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Unformatted text preview: SOLUTION (8.16)
Known: Four standard R.R. Moore specimens of known materials are given. Find: Estimate the longlife fatigue strength for reversed torsional loading. (State
whether it is for 108 or 5 X 108 cycles.) Schematic and Given Data: Wrought aluminum
Su = 29 ksi Wrought aluminum
Su = 73 ksi High grade
cast aluminum High grade
forged magnesium Assumption: Figs. 8.8, 8.9, and 8.10 can be used to estimate long life fatigue for
reversed loading. Analysis:
1. From Fig. 8.9, for the wrought aluminum having Su = 29 ksi, the rotating bending fatigue strength at 5 X 108 cycles is Sn’ = 12 ksi. Since, for reversed torsional loading Sn = 0.58 Sn’ , n = 0.58(12) = 7 ksi I
2. From Fig. 8.9, for the wrought aluminum having Su = 73 ksi, the rotating bending fatigue strength at 5 X 108 cycles is Sn’ = 19 ksi. Thus, for reversed torsional loading, Sn = 0.58(130) = 75 MPa I
3. From Fig. 8.8, for high grade cast aluminum, the rotating bending fatigue strength at 5 x 108 cycles is 11 ksi for sand cast and 15 ksi for permanent mold cast. Thus, for reversed torsional loading,
Sn 2 0.58(11) = 6.48 ksi for sand cast Sn = 0.58(15) = 8.7 ksi for permanent mold cast I
4. From Fig. 8.10, for high grade forged magnesium, the rotating bending fatigue strength at 108 cycles is 22 ksi. Thus, for reversed torsional loading, Sn =
0.58(22) = 12.76 ksi SOLUION (8.17)
Known: A steel bar having known Su and Sy has average machined surfaces. Find: Plot on loglog coordinates estimated S—N curves for (a) bending, (b) axial, and
(c) torsional loading. For each of the three types of loading, determine the fatigue strength corresponding to (1) 106 or more cycles and (2) 5 x 104 cycles. 8—12 Schematic and Given Data: Machined surface Assumptions:
1. Actual fatigue data is not available for this material.
2. The estimated S—N curves constructed using Table 8.1 are adequate. 3. Fig. 8.13 can be used to estimate surface factor, Cs.
4. The gradient factor, CG = 0.9, for axial and torsional loading. Analysis:
1. Endurance limits: (106 cycle strength)
Sn = Sn, CLCGCS For bending,
Sn’ = 0.5 Su = 0.5(97) = 48.5 ksi (Fig. 8.5)
CL = 1 (Table 8.1) CG = 0.9 (Table 8.1) Cs = 0.76 (Fig. 8.13) Sn = (48.5)(1)(0.9)(O.76) = 33.2 ksi
For axial, Sn’ = 48.5 ksi CL = 1 CG = 0.8 (between 0.7 and 0.9) C5 = 0.76 Sn = 48.5(1)(O.8)(0.76) = 29.5 ksi
For torsion, . Sn’ = 48.5 ksi CL = 0.58 CG = 0.9 C3 = 0.76 Sn = 48.5(0.58)(0.9)(0.76) = 19.2 ksi 2. 1_Q3 cycle strength
For bending, 0.9Su = 0.9(97) = 87.3 ksi (Table 8.1) For axial,
0.75Su = 0.75(97) = 72.8 ksi For torsion,
0.9Sus = 0.9(0.8)(97) = 69.8 ksi 8—13 3 . S—N curves A
GD
0 C —1
m .54
m"
U}
Q) :5 m Torsion 105 106
Cycles (log) 4. 5 x 104 cycle strength
Bending: 50.5 ksi Axial: 43.6 ksi
Torsion: 33.6 ksi Comments: 1 . The surface factor, Cs is not used for correcting the 103—cycle strength because for ductile parts the 103 strength which is close to the static strength, is unaffected by
surface ﬁnish. 2. For critical designs, pertinent test data should be used rather than the preceding
rough approximation. SOLUTION (8.18)
Known: A steel bar having known Su and Sy has a hot rolled surface ﬁnish. Find: Determine the fatigue strength at 2 x 105 cycles for reversed axial loading. Schematic and Given Data: Assumptions: 1. Actual fatigue data is not available for this material. 2. The estimated SN curves constructed using Table 8.1 are adequate.
