1 / 38

Failure Theories (5.3-5.8, 5.14)

Failure Theories (5.3-5.8, 5.14). MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering. Example. Static Loading. Failure theories

damian-clements
Download Presentation

Failure Theories (5.3-5.8, 5.14)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Failure Theories(5.3-5.8, 5.14) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering Failure Theory

  2. Example Failure theory

  3. Static Loading • Failure theories • For a given stress state (σ1, σ2, σ3) use properties from a simple tension test (Sy, Sut) to assess the strength. y y T F x A x A diameter = d Failure Theory

  4. Static Loading • For what values of T and F will the material fail if the yield strength is Sy? • If T = 0 • For T and F non-zero, how do we calculate an equivalent or effective stress to assess the strength? Failure Theory

  5. Ductile vs Brittle Column Design

  6. Maximum Normal Stress Theory (5.8) • Failure occurs when maximum principal stress exceeds the ultimate strength. • Primarily applies to brittle materials • Principal stresses: σ1, σ2, σ3 • Define σa, σb, σc where • σa= max (σ1, σ2, σ3) • σc = min (σ1, σ2, σ3) • σbis the value in between σy y τxy x σx Failure Theory

  7. Maximum Normal Stress Theory (5.8) • For failure • σa= Sut if σa > 0 (where Sut is ultimate strength in tension) • σc= -Suc if σc < 0 (where Suc is ultimate strength in compression) • Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy = 0) • σ1 = σo,σ2 = 0 & σ3 = 0 (plane stress) • σa = σo, σb = 0, σc = 0 • Failure when σo= Sut Failure Theory

  8. Maximum Normal Stress Theory (5.8) • Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0) • σ1 = +τxy,σ2 = -τxy & σ3 = 0 (plane stress) • σa = +τxy, σb = 0, σc = -τxy • Failure when τ xy= Sut or τ xy= Suc • Does not agree with experimental data. • Experimental data would show that failure occurs when τ ≈ 0.6 Sut. Failure Theory

  9. Maximum Shear Stress Theory (5.4) • Tresca yield criterion • Failure occurs when the maximum shear stress exceeds the yield strength (max shear stress in a tension test is Sy/2). • Applies to ductile materials. Failure Theory

  10. Maximum Shear Stress Theory (5.4) • Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy = 0) • σ1 = σo,σ2 = 0 & σ3 = 0 (plane stress) • σa = σo, σb = 0, σc = 0 • τmax = σo/2 = Sy/2 • Failure when σo= Sy • Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0) • σ1 = +τxy,σ2 = -τxy & σ3 = 0 (plane stress) • σa = +τxy, σb = 0, σc = -τxy • τmax = τxy • Failure when τ xy= Sy/2  Failure Theory

  11. Distortion Energy Theory (5.5) • von Mises theory • Failure occurs when • Applies to ductile materials. Failure Theory

  12. Distortion Energy Theory (5.5) • Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy = 0) • σ1 = σo,σ2 = 0 & σ3 = 0 (plane stress) • σo2 + σo2 = 2Sy2 • Failure when σo= Sy • Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0) • σ1 = +τxy,σ2 = -τxy & σ3 = 0 (plane stress) • (τxy +τxy)2 + τxy2 + τxy2 = 2Sy2 • 6 τxy2 = 2Sy2 • Failure when τ xy= 0.577Sy • Agrees very closely with experiments! • Section 5.14 in the textbook summarizes failure theories. Failure Theory

  13. Example Find the minimum allowable diameter, with a factor of safety of 2, using both Tresca and von Mises formulas. Assume Sy = 50,000 psi, P = 500 lbs, T = 1000 in-lb, and L = 5 in. y T Stress point x P + d Failure Theory

  14. Stress Concentration Factor(3.13) MAE 316 – Strength of Mechanical Components NC State Department of Mechanical and Aerospace Engineering Stress Concentration Factor

  15. Examples Stress Concentration Factor

  16. Stress Concentration Factor (3.13) • Consider the following two stress analysis problems: h h b b P P P P d Stress Concentration Factor

  17. Stress Concentration Factor (3.13) • For a plate with a hole, the maximum stress occurs around the hole. Stress Concentration Factor

  18. Stress Concentration Factor (3.13) • Maximum stress is defined using a stress concentration factor, Kt. Stress Concentration Factor

  19. Example For a plate with w = 2.0 in. and t = 1.0 in. subject to a 50,000 lb axial load, find the maximum stress for d = 0, 0.1 in., 0.5 in., and 1.0 in. Stress Concentration Factor

  20. Stress Concentration Factor (3.13) • What if the hole is elliptical? σo σmax 2a (hole) 2b (crack) σo Stress Concentration Factor

  21. Stress Concentration Factor (3.13) • This suggests that structures with sharp cracks could not sustain any level of applied stress without failure. • This cannot be correct – fracture mechanics analysis will resolve this. σo σo Stress Concentration Factor

  22. Stress Concentration Factor (3.13) • Other types of stress concentrations (Appendix A-15) • Plate with fillet • Plate with notch • Shaft with fillet • Grooved shaft Stress Concentration Factor

  23. Fracture Mechanics(5.12, 5.14) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering Fracture Mechanics

  24. Fracture Mechanics (5.12) P y h b x L Fracture Mechanics

  25. Fracture Mechanics (5.12) • For a sharp crack, Kt → ∞, σmax → ∞. • The conclusion is that P > 0 will lead to failure, but this is not reasonable. P crack a y h b x L Fracture Mechanics

  26. Stress Intensity Factor (5.12) Figure 5-23 Crack deformation types: (a) mode I, opening; (b) mode 2, sliding; (c) mode III, tearing Fracture Mechanics

  27. Stress Intensity Factor (5.12) • In fracture mechanics, design analysis is based not on stress, but stress intensity factor. • Stress intensity modification factors vary depending on load and geometry. • Refer to Figures 5-25 through 5-29 in the textbook. Fracture Mechanics

  28. Stress Intensity Factor (5.12) • So, for the cracked plate shown previously Figure 5-26 Fracture Mechanics

  29. Fracture Toughness (5.12) • Failure will occur when K ≥ KIc (KIc is fracture toughness, a material property). • “Failure” means the crack extends unstably and the structure fractures (i.e. breaks). Fracture Mechanics

  30. Fracture Toughness (5.12) • In fracture mechanics, factor of safety can also be expressed as Fracture Mechanics

  31. Fracture Toughness (5.12) • Two different analysis methods • To design conservatively for safety, we must do both analyses. Fracture Mechanics

  32. Example • The cracked plate shown below is made of 4340 steel has Sy = 240 ksi and KIc = 50 ksi(in)1/2. Find the maximum allowable load, P, that can be applied to the beam without failure. • Given: b = 1 in, h = 2 in, L = 24 in, a = 0.25 in P crack a y h b x L Fracture Mechanics

  33. Example • Find the stress intensity factor for a plate with a center crack if the average normal stress in the plate is 10 ksi. • Given: 2a = 3 in and 2b = 10 in, d=4 in Fracture Mechanics

  34. Example • Find the stress intensity factor for a plate with an edge crack if the average normal stress in the plate is 10 ksi. • Given: a = 3 in and b = 10 in Fracture Mechanics

  35. Fracture Mechanics

  36. Fracture Mechanics

  37. A Real-Life Example of Fracture Fracture Mechanics

  38. A Real-Life Example of Fracture Fracture Mechanics

More Related