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Question 1
ప్రశ్న 1
Which description best explains diagonal-tension cracking in a reinforced-concrete beam subjected to shear?
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Explanation:
Shear in a reinforced-concrete beam produces principal tensile stresses in addition to the stresses caused by flexure. Flexural cracks that begin near the tension face may become inclined as they extend into regions where shear is significant. Concrete has relatively low tensile strength, so these principal tensile stresses can produce diagonal cracking. Shear reinforcement is provided to cross such potential inclined cracks and contribute to shear resistance. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
For a reinforced-concrete beam of uniform depth, which expression gives the nominal shear stress τv under Limit State design?
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Explanation:
IS 456 defines the nominal shear stress in a beam of uniform depth as Vu divided by bd. Here Vu is the factored shear force at the section under consideration. The quantity b is the beam breadth, or the web breadth for a flanged beam, and d is the effective depth. The resulting nominal shear stress is compared with the design shear strength of concrete and the maximum permissible shear stress. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
A rectangular RCC beam has b = 300 mm, effective depth d = 500 mm and factored shear force Vu = 180 kN. What is the nominal shear stress τv?
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Explanation:
Nominal shear stress is τv = Vu/(bd). Substituting Vu = 180000 N, b = 300 mm and d = 500 mm gives 180000/(300 × 500). This evaluates to 1.20 N/mm². This value must subsequently be compared with τc and τc,max for the applicable concrete and reinforcement conditions. Therefore Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
A T-beam has effective flange width 1200 mm, web width bw = 250 mm, effective depth d = 550 mm and factored shear Vu = 200 kN. What nominal shear stress should be used for beam shear design?
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Explanation:
For shear design of a flanged beam, b is taken as the breadth of the web rather than the effective flange width. Thus τv = 200000/(250 × 550). The result is approximately 1.455 N/mm². Using the full flange width would substantially underestimate the nominal shear stress. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
For determining the design shear strength τc from IS 456 Table 19, how is the percentage of longitudinal tension reinforcement p calculated?
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Explanation:
Table 19 expresses the longitudinal tension reinforcement quantity as 100Ast/(bd). The Ast used is the relevant longitudinal tension reinforcement associated with the section being checked. The code note requires this reinforcement to continue at least one effective depth beyond the section, with the stated detailing qualification at supports. This percentage is one of the parameters governing the tabulated concrete shear strength τc. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
According to IS 456 Table 19, what is the design shear strength τc of M25 concrete when the longitudinal tension reinforcement percentage p is 1.00?
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Explanation:
IS 456 Table 19 gives design shear strengths of concrete for specified concrete grades and tension-reinforcement percentages. For M25 concrete at p = 1.00 percent, the tabulated value is 0.64 N/mm². This value represents the concrete contribution used in ordinary beam shear design. It is distinct from τc,max, which is the absolute maximum nominal shear stress permitted for the section. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
For M25 concrete, how does the IS 456 Table 19 design shear strength τc change when p increases from 0.50 percent to 1.50 percent?
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Explanation:
For M25 concrete, Table 19 gives τc = 0.49 N/mm² at p = 0.50 percent. At p = 1.50 percent, the corresponding value is 0.74 N/mm². Thus the tabulated concrete shear strength increases as the specified longitudinal tension reinforcement percentage increases over this range. This behaviour reflects the influence of longitudinal reinforcement on shear-transfer mechanisms after cracking. Therefore Option C is correct.
వివరణ:
Question 8
ప్రశ్న 8
What is the maximum nominal shear stress τc,max permitted for an M25 reinforced-concrete beam under IS 456?
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Explanation:
IS 456 Table 20 specifies the maximum shear stress permitted in beams. For M25 concrete, τc,max is 3.1 N/mm². This limit applies even when shear reinforcement is provided. It prevents a design from relying on arbitrarily large quantities of stirrups when the concrete diagonal-compression capacity is inadequate. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
An M25 beam has b = 300 mm, d = 500 mm and factored shear Vu = 480 kN. What is the appropriate design conclusion?
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Explanation:
The nominal shear stress is 480000/(300 × 500) = 3.20 N/mm². For M25 concrete, τc,max is 3.10 N/mm². IS 456 does not permit the nominal beam shear stress to exceed this maximum even when shear reinforcement is provided. The beam dimensions, concrete grade, loading arrangement, or another relevant design parameter must therefore be revised. Hence Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which set correctly lists forms of shear reinforcement recognized by IS 456 for beams when design shear reinforcement is required?
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Explanation:
IS 456 recognizes vertical stirrups as a standard form of beam shear reinforcement. It also permits bent-up bars used along with stirrups. Inclined stirrups are another recognized form. The chosen system must satisfy the relevant strength and detailing provisions. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
Where bent-up bars are used as part of the shear reinforcement of an RCC beam, what limitation applies to their contribution to shear resistance?
