Thin Walled Vessels

Data: $d = 1000$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 2\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{2(1000)}{2(12)} = 83.33\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 41.67\ \text{N/mm}^2$$

Maximum shear stress

The principal stresses are $\sigma_c$ and $\sigma_l$ (the radial stress is neglected), so

$$\tau_{max} = \frac{\sigma_c - \sigma_l}{2} = \frac{83.33 - 41.67}{2} = 20.83\ \text{N/mm}^2$$

(this is the maximum in the plane of the wall; it equals $\sigma_l/2$ and $\sigma_c/4$).

Change in dimensions

$$\varepsilon_c = \frac{\sigma_c - \nu\sigma_l}{E} = \frac{83.33 - 0.3(41.67)}{2\times10^5} = 3.541667e-04$$ $$\varepsilon_l = \frac{\sigma_l - \nu\sigma_c}{E} = \frac{41.67 - 0.3(83.33)}{2\times10^5} = 8.333333e-05$$ $$\delta d = \varepsilon_c\,d = 0.3542\ \text{mm}, \qquad \delta L = \varepsilon_l\,L = 0.3333\ \text{mm}$$

Volumetric strain $\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 7.916667e-04$, with $V = \frac{\pi}{4}d^2L = 3.1416e+09\ \text{mm}^3$:

$$\delta V = \varepsilon_v V = 2.4871e+06\ \text{mm}^3 \;(2487.1\ \text{cm}^3)$$

Answer: $\tau_{max} = 20.83$ N/mm²; $\delta d = 0.354$ mm (increase), $\delta L = 0.333$ mm (increase), $\delta V = 2487$ cm³.

To show: $\varepsilon_v = 2\varepsilon_c + \varepsilon_l$, where $\varepsilon_c$ is the circumferential (hoop) strain and $\varepsilon_l$ the longitudinal strain.

Consider a thin cylinder of internal diameter $d$ and length $L$. Its volume is

$$V = \frac{\pi}{4}d^2L$$

Derivation

Under internal pressure the diameter changes to $d + \delta d$ and the length to $L + \delta L$. The strains are

$$\varepsilon_c = \frac{\pi(d+\delta d) - \pi d}{\pi d} = \frac{\delta d}{d}, \qquad \varepsilon_l = \frac{\delta L}{L}$$

(circumferential strain equals diametral strain, since circumference $= \pi d$). The new volume is

$$V' = \frac{\pi}{4}(d+\delta d)^2(L+\delta L) = V\left(1+\varepsilon_c\right)^2\left(1+\varepsilon_l\right)$$

Expand and neglect products of small strains:

$$V' = V\left(1 + 2\varepsilon_c + \varepsilon_l + \varepsilon_c^2 + 2\varepsilon_c\varepsilon_l + \varepsilon_c^2\varepsilon_l\right) \approx V\left(1 + 2\varepsilon_c + \varepsilon_l\right)$$ $$\varepsilon_v = \frac{V'-V}{V} = 2\varepsilon_c + \varepsilon_l$$

Alternative (differentiation): from $V = \frac{\pi}{4}d^2L$,

$$\frac{dV}{V} = 2\frac{dd}{d} + \frac{dL}{L} = 2\varepsilon_c + \varepsilon_l$$

In terms of pressure

With $\sigma_c = \dfrac{pd}{2t}$ and $\sigma_l = \dfrac{pd}{4t}$:

$$\varepsilon_c = \frac{pd}{2tE}\left(1-\frac{\nu}{2}\right), \qquad \varepsilon_l = \frac{pd}{2tE}\left(\frac{1}{2}-\nu\right)$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{2tE}\left(\frac{5}{2} - 2\nu\right) = \frac{pd}{4tE}\left(5-4\nu\right)$$

A thin cylinder of internal diameter $d$, wall thickness $t$ and internal pressure $p$ is subjected to two stresses: circumferential (hoop) stress $\sigma_c$ acting on longitudinal sections, and longitudinal stress $\sigma_l$ acting on circular cross-sections. We show that $\sigma_l = \tfrac12\sigma_c$.

