Theory of Structures Principal planes are subjected to None of these Normal stresses only Tangential stresses only Normal stresses as well as tangential stresses None of these Normal stresses only Tangential stresses only Normal stresses as well as tangential stresses 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 + mbd³)/4D] BD³ + mbd³)/6D] BD² + mbd²)/6D] BD² + mbd²)/4D] BD² + mbd³)/4D] BD³ + mbd³)/6D] BD² + mbd²)/6D] 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 3p/a² 2p/a² p/a² 4p/a² 3p/a² 2p/a² p/a² 4p/a² ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures If Q is load factor, S is shape factor and F is factor of safety in elastic design, the following: Q = F – S Q = S – F Q = S × F Q = S + F Q = F – S Q = S – F Q = S × F Q = S + F ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures At any point of a beam, the section modulus may be obtained by dividing the moment of inertia of the section by Depth of the section Depth of the neutral axis Maximum tensile stress at the section Maximum compressive stress at the section Depth of the section Depth of the neutral axis Maximum tensile stress at the section Maximum compressive stress at the section 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 = E/R = F/Y M/I = E/R = Y/F I/M = R/E = F/Y M/I = R/E = F/Y M/I = E/R = F/Y M/I = E/R = Y/F I/M = R/E = F/Y M/I = R/E = F/Y ANSWER DOWNLOAD EXAMIANS APP
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