A. $$\frac{1}{7}$$
B. $$\frac{2}{7}$$
C. $$\frac{3}{7}$$
D. $$\frac{2}{5}$$
Answer: Option B
This Question Belongs to Civil Engineering >> Theory Of Structures
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
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. $$\frac{{\text{M}}}{{\text{T}}}$$
B. $$\frac{{\text{T}}}{{\text{M}}}$$
C. $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D. $$\frac{{2{\text{T}}}}{{\text{M}}}$$
For deflection at any point a from the fixed end is given as: δ= (Pa^2(3l-a))/(6EI). Now you can put the value a*(1/3) and a*(2/3) in the formula and work out for the ratio.
deflection at free end in a cantilever due to concentreted load=WL^3/3EI
then deflection at free end due to pt. load at any pt X (measured at a point X from the fixed end) is given by as below=(WX^3/3EI)+(WX^2(L-X)/2EI)
Now putting the value L/3 and 2L/3 in the above equation we can get the answer
using differtial equation
EI(d2y/dx2) = M = wx , x = distance from free end of cantilliver
EIy= 1/6wx3-1/2wL2x+1/3wL3
put values of x = 2/3L and 1/3L
Please upload the solution of this.