   Longitudinal section        Transverse section
   (hoop stress)               (longitudinal stress)

     ______________             _________
    |<----- d ---->|           /  p  p   \
  sc|   p  p  p    |sc        | p  p  p   |-> sl
    |______________|           \_________/
       length L                 area = pi*d*t

Circumferential stress (cut along a diameter)

Cut the cylinder by a longitudinal plane through its axis. The pressure on the projected area $d\times L$ is balanced by the tension in the two walls:

$$\text{Bursting force} = p\times d\times L, \qquad \text{Resisting force} = 2\times\sigma_c\times L\times t$$ $$p\,d\,L = 2\sigma_c\,L\,t \Rightarrow \sigma_c = \frac{pd}{2t}$$

Longitudinal stress (cut across the cylinder)

Cut the cylinder by a transverse plane. The pressure acts on the end area $\frac{\pi}{4}d^2$. This force is resisted by the metal ring of area $\pi d t$ (thin wall):

$$p\times\frac{\pi}{4}d^2 = \sigma_l\times\pi d\,t \Rightarrow \sigma_l = \frac{pd}{4t}$$

Relation

$$\frac{\sigma_l}{\sigma_c} = \frac{pd/4t}{pd/2t} = \frac{1}{2} \quad\Rightarrow\quad \sigma_l = \frac{\sigma_c}{2}$$

So the longitudinal stress is half of the circumferential stress. This is why a boiler shell fails along a longitudinal seam first, and why the longitudinal joint needs more strength than the circumferential joint.

Data: $d = 2500$ mm, $t = 8$ mm, $\delta V = 300\ \text{cm}^3 = 3\times10^5\ \text{mm}^3$, $E = 2.1\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Method

For a thin sphere the stress in every direction is $\sigma = \dfrac{pd}{4t}$. The strain in any direction is

$$\varepsilon = \frac{\sigma}{E}(1-\nu) = \frac{pd}{4tE}(1-\nu)$$

and the volumetric strain is three times this:

$$\frac{\delta V}{V} = 3\varepsilon = \frac{3pd}{4tE}(1-\nu)$$

Calculation

$$V = \frac{\pi d^3}{6} = \frac{\pi(2500)^3}{6} = 8.1812e+09\ \text{mm}^3$$ $$\frac{3\times10^5}{8.1812e+09} = \frac{3p(2500)}{4(8)(2.1\times10^5)}(1-0.3)$$ $$3.6669e-05 = 7.8125e-04\,p \Rightarrow p = 0.0469\ \text{N/mm}^2$$

Answer: $p \approx 0.047\ \text{N/mm}^2$ (0.047 MPa).

Data: $d = 1900$ mm, $t = 6$ mm, $\delta V = 400\ \text{cm}^3 = 4\times10^5\ \text{mm}^3$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

For a thin sphere, $\sigma = \dfrac{pd}{4t}$ in all directions, the linear strain is $\varepsilon = \dfrac{\sigma}{E}(1-\nu)$ and

$$\frac{\delta V}{V} = 3\varepsilon = \frac{3pd}{4tE}(1-\nu)$$

Calculation

$$V = \frac{\pi d^3}{6} = \frac{\pi(1900)^3}{6} = 3.5914e+09\ \text{mm}^3$$ $$\frac{4\times10^5}{3.5914e+09} = 1.1138e-04 = \frac{3p(1900)(0.75)}{4(6)(2\times10^5)} = 8.9063e-04\,p$$ $$p = 0.1251\ \text{N/mm}^2$$

Answer: $p \approx 0.125\ \text{N/mm}^2$.

The extra liquid fills the space created by the expansion of the shell. The liquid is assumed incompressible (no bulk modulus is given), so $\delta V_{shell} = 80\ \text{cm}^3$.

Data: $L = 300$ cm, $d = 30$ cm, $t = 1.4$ cm, $\nu = 0.36$, $E = 10^6\ \text{kg/cm}^2$.

