FME461 Engineering Design II

Transcription

1 FME461 Engineering Design II Dr.Hussein Jama Office 414 Lecture: Mon 8am -10am Tutorial Tue 3pm - 5pm 10/1/2013 1

2 Semester outline Date Week Topics Reference Reading 9 th Sept 1 House keeping issues Introduction to mechanical design Assignment 1 is given out Ch 1 - Norton, Shigley 16 th Sept 23 rd Sept 2 Ethics & safety Various 3 Assignment 1 is due Assignment 2 is given out Static & Fatigue Failure Various Ch 5 Shigley Ch 6 Shigley 30 th 4 FMEA Various Sept 7 th Oct 5 Continuous Assessment Test 1 (15%) Assignment 2 is due Assignment 3 is given out 14 th Oct 6 Shafts and shaft components Ch 7 Shigley 21 st Oct 7 Welding and permanent joints Ch 9 Shigley 28 th Oct 8 Mechanical springs Ch 10 Shigley 4 th Nov 9 Clutches & brakes Ch 16 Shigley 11 th Nov 10 Belts and chains Ch 17 Shigley 18 th Nov 11 Statistical consideration Ch 20 Shigley 25 th Nov 12 Continuous Assessment Test 2 (15%) 10/1/ nd Dec 13 Presentation of assignment 2 Assignment 2 is due

3 Discussion Shigley Chapter 5 - Static failure criteria Ductile materials Brittle materials Shigley Chapter 6 Fatigue failure criteria 10/1/2013 3

4 Static & Fatigue Failure Static load a stationary load that is gradually applied having an unchanging magnitude and direction Failure A part is permanently distorted and will not function properly A part has been separated into two or more pieces. 10/1/2013 4

5 Static failure theories Maximum shear Stress theory Distortion energy theory Ductile Coulomb-Mohr theory 10/1/2013 5

6 Definitions Material Strength S y = Yield strength in tension, S yt = S yc S ys = Yield strength in shear S u = Ultimate strength in tension, S ut S uc = Ultimate strength in compression S us = Ultimate strength in shear =.67 S u 10/1/2013 6

7 Ductile and brittle materials A ductile material deforms significantly before fracturing. Ductility is measured by % elongation at the fracture point. Materials with 5% or more elongation are considered ductile. Brittle material yields very little before fracturing, the yield strength is approximately the same as the ultimate strength in tension. The ultimate strength in compression is much larger than the ultimate strength in tension. 10/1/2013 7

8 Failure theories Ductile materials Maximum shear stress theory (Tresca 1886) ( ma specimen of the same material when that specimen x ) component > ( ) obtained from a tension test at the yield point Failure = S y To avoid failure = S y 2 ( max ) component < S y 2 = S y max = S y 2 n n = Safety factor 10/1/ =S y Design equation

9 Max. Shear Theory The maximum-shear-stress theory predicts that yielding begins whenever the maximum shear stress in any element equals or exceeds the maximum shear stress in a tension test. 10/1/2013 9

10 Failure theories Ductile materials Distortion energy theory (von Mises-Hencky) Simple tension test (S y ) t Hydrostatic state of stress (S y ) h t (S y ) h >> (S y ) t h Distortion contributes to failure much more than change in volume. h h t (total strain energy) (strain energy due to hydrostatic stress) = strain energy due to angular distortion > strain energy obtained from a tension test at the yield point failure 10/1/

11 Plane stress problems 10/1/

12 Stress components 10/1/

13 Failure theories ductile materials The area under the curve in the elastic region is called the Elastic Strain Energy. U = ½ ε 3D case U T = ½ 1 ε 1 + ½ 2 ε 2 + ½ 3 ε 3 Stress-strain relationship ε 1 = 1 E v 2 E v 3 E ε 2 = 2 E v 1 E v 3 E ε 3 = E v 3 1 E v 2 E 1 U T = ( ) - 2v ( ) 2E 10/1/

14 Failure theories Ductile materials Distortion strain energy = total strain energy hydrostatic strain energy 1 U d = U T U h U T = ( ) - 2v ( ) 2E (1) Substitute 1 = 2 = 3 = h U h = ( h + h + 2 h ) - 2v ( h h + h h+ h h) 2E Simplify and substitute = 3 h into the above equation 2 3 h U h = (1 2v) = 2E U d = U T U h = 6E ( ) 2 (1 2v) Subtract the hydrostatic strain energy from the total energy to obtain the distortion energy 1 + v 10/1/ E ( 1 2 ) 2 + ( 1 3 ) 2 + ( 2 3 ) 2 (2)

