Thin-walled pressure vessels and strain measurement

Pressure vessels hold fluids or gases at a pressure different from the surroundings. Civil engineers meet them as water storage tanks, penstocks and pressure pipelines, wastewater digesters, compressed-air receivers and the pipes of water distribution systems. Their analysis is a direct and satisfying application of equilibrium, and it produces a biaxial state of stress that makes good use of lesson 8. The second half of this lesson turns to measurement: how strain gauges let engineers find the stresses in a real structure, and how a three-gauge rosette gives the full plane strain state at a point.

When is a vessel thin-walled?

A vessel is treated as thin-walled when its wall thickness \(t\) is small compared with its radius \(r\), usually when \(r/t \ge 10\). Under that condition the normal stress through the thickness is nearly uniform, and simple equilibrium of a cut section gives accurate results. Thick-walled cylinders, such as hydraulic rams and gun barrels, have stresses that vary strongly through the wall and need the Lamé equations from a more advanced course.

Stresses in a cylindrical vessel

Cut a cylinder of internal pressure \(p\), radius \(r\) and wall thickness \(t\) with a plane through its axis, and consider a length \(L\) of one half. The pressure pushes on the projected area \(2r \times L\), giving a force \(2prL\). It is resisted by the circumferential (hoop) stress \(\sigma_h\) acting on the two cut wall strips, of total area \(2tL\). Equilibrium gives the hoop stress:

Now cut the cylinder across its axis. The pressure acts on the circular end area \(\pi r^2\) and is resisted by the longitudinal stress \(\sigma_l\) acting on the ring of wall, of area approximately \(2\pi r t\):

The hoop stress is twice the longitudinal stress. That is why an overpressurized pipe or a frozen water pipe splits lengthwise, along a line parallel to its axis: the crack opens against the larger hoop stress. It is also why longitudinal welded seams are more critical than circumferential ones, and why design codes apply joint efficiency factors to them.

Use the inner radius for \(r\) for a slightly conservative result; some codes use the mean radius. With \(r/t\) of 10, the two choices differ by about 5 %, and the difference shrinks for thinner walls.

Spherical vessels

A sphere is the same in every direction, so any cut through the center gives the same result as the longitudinal cut of a cylinder:

in every tangential direction. For the same pressure, radius and allowable stress, a sphere needs only half the wall thickness of a cylinder. That is why large gas storage tanks and some digesters are spherical or egg-shaped, even though they are harder to fabricate.

The state of stress in the wall

At a point on the outer surface of a cylinder, the hoop and longitudinal directions are principal directions: there is no shear stress on those planes, by symmetry. So \(\sigma_1 = pr/t\) and \(\sigma_2 = pr/(2t)\), both tensile. The maximum in-plane shear stress is \((\sigma_1 - \sigma_2)/2 = pr/(4t)\). But because both principal stresses have the same sign, the absolute maximum shear involves the third, radial direction, where the stress is zero on the outside surface: \(\tau_{\text{abs}} = \sigma_1/2 = pr/(2t)\), twice the in-plane value. This is exactly the case flagged in lesson 8.

On the inner surface the radial stress equals \(-p\). For thin walls this is small compared with the hoop stress, because \(p\) is much smaller than \(pr/t\), and it is usually neglected.

In the United States, boilers and pressure vessels are designed to the ASME Boiler and Pressure Vessel Code (Section VIII for most unfired vessels), with registration through the National Board and state or local jurisdictions. Large water pipelines and penstocks are usually designed to AWWA manuals and the owner's standards. Codes add joint efficiencies, corrosion allowances, minimum thicknesses and rules for heads, nozzles and supports that go well beyond the membrane formulas here.

Stress on an inclined weld

Large pipes are often made from steel strip wound in a helix and welded along a spiral seam. The stress normal to the seam, and the shear stress along it, follow from the transformation equations of lesson 8 with \(\sigma_x = \sigma_l\), \(\sigma_y = \sigma_h\) and \(\tau_{xy} = 0\).