Volumetric strain of a thin cylinder

$$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{2tE}\left(\frac{5}{2} - 2\nu\right) = \frac{pd}{4tE}(5-4\nu)$$

Volume and pressure

$$V = \frac{\pi}{4}(30)^2(300) = 212057.50\ \text{cm}^3$$ $$\frac{\delta V}{V} = \frac{80}{212057.50} = 3.7726e-04 = \frac{p(30)}{4(1.4)(10^6)}(5 - 4\times0.36) = 1.9071e-05\,p$$ $$p = 19.781\ \text{kg/cm}^2 \approx 1.940\ \text{N/mm}^2$$

Answer: pressure developed $\approx 19.78\ \text{kg/cm}^2$ (1.94 MPa).

The extra liquid occupies the additional space made by the expansion of the shell (liquid taken as incompressible), so $\delta V = 50\ \text{cm}^3$.

Data: $L = 200$ cm, $d = 25$ cm, $t = 1.2$ cm, $\nu = 0.34$, $E = 10^6\ \text{kg/cm}^2$.

Relation between pressure and volume change

For a thin cylinder, $\sigma_c = \dfrac{pd}{2t}$, $\sigma_l = \dfrac{pd}{4t}$, and

$$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = \frac{pd}{4tE}(5-4\nu)$$ $$V = \frac{\pi}{4}(25)^2(200) = 98174.77\ \text{cm}^3$$ $$\frac{\delta V}{V} = \frac{50}{98174.77} = 5.0930e-04 = \frac{p(25)}{4(1.2)(10^6)}(5-4\times0.34) = 1.8958e-05\,p$$ $$p = 26.864\ \text{kg/cm}^2 \approx 2.634\ \text{N/mm}^2$$

Answer: pressure developed $\approx 26.86\ \text{kg/cm}^2$ (2.63 MPa).

Assumption: the 1 m length is the cylindrical portion; the two spherical (hemispherical) ends have the same diameter 200 mm and thickness 6 mm. Total change in volume = change in the cylindrical part + change in the two ends (together one full sphere).

Data: $d = 200$ mm, $t = 6$ mm, $L = 1000$ mm, $p = 1.5\ \text{N/mm}^2$, $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.25$.

Cylindrical part

$$\sigma_c = \frac{pd}{2t} = 25.00\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 12.50\ \text{N/mm}^2$$ $$\varepsilon_c = \frac{25.00 - 0.25(12.50)}{2\times10^5} = 1.0938e-04, \qquad \varepsilon_l = \frac{12.50 - 0.25(25.00)}{2\times10^5} = 3.1250e-05$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 2.5000e-04, \quad V_{cyl} = \frac{\pi}{4}(200)^2(1000) = 3.1416e+07\ \text{mm}^3$$ $$\delta V_{cyl} = 2.5000e-04\times3.1416e+07 = 7854.0\ \text{mm}^3$$

Spherical ends (one complete sphere)

$$\sigma = \frac{pd}{4t} = 12.50\ \text{N/mm}^2, \qquad \varepsilon = \frac{\sigma}{E}(1-\nu) = 4.6875e-05$$ $$V_s = \frac{\pi d^3}{6} = 4.1888e+06\ \text{mm}^3, \qquad \delta V_s = 3\varepsilon V_s = 589.0\ \text{mm}^3$$

Total

$$\delta V = 7854.0 + 589.0 = 8443.0\ \text{mm}^3 = 8.44\ \text{cm}^3$$

Answer: increase in volume $\approx 8.44\ \text{cm}^3$.

A standpipe is a vertical pipe filled with water. The pressure is greatest at the base, so the wall must be thickest there; the hoop stress governs because it is twice the longitudinal stress.

Data: $d = 4$ m $= 4000$ mm, $h = 30$ m, $\sigma_{allow} = 120\ \text{N/mm}^2$, $\rho_w = 1000\ \text{kg/m}^3$, $g = 9.81\ \text{m/s}^2$.

Water pressure at the base

$$p = \rho g h = 1000(9.81)(30) = 294\,300\ \text{N/m}^2 = 0.2943\ \text{N/mm}^2$$

Thickness (hoop stress)

$$\sigma_c = \frac{pd}{2t} \Rightarrow t = \frac{pd}{2\sigma_c} = \frac{0.2943\times4000}{2\times120} = 4.91\ \text{mm}$$

Answer: required thickness $\approx 4.91\ \text{mm}$; provide about 5 mm (the thickness could be reduced higher up where the pressure is less).

