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Question 1
ప్రశ్న 1
A rectangular slab panel is supported on all four sides. Under the usual IS 456 classification, when is it generally designed as a one-way slab?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A slab supported on four sides may still behave predominantly as a one-way slab when the longer dimension is sufficiently large compared with the shorter dimension. IS 456 treats a slab as spanning one way when the ratio of longer span to shorter span exceeds 2. In that situation, most of the load is carried by flexural action across the shorter span. Main tension reinforcement is therefore provided in the shorter-span direction. Hence Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
In a simply supported one-way rectangular slab, in which direction is the principal flexural tension reinforcement provided?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A one-way slab carries most of its transverse load by bending across its shorter span. The tensile bending stresses corresponding to this action require main flexural reinforcement parallel to the shorter span. Reinforcement is also provided in the transverse direction for distribution, shrinkage and temperature effects. The transverse reinforcement is not normally the principal flexural steel in an ordinary one-way slab. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a simply supported slab that is not built integrally with its supports, how is the effective span determined under IS 456?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 specifies an effective-span rule for simply supported beams and slabs that are not built integrally with their supports. The effective span is taken as the clear span plus the effective depth or the centre-to-centre support distance, whichever is smaller. This prevents the design span from being overestimated when the support width is relatively large. The effective depth rather than overall depth is used in the first expression. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A simply supported one-way slab has a clear span of 3.50 m, effective depth of 125 mm and support width of 230 mm at each end. What effective span should be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The clear span plus effective depth is 3.50 + 0.125 = 3.625 m. The centre-to-centre support distance is 3.50 + 0.115 + 0.115 = 3.730 m. The effective span is the smaller of these two values. Hence the design span is 3.625 m. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
A 150 mm thick RCC slab is made with normal-weight concrete of unit weight 25 kN/m³. What is its self-weight per square metre?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The self-weight of a slab is its thickness multiplied by the unit weight of concrete. The thickness is 150 mm, which is 0.15 m. Therefore the self-weight is 0.15 × 25 = 3.75 kN/m². This load acts as a characteristic dead load before application of the Limit State load factor. Therefore Option C is correct.
వివరణ:
Question 6
ప్రశ్న 6
A 150 mm thick slab has self-weight 3.75 kN/m², floor finish 1.0 kN/m² and imposed load 3.0 kN/m². What is the total characteristic gravity load?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The characteristic gravity load is the sum of the characteristic dead and imposed loads. The slab self-weight contributes 3.75 kN/m² and the floor finish contributes another 1.0 kN/m². Adding the imposed load of 3.0 kN/m² gives 7.75 kN/m². This value has not yet been multiplied by the collapse load factor. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
For the slab in the previous question, what factored load is obtained for the collapse combination DL + IL using a factor of 1.5?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total characteristic gravity load is 7.75 kN/m². For the stated collapse combination, both dead and imposed loads are included with a factor of 1.5. Thus the factored load is 1.5 × 7.75 = 11.625 kN/m². For a 1 m wide design strip, this is numerically 11.625 kN/m. Therefore Option D is correct.
వివరణ:
Question 8
ప్రశ్న 8
A simply supported one-way slab has effective span 3.625 m and carries a factored uniformly distributed load of 11.625 kN/m². What is the design bending moment per metre width?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A 1 m wide strip of the simply supported slab carries a factored line load of 11.625 kN/m. For a simply supported strip under uniformly distributed load, the maximum bending moment is wuL²/8. Substitution gives 11.625 × 3.625²/8. The resulting design moment is approximately 19.09 kN·m per metre width. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
An M25 one-way slab strip of width 1000 mm and effective depth 125 mm must resist Mu = 19.09 kN·m using Fe500 tension steel. Using the singly reinforced LSM equations, what tension-steel area is required approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the 1 m strip, force equilibrium gives xu = 0.87fyAst/(0.36fckb). The moment equation is Mu = 0.87fyAst(d - 0.42xu). Solving these equations for Mu = 19.09 kN·m, b = 1000 mm, d = 125 mm, fck = 25 N/mm² and fy = 500 N/mm² gives Ast approximately 374 mm². This is greater than the minimum slab reinforcement for the stated slab thickness. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
For the slab requiring about 374 mm²/m of main reinforcement, what area is provided by 10 mm diameter bars at 200 mm centres?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The area of one 10 mm diameter bar is π(10²)/4 = approximately 78.54 mm². At 200 mm spacing, five such bars occur in one metre width. The steel area provided is therefore 78.54 × 1000/200 = approximately 392.7 mm²/m. This exceeds the required area of about 374 mm²/m. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
