RCC Structures Design For initial estimate for a beam design, the width is assumed 1/25th of span 1/30th of span 1/15th of span 1/20th of span 1/25th of span 1/30th of span 1/15th of span 1/20th of span ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The maximum ratio of span to depth of a slab simply supported and spanning in two directions, is 25 40 35 30 25 40 35 30 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The width of the flange of a T-beam, which may be considered to act effectively with the rib depends upon Breadth of the rib All of the listed here Overall thickness of the rib Centre to centre distance between T-beams Breadth of the rib All of the listed here Overall thickness of the rib Centre to centre distance between T-beams ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The thickness of base slab of a retaining wall generally provided, is One half of the width of the stem at the bottom One fourth of the width of the steam at the bottom Width of the stem at the bottom One-third of the width of the stem at the bottom One half of the width of the stem at the bottom One fourth of the width of the steam at the bottom Width of the stem at the bottom One-third of the width of the stem at the bottom ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design ‘P’ is the pre-stressed force applied to tendon of a rectangular pre-stressed beam whose area of cross section is ‘A’ and sectional modulus is ‘Z’. The minimum stress ‘f’ on the beam subjected to a maximum bending moment ‘M’ is f = (P/A) - (M/6Z) f = (A/P) - (M/Z) f = (P/A) - (M/Z) f = (P/'- (Z/M) f = (P/A) - (M/6Z) f = (A/P) - (M/Z) f = (P/A) - (M/Z) f = (P/'- (Z/M) 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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