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Question 1
ప్రశ్న 1
A 5.0 m span RCC floor slab has a calculated final deflection of 18 mm and an additional deflection of 13 mm occurring after erection of partitions and application of finishes. Which assessment is correct under the usual IS 456 serviceability limits?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a 5.0 m span, the usual limit on final deflection due to all loads including long-term effects is span/250, which is 5000/250 = 20 mm. The calculated final deflection of 18 mm is therefore within that limit. The additional deflection after partitions and finishes is limited to the lesser of span/350 and 20 mm. Here span/350 is about 14.29 mm, so the governing post-finish limit is 14.29 mm. Since 13 mm is also within this limit, both checks are satisfied and Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
A simply supported RCC slab of effective span 4.8 m is checked by the span-to-effective-depth method. If the applicable tension-reinforcement modification factor is 1.30 and no other modification is applicable, what effective depth is required approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a simply supported flexural member of span up to 10 m, the basic span-to-effective-depth ratio is 20. Multiplying by the stated modification factor gives an allowable ratio of 20 × 1.30 = 26. The required effective depth is therefore 4800/26 = 184.6 mm. This is rounded to approximately 185 mm for the purpose of the question. Hence Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
A continuous RCC slab has an effective span of 5.2 m. If the applicable tension-reinforcement modification factor is 1.25 and no other modification factor is used, what effective depth is indicated by the span-to-depth method?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The basic span-to-effective-depth ratio for a continuous flexural member with span up to 10 m is 26. Applying the given modification factor gives 26 × 1.25 = 32.5. The effective depth required is therefore 5200/32.5 = 160 mm. This check is a serviceability sizing procedure rather than a substitute for strength design. Therefore Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
A two-way slab panel has effective spans of 4.0 m and 6.0 m and an effective depth of 150 mm. For the ordinary span-to-depth deflection check, which span and corresponding unmodified span/depth ratio should be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For slabs spanning in two directions, IS 456 directs that the shorter of the two spans be used for the span-to-effective-depth ratio. The shorter span here is 4.0 m. Dividing 4000 mm by the effective depth of 150 mm gives about 26.67. The 6.0 m span is therefore not used as the governing span in this particular serviceability ratio check. Hence Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A simply supported slab has an effective span of 12 m. The applicable tension-reinforcement modification factor is 1.20 and no other factor is applicable. Using the IS 456 span-to-depth procedure for spans above 10 m, what effective depth is indicated approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The basic simply supported ratio is 20 for spans up to 10 m. For a non-cantilever span above 10 m, this value is multiplied by 10/span in metres, giving 20 × 10/12 = 16.67. Applying the stated tension-reinforcement modification factor of 1.20 gives an allowable ratio of about 20. The corresponding effective depth is 12000/20 = 600 mm. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
A cantilever slab has a span greater than 10 m. Which approach is consistent with the IS 456 deflection-control provision?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 gives basic span-to-depth ratios and a special adjustment for spans greater than 10 m. For simply supported and continuous members above 10 m, the basic values are reduced by the factor 10/span in metres. The code specifically excludes cantilevers from that simplified long-span adjustment. For a cantilever exceeding 10 m, deflection calculations should instead be made. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
Which statement correctly describes the modification-factor concept used with the basic span-to-effective-depth ratios for ordinary RCC slab serviceability checks?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The basic span-to-effective-depth ratios are not always used without adjustment. IS 456 permits modification based on the amount and service stress of tension reinforcement through the relevant tension-reinforcement modification factor. A further modification can be applied depending on the amount of compression reinforcement. These adjustments recognize the influence of reinforcement on flexural stiffness and therefore on deflection. Hence Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
A 175 mm thick slab uses Fe500 deformed bars. A drawing shows 8 mm bars at 250 mm centres in a direction where only minimum slab reinforcement is required. If the spacing itself is otherwise permissible, what is the correct assessment?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For high-strength deformed bars, the minimum slab reinforcement is 0.12 percent of the gross cross-sectional area in each required direction. For a one-metre strip of a 175 mm slab, the minimum area is 0.0012 × 1000 × 175 = 210 mm²/m. An 8 mm bar has an area of about 50.27 mm², and at 250 mm centres the provided steel is about 201.1 mm²/m. Thus the arrangement can satisfy a spacing limit yet still fail the minimum-area requirement. Therefore Option B is correct.
