RCC Structures Design A flat slab is supported On columns On beams On beams and columns On columns monolithically built with slab On columns On beams On beams and columns On columns monolithically built with slab ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a beam the local bond stress Sb, is equal to Leaver arm/(Bending moment × Total perimeter of reinforcement) Shear force/(Leaver arm × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Leaver arm/(Shear force × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) Shear force/(Leaver arm × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Leaver arm/(Shear force × Total perimeter of reinforcement) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If a rectangular pre-stressed beam of an effective span of 5 meters and carrying a total load 3840 kg/m, is designed by the load balancing method, the central dip of the parabolic tendon should be 5 cm 15 cm 20 cm 10 cm 5 cm 15 cm 20 cm 10 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design For a circular slab carrying a uniformly distributed load, the ratio of the maximum negative to maximum positive radial moment, is 1 2 5 3 1 2 5 3 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If Ac, Asc and A are areas of concrete, longitudinal steel and section of a R.C.C. column and m and σc are the modular ratio and maximum stress in the configuration of concrete, the strength of column is σc[A + (m - 1)Asc] σc(A - Asc) + m σcAsc All listed here σcAc + m σcAsc σc[A + (m - 1)Asc] σc(A - Asc) + m σcAsc All listed here σcAc + m σcAsc ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The transverse reinforcements provided at right angles to the main reinforcement Resist the shrinkage stress Distribute the load Resist the temperature stresses All of these Resist the shrinkage stress Distribute the load Resist the temperature stresses All of these ANSWER DOWNLOAD EXAMIANS APP
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