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Prestressed Concrete Structures MCQ Set 1

1. The shear stress is a function of:
a) Shear force and Cross section
b) Principle stresses and elevation
c) Strain & Compatibility
d) Axial prestress & tension

View Answer

Answer: a [Reason:] The shear distribution in un cracked structural concrete members linear deformations are assumed to be developed due to shear distribution and the shear stress in a function of shear force and cross section of the members which is given by the equations: τv =VS/IB, τv =shear stress, V = shear force, S = first moment of inertia, I = moment of inertia, B = width of the beam section.

2. The strength of concrete subjected to pure shear being nearly twice that in:
a) Compression
b) Tension
c) Bond
d) Anchorage

View Answer

Answer: b [Reason:] The effect of this shear stress is to induce principal tensile stresses on diagonal planes and in pure shear, the strength of the concrete is twice that of the strength in the tension local failures first appear in the form of diagonal tension cracks in legions of height shear stress.

3. The effect of maximum shear stress (τ v) produces:
a) Principal tensile stresses
b) Principal compression stresses
c) Principal strain stresses
d) Principal span stresses

View Answer

Answer: a [Reason:] cracks are observed at the point of the development of maximum shear stresses diagonally. The effect of this maximum shear stress (τ v) also produces principle tensile stresses on diagonal plane, the calculation of principle tensile stress resulting from direct at critical sections with or without bending and shear combined shall be carried out it is also done at the material change in width of section and should be less than 0.126(fc)1/2.

4. In a prestressed concrete member, the shear stress is generally accompanied by:
a) Zone stresses
b) Anchorage stresses
c) Direct stresses
d) Bondage stresses

View Answer

Answer: c [Reason:] In a prestressed concrete member, the shear stresses is generally accompanied by a direct stress in the axial direction of the member, and if transverse, vertical prestressing is adopted, compressive stresses in perpendicular to the axis of the moment will present in addition to the axial pre stresses.

5. The major principal stresses produced on diagonal plane is expressed as:
a) fx + fy/2
b) fx + fy/2 – 1/2 ((fx – fy )2 +4τ v2 )1/2
c) fx + fy/2 + 1/2 ((fx – fy)2 +4τ v2 )1/2
d) fx – fy/2

View Answer

Answer: b [Reason:] The major principal stress Fmax = fx + fy/2 + 1/2 ((fx – fy )2 +4τv2 )1/2 Minor principal stress Fmin = fx + fy/2 + 1/2 ((fx – fy )2 +4τ v2 )1/2 Fx, Fy are the direct stresses in horizontal & vertical directions respectively.

6. If the direct stresses are compressive, then the magnitude of principal stresses in prestressed concrete member gets:
a) Increased
b) Decreased
c) Constant
d) Zero

View Answer

Answer: a [Reason:] If the direct stresses are compressive, then the magnitude of principal stresses in prestressed concrete member acts reduced considerably and therefore under working loads, these principal stresses have to be compressive in nature in order to eliminate diagonal cracks in concrete.

7. How many ways are there for improving the shear resistance of structural concrete members by prestressing techniques?
a) 4
b) 6
c) 3
d) 2

View Answer

Answer: c [Reason:] In general three ways of improving the shear resistance of structural concrete members by prestressing techniques: Horizontal or axial prestressing by inclined or sloping cables and vertical or transverse prestressing.

8. A prestressed concrete beam span 10mm of rectangular section, 120mm wide & 300mm deep is axially prestressed on effective force of 180kn, uniformly distributed load of 5kn/m include the self weight of member. Find maximum shear stress at support?
a) 20.5n/mm2
b) 1.05n/mm2
c) 15.08n/mm2
d) 4.05n/mm2

View Answer

Answer: b [Reason:] A= (120×300) = 36×103 mm2, I = 27×107 mm4, Wd = 5kn/m Shear force at support V = (5×10/2) =25kn Maximum shear stress at support, τv = (3v/2bh) = (3/2)x(25×103 /120×300) = 1.05n/mm2.

