A thin-walled cylindrical pressure vessel has mean wall thickness of 𝑡 and nominal radius of 𝑟. The Poisson’s ratio of the wall material is 1/3. When it was subjected to some internal pressure, its nominal perimeter in the cylindrical portion increased by 0.1% and the corresponding wall thickness became 𝑡 . The corresponding change in the wall thickness of the cylindrical portion, i.e. 100 × (𝑡 − 𝑡)/𝑡, is ______%. (round off to 3 decimal places).

The correct answer is -0.06.

1. Understanding the Parameters and Strains in a Thin-Cylindrical Pressure Vessel
Let:
- t be the initial wall thickness, and t̄ be the final wall thickness.
- r be the nominal radius, and d = 2r be the nominal diameter of the cylindrical portion.
- μ be the Poisson's ratio of the wall material, given as:
$$\mu =\frac{1}{3}$$
- E be the Young's modulus of the material.
- p be the internal pressure acting on the vessel.

The circumferential (hoop) perimeter of the cylindrical portion is given by $P=2\pi r$. Therefore, the circumferential (hoop) strain ${ϵ}_h$ is equal to the fractional change in the perimeter:
$${ϵ}_h=\frac{\delta P}{P}=0.1\%=0.001$$

2. Stresses in the Pressure Vessel
For a thin-walled cylindrical pressure vessel, the three principal stresses are:
- Hoop stress (circumferential stress), ${\sigma }_1$:
$${\sigma }_1=\frac{pd}{2t}$$
- Longitudinal stress (axial stress), ${\sigma }_2$:
$${\sigma }_2=\frac{pd}{4t}=\frac{{\sigma }_1}{2}$$
- Radial stress, ${\sigma }_3$:
Since the vessel is thin-walled, the radial stress across the thickness is negligible compared to the hoop and longitudinal stresses, so we assume:
$${\sigma }_3\approx 0$$

3. Formulation of Hoop Strain
Using Hooke's law in three dimensions, the hoop strain is:
$${ϵ}_h=\frac{1}{E}[{\sigma }_1-\mu ({\sigma }_2+{\sigma }_3)]$$
Substituting ${\sigma }_2=\frac{{\sigma }_1}{2}$ and ${\sigma }_3=0$:
$${ϵ}_h=\frac{1}{E}[{\sigma }_1-\mu \left(\frac{{\sigma }_1}{2}\right)]=\frac{{\sigma }_1}{E}(1-\frac{\mu }{2})$$
Substituting ${\sigma }_1=\frac{pd}{2t}$:
$${ϵ}_h=\frac{pd}{4tE}(2-\mu )$$
Using this relationship, we can express the term $\frac{pd}{4tE}$ as:
$$\frac{pd}{4tE}=\frac{{ϵ}_h}{2-\mu }=\frac{0.001}{2-\frac{1}{3}}$$

4. Calculation of Radial Strain (Wall Thickness Strain)
The strain in the thickness direction (radial strain) is given by:
$$\frac{\delta t}{t}=\frac{\overline{t}-t}{t}=\frac{1}{E}[{\sigma }_3-\mu ({\sigma }_1+{\sigma }_2)]$$
Since ${\sigma }_3=0$ and ${\sigma }_2=\frac{{\sigma }_1}{2}$:
$$\frac{\delta t}{t}=-\frac{\mu }{E}({\sigma }_1+\frac{{\sigma }_1}{2})=-\frac{\mu }{E}\left(\frac{3{\sigma }_1}{2}\right)$$
Substitute ${\sigma }_1=\frac{pd}{2t}$:
$$\frac{\delta t}{t}=-\frac{3\mu }{2E}\left(\frac{pd}{2t}\right)=-3\mu \frac{pd}{4tE}$$

Now, substitute the value of $\frac{pd}{4tE}$ derived from the hoop strain:
$$\frac{\delta t}{t}=-3\times \frac{1}{3}\times \frac{0.001}{2-\frac{1}{3}}$$
Simplifying the expression:
$$\frac{\delta t}{t}=-1\times \frac{0.001}{5/3}=-0.001\times \frac{3}{5}=-0.0006$$

5. Computing the Percentage Change in Wall Thickness
The percentage change is given by:
$$100\times \frac{\overline{t}-t}{t}=100\times (-0.0006)=-0.06\%$$

Thus, the percentage change in the wall thickness of the cylindrical portion is -0.06%.