Conical Bar
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| 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.
Books and e-Books I recommend for Strength of Materials:
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