Elongation of a Conical Bar due to it's Self-Weight | QnA | Strength of Material | Solid Mechanics | By Akhand Dutta

Akhand Dutta

    Conical Bar 


    Elongation of a Conical Bar due to it's Self-Weight | QnA | Strength of Material | Solid Mechanics | By Akhand Dutta
    Elongation of a Conical Bar due to it's Self-Weight 





    Derive the relation for the elongation of a conical bar due to its self-weight.

    or

    Derive the expression for the elongation of a conical bar due to its self-weight.

    or

    Find out the formula for the elongation of a conical bar due to its self-weight.


    Let MNH be a conical bar of length 'l' m

    The diameter of the wider end = 'D' m.
    (fixed rigidly at MN)

    Density of material = 'P' kg/m^3

    Let us consider a small strip of thickness 'dy' and at a distance 'y' from the lower end.

    So, Weight of UHV = P * Area of UHV * y * g

    = (P𝝅d^2yg) / (3*4) 

    = (P𝝅d^2yg) / 12      ............................................................................................................ (1)


    On using properties of similar triangles in MNH and UVH we get,

    MN / UV = l / y

    or, UV = MNy / l

    or, d = Dy / l  ................................................................................................................... (2)


    Substituting the values from (2) in (1)

    Weight of UHV = [P𝝅 (Dy / l)^2 yg] / 12

    = (P𝝅 D^2 y^3 g) / 12 l^2 ....................................................................................... (3)


    Now, Stress at section UV = Force at UV / Cross-Sectional Area of UV

    = Weight of UHV / Cross-Sectional Area of UV

    = {(P𝝅 D^2 y^3 g) / (12 l^2)} / {(𝝅 D^2 y^2) / (4 l^2)}

    = Pyg/3 ................................................................................................................ (4)


    Extension in dy = (Pyg dy) / 3E
    (E = Young's Modulus)

    For total extension integrating within limits:

    dl = ∫ l → 0 {(Pyg dy) / 3E}

    dl = Pg / 3E ∫ l → 0 (y) dy

    dl = (Pgl^2) / (3E * 2)

    dl = Pgl^2 / 6E.


    A vertical tie of uniform strength is 20 metres long. If the area of the bar at the lower end is 600 m^2, find the area at the upper end when the tie is to carry a load of 800 kN. The material of the tie weighs 80 kN/m^3. (SOM, By Er. R. K. Rajput)


    The area at the bottom of the given tie (say A1) = 600 mm^2 = 600 × 10^-6 m^2


    Stress (say S) = (800 × 1000) / (600 × 10^-6)

    = 1333.3 MN/m^2


    Density (say P) = (80 × 1000) / 9.81 

    = 8155 kg/m^3


    By using A2 = A1 e^(Pgl/S) we get,


    Upper end's area (say A2) = 600 × 10^-6 × e^{(9.81 × 8155 × 20) / (1333.3 × 10^-6 )}

    A2= 600.036 × 10^-6 m^2

    or, A2 = 600.036 mm^2.



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    R.S. Khurmi- https://amzn.to/2Ut6VaX
    U.C. Jindal- https://amzn.to/3zKMXIQ




    *Self Typed
    *Source- Books, Internet, Self-Analysis

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