RCC Structures Design If the bearing capacity of soil is 10 tonnes/cm² and the projection of plain concrete footing from walls, is a cm, the depth D of footing is D = 0.775 a D = 0.775 √a D = 0.775 a² D = 0.0775 a D = 0.775 a D = 0.775 √a D = 0.775 a² D = 0.0775 a ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design For a continuous floor slab supported on beams, the ratio of end span length and intermediate span length, is 0.7 0.6 0.9 0.8 0.7 0.6 0.9 0.8 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A flat slab is supported On columns On beams and columns On beams On columns monolithically built with slab On columns On beams and columns On beams On columns monolithically built with slab ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Pick up the assumption for the design of a pre-stressed concrete member from the following: All listed here Variation of stress in reinforcement due to changes in external loading is negligible A transverse plane section remains a plane after bending During deformation limits, Hook's law is equally applicable to concrete as well as to steel All listed here Variation of stress in reinforcement due to changes in external loading is negligible A transverse plane section remains a plane after bending During deformation limits, Hook's law is equally applicable to concrete as well as to steel ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a beam the local bond stress Sb, is equal to Leaver arm/(Shear force × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Shear force/(Leaver arm × Total perimeter of reinforcement) Leaver arm/(Shear force × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Shear force/(Leaver arm × Total perimeter of reinforcement) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The load stress of a section can be reduced by Decreasing the lever arm Increasing the total perimeter of bars Replacing larger bars by greater number of small bars Replacing smaller bars by greater number of greater bars Decreasing the lever arm Increasing the total perimeter of bars Replacing larger bars by greater number of small bars Replacing smaller bars by greater number of greater bars ANSWER DOWNLOAD EXAMIANS APP
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