300+ REAL TIME Prestressed Concrete Structures Objective Questions & Answers 2023

250+ TOP MCQs on Loss Due to Elastic Deformation and Answers

Basic Prestressed Concrete Structures Questions and Answers on “Loss Due to Elastic Deformation”.

1. The loss of prestress due to elastic deformation of concrete depends on ____________
a) Modular ratio and average stress
b) Modular elasticity and shear
c) Prestress in concrete
d) Modulus of elasticity of steel
Answer: a
Clarification: The loss due to elastic deformation of concrete depends on the modular ratio and the average stress in concrete at the level of steel, consider a post tensioned member which is prestressed by a single tendon and the shortening of concrete occurs till the tendon is jacked and no shortening of concrete is observed after it.

2. The term E_c in losses developed by elastic deformation is expressed as ____________
a) P_e/A
b) P_c/E_a
c) P/A_cE_c
d) E_a/E_l
Answer: c
Clarification: The term E_c is defined as strain in concrete and the equation for loss due to elastic deformation is given as E_c = P_c/E_c = P/A_c×1/E_c, the tension in the tendon is obtained after the elastic shortening of concrete and therefore, there will not be losses due to elastic shortening.

3. The term E_s in losses developed by elastic deformation is defined as ____________
a) Shear in steel
b) Torsion in steel
c) Strain loss in steel
d) Loading in steel
Answer: c
Clarification: The term Es is defined as strain loss in steel, E_s = Δf_s/E_s,
Δf_s = Loss of stress in steel, Es = strain loss in steel.

4. The loss of stress in steel due to elastic shortening or deformation is ____________
a) α_ef_c
b) α_cf_c
c) α_c/f_c
d) α_e/f_c
Answer: a
Clarification: Loss of stress in steel due to elastic shortening is α_ef_c,
α_e = E_s/E_c = modular ratio, f_c = prestress in concrete at the level of steel, E_s = modulus of elasticity of steel, E_c = modulus of elasticity of concrete.

5. A pretensioned concrete beam, 100mm wide and 300mm deep in prestressed by straight wires and modulus of elasticity of steel and concrete are 210 and 35n/mm2. Find modular ratio?
a) 14
b) 7
c) 6
d) 10
Answer: c
Clarification: b = 100mm, d = 300mm, E_s = 210kn/mm², E_c = 35n/mm²
α_e = E_s/E_c = (210/35) = 6n/mm².

6. A pretensioned concrete beam 200mm wide and 300mm deep, is prestressed by straight wires carrying an initial force of 150kn at eccentricity of 50mm, area of steel wires is 188mm². Find initial stress in steel?
a) 1400
b) 800
c) 200
d) 100
Answer:b
Clarification: b = 200mm, d = 300mm, p = 150kn = 150×103, e = 50mm, a = 188n/mm²,
Initial stress in steel = (150×10³/188) = 800n/mm².

7. A pre tensioned concrete beam 100mm wide and 300mm deep, initial force of 150kn at an eccentricity of 50mm, moment of inertia is 225×10⁶mm⁴, initial stress in steel is 400n/mm², modular ratio is 8. Estimate the percentage loss?
a) 10%
b) 5%
c) 14%
d) 20%
Answer: c
Clarification: P = 150kn, y = d/6 = 300/6 = 50mm, a = (100×300) = 3×10⁴, I = 225×10⁶, α_e = 8, initial stress = 400n/mm², Stress in concrete, f_c = (150×10³/3×10⁴)+(150×10³×50×50/225×10⁶) = 6.66n/mm²,
Loss of stress due to elastic deformation of concrete = α_ef_c = (8×6.66) = 53n/mm²,
Percentage of loss of stress in steel = (53×100/400) = 13.25% = 14%.

