, Mu(required) = 16865knm, Dt = 400mm, d = 1750mm,
Mu = 0.9dApfp = (0.9x1750x7980x1862) = 2340x 106 nmm = 23402 knm.
10. Calculate failure by crushing of concrete such that bw is 200, d is 1750, fck is 60n/mm2, b is800, Df is 400?
Answer: c [Reason:] bw is 200, d is 1750, fck is 60n/mm2, b is800, Df is 400,
Mu = 0.176bwd2fck+2/3 x 0.8(b-bw)(d-0.5Df)Dffck = ((0.176x200x17502x60)+0.67×0.8(800-200)(1750-0.5×400) 400×60)) = 18253×106nmm.
11. Check for ultimate shear strength such that shear force, Vu = 2834 kn, According to IRC: 18-2000, the ultimate shear resistance of the support section uncracked in flexure loss ratio is 0.8, p is 12000×103, area is 0.88 x 106mm2?
Answer: d [Reason:] Flexure loss ratio is 0.8, P = 12000×103, Area = 0.88 x 106mm2,
fcp = (ɳp/A) = (0.8x12000x103/0.88×106) = 10.9n/mm2.
12. Calculate the slope angle such that eccentricity is 750, length is 40m and stress induced is 1000n/m2?
Answer: b [Reason:] Here, e = 750, l = 40m and stress induced is 1000n/m2,
θ = (4e/L) = (4×750/40×1000) = 0.075.
13. Calculate spacing of a prestressed concrete T slab using 12mm diameter of two legged stirrups such that fc is 415, vcw = 1900kn, balance shear is 934×103?
Answer: a [Reason:] fc is 415, vcw = 1900kn, balance shear = 934kn, using 12mm diameter two legged stirrups spacing,
sv = (0.87x415x2x113x1900/934×103) = 165mm.
Prestressed Concrete Structures MCQ Set 3
1. The prestressed concrete slab systems are ideally suited for:
Answer: b [Reason:] Prestressed concrete slabs systems are ideally suited for floor and roof construction of industrial buildings where the live loads to base supported are of a higher order and the uninterrupted floor space is desirable for which reason longer span between the supporting elements are required.
2. The precast prestressed hollow core slabs, with or without topping is important structural element in:
Answer: b [Reason:] Precast prestressed hollow core slabs, with or without topping are important structural elements in industrialized and large panel building construction and the slabs, produced on long casting beds using the pretensioning systems and cut to shorter specified span lengths, are mainly used in one way floors which are freely supported by transverse walls or base.
3. Prestressed pretensioned cored slabs with differ types of cavities are widely used as:
a) Floor panels
c) Wall coatings
Answer: a [Reason:] Prestressed pretensioned cored slabs with differ types of cavities are widely used as Floor panels of civil and industrial buildings in erstwhile U.S.S.R Graduck reports that these panels are produced in multiples of 200mm nominal width and lengths from 3.6 to 6.4m and hollow panels of oval cavity type are most economical for larger spans since they contain the least volume of concrete as compared to round cavity panels and prestressed concrete ribbons have been used as reinforcement for hollow-cored slabs and these consist of tensioned wires or strands embedded in high grade concrete of star of rectangular cross section.
4. One way slabs may be supported across the entire width of the slab by means of:
Answer: b [Reason:] One way slabs may be supported across the entire width of the slab by beams, piers or abutments or bearing walls, which are positioned perpendicular to the longitudinal axis of the span or the supports may be at an angle of the span directions and one way slabs may be continuous over one or several support.
5. The simple or continuous slabs are analyzed for:
a) Design foundation
b) Design reinforcement
c) Design moments
d) Design slab
Answer: c [Reason:] The simple or continuous slabs are analyzed for design moments by considering a unit width of the slab and the prestressing force and the eccentricity of the cable required at prominent sections to resist the dead and live load moments are determined and the spacing of the cables or wires fixed based on the availability of type of tendon.
6. The design of a two-way-slab supported on all four sides involves the computation of:
b) Cross sections
c) Bending moment
Answer: c [Reason:] The design of a two way slab supported on all four sides involves the computation of bending moment in the principal directions of the slab and the slab may be supported on masonry walls or beams and mayor may not be continuous over the supports and transverse loads are resisted by the development of two way slab action, resulting in moments in longer and shorter span directions.
7. The moment coefficients derived from the ultimate load method are generally lower in:
Answer: d [Reason:] The moment coefficients derived from the ultimate load method are generally lower in magnitude than those evaluated from elastic theories thus naturally resulting in savings in reinforcement and however slabs designed by the ultimate load method should be checked service loads according to the principle of limit state design.
8. A simple prestressed flat slab is generally supported by a network of:
Answer: b [Reason:] A simple prestressed flat slab is generally supported by a network of columns without beams and prestressed in two perpendicular directions and the design of typical simple flat slab involves the analysis of moments in the two principal directions so that cables may be arranged to resist these moments and the slab is analyzed as one way slab and the total number of cables required to resist the moments in each of two principal directions are determined.
9. The proportioning of tendons in design of prestressed concrete simple flat slab between the column and middle strips may be based on:
Answer: b [Reason:] The column strips being stiffer than the middle strips, a greater percentage of the tendons are housed in the column strips and the proportioning of the tendons between the column and middle strips may be based on the provisions of codes such as IS:456 and BS:8110, where column strips share a higher proportion of total moment.
