RCC Structures Design Steel beam theory is used for Design of simple steel beams Beams if shear exceeds 4 times allowable shear stress Doubly reinforced beams ignoring compressive stress in concrete Steel beams encased in concrete Design of simple steel beams Beams if shear exceeds 4 times allowable shear stress Doubly reinforced beams ignoring compressive stress in concrete Steel beams encased in concrete ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If longitudinally spanning stairs are casted along with their landings, the maximum bending moment per metre width, is taken as wl24 wl28 wl210 wl212 wl24 wl28 wl210 wl212 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the tendon is placed at an eccentricity e below the centroidal axis of the longitudinal axis of a rectangular beam (sectional modulus Z and stressed load P in tendon) the stress at the extreme top edge Remains unchanged Is increased by PZ/e Is increased by Pe/Z Is decreased by Pe/Z Remains unchanged Is increased by PZ/e Is increased by Pe/Z Is decreased by Pe/Z ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design An R.C.C. column of 30 cm diameter is reinforced with 6 bars 12 mm φ placed symmetrically along the circumference. If it carries a load of 40,000 kg axially, the stress is 250 kg/cm² 100 kg/cm² 49.9 kg/cm² 175 kg/cm² 250 kg/cm² 100 kg/cm² 49.9 kg/cm² 175 kg/cm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A pile of length ‘L’ carrying a uniformly distributed load ‘W’ per metre length is suspended at the centre and from other two points 0.15 L from either end ; the maximum hogging moment will be WL²/90 WL²/15 WL²/60 WL²/30 WL²/90 WL²/15 WL²/60 WL²/30 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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