RCC Structures Design The diameter of main bars in R.C.C. columns, shall not be less than 15 mm 10 mm 12 mm 11 mm 15 mm 10 mm 12 mm 11 mm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design An R.C.C. beam of 25 cm width and 50 cm effective depth has a clear span of 6 meters and carries a U.D.L. of 3000 kg/m inclusive of its self weight. If the lever arm constant for the section is 0.865, the maximum intensity of shear stress, is 21.5 kg/cm² 7.6 kg/cm² 21.5 kg/cm² 8.3 kg/cm² 21.5 kg/cm² 7.6 kg/cm² 21.5 kg/cm² 8.3 kg/cm² ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design In a pre-stressed member it is advisable to use Low strength concrete but high tensile steel Low strength concrete only High strength concrete and high tensile steel High strength concrete only Low strength concrete but high tensile steel Low strength concrete only High strength concrete and high tensile steel High strength concrete only ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If L is the effective span of a R.C.C. beam which is subjected to maximum shear qmax at the ends, the distance from either end over which stirrups for the shear, are provided, is (L/3) (1 - 5/qmax) (L/2) (1 - 5/qmax) (L/2) (1 - 2/qmax) (L/2) (1 - 3/qmax) (L/3) (1 - 5/qmax) (L/2) (1 - 5/qmax) (L/2) (1 - 2/qmax) (L/2) (1 - 3/qmax) ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The maximum shear stress (q) in concrete of a reinforced cement concrete beam is Shear force/(Lever arm × Width) (Shear force × Width)/Lever arm Width/(Lever arm × Shear force) Lever arm/(Shear force × Width) Shear force/(Lever arm × Width) (Shear force × Width)/Lever arm Width/(Lever arm × Shear force) Lever arm/(Shear force × Width) 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/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) y = L - (l - x̅) y = L/2 - (l + x̅) ANSWER DOWNLOAD EXAMIANS APP
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