3. Fig. 8.13 can be used to estimate surface factor, Cs. 8—14 Analysis:
1. Endurance limit (106 cycle strength)
Sn = SnICLCGCs
For axial,
n’ = 0.5811 = 0.5(950) = 475 MPa
CL = 1
CG = 0.8 (between 0.7 and 0.9)
C5 = 0475
Sn = (475)(1)(0.8)(0.475) = 180.5 MPa I 2. 1_03 cycle strength
For axial, 0.75811 = 075(950) = 712.5 MPa
3. SN curves A
an
O C}
c: E
m"
5/:
:u 2: {/3 2 5 10_5 cycles 105 106
Cycles (log) 4. 2 X 105 cycle strength
Axial: 248.7 MPa I Comments:
1. The surface factor, Cs is not used for correcting the 103cycle strength because for ductile parts the 103 strength is relatively unaffected by surface ﬁnish.
2. For critical designs, pertinent test data should be used rather than the preceding rough approximation.
3. Analytically the 200,000 cycle strength for reverse axial loading may be determined by solving
[log (712.5) — 10g (180.5)]/(6  3) = [log (S)  log (180.5)]/(6  log (200,000)). SOLUTION (8.19)
Known: A steel bar having known Su and Sy has a ﬁne ground surface. Find: Determine the fatigue strength for bending corresponding to (1) 106 or more
cycles and (2) 2 X 105 cycles. 815 lit
1: n =43,500 T=n—(4—%—65—00—)=85401b in. I 2. For fatigue failure, the appropriate endurance limit is:
Sn = Sn, CLCGCS where Sn’ = 0.5811 2 0.5(90)
CL = 0.58 (Table 8.1)
CG = 0.9 (Table 8.1)
Cs = 0.77 (Fig. 8.13)
Sn = 0.5(90)(0.58)(0.9)(0.77) = 18.1 ksi I
3. From Fig. 4.35, K = 1.72
From Fig. 8.24, q = 0.78
Thus, Kf = 1 + (K — 1)q [Eq. (8.2)]
= 1 + (0.72)(O.78) = 1.56
4. For fatigue failure, 16T 18,1001: KfT=18,100;T=m=2,2801b in. I Comment: For static loading of a ductile material, the very ﬁrst yielding at the notch
root is not signiﬁcant; hence, ignore stress concentration. SOLUTION (8.27)
Known: A shaft rotates at high speed while the imposed loads remain static. The shaft is machined from AISI 1040 steel, oil quenched and tempered at 1000 0F. The
loading is sufﬁciently great to produce a fatigue failure (after perhaps 106 cycles). Find: Determine where the failure would most likely occur. Schematic and Given Data: Assumption: The shaft is manufactured as speciﬁed. 830 Analysis: 2. Only locations A, B, and C need to be investigated.
From Appendix C—Sb, Su z 107 ksi 3. Since, 0'— ' %K1°‘= MgKi
(13 therefore, failure will occur at the highest value of $K, q Kf
Mm
Point Relative M d3 Kt (Fig. 8.24) [511. (8.2)] d3
A 1 1 Fig. 4.35a 1.70 1.70
1.35 A fatigue failure should occur at C 8—31 SOLUTION (8.28) Known: A machined shaft having a known hardness experiences completely reversed
torsion. Find: With a safety factor of 2, estimate the value of reversed torque that can be
applied without causing eventual fatigue failure. Schematic and Given Data: 150 Bhn Assumption: The shaft is manufactured as speciﬁed with regard to the critical shaft
geometry. Analysis:
1. For steel,
Su = 0.5 Bhn = 0.5(150) = 75 ksi or, 311 = 75 ksi (QQEEMIE) = 517 MPa 2. Sn = Sn, CLCGCS 311' = 0.5811 = 0.5(517) (Fig. 8.5)
CL = 0.58 (Table 8.1)
CG = 0.9 (Table 8.1)
Cs = 0.78 (Fig. 8.13)
Sn = 0.5(5l7)(0.58)(0.9)(0.78) = 105.3 MPa I
3. At the critical point (0.8 mm radius), r/d = 0.04 and D/d = 1.2
From Fig. 4.35(c), Kt = 1.65
From Fig. 8.23, q = 0.74
Hence, Kf= 1 + (K 1)q [Eq.(8.2)]
= 1 + (0.65)(0.74) = 1.48 4. Therefore, the nominal value of reversed torsional stress can be 1: = 105.3/1.48 =
