Diffrence between external diameter and internal diameter:
A compound tube is built by shrinking one tube on another, the ultimate dimension being, 80 mm internal diameter, 160 mm external diameter and 120 mm being the diameter at the junction.
If the radial pressure at the junction because of shrinking is 15 N/mm^{2}, discover the greatest tensile and compressive stresses induced in the material of the cylinder. What difference should there be in the external diameter of the inner cylinder & the internal diameter of the outer cylinder before shrinking?
Take Young's Modulus as 200 GPa.
Solution
For outer cylinder, at r = 60, σ_{r} = 15, and at r = 80, σ_{r} = 0.
We get, and
15 = (b_{1} /60^{2}) - a_{1}
0 = (b_{1 }/80^{2}) - a_{1}
135
On solving, we get b1 =864000/ 7 and a_{1} = 7 .
Circumferential stress at the inner surface,
σ _{h} =( 864000/7 × 60^{2} )+ (135/7) = 53.6 N/mm^{2 } (tensile)
For inner cylinder, at r = 60, σ_{r} = 15, and r = 40, σ_{r} = 0.
We get, 15 = ( b_{ 2}/60^{2} )- a_{2}
and
0 = b_{2}/40 - a_{2}
On solving out, we get b_{2} = - 43200 and a_{2} = - 27.
Circumferential stress at the inner surface,
σ_{h} = -( 43200/40^{2}) + (- 27) = - 54 N/mm^{2 } (compressive)
Circumferential stress at the junction,
σ_{ h} = - (43200 /60^{2})+ (- 27) = - 39 N/mm^{2} (compressive)
Required difference in diameter,
= (Junction diameter / E) × (Algebraic difference between hoop stresses at the junction)
= (120/2 × 10^{5}) (53.6 + 39) = 0.0556 mm