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Question 1
ప్రశ్న 1
Which statement most accurately distinguishes a T-beam from an L-beam in a monolithic RCC floor system?
Explanation:
In a monolithic floor system, part of the slab can participate with the beam web when the slab lies in the compression zone. An interior beam normally has slab available on both sides of its web and is therefore idealized as a T-section. An edge beam generally has participating slab mainly on one side and is idealized as an L-section. The distinction is based on cross-sectional geometry and effective slab participation rather than material grade. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
What is the principal meaning of the effective flange width used in RCC flanged-beam design?
Explanation:
Longitudinal compressive stress in a wide slab connected to a beam is not uniformly distributed across the complete slab width. The slab region close to the web generally participates more effectively than regions far away from it. Structural design therefore replaces the actual nonuniform stress distribution by an effective flange width. This effective width provides a practical representation of the portion of slab acting with the web. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which condition is required before a slab may be assumed to act as the compression flange of a T-beam or L-beam?
Explanation:
Flanged-beam action requires force transfer between the slab and beam web. Monolithic casting naturally provides the necessary connection when properly constructed. IS 456 also permits effective bonding by other suitable means where composite action is reliably developed. The slab cannot be counted as a compression flange if it is structurally disconnected from the web. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
In the absence of a more accurate determination, which expression gives the effective flange width bf of a conventional T-beam according to IS 456, subject to the applicable width limitation?
Explanation:
For a conventional T-beam, IS 456 gives an effective-width expression involving the distance between points of zero moment, web width and flange thickness. The expression is bf = l0/6 + bw + 6Df. The value is additionally subject to the geometric limitation specified for the available slab width. The L-beam expression contains smaller l0 and flange-thickness contributions because slab participation occurs mainly on one side. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
In the absence of a more accurate determination, which expression gives the effective flange width bf of a conventional L-beam according to IS 456, subject to the applicable width limitation?
Explanation:
An L-beam generally has effective slab participation predominantly on one side of the web. IS 456 therefore uses bf = l0/12 + bw + 3Df for the conventional L-beam expression. The terms involving l0 and Df are one-half of the corresponding additions used for a T-beam. The calculated effective width must still satisfy the applicable geometric limitation. Therefore Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
For estimating the effective flange width of a continuous beam or frame, what value may be assumed for l0, the distance between points of zero moment?
Explanation:
The effective-width expressions use l0, which represents the distance between points of zero moment. Determination of exact zero-moment locations would ordinarily require structural analysis. For continuous beams and frames, IS 456 permits l0 to be assumed as 0.7 times the effective span for this purpose. This provides a convenient approximation for effective flange-width calculations. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
A continuous T-beam has an effective span of 8 m, web width bw = 300 mm and flange thickness Df = 100 mm. Assume l0 = 0.7 times the effective span and that the geometric upper limit does not govern. What is bf approximately?
Explanation:
For the continuous beam, l0 = 0.7 × 8000 = 5600 mm. The T-beam expression is bf = l0/6 + bw + 6Df. Hence bf = 5600/6 + 300 + 600 = 1833.3 mm approximately. The problem states that the separate geometric upper limit does not control the result. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
An L-beam has l0 = 6 m, web width bw = 300 mm and flange thickness Df = 100 mm. Assuming the geometric upper limit does not govern, what effective flange width is obtained?
Explanation:
For an L-beam, bf = l0/12 + bw + 3Df. Using l0 = 6000 mm gives a span contribution of 500 mm. Adding the 300 mm web width and 300 mm flange-thickness contribution gives bf = 1100 mm. This calculation assumes that the available slab width does not impose a smaller value. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
For a conventional T-beam or L-beam, the effective flange width calculated from the IS 456 expression is additionally limited so that it is not greater than:
Explanation:
The effective-width equation cannot create a compression flange wider than the slab region physically available to the beam. IS 456 therefore places a geometric upper limit involving the web width and the clear distances to adjacent beams. This prevents overlapping or physically impossible effective flange widths in floor systems. The restriction is independent of the reinforcement area. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
If the neutral-axis depth xu of a flanged beam lies entirely within the flange thickness Df, how may the section be analysed for ultimate flexural resistance?
