Variable stresses in machine parts
- 1. Contents 1 Notations 2 2 Completely Reversed or Cyclic Stresses 3 3 Fatigue and Endurance Limit 3 4 Effect of Loading on Endurance Limit and Load Factor 4 5 Effect of Surface Finish on Endurance Limit and Surface Finish Factor 4 6 Effect of Size on Endurance Limit and Size Factor 4 7 Effect of Miscellaneous Factors on Endurance Limit 5 8 Relation Between Endurance Limit and Ultimate Tensile Strength 5 9 Factor of Safety for Fatigue Loading 5 10 Stress Concentration 5 11 Theoretical or Form Stress Concentration Factor 6 12 Stress Concentration due to Holes and Notches 6 13 Methods of Reducing Stress Concentration 7 14 Factors to be Considered while Designing Machine Parts to Avoid Fatigue Failure 8 15 Stress Concentration Factor for Various Machine Members 8 16 Fatigue Stress Concentration Factor 12 17 Notch Sensitivity 12 18 Combined Steady and Variable Stress 12 19 Gerber Method for Combination of Stresses 12 20 Goodman Method for Combination of Stresses 13 21 Soderberg Method for Combination of Stresses 14 22 Combined Variable Normal Stress and Variable Shear Stress 14 23 Application of Soderbergs Equation 15 24 Examples 17 25 References 31 26 Contacts 31
- 2. 1 Notations • σm = Mean or average stress (tensile or compres- sive). • σmax = Maximum applied stress. • σmin = Minimum applied stress. • σv = Reversed stress component or alternating or variable stress. • σe = Endurance limit for any stress range • σe = Endurance limit for completely reversed stresses • R = Stress ratio. • Kb = Load correction factor for the reversed or rotating bending load. Its value is usually taken as unity. • Ka = Load correction factor for the reversed axial load. Its value may be taken as 0.8. • Ks = Load correction factor for the reversed tor- sional or shear load. Its value may be taken as 0.55 for ductile materials and 0.8 for brittle materials • σeb = Endurance limit for reversed bending load. • σea = Endurance limit for reversed axial load. • τe = Endurance limit for reversed torsional or shear load. • Ksur = Surface finish factor. • σel = Surface finish endurance limit. • Ksz = Size factor. • σe2 = Size endurance limit. • Kr = Reliability factor. • KT = Temperature factor. • Ki = Impact factor. • F.S. = Factor of safety. • σy = Yield point stress. • Kt = Theoretical stress concentration factor for axial or bending loading. • Kts = Theoretical stress concentration factor for torsional or shear loading. • σu = Ultimate point stress. • Kf = Fatigue stress concentration factor. • τy = Yield point shear. • τu = Ultimate point shear. • σneb = Equivalent normal stress due to reversed bending loading. • σnea = Equivalent normal stress due to reversed axial loading. • σne = Total equivalent normal stress. • τes = Equivalent shear stress due to reversed tor- sional or shear loading • Wm = Mean or average load. • Wv = Variable load. • A = Cross-sectional area. • Mm = Mean bending moment. • Mv = Variable bending moment. • Z = Section modulus. • σb = Working or design bending stress. • Tm = Mean or average torque. • Tv = Variable torque. • d = Diameter of the shaft. • τm = Mean or average shear stress. • τv = Variable shear stress. • τ = Working or design shear stress. • Kfs = Fatigue stress concentration factor for tor- sional or shear loading. • k = Ratio of inner diameter to outer diameter.
