Chapter 5. Failures Resulting from Static Loading

Transcription

1 Chapter 5 Failures Resulting from Static Loading

2 2 Chapter Outline Stress Concentration Failure Theories for Ductile Materials Maximum-Shear-Stress Theory Distortion-Energy Theory Coulomb-Mohr Theory Failure Theories for Brittle Materials Maximum-Normal-Stress Theory Modifications of the Mohr Theory Selection of Failure Criteria

3 3 Stress Concentration Localized increase of stress near discontinuities K t = Theoretical (Geometric) Stress Concentration Factor (3.48)

4 4 Theoretical Stress Concentration Factor Appendix A 15 & A 16 Peterson s Stress-Concentration Factors Figure A 15 1: Bar in tension or simple compression with a transverse hole. σ 0 = F/A, A = (w d)t and t is the thickness.

5 5 Theoretical Stress Concentration Factor Figure A 15 9: Round shaft with shoulder fillet in bending. σ 0 = Mc/I, c = d/2 & I = πd 4 /64.

6 6 Stress Concentration for Static and Ductile Conditions With static loads and ductile materials: Highest stressed fibers yield (cold work) Load is shared with next fibers Cold working is localized Overall part does not see damage unless ultimate strength is exceeded SC effect is commonly ignored for static loads on ductile materials

7 7 Stress Concentration for Static and Ductile Conditions SC must be included for dynamic loading SC must be included for brittle materials, since localized yielding may reach brittle failure rather than coldworking and sharing the load.

8 8 Need for Static Failure Theories Uniaxial stress element (tension test) Strength S n Stress Multi-axial stress element One strength, multiple stresses How to compare stress state to single strength?

9 9 Need for Static Failure Theories Failure theories propose appropriate means of comparing multi-axial stress states to single strength

10 10 Maximum Normal Stress Theory (MNS) Yielding begins when max principal stress in a stress element exceeds the yield strength. Use Mohr s circle to find principal stresses Compare largest principal stress to yield strength

11 11 Max Normal Stress Theory (MNS) Theory is unsafe in 4 th quadrant Not safe to use for ductile materials

12 12 Max Shear Stress Theory (MSS) Yielding begins when t max in a stress element exceeds t max in a tension test specimen of same material when that specimen begins to yield. For a tension test specimen, t max = 1 /2. At yielding, 1 = S y and t max = S y /2 Yielding begins when max shear stress in a stress element exceeds S y /2

13 13 Max Shear Stress Theory (MSS) Mohr s circle: find max shear stress. Compare max shear stress to S y /2. Order principal stresses: Incorporate a design factor n n S y t /2 max

14 14 Max Shear Stress Theory (MSS) Consider a plane stress state A & B = two non-zero principal stresses Order them with zero principal stress such that Assuming A B, three cases to consider: Case 1: A B 0 Case 2: A 0 B Case 3: 0 A B

15 15 Max Shear Stress Theory (MSS) Case 1: A B 0 1 = A and 3 = 0 Eq. (5 1) reduces to A S y Case 2: A 0 B 1 = A and 3 = B Eq. (5 1) reduces to A B S y Case 3: 0 A B 1 = 0 and 3 = B Eq. (5 1) reduces to B S y

16 16 Max Shear Stress Theory (MSS) 1. A B 0, A S y 2. A 0 B, A B S y 3. 0 A B, B S y Fig. 5 7

17 17 Max Shear Stress Theory (MSS) Conservative in all quadrants Used for design situations

18 18 Distortion Energy (DE) Failure Theory Other Names: Octahedral Shear Stress Shear Energy Von Mises Von Mises Hencky

19 19 Distortion Energy (DE) Failure Theory Ductile materials stressed hydrostatically exhibited yield strengths greatly in excess of expected values. Strain energy is divided into: 1. hydrostatic volume changing energy 2. angular distortion energy Yielding is primarily affected by distortion energy

20 20 Distortion Energy (DE) Failure Theory Yielding occurs when distortion strain energy per unit volume reaches distortion strain energy per unit volume for yield in simple tension or compression of same material. Fig. 5 8

21 21 Distortion Energy (DE) Failure Theory Hydrostatic stress is average of principal stresses Strain energy per unit volume,

22 22 Distortion Energy (DE) Failure Theory Substituting av for 1, 2, and 3, volume Strain energy is obtained:

23 23 Distortion Energy (DE) Failure Theory Tension test specimen at yield has 1 = S y and 2 = 3 =0 Applying to Eq. (5 8), distortion energy for tension test specimen is DE theory predicts failure when DE (Eq. 5.8) exceeds DE of tension test specimen (Eq. 5.9)

24 24 Von Mises Stress (VMS) Von Mises stress For plane stress, simplifies to In terms of xyz components, in 3D

25 25 Distortion Energy Theory With VMS Von Mises Stress can be thought of as a single, effective stress for the entire general state of stress in a stress element. Distortion Energy failure theory simply compares von Mises stress to yield strength. Introducing a design factor, n S y

