RCC Structures Design A foundation is called shallow if its depth, is Half of its width Three-fourth of its width Equal to its width One-fourth of its width Half of its width Three-fourth of its width Equal to its width One-fourth of its width ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If l₁ and l₂ are the lengths of long and short spans of a two way slab simply supported on four edges and carrying a load w per unit area, the ratio of the loads split into w₁ and w₂ acting on strips parallel to l₂ and l₁ is w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)³ w₁/w₂ = (l₂/l₁)² w₁/w₂ = (l₂/l₁)⁴ w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)³ w₁/w₂ = (l₂/l₁)² w₁/w₂ = (l₂/l₁)⁴ 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
RCC Structures Design In a simply supported slab, alternate bars are curtailed at 1/6th of the span 1/7th of the span 1/4th of the span 1/5th of the span 1/6th of the span 1/7th of the span 1/4th of the span 1/5th of the span ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If R and T are rise and tread of a stair spanning horizontally, the steps are supported by a wall on one side and by a stringer beam on the other side, the steps are designed as beams of width R + T R - T √(R² + T²) T - R R + T R - T √(R² + T²) T - R 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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