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 Distribution reinforcement in a simply supported slab, is provided to distribute Shrinkage stress Load Temperature stress All listed here Shrinkage stress Load Temperature stress All listed here ANSWER DOWNLOAD EXAMIANS APP
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 sin θ W - 2P cos θ W - P sin θ W - P cos θ W - 2P sin θ W - 2P cos θ W - P sin θ W - P cos θ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The Young's modulus of elasticity of steel, is 275 KN/mm² 150 KN/mm² 250 KN/mm² 200 KN/mm² 275 KN/mm² 150 KN/mm² 250 KN/mm² 200 KN/mm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design With usual notations the depth of the neutral axis of a balanced section, is given by mc/t = (d - n)/n mc/t = n/(d - n) t/mc = (d - n)/n t/mc = (d + n)/n mc/t = (d - n)/n mc/t = n/(d - n) t/mc = (d - n)/n t/mc = (d + n)/n 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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