Explanation:
When xu is less than Df, the entire concrete compression block lies within the flange. The web below the flange is then on the tension side of the neutral axis and does not change the shape of the compression block. The flexural compression region can therefore be treated exactly like a rectangular section of width bf. The usual singly reinforced rectangular-section equations apply using bf as the compression-face width. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
An M25 T-beam has bf = 1200 mm, Df = 120 mm, effective depth d = 500 mm and Ast = 1500 mm² of Fe415 steel. Assuming the tensile steel develops 0.87fy, what is the neutral-axis depth xu approximately?
Explanation:
Assuming the neutral axis lies in the flange, force equilibrium is 0.36fck bf xu = 0.87fyAst. Substituting the given values gives xu = 0.87(415)(1500)/[0.36(25)(1200)]. The result is approximately 50.15 mm. Since this value is less than Df = 120 mm, the rectangular-flange assumption is verified. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the T-beam in the previous question with xu = 50.15 mm, what is its ultimate moment capacity approximately?
Explanation:
Because the neutral axis lies inside the flange, the section behaves as a rectangular section of width bf for flexure. The lever arm is d - 0.42xu = 500 - 0.42(50.15), or approximately 478.94 mm. The tensile force is 0.87fyAst. Multiplying the tensile force by the lever arm gives approximately 259.4 kN·m. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
For an under-reinforced flanged beam with the neutral axis in the web and with Df/xu sufficiently small that the full flange thickness may be represented at approximately uniform design compression, which compression-force expression is appropriate?
Explanation:
When the neutral axis lies below the flange, the compression zone contains both the web and the projecting flange portions. The web contribution is represented by 0.36fck bw xu. The additional flange outside the web contributes approximately 0.45fck(bf - bw)Df in the applicable case. The two concrete compression components together balance the tensile reinforcement force. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
An M25 flanged beam has bf = 1000 mm, bw = 300 mm, Df = 100 mm, d = 600 mm and Ast = 4000 mm² of Fe415 steel. Using C = 0.36fck bw xu + 0.45fck(bf - bw)Df and T = 0.87fyAst, what is xu approximately?
Explanation:
The tensile force is 0.87(415)(4000) = 1444.2 kN. The projecting flange contributes 0.45(25)(1000 - 300)(100) = 787.5 kN of compression. The remaining compression is provided by the web term 0.36(25)(300)xu. Solving gives xu approximately 243.22 mm, and Df/xu is about 0.411, which is consistent with the stated full-flange representation. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For the flanged beam in the previous question, what is the ultimate moment capacity approximately?
Explanation:
The web compression force is 0.36fck bw xu, which is approximately 656.7 kN. Its lever arm is d - 0.42xu, while the projecting-flange force of 787.5 kN acts approximately at Df/2 from the compression face. The moment capacity is therefore the sum of the web-block moment and the additional flange moment. Substitution gives approximately 760.1 kN·m. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
When the neutral axis lies in the web and the flange depth is relatively large so that the complete projecting flange cannot be represented by a constant 0.45fck stress over Df, what effective depth yf is used in the IS 456 flanged-section treatment?
Explanation:
For the relevant thick-flange case, the compression stress in the projecting flange is represented through an effective depth yf. IS 456 gives yf = 0.15xu + 0.65Df. The calculated yf is not permitted to exceed the actual flange thickness Df. This expression allows the equivalent flange compression to reflect the position of the neutral axis more realistically. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
An M20 flanged beam has bf = 900 mm, bw = 300 mm, Df = 100 mm, d = 450 mm and Ast = 2400 mm² of Fe415 steel. Using yf = 0.15xu + 0.65Df and force equilibrium 0.36fck bw xu + 0.45fck(bf - bw)yf = 0.87fyAst, which pair is obtained approximately?
