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Question 1
ప్రశ్న 1
In a singly reinforced rectangular RCC beam, what does the effective depth d represent?
Explanation:
Effective depth is measured from the extreme compression face to the centroid of the main tensile reinforcement. It is therefore generally smaller than the overall beam depth. The effective depth is used directly in the flexural strength equations because it defines the position of the tensile force. Its value depends on the overall depth, cover, stirrup diameter and main-bar diameter. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
A beam has an overall depth of 550 mm, nominal cover of 25 mm to the outer surface of an 8 mm stirrup, and 20 mm diameter main tension bars placed immediately inside the stirrup. What is the effective depth approximately?
Explanation:
The centroid of the main tension bar lies beyond the nominal cover by the stirrup diameter plus half the main-bar diameter. The effective cover is therefore 25 + 8 + 10 = 43 mm. Effective depth is the overall depth minus this effective cover. Hence d = 550 - 43 = 507 mm. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a singly reinforced rectangular beam in which the tension steel develops its design stress, which ultimate-force equilibrium equation is correct?
Explanation:
At ultimate flexure, the resultant concrete compression must equal the tensile force in the reinforcement. The equivalent concrete compression is 0.36fck b xu. When the tensile steel has sufficient strain to develop its design stress, its force is 0.87fyAst. Equating these two forces gives the neutral-axis depth for an under-reinforced or limiting singly reinforced section. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
An M25 rectangular beam has b = 300 mm, d = 500 mm and Ast = 1200 mm² of Fe415 tension steel. Assuming the steel develops 0.87fy, what is the neutral-axis depth xu approximately?
Explanation:
The neutral-axis depth is xu = 0.87fyAst/(0.36fckb). Substituting fy = 415 N/mm², Ast = 1200 mm², fck = 25 N/mm² and b = 300 mm gives xu approximately 160.47 mm. The corresponding ratio xu/d is approximately 0.321. This is below the Fe415 limiting ratio of 0.48. Therefore Option B is correct.
వివరణ:
Question 5
ప్రశ్న 5
For the beam in the previous question, how should the section be classified with respect to the Fe415 limiting neutral-axis depth?
Explanation:
For Fe415 reinforcement, xu,max/d is 0.48. With d = 500 mm, the limiting neutral-axis depth is therefore 240 mm. The calculated neutral-axis depth is only about 160.5 mm. Since xu is less than xu,max, the section satisfies the under-reinforced condition. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For the M25-Fe415 beam with d = 500 mm and xu = 160.47 mm, what is the ultimate lever arm z = d - 0.42xu approximately?
Explanation:
The equivalent concrete compression resultant acts at a depth of 0.42xu from the compression face. The lever arm between compression and tensile reinforcement is therefore d - 0.42xu. Substituting the stated values gives 500 - 0.42(160.47) = 432.60 mm approximately. This lever arm is then used with the tensile or compressive resultant to obtain ultimate moment capacity. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
What is the ultimate moment capacity of the M25 beam with b = 300 mm, d = 500 mm and Ast = 1200 mm² of Fe415 steel described above?
Explanation:
The tensile force is 0.87fyAst. For Fe415 steel and Ast = 1200 mm², this force is multiplied by the lever arm of approximately 432.6 mm. Thus Mu = 0.87(415)(1200)(432.6) N·mm. The result is approximately 187.43 kN·m. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
For Fe415 reinforcement, what is the limiting neutral-axis depth ratio xu,max/d used in IS 456 Limit State flexural design?
Explanation:
The limiting neutral-axis ratio is related to the minimum tensile-steel strain required at flexural failure. For Fe415 reinforcement, the specified limiting ratio is 0.48. Fe250 has a larger limiting ratio, while Fe500 has a smaller one. These different ratios reflect the different strain requirements associated with the steel grades. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
An M25 beam has b = 300 mm and d = 500 mm and uses Fe415 reinforcement. What is its limiting tensile-steel area Ast,lim approximately?
Explanation:
For Fe415 steel, xu,max = 0.48d = 240 mm. At the limiting condition, 0.36fck b xu,max = 0.87fyAst,lim. Substituting fck = 25 N/mm², b = 300 mm, xu,max = 240 mm and fy = 415 N/mm² gives approximately 1794.8 mm². This is the tensile-steel quantity corresponding to the limiting neutral-axis depth. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
What is the limiting ultimate moment capacity of an M25 rectangular beam with b = 300 mm, d = 500 mm and Fe415 reinforcement?
Explanation:
For Fe415 steel, the limiting moment coefficient is approximately 0.138. Thus Mu,lim is approximately 0.138fckbd². Substitution gives 0.138 × 25 × 300 × 500² N·mm. Using the exact stress-block relationship gives approximately 258.68 kN·m. Therefore Option D is correct.
వివరణ:
Question 11
ప్రశ్న 11
The M25, 300 mm × 500 mm effective-depth beam with Fe415 steel is provided with Ast = 1800 mm². Assuming 0.87fy initially gives xu approximately 240.7 mm, while xu,max = 240 mm. What is the correct interpretation?
