Theory of Structures The equivalent length of a column of length L having one end fixed and the other end free, is 2L L L L/2 2L L L L/2 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 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 Depth of the section Depth of the neutral axis Maximum tensile stress at the section ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures For determining the support reactions at A and B of a three hinged arch, points B and Care joined and produced to intersect the load line at D and a line parallel to the load line through A at D’. Distances AD, DD’ and AD’ when measured were 4 cm, 3 cm and 5 cm respectively. The angle between the reactions at A and B is 45° 90° 30° 60° 45° 90° 30° 60° ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures In case of a simply supported I-section beam of span L and loaded with a central load W, the length of elasto-plastic zone of the plastic hinge, is L/4 L/2 L/3 L/5 L/4 L/2 L/3 L/5 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 = R/E = F/Y M/I = E/R = Y/F M/I = E/R = F/Y I/M = R/E = F/Y M/I = R/E = F/Y M/I = E/R = Y/F M/I = E/R = F/Y ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures The yield moment of a cross section is defined as the moment that will just produce the yield stress in The neutral fibre of the section The inner most fibre of the section The outer most fibre of the section The fibre everywhere The neutral fibre of the section The inner most fibre of the section The outer most fibre of the section The fibre everywhere ANSWER DOWNLOAD EXAMIANS APP
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