Prestressed Concrete Structures MCQ Set 1
1. In the anchorage zone or the end block of a post tensioned prestressed concrete element, the state of stress distribution is considered as:
Answer: b [Reason:] In the anchorage zone or the end block of a post tensioned prestressed concrete element, the state of stress distribution is complex and three dimensional in nature, in most post tensioned members the prestressing wires are introduced in cable holes or ducts, preformed in the members, and then stressed and anchored at the end faces.
2. The stress distribution at a distance far away from the loaded face is computed from:
a) Simple bending theory
b) Complex bending theory
c) Axial bending theory
d) Parallel bending theory
Answer: a [Reason:] According to st.venant’s principle, the stress distribution at a distance far away from the loaded face (normally at a distance equal to or greater than the depth of the beam) can be computed from the simple bending theory.
3. The term anchorage zone or end block is termed as:
a) Zone between end and section
b) Zone between middle and section
c) Zone between edge and section
d) Zone between tension and section
Answer: c [Reason:] The zone between the end of the beam and the section where only longitudinal stress exists is generally referred to as anchorage zone or end block, as a result of this, large forces, concentrated over relatively small areas are applied on the end blocks, the highly discontinuous which are applied at the end while changing progressively to continuous linear distribution, develop transverse and shear stresses.
4. The transverse stresses developed in the anchorage zone are:
Answer: a [Reason:] The transverse stresses developed in the anchorages zone are tensile in nature over a large length and since concrete is weak in tension, adequate reinforcement should be provided to resist this tension.
5. Which knowledge is essential at the anchorage zone according to view of designer?
a) Transverse stresses
b) Distribution of stresses
c) Zone stresses
d) Anchorage stresses
Answer: b [Reason:] From the point of view of the designer, it is essential to have a good knowledge of the distribution of stresses in the anchorage zone, so that he can provide an adequate amount of steel, properly distributed to sustain the transverse tensile stresses.
6. The investigations studied the stress distribution in the anchorage zone stresses using:
a) Theoretical solutions
b) Linear equations
c) Pie diagrams
d) Charts and blocks
Answer: a [Reason:] A number of investigations have studied the stress distribution in the anchorage zone using empirical equations or theoretical equations or theoretical solutions based on two or three dimensional elasticity or experimental techniques.
7. Who among the one has done investigations on anchorage zone stress:
Answer: c [Reason:] The important investigations were those done by Magnel, Guyon, Iyengar, Zielinski, Rowe, Yettram, Robbins and Chandra sekhara, since the lines of force follow the same pattern with half the radius of curvature the length of the anchorage zone is halved, the transverse tension developed is also proportionately reduces, in similar way the greater the number of points of application of the prestressing force on the end block, the more uniform is the stress distribution.
8. The main aim of stress analysis in the anchorage zone is to obtain:
a) Zone tensile stress
b) Transverse tensile stress
c) Longitudinal tensile stress
d) Jack tensile stress
Answer: b [Reason:] The main aim of stress analysis in the anchorage zone is to obtain the transverse tensile stress distribution in the end block from which the total transverse bursting tension could be computed, the effect of transverse tensile stresses is to develop a zone of bursting tension in a direction perpendicular to the anchorage forces.
9. During anchorage of the end faces results in application of:
a) Small forces
b) Neutral forces
c) Large forces
d) Tensile forces
Answer: c [Reason:] In most post tensioned members, the prestressing wires are introduced in cable holes or ducts, performed in the members and then stressed and anchorage at end faces and as a result of this, large forces, concentrated over relatively small areas, are applied on end blocks.
10. The discontinuous forces applied at the end while changing progressively develop:
a) Transverse and shear
b) Principle and shear
c) Bursting and shear
d) Longitudinal and shear
Answer: a [Reason:] The highly discontinuous forces which are applied at the end, while changing progressively to continuous linear distribution, develop transverse and shear stresses, the ratio of transverse tensile stresses to the average compressive stress gradually decreases with the increase in the ratio of the depth of anchor plate to that of the end block.
Prestressed Concrete Structures MCQ Set 2
1. The deformation of prestressed members change with time as a result of:
a) Creep and shrinkage
b) Friction and torsion
c) Deformation and flexibility
d) Cracking moment
Answer: a [Reason:] The deformation of prestressed members change with time as a result of creep and shrinkage of concrete and relaxation of stress in steel, practically the change in stress obtained is relatively very small and hence it may be assumed that at constant stress the formation of creep occurs in concrete.