Data: $d = 600$ mm, $L = 3000$ mm, $t = 10$ mm, $p = 3\ \text{MPa}$, $E = 200\ \text{GPa} = 2\times10^5\ \text{MPa}$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{3(600)}{2(10)} = 90.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 45.0\ \text{MPa}$$

Strains

$$\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{90.0 - 0.3(45.0)}{2\times10^5} = 3.8250e-04, \qquad \varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = 9.0000e-05$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 8.5500e-04$$

Change in volume

$$V = \frac{\pi}{4}(600)^2(3000) = 8.4823e+08\ \text{mm}^3, \qquad \delta V = \varepsilon_vV = 7.2524e+05\ \text{mm}^3$$

Answer: $\delta V \approx 725.2\ \text{cm}^3$ (725237 mm³, increase).

The extra water to be pumped in must fill two things: the increase in the volume of the shell (it expands) and the reduction in the volume of the water itself (water is compressed).

Data: $d = 300$ mm, $L = 4000$ mm, $t = 12$ mm, $p = 6\ \text{N/mm}^2$, $E = 2.1\times10^5\ \text{N/mm}^2$, $\nu = 0.3$, $K = 2100\ \text{N/mm}^2$.

Increase in shell volume

$$\sigma_c = \frac{pd}{2t} = 75.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 37.50\ \text{N/mm}^2$$ $$\varepsilon_v = \frac{pd}{4tE}(5-4\nu) = \frac{6(300)}{4(12)(2.1\times10^5)}(5-1.2) = 6.7857e-04$$ $$V = \frac{\pi}{4}(300)^2(4000) = 2.8274e+08\ \text{mm}^3, \qquad \delta V_{shell} = 1.9186e+05\ \text{mm}^3$$

Compression of the water

$$K = \frac{p}{\delta V_w/V} \Rightarrow \delta V_w = \frac{pV}{K} = \frac{6\times2.8274e+08}{2100} = 8.0784e+05\ \text{mm}^3$$

Additional water required

$$\delta V = \delta V_{shell} + \delta V_w = 1.9186e+05 + 8.0784e+05 = 9.9970e+05\ \text{mm}^3$$

Answer: additional volume of water $\approx 999.7\ \text{cm}^3$ (of which 191.9 cm³ is the expansion of the shell).

Data: $d = 1000$ mm, $L = 5000$ mm, $t = 20$ mm, $p = 5\ \text{N/mm}^2$, $E = 200\ \text{GPa} = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{5(1000)}{2(20)} = 125.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 62.5\ \text{N/mm}^2$$ $$\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = 31.25\ \text{N/mm}^2$$

(maximum shear stress in the plane of the wall; it equals half the longitudinal stress).

Strains

$$\varepsilon_c = \frac{125.0-0.3(62.5)}{2\times10^5} = 5.3125e-04, \qquad \varepsilon_l = \frac{62.5-0.3(125.0)}{2\times10^5} = 1.2500e-04$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 1.1875e-03$$

Changes in dimensions

$$\delta d = \varepsilon_c d = 0.5312\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.6250\ \text{mm}$$ $$V = \frac{\pi}{4}(1000)^2(5000) = 3.9270e+09\ \text{mm}^3, \qquad \delta V = \varepsilon_v V = 4.6633e+06\ \text{mm}^3$$

Answer: $\sigma_c = 125.0$, $\sigma_l = 62.5$, $\tau_{max} = 31.25$ N/mm²; $\delta d = 0.531$ mm, $\delta L = 0.625$ mm, $\delta V = 4.663\times10^6$ mm³ ($\approx 4663$ cm³).

Data: external diameter 260 mm, $t = 5$ mm, so internal diameter $d = 260 - 2(5) = 250$ mm; $L = 2500$ mm, $p = 1.6\ \text{MPa}$, $E = 2\times10^5\ \text{MPa}$, $\nu = 0.3$, longitudinal joint efficiency $\eta_l = 0.85$, circumferential joint efficiency $\eta_c = 0.65$.

