Failure as a Design      Criterion

   Fracture Mechanics

   Failure Analaysis


Wire Rope Failure


Undercarriage Leg Failure
- Part 1
- Part 2
- Part 3
- Part 4
- Activity 1 - Fractography A
- Activity 2 - Fractography B
- Activity 3 - Fracture Stress


Aircraft Towbar Failure


Hail Damage


Insulator Caps


Fractography Resource

Estimation of Tensile and Shear Loads at Failure

In order to validate the hypothesis derived from the fractographic examination, i.e. that an impact load was experienced by the undercarriage leg either immediately prior to, or during, the accident, it is very useful to calculate first order estimates of the tensile and shear loads that the component may have experienced. We can obtain the tensile stress at failure from fracture mechanics (FM), which is the branch of applied mechanics that deals with cracked bodies under load. In particular, linear elastic fracture mechanics is fairly straightforward and usually works well for cases of fatigue crack growth. A comprehensive tutorial on fracture mechanics exists on the internet, which allows students to progress through carefully constructed and fully explained examples in LEFM, at their own speed. The shear stress can be estimated from the area that, fractographically, appears to have failed in shear.

The fundamental equation used in fracture mechanics states that fracture occurs when a combination of a particular crack length and specific applied tensile stress causes the fracture toughness of the material to be exceeded. For a high strength material like 4340 steel, linear elastic fracture mechanics (LEFM) should characterise the fracture, and the relevant equation linking these three parameters can be expressed as:

where KC is the fracture toughness, sigma is the global applied stress, a is crack length, and Y is a finite geometry correction factor. A first order estimate of the fracture toughness for 4340 steel can be obtained in the literature, with typical values ranging from 110 MPam1/2 at a tensile strength of 1290 MPa [1] to about 87 MPam1/2 at a tensile strength of 1060 MPa (University of Western Australia website). These values can be applied to the alloy used in this saddle clamp (tensile strength 1060 MPa), noting that fracture toughness generally decreases with increase in tensile strength in a particular class of alloys. The maximum crack length, measured from the fracture surface is 2.0 mm, and the geometry can be modelled by assuming the simplified case of a semi-elliptic crack in a flat plate subject to tension loading. Activity 3 allows you to calculate the fracture stress for the saddle clamp and to conduct a simple sensitivity analysis of the result to parameters such as crack shape, crack depth and assumed toughness.

A slightly more sophisticated analysis indicated that plastic collapse of this component would occur (rather than fracture) at a stress of 660 MPa. This was performed to Level 1 in PD 6493 : 1980 [5], now superseded by later versions, and contains an inherent 'safety' factor of about 2 (thus the 'actual' magnitude of fracture stress is similar to the predictions obtained with the simple calculations of Activity 3). . Whilst acknowledging the approximations in this analysis, it probably gives a reasonable order of magnitude for the failure stress. The results also agree broadly with the fractographic observations of strongly ductile fracture overtaken by shear, i.e. a situation that is on the border between failure by plastic collapse and failure by fracture. This analysis takes no account of the geometric stress concentration factor at the crack plane, but this is likely to be a second order effect because of the crack is relatively large compared with the section thickness. Even in the presence of a fairly large stress concentration factor, the failure stress would still be of the order of several hundred MPa, implying a large impact loading on the undercarriage legs.

The shear load can be estimated from simple mechanics concepts, as follows. For this 4340 steel the von Mises shear yield stress would be around 0.577 of the tensile yield strength, or perhaps around 0.5 of the tensile strength, i.e. some 530 MPa. The sheared region of the clamp was measured as approximately 3 mm by 38 mm and hence, as no stress concentration acts on the shear stress, a simple calculation indicates that the shear load would have been some 60.4 kN. Note that the order of magnitude of the shear stress and fracture stress are in agreement, which supports the results from the fracture analysis.

These results were also supported by a finite element analysis of the loads on the landing gear, due to a 10 nose-down impact. This work indicated the presence of a rearwards stress at the fracture plane, and a downwards moment. The rearwards stress would initiate fast ductile fracture of the clamp strap, while the downwards stress caused by the moment would tend to shear it. The magnitude of the shear load found by FE analysis was very close to the value calculated above. Thus the interpretation of the fractographic evidence is powerfully supported by this hypothesis of rotational impact to the undercarriage legs in that there is both ductile fast fracture and shear on the fracture surface in sequence after the fatigue region. The inference to be drawn is that during the accident, fast fracture initiated due to tensile stresses on the strap but was overtaken by shear failure. The shear failure was not a concomitant of the fast fracture process, as it was not in the correct plane for such an event (45 to the plane of the initial crack).

Proceed to summary and conclusions of case study.


Failure Analysis  -  Fracture Mechanics  -  Failure As A Design Criterion