2. The deflection of prestressed members can be computed relative to given:
a) Bending moment
b) Strain diagram
Answer: d [Reason:] The deflection of prestressed members can be computed relative to a given datum, if the magnitude and longitudinal distribution of curvatures for the beam span are known based on load history including prestressing forces and live loads.
3. The prestressed concrete member develops deformation under the influence of:
a) Flexural moments
b) Stress strain diagram
c) Prestress and transverse loads
d) Self weight
Answer: c [Reason:] The prestressed concrete members develop deformation under the influence of two usually opposing effects, which are the prestress effects and transverse loads and the deflections caused are to be changed first because the loss incurred due to prestress which in turn decreases the deflection and effects of creep which increases the deflection are suggested in the method for long term deflections.
4. The net curvature ϕtat a section at any given stage is given as:
a) ϕt = ϕw + ϕe
b) ϕt = ϕm + ϕn
c) ϕt = ϕp + ϕs
d) ϕt = ϕmt + ϕpt
Answer: d [Reason:] The net curvature ϕt at a section at any given stage is obtained
ϕt = ϕmt + ϕpt, ϕmt = change of curvature caused by transverse loads,
ϕpt = change of curvature caused by prestress, as the time changes the compressive stress distribution in the concrete also changes under sustained transverse loads.
5. The section of sustained transverse loads under compressive stress distribution in the concrete changes with:
c) Bending moment
Answer: a [Reason:] Under the section of sustained transverse loads, the compressive stress distribution in the concrete changes with time, axial force is a measurement of the forces required to pull something such as rope, wire or structural beam to the point where it breaks compression force is the application of power, pressure and erection against an object.
6. The creep strain due to the transverse loads is directly computed as a function of:
a) Strain coefficient
b) Creep coefficient
c) Stress coefficient
Answer: c [Reason:] The creep strain due to transverse loads is directly computed as a function of the creep coefficient so that the change of curvature can be estimated by the expression,
Φmt = (1+ϕ)ϕi,
ϕ = creep coefficient, ϕi = initial curvature immediately after the application of transverse loads.
7. Which of the following person made attributions to evaluate the curvature under simplified assumptions?
Answer: a [Reason:] Several methods have been proposed to evaluate the curvature under simplified assumptions and important ones are attributed by Buseman, Mchenry, Douglas, Corley, Sozen and Siessand and the numerical solutions developed ignore the influence of the tensile concrete zone on the strain distribution in the section, which considerably effect deflection the equation for long term deflection of cracked members.
8. The creep curvature due to prestress is obtained on the simplified assumption that creep is induced by the average prestress acting over the given time is according to:
Answer: d [Reason:] According to Neville and the ACI committee report, the creep curvature is obtained due to prestress based on a very simplified assumption that the creep is induced into the concrete by the average prestress acting over with respect to the given time.
9. A simplified but an approximate procedure for computing long time deflections is given by:
Answer: c [Reason:] “Lin” suggested a procedure which is not exactly accurate but it helps in calculating long term deflections in a very simplified manner, this procedure helped in calculating the long term deflections.
10. The long time deflections are expressed as:
a) af = (ail-aipxpt/pi) (1+ϕ)
b) af = (ail-aipxpt/pi)
c) af = (ail-aipxpt/pi)
d) af = (ail-aipxpt/pi)
Answer: a [Reason:] The principle of reduced modulus involving the creep coefficient is used to amplify the initial deflections and according to this method, the final long time deflections are expressed as:
af = (ail – aipxpt / pi) (1+ϕ).
Prestressed Concrete Structures MCQ Set 3
1. The term anchorage slip means:
a) Distance moved by friction wedges
b) Radius by friction wedges
c) Rotation by friction wedges
d) Twisting movement by friction wedges
Answer: a [Reason:] Anchorage slip is the distance moved by the friction wedges (in post tensioned members) after releasing the jacks at the ends of the member and before the wires get fixed perfectly in wedges, the loss during anchoring which occurs with wedge type grips is normally allowed for on the site by over-extending the tendon in the prestressing operation by the amount of the draw in before anchoring.
2. The anchorage slip is observed in:
a) Post tensioned members
b) Pre tensioned members
c) Anchorage members
d) Tensioned members
Answers: b [Reason:] The anchorage slip is observed in post tensioned members at the time of transfer of prestress to the concrete and the friction wedges employed to grip the wires, slip over a small distance before the wires are firmly housed between the wedges.
3. The amount of anchorage slip generally depends upon:
a) Type of wedge
b) Type of tendon
c) Type of anchor
d) Type of cement
Answer: a [Reason:] The amount of anchorage slip generally depends upon the type of wedge used at the ends and the magnitude of stress in the wires, since the loss of stress is caused by a definite total amount of shortening the percentage loss is higher for short members than for long ones.
