For calculating the allowable stress of long columns $${\sigma _0} = \frac{{{\sigma _y}}}{n}\left[ {1 - a{{\left( {\frac{1}{r}} \right)}^2}} \right]$$    is the empirical formula, known as

A simply supported beam A carries a point load at its mid span. Another identical beam B carries the same load but uniformly distributed over the entire span. The ratio of the maximum deflections of the beams A and B, will be

A. $$\frac{2}{3}$$

B. $$\frac{3}{2}$$

C. $$\frac{5}{8}$$

D. $$\frac{8}{5}$$

Principal planes are subjected to

A. Normal stresses only

B. Tangential stresses only

C. Normal stresses as well as tangential stresses

D. None of these

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

A. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$

B. $$\frac{{\text{I}}}{{\text{M}}} = \frac{{\text{R}}}{{\text{E}}} = \frac{{\text{F}}}{{\text{Y}}}$$

C. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{F}}}{{\text{Y}}}$$

D. $$\frac{{\text{M}}}{{\text{I}}} = \frac{{\text{E}}}{{\text{R}}} = \frac{{\text{Y}}}{{\text{F}}}$$

A shaft is subjected to bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in the shaft, is

A. $$\frac{{\text{M}}}{{\text{T}}}$$

B. $$\frac{{\text{T}}}{{\text{M}}}$$

C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$

D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$