వివరణ:
Question 9
ప్రశ్న 9
A solid slab has an effective depth of 85 mm. Without relying on a separate crack-width calculation to justify wider spacing, what is the maximum ordinary spacing of parallel main reinforcement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ordinary slab detailing rule limits the spacing of parallel main reinforcement to the smaller of 3d and 300 mm. With d = 85 mm, three times the effective depth is 255 mm. Since 255 mm is less than the absolute limit of 300 mm, it governs. This limit is a crack-control and distribution-related detailing requirement in addition to providing the required steel area. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
A slab has an effective depth of 70 mm and shrinkage-and-temperature reinforcement is shown at 375 mm centres. Under the ordinary prescriptive spacing rule, which statement is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Reinforcement provided against shrinkage and temperature is limited to the smaller of 5d and 450 mm under the ordinary slab rule. Five times the effective depth is 5 × 70 = 350 mm. The proposed spacing of 375 mm exceeds this governing value even though it is less than 450 mm. The smaller of the two limits must be used. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
For durability alone, what nominal cover may be used for main reinforcement of 12 mm diameter in a slab under mild exposure, assuming no greater fire-resistance requirement governs?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Table 16 gives 20 mm nominal cover for mild exposure. Its note permits a reduction of 5 mm for main reinforcement up to 12 mm diameter under mild exposure. This gives 15 mm. The general requirement that nominal cover not be less than the bar diameter is also satisfied because 15 mm exceeds 12 mm. Therefore, subject to fire and other project-specific requirements not governing, Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a normal-weight RCC slab in moderate exposure, what minimum nominal cover is specified by the ordinary durability table, assuming no fire requirement or other special provision governs?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 Table 16 specifies nominal cover values according to exposure condition. The value for moderate exposure is 30 mm. The special 5 mm reduction for bars up to 12 mm applies to mild exposure, not generally to moderate exposure. A larger cover can still be required by fire resistance or other project conditions. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A 250 mm thick slab in mild exposure uses a 25 mm diameter reinforcing bar. Ignoring fire requirements, what minimum nominal cover follows from the combined durability and bar-diameter provisions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ordinary durability cover for mild exposure is 20 mm. However, IS 456 also states that nominal cover shall not be less than the diameter of the reinforcing bar. For a 25 mm bar, the bar-diameter requirement therefore raises the minimum to 25 mm. The special mild-exposure reduction for bars up to 12 mm does not apply to a 25 mm bar. Hence Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
A slab uses M35 concrete and is exposed to a severe environment. Ignoring fire requirements, what nominal cover may be adopted under the Table 16 note permitting a reduction for higher-grade concrete?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The basic Table 16 nominal cover for severe exposure is 45 mm. A note to the table permits a reduction of 5 mm for severe and very severe exposure when concrete grade M35 or higher is used. The reduced durability cover is therefore 45 - 5 = 40 mm. This does not remove the need to satisfy any larger cover required for fire resistance or other conditions. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
Two equal 12 mm diameter parallel main bars are placed in a slab made with 20 mm nominal maximum size coarse aggregate. What minimum clear horizontal distance between the bars is ordinarily required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The minimum clear horizontal distance between parallel main bars is ordinarily the greatest of the applicable spacing criteria. For equal bars, one criterion is the bar diameter, which is 12 mm. Another criterion is 5 mm more than the nominal maximum size of coarse aggregate, giving 20 + 5 = 25 mm. The greater value is therefore 25 mm. Hence Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Which expression gives the development length Ld of a reinforcing bar according to IS 456?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Development length is the length required to develop the stress in a reinforcing bar through bond with the surrounding concrete. IS 456 expresses it as Ld = φσs/(4τbd). Here φ is the nominal bar diameter, σs is the stress in the bar at the section considered at design load, and τbd is the design bond stress. The required length therefore varies with bar size, steel stress, concrete bond capacity and bar type. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
A 12 mm Fe415 deformed bar in tension is embedded in M20 concrete and is stressed to 0.87fy at the design load. Using τbd = 1.2 N/mm² for a plain bar in M20 and the 60 percent increase for a deformed bar, what is the development length approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For M20 concrete, the design bond stress for a plain bar in tension is 1.2 N/mm². For a deformed bar this is increased by 60 percent, giving τbd = 1.92 N/mm². The steel stress is 0.87 × 415 = 361.05 N/mm². Substituting into Ld = φσs/(4τbd) gives 12 × 361.05/(4 × 1.92) ≈ 564 mm. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
For the same concrete and the same bar stress, how does using a deformed bar instead of a plain bar affect development length when the code's 60 percent increase in design bond stress applies?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Development length is inversely proportional to the design bond stress when bar diameter and bar stress remain unchanged. The design bond stress for a qualifying deformed bar is 1.60 times the corresponding plain-bar value. Therefore its development length becomes 1/1.60 = 0.625 times the plain-bar development length. This is one reason deformed bars can develop force over a shorter length than otherwise identical plain bars. Hence Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
