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Question 1
ప్రశ్న 1
Which assumption regarding strain distribution is used in Limit State flexural analysis of reinforced-concrete sections?
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Explanation:
Limit State flexural analysis assumes that plane sections remain plane after bending. This produces a linear strain distribution over the depth of the section. Perfect bond allows the strain in reinforcement to be related to the concrete strain at the same level. The neutral-axis position is determined from equilibrium and compatibility rather than being fixed at mid-depth. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
What maximum compressive strain is taken in concrete at the outermost compression fibre at the limit state of collapse in bending?
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Explanation:
The maximum concrete compressive strain at the extreme compression fibre in bending is taken as 0.0035. This value defines the upper end of the ultimate flexural strain distribution used in IS 456. It is distinct from the strain value of 0.002 associated with the beginning of the plateau in the idealized concrete stress-strain curve. The value is also different from the maximum strain adopted for pure axial compression. Therefore Option C is correct.
వివరణ:
Question 3
ప్రశ్న 3
For the IS 456 design stress-strain curve in flexure, concrete strength in the structure is taken as 0.67fck and the material partial safety factor is 1.5. What is the corresponding maximum design compressive stress approximately?
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Explanation:
For design, the concrete strength in the structure is represented as 0.67fck. Applying the material partial safety factor of 1.5 gives 0.67fck/1.5. This is approximately 0.4467fck, commonly rounded to about 0.447fck. This peak design stress should not be confused with the equivalent stress-block resultant coefficient of 0.36. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
For a singly reinforced rectangular section of width b and neutral-axis depth xu, what is the resultant concrete compressive force represented by the IS 456 equivalent design stress block?
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Explanation:
The actual design compression distribution is represented by an equivalent resultant for flexural design. For a rectangular section, this resultant concrete compression is 0.36fck b xu. The coefficient already reflects integration of the idealized concrete stress block. It is not obtained by simply multiplying the maximum concrete stress by the entire neutral-axis depth. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
At what distance from the extreme compression fibre does the resultant concrete compression act in the IS 456 equivalent flexural stress block?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The equivalent concrete compressive force acts at a specified centroidal depth from the extreme compression fibre. IS 456 gives this distance as 0.42xu. Consequently the lever arm to tensile reinforcement at effective depth d is d - 0.42xu. This relationship is central to the ultimate moment equation for singly reinforced rectangular sections. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
How is the tensile strength of concrete treated in the IS 456 Limit State flexural strength calculation?
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Explanation:
Concrete is weak in tension and flexural cracking is expected on the tension side of an ordinary reinforced-concrete member. The Limit State flexural-strength assumptions therefore ignore the tensile strength of concrete. Tensile resistance is provided primarily by the reinforcement crossing the cracked region. This assumption simplifies equilibrium without relying on uncertain post-cracking tensile concrete resistance. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
According to the IS 456 flexural assumptions, the strain in tensile reinforcement at failure shall not be less than which expression?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 prescribes a minimum tensile-reinforcement strain at flexural failure to ensure adequate ductile behaviour. The required minimum is fy/(1.15Es) + 0.002. The first term corresponds to the design stress level in steel expressed as an elastic strain component. The additional strain of 0.002 provides a ductility margin beyond that level. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For Fe415 reinforcement with Es = 200000 N/mm², what minimum tensile-steel strain is required at flexural failure approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The minimum strain is fy/(1.15Es) + 0.002. Substitution gives 415/[1.15(200000)] + 0.002. The first term is approximately 0.001804, giving a total of approximately 0.003804. Rounded suitably, the minimum required strain is about 0.00380. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
Using the extreme concrete strain of 0.0035 and the minimum tensile-steel strain condition for Fe415 steel, what limiting neutral-axis ratio xu,max/d is obtained approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Strain compatibility gives steel strain as 0.0035(d - xu)/xu. At the limiting condition for Fe415 steel, this equals approximately 0.003804. Solving gives xu/d approximately 0.479. IS 456 adopts the practical limiting ratio of 0.48 for Fe415 reinforcement. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which set gives the limiting neutral-axis depth ratios xu,max/d for Fe250, Fe415 and Fe500 reinforcement respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The limiting neutral-axis ratio depends on the steel grade because the required tensile-steel strain depends on fy. For Fe250, the limiting ratio is 0.53. For Fe415, it is 0.48, while for Fe500 it is 0.46. Higher-strength reinforcement requires greater tensile strain to satisfy the prescribed ductility condition, which leads to a shallower limiting neutral axis. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
