RCC Structures Design In a pre-stressed beam carrying an external load W with a bent tendon is having angle of inclination ? and pre-stressed load P. The net downward load at the centre is W - 2P cos θ W - P sin θ W - 2P sin θ W - P cos θ W - 2P cos θ W - P sin θ W - 2P sin θ W - P cos θ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The effective span of a simply supported slab, is Clear span plus effective depth of the slab Clear distance between the inner faces of the walls plus twice the thickness of the wall None of these Distance between the centers of the bearings Clear span plus effective depth of the slab Clear distance between the inner faces of the walls plus twice the thickness of the wall None of these Distance between the centers of the bearings ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A pre-stressed rectangular beam which carries two concentrated loads W at L/3 from either end, is provided with a bent tendon with tension P such that central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip h is WL/3P WL/P WL/2P WL/4P WL/3P WL/P WL/2P WL/4P ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The section of a reinforced beam where most distant concrete fibre in compression and tension in steel attains permissible stresses simultaneously, is called Critical section Economic section All listed here Balanced section Critical section Economic section All listed here Balanced section ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A column is regarded as long column if the ratio of its effective length and lateral dimension, exceeds 20 25 10 15 20 25 10 15 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 - (l - x̅) 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) ANSWER DOWNLOAD EXAMIANS APP
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