Category: Prestressed Concrete Structures Objective Questions

Prestressed Concrete Structures Multiple Choice Questions on “Pretensioning Systems”.

1. The tendons in the pretensioning system are tensioned between ___________
a) Rigid anchorages
b) Hydraulic jacks
c) Concrete beds
d) Variable beams
Answer: a
Clarification: In the prestressing system, the tendons are first tensioned between rigid anchorage blocks cast on the ground or a column or a unit mould type pretensioning bed, prior to the casting of concrete in the moulds.

2. When the concrete attains sufficient strength, which elements are released?
a) Jacks
b) Casting bed
c) Tendons
d) Beams
Answer: a
Clarification: High early strength concrete is often used in factory to facilitate early stripping the reuse of moulds and when the concrete attains sufficient strength, the jacking pressure is released and the edge of tendon at its either side is formed to an abutment and its other edge is to be pulled with the application of jack.

3. Which is one of the systems used for pretensioning?
a) Magnel-Balton system
b) Freyssinet system
c) Gifford-Udall system
d) Hoyer’s long line method
Answer: d
Clarification: Hoyer’s long line method is the system used in pretensioning and the other systems like Freyssinet, Gifford-Udall, and Magnel-Balton are post tensioning systems and large numbers of beams are produced in an individual alignment.

4. Hoyer’s system of pre tensioning is generally adopted for ___________
a) Small scale members
b) Large scale members
c) Middle span members
d) End members
Answer: b
Clarification: Hoyer’s system is generally recommended when the production of pretensioned members is required on a large scale is the principle which are used to precast the beams the post tensioning system was considered to pretensioning system when used for large spans, due to this reason the pre tensioning system was replaced by post tensioning system.

5. The transfer of prestress of concrete is achieved by ___________
a) Plates
b) Rings
c) Steel bars
d) Jacks
Answer: d
Clarification: The transfer of prestress to concrete is usually achieved by large hydraulic or screw jacks by which all the wires are simultaneously released after the concrete attains the required compressive strength generally strands of up to 18mm diameter and high tensile wires of up to 7mm diameter anchor themselves satisfactorily with the help of surface bond and the interlocking of the surrounding matrix in the micro indentations on the wires.

6. The bond of prestressing wires in Hoyer’s system can be formed by ___________
a) Helical crimping
b) Tangential crimping
c) Circular crimping
d) Diode crimping
Answer: a
Clarification: Bond of prestressing wires may be considerably improved by forming surface indentations and by helical crimping of wires in Hoyer’s system, strands have considerable better bond characteristics than plain wires of equal cross sectional area supplementary anchoring devices are required when single wires of larger diameter are used in the pretensioned units.

7. The Hoyer’s method of prestressing is done by ___________
a) Pulling out of wires
b) Pushing wires
c) Heating of wires
d) Stressing of wires
Answer: a
Clarification: Hoyer’s system of prestressing involves pulling out of wires between two bulkheads which are separated at large distances in order to produce a larger number of beams in an individual alignment, the concrete is to be poured by providing appropriate shuttering between the beams and then the wires are united after hardening of concrete and cutoff.

8. Hoyer’s system of pretensioning can be done for beams.
a) 1
b) More than 2
c) Less than 2
d) 2
Answer: b
Clarification: United beams more than 2 are hardened in Hoyer’s method of pretensiong in case of large distances between each beam, the most commonly used devices are the Weinberg clip developed in France and the Dorland clip developed in the united states these clips are clamped on to the tensioned wires close on the end diaphragms of the units before concreting operations.

9. The Hoyer’s system of prestressing proves to be economical for ___________
a) Pre tensioning system
b) Post tensioning system
c) Beam casting
d) Bed casting
Answer: b
Clarification: The post tensioning system was considered as economical when compared to pre tensioning system, when used for larger spans due to this reason pre tensioning system was replaced by post tensioning system, for mass production of pretensioned elements, the long line process developed by Hoyer is generally used in the factory and in this method the tendons are stretched several hundred meters apart so that a number of similar units may be cast along the same group of tensioned wires.

