Theory of Structures The point of contraflexure is the point where S.F. is zero M. is maximum M. changes sign M. is minimum S.F. is zero M. is maximum M. changes sign M. is minimum ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures Y are the bending moment, moment of inertia, radius of curvature, modulus of If M, I, R, E, F, and elasticity stress and the depth of the neutral axis at section, then M/I = R/E = F/Y M/I = E/R = Y/F I/M = R/E = F/Y M/I = E/R = F/Y M/I = R/E = F/Y M/I = E/R = Y/F I/M = R/E = F/Y M/I = E/R = F/Y ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures Principal planes are subjected to Normal stresses as well as tangential stresses Tangential stresses only Normal stresses only None of these Normal stresses as well as tangential stresses Tangential stresses only Normal stresses only None of these ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures The maximum bending moment for a simply supported beam with a uniformly distributed load w/unit length, is WI²/4 WI/2 WI²/8 WI²/12 WI²/4 WI/2 WI²/8 WI²/12 ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures For a strongest rectangular beam cut from a circular log, the ratio of the width and depth, is 0.404 0.303 0.707 0.505 0.404 0.303 0.707 0.505 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 All of these E = 3K (1 – 2/m) E = 2N (1 + 1/m) (3/2)K (1 – 2/m) = N (1 + 1/m) All of these E = 3K (1 – 2/m) E = 2N (1 + 1/m) (3/2)K (1 – 2/m) = N (1 + 1/m) ANSWER DOWNLOAD EXAMIANS APP
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