9. A prestressed concrete beam of span 10m of rectangular section, 120mm wide & 300mm deep a curved cable having an eccentricity of 100mm at the centre of span. Find the slope of cable of support:
a) 0.08 radians
b) 0.01 radians
c) 0.04 radians
d) 0.12 radians

View Answer

Answer: c [Reason:] I = 10mm, l=100mm Slope of cable at support = (4e/l) = (4×100/10×100) = 0.04 radians.

10. Which type of tensioning is generally uneconomical for vertical prestressing?
a) Post tensioning
b) Pre tensioning
c) Chemical tension
d) Thermal tension

View Answer

Answer: a [Reason:] Vertical prestressing is not generally adopted because the length of the cables being short, the loss of prestress due to anchorage slip is excessively large post tensioning is generally uneconomical for vertical prestressing due to losses of prestress encountered.

Prestressed Concrete Structures MCQ Set 2

1. The short term deflections are also known as:
a) Cracked
b) Un cracked
c) Instantaneous
d) Non instantaneous

View Answer

Answer: c [Reason:] Short term deflections of prestressed members are also known as instantaneous deflections governed by distribution of bending moment throughout the span and flexural rigidity of member, these theorems are applied for determining the deflections due to prestressing force, imposed loads and self weight.

2. Which of the following is the equation given Mohr’s first theorem?
a) Area of bending moment deflection/flexural rigidity
b) Moment/flexural rigidity
c) Deflection/flexural rigidity
d) Loads/flexural rigidity

View Answer

Answer: a [Reason:] When the beam AB is subjected to a bending moment distribution due to prestressing force or self weight or imposed loads, ACB is the centre line of the deformed structure under the system of given loads, According to Mohr’s first theorem Slope = area of bending moment deflection/flexural rigidity, θ = A/EI.

3. Which of the following is the equation given by Mohr’s second theorem?
a) Mid span/flexural rigidity
b) Moment of area of bending moment diagram/flexural rigidity
c) End span/flexural rigidity
d) Thickness/flexural rigidity

View Answer

Answer: b [Reason:] Mohr’s second theorem states that Intercept, a = (moment of the area of bending moment deflections/flexural rigidity), a = AX/EI, a = deflection at the centre for symmetrically loaded, simply supported beam (since the tangent is horizontal for such cases), A = area of bending moment deflection between A and C, x = distance of the centroid of the bending moment deflection between A and C from the left support, EI = flexural rigidity of beam.

4. Which of the following deflections are directly obtained by Mohr’s second area theorem?
a) Simply supported beam
b) Uniformly distributed load
c) Point beams
d) Fixed beams

View Answer

Answer: a [Reason:] The deflections of symmetrically loaded and simply supported beam at the mid span point are directly obtained from the second moment area theorem since the tangent is horizontal at this span, In most of cases of prestressed beams tendons are located with eccentricities towards the soffit of the beam to counteract the sagging bending moments due to transverse loads.

5. The problems involving unsymmetrical loading can be solved by:
a) Mohr’s theorem
b) Kennedy’s theorem
c) Row’s theorem
d) Casagrande’s theorem

View Answer

Answer: a [Reason:] More complicated problems involving unsymmetrical loading may be solved by combining both the moment area theorems Mohr’s first theorem and second theorem, since the bending moment at every section is the product of prestressing force and eccentricity the tendon profile itself will represent the shape of the bending moment diagram.

6. A straight tendon at a uniform eccentricity below the centroidal axis is given as:
a) –PeL2/4EI
b) –PeL2/8EI
c) –PeL2/14EI
d) –PeL2/16EI

View Answer

Answer: b [Reason:] A straight tendon at a uniform eccentricity below the centroidal axis is given as: If the camber of beam with straight tendons upward deflections are considered as negative and a = -(PeL) (L/4)/EI = -PeL2/8EI, P = effective prestressing force, e = eccentricity, L = length of beam.