15 Failure theories Ductile materials Strain energy from a tension test at the yield point 1 = S y and 2 = 3 = 0 Substitute in equation (2) U d = U T U h = 1 + v 6E ( 1 2 ) 2 + ( 1 3 ) 2 + ( 2 3 ) 2 U test = (S y ) v 3E To avoid failure, U d < U test ( 1 2 ) 2 + ( 1 3 ) 2 + ( 2 3 ) 2 2 ½ < S y 10/1/

16 Failure theories ductile materials ( 1 2 ) 2 + ( 1 3 ) 2 + ( 2 3 ) 2 2 ½ < S y 2D case, 3 = 0 ( 1 ½ ) < S y = Where is von Mises stress = S y n Design equation 10/1/

17 Failure theories -Ductile materials Pure torsion, = 1 = 2 ( ) = S y = S y 2 S ys = S y / 3 S ys =.577 S y Relationship between yield strength in tension and shear If y = 0, then 1, 2 = x/2 [( x )/2] 2 + ( xy ) 2 the design equation can be written in terms of the dominant component stresses (due to bending and torsion) ( x) 2 + 3( xy ) 2 S y = n 10/1/ /2

18 Summary Ductile materials 10/1/

19 Design process Distortion energy theory Maximum shear stress theory S y S y = max = n 2n Select material: consider environment, density, availability S y, S u Choose a safety factor n Size Weight Cost The selection of an appropriate safety factor should be based on the following: Degree of uncertainty about loading (type, magnitude and direction) Degree of uncertainty about material strength Uncertainties related to stress analysis Consequence of failure; human safety and economics Type of manufacturing process Codes and standards 10/1/

20 Flow process 10/1/

21 Design Process Use n = 1.2 to 1.5 for reliable materials subjected to loads that can be determined with certainty. Use n = 1.5 to 2.5 for average materials subjected to loads that can be determined. Also, human safety and economics are not an issue. Use n = 3.0 to 4.0 for well known materials subjected to uncertain loads. 10/1/

22 Design Process Select material, consider environment, density, availability S y, S u Choose a safety factor Formulate the von Mises or maximum shear stress in terms of size. Use appropriate failure theory to calculate the size. = S y n max = S y 2n Optimize for weight, size, or cost. 10/1/

23 Example from Shigley 10/1/

24 Solution 10/1/

25 Failure theories- brittle materials One of the characteristics of a brittle material is that the ultimate strength in compression is much larger than ultimate strength in tension. S uc >> S ut Mohr s circles for compression and tension tests. S uc 3 Stress 1 state S ut Compression test Tension test Failure envelope The component is safe if the state of stress falls inside the failure envelope. 1 > 3 and 2 = 0 10/1/

26 Failure theories brittle materials Modified Coulomb-Mohr theory 3 or 2 3 or 2 S ut S ut S uc Safe Safe Safe S ut 1 I II S ut 1 Safe -S ut -S ut III S uc S uc Cast iron data Three design zones 10/1/

27 Failure theories brittle materials Zone I 3 1 > 0, 2 > 0 and 1 > 2 1 = S ut n Design equation S ut I II 1 Zone II -S ut III 1 > 0, 2 < 0 and 2 < S ut 1 = S ut n Design equation S uc Zone III 1 > 0, 2 < 0 and 2 > S ut 1 ( 1 1 S ut ) S uc 2 = S uc 1 n 10/1/ Design equation

28 Summary Brittle materials 10/1/

29 Example 10/1/

30 Solution 10/1/

31 Solution continued 10/1/

32 Static failure summary - Ductile 10/1/

33 Summary Brittle materials 10/1/

34 Failure theories - Fatigue It is recognised that a metal subjected to a repetitive or fluctuating stress will fail at a stress much lower than that required to cause failure on a single application of load. Failures occurring under conditions of dynamic loading are called fatigue failures. Fatigue failure is characterized by three stages Crack Initiation Crack Propagation Final Fracture 10/1/

35 Jack hammer component, shows no yielding before fracture. Crack initiation site Propagation zone, striation Fracture zone 10/1/

36 Example VW crank shaft fatigue failure due to cyclic bending and torsional stresses Propagation zone, striations Crack initiation site Fracture area 10/1/

