Theory of Structures Gradually applied static loads do not change with time their Magnitude Point of application All of these Direction Magnitude Point of application All of these Direction ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures The point of contraflexure is the point where M. is minimum S.F. is zero M. is maximum M. changes sign M. is minimum S.F. is zero M. is maximum M. changes sign 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 tensile stress at the section Depth of the neutral axis Maximum compressive stress at the section Depth of the section Maximum tensile stress at the section Depth of the neutral axis Maximum compressive stress at the section Depth of the section ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures constant, depth of a cantilever of length of uniform strength loaded with Keeping breadth uniformly distributed load varies from zero at the free end and w l) at the fixed end l) at the fixed end 3w l at the fixed end 2w w l at the fixed end w l) at the fixed end l) at the fixed end 3w l at the fixed end 2w w l at the fixed end ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures A shaft subjected to a bending moment M and a torque T, experiences Maximum bending stress = 32M/πd³ Neither A nor B Maximum shear stress = 16 T/πd³ Both A and B Maximum bending stress = 32M/πd³ Neither A nor B Maximum shear stress = 16 T/πd³ Both A and B ANSWER DOWNLOAD EXAMIANS APP
Theory of Structures parabolic arch of span and rise , is given by The equation of a y = 2h/l² × (1 – x) y = h/l² × (1 – x ) y = 3h/l² × (1 – x) y = 4h/l² × (1 – x) y = 2h/l² × (1 – x) y = h/l² × (1 – x ) y = 3h/l² × (1 – x) y = 4h/l² × (1 – x) ANSWER DOWNLOAD EXAMIANS APP
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