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 σp and σw 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/ Ec - v (σl / Ec) = σ′w /Es
σ′p / Ec - 0.3 × (11.67 / Ec) = σ′w / (1.6 Es)
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 - σp = 7.
Using the relation between initial stresses, σp = 8.44 N/mm2.
∴ Tension in the wire = 20.15 N/mm2.