Failures Resulting from Static Loading Ali Vatankhah

Static Strength A static load is a stationary force or couple applied to a member. To be stationary, the force or couple must be unchanging in magnitude, point or points of application, and direction. A static load can produce axial tension or compression, a shear load, a bending load, a torsional load, or any combination of these. To be considered static, the load cannot change in any manner. You can now appreciate the following four design categories: Failure of the part would endanger human life, or the part is made in extremely large quantities; consequently, an elaborate testing program is justified during design. The part is made in large enough quantities that a moderate series of tests is feasible. The part is made in such small quantities that testing is not justified at all; or the design must be completed so rapidly that there is not enough time for testing. The part has already been designed, manufactured, and tested and found to be unsatisfactory. Analysis is required to understand why the part is unsatisfactory and what to do to improve it. Failure in machine design means that a part become permanently distorted(i.e. yielded) thus its function is compromised. Therefore, the failure strength we refer to is the yield strength (or if the material does not yield such as some brittle materials failure will mean fracture “ultimate strength”).

Stress concentration A stress concentration (often called stress raisers or stress risers) is a location in an object where stress is concentrated. An object is strongest when force is evenly distributed over its area, so a reduction in area, e.g., caused by a crack, results in a localized increase in stress. A material can fail, via a propagating crack, when a concentrated stress exceeds the material's theoretical cohesive strength. The real fracture strength of a material is always lower than the theoretical value because most materials contain small cracks or contaminants (especially foreign particles) that concentrate stress

Failure Theories Failure theory is the science of predicting the conditions under which solid materials fail under the action of external loads. The failure of a material is usually classified into brittle failure (fracture) or ductile failure (yield). 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. Static failure can be classified (as shown in the following sketch) into: 1.Ductile 2. Brittle

Cont Ductile: and have an identifiable yield strength that is often the same in compression as in tension. Significant plastic deformation between yield and fracture { 𝜀 𝑓 ≥ 0.05}. The generally accepted theories for ductile materials (yield criteria) are: a) Maximum shear stress (MSS). b) Distorsion energy (DE). c) Ductile Coulomb-Mohr (DCM).

Cont 2. Brittle: A material is brittle if, when subjected to stress, it breaks without significant deformation. Do not exhibit an identifiable yield strength, and are typically classified by ultimate tensile and compressive strengths. Yield ~= fracture { ε_f<0.05}. The generally accepted theories for brittle materials ( fracture criteria) are: Maximum normal stress(MNS) Brittle Coulomb- Mohr(BCM) Modified Mohr(MM)

Maximum-Shear-Stress Theory for Ductile Materials It predicts that yielding begins whenever the maximum shear stress (MSS) 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. Thus, for a general state of stress, the maximum-shear-stress theory predicts yielding when: 𝜏 𝑚𝑎𝑥 = 𝜎 1 − 𝜎 3 2 ≤ 𝑆 𝑦 2 𝑜𝑟 𝜎 1 − 𝜎 3 ≤ 𝑆 𝑦 5−1 Note that this implies that the yield strength in shear is given by: 𝑆 𝑠𝑦 =0.5 𝑆 𝑦 5−2 , which, as we will see later is about 15 percent low (conservative). For design purposes, Eq. (5–1) can be modified to incorporate a factor of safety, n. Thus, 𝜏 𝑚𝑎𝑥 = 𝑆 𝑦 2𝑛 𝑜𝑟 𝜎 1 − 𝜎 3 = 𝑆 𝑦 2𝑛 5−3 The MSS theory came from the observation that for ductile materials during tension test ,the slip lines (which occur after yielding) and fracture surfaces occur at 45 angle, which is the angle of the maximum shear stress.

Cont For plane stress (where one of the principal stresses is zero), Assuming that 𝜎 𝐴 ≥ 𝜎 𝐵 , there are three cases to consider in using Eq. (5–1): Case 1: 𝜎 𝐴 ≥ 𝜎 𝐵 ≥0. For this case, 𝜎 1 = 𝜎 𝐴 𝑎𝑛𝑑 𝜎 3 = 0. Equation (5–1) reduces to a yield condition of: 𝜎 𝐴 ≥ 𝑆 𝑦 5−4 Case 2: 𝜎 𝐴 ≥ 0≥ 𝜎 𝐵 . For this case, 𝜎 1 = 𝜎 𝐴 𝑎𝑛𝑑 𝜎 3 = 𝜎 𝐵 . Equation (5–1) becomes: 𝜎 𝐴 − 𝜎 𝐵 ≥ 𝑆 𝑦 5−5 Case 3: 0≥ 𝜎 𝐴 ≥ 𝜎 𝐵 . For this case, 𝜎 1 = 0 𝑎𝑛𝑑 𝜎 3 = 𝜎 𝐵 . Equation (5–1) reduces to a yield condition of: 𝜎 𝐵 ≤− 𝑆 𝑦 5−6 Equations (5–4) to (5–6) are represented in Fig. 5–7 by the three lines indicated in the 𝜎 𝐴 , 𝜎 𝐵 plane. Plane stress is defined to be a state of stress in which the normalstress, Oz, and the shear stresses, O'x z and 0'}, z, directed perpendicular to the. x-y plane are assumed to be zero. The geometry of the body is essentially that of a plate with one dimension

