RCC Structures Design Minimum spacing between horizontal parallel reinforcement of the same size should not be less than 3.5 diameters 3 diameters 2.5 diameters One diameter 3.5 diameters 3 diameters 2.5 diameters One diameter ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A simply supported beam 6 m long and of effective depth 50 cm, carries a uniformly distributed load 2400 kg/m including its self weight. If the lever arm factor is 0.85 and permissible tensile stress of steel is 1400 kg/cm², the area of steel required, is 14 cm² 15 cm² 17 cm² 16 cm² 14 cm² 15 cm² 17 cm² 16 cm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If W is total load per unit area on a panel, D is the diameter of the column head, L is the span in two directions, then the sum of the maximum positive bending moment and average of the negative bending moment for the design of the span of a square flat slab, should not be less than WL/12 (L - 2D/3)² WL/10 (L + 2D/3)² WL/12 (L - D/3)² WL/10 (L - 2D/3)² WL/12 (L - 2D/3)² WL/10 (L + 2D/3)² WL/12 (L - D/3)² WL/10 (L - 2D/3)² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a cantilever retaining wall without a heel slab Width of the base slab is kept 0.7 time the total height of the wall All listed here Thickness of the stem is kept same throughout Base slab is made 10 cm thicker than the stem Width of the base slab is kept 0.7 time the total height of the wall All listed here Thickness of the stem is kept same throughout Base slab is made 10 cm thicker than the stem ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If jd is the lever arm and ΣO is the total perimeter of reinforcement of an R.C.C. beam, the bond stress at the section having Q shear force, is Q/jdƩO 2 × Q/jdƩO Q/2jdƩO Q/3jdƩO Q/jdƩO 2 × Q/jdƩO Q/2jdƩO Q/3jdƩO ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If q is the punching shear resistance per unit area a, is the side of a square footing for a column of side b, carrying a weight W including the weight of the footing, the depth (D) of the footing from punching shear consideration, is D = W (a - b)/4a²bq D = W (a² - b²)/4abq D = W (a² - b²)/8a²bq D = W (a² - b²)/4a²bq D = W (a - b)/4a²bq D = W (a² - b²)/4abq D = W (a² - b²)/8a²bq D = W (a² - b²)/4a²bq ANSWER DOWNLOAD EXAMIANS APP
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