RCC Structures Design If the sides of a slab simply supported on edges and spanning in two directions are equal, the maximum bending moment is multiplied by 0.5 0.6 0.7 0.4 0.5 0.6 0.7 0.4 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The diameter of longitudinal bars of a column should never be less than 10 mm 6 mm 8 mm 12 mm 10 mm 6 mm 8 mm 12 mm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Though the effective depth of a T-beam is the distance between the top compression edge to the centre of the tensile reinforcement, for heavy loads, it is taken as 1/10th of the span 1/8th of the span 1/12th of the span 1/16th of the span 1/10th of the span 1/8th of the span 1/12th of the span 1/16th of the span ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If l₁ and l₂ are the lengths of long and short spans of a two way slab simply supported on four edges and carrying a load w per unit area, the ratio of the loads split into w₁ and w₂ acting on strips parallel to l₂ and l₁ is w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)³ w₁/w₂ = (l₂/l₁)² w₁/w₂ = (l₂/l₁)⁴ w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)³ w₁/w₂ = (l₂/l₁)² w₁/w₂ = (l₂/l₁)⁴ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a beam the local bond stress Sb, is equal to 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) Total perimeter of reinforcement/(Leaver arm × Shear force) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Distribution reinforcement in a simply supported slab, is provided to distribute Shrinkage stress All listed here Load Temperature stress Shrinkage stress All listed here Load Temperature stress ANSWER DOWNLOAD EXAMIANS APP
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