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The ratio of the deflections of the free end of a cantilever due to an isolated load at $${\frac{1}{3}^{{\text{rd}}}}$$ and $${\frac{2}{3}^{{\text{rd}}}}$$ of the span, is

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

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Comments ( 3 )

  1. Munna Bhaiya
    Munna Bhaiya :
    3 years ago

    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

  2. Sanjay Deka
    Sanjay Deka :
    3 years ago

    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

  3. Karma Yangzom
    Karma Yangzom :
    4 years ago

    Please upload the solution of this.

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

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

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