7. A tendon with a trapezoidal profile considering the bending moment and deflection at the centre of the beam is obtained by:
a) –Pe/6EI(2l12+6l1l2+3l22)
b) –Pe/6EI(2l12+6l1l2+3l22)
c) –Pe/6EI(2l12+6l1l2+3l22)
d) –Pe/6EI(2l12+6l1l2+3l22)

View Answer

Answer: b [Reason:] A draped tendon with a trapezoidal profile considering the bending moment diagram the deflection at the centre of the beam is obtained by taking the moment of the area of the bending moment diagram over one half of the span A = –Pe/6EI(2l12+6l1l2+3l22).

8. The deflection of a beam with parabolic tendon is given as:
a) –5PeL2/48EI
b) –10PeL2/48EI
c) –15PeL2/48EI
d) –3PeL2/48EI

View Answer

Answer: a [Reason:] The deflection of the beam with parabolic tendons having an eccentricity e at the center and zero at the supports is given by a = –5PeL2/48EI, a beam with a parabolic tendon having an eccentricity e1 at the centre of span and e2 at the support sections and the resultant deflection at the centre is obtained as the sum of the upward deflection of a beam with a parabolic tendon of eccentricity e1+e2 at the centre and zero at the supports and the downward deflection of a beam subjected to a uniform sagging bending moment of intensity pe2 throughout the length, the resultant stress becomes a = PL2/48EI(-5e1+e2).

9. The deflection is computed in a way similar to sloping tendon is given as:
a) 2PL2/24EI
b) 4PL2/24EI
c) PL3/24EI (-2e1+e2)
d) PL2/24EI (e1+e2)

View Answer

Answer: c [Reason:] The deflection in sloping tendon is computed in a way similar to: A = (-PL2/12EI(e1+e2)) + (Pe2L2/8EI) A = (PL3/24EI (-2e1+e2)).

10. The deflection due to self weight and imposed loads are:
a) 5(g+q)L4/384EI
b) 5(g+q)L4/384EI
c) 5(g+q)L4/384EI
d) 5(g+q)L4/384EI

View Answer

Answer: a [Reason:] At the time of transfer of prestress, the beam hogs up due to the effect of prestressing, at this stage the self weight of the beam includes downward deflections, which further increases due to the effect of imposed loads on the beam a = 5(g+q)L4/384EI and deflections due to concentrated live loads can be directly computed by using Mohr’s theorem.

Prestressed Concrete Structures MCQ Set 3

1. The estimation based on compatibility of strain of prestressed concrete involves:
a) Flexural strength
b) Tensile strength
c) Compressive strength
d) Bulking strength

View Answer

Answer: a [Reason:] The method by which the flexural strength of prestressed concrete is estimated based on the compatibility of strains and equilibrium of forces acting on the section at the stage of failure is known as strain compatibility method.

2. The distribution of concrete in strain compatibility method is:
a) Aligned
b) Curved
c) Linear
d) Bent

View Answer

Answer: c [Reason:] The distribution of concrete strain is linear (plane section normal to axis remains plane even after bending), the basic theory is applicable to all structural concrete sections whether reinforced or prestressed and some assumptions are made.

3. The resistance of concrete is neglected at:
a) Tension
b) Compression
c) Shear
d) Breakage

View Answer

Answer: a [Reason:] The resistance of concrete in tension is neglected in strain compatibility method at the end of the section, the maximum compressive strain in concrete at failure reaches a particular value are the assumptions made in strain compatibility method.

4. The stress distribution in the compression zone of concrete can be defined by means of:
a) Specific gravity
b) Coefficient
c) Modulus of elasticity
d) Span moment

View Answer

Answer: b [Reason:] The stress distribution in the compression zone of concrete can be defined by means of coefficients applied to the characteristic compressive strength and the average compressive stress and the position can be assumed.

5. The flexural compression stress in the compressive zone follows the:
a) Block curve
b) Anchorage curve
c) Mid span
d) Stress strain curve

View Answer

Answer: d [Reason:] The flexural compressive stress in the compressive zone closely follows the stress strain curve of concrete and the properties of the concrete stress block can be expressed in terms of the characteristic ratios of k1 & k2.

