RCC Structures Design Spacing of stirrups in a rectangular beam, is Kept constant throughout the length Increased at the centre of the beam Increased at the ends Decreased towards the centre of the beam Kept constant throughout the length Increased at the centre of the beam Increased at the ends Decreased towards the centre of the beam ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design By over-reinforcing a beam, the moment of resistance can be increased not more than 25 % 20 % 10 % 15 % 25 % 20 % 10 % 15 % 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 As per IS : 456, the reinforcement in a column should not be less than 0.8% and not more than 8% of cross-sectional area 0.7% and not more than 7% of cross-sectional area 0.6% and not more than 6% of cross-sectional area 0.5% and not more than 5% of cross-sectional area 0.8% and not more than 8% of cross-sectional area 0.7% and not more than 7% of cross-sectional area 0.6% and not more than 6% of cross-sectional area 0.5% and not more than 5% of cross-sectional area ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A foundation rests on Sub-grade Both B and C Foundation soil Base of the foundation Sub-grade Both B and C Foundation soil Base of the foundation 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/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) y = L - (l - x̅) ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS