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) (r/1 + r²) M (5/6) (r²/1 + r³) M (5/6) (r²/1 + r²) M (5/6) (r²/1 + r⁴) M (5/6) (r/1 + r²) M (5/6) (r²/1 + r³) M (5/6) (r²/1 + r²) M (5/6) (r²/1 + r⁴) M ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A pre-stressed concrete member is preferred because All listed here Removal of cracks in the members due to shrinkage Its dimensions are not decided from the diagonal tensile stress Large size of long beams carrying large shear force need not be adopted All listed here Removal of cracks in the members due to shrinkage Its dimensions are not decided from the diagonal tensile stress Large size of long beams carrying large shear force need not be adopted ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The floor slab of a building is supported on reinforced cement floor beams. The ratio of the end and intermediate spans is kept 0.2 0.7 0.8 0.9 0.2 0.7 0.8 0.9 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The diameter of the column head support a flat slab, is generally kept 0.25 times the span length 0.25 times the diameter of the column 4.0 cm larger than the diameter of the column 5.0 cm larger than the diameter of the column 0.25 times the span length 0.25 times the diameter of the column 4.0 cm larger than the diameter of the column 5.0 cm larger than the diameter of the column ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In favourable circumstances a 15 cm concrete cube after 28 days, attains a maximum crushing strength 100 kg/cm² 400 kg/cm² 300 kg/cm² 200 kg/cm² 100 kg/cm² 400 kg/cm² 300 kg/cm² 200 kg/cm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The load stress of a section can be reduced by Replacing larger bars by greater number of small bars Decreasing the lever arm Increasing the total perimeter of bars Replacing smaller bars by greater number of greater bars Replacing larger bars by greater number of small bars Decreasing the lever arm Increasing the total perimeter of bars Replacing smaller bars by greater number of greater bars ANSWER DOWNLOAD EXAMIANS APP
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