Introduction to forensic engineering
Introduction to forensic engineering

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Introduction to forensic engineering

5.7 Mechanical analysis

What stresses is a radiator reservoir subjected to in service? They can be calculated from the internal pressure of the system using standard formulae (Box 19). The water in the cooling system is under pressure, and one value suggested by the manufacturers was 25 psi. This is equivalent to a pressure of about twice atmospheric pressure, or 0.1725 MN m−2(1 psi = 6.9 kN m−2). When values of r = 22.5 mm and t = 2.5 mm are inserted in Equation (7), Box 19, it produces a hoop stress of about 1.55 MN m−2, a value well below the measured strength of the worst sample of about 55 MN m−2. What could cause the material to fail? In the discussion in Paper 3 it is suggested that four kinds of stress raisers that could initiate a brittle crack (page 193).

  1. Geometric stress raiser at the buttress corner.

  2. Internal voids.

  3. Cold slug fragments.

  4. Frozen-in strain.

Estimating the net effect of all these factors working together is difficult. Using the known dimensions of the buttress corner, it was possible to use a stress concentration diagram (Figure 40, Box 14) to estimate a Kt value of about 4.2. The model chosen to represent a void at a cold slug, for example, was that for a penny-shaped crack (Figure 71). However, the evidence for the existence of such a shape in the fracture surface was rather weak, because distinguishing a void from a crack growth region was difficult from the micrographs. Nevertheless, a value of Kt of about 6 emerged from the analysis, so that a net stress concentration of about 25 (6 × 4.2) could have been working to weaken the wall near the buttress corner.

Figure 71: Standard reference for calculating stress concentration of cavities under tension (Peterson, 1997)

The final factor was the distortion in the reservoir noted earlier. It was possible to calculate the effect it could have produced from a tensile curve made during mechanical testing, together with the observation of the widening of the crack in the whole failed sample reservoir. The analysis gave the surprisingly large value of about 20 MN m−2 (page 194, Paper 3). Thus, the net effect on the potential initiator was for a total stress of about 59 MN m 2. This value is now comparable with the strength of the weakest sample tested, so it seemed reasonable to conclude that the combination of defects initiated one, or more likely, a series of cracks, which created the final leak.

Box 19 Pressure pipes and vessels

A pressurised pipe can be analysed relatively easily for the stresses on the pipe wall, and this is surprisingly useful in a range of practical problems. There are two stresses at work in a thin-walled pipe: the hoop stress σH around the circumference, and the longitudinal stress σL acting along the main axis of the pipe (Figure 72). (A thin-walled pipe is defined as a pipe where the wall of thickness t, is less than about a tenth of the radius r, so t is small compared to r. For simplicity, consider there to be an internal pressurep that is constant at all places.)

Analysis shows the hoop stress is the largest stress in the wall and is given by the equation

The longitudinal stress is half the hoop stress and is given by the equation:

When the wall is more than a tenth of the radius, separate equations must be used, although they have a similar form to the above equations. An important inference from the above analysis is that longitudinal cracks grow preferentially to hoop or circumferential cracks. As cracks grow at right angles to the load, longitudinal cracks grow from the hoop stress, while hoop cracks grow from the longitudinal load on a pipe. If hoop cracks are found in a pipe, it therefore implies they have grown under a different load regime than simple hydrostatic pressure.

Figure 72: Wall stresses in pressurized pipe

SAQ 18

Draw a diagram to show the most likely sequence of events that ultimately led to the failure of the reservoir. Indicate on the diagram the evidence on which the sequence is based. Show the original cause of the failure, and discuss how the manufacturer of the radiator assembly could tackle the problem with the moulder of the reservoir.

Answer

Based on the comprehensive analysis of potential and actual defects found in the fracture surface, it is possible to construct the sequence of events in Figure 73. To the right is shown the evidence that pointed to each event.

Figure 73: Sequence of events and supporting evidence leading to car engine seizure

The cause of failure was faulty moulding of the radiator reservoir. The most likely origin being when an early moulding was placed by the operator in the tub intended for delivery to the assembler of the parts intended for the new car. As several cars were being tested, and only one radiator failed, it is unlikely the other reservoirs were poorly moulded. The overall moulding procedure was probably correct, the fault lying with the operator who mis-identified the sample.

The problem could be tackled by informing the moulder of the cause of the problem, and asking for a review of quality control procedures. This could include both discussion of instructions given to operators and sampling procedures used after moulding.

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