71.1 MPa. 3
16T __’C1td
Butt—““13 or T— 16 _ (71.1 MPa)1t(20 mm)3
‘ 16 111.7 Nm
2 T = 111,700 N‘mm with SF 2 2, T = = 55.8 Nm I 832 120 140 6. From the graph, (Sm = 6;, = 32 ksi.
With SF = 1.2,
(1.2)(116.3)F = 32,000
Therefore, F = 229 1b I SOLUTION (8.38) Known: A shaft is subjected to a ﬂuctuating nominal stress. The shaft is made of
steel having known Su and Sy. Find: Estimate the safety factor with respect to eventual fatigue failure if:
(a) the stresses are bending, (b) the stresses are torsional. Schematic and Given Data: Assumption: The shaft is manufactured as speciﬁed with regard to surface ﬁnish and
critical ﬁllet radii. 845 Analysis:
1. For bending stresses, Sn = Sn, CLCGCs [130 (81)]
Sn, = 0.5811 (Flg 8.5)
CL = 1 (Table 8.1)
CG = 0.9 (Table 8.1)
Cs = 0.77 (Fig. 8.13) Sn 2 0.5(600)(1)(0.9)(0.77) = 208 MPa
2. Highest stress is at the 1.5 mm ﬁllet where
D/d = 1.2 and r/d = 0.03
From Fig. 4.35, Kt = 2.3
From Fig. 8.24, q = 0.78
From Fig. (8.2), Kf = 1 +(Kt —1)q
Kr = 1 + (1.3)(0.78) = 2.01
3. At the ﬁllet om = 201(80516) = 64 MPa ca = 201(80; 16) = 96 MPa 4. Thus, for the bending stresses, Overload causing
eventual failure SF = 168/96 = 1.8
5. For torsional stresses, Sn = Sn, CLCGCS Sn, 3 0.5811
CL = 0.58
CG = 0.9
Cs = 0.77
n = 0.5(600)(0.58)(0.9)(0.77) = 121 MPa
6. From Fig. 4.35, K = 1.78
From Fig. 8.24, q = 0.81
Kf=1+(1.78  1)(0.81) = 1.63 8—46 7. At critical ﬁllet, rm =1.63(80516)= 52 MPa ca = 1.63(80;16)= 78 MPa 8. Thus, for torsional stresses, SF = 105/78 = 1.3 I SOLUTION (8.39)
Known: A round shaft made of steel having known Su and Sy is subjected to a torque ﬂuctuation. All critical surfaces are ground. Find: Estimate the safety factor for inﬁnite fatigue life with respect to an overload that
(a) increases both mean and alternating torque by the same factor,
(b) an overload that increases only the alternating torque. Schematic and Given Data: 847 Assumption: The shaft is manufactured as speciﬁed with regard to critical radii, hole
geometry, and surface ﬁnish. Analysis:
1. Sus = 0.8(162) = 130 ksi
Sys = 0.58(138) = 80 ksi
2. At the hole,
from Fig. 4.37, K; = 1.75
from Fig. 8.24, q = 0.88
Kf= 1 + (K:  1)q [Eq. (8.2)]
Kf= 1 + (0.75)(0.88) = 1.66
(At ﬁllet, K = 1.33; hence, not as critical as hole)
3. Using the equation in Fig. 4.37, Tm T = —————————K
m (1tD3/16)  (dD2/6) ‘
_ 5000 _ .
Tm ‘ 7120/16) — (1/16)(1/6)(1'66) ‘ 44600 P“
2000 _ .
