RCC Structures Design The anchorage value of a hook is assumed sixteen times the diameter of the bar if the angle of the bend, is 30° 40° 45° All listed here 30° 40° 45° All listed here ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A T-beam behaves as a rectangular beam of a width equal to its flange if its neutral axis Remains within the flange Coincides the geometrical centre of the beam None of these Remains below the slab Remains within the flange Coincides the geometrical centre of the beam None of these Remains below the slab ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design High strength concrete is used in pre-stressed member All listed here To overcome high bearing stresses developed at the ends To overcome bursting stresses at the ends To provide high bond stresses All listed here To overcome high bearing stresses developed at the ends To overcome bursting stresses at the ends To provide high bond stresses ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a combined footing for two columns carrying unequal loads, the maximum hogging bending moment occurs at Less loaded column More loaded column A point of the maximum shear force A point of zero shear force Less loaded column More loaded column A point of the maximum shear force A point of zero shear force ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The system in which high tensile alloy steel bars (silica manganese steel) are used as pre-stressing tendons, is known as C.L. standard system Magnel-Blaton system Lee-McCall system Freyssinet system C.L. standard system Magnel-Blaton system Lee-McCall system Freyssinet system 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 - (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) y = L/2 - (l + x̅) ANSWER DOWNLOAD EXAMIANS APP
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