Analysis of Uniformly Tapering Circular Rod

Analysis of Uniformly Tapering Circular Rod

A circular bar whose diameter gradually reduces from D₁ at one end to D₂ at the other end is shown in Fig. 1.13. Such a bar is said to be uniformly tapering.

Let:
P = Axial tensile load acting on the bar
L = Total length of the bar
E = Young’s Modulus of the material

Diameter at any Section

Consider a very small element of the rod at a distance x from the left end. The diameter at this section may be written as:

Dₓ = D₁ − kx

where

k = (D₁ − D₂) / L

Area of Cross-Section at Distance x

Aₓ = (π/4) × (D₁ − kx)²

Stress at Distance x

The stress in the infinitesimal section is:

σₓ = P / Aₓ  
σₓ = 4P / [π (D₁ − kx)²]

Strain at Distance x

eₓ = σₓ / E  
eₓ = 4P / [πE (D₁ − kx)²]

Extension of a Small Element dx

dL = eₓ × dx  
dL = [4P dx] / [πE (D₁ − kx)²]

Total Extension of the Rod

To get the overall elongation, integrate from x = 0 to x = L:

L_total = ∫ (4P / πE (D₁ − kx)² ) dx  

Performing the integration:

L_total = (4P / πEk) 
          × [ 1 / (D₁ − kx) ] from 0 to L

Substituting limits:

L_total = (4P / πEk) × [ (1/D₂) − (1/D₁) ]

Final Expression

Total extension,  ΔL = (4P / πE (D₁ − D₂)) × ( 1/D₂ − 1/D₁ )

Special Case – If Rod Has Uniform Diameter

If D₁ = D₂ = D, the extension reduces to:

ΔL = 4PL / (π E D²)

Problem 1.15 – Extension of a Uniformly Tapering Rod

Question

A circular rod tapers uniformly from a diameter of 40 mm at one end to 20 mm at the other end over a length of 400 mm. The rod is subjected to an axial tensile load of 5000 N. If E = 2.1 × 10⁵ N/mm², determine the extension of the rod.

Given

Larger diameter, D₁ = 40 mm
Smaller diameter, D₂ = 20 mm
Length of rod, L = 400 mm
Axial load, P = 5000 N
Young’s modulus, E = 2.1 × 10⁵ N/mm²

Formula for a Uniformly Tapering Circular Bar

For a bar tapering linearly from D₁ to D₂, the total extension is:

ΔL = (4 P L) / (π E D₁ D₂)

Calculation

ΔL = (4 × 5000 × 400) / (π × 2.1 × 10⁵ × 40 × 20)

ΔL ≈ 0.01515 mm

Final Answer

Extension of the tapering rod ≈ 0.0152 mm.

Problem 1.16 – Finding Modulus of Elasticity of a Tapering Rod

Question

A circular rod tapers uniformly from 30 mm diameter at one end to 15 mm diameter at the other end over a length of 350 mm. The rod carries an axial load of 5.5 kN, and the measured extension is 0.025 mm. Determine the modulus of elasticity E of the material.

Given

Larger diameter, D₁ = 30 mm
Smaller diameter, D₂ = 15 mm
Length of rod, L = 350 mm
Axial load, P = 5.5 kN = 5500 N
Measured extension, ΔL = 0.025 mm

Using the Tapered Bar Formula

Again,

ΔL = (4 P L) / (π E D₁ D₂)

Rearrange to solve for E:

E = (4 P L) / (π D₁ D₂ ΔL)

Calculation

E = (4 × 5500 × 350) / (π × 30 × 15 × 0.025)

E ≈ 217865 N/mm² = 2.17865 × 10⁵ N/mm²

Final Answer

Modulus of elasticity, E ≈ 2.18 × 10⁵ N/mm².