Concrete Technology and Design of Concrete Structures The compressive strength of 100 mm cube as compared to 150 mm cube is always None of these equal more less None of these equal more less ANSWER DOWNLOAD EXAMIANS APP
Concrete Technology and Design of Concrete Structures In a ring beam subjected to uniformly distributed loadi) shear force at mid span is zeroii) shear force at mid span is maximumiii) torsion at mid span is zeroiv) torsion at mid span is maximum The correct answer is (i) and (iv) (ii) and (iv) (ii) and (iii) (i) and (iii) (i) and (iv) (ii) and (iv) (ii) and (iii) (i) and (iii) ANSWER DOWNLOAD EXAMIANS APP
Concrete Technology and Design of Concrete Structures Minimum pitch of transverse reinforcement in a column is sixteen times the smallest diameter of longitudinal reinforcement bar to be tied forty-eight times the diameter of transverse reinforcement lesser of the above three values the least lateral dimension of the member sixteen times the smallest diameter of longitudinal reinforcement bar to be tied forty-eight times the diameter of transverse reinforcement lesser of the above three values the least lateral dimension of the member ANSWER DOWNLOAD EXAMIANS APP
Concrete Technology and Design of Concrete Structures Shrinkage of concrete depends uponi) humidity of atmosphereii) passage of timeiii) stress The correct answer is All (i), (ii) and (iii) (i) and (ii) (ii) and (iii) only (iii) All (i), (ii) and (iii) (i) and (ii) (ii) and (iii) only (iii) ANSWER DOWNLOAD EXAMIANS APP
Concrete Technology and Design of Concrete Structures Sinking of an intermediate support of a continuous beam i) reduces the negative moment at support ii) increases the negative moment at support iii) reduces the positive moment at center of span iv) increases the positive moment at center of span The correct answer is (ii) and (iv) (i) and (iv) (i) and (iii) (ii) and (iii) (ii) and (iv) (i) and (iv) (i) and (iii) (ii) and (iii) ANSWER DOWNLOAD EXAMIANS APP
Concrete Technology and Design of Concrete Structures The relation between modulus of rupture fcr, splitting strength fcs and direct tensile strength fcl is given by fc5>fcr>fC fcr fcr>fcs>fc. tcr – rcs = tct fc5>fcr>fC fcr fcr>fcs>fc. tcr – rcs = tct ANSWER DOWNLOAD EXAMIANS APP
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