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On the base of engineering mechanics, this chapter will develop more specific understanding in failure theories and material strength for machinery design. The failure usually reflect one of the most important perspectives ([pE5spektiv]) in assessing the safety of mechanical system and components. 3.1 THEORIES OF FATIGUE The maximum-normal-stress theory (the first strength theory) The maximum-normal-strain theory (the second strength theory) The maximum-shear-stress theory (the third strength theory) The distortion-energy theory (the fourth strength theory) 3.2 BULK STRENGTHS OF THE MACHINE COMPONENTWS 3.2.1 Load and Stress static load static stress Load stress varying load varying (fluctuating) stress Load: Loads are the external action between two bodies. In terms of the behaviour, load can be presented in a fashion of force, bending moment and torque. Static load: The magnitudes or directions of the load remain unchanged or change a little and slowly within a given time. Varying load: The magnitudes or direction of the load is continuously time-varying. Stress: Stresses are the external resistances or forces, which are set up in the material when a load acts on a component. Static stress: static stresses are the stresses whose magnitudes and directions remain unchanged or change a little and slowly within a given time.
σ σ=constant o t
Varying (fluctuating) stress: A varying stress varies its magnitude and directions all the time. Static or varying load both can cause varying stress.
3. limit of fatigue. (e) Repeated alternating (pulsant) stress. Allowable stress: Calculate the allowable maximum of stress Safety factor : S= σ lim [σ ] Calculated value of safety factor . calculate the stress that is equal to simple tension Ultimate stress: Some utmost of mechanical characters of material Strength utmost: off set limit. (d) Sinusoidal fluctuating stress. (f) Completely reversed (symmetrical) stress Stress Working stress: Using the formula in mechanics of materials calculate the stress on section plane of element Calculated stress: Followed the theory of strength.(b) and (c) Non-symmetrical (or nonsinusoidal) fluctuating stresses.Fig.1 Stress-time relationships for some typical stress (a) Static stress.
3.3.2 Stress-Strength Design Method Stress-strength method of factor-of-safety method of design is a method as old as engineering design itself. 3.2.3. We just need to pay attention to the fatigue theories.1 The Strength Calculation of Machine Elements Under the Static Stresses σ lim σ ca ≤ = [σ ] S Where σ ca ——calculated stress σ lim ——ultimate stress (strength utmost. off set limit) S——safety factor [σ]——allowable stress 3. ② The discrepancy of mechanical model and actual state； ③ The nonhomogeneity of mechanical characters ； ④ The importance of occasion that the elements apply in. That is why it is often called the classical method of design.2 The Strength Calculation of Machine Elements Under the Static Stresses 3. The maximum stress condition: σ max ≤ [σ ] or σ max ≤ τ max ≤ [τ ] or τ max ≤ The safety factor condition: σ lim [ Sσ ] τ lim [ Sτ ] σ lim ≥ [ Sσ ] σ τ Sτ = lim ≥ [ Sτ ] τ Sσ = 3.3.1 Cycle Performance and Fatigue Limit .2.S ca = σ lim σ ca The reason of introduce the safety factor: ① The inexactness in calculation of stress.3 STRENGTH DESIGN UNDER STATIC VARYING STRESSES Static strength: Both the design for ductile materials and brittle materials are not so difficult under static stresses. It mainly includes two parts.
1. static stress σ σa σmx a o T σa σ σmx a to σm σm in Sinusoidal fluctuating stress σm in r =+1 r =-1 σa σ σmx a σa to σm in r =0 σa σm σa t Completely reversed stress Repeated alternating stress 2. It taken the limit of fatigue of material by the parameter.3.Cycle Performance The parameters used to describing the stress condition： σmin=minimum stress σmax=maximum stress σa =alternating stress (or stress amplitude) σm=mean stress σr=stress range σs=steady . completely reversed stress σ min r= = 0. .or static stress From fig. based on the experiment we can get the curve from. the mean stress and alternating stress can be calculated by σ max + σ min 2 σ max − σ min σa = 2 σm = Cycle Performance : -1. repeated alternating stress σ max 1. σ-N fatigue curve (material fatigue curve Ⅰ) The σ-N fatigue curve is shown as follow.1.