26 26 DE Theory Compared to Experimental Data Plot VMS on principal stress axes to compare to experimental data DE curve is typical of data Fig. 5 15

27 27 DE Theory Compared to Experimental Data Typically equates to a 50% reliability from a design perspective Commonly used for analysis situations MSS theory useful for design situations where higher reliability is desired

28 28 Shear Strength Predictions Fig. 5 9 For pure shear loading, Mohr s circle shows that A = B = t Intersection of pure shear load line with failure curve indicates shear strength has been reached

29 29 Shear Strength Predictions For MSS theory, intersecting pure shear load line with failure line [Eq. (5 5)] results in Fig. 5 9

30 30 Shear Strength Predictions For DE theory, intersection pure shear load line with failure curve [Eq. (5 11)] gives Fig. 5 9

31 31 Selection of Failure Criteria

32 Example 5-1 A hot-rolled steel has a yield strength of S yt = S yc = 100 kpsi and a true strain at fracture of ε f = Estimate the factor of safety for the following principal stress states: a. σ x = 70 kpsi, σ y = 70 kpsi, τ xy = 0 kpsi b. σ x = 60 kpsi, σ y = 40 kpsi, τ xy = 15 kpsi c. σ x = 0 kpsi, σ y = 40 kpsi, τ xy = 45 kpsi d. σ x = 40 kpsi, σ y = 60 kpsi, τ xy = 15 kpsi e. σ 1 = 30 kpsi, σ 2 = 30 kpsi, σ 3 = 30 kpsi

33 33 Mohr Theory Some materials have compressive strengths different from tensile strengths Mohr theory is based on three simple tests: 1. Tension 2. Compression 3. shear Fig. 5 12

34 34 Coulomb-Mohr Theory Coulomb-Mohr theory simplifies to linear failure envelope using only tension and compression tests Fig. 5 13

35 35 Coulomb-Mohr Theory Fig. 5 13

36 36 Coulomb-Mohr Theory Incorporating factor of safety For ductile material, use tensile and compressive yield strengths For brittle material, use tensile and compressive ultimate strengths

37 37 Coulomb-Mohr Theory Consider three cases Case 1: A B 0, 1 = A and 3 = 0 Eq. (5 22) reduces to Case 2: A 0 B, 1 = A and 3 = B Eq. (5-22) reduces to Case 3: 0 A B, 1 = 0 and 3 = B Eq. (5 22) reduces to

38 38 Coulomb-Mohr Theory Fig. 5 14

39 39 Coulomb-Mohr Theory Intersect pure shear load line with the failure line to determine the shear strength

40 Example 5-2 A 25-mm-diameter shaft is statically torqued to 230 N m. It is made of cast 195-T6 aluminum, with a yield strength in tension of 160 MPa and a yield strength in compression of 170 MPa. It is machined to final diameter. Estimate the factor of safety of the shaft.

41 Example 5-3 A certain force F applied at D near the end of the 15-in lever shown in Fig. 5 16, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. This bar (OABC) is of AISI 1035 steel, forged and heat-treated so that it has a minimum (ASTM) yield strength of 81 kpsi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force.

42 Example 5-3

43 Example 5-4 The cantilevered tube shown in Fig is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tube from Table A 8 using a design factor n d = 4. The bending load is F = 1.75 kn, the axial tension is P = 9.0 kn, and the torsion is T = 72 N m. What is the realized factor of safety?

44 Example 5-4 Fig. 5 17

45 45 Failure Theories for Brittle Materials Experimental data indicates some differences in failure for brittle materials. Failure criteria is generally ultimate fracture rather than yielding Compressive strengths are usually larger than tensile strengths

46 46 Max Normal Stress Theory (MNS) Failure occurs when max principal stress in a stress element exceeds the strength. Predicts failure when For plane stress, Incorporating design factor,

47 47 Max Normal Stress Theory (MNS) Unsafe in part of fourth quadrant Not recommended for use historical comparison Fig. 5 18

48 48 Brittle Coulomb-Mohr Failure equations dependent on quadrant Quadrant condition Failure criteria Equation No.

49 49 Brittle Coulomb-Mohr Failure equations dependent on quadrant Fig. 5 14

50 50 Brittle Failure Experimental Data Coulomb-Mohr is conservative in 4 th quadrant Modified Mohr criteria adjusts to better fit the data in the 4 th quadrant Fig. 5 19

51 51 Modified-Mohr Quadrant condition Failure criteria Equation No.

52 Example 5-5 Consider the wrench in Ex. 5 3, Fig. 5 16, as made of cast iron, machined to dimension. The force F required to fracture this part can be regarded as the strength of the component part. If the material is ASTM grade 30 cast iron, find the force F with a. Coulomb-Mohr failure model. b. Modified Mohr failure model.

53 Example 5-5 Fig. 5 16

54 54 Selection of Failure Criteria