What is the minimum reinforcement in either direction of an RCC slab when high-strength deformed bars are used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 specifies minimum slab reinforcement in each direction. When high-strength deformed bars or welded wire fabric are used, the minimum is 0.12 percent of the total cross-sectional area. This minimum applies even in the direction that is not the principal flexural direction. It provides reinforcement for shrinkage, temperature effects and general crack control. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a 150 mm thick slab reinforced with high-strength deformed bars, what minimum steel area is required in each direction per metre width?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The gross concrete area of a 1 m wide strip is 1000 × 150 = 150000 mm². The minimum reinforcement with high-strength deformed bars is 0.12 percent of this area. Thus Ast,min = 0.0012 × 150000 = 180 mm². This minimum applies separately in each slab direction. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
What minimum reinforcement would be required per metre width in a 150 mm thick slab if mild steel reinforcement were used instead of high-strength deformed bars?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For mild steel reinforcement, IS 456 specifies a slab minimum of 0.15 percent of the total cross-sectional area. A 1 m wide, 150 mm thick slab has gross area 150000 mm². Multiplying by 0.0015 gives 225 mm². This is higher than the 180 mm² required when high-strength deformed bars are used. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a solid RCC slab of overall thickness 150 mm, what is the maximum permitted reinforcing-bar diameter according to the slab detailing rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 limits the diameter of reinforcing bars in a slab to one-eighth of the total slab thickness. For a 150 mm thick slab, the limiting diameter is 150/8. This gives 18.75 mm. Thus a 16 mm bar satisfies this particular limit, whereas a 20 mm bar would exceed it. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
A one-way slab has effective depth d = 125 mm. What is the maximum permitted spacing of the main flexural reinforcement under the normal slab-spacing rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The spacing of main reinforcement in a solid slab shall not exceed three times the effective depth or 300 mm, whichever is smaller. Three times 125 mm is 375 mm. Comparing 375 mm with the absolute limit of 300 mm shows that 300 mm governs. A smaller spacing may still be required by strength or crack-control considerations. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
A solid slab has effective depth d = 80 mm. What maximum spacing is permitted for its main flexural reinforcement under the general slab-spacing rule?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The main-bar spacing limit is the smaller of 3d and 300 mm. For d = 80 mm, 3d equals 240 mm. Since 240 mm is less than 300 mm, it controls. The reinforcement must therefore be spaced at not more than 240 mm under this rule. Therefore Option B is correct.
వివరణ:
Question 17
ప్రశ్న 17
A one-way solid slab has effective depth d = 125 mm. What maximum spacing is permitted for reinforcement provided primarily against shrinkage and temperature effects?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Shrinkage and temperature reinforcement in a solid slab is permitted a larger spacing than main flexural reinforcement. Its spacing shall not exceed five times the effective depth or 450 mm, whichever is smaller. Five times 125 mm is 625 mm. The absolute limit of 450 mm is therefore smaller and controls. Hence Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which statement correctly describes the transverse distribution reinforcement in an ordinary one-way slab?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Although flexural action is predominantly one way, reinforcement is still required in the perpendicular direction. This transverse steel helps distribute local effects and controls shrinkage and temperature cracking. It must satisfy the applicable minimum slab-reinforcement percentage. Its spacing is governed by the rules for shrinkage and temperature reinforcement unless another design requirement is more severe. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
For mild exposure, what nominal cover may be used for slab main reinforcement of diameter not exceeding 12 mm, subject to satisfying the applicable fire-resistance requirement?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The general durability cover for mild exposure is 20 mm. IS 456 permits a reduction of 5 mm for main reinforcing bars not exceeding 12 mm diameter under mild exposure. This gives a nominal cover of 15 mm for the stated condition. The selected cover must still satisfy any larger requirement arising from fire resistance or other applicable provisions. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which set gives the basic span-to-effective-depth ratios used for preliminary deflection control for spans up to 10 m?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The deflection-control provisions used for beams also apply to solid slabs. The basic span-to-effective-depth ratios for spans up to 10 m are 7 for cantilevers, 20 for simply supported members and 26 for continuous members. These are basic values rather than automatic final allowable ratios. Appropriate modification factors and other serviceability considerations may alter the permissible value for a particular reinforced-concrete member. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