Ta — F16 _1/96(1.66)— 17,900 ps1
4. Sn = Sn] CLCGCS [Eq. (8.1)]
Sn, = 0.55“ (Flg. 8.5)
CL = 0.58 (Table 8.1)
CG = 0.9 (Table 8.1)
CS = 0.89 (Fig. 8.13)
Sn = 0.5(162)(0.58)(0.9)(0.89) = 38 ksi I
5. N ormal
operation 6. For an overload that increases both the mean and the alternating torque by the same factor, SF = 22/179 = 1.2 I
For an overload that increases only the alternating torque, SF = 25/179 = 1.4 I 8—48 For zerotomaximum ﬂuctuation 6;, = O'm = 121 MPa, or cmx = 242 MPa.
From Fig. 4.38(a), H/h = 1.4, r/h = 0.08, K( = 1.87
From Fig. 8.24, q = 0.78
Thus, Kf= 1 + (K 1)q [Eq. (8.2)]
Kf = 1 + (0.87)(0.78) = 1.68
6. Using the equation from Fig. 4.38(a), Gmax = MHImXCKf = 6Mn5ath
bh 242 MPa = —ﬂm—al—§(1.68) (20 mm)(25 mm)
Therefore, Mum = 300,000 Nmm = 300 Nm I Ulb SOLUTION (8.42)
Known: A solid round shaft has a shoulder with known D and d. The shaft is made
of steel having known values of Sn and Sy. All surfaces are machined. In service the
shaft is subjected to a ﬂuctuating torsional load and is to have an inﬁnite life (with
safety factor =1). Find: Estimate the smallest ﬁllet radius. Schematic and Given Data: Assumption: The shaft is manufactured as speciﬁed with regard to the critical ﬁllet
geomeny and the shaft surface ﬁnish. Analysis:
.171 _ Ta _ 312113.222 _ 20_5 _
1' rm ‘ “T; ‘ (123 + 82)/2 “ 102.5 “ 0'20
2. 3n = 310' CLCGCS [Eq. (8.1)]
Sn, = 0.5811 (Flg. 8.5)
CL = 0.58 (Table 8.1)
CG = 0.9 (Table 8.1)
CS = 0.69 (Fig. 8.13) Sn = 0.5(150)(0.58)(0.9)(0.69) = 27 ksi 8—52 0U] s“, = 0.8311 = 0.8(150) = 120 ksi
sys = 0.583y = 0.58020) = 70 ksi From ”Cm”Ca plot, 13 = 14.5 ksi
Using the equation in Fig. 435(0), 16Ta
Ta = 3 K1
1rd
16 20.5 12
14,500 psi = #Ki
1t(0.5)
Kf = 1.45 From Fig. 8.24, estimate q z 0.85.
Then, Kf =1 +(Kt1)q (Kr 1)
q _ (1.45 — 1) ‘ 0.85 From Fig. 4.35, for D/d = 2 and K = 1.53,
r/d = 0.08 ; then, r = (0.08)(O.5) = 0.04 in. for which q z 0.88. Hence, I is slightly greater than 0.04 in. Kt: +1 +1 =1.53 853 SOLUTION (8.43) Known: A steel shaft used in a spur gear reducer is subjected to a constant torque
together with lateral forces that tend always to bend it downward in the center. The
stresses are known, but these values do not take into account stress concentration
caused by a shoulder with known dimensions. All surfaces are machined and the
strength values and hardness of the steel are known. Find: Estimate the safety factor with respect to inﬁnite life. Schematic and Given Data: Assumption: The shaft is manufactured as speciﬁed with regard to the critical ﬁllet
and shaft surface ﬁnish. Analysis:
1. We use the Fig. 8.16 relationship for "general biaxial loads":
0 Bending provides an alternating stress: Ga 2 Oea = 60 Kf W3
From Fig. 4.35(a), K = 1.63
From Fig. 8.24, q = 0.84 .
From Eq. 8.2, Kf = 1 + (0.63)(0.84) = 1.53
 Gea = 60(1.53) = 91.8 MP3
 Torsion provides a mean stress:
Tm = Gem = 80 Kf W3.
