Failures Resulting From Static Loading PDF
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A.K.Ebenezer
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Summary
This document explores various failure theories in static loading, including maximum shear stress (MSS) and distortion energy (DET) theories, applied to mechanical engineering analysis. It details these concepts which are important tools for effective design and engineering practice. The document also examines conditions for yielding and provides useful insights for designers to anticipate material failures.
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FAILURES RESULTING FROM STATIC LOADING A.K.EBEN Failure Theories Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading...
FAILURES RESULTING FROM STATIC LOADING A.K.EBEN Failure Theories Events such as distortion, permanent set, cracking, and rupturing are among the ways that a machine element fails. Testing machines appeared in the 1700s, and specimens were pulled, bent, and twisted in simple loading processes If the failure mechanism is simple, then simple tests can give clues. Just what is simple? The tension test is uniaxial (that’s simple) and elongations are largest in the axial direction, so strains can be measured and stresses inferred up to “failure.” Just what is important: a critical stress, a critical strain, a critical energy? In the next several sections, we shall show failure theories that have helped answer some of these questions. Failure Theories Unfortunately, there is no universal theory of failure for the general case of material properties and stress state. Instead, over the years several hypotheses have been formulated and tested, leading to today’s accepted practices. Being accepted, we will characterize these “practices” as theories as most designers do Structural metal behavior is typically classified as being ductile or brittle, although under special situations, a material normally considered ductile can fail in a brittle manner (see Sec. 5–12) Failure Theories Ductile materials are normally classified such that ε ≥ 0.05 and f have an identifiable yield strength that is often the same in compression as in tension (Syt = Syc = Sy). Brittle materials, εf < 0.05, do not exhibit an identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths, Sut and Suc, respectively (where Suc is given as a positive quantity). Failure Theories The generally accepted theories are: Ductile materials (yield criteria) Maximum shear stress (MSS), Sec. 5–4 Distortion energy (DE), Sec. 5–5 Ductile Coulomb-Mohr (DCM), Sec. 5–6 Brittle materials (fracture criteria) Maximum normal stress (MNS), Sec. 5–8 Brittle Coulomb-Mohr (BCM), Sec. 5–9 Modified Mohr (MM), Sec. 5–9 Failure Theories It would be inviting if we had one universally accepted theory for each material type, but for one reason or another, they are all used. Later, we will provide rationales for selecting a particular theory. First, we will describe the bases of these theories and apply them to some examples. Failure Theories- Maximum-Shear-Stress Theory for Ductile Materials The maximum-shear-stress (MSS) theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension-test specimen of the same material when that specimen begins to yield. The MSS theory is also referred to as the Tresca or Guest theory. Many theories are postulated on the basis of the consequences seen from tensile tests. As a strip of a ductile material is subjected to tension, slip lines (called Lüder lines) form at approximately 45° with the axis of the strip. These slip lines are the beginning of yield, and when loaded to fracture, fracture lines are also seen at angles approximately 45° with the axis of tension. Failure Theories Since the shear stress is maximum at 45° from the axis of tension, it makes sense to think that this is the mechanism of failure. It will be shown in the next section, that there is a little more going on than this. However, it turns out the MSS theory is an acceptable but conservative predictor of failure; and since engineers are conservative by nature, it is quite often used. Failure Theories Recall that for simple tensile stress, σ = P/A, and the maximum shear stress occurs on a surface 45° from the tensile surface with a magnitude of τmax = σ/2. So the maximum shear stress at yield is τmax = Sy/2. For a general state of stress, three principal stresses can be determined and ordered such that σ1 ≥ σ2≥ σ3. The maximum shear stress is then τmax = (σ1 - σ3)/2 (see Fig. 3–12). Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when Failure Theories Failure Theories we cannot arbitrarily call the in-plane principal stresses σ1 and σ2 until we relate them with the third principal stress of zero. To illustrate the MSS theory graphically for plane stress, we will first label the principal stresses given by Eq. (3–13) as σA and σB, and then order them with the zero principal stress according to the convention σ1 ≥ σ2 ≥ σ3. Assuming that σA ≥ σB, there are three cases to consider when using Eq. (5–1) for plane stress: Failure Theories Failure Theories Failure Theories Point a represents the stress state of a critical stress element The factor of safety guarding against yield at point a is given by the ratio of strength (distance to failure at point b) to stress (distance to stress at point a), that is n = Ob/Oa. Failure Theories Note that the first part of Eq. (5–3), τmax = Sy/2n, is sufficient for design purposes provided the designer is careful in determining τmax. However, consider the special case when one normal stress is zero in the plane, say σx and τxy have values and σy = 0. It can be easily shown that this is a Case 2 problem, and the shear stress determined by Eq. (3–14) is τmax. Failure Theories Shaft design problems typically fall into this category where a normal stress exists from bending and/or axial loading, and a shear stress arises from torsion. Example The principal stresses at a point in an elastic material are 200 N/mm2 (tensile), 100 N/mm2 (tensile) and 50 N/mm2 (compressive). If the stress at the elastic limit in simple tension is 200 N/mm2, a. Determine whether the failure of the material will occur according to Maximum Shear Stress theory. Take Poisson’s ratio = 0.3. b. Given a safety factor of 5, determine if the material will fail Failure Theories Distortion-Energy Theory for Ductile Materials The distortion-energy theory predicts that yielding occurs when the distortion strain energy per unit volume reaches or exceeds the distortion strain energy per unit volume for yield in simple tension or compression of the same material. Ductile materials stressed hydrostatically (equal principal stresses) exhibited yield strengths greatly in excess of the values given by the simple tension test. It was postulated that yielding was not a simple tensile or compressive phenomenon at all, but, rather, that it was related somehow to the angular distortion of the stressed element. Distortion-Energy Theory for Ductile Materials Failure Theories DET Failure Theories - DET Failure Theories DET VS MSS The model for the MSS theory ignores the contribution of the normal stresses on the 45° surfaces of the tensile specimen. However, these stresses are P/2A, and not the hydrostatic stresses which are P/3A. Herein lies the difference between the MSS and DE theories. THE END OF THE MATTER Von Mises represents complicated stress situations. Von Misses compares to yield strength of the material through Eq. (5- 11) Application of DET The distortion-energy theory predicts no failure under hydrostatic stress and agrees well with all data for ductile behavior. It is the most widely used theory for ductile materials and is recommended for design problems unless otherwise specified. DET & Shear Stresses Consider a case of pure shear τxy, where for plane stress σx = σy = 0. For yield, Eq. (5–11) with Eq. (5–15) gives which as stated earlier, is about 15 percent greater than the 0.5 Sy predicted by the MSS theory. Little Example A hot-rolled steel has a yield strength of Syt = Syc = 100 kpsi and a true strain at fracture of εf = 0.55. Estimate the factor of safety for the following principal stress states using DET & MSST: (a) σx = 70 kpsi, σy = 70 kpsi, τxy = 0 kpsi (b) σx = 60 kpsi, σy = 40 kpsi, τxy = −15 kpsi