RCC Structures Design If the permissible compressive and tensile stresses in a singly reinforced beam are 50 kg/cm² and 1400 kg/cm² respectively and the modular ratio is 18, the percentage area At of the steel required for an economic section, is 0.696 % 0.796 % 0.496 % 0.596 % 0.696 % 0.796 % 0.496 % 0.596 % ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Pick up the incorrect statement from the following. The intensity of horizontal shear stress at the elemental part of a beam section, is directly proportional to Moment of the beam section about its neutral axis Shear force Distance of the C.G. of the area from its neutral axis Area of the section Moment of the beam section about its neutral axis Shear force Distance of the C.G. of the area from its neutral axis Area of the section ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If a rectangular pre-stressed beam of an effective span of 5 meters and carrying a total load 3840 kg/m, is designed by the load balancing method, the central dip of the parabolic tendon should be 15 cm 5 cm 10 cm 20 cm 15 cm 5 cm 10 cm 20 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Bottom bars under the columns are extended into the interior of the footing slab to a distance greater than 42 diameters from the outer edge of the column 42 diameters from the inner edge of the column 24 diameters from the centre of the column 42 diameters from the centre of the column 42 diameters from the outer edge of the column 42 diameters from the inner edge of the column 24 diameters from the centre of the column 42 diameters from the centre of the column ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The advantage of a concrete pile over a timber pile, is No decay due to termites All of these Higher bearing capacity No restriction on length No decay due to termites All of these Higher bearing capacity No restriction on length 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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