10. In Hoyer’s system the projection of plugs left in concrete exceeds beyond ___________
a) Middle of member
b) End of member
c) First of member
d) Transfer part of member
Answer: b
Clarification: One of the disadvantages of Hoyer’s system is the projection of plugs which are left in concrete exceeds beyond the end of the member, application of heavy jacks which are uneconomical, projection of plugs which are left in concrete exceeds beyond the ends of the member, additional reinforcement is required to prevent failure of shear.

Prestressed Concrete Structures Multiple Choice Questions on “Loss Due to Anchorage Slip”.

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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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
Clarification: 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 __________
a) E_sA/L
b) E_aA/L
c) E_cA/L
d) E_wA/L
Answer: a
Clarification: 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/AE_s, But moment Loss of stress, (Δf)_a = P/A,
By considering both equations (Δf)_a = E_sΔ/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?
a) 3.5%
b) 4.5%
c) 5.5%
d) 8.7%
Answer: a
Clarification: Loss of stress due to anchorage slip = (E_sΔ/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, E_s 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?
a) 39.8kn
b) 40.2kn
c) 37.8kn
d) 48.8kn
Answer: c
Clarification: 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 __________
a) Lin
b) Marks
c) Keifer
d) Neville
Answer: a
Clarification: 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?(E_s = 210kn/mm², E_c = 35kn/mm²).
a) 21 and 15
b) 35 and 25
c) 15 and 20
d) 5 and 10
Answer: a
Clarification: A = 320mm², b = 200mm, d = 300mm, e = 50mm, p = 1000n/mm², l = 10m, E_s = 210kn/mm², E_c = 35kn/mm², Slip at anchorage = (1×210×10³/10×1000) = 21, Friction effect = (1000×0.0015×10) = 15.

Prestressed Concrete Structures Multiple Choice Questions on “Bond Stresses”.

1. The magnitude of bond stresses is developed between ____________
a) Concrete and steel
b) Aggregates and steel
c) Water and steel
d) Bricks and steel
Answer: a
Clarification: The magnitude of bond stresses developed between concrete and steel and its variation in the transfer zone of pretensioned beam, the deformation of concrete shrinkage depends on environmental conditions, age of concrete, size of concrete, concrete composition, etc.

2. The bond stress is zero at the distance equal to the ____________
a) Tensile length
b) Span length
c) Transmission length
d) Anchorage length
Answer: c
Clarification: The bond stress is zero at the end but builds up rapidly to a maximum over a very short length and this value decreases as the stress in the wire builds up at a distance equal to the transmission length, the bond stress is almost zero while the stress in steel and concrete reach their maximum values.

3. The relations proposed by Marshall for bond stress are given as ____________
a) (τ _bp)x = (τ _bp)_maxe^-4ψx/ϕ
b) (τ _bp)x = (τ _bp)_maxe^-4ψx
c) (τ _bp)x = (τ _bp)_maxe^-4ψ
d) (τ _bp)x = (τ _bp)_maxe^-4
Answer: a
Clarification: Based on tests conducted at the university of leeds, the following relations have been proposed by marshall (τ _bp)x = (τ _bp)_maxe^-4ψx/ϕ, fx = fsc(1-e^-4ψx/ϕ), (τ _bp)_x = maximum value of bond stress, (τ_bp)_max = bond stress of a distance x from the free end, ϕ = diameter of wire, f_x = stress in steel a distance x from the free end, f_se = effective stress in steel at the ends of transfer zone, ψ = constant, x = distance measured from the free end in mm.

4. According to Marshall the wires of 2 and 5mm diameter are stressed to ____________
a) 1500 and 1700n/mm²
b) 1575 and 1100n/mm²
c) 1400 and 1800n/mm²
d) 1250 and 1600n/mm²
Answer: b
Clarification: Based on tests conducted according to Marshall using wires of 2 and 5mm diameter stressed to 1575 and 1100n/mm² respectively in conjunction with a concrete having cube strength of 80n/mm², the values of maximum bond stress and constant ψ found to be 7.42n/mm² and 0.00725 respectively.