6. The stress strain characteristics of steel used as prestressing tendons is necessary for:
a) Principal computation
b) Stress computation
c) Flexural computation
d) Strain computation

View Answer

Answer: c [Reason:] Knowledge of the stress–strain characteristics of steel that is used as prestressing tendons is necessary for flexural strength computations by the strain compatibility method and a typical short- term design stress-strain curve for concrete recommended in the British & Indian standard codes.

7. The number of steps to be followed in the strain compatibility method is:
a) 4
b) 7
c) 10
d) 6

View Answer

Answer: d [Reason:] The major steps to be followed in the strain compatibility method are summarized below compute the effective strain, assume a trail value for the neutral axis, using the stress-strain curve for steel compute the total compression Lu & tension Tu, if the compressive and tensile forces are equal, evaluate the ultimate moment Mu.

8. Who suggested a graphical version compatibility method?
a) Cornd
b) Morsch
c) Lin
d) Musy

View Answer

Answer: b [Reason:] Morsch has suggested a graphical version of the strain compatibility method in which the failure of the compressive zone is assumed when the extreme compressive fiber reaches a strain limit of 0.2%, many codes have recommended simplified procedures for calculating the flexural strength of concrete sections which are reinforced with high tensile steel in the tension zone.

9. How much percentage of tensile strain is assumed at the failure of under reinforced sections?
a) 0.7%
b) 0.5%
c) 0.4%
d) 0.2%

View Answer

Answer: c [Reason:] For under reinforced sections, the failure of the prestressing steel is assumed to take place at a maximum tensile strain of 0.5% however the method can be considerably simplified under reinforced sections, in which the stress in tensile stress at the collapse stage is more of less equal to the characteristic tensile strength of tendons.

10. The strain compatibility method is generally applicable for:
a) Under & over reinforcement sections
b) Partially prestressed sections
c) Mid span sections
d) Fully prestressed sections

View Answer

Answer: a [Reason:] Many codes have recommended simplified procedures for calculating the flexural strength of concrete sections which are reinforced with high tensile steel in the tension zone; the use of the strain compatibility method, generally applicable for both under & over reinforced sections is illustrated.

Prestressed Concrete Structures MCQ Set 4

1. During stress distribution in end blocks the prestressing force is applied as:
a) Concentrated force
b) Deviated force
c) Tension force
d) Torsion force

View Answer

Answer: a [Reason:] The prestressing force is applied as a concentrated force in a tendon along the anchorages during stress distribution in end blocks, the compressive stresses of concrete ahead of the anchorage devices, location and magnitude of the bursting force and the edge tension forces can be estimated with approximate method of design of end blocks when the accurate analysis are not available.

2. The stress distribution in concrete member which is away from the anchorage and in the region of the anchorage will be:
a) Non uniform
b) Zero
c) Constant
d) Uniform

View Answer

Answer: d [Reason:] The stress distribution in concrete member which is away from the anchorage and in the region of the anchorage will be uniform reasonably and complex respectively this is stated by st.venant’s principle, a product of the correction factor in each direction is used, if a group of anchorage are closely spaced in two directions in case of computation of compressive stresses.

3. The tensile stresses which tend to split the concrete are placed in the transverse direction to the:
a) Edge of member
b) Span of member
c) Axis of member
d) End of member

View Answer

Answer: c [Reason:] The tensile stresses which tend to split the concrete are placed in the transverse direction to the axis of the member which is the more important effect for the design, the spalling and longitudinal edge tension forces are induced when the location of centroid of all the tendons is considered to be outside the keen of the section.