A simply supported one-way slab has effective span 3.625 m and effective depth 125 mm. Its actual span-to-effective-depth ratio is 29.0. What is the correct serviceability interpretation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The actual ratio is 3625/125 = 29.0. The basic value for a simply supported member is 20. The code permits the basic ratio to be modified using the applicable reinforcement-dependent factors, so exceeding 20 does not automatically mean failure. If the modified permissible ratio is still inadequate, the slab depth or reinforcement arrangement must be revised or deflection calculated directly. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Ignoring all modification factors, what effective depth would correspond to the basic span-to-depth ratio of 20 for a simply supported slab of effective span 3.60 m?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the basic simply supported criterion, L/d = 20. Rearranging gives d = L/20. With L = 3.60 m = 3600 mm, d = 3600/20 = 180 mm. This is only a basic preliminary value because applicable modification factors may permit a different effective depth. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A simply supported one-way slab is 125 mm thick and uses 8 mm main bars with 15 mm nominal cover. Its clear span is 2.50 m and each support is 230 mm wide. What are the effective depth and effective span approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The effective depth is measured to the centroid of the tensile bar. Thus d = 125 - 15 - 8/2 = 106 mm. Clear span plus effective depth is 2.50 + 0.106 = 2.606 m. The centre-to-centre support distance is 2.50 + 0.115 + 0.115 = 2.730 m, so the smaller value 2.606 m governs. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
For the slab in the previous question, take concrete unit weight as 25 kN/m³, floor finish as 1.0 kN/m² and imposed load as 2.0 kN/m². What factored load and simply supported design moment per metre width are obtained?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The slab self-weight is 0.125 × 25 = 3.125 kN/m². Adding 1.0 kN/m² finish and 2.0 kN/m² imposed load gives a characteristic load of 6.125 kN/m². Applying the factor 1.5 gives wu = 9.1875 kN/m². Using Leff = 2.606 m, Mu = wuL²/8 = approximately 7.80 kN·m per metre width. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
For the 125 mm slab strip with d = 106 mm, M20 concrete and Fe415 steel carrying Mu = 7.80 kN·m/m, the required main steel is approximately 213 mm²/m. Which reinforcement is the most suitable among the following, considering both required steel and the main-bar spacing limit?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The required flexural steel is approximately 213 mm²/m. Eight millimetre bars at 200 mm centres provide approximately 251 mm²/m, which exceeds the required area. At 250 mm centres the same bars provide only about 201 mm²/m and are therefore inadequate for the stated moment. The main-bar spacing limit is the smaller of 3d = 318 mm and 300 mm, so 200 mm spacing also satisfies the spacing rule. Therefore Option C is correct.
వివరణ:
Question 26
ప్రశ్న 26
Consider the following statements about one-way RCC slabs:
1. Main flexural reinforcement is generally placed along the shorter span.
2. With high-strength deformed bars, minimum reinforcement in each direction is 0.12 percent of the gross slab area.
3. Main reinforcement spacing shall not exceed 3d or 300 mm, whichever is smaller.
4. Distribution reinforcement may be omitted completely because the slab spans only in one direction.
Which statements are correct?
1. Main flexural reinforcement is generally placed along the shorter span.
2. With high-strength deformed bars, minimum reinforcement in each direction is 0.12 percent of the gross slab area.
3. Main reinforcement spacing shall not exceed 3d or 300 mm, whichever is smaller.
4. Distribution reinforcement may be omitted completely because the slab spans only in one direction.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 correctly describes the principal flexural reinforcement direction in an ordinary one-way slab. Statement 2 correctly gives the minimum reinforcement requirement when high-strength deformed bars are used. Statement 3 correctly states the maximum spacing rule for main reinforcement in solid slabs. Statement 4 is incorrect because reinforcement is still required in the transverse direction for shrinkage, temperature effects and general distribution. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. When the ratio of longer span to shorter span is greater than 2 సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. Along the shorter span సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. The lesser of clear span plus effective depth and centre-to-centre distance between supports సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. 3.625 m సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: C. 3.75 kN/m² సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: C. 7.75 kN/m² సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: D. 11.625 kN/m² సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 19.09 kN·m/m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: C. 374 mm²/m సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 393 mm²/m సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. 0.12 percent of the gross cross-sectional area సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 180 mm² సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 225 mm² సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 18.75 mm సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 300 mm సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. 240 mm సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 450 mm సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. It is provided perpendicular to the main flexural reinforcement and must satisfy the minimum slab-reinforcement requirement. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. 15 mm సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: A. Cantilever 7, simply supported 20, continuous 26 సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: A. The ratio exceeds the basic simply supported value of 20, so a suitable modification factor or a more detailed deflection check is required before accepting the depth. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 180 mm సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. d = 106 mm and Leff = 2.606 m సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: B. wu = 9.188 kN/m² and Mu = 7.80 kN·m/m సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: C. 8 mm bars at 200 mm centres సమాధానం: C.
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Question 26 ప్రశ్న 26Answer: B. 1, 2 and 3 only సమాధానం: B.