Theory of Structures In case of a simply supported rectangular beam of span L and loaded with a central load W, the length of elasto-plastic zone of the plastic hinge, is L/2 L/3 L/5 L/4 L/2 L/3 L/5 L/4 ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures The moment of inertia of a triangular section (height h, base b) about its base, is bh²/12 bh³/12 b³h/12 b²h/12 bh²/12 bh³/12 b³h/12 b²h/12 ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A steel bar 5 m × 50 mm is loaded with 250,000 N. If the modulus of elasticity of the material is 0.2 MN/mm² and Poisson’s ratio is 0.25, the change in the volume of the bar is: 1.125 cm³ 2.125 cm³ 3.125 cm³ 4.125 cm² 1.125 cm³ 2.125 cm³ 3.125 cm³ 4.125 cm² ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A compound bar consists of two bars of equal length. Steel bar cross -section is 3500 mm²and that of brass bar is 3000 mm². These are subjected to a compressive load 100,000 N. If Eb = 0.2 MN/mm² and Eb = 0.1 MN/mm², the stresses developed are: b = 8 N/mm² s = 16 N/mm² b = 5 N/mm² s = 10 N/mm² b = 6 N/mm² s = 12 N/mm² b = 10 N/mm² s = 20 N/mm 2 b = 8 N/mm² s = 16 N/mm² b = 5 N/mm² s = 10 N/mm² b = 6 N/mm² s = 12 N/mm² b = 10 N/mm² s = 20 N/mm 2 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 = 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) All of these E = 3K (1 – 2/m) 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 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 M/I = R/E = F/Y M/I = E/R = Y/F ANSWER DOWNLOAD EXAMIANS APP
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