Which statement correctly describes under-reinforced, limiting and over-reinforced singly reinforced rectangular sections in Limit State flexure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The limiting neutral-axis depth defines the boundary associated with the minimum required tensile-steel strain. An under-reinforced section has xu less than xu,max and therefore provides greater tensile-steel strain and desirable ductile behaviour. A limiting section has xu equal to xu,max. If xu exceeds xu,max, the section is over-reinforced relative to the code ductility requirement and should not be accepted as an ordinary singly reinforced flexural design. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a singly reinforced rectangular section in which tensile steel develops the design stress 0.87fy, which equilibrium relation is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At ultimate flexural equilibrium, the resultant concrete compression must equal the tensile force in the reinforcement. The concrete compression is 0.36fck b xu. When the tensile steel has sufficient strain to develop its design stress, its tensile force is 0.87fyAst. Equating the two forces gives 0.36fck b xu = 0.87fyAst. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
An M20 singly reinforced rectangular beam has b = 300 mm, d = 500 mm and Ast = 1200 mm² of Fe415 steel. Assuming the steel develops 0.87fy, what neutral-axis depth xu is obtained approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Force equilibrium gives xu = 0.87fyAst/(0.36fck b). Substituting fy = 415 N/mm², Ast = 1200 mm², fck = 20 N/mm² and b = 300 mm gives approximately 200.6 mm. The corresponding xu/d ratio is about 0.401. Since this is below the Fe415 limiting ratio of 0.48, the assumed tensile-steel design stress is consistent with an under-reinforced section. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
For the M20-Fe415 beam in the previous question with xu = 200.6 mm and d = 500 mm, what is the ultimate lever arm approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The resultant concrete compression acts at 0.42xu from the compression face. The lever arm is therefore z = d - 0.42xu. Substituting d = 500 mm and xu = 200.6 mm gives about 500 - 84.25 = 415.75 mm. Rounding to the available value gives approximately 415.7 mm. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For the same M20 beam with b = 300 mm, d = 500 mm and Ast = 1200 mm² of Fe415 steel, what is its ultimate moment of resistance approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The tensile force is 0.87fyAst and the lever arm is approximately 415.75 mm. Thus Mu = 0.87(415)(1200)(415.75) N·mm. This evaluates to approximately 180.13 × 10^6 N·mm. The result is about 180.1 kN·m and is lower than the limiting capacity of the same section. Therefore Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
For an M20 rectangular section with b = 300 mm, d = 500 mm and Fe415 steel, what are the limiting neutral-axis depth and corresponding limiting tensile-steel area approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For Fe415 steel, xu,max/d is 0.48. With d = 500 mm, xu,max is therefore 240 mm. The limiting steel area follows from 0.36fck b xu,max = 0.87fyAst,lim. Substitution gives Ast,lim approximately 1435.8 mm², which rounds to 1436 mm². Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
What is the limiting ultimate moment of resistance of the M20, b = 300 mm, d = 500 mm section reinforced with Fe415 steel?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
At the limiting condition, xu,max = 0.48d = 240 mm. The limiting moment is 0.36fck b xu,max(d - 0.42xu,max). Substitution gives 0.36(20)(300)(240)[500 - 0.42(240)] N·mm. This evaluates to approximately 206.95 kN·m. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
For a singly reinforced rectangular section with Fe415 steel, which approximate limiting-moment expression follows from xu,max/d = 0.48?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The general limiting expression is 0.36fckb xu,max(d - 0.42xu,max). Substituting xu,max = 0.48d produces a coefficient of 0.36(0.48)[1 - 0.42(0.48)]. This evaluates to approximately 0.13796. Rounded suitably, Mu,lim is approximately 0.138fckbd² for Fe415 reinforcement. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
Using the IS 456 limiting neutral-axis ratios, which set most closely gives the limiting moment coefficients multiplying fckbd² for Fe250, Fe415 and Fe500 steel respectively?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The limiting coefficient is 0.36r(1 - 0.42r), where r = xu,max/d. Using r = 0.53 for Fe250 gives approximately 0.148. Using 0.48 for Fe415 gives approximately 0.138, while 0.46 for Fe500 gives approximately 0.134. Thus the limiting coefficient decreases slightly as the specified steel grade increases because the permitted limiting neutral axis becomes shallower. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
For an M25 rectangular compression zone with b = 250 mm and xu = 180 mm, what are the resultant concrete compression and its distance from the extreme compression fibre?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The concrete compression is C = 0.36fck b xu. Substituting 25 N/mm², 250 mm and 180 mm gives 405000 N, or 405 kN. Its line of action is at 0.42xu from the compression face. This distance is 0.42 × 180 = 75.6 mm. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
