RCC Design (IS 456) interview questions

1. A.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
2. B.nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
3. C.Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
4. D.As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area

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1. A.Span-to-effective-depth control
2. B.Partial safety factors for materials
3. C.Development length
4. D.Minimum eccentricity for columns

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1. A.Partial safety factors for materials — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
2. B.Partial safety factors for materials — Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
3. C.Partial safety factors for materials — nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
4. D.Partial safety factors for materials — γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement

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1. A.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
2. B.Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
3. C.Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
4. D.As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area

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1. A.Partial safety factors for loads
2. B.Development length
3. C.Partial safety factors for materials
4. D.Span-to-effective-depth control

Answer + AI explanation with Pro

1. A.Partial safety factors for loads — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
2. B.Partial safety factors for loads — As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area
3. C.Partial safety factors for loads — Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
4. D.Partial safety factors for loads — basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages

Answer + AI explanation with Pro

1. A.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
2. B.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
3. C.every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
4. D.Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm

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1. A.Span-to-effective-depth control
2. B.Partial safety factors for materials
3. C.Crack width limit
4. D.Nominal cover by exposure

Answer + AI explanation with Pro

1. A.Nominal cover by exposure — surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
2. B.Nominal cover by exposure — Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
3. C.Nominal cover by exposure — Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
4. D.Nominal cover by exposure — γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement

Answer + AI explanation with Pro

1. A.As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area
2. B.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
3. C.every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
4. D.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F

Answer + AI explanation with Pro

1. A.Span-to-effective-depth control
2. B.Minimum tension steel in beams
3. C.Partial safety factors for loads
4. D.Crack width limit

Answer + AI explanation with Pro

1. A.Minimum tension steel in beams — Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
2. B.Minimum tension steel in beams — As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area
3. C.Minimum tension steel in beams — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
4. D.Minimum tension steel in beams — Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm

Answer + AI explanation with Pro

1. A.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
2. B.Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
3. C.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
4. D.Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression

Answer + AI explanation with Pro

1. A.Minimum eccentricity for columns
2. B.Development length
3. C.Partial safety factors for materials
4. D.Minimum tension steel in beams

Answer + AI explanation with Pro

1. A.Development length — every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
2. B.Development length — nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
3. C.Development length — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
4. D.Development length — surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F

Answer + AI explanation with Pro

1. A.Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
2. B.every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
3. C.nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
4. D.basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages

Answer + AI explanation with Pro

1. A.Span-to-effective-depth control
2. B.Minimum eccentricity for columns
3. C.Partial safety factors for loads
4. D.Minimum tension steel in beams

Answer + AI explanation with Pro

1. A.Span-to-effective-depth control — Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
2. B.Span-to-effective-depth control — nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
3. C.Span-to-effective-depth control — Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
4. D.Span-to-effective-depth control — basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages

Answer + AI explanation with Pro

1. A.Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
2. B.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
3. C.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
4. D.nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess

Answer + AI explanation with Pro

1. A.Minimum tension steel in beams
2. B.Nominal cover by exposure
3. C.Shear design procedure
4. D.Development length

Answer + AI explanation with Pro

1. A.Shear design procedure — nominal shear Vu/bd is compared with τc from Table 19 for the steel percentage, must not exceed τc,max of Table 20, and stirrups carry the excess
2. B.Shear design procedure — Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
3. C.Shear design procedure — every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
4. D.Shear design procedure — γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement

Answer + AI explanation with Pro

1. A.As/bd must be at least 0.85/fy, about 0.20 percent for Fe415, and total steel on either face may not exceed 4 percent of gross area
2. B.Table 18 uses 1.5 on dead plus live load, 1.2 when dead, live and wind or earthquake act together, and 0.9 on dead load when it helps stability
3. C.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
4. D.surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F

Answer + AI explanation with Pro

1. A.Development length
2. B.Crack width limit
3. C.Partial safety factors for materials
4. D.Partial safety factors for loads

Answer + AI explanation with Pro

1. A.Crack width limit — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression
2. B.Crack width limit — γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
3. C.Crack width limit — surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
4. D.Crack width limit — basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages

Answer + AI explanation with Pro

1. A.Table 16 requires 20 mm for mild, 30 mm moderate, 45 mm severe, 50 mm very severe and 75 mm extreme exposure, with footings never below 50 mm
2. B.γm is 1.5 for concrete and 1.15 for steel, so design strengths become 0.446 fck in the stress block and 0.87 fy for reinforcement
3. C.every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
4. D.basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages

Answer + AI explanation with Pro

1. A.Crack width limit
2. B.Partial safety factors for materials
3. C.Minimum eccentricity for columns
4. D.Partial safety factors for loads

Answer + AI explanation with Pro

1. A.Minimum eccentricity for columns — surface crack width is limited to 0.3 mm generally, 0.2 mm where aggressive exposure exists and 0.1 mm for very severe conditions, checked by Annex F
2. B.Minimum eccentricity for columns — basic ratios of 7 for cantilevers, 20 for simply supported and 26 for continuous spans up to 10 m, modified for tension and compression steel percentages
3. C.Minimum eccentricity for columns — every column is designed for at least L/500 plus D/30 or 20 mm eccentricity, whichever is more, about each axis separately
4. D.Minimum eccentricity for columns — Ld = φ σs / (4 τbd), with design bond stress from Table for plain bars increased 60 percent for deformed bars and 25 percent more in compression

Answer + AI explanation with Pro