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## Prestressed Concrete Structures MCQ Set 1

1.Calculate ultimate moment and shear of effective span is 30m, live load is 9kn/m, dead load excluding self weight is 2kn/m, load factors for dead load is 1.4 for live load is 1.6 cube strength of concrete fcu is 50n/mm2 cube strength at transfer is fci is 35n/mm2, tensile strength of concrete Ec is 34kn/mm2 loss ratio ɳ is 0.85 and 8mm diameter high tensile strength fpu is 1500n/mm2 are available for use and the modulus of elasticity of high tensile wires is 200kn/mm2?
a) 340 and 450kn
b) 240 and 340kn
c) 140 and 240kn
d) 100 and 200kn

Answer: a [Reason:] Wmin/Wud = (50x2400x9.81×0.125x25x30/50×106x0.852) = 0.31 Ultimate load excluding the factored selfweight = (1.4×2)+(1.6×9) = 17.2kn/m, Wud = 17.2/1-1.4×0.31) = 30KN/M, Wmin = (0.31×30) = 9.3kn/m, Ultimate moment , Mu = (0.125x30x302) = 3400knm, Ultimate shear, Vu = (0.5x30x30) = 450kn.

2. Find cross-sectional dimensions thickness of web if hf/d ratio is 0.23 and bw/b ratio is 0.25 and b is 0.5d?
a) 100mm
b) 110mm
c) 120mm
d) 30mm

Answer: c [Reason:] hf/d =0.23 and bw/b = 0.25 and b = 0.5d, Mu = 0.10fcubd2 d = (3400×106/0.10x50x0.5)1/3 = 1130mm, h = (1130/0.85) = 1300, b = 600mm, hf = (0.2×1130) = 250mm, adopt an effective depth, d = 1150mm, thickness of web, bw = (0.6vu/fth) = (0.6x450x103/1.7×1300) = 120mm.

3. Calculate working moment if design working load is 19.8kn/m covered over a span of 30m( actual self weight of girder is 8.8kn/m)?
a) 3000
b) 2000
c) 4340
d) 2230

Answer: d [Reason:] Actual self weight of the beam and the girder = 8.8kn/m, span = 30m Minimum moment Mmin = 990knm, Design working load = 19.8kn/m, Working moment Md = (0.125×19.8×302) = 2230knm.

4. Find the Permissible stresses and range of stresses for class 1 structure fcu = 50n/mm2,fck = 35n/mm2 according to BS: 8110 recommendations for fcu = 50n/mm2 and fci = 35n/mm2,fct = 0.5fci= 17.5n/mm2?
a) 16.5n/mm2
b) 12.56n/mm2
c) 13.56n/mm2
d) 12.00n/mm2

Answer: a [Reason:] fcu = 50n/mm2,fck = 35n/mm2 according to BS: 8110 recommendations for fcu = 50n/mm2 and fci = 35n/mm2, fct = 0.5fci= 17.5n/mm2For class 1 structure fu = htw = 0, fbr = (ɳfct-ftw) = (0.85×17.5) = 15n/mm2, fcw = 0.33fcu = (0.33×50) = 16.5n/mm2, fcu = (fcw-ɳfu) = 16.5n/mm2.

5. Find prestressing force if area is 36.75mm2 of eccentricity 580given finf is 26.5kn/m and zb is 99×106?
a) 405
b) 308
c) 453
d) 206

Answer: b [Reason:] Area = 36.75mm2, e = 580, finf = 26.5kn/m, zb = 99×106 p =(AfinfZb/Zb+Ae) =(367500×26.5x99x106/(99×106)+(367500×580)) = 308x104kn/m2.

6. Find force in cable using Freyssinet cables 12-8mm diameter and stressed to 1100n/mm2 of eccentricity 50 and the prestressing force is given as 1000n/mm2?
a) 660kn
b) 234kn
c) 300kn
d) 230kn

Answer: a [Reason:] 12 diameter, stress = 1100n/mm2, e = 50, prestressing force =1000n/mm2 Force in each cable = ( (12x50x1100)/1000)) = 660kn.

