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The deflection of a uniform circular bar of diameter d and length $$l$$, which extends by an amount under a tensile pull W, when it carries the same load at its mid-span, is

A. $$\frac{{{\text{e}}l}}{{2{\text{d}}}}$$

B. $$\frac{{{{\text{e}}^2}l}}{{3{{\text{d}}^2}}}$$

C. $$\frac{{{\text{e}}{l^2}}}{{3{{\text{d}}^2}}}$$

D. $$\frac{{{{\text{e}}^{\frac{1}{2}}}}}{{3{{\text{d}}^2}}}$$

Answer: Option C


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 5 )

  1. Mudassar Baghdad
    Mudassar Baghdad :
    2 years ago

    The deflection of a uniform circular bar of diameter d and length l, which extends by an amount e under a tensile pull W, when it carries the same load at its mid-span, is:

    This is the correct question

  2. Dhaval Shanti
    Dhaval Shanti :
    3 years ago

    it should be length cube l³ in the answer which is wrongly given hers

  3. Sabeeh Swati
    Sabeeh Swati :
    3 years ago

    For 1st case; e = WL/AE.
    For the 2nd case; deflection= WL³/48EI.
    I = π d^4/64 = A*d²/16 where A = πd²/4.
    Def=WL³/3EAd², replacing by "e"
    Now Def=eL^2/3d^2.

  4. Ashna Aliyar
    Ashna Aliyar :
    4 years ago

    Qstn is not clear

  5. Ashna Aliyar
    Ashna Aliyar :
    4 years ago

    Pls xplain the answr

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