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Explanation:
IS 456 limits the amount of shear resistance that may be credited to bent-up bars. Their contribution is not permitted to exceed one-half of the total shear resistance provided by the shear reinforcement. This provision avoids excessive dependence on bent-up longitudinal reinforcement for shear resistance. Stirrups consequently remain an important component when bent-up bars are used. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
A beam has Vu = 180 kN, b = 300 mm and d = 500 mm. If τc = 0.64 N/mm², what shear force must be resisted by designed shear reinforcement?
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Explanation:
The concrete shear resistance represented by τcbd is 0.64 × 300 × 500 N. This equals 96000 N, or 96 kN. The shear reinforcement is designed for Vu - τcbd. Therefore Vus = 180 - 96 = 84 kN. Hence Option B is correct.
వివరణ:
Question 13
ప్రశ్న 13
Which expression gives the shear resistance Vus of vertical stirrups under the IS 456 Limit State method?
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Explanation:
For vertical stirrups, IS 456 gives Vus = 0.87fyAsvd/sv. Asv is the total cross-sectional area of the stirrup legs effective in resisting shear. The parameter sv is the spacing of stirrups measured along the member. For this calculation, the characteristic strength of shear reinforcement is not taken greater than the code-prescribed limit. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
The beam in Question 12 requires Vus = 84 kN. If two-legged 8 mm diameter Fe415 vertical stirrups are used with d = 500 mm, what stirrup spacing sv is required approximately?
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Explanation:
A two-legged 8 mm stirrup has Asv = 2 × π(8²)/4 = approximately 100.53 mm². For vertical stirrups, sv = 0.87fyAsvd/Vus. Substituting fy = 415 N/mm², d = 500 mm and Vus = 84000 N gives approximately 216.1 mm. This value is also below the general maximum spacing permitted for vertical stirrups. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
Which expression states the IS 456 minimum shear reinforcement requirement for stirrups in a beam?
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Explanation:
IS 456 specifies a minimum quantity of beam shear reinforcement in the form of stirrups. The requirement is Asv/(bsv) not less than 0.4/(0.87fy). The characteristic strength fy used in this expression is not taken greater than 415 N/mm². This minimum reinforcement improves post-cracking behaviour and provides reserve shear resistance even when calculated shear demand is relatively low. Therefore Option A is correct.
వివరణ:
Question 16
ప్రశ్న 16
A 300 mm wide beam with d = 500 mm uses two-legged 8 mm Fe415 vertical stirrups. Considering both the minimum-shear-reinforcement equation and the general maximum-spacing rule, what is the largest permissible spacing approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The two-legged 8 mm stirrup area is approximately 100.53 mm². The minimum-reinforcement equation by itself gives a limiting spacing of about 302.5 mm. For vertical stirrups, however, the general spacing limit is the smaller of 0.75d and 300 mm. Here 0.75d = 375 mm, so the absolute 300 mm limit governs. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
A beam has effective depth d = 320 mm and uses vertical stirrups. Ignoring any smaller spacing required by shear strength or minimum reinforcement, what is the maximum spacing permitted along the beam?
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Explanation:
The maximum spacing of vertical shear reinforcement is 0.75d, subject also to an absolute maximum of 300 mm. For d = 320 mm, 0.75d equals 240 mm. Since 240 mm is smaller than 300 mm, it governs. A strength calculation could require a still smaller spacing, but it cannot justify a larger one. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which expression gives the shear resistance of inclined stirrups making an angle α with the axis of a beam, where α is not less than 45 degrees?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For inclined stirrups or a series of bars bent up at different cross-sections, IS 456 includes both sine and cosine components in the shear-resistance expression. The relation is Vus = [0.87fyAsvd/sv](sin α + cos α). The angle α is measured between the inclined reinforcement and the member axis. The code requires this angle not to be less than 45 degrees for the stated provision. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
An M30 beam has b = 300 mm, d = 500 mm and Ast = 2250 mm² of longitudinal tension reinforcement. What are the percentage tension reinforcement p and the corresponding τc from IS 456 Table 19?
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Explanation:
The reinforcement percentage is p = 100Ast/(bd). Substituting the given values gives 100 × 2250/(300 × 500) = 1.50 percent. For M30 concrete at p = 1.50 percent, IS 456 Table 19 gives τc = 0.76 N/mm². The Table 20 maximum value of 3.5 N/mm² for M30 is a different limit and should not be confused with τc. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
For the M30 beam in the previous question, the factored shear force is Vu = 240 kN. How much shear force must be carried by shear reinforcement?
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Explanation:
The concrete shear contribution is τcbd. Using τc = 0.76 N/mm² gives 0.76 × 300 × 500 = 114000 N, or 114 kN. The shear reinforcement must therefore resist 240 - 114 = 126 kN. The nominal shear stress is 1.60 N/mm², which is below the M30 maximum of 3.5 N/mm². Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
For the M30 beam requiring Vus = 126 kN, two-legged 10 mm Fe415 vertical stirrups are proposed. What spacing is required approximately?