8. A rectangular concrete beam 360mm deep and 200mm wide, is prestressed by means of fifteen 5mm diameter wires located 65mm from the bottom of the beam and three 5mm wires, located 275mm from top of the beam, initial tension stress is 840n/mm². Calculate prestressing force?
a) 504×10²kn
b) 500×10²kn
c) 620×10²kn
d) 400×10²kn
Answer: a
Clarification: Position of the centroid of wires from soffit of the beam y = ((15×65)+(3×25)/(15+3)) = 100mm, e = (150-100) = 50mm, area of concrete A = (200×300) = 6×10⁴mm², I = (200×300³)/12 = 45×10⁷mm⁴, Prestressing force = initial stress×area = 840×6×10⁴ = 504×10⁵N = 500×10²kn.

9. A post tensioned concrete beam, 100mm wide and 400mm deep is prestressed by three cables, each with a cross sectional area of 50mm², initial stress of 1200n/mm². Calculate the stress in concrete at level of steel?
a) 2.4n/mm²
b) 2.0n/mm²
c) 2.7n/mm²
d) 1.5n/mm²
Answer: c
Clarification: Force in each cable, p = (50×1200) = 60×10³n = 60kn, A = 3×10⁴mm², I = 225×10⁶mm⁴, e = 50mm, y = 50mm stress in concrete at the level of steel f_c = (60×10³/3×10⁴)+(60×10³×50×50/225×10⁶) = 2.7n/mm².

10. The loss of stress due to successive tensioning of curved cables in elastic deformation of concrete is estimated by considering ___________
a) Initial stress
b) Average stress
c) Bondage stress
d) Anchorage stress
Answer: b
Clarification: In most bridge girders, the cables are curved with maximum eccentricity in center of the span in such cases loss of stress due to elastic deformation of concrete is estimated by considering stress in concrete at the level of steel.

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250+ TOP MCQs on Ultimate Shear Resistance and Answers

Prestressed Concrete Structures Multiple Choice Questions on “Ultimate Shear Resistance “.

1. How many types of modes of shear cracking are present in structural concrete beams?
a) 4
b) 2
c) 5
d) 8
Answer: b
Clarification: Research over the years have shown that there are two major modes of shear cracking in structural concrete beams, web shear cracks, flexure shear cracks and different modes of shear failure patterns are also considered like diagonal tension failure, shear compression failure, web compression failure.

2. The web shear cracks generally start from __________
a) Interior point
b) Exterior point
c) Edge
d) Mid span
Answer: a
Clarification: Web shear cracks generally start from an interior point, when the local principal tensile stress exceeds the tensile strength of concrete, the British code (BS:8110:-1985) and the Indian code (IS:1343-1980) specify a modified version of this relation given by: V_cw = 0.67b_wh(f₁²+0.8f_cpf_t)^1/2 .

3. The web shear cracks are developed when the beam is subjected to __________
a) Point load
b) Concentrated loads
c) Shear loads
d) Uniformly distributed load
Answer: b
Clarification: Web shear cracks are likely to develop in highly prestressed beams with thin weds, particularly when the beam is subjected to large concentrated loads near a simple support, in which the value of 0.67 h is somewhat lower for flanges sections this together with reduced value of 0.8f_cp results in conservation estimates of the shear resistance of flanges sections and if there are inclined cables, the sheaing force v_cw is increased by an amount equal to the vertical component of the prestressing force, in the above expression for computing v_cw the tensile strength of concrete may be assumed as f_t = 0.24(f_ck).

4. In which direction the flexural cracks are first initiated?
a) Straight
b) Inclined
c) Bended
d) Dotted
Answer: c
Clarification: Flexure – shear cracks are first initiated by flexural cracks in the inclined direction and they are developed when the combined shear & flexural tensile stresses produce a principal tensile stress exceeding the tensile strength of concrete.

5. What is provided in order to protect a member from collapsing suddenly after the development of shear cracks?
a) Edge reinforcement
b) Principal reinforcement
c) Span reinforcement
d) Shear reinforcement
Answer: d
Clarification: In order to protect a member from collapsing suddenly after the development of shear crack, a minimum shear reinforcement is provided (A_sv) is provided in the form of stirrups which is obtained by satisfying the following condition:
A_sv/b_sv = 0.4/0.87f_y
In member without shear reinforcement, the inclined shear cracks extend to the compression face resulting in sudden explosive failure this is sometimes referred to as the diagonal tensile modes of failure.