Generalized Hooke's law and the change in size

In a biaxial stress state, each stress causes strain in its own direction and Poisson contraction in the other. For an isotropic material in plane stress:

For the penstock example, the hoop strain is \(\varepsilon_h = (108 - 0.3 \times 54)/200\,000 = 4.59 \times 10^{-4}\), so the diameter grows by \(4.59 \times 10^{-4} \times 1800 = 0.83\ \text{mm}\). The longitudinal strain is \((54 - 0.3 \times 108)/200\,000 = 1.08 \times 10^{-4}\). See how Poisson's effect reduces the longitudinal strain to less than a quarter of the hoop strain even though the longitudinal stress is half the hoop stress.

Measuring strain

Stress cannot be measured directly; strain can. The electrical resistance strain gauge, a fine metal foil grid bonded to the surface, changes resistance in proportion to the strain along its grid. Connected in a Wheatstone bridge, it can resolve strains of a few microstrain. Engineers use strain gauges to verify designs, monitor bridges and dams, load-test structures and find the stresses in components too complex to analyze by hand. Vibrating-wire gauges, which measure the change in frequency of a tensioned wire, are preferred for long-term monitoring of concrete and soil structures because they are very stable over years.

A single gauge measures strain in only one direction. At a point where the principal directions are unknown, three gauges at different angles are needed, because plane strain has three unknowns: \(\varepsilon_x\), \(\varepsilon_y\) and \(\gamma_{xy}\). Such a group is called a strain rosette.

Normal strain transforms with angle in the same way as normal stress, with \(\gamma_{xy}/2\) playing the role of \(\tau_{xy}\):

For a rectangular rosette with gauges a, b and c at 0°, 45° and 90°, this gives \(\varepsilon_x = \varepsilon_a\), \(\varepsilon_y = \varepsilon_c\) and

The principal strains then follow from the same formulas as principal stresses, with \(\gamma_{xy}/2\) in place of \(\tau_{xy}\). Finally, the principal stresses come from the generalized Hooke's law solved for stress:

Rosette reduction by code

Monitoring systems log thousands of rosette readings, so the reduction is always automated. The function below does the full chain, from three gauge readings to principal stresses, their direction and the von Mises stress.

# Rectangular (0-45-90) strain rosette reduction
import math

def rosette(ea, eb, ec, E=200000.0, nu=0.3):
    """Gauge readings in microstrain; E in MPa. Returns principal stresses in MPa."""
    ex, ey = ea * 1e-6, ec * 1e-6
    gxy = (2 * eb - ea - ec) * 1e-6
    C = (ex + ey) / 2
    R = math.hypot((ex - ey) / 2, gxy / 2)
    e1, e2 = C + R, C - R
    th = 0.5 * math.degrees(math.atan2(gxy, ex - ey))       # from gauge a to the e1 direction
    k = E / (1 - nu**2)
    s1, s2 = k * (e1 + nu * e2), k * (e2 + nu * e1)
    vm = math.sqrt(s1**2 - s1 * s2 + s2**2)
    return e1 * 1e6, e2 * 1e6, th, s1, s2, vm

for name, r in [("girder web", (420, -60, -180)),
                ("pipe wall", (110, 300, 460)),
                ("pure shear", (0, 250, 0))]:
    e1, e2, th, s1, s2, vm = rosette(*r)
    print(f"{name:11s}: e1 = {e1:6.1f}, e2 = {e2:6.1f} microstrain at {th:6.1f} deg; "
          f"s1 = {s1:6.1f}, s2 = {s2:6.1f} MPa; von Mises = {vm:5.1f} MPa")

girder web : e1 =  469.9, e2 = -229.9 microstrain at  -15.5 deg; s1 =   88.1, s2 =  -19.5 MPa; von Mises =  99.3 MPa
pipe wall  : e1 =  460.6, e2 =  109.4 microstrain at   87.6 deg; s1 =  108.5, s2 =   54.4 MPa; von Mises =  93.9 MPa
pure shear : e1 =  250.0, e2 = -250.0 microstrain at   45.0 deg; s1 =   38.5, s2 =  -38.5 MPa; von Mises =  66.6 MPa

The pipe-wall case is close to the 2:1 hoop-to-longitudinal pattern of a pressurized cylinder (gauge c along the hoop direction), and the pure-shear case gives equal and opposite principal stresses at 45°, both just as theory predicts.