6.9 Burst Calculation of a Compressor

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Plasticity is an important phenomenon in the deformation of metallic materials. Because

of plasticity, high linear elastic stresses at notches and other geometric discontinuities

are relaxed and redistributed. In the present application, the rotational speed of a radial

compressor is increased till burst. The compressor is made of an aluminum alloy with

Young’s modulus E = 75 000 MPa and a Poisson coefficient ν = 0.3. The isotropic hardening

curve is bilinear and described in Table 6.1. The geometry of the compressor can

be downloaded from the CalculiXHomepage (CalculiX GraphiX examples, (CalculiX

2003)).

Figure 6.6 shows the equivalent plastic strain at a location at the bore (inner radius) and

the rim (outer radius) of the disk. It is well known from the theory of elasticity (Timoshenko

and Goodier 1970) that the stresses are highest in the bore region and that is where plastic

flow starts from. At a rotational speed of about 170 000 cycles/min, the disk collapses. This

Table 6.1 Isotropic hardening curve.

von Mises stress Equivalent plastic strain

(MPa) (%)

290 0

347 6

347 100

FINITE STRAIN ELASTOPLASTICITY 303

−2

0

2

2

4

4

6

6

8

8

10

10

12

12

14

14

16

16

18

18

Rotational speed (104 cycles/min)

Equivalent plastic strain (%)

Bore

Rim

Figure 6.6 Equivalent plastic strain in the disk

0

0.5

1

2 4 6 8 10 12 14 16 18

0.1

0.2

0.3

0.4

0.6

0.7

0.8

0.9

Rotational speed (104 cycles/min)

Radial displacement (mm)

Bore

Rim

Figure 6.7 Radial displacements in the disk

is clear from the asymptotic increase of the plastic flow in the bore region. At the same

time, the inner and outer radii also increase significantly (Figure 6.7). The bore radius at

rest is about 3.5 mm, the rim radius is 43.5 mm. The calculation allows us to determine

the burst margin for a given operation point.

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