6. The ultimate shear resistance of prestressed concrete with web shear cracking but without flexural crakes & mainly governed by __________
a) Limiting value
b) Constant value
c) Zero
d) Infinity
Answer: a
Clarification: The ultimate shear resistance of prestressed concrete sections with web shear cracking but without flexural cracks, is mainly governed by the limiting value of the principal tensile stress developed in concrete, the failure is assumed to take place when the principal tension exceeds the tensile strength of concrete.

7. The relation for computing the ultimate shear force is given as __________
a) V_cw = b_w d_p((0.3λ)f_cy^1/2 – 0.3f_cp) + v_p
b) V_cw = b_w d_p((0.3λ)f_cy^1/2 + 0.3f_cp) + v_p
c) V_cw = b_w d_p(0.3λ + v_p)
d) V_cw = b_w d_p
Answer: b
Clarification: The relation for computing the ultimate shear force, which includes a multiplying factor to multiplying factor to account for the type of concrete is given as V_cw = b_w d_p((0.3λ)f_cy^1/2 + 0.3f_cp)+ v_p
B_w = breadth of web, d_p = effective depth to tender or 0.8h, λ = 1.0 for nominal weight concrete and less for light weight concrete, f_cy = cylinder compressive strength of concrete(n/mm²), f_cp = compressive prestress at centroid of a section, v_p = the vertical component of the effective prestressing force at section.

8. The ultimate shear resistance vcf of section cracked in flexure is expressed as __________
a) v_cf = (1-0.55f_pc/f_p)τ _cb_wd+(m₀/m)v
b) v_cf = (1-0.55f_pc/f_p)τ _cb_wd
c) v_cf= (1-0.55f_pc/f_p)
d) v_cf= (1-0.55f_pc)
Answer: a
Clarification: v_cf = (1-0.55f_pc/f_p)τ _cb_wd+(m₀/m)v less than or equal to 0.1b_wdf_ck^1/2, f_pc= effective prestress, f_p = characteristic strength of prestressing steel, τ_c = ultimate shear stress capacity of concrete, b_w = breadth of the member, d = effective depth of tendons, m₀= moment necessary to produce zero, v and m = shear force and bending moment at section.

9. The flexure shear cracking load in a prestressed member is given by __________
a) ACI
b) Indian
c) British
d) Canada
Answer: a
Clarification: The American code (ACI: 318-1989) recommendations for the flexure shear cracking load in a prestressed member are based on experimental observations that flexure shear cracking initiates at the flexural cracking load plus an additional shear which is a flexural cracking load plus an additional shear which is a function of the strength and density of concrete & the dimensions of the section.

10. According to Mast, a complete shear analysis is necessary only in beams of __________
a) High span
b) Mid span
c) Low span
d) Edge span
Answer: c
Clarification: In general, the web shear and flexural shear resistance of the beam at important sections are compared with the ultimate shear requirements to identify zones where web reinforcement are required, According to Mast, a complete shear analysis is necessary only in beams of low span/depth ratio and in other cases the specified minimum proportions of web reinforcement is sufficient.

250+ TOP MCQs on Crack Widths in Members and Answers

Prestressed Concrete Structures Multiple Choice Questions on “Crack Widths in Members”.

1. The width of cracks that developed in prestressed members is governed by how many factors?
a) 3
b) 4
c) 5
d) 7
Answer: b
Clarification: ACI committee and Nawy have indicated that the width of cracks that develops in prestressed members is governed by: the average strain at the level at which cracks are considered, the minimum cover to the tension steel, the overall depth of the member in the neutral axis depth.