In the origin. so the value is the strength limit σ B . . It means the material will be snapped when it is loaded to the maximum. the number of stress cycles is N=1/4.
Since point D: Curve goes to be horizontal.so this kind of destroy is called low-cycle fatigue or strain fatigue. σ max declines. σ r ∞ .AB ： N< 103 . it is called infinite life. σ rN N = σ rm N0 = C σB σrN σr σmx a AB C D N N N0≈107 N=1/4 103 104 . N ≈ 104 . Sometimes it is necessary to do the strength check of static stress for the element which have prodigious peak value but little action. We use N O and its fatigue limit σ r m approximatively stand for N D . at the point C. σ max has little changed. BC ： As the N augments. CD: As the N augments. σ max declines rapidly. it means as the N augments no longer declines. so it can be approximated to static stress strength. when doing the fatigue experiment we always ordain a cycle index N O (cycle radix). m (N Describe as follow: σ rN N = C C ≤ N ≤ ND ) Explanation: machinery design mainly discuss the high cycle fatigue. N ≈ 106 : 25 ×107 The equation is: σ rN = σ r∞ （N > ND ) Because of the value of N D is large. Most fatigue of machine element happen in this phase. this phase is called high cycle fatigue or finite lifetime.
High circle fatigue: CD-----fatigue for finite life After D----fatigue for infinite life σ rN = σ r m Kn = m N0 N N0 life factor N Eg: In the strength calculation of gear and worm. the value of is definited by material testing. the material will never be destroyed. Two reduced method in practical application . use the following equation to calculate the allowable stress. In that way no matter how many times of the circles. it is called constant life fatigue curve. But since the point D. σr [σ ] = Kn S 3. Constant Life Fatigue Curve (Material Fatigue Curve Ⅱ) Material fatigue curve can also be shown by using the relationship among limit stress amplitudes in the given N. As the testing result shown in interval CD. after relevant numbers of varying stresses the check bar will happen endurance failure.interval CD: the relationship between N and fatigue limit σ rN ： σ rN = σ r m N0 N m σ N = r N0 σ rN Among the equation above. if the maximum stress less than the stress of the point D.
18.104.22.168.Stress Limit Curve( σ m − σ a Curve) Allowable fatigue design diagram: 1.The stress limit of material .
The explanation of stress limit： (1)Every point on the curve stands for limiting fatigue stress of material under the different cycle performance. )→D( . (2)Point A’ correspond the cycle of r=-1. Every point on line CG’ stands for the varying static condition that σ max = σ m + σ a = σ s . (4)The area inside the broken line A ' D ' G ' C ' is yielding and fatigue safety zone. size dimension. σ a = σ −1 = σ max Point D’ correspond the cycle of r=0. σ a = 0 . σ min = 0. (5)For simplifying the calculation. ) 2 2 2 2Kσ The equation of AD： σ −1 = K σσ ae '+ψσσ me ' Where ψ σ is a constant with regarding to the material and can be determined by fatigue tests of following equation: . the fatigue value of material is above and beyond the Fatigue value of element. use the equation but not the curve.σ-1)→A（0， ）or（0， σ −1 ） D’( σ0 σ0 σ0 σ0 . 2. σ m = 0.The stress limit of element ： Because of the geometrical shape. σ m = σ a = σ 0 / 2 = σ max / 2 Line OC correspond the static stress r=1. (3) A straight line is drawn from point C with an angle of 45°to the mean-stress axis and stops at point G.choose the coordinate values of point C be equal to the off set limit σ s of material.Surface texture and intensifying factor. fatigue value of material fatigue limit fatigue value of element influence coefficient K σ——bending fatigue limit influential factors K Kτ——shear fatigue limit influential factors = eg： σ −1 σ − 1e σ-1e——symmetric cycle bending fatigue limit of element σ-1 ——symmetric cycle bending fatigue limit of material Amending the stress limit of material diagram A’G’: The coordinate value changed: Kσ = A’(0.
