E311 achine Design Lecture 6: Fluctuating Fatigue and the Goodman Diagram; Impact W Dornfeld 19Oct17 Fairfield University chool of Engineering Fluctuating Fatigue o far we have discussed loading that ernately went from tension to compression with the extremes equal and opposite. Now we look at the more general case where there could be a value. This is called Fluctuating Fatigue, and is characterized by both a and an ernating component. + min min verage or range tress mplitude teady tress Hamrock ection 7. 1
+ min min Fluctuating or Fully Reversing? 8 4 4-4 4 Effect of ean tress This axis is the fraction that is of the Endurance trength, e Circles represent failures s the ean stress increases, the material breaks at a lower lternating stress amplitude. This axis is the fraction that is of the Ultimate trength,
Fluctuating Fatigue Diagrams everal ways are available to characterize the Fluctuating Fatigue behavior. Two common ones are: Goodman line and Gerber line. Both approximate the material behavior. We will use the Goodman line because it is simpler and conservative. The odified Goodman Diagram is the red line and is the Goodman line truncated by the Yield line. odified Goodman Diagram Hamrock ection 7.1 The diagram is based on material properties, y, and e. s with the -N curve, e should be derated to reflect your part: e k f k s k r k t k m e (y Part) (Test pecimen) Drawing the odified Goodman Diagram For torsional (shear) loading, use sy.577 y, us.67, and e for Torsion 3
Drawing the odified Goodman Diagram Then plot your ernating and stress. If your point is below the od Goodman line, the part should have unlimited life. Note: This is very different from the complete odified Goodman Diagram that Hamrock details on P. 178 179. We will not use that version it is pretty confusing. Factors of afety The Factor of afety depends on how the stresses behave. They might: 1. Grow proportionately. Only grow in 3. Only grow in ernating How they behave depends on the actual hardware and loading involved. lt. Only Proportional ean Only 4
If and Increase Proportionately To make the Goodman Line w/f.o.. go through (, ) : 1 e( ) n e 1 n 1 n e + e e /n Load Line /n Equations of the Goodman Line: e + e e ( For a Factor of afety of n : + 1) or 1 e ( n e (1 ) ) If and Increase Proportionately Lim e n 1 + 1 e e Equation to find ernating stress when operating point hits the FO line or Goodman line. e /n et n 1 to get this level /n me an 5
If Only Increases Equation of the Goodman Line: e(1 ) e a a e(1 Equation to find ernating stress when operating point hits the Goodman line. ) n a Fatigue Exercise Given a bar of steel with these properties Yield UltimateTensile Endurance 7 4 ksi 65 ksi 3 ksi. B. C. On a Goodman Diagram, predict fatigue for these loadings: in ax ean lt 36 ksi -7 37 ksi 14 3 ksi 6 lternating tress (ksi) 5 4 3 1 1 3 4 5 6 7 ean tress (ksi) 6
Fatigue Exercise Yield UltimateTensile Endurance 4 ksi 65 ksi 3 ksi. B. C. in ax ean lt 36 18 18 ksi -7 37 5 3 ksi -18 46 14 3 ksi 45 4 lternating tress (ksi) 35 3 5 15 B C 1 5 5 1 15 5 3 35 4 45 5 55 6 65 7 ean tress (ksi) Fatigue Diagram for Finite Life From EngRasp ETBX 7
nother Type of Fatigue Diagram Note Finite Life curves 5 ksi 7 ksi tress Ratios for This Diagram 5 ksi 7 ksi min tress ratio R mplitude ratio 8
ore Cantilever Beam Ignore tress Concentration e k f k s k r e (.63)(1)(.87)(1) 54.8 ksi Details 1 Gauge (.194 thick).75 in. wide 4 in. long High trength teel, with 45 ksi achined finish Room Temperature CE : Tip is flexed between.75 in and.5 in. What is life for 95% survival? By proportioning, the force now fluctuates between 8.631 lb and 3 x 8.631 5.893 lb. tresses go from +3.1 ksi to +69.3 ksi. Tip Deflection (in).5.15.75 69.3 46. 3.1 tress (ksi) Cantilever Beam, contd. Tip Deflection (in).5.15.75 69.3 46. 3.1 tress (ksi) + min 69.3 + 3.1 46. ksi min 69.3 3.1 3.1ksi lternating tress (ksi) 6 e 4 4 6 8 1 1 14 16 18 4 6 ean tress (ksi) 9
46. ksi 3.1ksi lternating tress (ksi) 6 e 4 Cantilever Beam FO 4 6 8 1 1 14 16 18 4 6 ean tress (ksi) What is the Factor of afety?. If both ernating and stresses increase proportionately: 1 3.1 46. + +.4 +.189.611 n n 54.8 45 e B. If only ernating stress increases: 46. a e(1 ) 54.8(1 ) 54.8(1.189) 44.4ksi 45 n a 44.4 1.9 3.1 1.611 1.64 The energy of the falling block is transferred to stored energy in the spring: 1 W ( h + δ ) ( k δ ) δ tatic displacement of Weight gently placed on pring of stiffness, k W δ st k Impact factor is I m δ P h 1+ 1 δ W + δ static st Impact k W h δ W Effect is dependent on pring tiffness. soft spring s large static deflection, which s smaller Impact factor. Impact significantly increases forces! 1
For this situation, what is: 1. The static deflection. The Impact factor 3. The deflection 4. The force Impact 1Lb 1 1Lb 1 Lb/in δ Impact For the case of a weight sliding horizontally with a velocity, V, and hitting the spring δ WV gk where g is the gravitational constant, 386 in/s or 9.8 m/s. Recognizing that W/k δ st δ δ g stv W V W δ k 11
Impact For this setup, what velocity gives the same force as the falling weight just did? 1Lb/in V 1Lb δ 1Lb Impact Read Hamrock s Example 7.11 of a diver landing on a diving board. Note that the spring here is a beam, whose stiffness is calculated as Force/Deflection. lso note that he deflects the end of the board 9mm [3.6 ], and sees a force of 13.5kN [335lb]! 1