2. The formula recommended by the British code BS: 8110 – 1985 for the estimation of surface crack width W_cr is?
a) W_cr = 3a_crε_m / 1+2(a_cr – C_min/h-x)
b) W_cr = 3a_crε_m
c) W_cr = 3a_crε_m / 1+2(a_cr + C_min/h-x)
d) W_cr = 5a_crε_m
Answer: a
Clarification: W_cr = 3a_crε_m / 1+2(a_cr – l_min/h-x) is recommended by the British code BS: 8110 – 1985 for the estimation of surface crack width based on the work of beedy, a_cr = distance from the point considered to the surface of the nearest longitudinal area, ε_m = average strain at the level where cracking is being considered, C_min = minimum cover to the tension steel, h = overall depth of member, x = neutral axis.

3. The reason for smaller crack widths in slabs under service loads is?
a) H – x is large
b) H + x is small
c) 0
d) Constant
Answer: b
Clarification: In structural members, ε_m is a maximum at the tension face if h-x is sufficiently small for the crack width at the tension face not to exceed the permissible limit of 0.3mm, it will not exceed the limit anywhere, this is the reason for smaller crack widths in slabs under service loads, provided the thickness does not exceed about 200mm.

4. The stabilized mean crack spacing acs can be expressed as ____________
a) C (A₁/Ʃ0)
b) C (A₁/Ʃ8)
c) C (A₁/Ʃ5)
d) C (A₁/Ʃ7)
Answer: a
Clarification: Nawy based on extensive research work, have developed empirical relations to predict the maximum width of cracks and their mean spacing in pre tensioned and post tensioned beams and according to the investigations the stabilized mean crack spacing, a_cs can be expressed as C (A₁/Ʃ0)
A_t = effective concrete area in tension, Ʃ0 = sum of the circumferences of the reinforcing elements, C = empirical constant.

5. The width of cracks is influenced by?
a) Hollow stress
b) Linear stress
c) Net stress
d) Shear stress
Answer: c
Clarification: The widths of cracks are influenced by the net stress in the tendons, their surface characteristics and the effective area in tension.
Width of crack of Case 1: directly over a bar, the distance a_cr is equal to the concrete cover C_min equation then reduces: W_cr = 3a_crε_m and Case 2: when distance a_cr is large: W_cr = 1.5(h-x) εmin.

6. The British and Indian standard codes on prestressed concrete maximum limiting crack for members exposed to aggressive environment is?
a) 0.1mm
b) 0.3mm
c) 0.6mm
d) 0.4mm
Answer: a
Clarification: The British and Indian standard codes on prestressed concrete prescribe maximum limiting crack widths of 0.1mm for members exposed to aggressive environment and 0.2mm for all other members.

7. A nano polymer range which meets wider expectations as aesthetics is known as ____________
a) Refit
b) Break
c) Struck
d) Collapse
Answer: a
Clarification: A nano polymer range which meets wider expectations as aesthetics, economics, durability and performance of manufactured concrete products and the application of nano technology in the production of nano polymers has revolutionized the concrete industry.

8. The construction joints should be located at points of ____________
a) Maximum shear
b) Minimum shear
c) Total shear
d) Average shear
Answer: b
Clarification: Construction joints should be planned in advance and preferably they should be located at points of minimum shear and they should be nearly perpendicular to the principal lines of stress and construction joints are generally either vertical or horizontal.

9. When the work has resumed the surfaces of the concrete previously placed should be cleaned of ____________
a) Ash
b) Mud
c) Dirt
d) Soil
Answer: c
Clarification: When the work is resumed the surface of the concrete previously placed should be thoroughly cleaned of dirt, scum, laitance, loosely projecting aggregates and other soft material using stiff wire brushes and the surface should then be thoroughly soaked with clean water for two to three hours before further concreting using a thin layer of cement slurry.

10. In roadway slabs, construction joints should be formed?
a) Vertical
b) Horizontal
c) Aligned
d) Loaded
Answer: a
Clarification: In roadway slabs, construction joints shall be formed vertical and in true alignment and shear layers in construction joints should be constructed as shown in working site plans and in the case of box girders webs, theses shear keys are normally shown on the plans to the full width.

