BIAXIAL LOADING

15.10.2013 | Автор: admin

In a borehole, the loads on casing due to internal or external pressure induce tangential stresses, which are always accompanied by greater or lesser superimposed axial stresses, predominantly tensile stresses. The effect of axial stresses on

^’compare with’

Fig. 2.18: Comparison of experimentally determined minimal collapse strength.

w’th ^at calculated by the methods of API (Pi) and Clinedinst (P4). (After Krug and Marx, 1980; courtesy of ITE-PU Clausthal.)

internal or external pressure was first recognized by Holmquist and Nadai (1939). According to classical distortion energy theory, the relationship for tlie effect of axial stress is given by the following equation:

(at — az)2 + (oy — oy)2 + (oy — oy)2 = ‘2a (2.150)

where oy = yield stress and ay. ay and ay (ay) are the radial, tangential and axial stresses, respectively, in the principal planes. Expanding and regrouping Eq. 2.150 yields:

(oy — o>)2 — (<Ta — ar) (oy — ay) + (ay — oy)2 = a2 (2.151)

I (<Tt — o-r)2 + (a. — ^^)2 — <72 = o (2.152)

Denoting oy — ay = x, and oy — i(oy — ay) = y. one can express the relationship between the principal stresses in the form of an equation of an ellipse, in this case the ellipse of plasticity:

Consider Eqs. 2.113 and 2.114 in conjunction with the free body diagram (Fig. 2.10). If the pipe is now subjected to an external pressure p0 and an internal pressure p, then for a given set of boundary conditions constants Кг and I2 can be determined. The tangential and radial stresses on the pipe body at any radius r are given by:

УУ + г2)

(2.154)

Pi rf (ro + ?’2) — P-

r2 (rf — rf)

(2.155)

-Pi rf {rf — r2) — p0 rf (r2 + rf) r2 (r2 _ ,.2)

Under the action of external and internal pressures, the pipe will experience the maximal stress at its inner surface. Letting r — r, in Eq. 2.155 yields the condition for equilibrium: ar = —pt. Substituting for <rr in Eq. 2.151 one obtains the quadratic equation:

(2.156)

<?t+P, Y (°a + Pi (<rt + Pi ((g„ + P<)2′ | _ j _ о

Solving the quadratic Eq. 2.156:

<7q + Pi

+ Pi

= ±

(2.157)

Equation 2.157 is regarded as the ellipse of plasticity. Denoting (a, + p,)/py as positive if the pipe is subjected to an internal pressure (burst), and negative if it is subjected to an external pressure (collapse), the equation of the ellipse of plasticity can be presented as in Fig. 2.19. From the plot it can be seen that the tensile force has a negative effect on the collapse pressure and a positive effect on the burst pressure. In contrast, axial compression has a negative effect on the burst pressure and positive effect on collapse pressure.

In practice, however, maximal burst pressure occurs at the surface, where casing is subjected to tensile load due to its own weight. The ellipse of plasticity, therefore, is usually applied when computing the additional effect of tensile force on collapse pressure rating.

EXAMPLE 2-11:

For the N-80, 9| in., 47 lb/ft piece of casing, compute its collapse pressure rating for: (i) <7a = 0, (ii) cra — 25,000 psi and p, — 5.400 psi. Assume a yield strength mode of failure. Compare the answer with the one obtained in Example 2-10.

/Oz+Pi

^)x100»

Oyietd

СГу)виу

d120 -100 -80 -60 -40 -20 0 20 40 60 80 100 120

TENSION

СО

СС

x 100%

/Gt + Pi yield )

COMPRESSION /Oz+R VOyieW J

0 10 20 30 40 50 60 70 80 90 100

x 100% 8

TENSION

SHAPE * MERGEFORMAT

Fig. 2.19: Ellipse of plasticity showing the effect of axial load on both the collapse and burst pressures.

Solution:

The term a ‘yield strength mode of failure’ implies that the casing fails at its inner surface first, i. e., r = r, in Eq. 2.154. Hence, substituting r with r, one can obtain:

PirKrl + rf) — РоГЦг] + rf)

1 rf(r20 — rf)

= Pib’o + rf) — 2 po rf (r2 _ r2)

Rearranging the above equation for direct substitution into the quadratic Eq. 2.156 yields:

‘ at + Pi ( ‘2ro

(i) For the first case where ua — 0 and p, = 0. Eq. 2.157 reduces to: ‘at + Pi

= 0

= ±VT, NOTE:

Substituting:

2 x (Э.625)2

ч9.6252 — 8.6812) 80,000;

Which yields pa = 7,462 psi (from Example 2-10. py = 7,461 psi)

(ii) In the case where <ra = 25,000 and p, = 5,400 psi. one can obtain: ‘at + P, 5,400 — p0

7,462

aa + Pi 25,000 — 5,400

= 0.245

80,000

Substituting these terms into Eq. 2.157 yields:

5,47°°462 P° = ±^l — 0.75 (0.245)2 + i(0.245)

= ±0.977 ± 0.123 = -0.855

Thus:

—p0 — —0.855 x (7,462) — 5.400 p0 = 11,778 psi

Note that in case (i) the pressure differential for collapse was Др = 7.462 psi. The effect of the combined stresses in case (ii) reduced the previously obtained
pressure differential to Ap = 11, 778 — 5,400 = 6. 378 psi. a 45.85 Vc reduction in collapse pressure rating.

The sample N-80 casing actually failed in the ‘plastic range’ in Example ‘2-10 and the collapse pressure value of py — 7.462 psi is well above the actual collapse pressure of pc — 4,760 psi for cra = 0. The lesson then is that the ellipse of plasticity cannot be haphazardly applied in casing design. The mode of failure must be known to be yield strength failure for a valid answer using the above approach, so first check is whether the d0/t ratio falls in the yield range? Here it does not. By using the API approach to biaxial loading as illustrated in Example 2-12, one automatically obtains the correct collapse pressure.

Рубрика: Casing design theory and practice