Causes of Casing Buckling

17.10.2013 | Автор: admin

Buckling of long sections of columns or cylindrical bars is generally caused by the change from a stable equilibrium to an unstable one. The concept of stability can

Fig. 2.24: Three states of equilibrium.

best be visualized by considering three possible states of a ball when it is placed on plane surfaces of different geometry as shown in Fig. 2.24 (Higdon et al.. 1978). The ball in Fig. 2.24(a) is in a stable equilibrium position at the bottom of the concave surface because gravity will cause it to return to the equilibrium position if disturbed. In other words, the potential energy at the bottom of the concave plane is a minimum. Similarly the ball in Fig. 2.24(b) is in a neutral equilibrium position on a horizontal plane because the potential energy is the same irrespective of its position on the plane. The ball in Fig. 2.24(c). however, is in an unstable equilibrium position at the top of the convex surface because the potential energy at this point is a maximum. If it is disturbed, gravity will cause it to move down the slope from its original position until it eventually reaches a point at which the potential energy is a minimum. Thus, equilibrium is stable if the potential energy is at a minimum, unstable if it is at a maximum, and neutral if it is constant.

Consider a situation where casing is freely suspended in air as shown in Fig. 2.25(a), the only stress that exists in the pipe body is the tensile stress due to its own weight which is a maximum at the surface and zero at the bottom of the string. The weight of the casing tends to keep the pipe straight and if the string is deflected it will, under the action of its own weight, return to the straight position. Thus, casing under tension is in a state of stable equilibrium.

If a force is applied to the casing shoe by slacking off below the weight of the string in air at the surface, i. e.. by partially standing the string on-end. the stress distribution will now consist of a compressive stress having the magnitude of slackoff weight at the shoe and a tensile stress at the surface, which is equal to the casing weight in air minus the slackoff weight. Along the axis of the pipe, there is a point where the value of the axial stress in the pipe is zero. Casing below this point O’ in Fig. 2.25(b) is under a compressive stress equivalent to the slackoff weight at that point. As the slackoff weight increases, the point of zero stress gradually moves upward along the pipe axis. If the slackoff weight is further increased, a critical value is reached at which the pipe equilibrium is on the verge of becoming unstable. Any additional slackoff leads to the pipe

COMPRESSION

(-)

NEUTRAL■ POINT

O’

TENSION R/lT

(b) SLACK-OFF NO FLUID

COMPRESSION

TENSION

HYDROSTATIC

PRESSURE

COMPRESSION	TENSION
(-)/	(+)/N’ ‘
HYDROSTATIC / PRESSURE i V/	i i i i I ZERO -«-/— AXIAL / STRESS
/ 1 t 1 1/ 1/ Q’f	‘STRESS DUE /TO BUOYANT / WEIGHT ‘ wt/ft /p /
NEUTRAL POINT

Fig. 2.25: Basic forces acting on casing under different bottom hole conditions. (After Hammerlindl, 1980.)

deflecting laterally and buckling.

When a pipe is freely suspended in fluid, it experiences a buoyancy force according to Archimedes’ Principle. The horizontal component of this force is evenly distributed over the entire length of the pipe and the vertical component is concentrated at the lower end. The stress distribution of the casing suspended in fluid is shown in Fig. 2.25(c). The differences between the stress distributions for the pipe in air and in the fluid can be summarized as follows:

1. The lower end of the pipe in Fig. ‘2.25(b) is under compressive stress due to the applied force, whereas the pipe in Fig. 2.25 (c) is under compression due to the hydrostatic pressure applied vertically upward.

2. The radial and tangential stresses of the pipe in Fig. 2.25 (c) are no longer equal to zero, but instead are equal to hydrostatic pressure of the fluid (line OQ’). At point Q’, the axial stress (compressive) is equal to the hydrostatic

pressure of the fluid.

The buoyancy force on the casing, which is partially cemented at the shoe, is due to the hydrostatic pressure applied vertically to the exposed shoulders and end areas of the pipe. Figure 2.25(d) represents the stress distribution of the casing submerged in fluid with slackoff of part of its own weight. The point of zero axial stress in Fig. 2.25 (d) has moved further upwards along the axis of the pipe. Lines R’S and OQ’ represent axial stress and hydrostatic pressure on the casing, respectively. The point Q1 in Fig. 2.25(c). at which the axial stress is equal to the hydrostatic pressure of the fluid, has also moved upwards to point T in Fig. 2.25 (d) as the force at the lower end increases.

Existence of the compressive stress at the lower end of the casing does not necessarily mean that the casing will buckle. In a discussion of a paper by Klinkenberg (1951), Wood (1951) proposed the following criteria based on the concept of potential energy. There exists a neutral point along the axis of the casing (point Q’ in Fig. 2.25(c) and point T in Fig. 2.25(d)). at which the axial stress is equal to the average of radial and tangential stresses. Below this point buckling will occur, whereas above this point buckling is unlikely.

Fig. 2.26: Pipe subjected to three principal forces: internal, external and axial.

A tubular member with different internal and external pressures is presented in Fig. 2.26 (Klinkenberg, 1951). The lower end of the tube is sealed and fastened to the bottom of the pressure chamber. At the top of the tube, the bore of the pressure chamber is reduced to make a sliding leak-proof fit between the pressure chamber and the outside of the tube. A plunger, which is an integral part of

the pressure chamber extends down into the end of the tube making a sliding leakproof fit with the bore of the tube. Thus, the pressure chamber and the tube comprises three chambers in which separate pressures p,, pQ and p can be

maintained. It is also assumed that the tube is weightless and that its length is

greater than its stiffness, i. e., it can be deflected.

For a small lateral displacement of the tube at its middle, the top of the tube will slide down a small distance Д/. The energy consumed in bending the tube is very small and hence the only changes in potential energy that need to be considered are the following:

1. The chamber at pressure p0 will decrease in volume by irrf Al and the work done is equal to 7тг20 A/ p0.

2. The chamber at pressure pt will increase in volume by nrf A/ and the work done is equal to irrf A/ p,.

3. The chamber at pressure p will increase in volume by ir(rf — rf) A/ and the

work done is equal to — rf) Alp.

where:

А/ = change in length.

The total change in potential energy. APe- is given by:

A PE = 7Г r] Alp0 — ж rf Alp, — 7Г (rf — rf) AI p (2.169)

It should be noted that the volume of each chamber is large so that any small

change in volume produces no change in pressure.

Inasmuch as the tube is weightless, the axial stress. cra, results entirely from the pressure, p. Thus, Eq. 2.169 can be rewritten as:

APE = nr20Alp0- irrf Alр{ + 7Г {rf — rf) Alcra (2.170)

If there is no change in potential energy, the equilibrium is neutral. Hence,

substituting APe = 0 in Eq. 2.170 gives:

РгА,-РоА0 /ОТ7М

<та = — —— — (A1<1)

where:

A0 — A, = ds, the cross-sectional area of steel pipe.

(2.173)

Oa <

Ад — Аг

Similarly the equilibrium is stable if A Pe > 0:

^ ptAt-p0A0 °a A0 — Ai

and the equilibrium is unstable if A Pe < 0:

p, A, ~ PoA0

(2.174)

(2.175)

Thus, for a tube of internal radius, r,, and external radius, r0, and having an internal pressure, p,, and an external pressure, the radial and tangential stresses can be expressed as follows (Graisse, 1965):