4. The bursting tensile forces in end bocks with proportion Pi is given as:
a) Fbst = Pi(0.32-0.3(ypo/yo))
b) Fbst = fi(0.32-0.3(ypo/yo))
c) Fbst = Ti(0.32-0.3(ypo/yo))
d) Fbst = πi(0.32-0.3(ypo/yo))

View Answer

Answer: a [Reason:] The bursting tensile force in end blocks with proportion pi is given as: Fbst = Pi(0.32-0.3(ypo/yo)), ypo/yo < or equal 0.3, Fbsr/Pi = 0.23, ypo/yo > or equal 0.3, Fbst/Pi = 0.11.

5. The longitudinal extent of the concrete member which is rectangular in cross section is:
a) Zero
b) Equal
c) Constant
d) Unity

View Answer

Answer: b [Reason:] The longitudinal extent of the concrete member which is rectangular in cross section is at least equal to the largest transverse dimension of the cross section; the axial flexural beam analysis at one half the depth of section away from the loaded area enables to determine the longitudinal edge tension force.

6. The member within or ahead of the anchorage zone will not have any:
a) Strain
b) Stress
c) Discontinous
d) Torsion

View Answer

Answer: c [Reason:] The member within or ahead of the anchorage zone will not have any discontinuities but they are subjected to stress during prestressing, the more detailed analysis is required such as strut-and-ties models or another analytical procedures to find out the spalling forces for larger spacing.

7. The main plate of the member in the anchorage zone has minimum edge distance of at least:
a) 2.0
b) 1.5
c) 1.8
d) 3.0

View Answer

Answer: b [Reason:] The main plate of the member in the anchorage zone has minimum edge of at least 1.5 times the corresponding lateral dimension compressive stresses of the anchorage devices, the basic anchorage devices which satisfies below condition will have no additional check of compressive stresses of concrete fb< 0.7ϕf”ci(A/Ag).

8. The anchorage zone consists of how many devices:
a) 5
b) 3
c) 2
d) 1

View Answer

Answer: d [Reason:] The anchorage zone consists of only one anchorage device or group of anchorages devices which are closely spaced, at the interface between the local zone and the general zone of the special anchorage devices ahead, the concrete compressive stresses.

9. If the centre to centre spacing of the anchorage devices will not exceed 1.5 times width then they are considered as:
a) Closely spaced
b) Gapely spaced
c) Farley spaced
d) Rectangular spaced

View Answer

Answer: a [Reason:] If the center spacing of the anchorage devices will not exceed 1.5 times the width of these devices in the direction in which it is considered is treated as closely spaced devices and if the centre to centre spacing for multiple anchorages is less than 0.4 times the depth of section, the spalling forces will not be less than 2% of the total factored tendon force in any case.

10. If the anchor force points are towards the centroid, the center line of the member will not be greater than:
a) 15o
b) 10o
c) 20o
d) 25o

View Answer

Answer: b [Reason:] If the anchorage force points are towards and away from the centroid of the section, the center line of the sectional member will not be greater than 20o and 5o respectively will be the angle of inclination of a tendon.

Prestressed Concrete Structures MCQ Set 5

1. The prestressed member undergoes deformation due to the action of:
a) Prestressing force and flexural loads
b) Prestressing force and combined loads
c) Prestressing force and transverse loads
d) Prestressing force and tangential loads

View Answer

Answer: c [Reason:] A prestressed member undergoes deformation due to the action of the prestressing force and transverse loads acting on the member and the curvature of the cable changes which results in a slight variation of stresses in the tendons.

2. The rotation equation obtained by applying Mohr’s theorem considering a concrete beam of span l, force p, eccentricity e is:
a) θp = PeL/2EI
b) θp = PeL/4EI
c) θp = PeL/16EI
d) θp = PeL/20EI

View Answer

Answer: a [Reason:] A concrete beam of span l is prestressed by a cable carrying an effective force p at an eccentricity e the rotation at the supports due to hogging of the beam is obtained by applying Mohr’s theorem as θp = Area of bending moment/flexural rigidity = PeL/2EI.

