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 The advantage of reinforced concrete, is due to All listed here Monolithic character Economy because of less maintenance cost Fire-resisting and durability All listed here Monolithic character Economy because of less maintenance cost Fire-resisting and durability ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If L is the effective span of a R.C.C. beam which is subjected to maximum shear qmax at the ends, the distance from either end over which stirrups for the shear, are provided, is (L/3) (1 - 5/qmax) (L/2) (1 - 5/qmax) (L/2) (1 - 3/qmax) (L/2) (1 - 2/qmax) (L/3) (1 - 5/qmax) (L/2) (1 - 5/qmax) (L/2) (1 - 3/qmax) (L/2) (1 - 2/qmax) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If diameter of a reinforcement bar is d, the anchorage value of the hook is 8d 4d 16d 12d 8d 4d 16d 12d ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Based on punching shear consideration, the overall depth of a combined footing under a column A, is (Area of the column A × Safe punching stress)/Load on column A (Perimeter of column A × Safe punching stress)/(Load on column A × Upward pressure × Area of the column) (Perimeter of column A × Safe punching stress)/(Load on column A + Upward pressure × Area of the column) None of these (Area of the column A × Safe punching stress)/Load on column A (Perimeter of column A × Safe punching stress)/(Load on column A × Upward pressure × Area of the column) (Perimeter of column A × Safe punching stress)/(Load on column A + Upward pressure × Area of the column) None of these ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the ratio of long and short spans of a two way slab with corners held down is r, the actual reduction of B.M. is given by (5/6) (r²/1 + r²) M (5/6) (r²/1 + r⁴) M (5/6) (r²/1 + r³) M (5/6) (r/1 + r²) M (5/6) (r²/1 + r²) M (5/6) (r²/1 + r⁴) M (5/6) (r²/1 + r³) M (5/6) (r/1 + r²) M ANSWER DOWNLOAD EXAMIANS APP