7. Find ratio for ultimate flexural strength at the centre – span section given that Aps = 3000mm2, d= 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm?
a) 9.5
b) 0.23
c) 6.7
d) 3.4

Answer: b [Reason:] Aps = 3000mm2, d= 1150mm, fcu = 50n/mm2, bw = 150mm, fpu = 1500n/mm2, b = 600mm, ht = 250mm, design ultimate moment mud = 3400knm, according to BS: 8110-1985, Aps = (Apw+Apf) = Apf = 0.45×50(600-150)(250/1500) = 0.45xfcu(b-bw)(hf/fpu) = 1680mm2, Apw = (1300-1680) = 1320mm2, ratio(fpuApw/fcubwd) = (1500×1320/50x150x1150) = 0.23.

8. Calculate the slope of cable section at support uncracked in flexure given that eccentricity is 410, length is 30m and stress induced is 1000?
a) 0.0547
b) 2.456
c) 0.0234
d) 0.0123

Answer: a [Reason:] e = 410, length = 30m, stress induced = 1000 Slope of cable θ = (4e/l) = ((4×410)/(30×1000)) = 0.0547.

9. Calculate the span section cracked in flexure (M=M0) Fcp = 23.4n/mm2, zb is 99×106 and stress induced is 1000?
a) 1200kn
b) 1850kn
c) 2300kn
d) 4300kn

Answer: b [Reason:] Fcp = 23.4n/mm2, zb is 99×106, stress is 1000 m0 = (0.8fcpZb) = (0.8 x 23.4 x (99×106/1000)) = 1850knm.

10. Find resultant maximum long term deflection if ϕ is 2.6, αy is 38.5mm, αg is 46mm, αp is 74.7mm?
a) 95mm
b) 35mm
c) 55mm
d) 20mm

Answer: a [Reason:] Ece = (Ec/1+ϕ) = (Ec/2.6), ϕ = 2.6, αy = 38.5mm, αg = 46mm, αp = 74.7mm, resultant maximum long term deflection = (2.6×46)+38.5-(0.85×74.7) = 95mm which is less than the code limit (span/250) = 120mm, ɳ = 0.85.

## Prestressed Concrete Structures MCQ Set 2

1. Design a pretensioned roof pull-in to suit the data Fcu, concrete cube strength = 50n/mm2, effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, Dc = 2400, and determine ultimate moment and shear?
a) 42 and 27.75
b) 54 and 27.75
c) 34 and 27.75
d) 20 and 28

Answer: a [Reason:] Wmin/Wud = KDcgβ(L/h)L/fcu(d/h)2 = 7.5×2400 x 9.81 x 0.125×25 x 6/50x 106x(0.85)2 = 0.094 Fcu, concrete cube strength = 50n/mm2, effective span = 6m, applied load = 5kn/m, dead load = 1.4, live load = 1.6, β = 0.125, k = 7.5, Dc = 2400, fcu = 50n/mm2, wmin = (0.094)(9.25) = 0.86kn/m, mu = (0.125×9.25×62) = 42knm, vu = (0.5×9.25×6) = 27.75kn.

2. Design cross sectional dimensions of pretensioned roof pull given that b is 0.5d?
a) 250
b) 260
c) 270
d) 280

Answer: c [Reason:] Mu = 0.10fcubd2 and if b = 0.5d D = (42×106x2/0.10×50)1/3 = 270mm.

3. Find the approximate thickness of web if b is 0.5d, d is 270mm, d/h ratio is 0.85, h is 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent?
a) 45mm
b) 43mm
c) 41mm
d) 42mm

Answer: b [Reason:] b = 0.5d, d = 270mm, d/h = 0.85, h = 315mm, adopt effective depth, d = 275mm overall depth , h is 320mm, width of flange of 160mm and Average thickness of flange is 70mm since sloping flanges are used, increases the flange thickness by 20 percent: Thickness of flange = (0.2×275) =55mm Approximate thickness of web = (0.85vu/fth) = (0.85×27.75×103/1.7×320) = 43mm.

4. Find minimum range of stresses if fct is 15n/mm2, fcw is 17, ftw is zero, fu is -1n/mm2, ɳ is 0.8?
a) 12 and 18n/mm2
b) 13 and 14n/mm2
c) 12 and 15n/mm2
d) 10 and 16n/mm2

Answer: a [Reason:] Range of stress fbr = (ɳfct-fcw) = (0.85×15-0) = 12n/mm2, ftr = (fcw – ɳfu) = (17-0.8x(-1)) = 17.8n/mm2, fct = 15n/mm2, fcw = 17, ftw = 0, fu = -1n/mm2, ɳ = 0.8.

5. Find minimum section modulus given data is mg is 3.86×106, mq is 22.50×106, fbr is given as 12 and the loss ratio is 0.8?
a) 134×104
b) 182×104
c) 123×104
d) 120×104

Answer: b [Reason:] mg = 3.86×106, mq = 22.50×106, fbr = 12, loss ratio = 0.8 Zb > or equal (mq+(1-ɳ)mg/fbr) > or equal ((22.50×106)+(1-0.8)3.86×106)/12) Greater than equal to 182x104mm3.

6. Find the supporting force if given characteristic strength is -1, moment of gravity is 3.86×106, zt = 230×104?
a) -2.68n/mm2
b) -3.45n/mm2
c) -1.23n/mm2
d) 13.56n/mm2

Answer: a [Reason:] p = (A(finfZb+fsubZt)/Zt+Zb) Finf = ((ftw/ɳ+(mq+mg)/ɳzb)) = ( 0+ (26.36×106/0.8x230x104)) Fsup = (fu – mg/zt) = (-1 – (3.86×106)/(230×104)) = -2.68n/mm2.

7. Check for ultimate flexural strength if given Aps is 154mm2, fpu is 1600n/mm2, b is 160mm, fcu is 50n/mm2and diameter is 265mm?
a) 9.65
b) 0.116
c) 3.442
d) 2.345

Answer: b [Reason:] Aps = (38.5xy) = 154mm2, fpu = 1600n/mm2, b = 160mm, fcu 50n/mm2,, d = 265mm (Apsfpu/bdfcu) = (154×1600/160x265x50) = 0.116.

8. Find ultimate shear strength (check it for safe against shear failure) if vu is 27.75kn, Loss ratio is 0.8, prestressing force is 182000, area is 31400, breadth is 50 where height is 320, prestressing force is 1.7, fcp = 4.65, ft is 1.7?
a) Safe
b) Unsafe
c) Zero
d) Collapse

Answer: a [Reason:] Fcp = (ɳp/A) = (0.8×182000/31400) = 4.65n/mm2 Vcw = 0.67bh(f12+0.8fcpft)1/2 = (0.67x50x320(1.72+0.8×4.65×1.7)1/2/103) = 33.2kn Vcw > Vu hence safe against shear failure.

9. Check for deflection due to prestressing force if given data is Prestressing force is 182×103 eccentricity of cable is 105, Length of the cable is 1000, elastic modulus of concrete is 34×103, Moment of inertia is 3200×105?
a) 9.4
b) 4.5
c) 6.8
d) 9.8

Answer: c [Reason:] P = 182×103 e = 105, L = 1000, elastic modulus of concrete = 34×103, I = 3200×105 PeL2/8EcI = (182×103x105x62x10002/8x34x103x3200x105) = 6.8mm.

10. Find the deflection due to self weight given that ϕ = 1.6, Ee = 2.6Ece, elastic modulus of concrete is 34×103, gravity is given as 6, self weight is 0.76, Length of the cable is 1000, elastic modulus of concrete is 34×103 , Moment of inertia is 3200×105?
a) 1.66mm
b) 5.3mm
c) 23.4mm
d) 1.02mm

Answer: d [Reason:] Ece = Ec/1+ϕ, ϕ = 1.6, Ee = 2.6Ece Deflection due to self weight g = (5gL4/384EcI) = (5×0.76×64x10004/384x34x103x3700x105) = 1.02mm.

