RCC Structures Design Cantilever retaining walls can safely be used for a height not more than 5 m 4 m 6 m 3 m 5 m 4 m 6 m 3 m ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the length of an intermediate span of a continuous slab is 5 m, the length of the end span is kept 4.7 m 4.5 m 4.1 m 4.0 m 4.7 m 4.5 m 4.1 m 4.0 m ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If a bent tendon is required to balance a concentrated load W at the centre of the span L, the central dip h must be at least WL/P WL/2P WL/3P WL/4P WL/P WL/2P WL/3P WL/4P ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Though the effective depth of a T-beam is the distance between the top compression edge to the centre of the tensile reinforcement, for heavy loads, it is taken as 1/10th of the span 1/12th of the span 1/16th of the span 1/8th of the span 1/10th of the span 1/12th of the span 1/16th of the span 1/8th of the span ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If p1 is the vertical intensity of pressure at a depth h on a block of earth weighing w per unit volume and the angle of repose φ, the lateral intensity of pressure p2 is wh (1 - cos φ)/(1 + sin φ) w (1 - cos φ)/h (1 + sin φ) wh (1 - sin φ)/(1 + sin φ) wh (1 - tan φ)/(1 + tan φ) wh (1 - cos φ)/(1 + sin φ) w (1 - cos φ)/h (1 + sin φ) wh (1 - sin φ)/(1 + sin φ) wh (1 - tan φ)/(1 + tan φ) 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 - (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) ANSWER DOWNLOAD EXAMIANS APP
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