RCC Structures Design A pre-stressed concrete member is preferred because Its dimensions are not decided from the diagonal tensile stress All listed here Removal of cracks in the members due to shrinkage Large size of long beams carrying large shear force need not be adopted Its dimensions are not decided from the diagonal tensile stress All listed here Removal of cracks in the members due to shrinkage Large size of long beams carrying large shear force need not be adopted ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If the maximum shear stress at the end of a simply supported R.C.C. beam of 16 m effective span is 10 kg/cm², the length of the beam having nominal reinforcement, is 12 cm 8 cm 6 cm 10 cm 12 cm 8 cm 6 cm 10 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The diameter of the column head support a flat slab, is generally kept 0.25 times the diameter of the column 4.0 cm larger than the diameter of the column 5.0 cm larger than the diameter of the column 0.25 times the span length 0.25 times the diameter of the column 4.0 cm larger than the diameter of the column 5.0 cm larger than the diameter of the column 0.25 times the span length ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Pick up the incorrect statement from the following: Tensile reinforcement bars of a rectangular beam Are bent up at suitable places to serve as shear reinforcement Are bent down at suitable places to serve as shear reinforcement Are maintained at bottom to provide at least local bond stress Are curtailed if not required to resist the bending moment Are bent up at suitable places to serve as shear reinforcement Are bent down at suitable places to serve as shear reinforcement Are maintained at bottom to provide at least local bond stress Are curtailed if not required to resist the bending moment ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If ‘p’ is the net upward pressure on a square footing of side ‘b’ for a square column of side ‘a’, the maximum bending moment is given by M = pb (b + a)/8 M = pb (b - a)²/4 M = pb (b - a)²/8 M = pb (b + a)/8 M = pb (c - a)/4 M = pb (b + a)/8 M = pb (b - a)²/4 M = pb (b - a)²/8 M = pb (b + a)/8 M = pb (c - a)/4 ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The amount of reinforcement for main bars in a slab, is based upon Minimum bending moment Maximum bending moment Minimum shear force Maximum shear force Minimum bending moment Maximum bending moment Minimum shear force Maximum shear force ANSWER DOWNLOAD EXAMIANS APP
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