For the section in the previous question, if the effective depth d is 450 mm, what ultimate moment is represented by the concrete stress block?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The concrete compression resultant is 405 kN. Its line of action is 75.6 mm from the compression face, while the tensile reinforcement is at an effective depth of 450 mm. The lever arm is therefore 450 - 75.6 = 374.4 mm, or 0.3744 m. Multiplying 405 kN by 0.3744 m gives approximately 151.6 kN·m. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
A singly reinforced section has xu/d = 0.40 at ultimate flexure. What is the tensile-steel strain from strain compatibility, and does it satisfy the minimum requirement for Fe415 steel?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The steel strain is 0.0035(d - xu)/xu. For xu/d = 0.40, this becomes 0.0035(0.60/0.40). The resulting strain is 0.00525. This exceeds the Fe415 minimum required value of approximately 0.00380, confirming adequate tensile strain and an under-reinforced condition. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
For Fe500 reinforcement with Es = 200000 N/mm², which pair most closely gives the minimum tensile-steel strain at failure and the corresponding limiting xu,max/d value?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The minimum tensile strain is fy/(1.15Es) + 0.002. For Fe500 steel, this gives 500/[1.15(200000)] + 0.002, or approximately 0.004174. Combining this with the extreme concrete strain of 0.0035 gives an exact compatibility ratio near 0.456. The code limiting ratio is taken as 0.46 for Fe500 reinforcement. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
An M20 rectangular beam has b = 300 mm, d = 500 mm and 1600 mm² of Fe415 tensile reinforcement. If force equilibrium is first evaluated assuming the tensile steel develops 0.87fy, the calculated xu is about 267.4 mm, whereas xu,max is 240 mm. What is the correct design interpretation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For Fe415 reinforcement, the limiting neutral-axis depth is 0.48d, which is 240 mm for this beam. The assumed yielding-steel equilibrium gives xu approximately 267.4 mm, exceeding that limit. Such a result indicates an over-reinforced condition relative to the code flexural ductility requirement. The simple singly reinforced yielding-steel solution should therefore not be accepted as the design state, and the section should be redesigned appropriately. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about the IS 456 Limit State flexural stress block:
1. The maximum concrete compressive strain in bending is 0.0035.
2. The equivalent concrete compressive resultant is 0.36fck b xu and acts at 0.42xu from the compression face.
3. Tensile concrete is ignored in calculating flexural strength.
4. A singly reinforced section with xu greater than xu,max is preferred because it provides greater ductility.
Which statements are correct?
1. The maximum concrete compressive strain in bending is 0.0035.
2. The equivalent concrete compressive resultant is 0.36fck b xu and acts at 0.42xu from the compression face.
3. Tensile concrete is ignored in calculating flexural strength.
4. A singly reinforced section with xu greater than xu,max is preferred because it provides greater ductility.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 correctly gives the ultimate concrete compression strain used in bending. Statement 2 correctly states both the magnitude and location of the equivalent concrete compression resultant. Statement 3 is correct because tensile concrete is ignored in the flexural-strength calculation. Statement 4 is false because xu greater than xu,max indicates inadequate tensile strain and an over-reinforced condition rather than desirable ductile behaviour. 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. Plane sections normal to the axis before bending remain plane after bending. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: C. 0.0035 సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: B. 0.447fck సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: A. 0.36fck b xu సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. 0.42xu సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: C. It is ignored in calculating flexural strength. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. fy/(1.15Es) + 0.002 సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. 0.00380 సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: C. 0.48 సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: B. 0.53, 0.48, 0.46 సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: A. Under-reinforced: xu < xu,max; limiting: xu = xu,max; over-reinforced: xu > xu,max. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. 0.36fck b xu = 0.87fy Ast సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. 200.6 mm సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. 415.7 mm సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. 180.1 kN·m సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. xu,max = 240 mm and Ast,lim = 1436 mm² సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: C. 206.9 kN·m సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: B. Mu,lim = 0.138fckbd² సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 0.148, 0.138, 0.134 సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. C = 405 kN at 75.6 mm సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: B. 151.6 kN·m సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 0.00525; it satisfies the requirement సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 0.00417 and 0.46 సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: C. The calculated neutral axis exceeds the permitted limiting depth, so the ordinary singly reinforced section should be redesigned rather than accepting the simple yielding-steel solution. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: B. 1, 2 and 3 only సమాధానం: B.