- 3. 2 Completely Reversed or Cyclic Stresses For each revolution of the beam, the stresses are re- versed from compressive to tensile. The stresses which vary from one value of compressive to the same value of tensile or vice versa, are known as completely re- versed or cyclic stresses. Notes: 1. The stresses which vary from a minimum value to a maximum value of the same nature, (i.e. tensile or compressive) are called fluctuating stresses. Figure 1: Reversed or cyclic stresses. 2. The stresses which vary from zero to a certain maximum value are called repeated stresses. 3. The stresses which vary from a minimum value to a maximum value of the opposite nature (i.e. from a certain minimum compressive to a certain maximum tensile or from a minimum tensile to a maximum compressive) are called alternating stresses. 3 Fatigue and Endurance Limit It has been found experimentally that when a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue. The failure is caused by means of a progressive crack formation which are usually fine and of microscopic size. The failure may occur even without any prior indication. The fatigue of material is effected by the size of the component, relative magnitude of static and fluctuating loads and the number of load reversals. Figure 2: Time-stress diagrams. Mean or average stress, σm = σmax + σmin 2 Reversed stress component or alternating or variable stress, σv = σmax − σmin 2 Stress ratio, R = σmax σmin For completely reversed stresses, R = −1 and for repeated stresses, R = 0. It may be noted that R cannot be greater than unity. Endurance limit, σe = 3σe 2 − R
- 4. 4 Effect of Loading on Endurance Limit and Load Factor Endurance limit for reversed bending load, σeb = σeKb = σe Endurance limit for reversed axial load, σea = σeKa Endurance limit for reversed torsional or shear load, τe = σeKs 5 Effect of Surface Finish on Endurance Limit and Surface Finish Factor Figure 3: Surface finish factor for various surface conditions. Endurance limit, σel = σebKsur = σeKbKsur = σeKsur ...(For reversed bending load) = σeaKsur = σeKaKsur ...(For reversed axial load) = τeKsur = σeKsKsur ...(For reversed torsional or shear load) Note: The surface finish factor for non-ferrous metals may be taken as unity. 6 Effect of Size on Endurance Limit and Size Factor Endurance limit, σe2 = σelKsz = σebKsurf Ksz = σeKbKsurf Ksz = σeKsurKsz ...(For reversed bending load) = σeaKsurKsz = σeKaKsurKsz ...(For reversed axial load) = τeKsurKsz = σeKsKsurKsz ...(For reversed torsional or shear load) Notes: 1. The value of size factor is taken as unity for the standard specimen having nominal diameter of 7.657 mm. 2. When the nominal diameter of the specimen is more than 7.657 mm but less than 50 mm, the value of size factor may be taken as 0.85. 3. When the nominal diameter of the specimen is more than 50 mm, then the value of size factor may be taken as 0.75.
- 5. 7 Effect of Miscellaneous Factors on Endurance Limit Endurance limit, σe = σebKsurKszKrKT Ki ...(For reversed bending load) = σeaKsurKszKrKT Ki ...(For reversed axial load) = τeKsurKszKrKT Ki ...(For reversed torsional or shear load) Note: In solving problems, if the value of any of the above factors is not known, it may be taken as unity. 8 Relation Between Endurance Limit and Ultimate Tensile Strength Figure 4: Endurance limit of steel corresponding to ultimate tensile strength. σe = 0.5σu ...(For steel) = 0.4σu ...(For cast steel) = 0.35σu ...(For cast iron) = 0.3σu ...(For non-ferrous metals and alloys) 9 Factor of Safety for Fatigue Loading F.S. = Endurance limit stress Design or working stress = σe σd Note: σe = 0.8 to 0.9 σy ...(For steel) 10 Stress Concentration Whenever a machine component changes the shape of its cross-section, the simple stress distribution no longer holds good and the neighborhood of the discon- tinuity is different. This irregularity in the stress dis- tribution caused by abrupt changes of form is called stress concentration. It occurs for all kinds of stresses in the presence of fillets, notches, holes, key- ways, splines, surface roughness or scratches etc. Figure 5: Stress concentration.
- 6. 11 Theoretical or Form Stress Concentration Factor The theoretical or form stress concentration factor is defined as the ratio of the maximum stress in a member (at a notch or a fillet) to the nominal stress at the same section based upon net area. Mathematically, theoretical or form stress concentration factor, Kt = Maximum stress Nominal stress The value of Kt depends upon the material and geometry of the part. Notes: 1. In static loading, stress concentration in ductile materials is not so serious as in brittle materials, because in ductile materials local deformation or yielding takes place which reduces the concentration. In brittle materials, cracks may appear at these local concentrations of stress which will increase the stress over the rest of the section. It is, therefore, necessary that in designing parts of brittle materials such as castings, care should be taken. In order to avoid failure due to stress concentration, fillets at the changes of section must be provided. 2. In cyclic loading, stress concentration in ductile materials is always serious because the ductility of the material is not effective in relieving the concentration of stress caused by cracks, flaws, surface roughness, or any sharp discontinuity in the geometrical form of the member. If the stress at any point in a member is above the endurance limit of the material, a crack may develop under the action of repeated load and the crack will lead to failure of the member. 12 Stress Concentration due to Holes and Notches The maximum stress is given by σmax = σ 1 + 2a b and the theoretical stress concentration factor, Kt = σmax σ = 1 + 2a b When a/b is large, the ellipse approaches a crack transverse to the load and the value of Kt becomes very large. When a/b is small, the ellipse approaches a longitudinal slit and the increase in stress is small. When the hole is circular, then a/b= 1 and the maximum stress is three times the nominal value. Figure 6: Stress concentration due to holes. The stress concentration in the notched tension mem- ber, is influenced by the depth a of the notch and ra- dius r at the bottom of the notch. The maximum stress, which applies to members having notches that are small in comparison with the width of the plate, may be ob- tained by the following equation, σmax = σ 1 + 2a r Figure 7: Stress concentration due to notches.