4. The anchorage slip is low in members with:
a) Small spans
b) Large spans
c) Middle spans
d) End spans
Answer: a [Reason:] The anchorage slip involves the definite total amount shortening of concrete and hence it is low in members with small spans compared to the members with large spans while prestressing a short member, due care should be taken to allow for the loss of stresses due to anchorage slip, which forms a major portion of total loss.
5. In case of long line pre tensioning system, anchorage slip is less than:
a) Magnitude of wires
b) Length of wires
c) Distance of wires
d) Radius of wires
Answer: b [Reason:] In case of long line pre tensioning system, anchorage slip is much less in comparison with the length of the tensioned wires and hence it can be ignored for calculation of stresses, slip of anchorages, length of cable, cross-sectional area of the cable, modulus of elasticity, prestressing force in cable are considered.
6. Which of the following system is similar to loss due to anchorage slip?
a) Freyssinet system
b) Magnel Balton system
c) Leonhardt-Baur system
d) Gifford Udall system
Answer: c [Reason:] The systems in which tendons are lopped around concrete anchorage blocks, as in the case of Leonhardt-Baur system, loss of stress may take place, the Baur Leonhardt system is included in the third type of anchorages which work under the principle of looping tendon wires at the ends of concrete member, in this method double tendons are wrapped around the end block.
7. The loss of stress due to anchorage slip of anchorages is given as:
Answer: a [Reason:] The method used for slip in anchorage by extending tendon is satisfactory provided by overstress does not exceed the prescribed limits of 80-85percent of the ultimate tensile strength of the wire and the magnitude of the loss of stress due to slip in anchorage is computed as
Anchorage slip Δ = PL/AEs, But moment Loss of stress, (Δf)a = P/A,
By considering both equations (Δf)a = EsΔ/l.
8. A concrete beam is post tensioned by a cable carrying an initial stress of 1000n/mm2, the slip at the jacking end was end was observed to be 5mm; modulus of elasticity of steel is 210kn/mm2. Estimate the percentage loss of stress due to anchorage if length of beam is 30m?
Answer: a [Reason:] Loss of stress due to anchorage slip = (EsΔ/l),
For a 30m long beam, loss of stress = (210×103×5)/(30×1000) = 35n/mm2,
Loss of stress = 35/1000×100 = 3.5%.
9. A post tensioned cable of beam 10m long is initially tensioned to a stress of 1000n/mm2 at one end, slope is 1 in 24 tendons curved at each end , area is 600mm2, Es is 210kn/mm2, coefficient of friction between duct and cable is 0.55, friction coefficient for wave effect is 0.0015perm. During anchorage, if there is a slip 3mm at the jacking end, calculate final force?
Answer: c [Reason:] Total change of slope from end to end α = (2×1/24) = (1/12),
μα = (0.55×1/12) = 0.046, kx = (0.0015×10) = 0.015,
P˳(μα+kx) = 1000(0.046+0.015) = 61n/mm2, Slip at the jacking end = 3 = (PL/AE)
P = (3×210×103×600/10×1000) = 37800kn = 37.8kn.
10. The total losses of stress that could be encountered under normal conditions of work were recommended by:
Answer: a [Reason:] Typical values of the total losses of stress that could be encountered under normal conditions of work were recommended by Lin, long term field studies on the loss of prestress in post tensioned concrete bridge girders have been carried out by marks and Keifer, Neville gave the losses in prestress considering various influencing parameters.
11. A prestressed concrete beam, 200mm wide and 300mm deep is prestressed with wires (area is 320mm2) located at a constant eccentricity of 50mm, initial stress of 1000n/mm2, span is 10m. Calculate loss of stress due to friction and slip anchorage of post tensioned beam?(Es = 210kn/mm2, Ec = 35kn/mm2)
a) 21 and 15
b) 35 and 25
c) 15 and 20
d) 5 and 10
Answer: a [Reason:] A = 320mm2, b = 200mm, d = 300mm, e = 50mm, p = 1000n/mm2, l = 10m, Es = 210kn/mm2, Ec = 35kn/mm2, Slip at anchorage = (1×210×103/10×1000) = 21, Friction effect = (1000×0.0015×10) = 15.
Prestressed Concrete Structures MCQ Set 4
1. Which of the following results in the reduction of stress in steel used for prestress?
d) Anchorage slip
Answer: c [Reason:] The sustained prestress in the concrete of a prestressed members results in creep of concrete which effectively reduces the stress in high tensile steel and the progressive inelastic strains due to creep in a concrete are likely to occur under the smallest sustained stresses at ambient temperatures, shrinkage and creep of concrete are basically similar in origin, being largely the result of migration of water in the capillaries of cement paste.
