RCC Structures Design The breadth of a ribbed slab containing two bars must be between 6 cm to 7.5 cm 8 cm to 10 cm 12 cm to 15 cm 10 cm to 12 cm 6 cm to 7.5 cm 8 cm to 10 cm 12 cm to 15 cm 10 cm to 12 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a combined footing if shear stress exceeds 5 kg/cm², the nominal stirrups provided are: 10 legged 12 legged 6 legged 8 legged 10 legged 12 legged 6 legged 8 legged ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The anchorage value of a hook is assumed sixteen times the diameter of the bar if the angle of the bend, is 30° All listed here 40° 45° 30° All listed here 40° 45° ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the depth of actual neutral axis of a doubly reinforced beam Is less than the depth of critical neutral axis, the steel in the tensile zone attains its maximum stress earlier Is greater than the depth of critical neutral axis, the concrete attains its maximum stress earlier Is equal to the depth of critical neutral axis; the concrete and steel attain their maximum stresses simultaneously All listed here Is less than the depth of critical neutral axis, the steel in the tensile zone attains its maximum stress earlier Is greater than the depth of critical neutral axis, the concrete attains its maximum stress earlier Is equal to the depth of critical neutral axis; the concrete and steel attain their maximum stresses simultaneously All listed here ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If ‘p’ is the net upward pressure on a square footing of side ‘b’ for a square column of side ‘a’, the maximum bending moment is given by M = pb (b + a)/8 M = pb (c - a)/4 M = pb (b - a)²/4 M = pb (b + a)/8 M = pb (b - a)²/8 M = pb (b + a)/8 M = pb (c - a)/4 M = pb (b - a)²/4 M = pb (b + a)/8 M = pb (b - a)²/8 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/2 + (l - x̅) 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) ANSWER DOWNLOAD EXAMIANS APP
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