RCC Structures Design If the diameter of longitudinal bars of a square column is 16 mm, the diameter of lateral ties should not be less than 6 mm 5 mm 7 mm 4 mm 6 mm 5 mm 7 mm 4 mm 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 Kx (1 - x/l) + bw Kx (1 + x/l) + bw All listed here K/x (1 + x/d) + bw Kx (1 - x/l) + bw Kx (1 + x/l) + bw All listed here K/x (1 + x/d) + bw ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In the zone of R.C.C. beam where shear stress is less than 5 kg/cm², nominal reinforcement is provided at a pitch of One and half lever arm of the section One-half lever arm of the section One-third lever arm of the section Lever arm of the section One and half lever arm of the section One-half lever arm of the section One-third lever arm of the section Lever arm of the section ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The diameter of transverse reinforcement of columns should be equal to one-fourth of the diameter of the main steel rods but not less than 7 mm 5 mm 4 mm 6 mm 7 mm 5 mm 4 mm 6 mm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a combined footing if shear stress does not exceed 5 kg/cm², the nominal stirrups provided are 10 legged 6 legged 8 legged 12 legged 10 legged 6 legged 8 legged 12 legged 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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