the similar form linear equation can get τ − 1 = K ττ ae '+ψττ me ' τ ae '+ τ me ' = τ s 3. 1.It is the stress condition of the majority of the rotating shaft. and then definite the operating point M on the limit stress curve of element . first calculate the maximum (minimum) stress on minimum life section.the cycle performance of the varying stress remains the same . σ min holds the value on minimum life section Or τ max .τ min holds the value on minimum life section.2. Pure static varying stress: one of the normal or tangential stress Doubleaction static varying stress: both normal and tangential stress.ψσ = ψ ----material characteristic 2σ −1 − σ 0 σ0 ψ σ ---. Which point is the value of the limit stress depends on the varying discipline of the element working stress.3.material shearing characteristic Whereψ σ is a constant with regarding to the material and can be determined by fatigue tests of following equation: 2σ −1 − σ 0 σ0 ψσ = The equation of line CG( considered by static stress ) σ ae '+ σ me ' = σ s In the condition of shearing stress . .3 The Strength Calculation of Machine Elements Under the Static Stresses The characteristic of static stress: σ max . The design method: theoretical design Use for calculation: safety factor checking σ lim Sca = ≥S σ Machine element’s fatigue value of pure static varying stress when calculating the endurance bending strength of machine element.In the case of r=C (a constant). second calculate the.material bending characteristic ψ τ --.
the mean stress of the varying stress remains the same.σ a σ max − σ min 1 − r = = =C' σ m σ max + σ min 1 + r The limit stress of the point M ' ' ' σ max = σ ae + σ me = σ −1 (σ m + σ a ) σ −1σ max = Kσ σ a + ψ σ σ m Kσ σ a + ψ σ σ m The calculation of safety factor and stress condition: Sca = Static stress condition: Sca = ' σ −1 σ lim σ max = ≥S = σ max Kσ σ a + ψ σ σ m σ σ σs σ lim = s = ≥S σ σ max σ a + σ m 2. It is the stress condition of the vibrating loaded spring. In the case of σ m = C . The limit stress of point M: ' σ max = σ −1e + σ m (1 − ψσ σ + ( Kσ −ψ σ )σ m ) = −1 Kσ Kσ The calculation of safety factor and stress condition: ' σ lim σ max σ −1 + ( Kσ −ψ σ )σ m Sca = = = ≥S σ σ max Kσ (σ a + σ m ) Static stress condition: .
etc. Sca = σ −1 Kσ σ a + ψ σ σ m = t he f at i gue l i mt of symet ri c cycl e i m ≥S working stress amplitude × stress amplitude influence coefficient Also .In the case of σ min = C theminimum stress of the varying stress remains the same. for the shear stress. . σ min = σ m − σ a = C The limit stress of point M: ' ' ' σ max = σ me + σ ae = 2σ −1 + ( Kσ −ψ σ )σ min Kσ + ψ σ The calculation of safety factor and stress condition: ' σ lim σ max 2σ −1 + ( Kσ −ψ σ )σ min Sca = = = ≥S σ σ max ( Kσ +ψ σ )(σ a + σ min ) illustration：when the law of how the stress varies is dubious.we always choose the calculation method of r=C. the only thing we have to do is using the τ instead of σ in the equation above. tyre and road sense. vehicle speed. Machine element’s fatigue value of pure unstable varying stress Eg: The stress condition of automobile leaf-spring is affected by the load. pavement.Sca = σ σs σ lim = s = ≥S σ σ max σ a + σ m 3. It is the stress condition of the bolt joining under the axial varying load.
the material damage ratio of stress σ 1 is 1/ N1 in every circle. so in n1 circles the material .Fig3-9 regular unstable varying stress diagram Fig3-10 unstable varying stress on the σ-N curve Adopt: fatigue damage accumulation hypothesis n1 n2 n3 + + =1 N1 N 2 N 3 Regular unstable varying stress diagram Unstable varying stress on the σ − N curve If every circle of stresses make the equicohesive destruct on the material.