Explanation:
Substitution of yf = 0.15xu + 65 mm into the compression-force equation gives a single equation in xu. Solving gives xu approximately 173.58 mm. The corresponding yf is 0.15(173.58) + 65 = approximately 91.04 mm. The neutral axis lies below the 100 mm flange, while Df/xu exceeds 0.43, making the yf treatment appropriate. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
For the M20 flanged beam in the previous question with xu = 173.58 mm and yf = 91.04 mm, what is the ultimate moment capacity approximately?
Explanation:
The web-block moment is 0.36fck bw xu(d - 0.42xu). The projecting-flange moment is 0.45fck(bf - bw)yf(d - yf/2). Both components must be included because the neutral axis lies below the flange. Substitution of the stated dimensions gives a total ultimate moment of approximately 340.2 kN·m. Therefore Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
An M20 T-beam has bf = 1000 mm, bw = 300 mm, Df = 80 mm, d = 500 mm and Fe415 steel. Since Df/d = 0.16, use the IS 456 limiting flanged-section expression with xu,max/d = 0.48. What is Mu,lim approximately?
Explanation:
For Fe415 steel, xu,max = 0.48d = 240 mm. The limiting capacity contains the web contribution plus the projecting-flange contribution over the full 80 mm flange thickness. The web term is 0.36fck bw xu,max(d - 0.42xu,max). The flange term is 0.45fck(bf - bw)Df(d - Df/2), and their sum is approximately 438.8 kN·m. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
An M20 T-beam has bf = 1000 mm, bw = 300 mm, Df = 120 mm, d = 500 mm and Fe415 reinforcement. At the limiting neutral axis xu,max = 0.48d, what value of yf = 0.15xu,max + 0.65Df is obtained?
Explanation:
For Fe415 steel, xu,max = 0.48 × 500 = 240 mm. The effective flange compression depth is yf = 0.15(240) + 0.65(120). This gives 36 + 78 = 114 mm. Since 114 mm is less than the actual flange thickness of 120 mm, no capping at Df is required. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
For the T-beam in the previous question with xu,max = 240 mm and yf = 114 mm, what limiting ultimate moment capacity is obtained approximately?
Explanation:
The limiting web contribution is 0.36fck bw xu,max(d - 0.42xu,max). The additional flange contribution is 0.45fck(bf - bw)yf(d - yf/2). Using fck = 20 N/mm², bf = 1000 mm, bw = 300 mm, d = 500 mm, xu,max = 240 mm and yf = 114 mm gives approximately 525.1 kN·m. The use of yf is necessary because Df/d exceeds 0.2 in this example. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
At an interior support of a continuous monolithic T-beam carrying negative bending moment, the top slab is generally in tension. Which flexural idealization is usually appropriate for the strength calculation at that region when no effective compression flange exists on the opposite face?
Explanation:
Flanged-beam enhancement is obtained when the slab lies in the compression zone and can participate as a flange. Over an interior support under negative bending, the top slab is normally on the tension side. Ordinary ultimate flexural design does not rely on tensile concrete to provide the compression flange action. The compression zone may therefore be confined mainly to the beam web unless another effective compression flange exists. Therefore Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
An M25 T-beam has bf = 1200 mm, bw = 300 mm, Df = 100 mm and d = 500 mm and uses Fe415 tension reinforcement. It must resist Mu = 200 kN·m. Assuming the neutral axis lies in the flange, solving the rectangular-section LSM equations gives what required Ast and corresponding xu approximately?
Explanation:
With the neutral axis in the flange, the section is analysed as a rectangular section of width bf = 1200 mm. Simultaneously solving xu = 0.87fyAst/(0.36fckbf) and Mu = 0.87fyAst(d - 0.42xu) gives the smaller physical root Ast approximately 1145 mm². The associated neutral-axis depth is approximately 38.3 mm. Since 38.3 mm is well below Df = 100 mm, the initial flange-only compression assumption is self-consistent. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
Consider the following statements about flanged RCC beams:
1. Effective flange width accounts for the nonuniform longitudinal compression distribution in the slab.
2. If xu lies within Df, the flexural section may be analysed as a rectangular section of width bf.
3. The conventional T-beam and L-beam effective-width expressions are different because their slab participation is different.
