Prestressed Concrete Structures MCQ Set 1
1. What are necessary to prevent failure at the end zones, in the transfer zone of pretensioned beams?
Answer: a [Reason:] In the transfer zone of pretensioned beams, transverse reinforcements are necessary to prevent the failure of the end zones due to cracking of concrete as a consequence of large transverse tensile stresses, which often exceed the tensile strength of concrete.
2. The designed anchorage zone in the main reinforcement using transverse stress distribution should withstand the:
a) Compression tension
b) Bursting tension
c) Anchorage tension
d) Principal tension
Answer: b [Reason:] By using the transverse stress distribution the designed anchorage zone in the main reinforcement should withstand the bursting tension and coincides with the line of action of the largest individual force on the critical axis.
3. The arrangement of reinforcement in end blocks like links, loops, helices, mats are placed in:
a) Linear directions
b) Parallel directions
c) Perpendicular directions
d) Angular directions
Answer: c [Reason:] The arrangement of reinforcement for Freyssinet type of anchorages in end blocks the links or loops, helices, mats are placed in the perpendicular directions of the reinforcements, the reinforcement is provided in the form of closed stirrups enclosing all the tendons, wherever single leg stirrups are used care should be taken to anchor the stirrups to bottom and top tendons with cross section.
4. Which is more efficient than mat reinforcement?
a) Helical reinforcements
b) Loops reinforcements
d) Hair pin bars
Answer: a [Reason:] When compared to mat reinforcements, the helical reinforcement is more efficient and it is tested by Zielinski and Rowe, the first stirrup should be placed at close to the end face as possible with due regard to the minimum cover reinforcements.
5. Which are necessary in case of short bond length, along with the deformed bars, loops?
a) Parallel bends
b) Right angle bends
c) Transverse bends
d) Tensile bends
Answer: b [Reason:] Along with the deformed bars, loops, hooks or right angle bends are necessary in the case of short bond lengths; about half of the total reinforcement is preferably located within a length equal to one third of transmission length from the end, the rest being distributed in the remaining distance.
6. What should be provided in case of end zone reinforcements to prevent failure of corner zones?
c) Hair pin bars
d) Transverse bars
Answer: c [Reason:] To prevent failure in case of end zone reinforcement of corner zones, where spalling or secondary tension develops at the corners, the hair pin bars of suitable steel should be provided, while designing the formwork, the use of cap cables must be considered and the space provided for fixing and handling the hydraulic jack must be sufficient especially at the soffits of beam when cap cables are used.
7. The secondary reinforcements can be bent if the suitable packets are provided behind the:
Answer: d [Reason:] The secondary reinforcements can be bent if the suitable packets are provided behind the anchorages and after the operations of prestressing are done, the pocket is filled with mortar, In case of uncracked members bond stresses are computed by considering the complete section, which is effective.
8. In case of end blocks, the steel case should be provided with bearing plates to overcome:
Answer: a [Reason:] In case of end blocks, the steel cage should be provided to overcome the overlapping of it with bearing plates when they are placed close to the edge of the block, in case of cracked flexural members, bond stresses change suddenly at the cracks due to the abrupt transfer of tension from concrete to steel in the vicinity of the cracks.
9. The cost of end anchorage steel compared to entire structural element is:
Answer: b [Reason:] The cost of end anchorage steel is low when compared to the entire structural members so, Morice, advices that it is better to provide extra reinforcement in doubtful situations in case of anchorage zone reinforcements and proper compaction of concrete in the end zones by vibration is essential to achieve dense concrete associated with high strengths.
10. In case of designing the formwork, the use of cab cables is:
b) Not considered
c) Made equal
d) Made Zero
Answer: a [Reason:] While designing the formwork, the use of cap cables must be considered and the space provided for fixing and handling of the hydraulic jack must be sufficient especially at the soffits of the beam when the cap cables are used.