## Prestressed Concrete Structures MCQ Set 3

1. The ultimate shear resistance for any given section, vc should be least of values:
a) vcw and vcf
b) vew and vef
c) vrw and vrf
d) vdw and vdf

Answer: a [Reason:] At any given section, the ultimate shear resistance, vc should be the least of values vcw and vcf where, vcw = ultimate shear resistance of section cracked in web, vcf = ultimate shear resistance of section cracked in flange.

2. What should be provided if the shear force due to ultimate load is less than the shear force of the concrete?
a) Tensile reinforcement
b) Shear reinforcement
c) Principle reinforcement
d) Compressive reinforcement

Answer: b [Reason:] When V the shear force due to ultimate loads is less than the shear force of the concrete vc, the shear force which can be carried by the concrete, a minimum shear reinforcement should be provided in the form of stirrups with spacing.

3. The spacing provided for shear reinforcement is given as:
a) Sv = (Asv0.87fy/0.4b)
b) (Asv0.91fy/0.4b)
c) (Asv0.12fy/0.4b)
d) (Asv0.23fy/0.4b)

Answer: a [Reason:] The spacing provided for shear reinforcement Sv = (Asv0.87fy/0.4b), Sv = spacing of stirrups along the length of member, Asv = cross sectional area of stirrup legs effective in shear, b = breadth (for T,I,L beams it is taken as breadth of rib, bw).

4. When the shear force due to ultimate loads is less than 0.5 times shear force of concrete then shear reinforcement is:
a) Provided
b) Not provided

Answer: b [Reason:] When the shear force due to ultimate loads V is less than 0.5 times the shear force of concrete: (v < 0.5vc) Then no shear reinforcement is provided in the members of low importance.

5. When v > vc condition exits, then the shear reinforcement provided includes which extra terms than actual spacing equation?
a) fy, dt, v, vc
b) ft, dt, v, vc
c) dt, v, vc
d) fe, dt, v, vc

Answer: a [Reason:] When V exceeds Vc shear reinforcement is required conforming to the relation (V > Vc) The extra terms used are fy , dt, v, vc these are included in the spacing equation Sv Sv = (Asv0.87fydt / V-Vc).

6. The term dt in the spacing equation is termed as:
a) Lowest value of depth
b) Neutral value of depth
c) Highest value of depth
d) Peak value of depth

Answer: c [Reason:] When v > vc, the spacing provided in that: dt is the highest value of depth from the extreme compression fiber to longitudinal bars and depth from extreme compression fiber to centroid of tendons.

7. The spacing of stirrups for maximum shear stress is:
a) 0.9dt
b) 0.10dt
c) 0.12dt
d) 0.7dt

Answer: d [Reason:] The spacing of stirrups should exceed neither 0.75dt nor 4 times the web thickness of flanged members, for maximum shear stress the spacing of stirrups: Sv > 0.75dt = > x web thickness.

8. The maximum spacing (Sv)max is provided for the condition:
a) V > 1.8Vc
b) V > 1.5Vc
c) V > 2.0Vc
d) V > 3.5Vc

Answer: a [Reason:] When V exceeds 1.8Vc, the maximum spacing should be reduced to 0.5dt, the lateral spacing of the individual legs of the stirrups provided at a cross section should not exceed 0.75dt, if V > 1.8Vc, the maximum spacing is: (Sv) max = 0.5 dt.

9. The maximum shear stress value for M35 and M50 is:
a) 3.5 and 4.0
b) 3.7 and 4.6
c) 3.8 and 6.0
d) 4.2 and 4.0

Answer: b [Reason:] The maximum shear stress value for M35 grade concrete is 3.7n/mm2, the maximum shear stress value for M50 grade concrete is 4.6n/mm2, M30 grade concrete is 3.5n/mm2, M40 grade concrete is 4.0n/mm2, M45 grade concrete is 4.3, M55 grade concrete is 4.8n/mm2.

10. The section should be redesigned if the nominal shear stresses:
a) Exceeds the given values
b) Equal
c) Zero
d) Constant

Answer: a [Reason:] When the nominal shear stress, v/bd exceeds the maximum permissible shear stress values for the particular grades of concrete like M-30,M-35,M-40,M-45,M-50,M-55 and above then the section should be redesigned( this is according to table of maximum shear stress(n/mm2) (IS: 1343-1980).