From Fig. 435(0), Kt = 1.33 From Fig. 8.24, q = 0.86
From Eq. 8.2, Kr = 1 + (0.33)(0.86) = 1.28 Gem = 800.28) = 102.4 MPa. 2. n = Sn, CLCGCS [Eq. (8.1)]
Sn’ = 0.5311 (Fig. 8.5)
CL = 1 (Table 8.1)
CG = 0.9 (Table 8.1)
C5 = 0.76 (Fig. 8.13) Sn = 0.5(700)(1)(0.9)(0.76) = 239 MPa 854 4. SF =175/91.8 = 1.9 I SOLUTION (8.44) Known: A pump is geardriven at uniform load and speed. The shaft is supported by
bearings mounted in the pump housing. The shaft is made of steel having known
values of Sn and Sy. The tangential, axial, and radial components of force applied to
the gear are known. The surface of the shaft ﬁllet has been shotpeened, which is
estimated to be equivalent to a laboratory mirror—polished surface. Fatigue stress
concentration factors for the ﬁllet have been determined. Find: Estimate the safety factor with respect to eventual fatigue failure at the ﬁllet. Schematic and Given Data: 25 mm solid
Bending Kr = 20
Torsional Kr: 1.5
Axial Kr: 1.8 855 2000 500 *750 1:9 Mv = 2000(50) MH = 500025) + 750(50)
= 100,000 Nmm = 100,000 Nmm M=VM%;+M%{=141,000Nmm Assumption: The shaft is manufactured as speciﬁed with regard to shaft geometry
and surface ﬁnish. Analysis:
1. MV = (2000)(50) = 100,000 Nmm, MH = 500(125) + 750(50) = 100,000 Nmm; M = V M% + Mﬁ = 141,000 Nmm
2. We use the Fig. 8.16 relationship for " general biaxial loads":
Alternating stress: 32MK—32(141’0300) 11:d31t(25)3
693‘ — 183.8 W8
Mean stresses: _16M _16(2000)(3125) Ga: (2) — 183. 8 MPa (1. 5) — 122.2 MPa 1r.d31C(25)3
500 4
0:319: (30.8): 1.83 MPa
A 1c(25)
0' 2 2
Gem=§+ T +(%) = .92 +1/122.22+ 0.922
cm = 123.1 MPa
3. sn 2 Sn’ CLCGCS [Eq. (8.1)]
Sn, = 0.58u (Flg. 8.5)
CL = 1 (Table 8.1)
CG = 0.9 (Table 8.1)
CS =1 (Fig. 8.13) Sn = 0.5(1000)(1)(0.9)(1) = 450 MPa 856 5. SF = 235/123 = 1.9 I 857 SOLUTION (8.45) Known: A countershaft has helical gear (B), bevel gear (D), and two supporting
bearings (A and C). Loads acting on the bevel gear are known. Forces on the helical
gears can be determined. Shaft dimensions are known. All shoulder ﬁllets have a
radius of 5 mm. Only bearing A takes thrust. The shaft is made of hardened steel
having known values of Sn and Sy. All important surfaces are ﬁnished by grinding. Find: (a) Draw load, shear force, and bending moment diagrams for the shaft in the xy
and x2 planes. Also draw diagrams showing the intensity of the axial force and
torque along the length of the shaft. (b) At points B, C, and E of the shaft, calculate the equivalent stresses in preparation for making a fatigue safety factor determination. (Note: Refer to Fig. 8.16.) (c) For a reliability of 99% (and assuming 6 = 0.08 Sn), estimate the safety factor of
the shaft at points B, C, and E. Schematic and Given Data: Forces act at
375 mm dia. Fz = 0.3675 Fy
Forces act at 500mmdia. {B
 K f = 1.6 for bend and torsion; _
1.0 for axial load. Use CS = 1 S“ “1069 Mpa
with these values. Sy = 896 MPa 858 Assumption: The shaft is manufactured as speciﬁed with regard to the critical shaft
geometry and surface ﬁnish. Analysis: 1. Load determination (a) Helical gear forces:
For EMX = O, the torque at the two gears must be equal. Therefore, Fy (250 mm)
= 5.33(187.5 mm). Hence, Fy = 4.00 kN.
