RCC Structures Design If jd is the lever arm and ΣO is the total perimeter of reinforcement of an R.C.C. beam, the bond stress at the section having Q shear force, is Q/jdƩO Q/3jdƩO 2 × Q/jdƩO Q/2jdƩO Q/jdƩO Q/3jdƩO 2 × Q/jdƩO Q/2jdƩO ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a doubly-reinforced beam if ‘c’ and ‘t’ are stresses in concrete and tension reinforcement, ‘d’ is the effective depth and ‘n’ is depth of critical neutral axis, the following relationship holds good mc/t = (d - n)/t (t + c)/n = (d + n)/n (m + c)/t = n/(d + n) mc/t = n/(d - n) mc/t = (d - n)/t (t + c)/n = (d + n)/n (m + c)/t = n/(d + n) mc/t = n/(d - n) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design For stairs spanning l meters longitudinally between supports at the bottom and top of a flight carrying a load w per unit horizontal area, the maximum bending moment per metre width, is ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a beam the local bond stress Sb, is equal to Total perimeter of reinforcement/(Leaver arm × Shear force) Leaver arm/(Shear force × Total perimeter of reinforcement) Shear force/(Leaver arm × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Leaver arm/(Shear force × Total perimeter of reinforcement) Shear force/(Leaver arm × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a pre-stressed beam carrying an external load W with a bent tendon is having angle of inclination ? and pre-stressed load P. The net downward load at the centre is W - 2P sin θ W - 2P cos θ W - P cos θ W - P sin θ W - 2P sin θ W - 2P cos θ W - P cos θ W - P sin θ 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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