For a bar having the same diameter and design stress, the development length in tension is calculated as 500 mm. If the bar is instead in compression and the IS 456 25 percent increase in bond stress for compression applies, what development length results approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For bars in compression, IS 456 increases the design bond stress by 25 percent relative to the corresponding tension value. Since development length is inversely proportional to design bond stress, the compression development length is the tension value divided by 1.25. Thus 500/1.25 = 400 mm. The calculation assumes the same bar diameter and stress so that the bond-stress change is the only variable. Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which statement correctly describes end anchorage of longitudinal tension reinforcement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 permits deformed bars in tension to be used without special end anchorage when the required development length is available. Their surface deformation provides enhanced bond with concrete. Plain bars in tension do not receive that same bond advantage, and hooks should normally be provided for them. In all cases, the design force in the reinforcement must be adequately developed by length, anchorage, or a combination. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
What anchorage value is assigned to a standard U-type hook on a 12 mm tension bar?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 assigns a standard U-type hook an anchorage value equal to 16 times the bar diameter. For a 12 mm bar, this is 16 × 12 = 192 mm. This anchorage value contributes toward satisfying the required anchorage or development condition. It should not be confused with the total development length calculated from bond stress. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
Away from a simple support or the end of a cantilever, a 16 mm tension bar in a slab is theoretically no longer required for flexure at a certain section. If the effective depth is 140 mm, how far beyond that theoretical cutoff point should the bar extend under the general curtailment rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The general curtailment rule requires tension reinforcement to extend beyond the point where it is theoretically no longer needed for flexure. The extension is the greater of the effective depth and 12 times the bar diameter. Here d = 140 mm, while 12φ = 12 × 16 = 192 mm. The larger value is therefore 192 mm. Hence Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A simply supported slab strip has 900 mm² of positive-moment reinforcement at mid-span, and the calculated development length of those bars is 600 mm. Under the general positive-moment support detailing rule, what minimum steel area and length should extend into the simple support along the same face?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a simple flexural member, at least one-third of the positive-moment reinforcement is to extend into the support along the same face. One-third of 900 mm² is 300 mm². The required extension into the support is Ld/3 under this provision. With Ld = 600 mm, Ld/3 is 200 mm. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
A continuous slab strip has 800 mm² of positive-moment reinforcement at a span region, and the relevant development length is 600 mm. Under the general positive-moment support rule, what minimum amount should extend into the support along the same face and for what length?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a continuous flexural member, at least one-fourth of the positive-moment reinforcement is required to extend into the support along the same face. One-fourth of 800 mm² is 200 mm². The prescribed extension associated with this rule is Ld/3. For Ld = 600 mm, this length is 200 mm. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
At a continuous support, a slab has negative-moment reinforcement consisting of 12 mm bars. The effective depth is 140 mm and the clear span is 4.8 m. At least one-third of the negative-moment reinforcement must extend beyond the point of inflection by not less than what distance?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For negative-moment reinforcement at a support, at least one-third of the total provided area must extend beyond the point of inflection. The required distance is not less than the greatest of the effective depth, 12 bar diameters, and one-sixteenth of the clear span. These values are 140 mm, 12 × 12 = 144 mm, and 4800/16 = 300 mm respectively. The governing distance is therefore 300 mm. Hence Option D is correct.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. Both deflection checks are satisfied సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. 185 mm సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: C. 160 mm సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: B. 4.0 m and about 26.7 సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 600 mm సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: C. Carry out deflection calculations rather than relying on the span-above-10-m reduction rule సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. The tension-reinforcement factor depends on the area and stress of tension steel, and a further factor may apply for compression reinforcement సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: B. The arrangement is inadequate because the minimum is 210 mm²/m while it provides only about 201 mm²/m సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: B. 255 mm సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: B. The spacing is unacceptable because the governing maximum is 350 mm సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. 15 mm సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 30 mm సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 25 mm సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 40 mm సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 25 mm సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. Ld = φσs/(4τbd) సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: C. 564 mm సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: B. The development length becomes 0.625 times the plain-bar value సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: A. 400 mm సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: B. Deformed bars may be used without end anchorage if the development-length requirement is satisfied, while hooks should normally be provided for plain tension bars సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. 192 mm సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: C. 192 mm సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 300 mm² for 200 mm సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. 200 mm² for 200 mm సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. 300 mm సమాధానం: D.