10. The design principles of continuous flat slab floors are similar to:
a) One way slab
b) Two way slab
Answer: b [Reason:] The design of continuous flat slab floors are similar to those of two way reinforced concrete slabs and a strip of slab of unit width continuous over supports is analyzed as continuous beam and its prestressing results in secondary moments.
Prestressed Concrete Structures MCQ Set 4
1. Prestressed concrete tanks have been widely used for the storage of:
Answer: c [Reason:] 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
Answer: a [Reason:] 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
Answer: b [Reason:] 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:
Answer: b [Reason:] 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:
Answer: c [Reason:] 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:
Answer: b [Reason:] 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:
Answer: a [Reason:] 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
Answer: d [Reason:] 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?
Answer: b [Reason:] 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)?
Answer: a [Reason:] 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?
Answer: b [Reason:] 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?
Answer: a [Reason:] 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)?
Answer: b [Reason:] 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?
Answer: c [Reason:] 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?
Answer: b [Reason:] fc = 8.2, stress = 1000, diameter = 150mm
F = (8.2 x 1000x 150)/1000 = 1230kn.
Prestressed Concrete Structures MCQ Set 5
1. The tendons in the pretensioning system are tensioned between:
a) Rigid anchorages
b) Hydraulic jacks
c) Concrete beds
d) Variable beams
Answer: a [Reason:] In the prestressing system, the tendons are first tensioned between rigid anchorage blocks cast on the ground or a column or a unit mould type pretensioning bed, prior to the casting of concrete in the moulds.
2. When the concrete attains sufficient strength, which elements are released?
b) Casting bed
Answer: a [Reason:] High early strength concrete is often used in factory to facilitate early stripping the reuse of moulds and when the concrete attains sufficient strength, the jacking pressure is released and the edge of tendon at its either side is formed to an abutment and its other edge is to be pulled with the application of jack.
3. Which is one of the systems used for pretensioning?
a) Magnel-Balton system
b) Freyssinet system
c) Gifford-Udall system
d) Hoyer’s long line method
Answer: d [Reason:] Hoyer’s long line method is the system used in pretensioning and the other systems like Freyssinet, Gifford-Udall, and Magnel-Balton are post tensioning systems and large numbers of beams are produced in an individual alignment.
4. Hoyer’s system of pre tensioning is generally adopted for:
a) Small scale members
b) Large scale members
c) Middle span members
d) End members
Answer: b [Reason:] Hoyer’s system is generally recommended when the production of pretensioned members is required on a large scale is the principle which are used to precast the beams the post tensioning system was considered to pretensioning system when used for large spans, due to this reason the pre tensioning system was replaced by post tensioning system.
5. The transfer of prestress of concrete is achieved by:
c) Steel bars
Answer: d [Reason:] The transfer of prestress to concrete is usually achieved by large hydraulic or screw jacks by which all the wires are simultaneously released after the concrete attains the required compressive strength generally strands of up to 18mm diameter and high tensile wires of up to 7mm diameter anchor themselves satisfactorily with the help of surface bond and the interlocking of the surrounding matrix in the micro indentations on the wires.
6. The bond of prestressing wires in Hoyer’s system can be formed by:
a) Helical crimping
b) Tangential crimping
c) Circular crimping
d) Diode crimping
Answer: a [Reason:] Bond of prestressing wires may be considerably improved by forming surface indentations and by helical crimping of wires in Hoyer’s system, strands have considerable better bond characteristics than plain wires of equal cross sectional area supplementary anchoring devices are required when single wires of larger diameter are used in the pretensioned units.
7. The Hoyer’s method of prestressing is done by:
a) Pulling out of wires
b) Pushing wires
c) Heating of wires
d) Stressing of wires
Answer: a [Reason:] Hoyer’s system of prestressing involves pulling out of wires between two bulkheads which are separated at large distances in order to produce a larger number of beams in an individual alignment, the concrete is to be poured by providing appropriate shuttering between the beams and then the wires are united after hardening of concrete and cutoff.
8. The Hoyer’s system of pretensioning can be done for beams:
b) More than 2
c) Less than 2
Answer: b [Reason:] United beams more than 2 are hardened in Hoyer’s method of pretensiong in case of large distances between each beam, the most commonly used devices are the Weinberg clip developed in France and the Dorland clip developed in the united states these clips are clamped on to the tensioned wires close on the end diaphragms of the units before concreting operations.
9. The Hoyer’s system of prestressing proves to be economical for:
a) Pre tensioning system
b) Post tensioning system
c) Beam casting
d) Bed casting
Answer: b [Reason:] The post tensioning system was considered as economical when compared to pre tensioning system, when used for larger spans due to this reason pre tensioning system was replaced by post tensioning system, for mass production of pretensioned elements, the long line process developed by Hoyer is generally used in the factory and in this method the tendons are stretched several hundred meters apart so that a number of similar units may be cast along the same group of tensioned wires.
10. In Hoyer’s system the projection of plugs left in concrete exceeds beyond:
a) Middle of member
b) End of member
c) First of member
d) Transfer part of member
Answer: b [Reason:] One of the disadvantages of Hoyer’s system is the projection of plugs which are left in concrete exceeds beyond the end of the member, application of heavy jacks which are uneconomical, projection of plugs which are left in concrete exceeds beyond the ends of the member, additional reinforcement is required to prevent failure of shear.