3. The cross section of a prestressed concrete beam is 100mm wide and 300mm deep and the initial stress in tendons are located at a eccentricity of 50mm is 1000n/mm2, the sectional area is 100mm2. Find rotation due to prestress (hogging moment)?
a) 0.00155
b) 0.00165
c) 0.00175
d) 0.00185

View Answer

Answer: d [Reason:] Moment of area I = (100×3003/12) = 225×106mm4 Prestressing force p = (1000×100) = 105 = 100kn Rotation due to prestressing force θp = PeL/2EI = (100×50×6×103/2×36×225×106) Hogging moment = 0.00185radians.

4. In the elastic range, any increase in prestressed member does not show any effect on:
a) Steel stress
b) Compressive stress
c) Bending stress
d) Flexural stress

View Answer

Answer: a [Reason:] In the elastic range any increase in loading on a prestressed member does not result in any significant change in the steel stress and in other words, the stress in steel is more or less constant in elastic range of prestressed concrete.

5. The rate of increase in stress in the tendons of a prestress concrete member depends upon:
a) Bond and breakage
b) Bond and surrounding concrete
c) Bond and elasticity
d) Bond and anchorage

View Answer

Answer: b [Reason:] The rate of increase of stress in the tendons of a prestressed concrete member under loads depends upon the degree of bond between the high tensile steel wires and the surrounding concrete, the increase of stress in steel depends on the average strain in concrete at the level of steel.

6. A prestressed concrete beam used over a span of 6m is 100mm wide and 300mm deep, live load of 4kn/m, density of concrete is 24kn/m3, modulus of elasticity of concrete is 36 and steel is 210kn/mm2. Find rotation due to loads(sagging moment)?
a) 0.005
b) 0.00525
c) 0.0024
d) 0.0045

View Answer

Answer: d [Reason:] E = 210kn/mm2, Self weight of the beam, g = (0.1×0.3×24) = 0.72kn/m Live load on the beam (q) = 4kn/m, Total load on the beam, Wd = (q+g) = (4+0.72) = 0.00472kn/mm Rotation due to prestressing force θp = PeL/24EI = (100×50×6×103/24×36×210×106) Sagging moment = 0.00525radians.

7. The stress in tendons of bonded beams is:
a) αe (My/I)
b) αe (My/R)
c) αe (My/L)
d) αe (My/20)

View Answer

Answer: a [Reason:] The stresses in tendons of bonded beams is αe (My/I), αe = Modular ratio of steel to concrete, y = vertical distance of a point from centroid of concrete section, M = bending moment, I = moment of area of the section, in cases of bonded members such as pretensioned elements or post tensioned grouted members, the composite action between steel and concrete prevails and the stresses in steel are computed using the theory of composite sections up to stage of cracking.

8. The rate of increase of stress is larger in case of:
a) Bonded beams
b) Un bonded beams
c) Tensioned beams
d) Anchorage beams

View Answer

Answer: a [Reason:] The rate of increase of stress is larger in case of bonded beams than in un bonded beams both in the pre cracking and post cracking stages, in case of unbounded beams, the tendons are free to elongate independently throughout their length under the action of transverse loads on the beam.

9. The development of crack widths is comparatively larger in:
a) Bonded beams
b) Un bonded beams
c) Localized beams
d) Strengthened beams

View Answer

Answer: a [Reason:] In post cracking stage, while the bonded beams are characterized by small cracks, which are well distributed in the zone of the larger moments, unbounded beams develop only a few cracks, which are localized at weaker sections and the crack widths are correspondingly larger in comparison with the bonded beams.

10. Which beams are preferred mostly due to their higher flexural strength?
a) Bonded beams
b) Un bonded beams
c) Exhaustive beams
d) Extended beams

View Answer

Answer: a [Reason:] In general bonded beams are preferable due to their higher flexural strength and predictable deformation characteristics, after the onset of cracking the stress in steel increases at a faster rate in both types of beams since steel does not reach its ultimate strength in case of unbounded beams the ultimate load supported by the beam is smaller than that of bonded beam in which the steel attains its ultimate strength at the failure stage of member.

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