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Explanation:
A two-legged 10 mm stirrup has Asv = 2 × π(10²)/4 = approximately 157.08 mm². The spacing follows from sv = 0.87fyAsvd/Vus. Substitution of fy = 415 N/mm², d = 500 mm and Vus = 126000 N gives approximately 225.1 mm. This spacing is below the general maximum of 300 mm for this effective depth. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
For a member subjected to axial compression Pu, IS 456 permits the Table 19 concrete shear strength to be multiplied by δ = 1 + 3Pu/(Agfck), subject to δ not exceeding 1.5. For Pu = 600 kN, Ag = 150000 mm² and fck = 25 N/mm², what is δ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The enhancement factor is δ = 1 + 3Pu/(Agfck). Substitution gives 1 + 3(600000)/[150000 × 25]. The added term is 0.48, so δ = 1.48. This is below the code maximum of 1.5 and therefore requires no further capping. Hence Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
If the unmodified Table 19 design shear strength for the axially compressed member in the previous question is τc = 0.64 N/mm², what enhanced design shear strength is obtained?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The applicable multiplier from the previous calculation is 1.48. The enhanced concrete shear strength is therefore 1.48 × 0.64 N/mm². This gives 0.9472 N/mm². Rounded suitably, the value is approximately 0.95 N/mm². Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
For a beam carrying generally uniformly distributed load, or where the principal load is farther than 2d from the face of a support, which simplified IS 456 shear-design provision may be used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
IS 456 contains provisions recognizing enhanced shear behaviour near supports. For beams carrying generally uniform load, or where the principal load lies farther than 2d from the support face, the shear stress may be calculated at a section d from the face of the support. This does not mean that shear at the support is physically zero. The accompanying requirements on concrete shear strength, anchorage and maximum shear stress remain relevant. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about Limit State shear design of RCC beams:
1. Nominal shear stress in a uniform-depth beam is τv = Vu/(bd).
2. Even with shear reinforcement, τv shall not exceed τc,max from Table 20.
3. The credited shear contribution of bent-up bars shall not exceed one-half of the total shear resistance provided by shear reinforcement.
4. The value of fy used for shear reinforcement calculations may always be taken greater than 415 N/mm² when higher-grade bars are physically provided.
Which statements are correct?
1. Nominal shear stress in a uniform-depth beam is τv = Vu/(bd).
2. Even with shear reinforcement, τv shall not exceed τc,max from Table 20.
3. The credited shear contribution of bent-up bars shall not exceed one-half of the total shear resistance provided by shear reinforcement.
4. The value of fy used for shear reinforcement calculations may always be taken greater than 415 N/mm² when higher-grade bars are physically provided.
Which statements are correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement 1 correctly gives the nominal shear-stress relation for a uniform-depth beam. Statement 2 correctly states the absolute maximum shear-stress restriction imposed by Table 20. Statement 3 correctly states the limitation placed on the credited contribution of bent-up bars. Statement 4 is incorrect because the characteristic strength used for the relevant shear-reinforcement design provisions is not taken greater than 415 N/mm². 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. Inclined cracking can develop when principal tensile stresses produced by the combined shear and flexural stress state exceed the tensile capacity of concrete. సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. τv = Vu/(bd) సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: C. 1.20 N/mm² సమాధానం: C.
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Question 4 ప్రశ్న 4Answer: C. 1.455 N/mm² సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. p = 100Ast/(bd), using the relevant longitudinal tension reinforcement continuing sufficiently beyond the section సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 0.64 N/mm² సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: C. It increases from 0.49 to 0.74 N/mm². సమాధానం: C.
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Question 8 ప్రశ్న 8Answer: C. 3.10 N/mm² సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. τv = 3.20 N/mm², which exceeds τc,max = 3.10 N/mm², so the section should be redesigned rather than made acceptable merely by adding shear reinforcement. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. Vertical stirrups, bent-up bars used along with stirrups, and inclined stirrups సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. Their contribution shall not exceed one-half of the total shear resistance provided by shear reinforcement. సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: B. 84 kN సమాధానం: B.
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Question 13 ప్రశ్న 13Answer: A. Vus = 0.87fyAsvd/sv సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. 216 mm సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: A. Asv/(bsv) ≥ 0.4/(0.87fy), with fy for this purpose not taken greater than 415 N/mm² సమాధానం: A.
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Question 16 ప్రశ్న 16Answer: C. 300 mm సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 240 mm సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. Vus = [0.87fyAsvd/sv](sin α + cos α) సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. p = 1.50 percent and τc = 0.76 N/mm² సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: C. 126 kN సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: C. 225 mm సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: C. 1.48 సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 0.95 N/mm² సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. The shear stress may be calculated at a section located a distance d from the face of the support. సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: B. 1, 2 and 3 only సమాధానం: B.