250+ TOP MCQs on Effects of Indeterminate structures and Answers

Prestressed Concrete Structures online quiz on “Effects of Indeterminate structures”.

1. The ultimate load capacity is higher in case of ____________
a) Statically indeterminate structures
b) Statically determinate structures
c) Prestressed structures
d) Reinforced structures
Answer: a
Clarification: The ultimate load carrying capacity is higher than in statically indeterminate structures than in determinate structure due to the phenomenon of redistribution of moments, reduction in the size of member’s results in lighter structures.

2. The bending moments are more evenly distributed between ____________
a) Ends and supports
b) Centre and supports
c) Edge and supports
d) Surface and supports
Answer: b
Clarification: The bending moments are more evenly distributed between the centre of span and the supports of member’s leads to increased stability and in continuous prestresssed structures, the deflections are comparatively small as compared to simply supported span.

3. In continuous post tensioned girders the curved cables can be positioned to resist ____________
a) Edge and supports
b) Beam and supports
c) Span and supports
d) Columns and supports
Answer: c
Clarification: In continuous post tensioned girders, the curved cables can be suitably positioned to resist the span and support moments, continuity of the members in framed structures leads to increase stability, in framed structures leads to increase stability continuous girders are formed by segmental construction using precast units connected by prestressed cables.

4. There is a reduction of anchorage in case of ____________
a) Precast prestressed beam
b) Pre tensioned prestressed beam
c) Continuous prestressed
d) Partially prestressed
Answer: d
Clarification: A reduction in the number of anchorages in a continuous prestressed beam in comparison with a series of simply supported beam only one pair of post tensioning anchorages and a single stressing operation can serve several members.

5. When an indeterminate structure is prestressed which reactions develop?
a) Extrusion
b) Redundant
c) Compressive
d) Deformation
Answer: b
Clarification: When an indeterminate structure is prestressed, reductant reactions will develop due to the redundancies exercising a restraint at the supports and the redundant reactions which develop as a consequence of prestressing an indeterminate structure result in secondary moments.

6. Which type of structure is free to deform?
a) Statically determinate
b) Continuous structure
c) Statically indeterminate
d) Partially prestressed structure
Answer: c
Clarification: While a statically determinate structure is free to deform when prestressed a continuous structure cannot deform freely and however the deflections should conform to the law of consistent deformation.

7. The formation of redundant reactions is examined with reference to ____________
a) One span continuous beam
b) Two span continuous beam
c) Three span continuous beam
d) Four span continuous beam
Answer: b
Clarification: The formation of redundant reactions and secondary moments are examined with reference to a two span continuous beam, prestressed by a straight cable local at a uniform eccentricity throughout the span.

8. The problem of excessive frictional losses can be tackled by reducing ____________
a) Eccentricity
b) Prestressing force
c) Curvature
d) Loads
Answer: c
Clarification: The problem of excessive frictional losses can be tackles by reducing the curvature of the cables housed in members of variable depth and also by temporarily overstressing the tendons from both ends.

9. The stresses due to secondary moments can be eliminated by selecting suitable ____________
a) Cable profile
b) Tendon profile
c) Anchorage profile
d) Wedge profile
Answer: b
Clarification: Stresses due to secondary moments can be eliminated by selecting suitable tendon profile which do not induce secondary moments and it is also possible to provide for secondary stresses in the design and if under reinforced sections are used, the redistribution of moments will be more or less complete resulting in higher collapse loads and these loads could be estimated by suing the well established plastic theory as applied to structural steel members.

10. The computation of collapse or ultimate load is influenced by ____________
a) Degree of compression
b) Degree of bending
c) Degree of redistribution
d) Degree of strain
Answer: c
Clarification: The computation of collapse or ultimate load is influenced by the degree of redistribution of moments in the continuous structure cables positioned to cater for secondary moments are not generally suitable to provide the required ultimate moment under a given system of loads.