Prestressed Concrete Structures MCQ Set 2
1. In partially prestressed members, to which extent tensile stresses are permitted:
Answer: b [Reason:] In partially prestressed members limited tensile stresses are permitted in concrete under service wads with controls on the maximum width of cracks and depending upon the type of prestressing and environmental conditions and the use of partial prestressing was first proposed by Emperger in 1939 and further progress in this field was mainly due to the sustained work of abeles, Goschy and others and the west German code DIN 42227 has provided for partial prestressing even before the C.E.B-FIP provisions were introduced in 1970.
2. Which type of reinforcement is required for the advantage of partial prestressing?
a) Un tensioned reinforcement
b) Tensioned reinforcement
c) Forced reinforcement
d) Embedded reinforcement
Answer: a [Reason:] The main point in favour of partial or limited prestressing is that required in the cross section of a prestressed member for various reactions such as to resist the differential shrinkage temperature effects and handling stresses.
3. The reinforcement can cater for which requirements:
a) Limited requirement
b) Serviceability requirements
c) Range requirements
d) Termed requirements
Answer: b [Reason:] The reinforcement can cater for the serviceability requirements such as control of cracking and partially for the ultimate limit state of collapse which can result in considerable reductions in the costlier high tensile steel and the saving in prestressing steel contributes to an overall saving in the cost of structure.
4. The fully prestressed members are prone to excessive:
a) Downward deflections
b) Upward deflections
c) Side deflections
d) Transverse deflections
Answer: c [Reason:] Fully prestressed members are prone to excessive upward deflections especially in bridge structures where dead loads form a major portion of the total service loads and these deflections may increase with time due to the effect of creep.
5. The energy capacity is comparatively low in case of:
Answer: a [Reason:] It is well established that fully prestressed members due to their higher rigidity have a lower energy absorption capacity in comparison with partially prestressed members, which exhibit a ductile behavior.
6. Which are used as untensioned reinforcement?
a) Yield steel and mild steel
b) High tensile steel and mild steel
c) Tensile steel and mid steel
d) Principle steel and mild steel
Answer: b [Reason:] High tensile steel and mild steel have been used as un tensioned reinforcement and the present practice is to use high yield strength deformed bars which are considerably cheaper than prestressing steel and at the same time have higher yield strength and better crack control characteristics by virtue of their surface configuration as compared to mild steel bars with plain surface.
7. The width of crack is influenced by:
b) Limit states
d) Factor of safety
Answer: a [Reason:] The method of calculating the crack width is of considerable importance in checking the limit state of cracking at service loads and it is well established that the width of crack primarily depending upon the stress in the reinforcement is also influenced by the cover and the type of reinforcement.
8. The calculation based on the stress in the reinforcement is obtained by:
a) Limit state theory
b) Convential theory
c) Retardation theory
d) Principle theory
Answer: b [Reason:] Several empirical relations have been developed to estimate the width of cracks but it is considered that a calculation based on the stress in the reinforcement obtained by the convential theory of cracked reinforcement concrete section is inherently more accurate than a computation based on the fictitious tensile stress in an uncracked section.
9. The tensile strength of the concrete below the neutral axis is neglected in:
a) Cracked section analysis
b) Uncracked section analysis
c) Beam section analysis
d) Strain analysis
Answer: a [Reason:] The cracked section analysis of a partially prestressed flange section with tensioned high tensile steel and un tensioned reinforcement is carried out under the following assumptions: The strain distribution across the section is linear, the tensile strength of the concrete below the neutral axis.
10. The stresses and strains developed and the forces acting on cracked prestressed concrete sections which are subjected to a moment Mcr is given as:
a) εpe = P/ApEp
b) εpe = P/AeEp
c) εpe = P/AsEp
d) εpe = P/AlEp
Answer: c [Reason:] In excess of the cracking moment Mcr just prior to the application of the moment the tensile strain in the prestressing steel is εpe and the compressive strain in the concrete at the tendon level is εce and these strains can be evaluated from the prestressing force acting at this stage εpe = p/ApEp , εce = P/Ec(e2/Ic+1/Ac), P = effective prestressing force, e = eccentricity of tendons, Ac = cross sectional area of the concrete section, Ic = second moment of area of the concrete section.