damage ratio of σ 1 is n1 / N1 and in n2 circles the material damage ratio of σ 2 is n2 / N 2 . So we obtain: n n n n ∑ Ni = N1 + N2 + N3 + . we can believe that the stress is not damaging. + nzσ z ) = i =1 m = 1 m N 0σ −1 N0 σ −1 ∑nσ i =1 i z m i m < N 0σ −1 If under these stress the material isn’t destroyed.. the calculation value . = 1 i =1 i 1 2 3 z σ w here：N i = N 0 −1 σi m As the experiment shown： 1)When the sequence of stress operation is increased and then decreased.. we obtain: ∑N i =1 ni i >1 In general case: ∑N i =1 z ni i = 0..the material happens to endurance failure. Eg: when the shaft is under the action of bending and torsion. we obtain: ∑N i =1 z z ni i <1 2) When the sequence of stress operation is decreased and then increased..7 ~ 2.2 z In limit case: ∑ niσ im 1 m ( n1σ 1m + n2σ 2m + ... If the stress is lower than σ −1∞ .. When the damage ratio is up to 100%. the strength condition is : 1 N0 ∑nσ i =1 i z m i Sca = σ −1 ≥S σ ca The fatigue value of doubleaction static varying stress The element which is under the alternate stress of bending and torsion is belong to this kind of fatigue value.we can obtain: Let the calculation stress of unstable stress to be: σ ca = m So σ ca < σ −1 .
3 the Measure for Increasing the Fatigue Value of Machine Element In the design phase the design measure should be taken to increasing the fatigue value of machine element : ①. When the working points are on fatigue safety zone: Sσ a = k N σ −1 ≥ [ Sσ a ] (kσ ) D σ a +ψ σ σ m Whereψ σ is a constant with regarding to the material and can be determined by fatigue tests of following equation: 2σ −1 − σ 0 σ0 ψσ = When the working points are on plastic safety zone: Sσ = σs ≥ [ Sσ ] σa +σs Illumination : Two other cases (σm =constant and σmin =constant) are not given in detail here for they are not very common. . the calculation value of safety factor Sτ = τ −1 K ττ a +ψττ m 3. ④. ⑤.of safety factor and strength condition is : Sca = Among this: Sσ ——only under the normal stress σ . Lower the stress concentration as much as possible； ②. Increase the surface texture of machine element. ③.Choose the heat treatment method and reinforced technics to increasing the fatigue value of machine element.Decrease or eliminate the original crack size of the element surface. For the case of constant ratio r. Choose the material of high fatigue value.3. the calculation value of safety factor Sσ = S σ Sτ 2 Sσ + Sτ2 ≥S σ −1 K σσ a +ψσσ m Sτ ——only under the shearing stress τ .
5. the critical number of cycles is N 0 = 107 .4 The material of shaft is 40Cr. Kτ = 1. can draw the simplified limit stress diagram by knowing some basic mechanical characters ( σ B . Can use the line graph and numerical tables in the appendix.index of life is m=9. the life factor K N andfatigue limit σ -1N . off set limit σ s = 355MPa. Grasp the checking method of doubleaction varying stress. and know the concept of equivalent stress. m.Estimate: when the number of cycles Nis 105 . its mechanical character is ψ 0 = 0. σ −1 .5 × 106 . limit influential factors Kσ = 2. the bending momentM=300Ng the torque T=800Ng fatigue m. Example3. The diameter of the its minimum life section is d=40mm.1 The fatigue limit of symmetric σ -1 = 275Mpa. σ S . . Example 3. Grasp the calculation method of strength under the pure varying stress.1. σ 0 )and the geometrical property.2. or 108 .MAIN CONTENTS OF THE CHAPTER Knowing the significance and use of the fatigue curve and limit stress diagram. σ −1 = 355MPa. thermal refining.5.ψ τ = 0.τ −1 = 205MPa. Know the significance and application mode of fatigue damage accumulation hypothesis (Miner law).
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