Explanation:
The code neutral-axis limit is a ductility requirement rather than merely a numerical approximation. Here the force-equilibrium calculation based on 0.87fy gives xu slightly greater than xu,max. This indicates that the assumed ordinary singly reinforced yielding-steel condition has crossed the limiting state. The design should therefore be modified rather than accepting the calculated capacity without qualification. Hence Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
An M25 rectangular beam of width 300 mm with Fe415 reinforcement must resist a factored bending moment of 200 kN·m. Using the limiting coefficient Mu,lim = 0.138fckbd², what minimum effective depth is required approximately if the section is designed at the limiting condition?
Explanation:
At the limiting condition, d = √[Mu/(0.138fckb)]. Substituting Mu = 200 × 10^6 N·mm, fck = 25 N/mm² and b = 300 mm gives approximately 439.6 mm. This rounds to about 440 mm. A larger adopted effective depth would generally permit the required moment to be resisted with an under-reinforced steel area. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
An M25 rectangular beam has b = 300 mm and d = 500 mm and must resist a factored moment of 180 kN·m using Fe415 tension steel. Solving the singly reinforced Limit State equations gives what required tensile-steel area approximately?
Explanation:
The design equations are xu = 0.87fyAst/(0.36fckb) and Mu = 0.87fyAst(d - 0.42xu). Substitution of the beam dimensions, material strengths and required moment produces a quadratic equation in Ast. The smaller physically acceptable root is approximately 1144 mm². Its corresponding neutral-axis depth is about 153.0 mm, which is well below the Fe415 limiting depth of 240 mm. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
An M25 beam has b = 300 mm, d = 500 mm and four 20 mm diameter Fe415 tension bars. Neglecting any difference between nominal and actual bar area beyond πd²/4, what is its ultimate moment capacity approximately?
Explanation:
Four 20 mm bars provide Ast = 4π(20²)/4 = approximately 1256.6 mm². The corresponding neutral-axis depth is approximately 168.0 mm, which is below the Fe415 limiting depth. The lever arm is about 429.4 mm. Substitution in Mu = 0.87fyAstz gives approximately 194.83 kN·m. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For a beam with b = 300 mm, d = 500 mm and Fe415 reinforcement, what is the minimum area of tension reinforcement according to Ast,min/(bd) = 0.85/fy?
Explanation:
The minimum tension reinforcement is Ast,min = 0.85bd/fy. Substituting b = 300 mm, d = 500 mm and fy = 415 N/mm² gives approximately 307.2 mm². This requirement prevents a beam from being provided with an inadequately small quantity of tension reinforcement. The minimum requirement applies even when a smaller area might result from a flexural-strength calculation. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A rectangular beam has b = 300 mm and overall depth D = 550 mm. What is the maximum permitted area of tension reinforcement based on the IS 456 limit of 0.04bD?
Explanation:
The maximum tension reinforcement in a beam is limited to 0.04bD. Here bD = 300 × 550 = 165000 mm². Four percent of this area is 6600 mm². The maximum-reinforcement rule is a detailing limitation and is distinct from the smaller steel quantity that may be permitted by the flexural neutral-axis limit. Therefore Option D is correct.
వివరణ:
Question 17
ప్రశ్న 17
A flexural calculation for a beam with b = 300 mm, d = 500 mm and Fe415 steel gives a required tension-steel area of only 250 mm². The minimum code area is approximately 307 mm². What should be provided as a minimum for the flexural tension reinforcement?
Explanation:
Design reinforcement must satisfy both strength requirements and minimum detailing requirements. In this case the strength calculation produces an area smaller than the prescribed minimum. The minimum code reinforcement therefore governs. At least approximately 307 mm² must be provided, subject also to practical bar arrangement and other detailing rules. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which section dimension is used in the IS 456 maximum tension-reinforcement limit Ast,max = 0.04bD?
Explanation:
The maximum tension reinforcement rule is expressed as 0.04bD. The symbol D represents the overall depth of the beam rather than its effective depth. This is different from the minimum reinforcement expression, which uses bd. Care is therefore needed not to substitute d for D when checking the maximum reinforcement limit. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
An M30 beam has b = 250 mm, d = 450 mm and Ast = 1000 mm² of Fe500 steel. Assuming the steel develops 0.87fy, which set most closely gives xu and the ultimate moment capacity?
Explanation:
Force equilibrium gives xu = 0.87(500)(1000)/[0.36(30)(250)]. This gives approximately 161.1 mm. The lever arm is 450 - 0.42(161.1) = approximately 382.3 mm. The resulting moment is 0.87(500)(1000)(382.3), or approximately 166.3 kN·m. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
For the M30, b = 250 mm, d = 450 mm beam with Fe500 steel in the previous question, what is the limiting moment capacity approximately?
Explanation:
For Fe500 steel, xu,max/d is 0.46. The corresponding limiting moment coefficient is approximately 0.134fckbd². Using the exact coefficient from the stress-block relationship gives about 0.1336. Substituting fck = 30 N/mm², b = 250 mm and d = 450 mm gives approximately 202.9 kN·m. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
For a fixed rectangular beam section and material grades, what is the general effect of increasing Ast from a low under-reinforced value while still remaining within the singly reinforced limiting condition?
