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The ratio of the maximum deflections of a simply supported beam with a central load W and of a cantilever of same length and with a load W at its free end, is

A. $$\frac{1}{8}$$

B. $$\frac{1}{{10}}$$

C. $$\frac{1}{{12}}$$

D. $$\frac{1}{{16}}$$

Answer: Option D


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 4 )

  1. Girishkumar Gange
    Girishkumar Gange :
    2 years ago

    Maximum deflection at centre of span for simply supported beam = wl^3/48EI
    And maximum deflection at the free end for cantilever beam of same span= wl^3/3EI
    Ratio=(Wl^3/48EI)÷(wl^3/3EI)
    =1/16 is the Correct Answer

  2. Muhammad Imran
    Muhammad Imran :
    4 years ago

    For Simply Supported beam deflection= PL^3/48EI
    For Cantilever beam deflection= PL^3/3EI
    ratio=PL^3/48EI * 3EI/PL^3=3/48=1/16

  3. Abhishek Singh
    Abhishek Singh :
    5 years ago

    WL^2/48EI *3EI/WL^3 = 1/16L

  4. Prakash Neupane
    Prakash Neupane :
    5 years ago

    Maximum deflection at centre of span for simply supported beam = wl^3/48EI
    And maximum deflection at the free end for cantilever beam of same span= wl^3/3EI
    Ratio=(Wl^3/48EI)÷(wl^3/3EI)
    =1/16

Related Questions on 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

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