Stresses

In the solid plate (away from the joints):

$$\sigma_c = \frac{pd}{2t} = \frac{1.6(250)}{2(5)} = 40.0\ \text{MPa}, \qquad \sigma_l = \frac{pd}{4t} = 20.0\ \text{MPa}$$

At the joints (for strength, the efficiency reduces the resisting area):

$$\sigma_{c,joint} = \frac{pd}{2t\eta_l} = \frac{40.0}{0.85} = 47.06\ \text{MPa}, \qquad \sigma_{l,joint} = \frac{pd}{4t\eta_c} = \frac{20.0}{0.65} = 30.77\ \text{MPa}$$

Changes in dimensions

The joint efficiencies affect only the local stress at the riveted or welded seams. The overall deformation of the shell is governed by the stresses in the solid plate, so the efficiencies are not used in the strain calculation.

$$\varepsilon_c = \frac{40.0-0.3(20.0)}{2\times10^5} = 1.7000e-04, \qquad \varepsilon_l = \frac{20.0-0.3(40.0)}{2\times10^5} = 4.0000e-05$$ $$\delta d = \varepsilon_c d = 0.04250\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.10000\ \text{mm}$$ $$\varepsilon_v = 2\varepsilon_c + \varepsilon_l = 3.8000e-04, \qquad V = \frac{\pi}{4}(250)^2(2500) = 1.2272e+08\ \text{mm}^3$$ $$\delta V = \varepsilon_v V = 46633.0\ \text{mm}^3 = 46.63\ \text{cm}^3$$

Answer: $\delta d = 0.0425$ mm, $\delta L = 0.1000$ mm, $\delta V = 46.63$ cm³ (all increases).

A cylinder is called thin walled when the wall thickness is small compared with the diameter ($t < d/20$). Under internal fluid pressure $p$ the wall is stressed in two principal directions; the radial stress is so small that it is neglected.

        hoop (circumferential)
          ^  ^  ^
   sl <--[=======]--> sl
        longitudinal

1. Circumferential (hoop) stress, $\sigma_c$

• Acts tangentially around the circumference, tending to split the cylinder along a longitudinal section.
• Considering half the cylinder: bursting force $p\,d\,L$ = resisting force $2\sigma_c\,t\,L$.

$$\sigma_c = \frac{pd}{2t}$$

2. Longitudinal (axial) stress, $\sigma_l$

• Acts along the axis of the cylinder, tending to separate it along a circular (transverse) section.
• Force on the ends $p\times\frac{\pi}{4}d^2$ = resisting force $\sigma_l\,\pi d\,t$.

$$\sigma_l = \frac{pd}{4t}$$

3. Radial stress

Varies from $-p$ at the inner surface to zero at the outer surface; it is very small compared with the other two and is ignored in thin cylinders.

4. Maximum shear stress

With $\sigma_c$ and $\sigma_l$ as principal stresses, the maximum in-plane shear stress is

$$\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = \frac{pd}{8t}$$

Notes

• $\sigma_c = 2\sigma_l$, so the hoop stress decides the thickness; a boiler shell fails along a longitudinal seam first.
• Both stresses are tensile, and both are uniform over the thickness.

Data: internal diameter $d = 500$ mm, pressure head $h = 100$ m of water, $\gamma_w = 10\ \text{kN/m}^3$, allowable stress $\sigma = 20\ \text{N/mm}^2$.

Pressure in the pipe

$$p = \gamma_w h = 10\ \text{kN/m}^3\times100\ \text{m} = 1000\ \text{kN/m}^2 = 1\ \text{N/mm}^2$$

Thickness

The hoop stress is the larger stress, so it governs the design:

$$\sigma_c = \frac{pd}{2t} \Rightarrow t = \frac{pd}{2\sigma_c} = \frac{1\times500}{2\times20} = 12.50\ \text{mm}$$

Check of the longitudinal stress: $\sigma_l = \dfrac{pd}{4t} = 10.0\ \text{N/mm}^2 < 20$, safe.

Answer: thickness $t = 12.5\ \text{mm}$ (provide 13 mm or more in practice).

