RCC Structures Design A foundation rests on Foundation soil Sub-grade Both B and C Base of the foundation Foundation soil Sub-grade Both B and C Base of the foundation ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A part of the slab may be considered as the flange of the T-beam if It is built integrally with the beam It is effectively bonded together with the beam Flange has adequate reinforcement transverse to beam All of the listed here It is built integrally with the beam It is effectively bonded together with the beam Flange has adequate reinforcement transverse to beam All of the listed here 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 6 cm 8 cm 10 cm 12 cm 6 cm 8 cm 10 cm ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Spacing of stirrups in a rectangular beam, is Kept constant throughout the length Decreased towards the centre of the beam Increased at the centre of the beam Increased at the ends Kept constant throughout the length Decreased towards the centre of the beam Increased at the centre of the beam Increased at the ends 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) Total perimeter of reinforcement/(Leaver arm × Shear force) Shear force/(Leaver arm × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) Leaver arm/(Shear force × Total perimeter of reinforcement) Total perimeter of reinforcement/(Leaver arm × Shear force) Shear force/(Leaver arm × Total perimeter of reinforcement) Leaver arm/(Bending moment × Total perimeter of reinforcement) 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 - (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) y = L/2 + (l - x̅) ANSWER DOWNLOAD EXAMIANS APP
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