Engineering, 27.12.2020 02:00 rileychas4114

A compound cylinder is formed by shrinking a tube of 250 mm internal diameter and 25 mm wall thickness onto another tube of 250 mm external diameter and 25 mm wall thickness, both tubes being made of the same material. The stress set up at the junction owing to shrinkage is

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Answer from: lydiadmanautou04

The question is incomplete. The complete question is :

A compound cylinder is formed by shrinking a tube of 250 mm internal diameter and 25 mm wall thickness onto another tube of 250 mm external diameter and 25 mm wall thickness, both tubes being made of the same material. The stress set up at the junction owing to shrinkage is 10 MN/m2. The compound tube is then subjected to an internal pressure of 80 MN/m2. Compare the hoop stress distribution now obtained with that of a single cylinder of 300 mm external diameter and 50 mm thickness subjected to the same internal pressure.

Solution :

Internal pressure = $80 \ MN/m^2$

Stress set up at the junction owing to shrinkage = $10 \ MN/m^2$

Therefore shrinkage at the outer tube at r = 0.15, $\sigma_{r} = 0$ and r = 0.125, $\sigma_{r} = -10 \ MN/m^2$

∴ $0=A-\frac{B}{(0.15)^2}=A-44.5B$ .....................(i)

$-10=A-\frac{B}{(0.125)^2}=A-64B$ .................(ii)

By solving the above equations we get A = 22.85 and B = 0.514

Now, hoop stress at radius = 0.15 m :

A + 44.5 B = 22.85+44.5 (0.514)

= 45.7 MPa

Hoop stress at radius 0.125 m :

A + 64 B = 22.85 + 64 (0.514)

= 55.74 MPa

Now shrinkage in the inner tubes

At r = 0.10, $\sigma_{r} = 0$ and r = 0.125, $\sigma_{r} = -10 \ MN/m^2$

$0=A-\frac{B}{(0.1)^2}=A-100B$ ....................(iii)

$-10=A-\frac{B}{(0.125)^2}=A-64B$ ...............(iv)

By solving the above equations,

A = -27.8 and B = -0.278

Now hoop stress at 0.125 m radius :

A + 64 B = -45.6 MPa

Hoop stress at 0.10 m radius:

A + 100 B = -55.6 MPa

Considering the internal pressure only on complete cylinder at r = 0.15, $\sigma_{r} = 0$ and r = 0.10 , $\sigma_{r} = -80$

$0=A-\frac{B}{(0.15)^2}=A-44.5B$ .............(v)

$-80=A-\frac{B}{(0.1)^2}=A-100B$ ...........(vi)

∴ A = 64.2 and B = 1.44

At r = 0.15 m,

$\sigma_{H} = A+44.5B$

$=128.4 \ MN/m^2$

At r = 0.125 m,

$\sigma_{H} = A+64B$

$=156.4 \ MN/m^2$

At r = 0.125 m,

$\sigma_{H} = A+100B$

$=208.2 \ MN/m^2$

The stress for the combined shrinkage and the internal pressure are :

Outer tube

r = 015

$\sigma_{H} = 128.4 + 45.1$

$=174.1 \ MN/m^2$

r = 0.125

$\sigma_{H} = 156.4 + 55.75$

$=212.15 \ MN/m^2$

For inner tubes,

r = 0.125

$\sigma_{H} = 156.4 - 45.6 = 110.8 \ MN/m^2$

r = 0.1

$\sigma_{H} = 208.2 - 55.6 = 152.6 \ MN/m^2$

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