Explanation:
The neutral-axis equation contains Ast directly in the tensile-force term. Increasing the tensile-steel area therefore increases the required concrete compression force and moves the neutral axis deeper. Over the valid under-reinforced range, the larger tensile force generally increases the ultimate moment capacity. This trend cannot be extended indefinitely because the neutral-axis depth must not exceed xu,max for an ordinary singly reinforced design. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Which statement correctly distinguishes nominal cover from effective cover for flexural calculations in a reinforced-concrete beam?
Explanation:
Nominal cover is the design cover specified to the reinforcement, including outer links or stirrups where they are the outermost steel. Effective cover is the distance from the tension or compression face to the centroid of the corresponding main reinforcement. Consequently effective cover can include nominal cover, stirrup diameter and part of the main-bar diameter. Effective depth is obtained by subtracting the effective cover on the tension side from the overall depth. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
Which statement correctly distinguishes analysis from design of a singly reinforced beam by the Limit State Method?
Explanation:
In an analysis problem, the member geometry, materials and reinforcement are normally given. The task is then to calculate quantities such as xu and ultimate moment capacity and check the resulting condition. In a design problem, the required factored actions are given and suitable dimensions and reinforcement must be selected. The completed design must satisfy flexural strength, neutral-axis limits and applicable detailing requirements. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about singly reinforced rectangular beams designed by LSM:
1. For Fe415 steel, xu,max/d = 0.48.
2. Minimum tension steel is given by Ast/(bd) = 0.85/fy.
3. Maximum tension reinforcement is limited to 0.04bD.
4. If a calculated yielding-steel neutral axis exceeds xu,max, the excess may simply be ignored while retaining the same singly reinforced strength calculation.
Which statements are correct?
1. For Fe415 steel, xu,max/d = 0.48.
2. Minimum tension steel is given by Ast/(bd) = 0.85/fy.
3. Maximum tension reinforcement is limited to 0.04bD.
4. If a calculated yielding-steel neutral axis exceeds xu,max, the excess may simply be ignored while retaining the same singly reinforced strength calculation.
Which statements are correct?
Explanation:
Statement 1 correctly gives the limiting neutral-axis ratio for Fe415 reinforcement. Statement 2 correctly states the minimum tension-reinforcement requirement for beams. Statement 3 correctly states the maximum tension-reinforcement detailing limit. Statement 4 is incorrect because exceeding xu,max indicates that the ordinary singly reinforced yielding-steel solution does not satisfy the required flexural ductility condition. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each singly reinforced LSM quantity with its correct expression:
1. Neutral-axis depth when tensile steel develops design stress
2. Ultimate lever arm
3. Ultimate moment of resistance
4. Minimum tension reinforcement
P. 0.85bd/fy
Q. d - 0.42xu
R. 0.87fyAst(d - 0.42xu)
S. 0.87fyAst/(0.36fckb)
Which matching is correct?
1. Neutral-axis depth when tensile steel develops design stress
2. Ultimate lever arm
3. Ultimate moment of resistance
4. Minimum tension reinforcement
P. 0.85bd/fy
Q. d - 0.42xu
R. 0.87fyAst(d - 0.42xu)
S. 0.87fyAst/(0.36fckb)
Which matching is correct?
Explanation:
The neutral-axis depth follows from force equilibrium and is represented by S. The ultimate lever arm is the distance between the tensile steel and the equivalent concrete compression resultant, so item 2 matches Q. Ultimate moment is tensile force multiplied by that lever arm, making item 3 match R. Minimum tension reinforcement is 0.85bd/fy, so item 4 matches P. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. The distance from the extreme compression face to the centroid of the tensile reinforcement సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. 507 mm సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. 0.36fck b xu = 0.87fy Ast సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: B. 160.5 mm సమాధానం: B.
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Question 5 ప్రశ్న 5Answer: A. Under-reinforced సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 432.6 mm సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. 187.4 kN·m సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 0.48 సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: C. 1795 mm² సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: D. 258.7 kN·m సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: B. The section is slightly beyond the limiting singly reinforced condition and should not be accepted using the ordinary yielding-steel solution. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 440 mm సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 1144 mm² సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 194.8 kN·m సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. 307 mm² సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: D. 6600 mm² సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. Approximately 307 mm² or more సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. The overall depth D సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. xu = 161 mm and Mu = 166 kN·m సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 202.9 kN·m సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: A. xu generally increases and the flexural capacity generally increases. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. Effective cover is the distance from the concrete surface to the centroid of the main reinforcement, whereas nominal cover is specified to the outermost reinforcement. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. Analysis generally starts with known section dimensions and reinforcement to determine strength, whereas design starts with required actions and selects dimensions and reinforcement satisfying the code. సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: B. 1, 2 and 3 only సమాధానం: B.
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Question 25 ప్రశ్న 25Answer: A. 1-S, 2-Q, 3-R, 4-P సమాధానం: A.
