RCC Structures Design The breadth of a ribbed slab containing two bars must be between 12 cm to 15 cm 8 cm to 10 cm 6 cm to 7.5 cm 10 cm to 12 cm 12 cm to 15 cm 8 cm to 10 cm 6 cm to 7.5 cm 10 cm to 12 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If A is the area of the foundation of a retaining wall carrying a load W and retaining earth of weight 'w' per unit volume, the minimum depth (h) of the foundation from the free surface of the earth, is h = (W/Aw) [(1 - sin φ)/(1 + sin φ)] h = (W/Aw) [(1 + sin φ)/(1 + sin φ)] h = √(W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 - sin φ)/(1 + sin φ)] h = (W/Aw) [(1 + sin φ)/(1 + sin φ)] h = √(W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 - sin φ)/(1 + sin φ)]² 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)/8a²bq D = W (a² - b²)/4a²bq D = W (a² - b²)/4abq D = W (a - b)/4a²bq D = W (a² - b²)/8a²bq D = W (a² - b²)/4a²bq D = W (a² - b²)/4abq ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Distribution reinforcement in a simply supported slab, is provided to distribute Shrinkage stress Temperature stress Load All listed here Shrinkage stress Temperature stress Load All listed here ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design An R.C.C beam of 25 cm width has a clear span of 5 metres and carries a U.D.L. of 2000 kg/m inclusive of its self weight. If the lever arm of the section is 45 cm., the beam is Is safe with stirrups Safe in shear Is safe with stirrups and inclined members Needs revision of the section Is safe with stirrups Safe in shear Is safe with stirrups and inclined members Needs revision of the section ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the length of a combined footing for two columns l meters apart is L and the projection on the left side of the exterior column is x, then the projection y on the right side of the exterior column, in order to have a uniformly distributed load, is (where x̅ is the distance of centre of gravity of column loads). y = L - (l - x̅) y = L/2 - (l - x̅) y = L/2 + (l - x̅) y = L/2 - (l + x̅) y = L - (l - x̅) y = L/2 - (l - x̅) y = L/2 + (l - x̅) y = L/2 - (l + x̅) ANSWER DOWNLOAD EXAMIANS APP
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