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The moment of inertia of a triangular section (height h, base b) about its base, is

A. $$\frac{{{\text{b}}{{\text{h}}^2}}}{{12}}$$

B. $$\frac{{{{\text{b}}^2}{\text{h}}}}{{12}}$$

C. $$\frac{{{\text{b}}{{\text{h}}^3}}}{{12}}$$

D. $$\frac{{{{\text{b}}^3}{\text{h}}}}{{12}}$$

Answer: Option C


This Question Belongs to Civil Engineering >> Theory Of Structures

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Comments ( 4 )

  1. MUhammad Iftikhar
    MUhammad Iftikhar :
    6 months ago

    I for triangle =bh^3/36
    At base
    I+Ay^2=bh^3/36+1/2*b*h(h/3)^2
    =bh^3/12

  2. Manoj Kumar
    Manoj Kumar :
    3 years ago

    About its centroid bh^3/36
    And
    Base bh^3/12

  3. Saurav S
    Saurav S :
    4 years ago

    Wrong

  4. Muhammad Saqib
    Muhammad Saqib :
    4 years ago

    BH^3/36 for triangular section

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