RCC Structures Design The thickness of base slab of a retaining wall generally provided, is One-third of the width of the stem at the bottom One fourth of the width of the steam at the bottom One half of the width of the stem at the bottom Width of the stem at the bottom One-third of the width of the stem at the bottom One fourth of the width of the steam at the bottom One half of the width of the stem at the bottom Width of the stem at the bottom ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design If Ac, Asc and A are areas of concrete, longitudinal steel and section of a R.C.C. column and m and σc are the modular ratio and maximum stress in the configuration of concrete, the strength of column is σcAc + m σcAsc σc[A + (m - 1)Asc] All listed here σc(A - Asc) + m σcAsc σcAc + m σcAsc σc[A + (m - 1)Asc] All listed here σc(A - Asc) + m σcAsc ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design The effective width of a column strip of a flat slab, is Half the width of the panel Radius of the column Diameter of the column One-fourth the width of the panel Half the width of the panel Radius of the column Diameter of the column One-fourth the width of the panel ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design High strength concrete is used in pre-stressed member To overcome bursting stresses at the ends To overcome high bearing stresses developed at the ends All listed here To provide high bond stresses To overcome bursting stresses at the ends To overcome high bearing stresses developed at the ends All listed here To provide high bond stresses ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design Design of a two way slab simply supported on edges and having no provision to prevent the corners from lifting, is made by Rankine Grashoff formula Marcus formula Rankine formula Grashoff formula Rankine Grashoff formula Marcus formula Rankine formula Grashoff formula ANSWER DOWNLOAD EXAMIANS APP
RCC Structures Design A reinforced concrete cantilever beam is 3.6 m long, 25 cm wide and has its lever arm 40 cm. It carries a load of 1200 kg at its free end and vertical stirrups can carry 1800 kg. Assuming concrete to carry one-third of the diagonal tension and ignoring the weight of the beam, the number of shear stirrups required, is 30 35 45 40 30 35 45 40 ANSWER DOWNLOAD EXAMIANS APP
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