- 7. 13 Methods of Reducing Stress Concentration Figure 8: Methods of reducing stress concentration. Figure 9: Methods of reducing stress concentration in cylindrical members with shoulders. Figure 10: Methods of reducing stress concentration in cylindrical members with holes. Figure 11: Methods of reducing stress concentration in cylindrical members with holes.
- 8. Figure 12: Methods of reducing stress concentration of a press fit. 14 Factors to be Considered while Designing Machine Parts to Avoid Fatigue Failure The following factors should be considered while designing machine parts to avoid fatigue failure: 1. The variation in the size of the component should be as gradual as possible. 2. The holes, notches and other stress raisers should be avoided. 3. The proper stress de-concentrators such as fillets and notches should be provided wherever necessary. 4. The parts should be protected from corrosive atmosphere. 5. A smooth finish of outer surface of the component increases the fatigue life. 6. The material with high fatigue strength should be selected. 7. The residual compressive stresses over the parts surface increases its fatigue strength. 15 Stress Concentration Factor for Various Machine Members
- 12. 16 Fatigue Stress Concentration Factor When a machine member is subjected to cyclic or fatigue loading, the value of fatigue stress concentration factor shall be applied instead of theoretical stress concentration factor. Since the determination of fatigue stress concentration factor is not an easy task, therefore from experimental tests it is defined as Kf = Endurance limit without stress concentration Endurance limit with stress concentration 17 Notch Sensitivity In cyclic loading, the effect of the notch or the fillet is usually less than predicted by the use of the theoreti- cal factors as discussed before. The difference depends upon the stress gradient in the region of the stress con- centration and on the hardness of the material. The term notch sensitivity is applied to this behavior. It may be defined as the degree to which the theoret- ical effect of stress concentration is actually reached. The stress gradient depends mainly on the radius of the notch, hole or fillet and on the grain size of the material. q = Kf − 1 Kt − 1 Kf = 1 + q (Kt − 1) ...(For tensile or bending stress) Kfs = 1 + q (Kts − 1) ...(For shear stress) 18 Combined Steady and Variable Stress The failure points from fatigue tests made with dif- ferent steels and combinations of mean and variable stresses are plotted in Fig.13 as functions of variable stress (σv) and mean stress (σm). The most significant observation is that, in general, the failure point is little related to the mean stress when it is compressive but is very much a function of the mean stress when it is tensile. In practice, this means that fatigue failures are rare when the mean stress is compressive (or negative). Therefore, the greater emphasis must be given to the combination of a variable stress and a steady (or mean) tensile stress. There are several ways in which problems involving this combination of stresses may be solved, but the follow- ing are important from the subject point of view : 1. Gerber method, 2. Goodman method, and 3. Soderberg method. Figure 13: Combined mean and variable stress. 19 Gerber Method for Combination of Stresses According to Gerber, variable stress, 1 F.S. = σm σu 2 F.S. + σv σe Considering the fatigue stress concentration factor (Kf ), 1 F.S. = σm σu 2 F.S. + σv Kf σe
- 13. 20 Goodman Method for Combination of Stresses According to Goodman, variable stress, 1 F.S. = σm σu + σv σe Considering the fatigue stress concentration factor (Kf ), 1 F.S. = σm σu + σv Kf σe Considering the load factor, surface finish factor and size factor, 1 F.S. = σm σu + σv Kf σeb Ksur Ksz ...(For ductile materials subjected to reversed bending loading) 1 F.S. = σm Kt σu + σv Kf σeb Ksur Ksz ...(For brittle materials subjected to reversed bending loading) 1 F.S. = σm σu + σv Kf σea Ksur Ksz ...(For ductile materials subjected to reversed axial loading) 1 F.S. = σm Kt σu + σv Kf σea Ksur Ksz ...(For brittle materials subjected to reversed axial loading) 1 F.S. = τm τu + τv Kf τe Ksur Ksz ...(For ductile materials subjected to reversed torsional or shear loading) 1 F.S. = τm Kts τu + τv Kf τe Ksur Ksz ...(For brittle materials subjected to reversed torsional or shear loading) where suffix ’s’ denotes for shear. Note: For reversed torsional or shear loading, the values of ultimate shear strength (τu) and endurance shear strength (τe) may be taken as follows: τu = 0.8σu; and τe = 0.8σe Figure 14: Goodman method.