Prestressed Concrete Structures MCQ Set 3
1. The restoration of structures by rectifying constructional deficiencies are as a result of:
a) Cross section
Answer: c [Reason:] The range of application of this strengthening technique is ideally suited for the following situations: restoration of structures by rectifying constructional deficiencies that impair the safety of the structure as a result of faulty dimensioning, corrosion of reinforcement, insufficient reinforcement, overloading etc.
2. The first attempt to strengthen concrete flexural elements by externally bonded steel plates was attempted in:
Answer: b [Reason:] The first reported attempts to strengthen concrete flexural elements by externally bonded steel plates were attempted in France around 1964-65 and practical applications date back to 1966-67 in France and South Africa followed by Japan and Russia.
3. The load bearing capacity of a structural element can be increased by changing:
Answer: c [Reason:] The strengthening of an existing structural element by increasing its load bearing capacity and altering the load supporting system by changing spans by shifting or removing of supports, conversion of continuous beam to single span beam and vice versa etc.
4. In Switzerland the externally bonded steel plates were used for:
Answer: a [Reason:] In Switzerland this method has been extensively used in both buildings and bridges and experiments conducted have shown that reinforced concrete beams, when epoxy bonded with steel plates on the tension face it exhibits significant increase up to 3% in the ultimate flexural strength in comparison with non plated beams.
5. Which grade of structural steel is suitable for bonded reinforcing plates?
c) Any grade
Answer: c [Reason:] Generally any grade of structural steel is suitable for bonded reinforcing plates and plate gauges below 3mm are not suitable because sand blasting can deform them and steel plates between 6-16mm thick were used in some strengthening works.
6. The pretreatment of concrete surface is generally carried out by:
a) Sand blasting
Answer: a [Reason:] Pretreatment of the concrete surface is generally carried out by sand blasting, shotcrete blasting, grinding or roughening with pneumatic needle gun or granulating hammer and the grain structure of the concrete must be exposed before the steel plates are fixed.
7. The tensile shear strength of the adhesive is initially proportional to the square root of:
Answer: b [Reason:] The adhesive joint is generally between 1-3mm thick and tests have shown that the tensile shear strength of the adhesive is initially proportional to the square root of the thickness however the tensile shear strength reaches a maximum and then starts decreasing as the adhesive thickness is further increased and hence the thinner layers prove stronger and have greater resistance than thick ones.
8. The repair procedure in severe damage includes:
Answer: c [Reason:] The detailed examination of the damage and review of calculations will help in selecting a cost effective and appropriate restoration technique for the damaged structure and the loss of prestress is excessive resulting in tensile cracks, preloading method should be seriously considered in making concrete repairs in order to restore the equivalent full or partial prestresss effect, as per original designs and the repair procedure may also include epoxy resin pressure injection, Shortcrete and additional welded fabric with drilled anchors and guniting.
9. Before patching the exposure strands should be coated with:
a) Thick cement grout
b) Slurry cement grout
c) Hardened cement grout
d) Watery cement grout
Answer: b [Reason:] In moderate damage it is recommended that welded wire fabrics be attached to drilled dowels placed at about 500mm spacing or to the existing reinforcement in the damaged area and if the prestressing strands are exposed, sufficient care must be taken so as not to damage the steel during the cleaning operation and the exposed strands should be coated with epoxy resin bonding compound or slurry cement grout before patching.
10. Elastomers have excellent adhesion to concrete and are not susceptible to:
Answer: d [Reason:] Elastomers have excellent adhesion to concrete and are not suspectable to softening within the normal range of ambient temperatures and normally Elastomers exhibit a higher degree of elongation of as much as 10 percent extension but in practice, this should be limited to 50 percent and the groove depth to width ratio should be 1.2, the material should be prevented from adhering to the bottom so that the crack remains free as a live crack.