2. The loss of stress due to creep of concrete can be estimated by:
a) Ultimate creep strain
b) Ultimate load
c) Ultimate creep stress
d) Ultimate creep tension
Answer: a [Reason:] The loss of stress in steel due to creep of concrete can be estimated if the magnitude of ultimate creep strain or creep coefficient is known and the values of creep coefficient which is the ratio of ultimate creep strain to the elastic strain is 2.2 at 7 days of loading, 1.6 at 28 days and 1.1 when the age at loading is 1 year.
3. The value of creep coefficient ‘f’ depends upon:
c) Uv rays
Answer: a [Reason:] The value of creep coefficient ‘f’ depends upon various factors such as humidity duration of load applied, age of loading and effective section thickness and the effective section thickness is defined for uniform sections as twice the cross sectional area divided by the exposed perimeter, it can be assumed about 4060 and 80 percent, respectively of the final creep develops during the first, six and 30 months under load when concrete is exposed to conditions of constant relative humidity.
4. The creep coefficient varies from a minimum value of:
Answer: b [Reason:] The creep coefficient varies from a minimum value of 1.5(for wet conditions) to a maximum value of 4.0(for dry condition), for design purposes it is convention to differentiate between deformation due to externally applied stress generally referred as creep and the deformation which occurs without externally applied stresses referred as shrinkage as the increase in strain under a sustained stress is several times larger than the strain on loading, it is of considerable importance in prestressed structural members.
5. The creep coefficient is high for:
b) Post tensioned
c) Chemical tensioned
d) Electrical tensioned
Answer: a [Reason:] Creep coefficient is generally low for post tensioned members and high for pre tensioned members, the various factors influencing creep of concrete are relative humidity, stress level, strength of concrete, age of concrete at loading, duration of stress, water cement ratio, and type of cement and aggregate in the concrete, for stress up to half of the crushing strength of concrete.
6. The creep coefficient equation is given as:
a) Creep strain/elastic strain
b) Creep stress/elastic stress
c) Elastic strain/creep strain
d) Elastic stress/creep stress
Answer: a [Reason:] The magnitude of the creep coefficient ϕ varies depending upon the humidity, concrete quality, duration of applied loading and the age of concrete when loaded,
Creep coefficient = (creep strain / elastic strain), ϕ = ℇc/ℇe, ℇc = ϕ ℇe = ϕ(fs/Ec),
Loss of stress in steel = ℇcEcϕEs = ϕ(fs/Ec)Es = ϕfcαe.
7. The loss of stress in steel due to creep of concrete is:
a) ℇcc fc Es
b) ℇc fc Ec
c) ℇc fe αe
d) ℇcc fc ϕ
Answer: a [Reason:] The loss of stress in steel due to creep of concrete is ℇcc fc Es,
ℇcc = ultimate creep strain for a sustained unit stress, fc = compressive stress in concrete at the level of steel, Es = modulus of elasticity of steel.
8. A concrete beam of rectangular section 100mm wide, area is 3×104, initial stress is 1200n/mm2 of 7mm diameter located at an eccentricity 50mm. Estimate the loss of stress due to creep of concrete using the ultimate creep strain method?
Answer: a [Reason:] Es = 210kn/mm2, I = 225×106mm4, Ec = 35n/mm2, a = 3×104mm2, p = (1200×38.5×5) = 23×104n, ℇcc = 41×10-6mm/mm per n/mm2, αe = (Es/Ec) = 6, ϕ = 1.6, fc = (23×104/3×104+23×104×50×50/225×106) = 10.2n/mm2, Ultimate creep strain method loss = ℇcc fc Es = (41×10-6)(10.2)(210×103) = 88n/mm2.
9. A post tensioned concrete beam of rectangular section 200mm wide and 300mm deep is stressed by a parabolic cable with eccentricity 50mm at centre of span, area is 3×104mm2, initial stress is 1200n/mm2 Find the stress in concrete at the level of steel?
Explanation: A = 3×104mm2, p = (200×1200) = 240000n, e = 50mm,
Stress in concrete at the level of steel: (240000/30000) = 8n/mm2.
10. The British code for structural concrete recommends design values of ultimate creep strain of pretensioning is:
Answer: a [Reason:] The British code for structural concrete recommends design values of ultimate creep strain of 48×10-6 for pretensioning and 36×10-6 for post tensioning and these values have to be increased in inverse proportion if the compressive strength of concrete at transfer is less than 40n/mm2.