- 14. 21 Soderberg Method for Combination of Stresses According to Soderberg, variable stress, 1 F.S. = σm σy + σv σe Considering the fatigue stress concentration factor (Kf ), 1 F.S. = σm σy + σv Kf σe Considering the load factor, surface finish factor and size factor, 1 F.S. = σm σy + σv Kf σeb Ksur Ksz ...(For ductile materials subjected to reversed bending loading) 1 F.S. = σm σy + σv Kf σea Ksur Ksz ...(For ductile materials subjected to reversed axial loading) 1 F.S. = τm τy + τv Kf τe Ksur Ksz ...(For ductile materials subjected to reversed torsional or shear loading) Note: The Soderberg method is particularly used for ductile materials. Figure 15: Soderberg method. 22 Combined Variable Normal Stress and Variable Shear Stress When a machine part is subjected to both variable normal stress and a variable shear stress; then it is designed by using the following two theories of combined stresses : 1. Maximum shear stress theory, and 2. Maximum normal stress theory.
- 15. Equivalent normal stress due to reversed bending, σneb = σm + σv σy Kfb σeb Ksur Ksz Equivalent normal stress due to reversed axial loading, σnea = σm + σv σy Kfa σea Ksur Ksz Total equivalent normal stress, σne = σneb + σnea = σy F.S. Equivalent shear stress due to reversed torsional or shear loading, τes = τm + τv τy Kfs τe Ksur Ksz The maximum shear stress theory is used in designing machine parts of ductile materials. According to this theory, maximum equivalent shear stress, τes(max) = 1 2 (σne)2 + 4 (τes)2 = τy F.S. The maximum normal stress theory is used in designing machine parts of brittle materials. According to this theory, maximum equivalent normal stress, σes(max) = 1 2 σne + 1 2 (σne)2 + 4 (τes)2 = σy F.S. 23 Application of Soderbergs Equation 1 F.S. = σm σy + σv Kf σe = σm σe + σv σy Kf σy σe ∴ F.S. = σy σe σm σe + σv σy Kf = σy σm σe + σv σe σy Kf Working or design stress, = σm σe + σv σe σy Kf Let us now consider the use of Soderberg’s equation to a ductile material under the following loading conditions. 1. Axial loading In case of axial loading, we know that the mean or average stress, σm = Wm A and variable stress, σv = Wv A Working or design stress, = Wm A + σy σe Wv A Kf = Wm + σy σe Wv Kf A ∴ F.S. = σy A Wm + σy σe Wv Kf 2. Simple bending In case of simple bending, we know that the bending stress, σb = My I = M Z
- 16. Mean or average bending stress, σm = Mm Z Variable bending stress, σv = Mv Z Working or design bending stress, σb = Mm Z + σy σe Mv Z Kf = Mm + σy σe Mv Kf Z ∴ F.S. = σy Z Mm + σy σe Mv Kf For circular shafts, Z = πd3 32 σb = 32 πd3 Mm + σy σe Mv Kf F.S. = σy 32 πd3 Mm + σy σe Mv Kf 3. Simple torsion of circular shafts In case of simple torsion, we know that the torque, T = π 16 τd3 or τ = 16T πd3 Mean or average shear stress, τm = 16Tm πd3 Variable shear stress, τm = 16Tm πd3 Working or design shear stress, τ = 16Tm πd3 + τy τe Kfs 16Tv πd3 = 16 πde Tm + τy τe Kfs Tv Note: For shafts made of ductile material, τy = 0.5σy , and τe = 0.5σe may be taken. 4. Combined bending and torsion of circular shafts In case of combined bending and torsion of circular shafts, the maximum shear stress theory may be used. According to this theory, maximum shear stress, τmax = τy F.S. = 1 2 (σb)2 + 4τ2 = 1 2 32 πd3 Mm + σy σe Kf Mv 2 + 4 16 πd3 Tm + τy τe Kfs Tv 2 = 16 πd3 Mm + σy σe Kf Mv 2 + Tm + τy τe Kfs Tv The majority of rotating shafts carry a steady torque and the loads remain fixed in space in both direction and magnitude. Thus during each revolution every fiber on the surface of the shaft under-goes a complete reversal of stress due to bending moment. Therefore for the usual case when Mm = 0, Mv = M, Tm = T and Tv = 0, the above equation may be written as τy F.S. = 16 πd3 σy σe Kf M 2 + T2 Note: The above relations apply to a solid shaft. For hollow shaft, the left hand side of the above equations must be multiplied by (1 − k4 ).
- 17. 24 Examples
- 31. 25 References 1. R.S. KHURMI, J.K. GUPTA, A Textbook Of Machine Design 26 Contacts mohamed.atyya94@eng-st.cu.edu.eg