Distortion-Energy (DE) Theory for Ductile Materials: It 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. The distortion-energy (DE) theory originated from the observation that ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of the values given by the simple tension test. Therefore 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. To develop the theory, note, in Fig. 5–8a, the unit volume subjected to any three-dimensional stress state designated by the stresses 𝜎 1 , 𝜎 2 , 𝑎𝑛𝑑 𝜎 3 . The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses 𝜎 𝑎𝑣 , acting in each of the same principal directions as in Fig. 5–8a. The formula for 𝜎 𝑎𝑣 , is simply:

The stress state shown in Fig The stress state shown in Fig. 5–8b is one of hydrostatic tension due to the stresses 𝜎 𝑎𝑣 , acting in each of the same principal directions as in Fig. 5–8a. The formula for 𝜎 𝑎𝑣 , is simply: 𝜎 𝑎𝑣𝑒 = 𝜎 1 + 𝜎 2 + 𝜎 3 3 𝑎 If we regard 𝜎 𝑎𝑣𝑒 as a component of 𝜎 1 , 𝜎 2 , 𝑎𝑛𝑑 𝜎 3 , then this component can be subtracted from them, resulting in the stress state shown in Fig. 5–8c. This element is subjected to pure angular distortion, that is, no volume change. The strain energy per unit volume for the element shown in Fig. 5–8a is: 𝑢= 1 2 𝜀 1 𝜎 1 + 𝜀 2 𝜎 2 + 𝜀 3 𝜎 3 𝑏 Using Eq. of Hooke's law with substituting Eq.(b) for the principal strains in gives: 𝑢= 1 2𝐸 𝜎 1 2 + 𝜎 2 2 + 𝜎 2 2 −2𝜐 𝜎 1 𝜎 2 + 𝜎 2 𝜎 3 + 𝜎 3 𝜎 1 𝑐 The strain energy for producing only volume change 𝑢 𝑣 can be obtained by substituting Eq. (a) in Eq. (c). The result is: 𝑢 𝑣 = 3 𝜎 𝑎𝑣 2 2𝐸 1−2𝜐 𝑑

𝑢 𝑣 = 1−2𝜐 6𝐸 𝜎 1 2 + 𝜎 2 2 + 𝜎 2 2 + 𝜎 1 𝜎 2 + 𝜎 2 𝜎 3 + 𝜎 3 𝜎 1 5−7 If we now substitute the square of Eq. (a) in Eq. (d) and simplify the expression, we get: 𝑢 𝑣 = 1−2𝜐 6𝐸 𝜎 1 2 + 𝜎 2 2 + 𝜎 2 2 + 𝜎 1 𝜎 2 + 𝜎 2 𝜎 3 + 𝜎 3 𝜎 1 5−7 Then the distortion energy is obtained by subtracting Eq. (5–7) from Eq. (c). This gives: 𝑢 𝑑 =𝑢− 𝑢 𝑣 = 1+𝜐 3𝐸 𝜎 1 − 𝜎 2 2 + 𝜎 2 − 𝜎 3 2 + 𝜎 3 − 𝜎 1 2 2 5−8 Note that the distortion energy is zero if 𝜎 1 = 𝜎 2 = 𝜎 3 =0 . For the simple tensile test, at yield, 𝜎 1 = 𝑆 𝑦 𝑎𝑛𝑑 𝜎 2 = 𝜎 3 =0 and from Eq. (5–8) the distortion energy is 𝑢 𝑑 = 1+𝜐 3𝐸 𝑆 𝑦 2 5−9 So for the general state of stress given by Eq. (5–8), yield is predicted if Eq. (5–8) equals or exceeds Eq. (5–9). This gives: 𝜎 1 − 𝜎 2 2 + 𝜎 2 − 𝜎 3 2 + 𝜎 3 − 𝜎 1 2 2 1 2 ≥ 𝑆 𝑦 5−10

Thus, the left of Eq. (5–10) can be thought of as a single, equivalent, or effective stress. This effective stress is usually called the Von Mises stress, σ′, named after Dr. R. Von Mises, who contributed to the theory. Thus Eq. (5–10), for yield, can be written as: σ ′ ≥ 𝑆 𝑦 5−11 , where the von Mises stress is: σ ′ = 𝜎 1 − 𝜎 2 2 + 𝜎 2 − 𝜎 3 2 + 𝜎 3 − 𝜎 1 2 2 1 2 5−12