Prestressed Concrete Structures MCQ Set 4
1. In case of composite sections, the percentage of tensioned reinforcement is less than:
a) Simple beams
b) Complex beams
c) Tee beams
d) Edge beams
Answer: a [Reason:] In case of composite sections the percentage of tensioned reinforcement is less than that in most simple beams, so that the section is invariably under reinforced and the compression zone generally consists entirely of insitu concrete of lower compressive strength, and the value of the cube strength of concrete to be used in flexural strength equations will obviously be that of insitu cast concrete.
2. If the compression zone contains part of precast element, the computation is done by considering:
a) Grades of concrete
b) Cross section
d) Strength characteristics
Answer: b [Reason:] If the compression zone contains a part of precast element, the average compressive strength computed by considering the cross sectional area of insitu and precast concrete is used in the computation of compressive force.
3. The effective bonding between the two parts of composite beams is developed by:
Answer: b [Reason:] Effective bending between the two parts of a composite beam may be developed by providing castellation in the precast unit or by roughening the contact surface of the precast unit before placing the insitu concrete or by projecting reinforcements from the precast unit which serve as ahead connectors.
4. The design of shear connections depends upon the strength of:
a) Ex situ cast concrete
b) In situ cast concrete
c) Tampered concrete
d) Prestressed concrete
Answer: b [Reason:] In the design of shear connection it is generally assumed that the natural bond at the interface contributes a part of the required shear resistance depending upon the strength of the insitu cast concrete and the roughness of the precast element and any extra shear resistance over and above this should be provided by shear connectors.
5. When ties are not provided the permissible values of the horizontal shear stress is:
Answer: a [Reason:] The permissible values of the horizontal shear stress for different types of contact surfaces is specified as: 0.6n/mm2 when ties are not provided and the contact surface of the precast element is free of laitance and intentionally roughened to an amplitude of 5mm and 25n/mm2 when minimum vertical ties are provided and the contact surface is roughened to an amplitude of 5mm.
6. The ties consisting of single bars and multiple leg stirrups should have a spacing not exceeding:
Answer: a [Reason:] The ties consisting single bars, multiple leg stirrups or vertical legs of welded face fabric should not have a spacing not exceeding four times the least dimension of the supported element nor 600mm whichever is less the Indian standard code IS: 1343 does not make any specific recommendations regarding the shear stresses in composite sections.
7. If the surface is rough tamped and without links to withstand a horizontal shear stress of 0.6n/mm2 and assuming the moduli of elasticity of precast and insitu cast concrete to be equal the centroid of the composite section is located 110mm from the top of the slab, second moment area of composite section is 2487×105mm4, calculate ultimate shearing resistance?
Answer: b [Reason:] vu = ultimate shearing force, τ = vus/ln , s= (350x30x95)mm3, b = 80mm, τ = 0.6n/mm2, vu= (0.6x2487x105x80/350x30x95) = 12000n = 12kn.
8. Design ultimate horizontal shear stress with nominal links and the contact surface are as cast (assume τ = 1.2n/mm2)?
Answer: d [Reason:] The ultimate shear resistance is expressed as vu= (τlb/s) = (1.2x2487x105x80/350x30x95) = 24000n = 24kn, τ = 1.2n/mm2.
Prestressed Concrete Structures MCQ Set 5
1. Which of failure is considered in the design of prestressed concrete members for the limit states of collapse?
a) Total failure
b) Shear failure
c) Ultimate failure
d) Collapse failure
Answer: c [Reason:] The distribution of compressive stress in the section at the ultimate failure stage is considered and the effective depth required is estimated by equating the total ultimate moment with the internal resisting angle.
2. The maximum design value of the moment of resistance of rectangular flanged section varies from:
a) 0.08fckbd2 to 0.2fckbd2
b) 0.11fcckd2 to 0.5fckbd2
c) 0.15fcckd2 to 0.8fckbd2
d) 0.21fcckd2 to 0.9fckbd2
Answer: a [Reason:] A comparative analysis of the various code recommendations indicate that the maximum design value of the moment of resistance of rectangular and flange of sections vary from 0.08fckbd2 to 0.2fckbd2, depending upon the recommendations of the stress block parameters.