37 928 Porsche timing pulley 10/1/ Crack started at the fillet

38 Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure. 25mm diameter steel pins from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel 10/1/

39 bicycle crank spider arm This long term fatigue crack in a high quality component took a considerable time to nucleate from a machining mark between the spider arms on this highly stressed surface. However once initiated propagation was rapid and accelerating as shown in the increased spacing of the 'beach marks' on the surface caused by the advancing fatigue crack. 10/1/

40 Crank shaft Gear tooth failure 10/1/

41 Hawaii, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin area rips off in mid-flight. Metal fatigue was the cause of the failure. 10/1/

42 Fracture surface characteristics Mode of fracture Ductile Typical surface characteristics Cup and Cone Dimples Dull Surface Inclusion at the bottom of the dimple Brittle Intergranular Brittle Transgranular Shiny Grain Boundary cracking Shiny Cleavage fractures Flat Fatigue Beachmarks Striations (SEM) Initiation sites Propagation zone Final fracture zone 10/1/

43 Fatigue failure type of fluctuating stresses Alternating stress min = 0 a = m = max / 2 10/1/ a = Mean stress = m max min 2 max + 2 min

44 Fatigue Failure, S-N Curve Typical testing apparatus, pure bending Test specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied. Motor Load Rotating beam machine applies fully reverse bending stress 10/1/

45 Fatigue Failure, S-N Curve Finite life Infinite life S e S e = endurance limit of the specimen 10/1/

46 Relationship Between Endurance Limit and Ultimate Strength Steel 0.5S ut S e = 100 ksi 700 MPa S ut 200 ksi (1400 MPa) S ut > 200 ksi S ut > 1400 MPa Cast iron Cast iron S e = 0.4S ut S ut < 60 ksi (400 MPa) S ut 60 ksi 24 ksi 160 MPa S ut < 400 MPa 10/1/

47 Relationship Between Endurance Limit and Ultimate Strength 0.4S ut S e = 19 ksi 130 MPa Aluminium S ut < 48 ksi (330 MPa) S ut 48 ksi S ut 330 MPa Copper alloys 0.4S ut S e = 14 ksi 100 MPa Copper alloys S ut < 40 ksi (280 MPa) S ut 40 ksi S ut 280 MPa For N = 5x10 8 cycle 10/1/

48 For materials exhibiting a knee in the S-N curve at 10 6 cycles S = endurance limit of the specimen (infinite life > 10 6 e ) S e = endurance limit of the actual component (infinite life > 10 6 ) S S e N For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x10 8 cycles S f S f = fatigue strength of the specimen (infinite life > 5x10 8 ) = fatigue strength of the actual component (infinite life > 5x10 8 ) S S f 10/1/ x10 8 N

49 Correction factor s for specimen s endurance limit S e = C load C size C surf C temp C rel (S e ) S f = C load C size C surf C temp C rel (S f ) Load factor, C load (page 326, Norton s 3 rd ed.) Pure bending C load = 1 Pure axial C load = 0.7 Pure torsion Combined loading C load = 1 C load = 1 if von Mises stress is used, use if von Mises stress is NOT used. 10/1/

50 Correction factor s for specimen s endurance limit Size factor, C size (p. 327, Norton s 3 rd ed.) Larger parts fail at lower stresses than smaller parts. This is mainly due to the higher probability of flaws being present in larger components. For rotating solid round cross section d 0.3 in. (8 mm) C size = in. < d 10 in. C size =.869(d) mm < d 250 mm C size = 1.189(d) If the component is larger than 10 in., use C size =.6 10/1/

51 Correction factor s for specimen s endurance limit For non rotating components, use the 95% area approach to calculate the equivalent diameter. Then use this equivalent diameter in the previous equations to calculate the size factor. d 95 =.95d d A 95 = (π/4)[d 2 (.95d) 2 ] =.0766 d 2 d equiv = ( A )1/2 Solid or hollow non-rotating parts Rectangular parts 10/1/ d equiv =.37d d equiv =.808 (bh) 1/2

52 Correction factor s for specimen s endurance limit I beams and C channels 10/1/

53 Correction factor s for specimen s endurance limit surface factor, C surf (p , Norton s 3 rd ed.) The rotating beam test specimen has a polished surface. Most components do not have a polished surface. Scratches and imperfections on the surface act like a stress raisers and reduce the fatigue life of a part. Use either the graph or the equation with the table shown below. C surf = A (S ut ) b 10/1/

