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/mm^{2} generates a tensile circumferential stress of 7 N/mm^{2} 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/mm^{2}
σ′_{w} = 19.11 N/mm^{2}
∴ Final tensile stress in the pipe, 15.44 - σ_{p} = 7.
Using the relation between initial stresses, σ_{p} = 8.44 N/mm^{2}.
∴ Tension in the wire = 20.15 N/mm^{2}.