From the given data, Fx = .2625Fy = 1.05 kN; Fz = .3675 Fy = 1.47 kN. (b) Determine shaft loads in the xy and xz planes xy or vertical plane xz or horizontal plan
550 450 400 Vertical forces:
4 550 1.37 187.5  1.37 1400
2MA=():CV=< )+ (1000) ( ) = 0.54 kN downward
2F = 0: v = 4  0.54  1.37 = 2.09 kN downward Horizontal forces:
1. 2 0 1.47 550 5.33 1400
2MA=01CH= 05(5) ( >+ ( > 1000
= 6.92 kN upward .47 + 6.92 5.33 2F = 0 : An 1
3.06 kN downward 8—59 2. Stress determination (a) At E, the loading is:
Compression of 1.37 kN, K = 2.2, q = .94,
Kf = 2.13. Axial stress (mean or constant) = 4 137 2.13
4PKf=——(————)(————) = 0.581MPa 2 2
nd 1r(80)
The tension stress is zero. M = «2.09 x 400)2 + (3.06 x 400)2 = 1482 kNmm
Kt = 1.9, q = .94. Therefore, Kr = 1.85 Bending stress (alternating) = 32¥Kt 1rd 32(1482 x 103)
= 1:030)3
From Eq. (a) and Eq. (b) in the ﬁgure caption of Fig. 8.16, O'em = 0;
(Sea = 54.5 MPa
(b) At B, the loading is:
Axial, P = 1.37 kN, Kr = 1.0, 0' = —0.27 MPa
Torsion = (4.0)(250) = 1000 kNmm Bendmg ; M = v (2.09 x 550)2 + (3.06 x 550)2 = 2038 kNmm Kf = 1.6 for bending and torsion
32M (1.85) = 54.5 MPa Bending stress (alternating) = 3 K1
1rd
 32(2038 x103) 1 6) _ 64 9 MP
‘ 1:030)3 ' “ ' a
6
Torsional stress (mean) = ‘11E6’d'rg—Kf2 17166815: (1.6) = 15.9 MPa
Gem = %1 (15.9)2+ (‘0227) = 15.76 MPa ; oea = 64.9 MPa
(c) At C, the loading is: Bending: M = 6/(533 x 400)2 +1.37 x (400  1875)]2 = 2152 kNmm . , 32(2152) x 103
Bending stress (altematlng) = ————————— = 42.8 MPa 1t(80)3
oea = 42.8 MPa
Torsional stress — same as (b) except no stress concentration factor; axial same as (b). 860 0.27+ (15_.9 +.27) _980MPa “em = 2 1.6 2
3. Strength and safety factor determination Su = 155 ksi = 1069 MPa; Sy = 130 ksi = 896 MPa For working with equivalent bending stress, Sn is (1_0_6_9) Sn = Sn’ CLCGCs= (1)(0 8)* (0 9) =385 MPaforCs=0.9
*(See note b, Table 8.1) Sn = Sn' CLCGCs — (1069)(1)(0. 8)(1. 0) =428lVfPaf0rCs= 1.0
But for 99% reliability, reduce this by 2.3 standard deviations, which amounts to
multiplying by a factor of (1  2.3 X .08) = .816
Thus, for 99% reliability,
Sn  385(. 816) — 314 MPa (for Cs: .9)
4=28(. 816) — 349 MPa (for Cs  1.0) 325 oolife with C _ .0 9, use for nonkeyway
E (0, 314) (points C and E) Gca
(MPa)
E (0, 54.5) 5. Safety factors: (B) SF = 325/649 = 5.0
(C) SF = 290/428 = 6.8
(E) SF = 314/54.5 = 5.8 SOLUTION (8.46)
Known: A stepped shaft having known dimensions was machined from AISI steel of known hardness. The loading is one of completely reversed torsion. During a typical
30 seconds of operation under overload conditions the nominal (Tc/J) stress in the 1— in.dia. section was measured. Find: Estimate the life of the shaft when operating continuously under these
conditions. 861 ...
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