Irodov 1.291: Curvature Doubles a Sphere's Strength – Amit & Deepali Sinha Foundation

Problem Irodov 1.291

A thin-walled glass vessel has wall thickness $\Delta r = 1.0$ mm and inner radius $r = 25$ mm, and holds an internal pressure with no external pressure acting on it. Find the greatest internal pressure the vessel can sustain before its wall fails in tension, (a) when the vessel is a long cylindrical tube, and (b) when the vessel is a spherical flask of the same radius and wall thickness.

Given

$\Delta r = 1.0$ mm
$r = 25$ mm
glass

Find

(a)

p

the tube can sustain, (b)

p

the spherical flask can sustain

Solution

Step 1Hoop stress in the tube’s wall

The tube, cut lengthwise through its axis; the upper half is the free body. The internal pressure p pushes radially outward on the inner face, and the tension σ Δ r per unit length acts straight down at each cut face. — The tube, cut lengthwise through its axis; the upper half is the free body. The internal pressure $p$ pushes radially outward on the inner face, and the tension $\sigma\,\Delta r$ per unit length acts straight down at each cut face.

Cut the tube lengthwise through its axis and take a slice of length $L$ ; keep the upper half as the free body. The internal pressure pushes on the projected area of the inner face, $2rL$ , and this is balanced by the wall tension acting across the two cut faces, each of cross section $\Delta r\,L$ .

p\cdot2rL = 2(\sigma\Delta r)L

Both sides carry $L$ , which cancels, leaving the stress in terms of $p$ , $r$ and $\Delta r$ alone.

\sigma = \frac{pr}{\Delta r}

(1)

Step 2The tube’s bursting pressure

The wall fails once $\sigma$ reaches glass’s tensile strength, $\sigma_m = 5.0\times10^{7}$ Pa. Setting $\sigma=\sigma_m$ in (1) and solving for $p$ gives the tube’s bursting pressure.

\begin{align*}\sigma_m &= \frac{p_a r}{\Delta r} \\ p_a &= \frac{\sigma_m\Delta r}{r}\end{align*}

Substituting that strength and the given dimensions evaluates it.

p_a = \frac{5.0\times10^{7}\times1.0\times10^{-3}}{25\times10^{-3}} = 2.0\times10^{6}\ \text{Pa} = 19.7\ \text{atm}

p_a = \frac{\sigma_m\Delta r}{r} \approx 20\ \text{atm}

(2)

Step 3Hoop stress in the sphere’s wall

The same half section for the spherical flask, cut through its equator (dashed ellipse). The pressure fan is as before, and the tension σ Δ r acts all the way around the circular equatorial cut. — The same half section for the spherical flask, cut through its equator (dashed ellipse). The pressure fan is as before, and the tension $\sigma\,\Delta r$ acts all the way around the circular equatorial cut.

Cut the flask through its equator and again keep the upper half as the free body. Now the pressure pushes on the full projected disc of radius $r$ , area $\pi r^2$ , and the wall tension acts all the way around the circular equatorial cut, of circumference $2\pi r$ and cross section $\Delta r$ per unit length of that circumference.

p\cdot\pi r^2 = (\sigma\Delta r)\cdot2\pi r

Dividing through by $\pi r$ isolates the stress.

\sigma = \frac{pr}{2\Delta r}

(3)

Step 4The sphere’s bursting pressure

Setting $\sigma=\sigma_m$ in (3) and solving for $p$ gives exactly twice the tube’s bursting pressure.

\begin{align*}\sigma_m &= \frac{p_b r}{2\Delta r} \\ p_b &= \frac{2\sigma_m\Delta r}{r}\end{align*}

The factor of two is the only change from the tube, so the number is twice the one found there.

p_b = 2\times2.0\times10^{6}\ \text{Pa} = 4.0\times10^{6}\ \text{Pa} = 39.5\ \text{atm}

p_b = \frac{2\sigma_m\Delta r}{r} \approx 40\ \text{atm}

(4)

Check

Both stresses carry units $[\text{Pa}][\text{m}]/[\text{m}] = \text{Pa}$ , correct. And the lengthwise cut behind (1) is the governing one for the tube: cutting it instead straight across its axis puts the pressure on the same projected disc $\pi r^2$ used for (3), now balanced by the whole annular wall of area $2\pi r\,\Delta r$ , giving an axial stress $pr/(2\Delta r)$ , exactly half the hoop stress. The wall is worked hardest along the lengthwise cut, which is why that cut, and not the transverse one, sets the bursting pressure.

Remark

The sphere sustains twice the tube’s pressure because it is curved in two independent directions at once, so its equatorial cut is reinforced by wall material running all the way around a closed circle, while the tube’s axial cut is reinforced along only two straight, finite-length edges. This is the reason pressure vessels built for high pressure (gas cylinders, spacecraft tanks) favor spherical or hemispherically-capped shapes over plain cylinders wherever weight or wall thickness is at a premium.