For plane stress, let σA and σB be the two nonzero principal stresses For plane stress, let σA and σB be the two nonzero principal stresses. Then from Eq. (5–12), we get: σ ′ = 𝜎 𝐴 2 − 𝜎 𝐴 𝜎 𝐵 + 𝜎 𝐵 2 1 2 5−13   Equation (5–13) is a rotated ellipse in the 𝜎 𝐴 , 𝜎 𝐵 plane, as shown in Fig. 5–9 with ′ = 𝑆 𝑦 . The dotted lines in the figure represent the MSS theory, which can be seen to be more restrictive, hence, more conservative. Using xyz components of three-dimensional stress, the von Mises stress can be written as: σ ′ = 1 2 𝜎 𝑥 − 𝜎 𝑦 2 + 𝜎 𝑦 − 𝜎 𝑧 2 + 𝜎 𝑧 − 𝜎 𝑥 2 +6 𝜏 𝑥𝑦 2 + 𝜏 𝑦𝑧 2 + 𝜏 𝑧𝑥 2 1 2 5−14 , and for plane stress, σ ′ = 1 2 𝜎 𝑥 2 − 𝜎 𝑥 𝜎 𝑦 + 𝜎 𝑦 2 +3 𝜏 𝑥𝑦 2 1 2 5−15 The distortion-energy theory is also called: 1.The von Mises or von Mises–Hencky theory 2.The shear-energy theory 3.The octahedral-shear-stress theory

Coulomb-Mohr Theory for Ductile Materials

𝐵 2 𝐶 2 − 𝐵 1 𝐶 1 𝑂𝐶 2 − 𝑂𝐶 1 = 𝐵 3 𝐶 3 − 𝐵 1 𝐶 1 𝑂𝐶 3 − 𝑂𝐶 1 𝐵 2 𝐶 2 − 𝐵 1 𝐶 1 𝑂𝐶 2 − 𝑂𝐶 1 = 𝐵 3 𝐶 3 − 𝐵 1 𝐶 1 𝑂𝐶 3 − 𝑂𝐶 1 𝜎 1 − 𝜎 3 2 − 𝑆 𝑡 2 𝑆 𝑡 2 − 𝜎 1 + 𝜎 3 2 = 𝑆 𝑐 2 − 𝑆 𝑡 2 𝑆 𝑐 2 + 𝑆 𝑡 2 Cross-multiplying and simplifying reduces this equation to: 𝜎 1 𝑆 𝑡 − 𝜎 3 𝑆 𝑐 =1 5−21 , where either yield strength or ultimate strength can be used.

For plane stress (where one of the principal stresses is zero), Assuming that 𝜎 𝐴 ≥ 𝜎 𝐵 , there are three cases to consider in using Eq. (5–1): Case 1: 𝜎 𝐴 ≥ 𝜎 𝐵 ≥0 . For this case, 𝜎 1 = 𝜎 𝐴 𝑎𝑛𝑑 𝜎 3 = 0. Equation (5–21) reduces to a yield condition of: 𝜎 𝐴 ≥ 𝑆 𝑡 5−22 Case 2: 𝜎 𝐴 ≥ 0≥ 𝜎 𝐵 . For this case, 𝜎 1 = 𝜎 𝐴 𝑎𝑛𝑑 𝜎 3 = 𝜎 𝐵 . Equation (5–21) becomes: 𝜎 𝐴 𝑆 𝑡 − 𝜎 𝐵 𝑆 𝑐 ≥1 5−23 Case 3: 0≥ 𝜎 𝐴 ≥ 𝜎 𝐵 . For this case, 𝜎 1 = 0 𝑎𝑛𝑑 𝜎 3 = 𝜎 𝐵 . Equation (5–1) reduces to a yield condition of: 𝜎 𝐵 ≤− 𝑆 𝑐 5−24

Maximum-Normal-Stress Theory for Brittle Materials:

, which is plotted in Fig. 5–18a , which is plotted in Fig. 5–18a. As before, the failure criteria equations can be converted to design equations as: The load lines are shown in Fig. 5–18b.

Modifications of the Mohr Theory for Brittle Materials: The equations provided for the theories will be restricted to plane stress and be of the design type incorporating the factor of safety. On the basis of observed data for the fourth quadrant, the modified Mohr theory expands the fourth quadrant as shown in Fig. 5–19. Brittle-Coulomb-Mohr: 5–12 Introduction to Fracture Mechanics: Self reading 5–13 Stochastic Analysis: 5–14 Important Design Equations:

Modified Mohr:

Selection of Failure Criteria: Figure 5–21 provides a summary flow-chart for the selection of an effective procedure for analyzing or predicting failures from static loading for brittle or ductile behavior.

Example1 Shaft of ASTM G25 cast iron subject to loading shown From Table A-24 Sut = 26 kpsi Suc = 97 kpsi Find: For a factor of safety of n = 2.8, what should the diameter of the shaft (d) be?

Example 2

Example 3

Example 4

Example 5 Two steel tubes are shrink-fitted together where the nominal diameters are 1.50, 1.75, and 2.00 in. Careful measurement before fitting revealed that the diametral interference between the tubes to be 0.00246 in. After the fit, the assembly is subjected to a torque of 8000 lbf · in and a bending-moment of 6000 lbf · in. Assuming no slipping between the cylinders, analyze the outer cylinder at the inner and outer radius. Determine the factor of safety using distortion energy with Sy = 60 kpsi.