## Prestressed Concrete Structures MCQ Set 4

1. Which type of cables are advantages in reducing the effective shear?
a) Straight
b) Curved
c) Trapezoidal
d) Longitudinal

Answer: b [Reason:] Curved cables are advantageous in reducing the effective shear and together with the horizontal compressive prestress, reduce the magnitude of the principal tension and the effect of shear is to induce tensile stresses on diagonal planes and prestressing is beneficial since it reduces the magnitude of the principal tensile stress in concrete.

2. The various codes recommend empirical relations to estimate:
a) Ultimate shear resistance
b) Ultimate torsional resistance
c) Ultimate bending resistance

Answer: a [Reason:] The various codes recommend empirical relations to estimate the ultimate shear resistance of the section by considering the flexure shear and web shear cracking modes and the design shear resistance should exceeded the ultimate shear due to the transverse loads and if not, suitable transverse reinforcements are designed to resist the balance shear force.

3. Which type of shear reinforcement should be provided for members with thin webs?
a) Maximum shear reinforcement
b) Minimum shear reinforcement
c) Nominal shear reinforcement
d) Tensile reinforcement

Answer: c [Reason:] In members with thin webs such as I and T sections nominal shear reinforcements have to be provided to prevent cracking due to variations in temperature the provisions for design of shear reinforcements prescribed in British, American and Indian standard have been dealt with in the case off structural concrete members subjected to torsion, shear stresses develop depending upon the type of cross section and magnitude of torque and the shear stresses in association with the flexural stresses may give rise to principal tensile stress the value of which when it exceeds tensile strength of the concrete results in the development of cracks on the surface of the member.

4. The pre and post tensioned members with bonded tendons bond stress between:
a) Steel and concrete
b) Steel and water
c) Steel and aggregates
d) Steel and plastic

Answer: a [Reason:] Pre tensioned or post tensioned members with bonded tendons develop bond stresses between steel and concrete when the sections are subjected to transverse shear forces due to the rate of change of moment along length of the beam and in the case of type 1 and 2 members, which are uncracked at service loads, the flexural bond stresses developed are computed by considering the complete section.

5. In case of pre tensioned member, the computations of transmission length is influenced by:
a) Bond
b) Flexure
c) Torsion
d) Tension

Answer: a [Reason:] In the case of pre tensioned members, the computations of the transmission length at the ends is of practical significance since the support positions are influenced by bond and transmission length.

6. When prestress is transferred to concrete by means of external anchorages which pressure is developed:
a) Bearing
b) Twisting
c) Torsion
d) Bent

Answer: a [Reason:] In the case of post tensioned members where prestress is transferred to concrete by means of external anchorages, the bearing pressures develop behind the anchorages have to be investigated and suitably controlled to prevent crushing failure of the end block zone.

7. The bearing pressure on the concrete is given as:
a) 0.4fci
b) 0.8fci
c) 0.12fci
d) 0.2fci

Answer: b [Reason:] According to the Indian standard code IS: 1343-1980, the permissible unit bearing pressure on the concrete after allowing for all losses is limited to: 0.48fci (Abr/Apan) or 0.8 fci, fci = cube strength of concrete at transfer, Abr= bearing area, Apan = punching area.

8. The effective punching area Apun is generally the contact area of:
a) Tendon device
b) Anchorage device
c) Stress device
d) Strain device

Answer: b [Reason:] The effective punching area is generally the contact area of the anchoring device and the bearing area is taken as the maximum area of that portion of the member which is geometrically similar and concentric to the effective punching area.

9. The end block of a beam has a rectangular section 100mm wide by 200mm deep, the force of cable is 200kn. Find the actual bearing pressure?
a) 30n/mm2
b) 40n/mm2
c) 20n/mm2
d) 10n/mm2

Answer: c [Reason:] force of cable = 200kn, b = 100mm,d = 200mm Bearing area Abr = (100×100) = 10000mm2, Actual bearing pressure = (200×103) / 10000 = 20n/mm2.