54 Correction factor s for specimen s endurance limit Temperature factor, C temp (p.331, Norton s 3 rd ed.) High temperatures reduce the fatigue life of a component. For accurate results, use an environmental chamber and obtain the endurance limit experimentally at the desired temperature. For operating temperature below 450 o C (840 o F) the temperature factor should be taken as one. C temp = 1 for T 450 o C (840 o F) 10/1/

55 Correction factor s for specimen s endurance limit Reliability factor, C rel (p. 331, Norton s 3 rd ed.) The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit). 10/1/

56 Fatigue Stress Concentration Factor, K f Experimental data shows that the actual stress concentration factor is not as high as indicated by the theoretical value, K t. The stress concentration factor seems to be sensitive to the notch radius and the ultimate strength of the material. Fatigue stress concentration factor K f = 1 + (K t 1)q Notch sensitivity factor (p. 340, Norton s 3 rd ed.) Steel 10/1/

57 Fatigue Stress Concentration Factor, K f for Aluminum (p. 341, Norton s 3 rd ed.) 10/1/

58 Design process Determine the maximum alternating applied stress ( a ) in terms of the size and cross sectional profile Select material S y, S ut Choose a safety factor n Determine all modifying factors and calculate the endurance limit of the component S e Determine the fatigue stress concentration factor, K f Use the design equation to calculate the size S e K f a = n Investigate different cross sections (profiles), optimize for size or weight You may also assume a profile and size, calculate the alternating stress and determine the safety factor. Iterate until you obtain the desired safety factor 10/1/

59 Design for finite life A S S n = a (N) b equation of the fatigue line B S e N Point A S n =.9S ut N = 10 3 Point A S n =.9S ut N = 10 3 Point B S n = S e N = 10 6 Point B S n = S f N = 5x /1/

60 Design for finite life S n = a (N) b log S n = log a + b log N log.9s ut = log a + b log 10 3 Apply boundary conditions for point A and B to find the two constants a and b a = (.9S ut ) 2 S e log S e = log a + b log 10 6 b = 1 3 log.9s ut S e S n = S e ( N ) ⅓ log ( 10 6 S e.9s ut ) Calculate S n and replace S e in the design equation S n K f a = n Design equation 10/1/

61 The effect of mean stress on fatigue life Mean stress exist if the loading is of a repeating or fluctuating type. a Mean stress is not zero Alternating stress S e Gerber curve Goodman line Soderberg line Mean stress S y S ut m 10/1/

62 The effect of mean stress on fatigue life goodman diagram a Yield line Alternating stress Safe zone C Goodman line S y Mean stress S ut m 10/1/

63 The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram a S y Yield line Goodman line Safe zone Safe zone - m - S yc S ut S y + m 10/1/

64 The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram Fatigue, m 0 a Fatigue, m > 0 a m 1 + = S e S ut n f Infinite life Yield a + m = S yc n y a = S e n f Safe zone S e Safe zone a m S ut + = 1 S n C Yield Finite life a + m = S y n y - m - S yc 10/1/

65 Applying Stress Concentration factor to Alternating and Mean Components of Stress Determine the fatigue stress concentration factor, K f, apply directly to the alternating stress K f a If K f max < S y then there is no yielding at the notch, use K fm = K f and multiply the mean stress by K fm K fm m If K f max > S y then there is local yielding at the notch, material at the notch is strain-hardened. The effect of stress concentration is reduced. Calculate the stress concentration factor for the mean stress using the following equation, Fatigue design equation K fm = S y K f a K f a K fm m 1 + = Infinite life S e S ut n f 10/1/ m

66 Combined loading All four components of stress exist, xa xm xya alternating component of normal stress mean component of normal stress alternating component of shear stress xym mean component of shear stress Calculate the alternating and mean principal stresses, 1a, 2a = ( xa /2) ( xa /2) 2 + ( xya ) 2 1m, 2m = ( xm /2) ( xm /2) 2 + ( xym ) 2 10/1/

67 Combined loading Calculate the alternating and mean von Mises stresses, 2 2 a = ( 1a + 2a - 1a 2a) 1/2 2 2 m = ( 1m + 2m - 1m 2m) 1/2 Fatigue design equation a m 1 + = Infinite life S e n f S ut 10/1/

68 Example /1/

69 Example 6-9 Shaft is made from SAE1050 steel and is cold formed 10/1/

70 10/1/

71 Solution 10/1/

72 Solution 10/1/

73 10/1/