__Conical Bar __

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'**mThe diameter of the wider end =

**'D'**m.(fixed rigidly at MN)

Density of material =

**'P'**kg/m^3Let 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^2Stress (

**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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**S.S. Rattan- https://amzn.to/3cX8Mez**

**R.K. Bansal- https://amzn.to/2SRJUhi**

**S.S. Bhavikatti- https://amzn.to/35H7sZ0**

**Timoshenko- https://amzn.to/3cZg9lO**

**R.K. Rajput- https://amzn.to/3cYsPZR**

**R.S. Khurmi- https://amzn.to/2Ut6VaX**

**U.C. Jindal- https://amzn.to/3zKMXIQ**

*Self Typed

*Source- Books, Internet, Self-Analysis