4. If xu lies below the flange, the section should always be analysed as a rectangular section of width bf over the entire compression depth.
Which statements are correct?
1. Effective flange width accounts for the nonuniform longitudinal compression distribution in the slab.
2. If xu lies within Df, the flexural section may be analysed as a rectangular section of width bf.
3. The conventional T-beam and L-beam effective-width expressions are different because their slab participation is different.
4. If xu lies below the flange, the section should always be analysed as a rectangular section of width bf over the entire compression depth.
Which statements are correct?
Explanation:
Statement 1 correctly explains why an effective rather than complete slab width is used. Statement 2 is correct because a neutral axis within the flange produces a rectangular compression block of width bf. Statement 3 is correct because T-beams normally have slab participation on both sides while L-beams generally have it mainly on one side. Statement 4 is false because a neutral axis below the flange requires separate treatment of the web and projecting flange portions. Therefore Statements 1, 2 and 3 are correct and Option B is the unique answer.
వివరణ:
Question 25
ప్రశ్న 25
Match each flanged-beam concept with its correct expression or treatment:
1. Conventional T-beam effective flange width
2. Conventional L-beam effective flange width
3. Neutral axis within the flange
4. Effective flange-compression depth for the applicable thick-flange case
P. bf = l0/12 + bw + 3Df
Q. Treat as rectangular section of width bf
R. yf = 0.15xu + 0.65Df, not greater than Df
S. bf = l0/6 + bw + 6Df
Which matching is correct?
1. Conventional T-beam effective flange width
2. Conventional L-beam effective flange width
3. Neutral axis within the flange
4. Effective flange-compression depth for the applicable thick-flange case
P. bf = l0/12 + bw + 3Df
Q. Treat as rectangular section of width bf
R. yf = 0.15xu + 0.65Df, not greater than Df
S. bf = l0/6 + bw + 6Df
Which matching is correct?
Explanation:
The conventional T-beam effective-width expression is l0/6 + bw + 6Df, so item 1 matches S. The corresponding L-beam expression is l0/12 + bw + 3Df, so item 2 matches P. If the neutral axis remains within the flange, the compression section is treated as a rectangle of width bf, making item 3 match Q. The applicable thick-flange treatment uses yf = 0.15xu + 0.65Df with the stated upper limit, so item 4 matches R. Therefore Option A gives the correct complete matching.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. A T-beam generally has effective slab flange on both sides of its web, while an L-beam generally occurs at an edge with effective flange mainly on one side. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. It is a hypothetical width over which the flange compression is represented so that the section can account for the nonuniform longitudinal stress distribution in the slab. సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. The slab and web should be cast integrally or otherwise effectively bonded so that composite action can develop. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. bf = l0/6 + bw + 6Df సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: B. bf = l0/12 + bw + 3Df సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: B. 0.7 times the effective span సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: C. 1833 mm సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 1100 mm సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. The web width plus half the sum of the clear distances to the adjacent beams on the two sides, as applicable సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. As a rectangular section of width bf సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. 50.1 mm సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 259.4 kN·m సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: A. C = 0.36fck bw xu + 0.45fck(bf - bw)Df సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. 243.2 mm సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 760 kN·m సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. yf = 0.15xu + 0.65Df, but not greater than Df సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. xu = 173.6 mm and yf = 91.0 mm సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: C. 340.2 kN·m సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: C. 438.8 kN·m సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: C. 114 mm సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: C. 525.1 kN·m సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: A. Treat the compression zone essentially as a rectangular section of web width rather than crediting the top slab as a compression flange. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: B. Ast = 1145 mm² and xu = 38.3 mm సమాధానం: B.
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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-P, 3-Q, 4-R సమాధానం: A.