5. The magnitude of the average bond stress compared to the maximum local bond stress is considerably ____________
a) Less
b) More
c) Zero
d) Constant
Answer: a
Clarification: The magnitude of the average bond stress is considerably less than the maximum local bond stress but according to the investigations of R_os, the average bond stress varied from 3.25 to 1n/mm² for round wires of 1.5 to 5mm diameter in the case of wires initially tensioned to a stress of 1200n/mm² and tensile stress shall not be allowed at any loading stage up to cracking in case of members assembled out of precast blocks.

6. The stress in a steel wire gradually increases from zero to the ____________
a) Middle
b) Bottom
c) End
d) Top
Answer: c
Clarification: The stress in a steel wire gradually increases from zero at the end of the beam to 100 percent of the effective stresses at the end of the transmission length, after the beam is placed the slab is casted and different rates of shrinkage imposed forces are developed because a moment is induced if the slab is not at the neutral axis of section.

7. The prestressing force of 90 to 95 percent is obtained at about ____________
a) Half of the transmission length
b) Three fourth of transmission length
c) Four third of transmission length
d) 1.5 of transmission length
Answer: b
Clarification: The effective prestressing force of 75 to 80 percent develops to about half of the transmission length, 90 to 95 percent of the prestressing force is attained at about three fourths of the transmission length from the end face of the beam.

8. A pre tensioned beam is prestressed with the cube strength of concrete at transfer is 30n/mm² (β = 0.0235) calculate the transmission length?
a) 645mm
b) 1100mm
c) 485mm
d) 1600mm
Answer: c
Clarification: β = 0.0235, cube strength of concrete is 30n/mm²,
Transmission length (L_t) = ((f_cu)^1/2 x 103/β)^1/2
= ((30)^1/2 x 10³/0.0235) = 485mm.

9. A pretensioned beam is prestressed using 5mm diameter wires, τ _bp = 7.42, ϕ = 0.00725, L_t = 485mm. Calculate the bond stress at l/4 and l/2 the transmission length from the end?
a) 3.7n/mm² and 1.82n/mm²
b) 2.2n/mm² and 5.8n/mm²
c) 4.8n/mm² and 10.5n/mm²
d) 1.5n/mm² and 7.4n/mm²
Answer: a
Clarification: Bond stress given by (τ _bp)_x = (τ _bp)_maxe^-4ψx/ϕ = 7.42 e-(4×0.00725xX)/5
τ _bp = 7.42e^-0.0058X
Bond stress at L/4 is given by: τ _bp = 7.42e^{-0.0058×121.25} = 8.7n/mm², τ _bp = 7.42e^{-0.0058×242.5} = 1.82n/mm².

10. A pre tensioned beam is prestressed using 5mm diameter wires with an initial stress of 80 percent of the ultimate strength of steel fpu = 1600n/mm². Calculate average bond stress?
a) 4.20n/mm²
b) 3.30n/mm²
c) 2.0n/mm²
d) 4.9n/mm²
Answer: b
Clarification: f_pu = 1600n/mm², d = 5mm, initial stress = 80% = 0.08,
Average bond stress = (τ _bp)_average
= (19.6×0.8×1600/πx5x485) = 3.30n/mm².

Prestressed Concrete Structures Questions and Answers for Campus interviews on “Estimation of Self Weight of Beams”.

1. The computation of total ultimate moment required for the design of prestressed beams, knowledge of is necessary?
a) Spans
b) Self weights
c) Loads
d) Deflection
Answer: b
Clarification: Generally, the self weight may be assumed on the basis of previous experience and the use of design chart containing dimensions of beams for various spans and applied loads as recommended by magnel is very useful in the regard.

2. The estimation of self weight is expressed as ____________
a) w_min/w_ud = KD_cgβ(L/h)L/f_cu(d/h)²
b) w_min/w_ud = KD_cgβ(L/h)L/f_cu
c) w_min/w_ud = KD_cgβ(L/h)
d) w_min/w_ud = KD_cgβ
Answer: a
Clarification: Bennelt has recently proposed a simple formula for estimating the self weight of the girder by considering several influencing parameters
w_min/w_ud = KD_cgβ(L/h)L/f_cu(d/h)²
w_min = Self weight or minimum load, L = effective span, K = numerical constant, D_c = density of the concrete member, g = acceleration, β = moment coefficient, h = overall depth of girder,
w_ud = Ultimate design load.

