Theory of Structures A simply supported beam which carries a uniformly distributed load has two equal overhangs. To have maximum B.M. produced in the beam least possible, the ratio of the length of the overhang to the total length of the beam, is 0.508 0.307 0.407 0.207 0.508 0.307 0.407 0.207 ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A material is said to be perfectly elastic if None of these It regains its original shape partially on removal of the load It regains its original shape on removal of the load It does not regain its original shape at all None of these It regains its original shape partially on removal of the load It regains its original shape on removal of the load It does not regain its original shape at all ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A square column carries a load P at the centroid of one of the quarters of the square. If a is the side of the main square, the combined bending stress will be p/a² 2p/a² 4p/a² 3p/a² p/a² 2p/a² 4p/a² 3p/a² ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A steel plate d × b is sandwiched rigidly between two timber joists each D × B/2 in section. The steel will be (where Young’s modulus of steel is m times that of the timber). BD² + mbd²)/4D] BD² + mbd²)/6D] BD³ + mbd³)/6D] BD² + mbd³)/4D] BD² + mbd²)/4D] BD² + mbd²)/6D] BD³ + mbd³)/6D] BD² + mbd³)/4D] ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures The general expression for the B.M. of a beam of length l is the beam carries M = (wl/2) x – (wx²/2) A load varying linearly from zero at one end to w at the other end A uniformly distributed load w/unit length An isolated load at mid span None of these A load varying linearly from zero at one end to w at the other end A uniformly distributed load w/unit length An isolated load at mid span None of these ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures If E, N, K and 1/m are modulus of elasticity, modulus of rigidity. Bulk modulus and Poisson ratio of the material, the following relationship holds good E = 3K (1 – 2/m) All of these (3/2)K (1 – 2/m) = N (1 + 1/m) E = 2N (1 + 1/m) E = 3K (1 – 2/m) All of these (3/2)K (1 – 2/m) = N (1 + 1/m) E = 2N (1 + 1/m) ANSWER DOWNLOAD EXAMIANS APP
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