Data (as given): $L = 4000$ mm, $t = 15$ mm, $d = 15$ mm, $p = 25\ \text{N/mm}^2$, $E = 2.05\times10^5\ \text{N/mm}^2$, $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{25(15)}{2(15)} = 12.50\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 6.25\ \text{N/mm}^2$$

Strains

$$\varepsilon_c = \frac{\sigma_c-\nu\sigma_l}{E} = \frac{12.50-0.3(6.25)}{2.05\times10^5} = 5.1829e-05$$ $$\varepsilon_l = \frac{\sigma_l-\nu\sigma_c}{E} = \frac{6.25-0.3(12.50)}{2.05\times10^5} = 1.2195e-05$$

Changes

$$\delta d = \varepsilon_c d = 0.00078\ \text{mm}, \qquad \delta L = \varepsilon_l L = 0.0488\ \text{mm}$$

Answer (data as given): $\delta L \approx 0.049\ \text{mm}$ and $\delta d \approx 0.0008\ \text{mm}$ (both increases).

Note: with $t = d$ the shell is not really thin, so the thin-shell formulas are only approximate here. If the internal diameter was meant to be 150 mm, the same method gives $\sigma_c = 125.0$, $\sigma_l = 62.5\ \text{N/mm}^2$, $\delta d = 0.078$ mm and $\delta L = 0.488$ mm.

Maximum shear stress is half of the longitudinal stress

In a thin cylinder of diameter $d$, thickness $t$ and pressure $p$, the two principal stresses are the hoop stress and the longitudinal stress (the radial stress is neglected):

$$\sigma_1 = \sigma_c = \frac{pd}{2t}, \qquad \sigma_2 = \sigma_l = \frac{pd}{4t}$$

The maximum shear stress (in the plane of the wall) is half the difference of the principal stresses:

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \frac{1}{2}\left(\frac{pd}{2t} - \frac{pd}{4t}\right) = \frac{pd}{8t}$$

Since $\sigma_l = \dfrac{pd}{4t}$:

$$\tau_{max} = \frac{pd}{8t} = \frac{1}{2}\,\sigma_l$$

Hence the maximum shear stress is half of the longitudinal stress (and one quarter of the hoop stress). It acts on planes at $45^\circ$ to the axis.

Volumetric strain

Volume $V = \dfrac{\pi}{4}d^2L$. Taking the differential,

$$\frac{\delta V}{V} = 2\frac{\delta d}{d} + \frac{\delta L}{L} = 2\varepsilon_c + \varepsilon_l$$

The strains (including Poisson's effect) are

$$\varepsilon_c = \frac{\sigma_c}{E} - \nu\frac{\sigma_l}{E} = \frac{pd}{2tE}\left(1-\frac{\nu}{2}\right), \qquad \varepsilon_l = \frac{\sigma_l}{E} - \nu\frac{\sigma_c}{E} = \frac{pd}{2tE}\left(\frac{1}{2}-\nu\right)$$ $$\varepsilon_v = \frac{pd}{2tE}\left[2\left(1-\frac{\nu}{2}\right) + \frac{1}{2} - \nu\right] = \frac{pd}{2tE}\left(\frac{5}{2}-2\nu\right) = \frac{pd}{4tE}\left(5-4\nu\right)$$

so the change in volume is $\delta V = \dfrac{pd}{4tE}(5-4\nu)\,V$.

For a thin cylinder (internal diameter $d$, length $L$, thickness $t$, internal pressure $p$, $t < d/20$) the radial stress is negligible and the wall is in a state of biaxial tension.