3. In the case of unsymmetrical I girders the range of values of h_f/d for economical designs is generally?
a) 0.15 to 0.10
b) 0.15 to 0.25
c) 0.8 to 1.0
d) 3.4 to 6.0
Answer: b
Clarification: In the case of unsymmetrical I girders the range of values of h_t/d and b_w/b for economical designs is generally 0.15 to 0.25 and 0.2 to 0.3 respectively, however the thickness of web, b_w is designed based on the dual criteria shear and housing the cables with adequate cover.

4. The breadth of the compression face may be assumed by considering the number of ____________
a) Anchorages
b) Splices
c) Girders
d) Ridges
Answer: c
Clarification: In dimensioning prestressed concrete flexural members the effective depth and breadth of the section at the compression face are determined solely on basis of the ultimate flexural strength requirements and The breadth of the compression face may be assumed by considering the number of covering a given width of bridge deck of a suitable ratio of b’d being in the range of 0.4 to 0.6.

5. The thickness of the web is generally determined on which of the basis?
a) Shear stress
b) Shear strength
c) Principle shear
d) Tensile shear
Answer: b
Clarification: The thickness of the web is generally determined on the basis of shear strength considerations discussed according to british code recommendations shear reinforcements are not required where V is less than 0.5v_c and in members of minor imporatance when the shear for V exceeds (V_c to 0.4b_wd), shear reinforcement are designed at spacing S_v = A_sv0.8f_y/0.4b_w.

6. The small span girders with straight tendons, b_w is?
a) 0.85v_u/f_th
b) 0.60v_u/f_th
c) 0.70v_u/f_th
d) 0.69v_u/f_th
Answer: a
Clarification: In the case of small span prestressed members, thinner webs of about 40 to 60mm may be used however in the case of long span, heavily loaded girders when large, curved cables have to pass through the webs a minimum thickness of 120 to 150mm is mandatory to accommodate the cables with adequate cover.

7. The condition that the principal tensile stress is not to exceed the tensile strength of concrete yields a criterion of the type ____________
a) b_w > (v_u/ (I/s) f_t(1-f_cp/f_t)^1/2
b) b_w > (v_u/ (I/s) f_t(1+f_cp/f_t)^1/2
c) b_w > (v_u/ (I/s)
d) b_w > (v_u/ (I/s) f_t
Answer: b
Clarification: The value of the shear moment arm I/S varies between 0.67 and 0.85h for I sections, the ratio f_cp/f_t generally varies between 2 and 3 for small span girders with straight tendons for long span girders with curved tendons, the ratio, f_cp/f_t can be taken between 3 and 4 and the effective shear as 0.8vu since the curved cables contribute to the ultimate shear resistance of the section.

8. The ultimate design load includes?
a) Partial factor of safety and live load
b) Ultimate load
c) Tensile load
d) Overloaded load
Answer: a
Clarification: The ultimate load includes the self weight enhanced by partial factor of safety
γ_f1q+γ_f2w_min, W_ud = γ_f1q/1-γ_f2(W_min/W_ud).

9. The value of numerical constant K is between ____________
a) 4 to 5
b) 6 to 7.5
c) 4 to 8
d) 5 to 9
Answer: b
Clarification: The value of numerical constant K is between 6 to 7.5 for rectangular sections and I section girders of short spans, while it takes a value between 4 and 5 for the flanged T or I section girders of long spans for self weight equation.

10. The load combination of dead and imposed has a beneficial dead load of ____________
a) 1.0
b) 1.5
c) 1.8
d) 2.0
Answer: a
Clarification: Load combinations: Dead and imposed (and earth and water pressure) – Dead beneficial is 1.0, Dead and wind (and earth and water pressure) – Dead beneficial is 1.0, Dead and wind and imposed (and earth and water pressure) – Dead beneficial is 1.2.

Prestressed Concrete Structures for Campus Interviews,