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250+ TOP MCQs on Construction Features and Answers

Prestressed Concrete Structures Multiple Choice Questions on “Construction Features”.

1. The chords and struts of trusses are designed for convenience in __________
a) Plastering
b) Fabricating
c) Rubbing
d) Forrowing
Answer: b
Clarification: The chords and struts of trusses are designed to have the same width for convenience in fabricating the truss in a horizontal position such that it is protected by external moments and also smoothen the members.

2. The precast roof slabs are used for __________
a) Roof coverings
b) Slab coverings
c) Column coverings
d) Beam coverings
Answer: a
Clarification: If precast roof slabs are used for roof covering the upper chord panels are made equal to the width of the precast slabs which is usually about 3m and the lower tension chord is prestressed using bunched high strength wires or cables running in perforated holes.

3. For spans in the range of 18-24m, the trusses are made in how many pieces?
a) 4 pieces
b) 2 pieces
c) 1 piece
d) 5 pieces
Answer: c
Clarification: For spans in the range of 18-24m, the trusses are made in one piece but when spans run from 24-30m, they are made in two pieces with a joint at mid span and beyond them they are divided into equal spans and joints are made.

4. The polygonal trusses with inclined top chords are generally made of __________
a) 8m Blocks
b) 6m blocks
c) 12m block
d) 65m block
Answer: b
Clarification: Polygonal trusses with inclined top chords are generally made of 6m blocks or half trusses with 3m panels and due to higher tensions developed in the diagonal members of large span trusses, prestressing them becomes inevitable.

5. The polygonal trusses are less economical than?
a) Circular
b) Bow type
c) Curved
d) Oval
Answer: a
Clarification: In general polygonal trusses are less economical than the bow type with regard to material and labour costs and various types of trusses like circular, cylindrical, hollow etc used for various structures are not much economical because of their shapes and freeness to work.

6. The steel bearing plates which are anchored serves as __________
a) Loading
b) Bearing
c) Stressing
d) Deforming
Answer: b
Clarification: At the ends of trusses near supports, 10-12mm steel bearing plates are anchored and embedded while casting, which serve as bearing and for fixing the trusses on neoprene pad bearings located on the columns.

7. Calculate the area of concrete section such that loss ratio is 0.18 and compressive strength is 15n/mm² (N_d = 377×10³)?
a) 9.43n/mm²
b) 6.54n/mm²
c) 8.5n/mm²
d) 9.34n/mm²
Answer: a
Clarification: N_d = 377×10³, ɳ = 0.18, f_ct = 15n/mm²
Area of concrete section = (N_d/ɳf_ct) = (377×10³)/0.8×15) = 3.316mm².

8. Calculate the number of wires which are subjected to prestressing force of 471.5kn, the section adopted is 50000 using 7mm diameter high tensile wires initially stressed at 1100n/mm²?
a) 13.2
b) 11.13
c) 24.5
d) 34.2
Answer: b
Clarification: Given section adopted = 50000, prestressing force 471.5kn, using 7mm diameter high tensile wires initially stressed so 1100n/mm²
N = (471.5×10³/38.5×1100) = 11.13.

9. Calculate the cracking load such that section adopted is 50000, the loss ratio is 0.8, compressive strength of concrete is 9.43 and minimum reinforcement is 4.0mm²?
a) 455kn
b) 324kn
c) 577.2kn
d) 456.6kn
Answer: c
Clarification: The cracking load = (50000 (0.8 x 9.43) + 4.0 / 1000),
Section adapted = 50000, ɳ = 0.8, f_ct = 9.43, A_s = 4.0 (A_s is the minimum reinforcement of 0.8 percent in the section).

10. Which type of analysis should be done which will lead to an optimal design of planning a structure?
a) Comparative analysis
b) Strength analysis
c) Transferred analysis
d) Global analysis
Answer: a
Clarification: It is important to note that there is no single form of design which would be most economical in a given situation and to arrive at an economical design, several alternatives using different materials and structural configurations should be examined and a comparative analysis made which will lead to an optimal design.