Hoop (circumferential) stress

Cut the cylinder by a plane along its axis. The pressure acts on the projected area $d\times L$; two wall sections of area $t\times L$ resist it.

$$p\,d\,L = 2\,\sigma_c\,t\,L \quad\Rightarrow\quad \sigma_c = \frac{pd}{2t}$$

Longitudinal stress

Cut across the cylinder. The pressure on the end area $\frac{\pi}{4}d^2$ is resisted by the ring area $\pi d\,t$.

$$p\,\frac{\pi}{4}d^2 = \sigma_l\,\pi d\,t \quad\Rightarrow\quad \sigma_l = \frac{pd}{4t}$$

Maximum shear stress

$\sigma_c$ and $\sigma_l$ are principal stresses (no shear on these planes), so

$$\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = \frac{pd}{8t}$$

Summary

Data: $L = 4000$ mm, $d = 1000$ mm, $p = 2\ \text{N/mm}^2$, $t = 8$ mm. For the change in volume, $E$ and $\nu$ are needed; they are not given, so take the usual values for steel, $E = 2\times10^5\ \text{N/mm}^2$ and $\nu = 0.3$.

Stresses

$$\sigma_c = \frac{pd}{2t} = \frac{2(1000)}{2(8)} = 125.0\ \text{N/mm}^2, \qquad \sigma_l = \frac{pd}{4t} = 62.5\ \text{N/mm}^2$$

Maximum shear stress

$$\tau_{max} = \frac{\sigma_c-\sigma_l}{2} = \frac{125.0-62.5}{2} = 31.25\ \text{N/mm}^2$$

Change in volume

$$\varepsilon_v = \frac{pd}{4tE}(5-4\nu) = \frac{2(1000)}{4(8)(2\times10^5)}(5-1.2) = 1.1875e-03$$ $$V = \frac{\pi}{4}(1000)^2(4000) = 3.1416e+09\ \text{mm}^3, \qquad \delta V = \varepsilon_v V = 3.7306e+06\ \text{mm}^3 = 3731\ \text{cm}^3$$

Answer: $\sigma_c = 125.0$ N/mm², $\sigma_l = 62.5$ N/mm², $\tau_{max} = 31.25$ N/mm², $\delta V \approx 3731$ cm³ (increase).

Data: $d = 1000$ mm, $t = 8$ mm, $p = 10\ \text{N/mm}^2$ (the pressure raised above atmospheric), $E = 2\times10^5\ \text{N/mm}^2$, $\nu = 0.3$, $K_{water} = 2100\ \text{N/mm}^2$. (E, ν and K are needed only for volume change; the stresses depend only on $p$, $d$, $t$.)

Principal stresses

$$\sigma_1 = \sigma_c = \frac{pd}{2t} = \frac{10(1000)}{2(8)} = 625.0\ \text{N/mm}^2, \qquad \sigma_2 = \sigma_l = \frac{pd}{4t} = 312.5\ \text{N/mm}^2$$

The radial stress (maximum $-p = -10$ N/mm² on the inner surface) is small compared with these and is neglected in thin-shell theory.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \frac{625.0-312.5}{2} = 156.25\ \text{N/mm}^2$$

(in the plane of the wall, on planes at $45^\circ$ to the axis). If the radial stress is taken as zero, the absolute maximum shear stress, $\sigma_1/2 = 312.5$ N/mm², occurs on planes at $45^\circ$ to the wall surface.

Answer: $\sigma_1 = 625.0$ N/mm², $\sigma_2 = 312.5$ N/mm², $\tau_{max} = 156.25$ N/mm².

Data: radius $r = 1000$ mm, so $d = 2000$ mm; $t = 10$ mm, $p = 1\ \text{MPa}$, $E = 200\ \text{GPa} = 2\times10^5\ \text{MPa}$, $\nu = 0.25$.

Stress

In a thin sphere the stress is the same in every direction in the wall (tangential stress): balancing the pressure on a diametral section $p\frac{\pi}{4}d^2$ with the wall force $\sigma\pi d t$:

$$\sigma = \frac{pd}{4t} = \frac{1(2000)}{4(10)} = 50.0\ \text{MPa}$$

Change in diameter

$$\varepsilon = \frac{\sigma}{E} - \nu\frac{\sigma}{E} = \frac{\sigma}{E}(1-\nu) = \frac{50.0(0.75)}{2\times10^5} = 1.8750e-04$$ $$\delta d = \varepsilon\,d = 1.8750e-04\times2000 = 0.3750\ \text{mm}$$

Answer: circumferential stress $= 50.0$ MPa; change in diameter $= 0.375$ mm (increase).