3. It is often preferable to use a larger section, because it means a saving on:
Answer: a [Reason:] The maximum ultimate moment of resistance of a resistance of a rectangular section rectangular section according to the Indian standard code IS: 1343-1980 is given by:
Mud = 0.21fckbd2, the dimension based on this expression are the minimum values and it often preferable to use a larger section because it means a saving on the costly prestressing tendons.
4. The area of high tensile and un tensioned reinforcement required to mobilize the desired flexural strength is computed using?
a) Tension equilibrium
b) Force equilibrium
c) Transverse equilibrium
d) Rotational equilibrium
Answer: b [Reason:] The area of high tensile and un tensioned reinforcement required to mobilize the desired flexural strength is computed using the force equilibrium at the limit state of collapse and the force equilibrium rather depends upon the pressures exerted in different conditions.
5. A pretensioned prestressed concrete beam of rectangular section is required to support a design ultimate moment of 100knm. Design the section if fck is 50n/mm2 and fp 1600n/mm2, xu/d is 0.5?
Answer: a [Reason:] Mu = 0.36fckbxu(d-0.42xu) = 0.14fckbd2, b = 0.5d, d3 = (mu/0.14fck0.5) = (100×106/0.15x50x0.5), d = 300mm, b = 150mm, xu = (0.5×300) = 150mm for (xu/d) = 0.5, fpu = 0.87fp, Ap = (mu/0.87fp(d-0.42xu) = (100×1006/0.87×1600(300-0.42×150)) = 300mm2, adopt a section 150mm wide by 350mm deep with 300mm2 of high tensile steel located at an effective depth of 300mm.
6. A post tensioned beam of unsymmetrical I section is required to support a design ultimate moment of 1200 knm and determine the overall depth and thickness of the compression flange required if fck is 35n/mm2 and fp is 1500n/mm2?
Answer: d [Reason:] For flanged sections, mud = 0.08fckbd2 assuming b = 0.5d and bw = 0.25b, d3 = ((1200×106)/0.10x35x0.5)), d = 1000mm, b = 500mm, thickness of top flange = hf = 0.2d = (0.2×10000) = 200mm, thickness of web = bw = 0.25b = (0.25×500) = 125mm
Assuming the neutral axis depth, xu = hf = 200mm, mu = 0.87fpAp (d-0.42xu), Ap = (mu/0.87fp (d-0.42xu) = (1200×106/0.87×1500(1000-0.42×200)) = 1003mm2.
7. Due to presence of precompression, prestressed concrete is ideally suited for the design of members subjected to:
a) Axial tension
b) Prestressed tension
c) Principle stress
d) Bonded stress
Answer: a [Reason:] Due to presence of precompression, prestressed concrete is ideally suited for the design of members subjected to axial tension and the axial tension in a member depends upon the direction of tension acting on a member and its reaction.
8. The design essentially considered for axial tension is to determine the:
c) Cross section
Answer: c [Reason:] The design essentially consists of determining the cross section area of the member and the required prestressing force to safely support the axial tensile load conforming to the limit state of serviceability and collapse.
9. According to Indian standard code IS: 3370 the load factors against cracking and collapse should not be less than:
Answer: b [Reason:] According to Indian standard code IS: 3370 1967 the load factors against cracking and collapse should be not less than 1.2 and 2 respectively and IS: 3370 code is used for design factors of load factors.
10. Design a suitable section for the tie member of a truss to support a maximum design tensile force of 500kn. The permissible compressive stress in concrete at transfer is 15n/mm2
Answer: a [Reason:] Design tensile load, nd = 500kn, fct = 15n/mm2, ftw = 0, ɳ = 0.80
Area of concrete section = (Nd/ ɳ fct) = (500×103/0.8×15) = 41500mm2.