10. The end block of a prestressed concrete beam has a rectangular section; a cable carrying a force of 200kn is to be anchored against the end block at the centre if the cube strength of concrete at transfer is 30n/mm2. Design the maximum permissible bearing pressure?(Abr = 10000mm2)
a) 24n/mm2
b) 48n/mm2
c) 54n/mm2
d) 12n/mm22

Answer: a [Reason:] Assuming an anchor plate of size 50mm by 50mm, punching area Apun = (50×50) = 2500mm2, Abr = 10000mm2, maximum permissible bearing pressure = 0.48fci(Abr/Apun) or 0.8fci whichever is smaller, fb = 0.48x30x(10000/2500)1/2 = 28.8n/mm2 or ( 0.8×30) = 24n/mm22.

## Prestressed Concrete Structures MCQ Set 5

1. The span of trusses generally lies in the range of:
a) 18-30m
b) 20-30m
c) 40-50m
d) 12-18m

Answer: a [Reason:] The height of reinforced concrete truss at midspan is in the range of 1/7 to 1/9 of its span length and the span of the trusses generally lies of the range of 18-30m and these spans are commonly used for bridge decks.

2. The width of various components is kept constant at:
a) 100-200mm
b) 200-350mm
c) 150-300mm
d) 1200-1400mm

Answer: b [Reason:] The width of the various compression and tension members is kept constant at 200-350mm depending upon the span of the truss and the depth of top bottom members which are in compression generally is the range of 200-300mm.

3. The bottom tie member should be sufficient size to:
a) Tension
b) Stress
c) House
d) Strain

Answer: c [Reason:] The bottom tie members should be of sufficient size to house the pretensioned wires or post tensioned cables and the depth is around 200mm for spans of 15m and increasing to 300mm for spans of 30m.

4. The depth of diagonal web members is at a range of:
a) 100-150
b) 200-250
c) 110-210
d) 114-115

Answer: a [Reason:] The depths of diagonal web members which are in comparison and tension generally vary in the narrow range of 100-150mm but they are not limited and they change the values from one to another diagonal web members.

5. The use of concrete trusses with modular coordination for spans of 6and 9 are common for countries like:
a) America
b) Poland
c) India
d) Nueziland

Answer: b [Reason:] The use of concrete trusses with modular coordinates for spans of 6,9,12,15,18,24,30 and 36m with a base module of 30m is most common for industrial buildings of east European countries such as Russia, Poland, Slovakia, Germany, Branko zezelji has reported the construction of reinforced concrete wit prestressed tie members for spans up to 60m in erstwhile Yugoslavia.

6. The precast pretensioned cored slabs having circular or elliptical cavities have been used in:
a) France
b) Russia
c) Phillipines
d) Losangles

Answer: b [Reason:] Prestressed concrete is ideally suited for office, industrial and commercial buildings when large column free open spaces are required and precast pretensioned cored slabs having circular or elliptical cavities have been widely used in Russia for floor panels of multistory buildings.

7. The prestressed concrete folded plate units are also widely used to cover roofs of:
a) Industrial buildings
b) Commercial buildings
c) Structural buildings
d) Regional buildings

Answer: a [Reason:] Folded plates prestressed with curved cables to cover the 60m span roof of a town hall in west phalia, germany and in comparison with types of roofs, folded plate roof provides the most economical solution with minimum maintenance costs.

8. Prestressed hollow inverted pyramids have been used as a transfer:
a) Girder system
b) Flange system
c) Block system
d) Chain system

Answer: a [Reason:] Prestressed hollow inverted pyramids have been used as a transfer: Girder system to support the four storey complex housing the administration building for engineering construction corporation limited at manapakkam in madras and the upper four storeys housing the administrative complex rest on four hollow prestressed pyramids which in turn are supported on hollow core walls.

9. Another example of application of prestressed concrete in buildings can be seen in construction of:
a) 11 – storied structure
b) 13 – storied structure
c) 10 – storied structure
d) 15 – storied structure