RCC Structures Design High strength concrete is used in pre-stressed member To overcome high bearing stresses developed at the ends All listed here To overcome bursting stresses at the ends To provide high bond stresses To overcome high bearing stresses developed at the ends All listed here To overcome bursting stresses at the ends To provide high bond stresses ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design For normal cases, stiffness of a simply supported beam is satisfied if the ratio of its span to its overall depth does not exceed 25 10 20 15 25 10 20 15 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If A is the area of the foundation of a retaining wall carrying a load W and retaining earth of weight 'w' per unit volume, the minimum depth (h) of the foundation from the free surface of the earth, is h = (W/Aw) [(1 + sin φ)/(1 + sin φ)] h = (W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 - sin φ)/(1 + sin φ)] h = √(W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 + sin φ)/(1 + sin φ)] h = (W/Aw) [(1 - sin φ)/(1 + sin φ)]² h = (W/Aw) [(1 - sin φ)/(1 + sin φ)] h = √(W/Aw) [(1 - sin φ)/(1 + sin φ)]² 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 6 legged 10 legged 8 legged 12 legged 6 legged 10 legged 8 legged 12 legged 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)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 w₁/w₂ = l₂/l₁ w₁/w₂ = (l₂/l₁)³ ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The percentage of minimum reinforcement of the gross sectional area in slabs, is 0.12 % 0.15 % 0.10 % 0.18 % 0.12 % 0.15 % 0.10 % 0.18 % ANSWER DOWNLOAD EXAMIANS APP
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