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 20 cm 15 cm 10 cm 5 cm 20 cm 15 cm 10 cm 5 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Pick up the incorrect statement from the following: Tensile reinforcement bars of a rectangular beam Are maintained at bottom to provide at least local bond stress Are curtailed if not required to resist the bending moment Are bent up at suitable places to serve as shear reinforcement Are bent down at suitable places to serve as shear reinforcement Are maintained at bottom to provide at least local bond stress Are curtailed if not required to resist the bending moment Are bent up at suitable places to serve as shear reinforcement Are bent down at suitable places to serve as shear reinforcement ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If K is a constant depending upon the ratio of the width of the slab to its effective span l, x is the distance of the concentrated load from the nearer support, bw is the width of the area of contact of the concentrated load measured parallel to the supported edge, the effective width of the slab be is All listed here Kx (1 + x/l) + bw Kx (1 - x/l) + bw K/x (1 + x/d) + bw All listed here Kx (1 + x/l) + bw Kx (1 - x/l) + bw K/x (1 + x/d) + bw 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 T - R R + T √(R² + T²) R - T T - R R + T √(R² + T²) R - T ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The minimum cube strength of concrete used for a pre-stressed member, is 350 kg/cm² 250 kg/cm² 150 kg/cm² 50 kg/cm² 350 kg/cm² 250 kg/cm² 150 kg/cm² 50 kg/cm² 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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