Prestressed Concrete Structures MCQ Set 5
1. The loss of prestress due to friction generally occurs in case of:
a) Post tensioned members
b) Pre tensioned members
c) Chemical members
d) Electrical members
Answer: a [Reason:] Loss of prestress due to friction occurs in the case of post tensioned members, the tendons are housed in the ducts performed in concrete and the ducts are either straight or follow a curved profile depending upon the design requirements.
2. The total loss of prestress due to friction is of:
a) 4 types
b) 2 types
c) 8 types
d) 3 types
Answer: b [Reason:] The total loss due to friction is divided into two types:
Loss of prestress due to effect of curvature, Loss of prestress due to wobble effect and frictional losses can be reduced by over tensioning the tendons by an amount equal to the maximum frictional loss and jacking the tendons from both ends of the beam adopted generally, when the tendons are long or when the angles of bearing are large.
3. The loss of stress due to curvature effect depends upon:
d) Exterior point
Answer: a [Reason:] The loss of stress due to the curvature effect, which depends upon the tendon form or alignment which generally follows a curved profile along the length of the beam, curvature coefficient is expressed as μ and wobble coefficient is expressed as k/m.
4. The wobble effect due to loss of stress is also known as:
a) Wave effect
b) Ray effect
c) Bubble effect
d) Bulb effect
Answer: a [Reason:] Loss of stress due to wobble effect, which depends upon the local deviations in the alignment of the cable and the wobble effect is also known as wave effect, the friction coefficient values for wave effect k are 0.15 per 100m for normal conditions, 1.5 per 100m for thin walled ducts where heavy vibrations are encountered and in other adverse conditions.
5. The wobble effect is the result of:
b) Extreme alignment
c) Tensile alignment
d) Anchorage alignment
Answer: a [Reason:] The wobble or wave effect is the result of accidental or unavoidable misalignment since ducts cannot be perfectly located to follow a predetermined profile throughout the length of the beam, the coefficient due to wobble effect may be reduced to zero where the clearance between the duct and cable is sufficiently large to eliminate wave effect so as the sheath is made up of heavy gauge steel tube with water tight joints, where a deformation of duct profile is prevented during the vibration of concrete.
6. The loss of stress due to friction is given as:
Answer: a [Reason:] The prestressing force at section is given by:
Px = p˳e-(μα+kx), Loss of stress (Δf)f = 1-(μα+Kx),
Px = Prestressing force at section x, P˳ = initial prestressing force, μ = coefficient of friction between the cable and concrete, k = wobble effect, d = cumulative angle.
7. The value of ‘μ’ in loss of stress equation depends upon:
a) Type of curing
b) Type of concrete
c) Type of steel
d) Type of aggregates
Answer: c [Reason:] The values of ‘μ’ (coefficient of curvature effect) depend upon the type of steel and concrete used in construction and are given in Indian standard codes of practice, coefficient of friction can be considerably reduced by variety of lubricants, particularly greases, oil, graphite mixtures, paraffin, the use of paraffin wax gives by far the coefficient of friction especially with high contact pressure.
8. A post tensioned concrete beam 200mm wide and 450mm deep, of span 10m, initial stress of 840n/mm2 is available in the un jacked end immediately after the anchoring. Find the angle between tangents to the cable at supports?
Answer: b [Reason:] b = 200mm, d = 450mm, l = 10m, r = 84m, d = 5m,
Angle between the horizontal tangent drawn to the cable at support sinα = (5/84) = 0.06radians,
Cumulative angle between tangents to the cable at supports = (2×0.06) = 0.12radians.
9. A concrete tank if has a minimum stress in wires 600n/mm2 immediately after tensioning and the coefficient of friction is 0.5. Calculate the maximum stress to be applied to the wires at the jack?
Answer: b [Reason:] Px = 600n/mm2, e = 2.7183, μ = 0.5
Px = P˳e-μα,
600 = P˳e-(0.5×π/2),
P˳ = (600) × (2.71830.79) = 1320n/mm2, Average stress in wires = (1320+600/2) = 960n/mm2.
10. A cylindrical concrete tank, 40m external diameter is to be prestressed circumferentially by means of a high strength steel wire (Es = 210kn/mm2) jacked at 4 points and 90 degrees apart. Find the expected extension at the jack?
Answer: c [Reason:] d = 40m, Es = 210kn/mm2, n = 4points, θ = 90˚,
Length of wires = (π×40×1000/4) = 104πmm,
Extension at the jack = (960/210×103×104π) = 144mm.