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)² w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)⁴ w₁/w₂ = (l₂/l₁)³ w₁/w₂ = (l₂/l₁)² w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)⁴ w₁/w₂ = (l₂/l₁)³ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The width of the flange of a T-beam should be less than Least of the above One-third of the effective span of the T-beam Breadth of the rib plus twelve times the thickness of the slab Distance between the centers of T-beam Least of the above One-third of the effective span of the T-beam Breadth of the rib plus twelve times the thickness of the slab Distance between the centers of T-beam ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the ratio of long and short spans of a two way slab with corners held down is r, the actual reduction of B.M. is given by (5/6) (r/1 + r²) M (5/6) (r²/1 + r³) M (5/6) (r²/1 + r⁴) M (5/6) (r²/1 + r²) M (5/6) (r/1 + r²) M (5/6) (r²/1 + r³) M (5/6) (r²/1 + r⁴) M (5/6) (r²/1 + r²) M ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If q is the punching shear resistance per unit area a, is the side of a square footing for a column of side b, carrying a weight W including the weight of the footing, the depth (D) of the footing from punching shear consideration, is D = W (a - b)/4a²bq D = W (a² - b²)/4abq D = W (a² - b²)/4a²bq D = W (a² - b²)/8a²bq D = W (a - b)/4a²bq D = W (a² - b²)/4abq D = W (a² - b²)/4a²bq D = W (a² - b²)/8a²bq ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The shear reinforcement in R.C.C. is provided to resist Vertical shear Horizontal shear Diagonal compression Diagonal tension Vertical shear Horizontal shear Diagonal compression Diagonal tension ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A pre-stressed concrete member is preferred because All listed here Large size of long beams carrying large shear force need not be adopted Removal of cracks in the members due to shrinkage Its dimensions are not decided from the diagonal tensile stress All listed here Large size of long beams carrying large shear force need not be adopted Removal of cracks in the members due to shrinkage Its dimensions are not decided from the diagonal tensile stress ANSWER DOWNLOAD EXAMIANS APP
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