Approximate the tension:

A copper tube 38 mm external diameter and 35 mm internal diameter is nearly wound with a steel wire of 0.8 mm diameter. Approximate the tension at which the wire should have been wound if an internal pressure of 2 N/mm² generates a tensile circumferential stress of 7 N/mm² in the tube. Young's Modulus of steel is equal to 1.6 times that of copper. Take Poisson's ratio of copper is equal to  0.3.

Solution

Assume σ_p and σ_w be the stresses in the pipe and the wire, respectively, before applying the internal pressure.

Assume for 1 mm length of pipe, we know, Compressive force in pipe = Tensile force in wire.

σ _p  × 2 × 1.5 = σ _w  × (1/0.8)  × 2 × (π /4)× 0.82

σ _p  = 0.419 σw

Because of internal pressure alone, let the stresses be σ′_p and σ′_w . Then, we know,

Tensile force in pipe + Tensile force in wire = Bursting force

σ′ p × 2 × 1.5 + σ′ w × (1/0.8)  × 2 × (π/4) × 0.82  = 2 × 35

 3σ′_p  + 1.257 σ′_w = 70

Longitudinal stress

pd /4t  = 2 × 35 / (4 × 1.5) = 11.67 N/mm2

Hoop strain in the pipe = Strain in wire at the junction

σ′_p/ E_c - v (σ_l / E_c) = σ′_w /E_s

σ′_p / E_c - 0.3 × (11.67 / E_c) = σ′_w / (1.6 E_s)

σ′_p  - 3.5 = 0.625 σ′_w

σ′_p  - 0.625 σ′_w = 3.5

Solving for σ′_p and σ′_w ,

σ′_p  = 15.44 N/mm²

σ′_w  = 19.11 N/mm²

∴  Final tensile stress in the pipe, 15.44 - σ_p = 7.

Using the relation between initial stresses, σ_p = 8.44 N/mm².

∴  Tension in the wire = 20.15 N/mm².