MECH 344 Problem Set 4-Chapter 8_Fatigue-Selected Problems Concordia University

MECH 344 Problem Set 4-Chapter 8_Fatigue-Selected Problems Concordia
University


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 ✕ 105 cycles.


Schematic and Given Data:

Su = 1200 MPa
10 mm
Sy = 950 MPa



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.

Analysis:
1. Endurance limits: (106 cycle strength)
Sn = Sn CLCGCsCTCR
For bending,
Sn = 0.5 Su = 0.5(1200) = 600 MPa (Fig. 8.5) CL
= CT = CR = 1 (Table 8.1)
CG = 0.9 (Table 8.1)
Cs = 0.86 (Fig. 8.13)
Sn = (600)(1)(0.9)(0.86)(1)(1) = 464.4 MPa ■
3
2. 10 cycle strength
For bending,
0.9Su = 0.9(1200) = 1080 MPa (Table 8.1)
3. S-N curves




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, 1080
Stress, MPa (log)


565.5

Bending




2 x 105 cycles
464.4




103 104 105 106 107
Cycles (log)


4. 2 ✕ 105 cycle strength
Bending: 565.5 MPa ■
Comments:
1. The surface factor, Cs is not used for correcting the 103-cycle 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 fatigue strength for bending may be determined
by solving
[log (1080) - log (565.5)]/(6 - 3) = [log (S) - log (565.5)]/(6 - log (200,000)).
SOLUTION (8.20)
Known: A steel bar having known Su and Sy has a hot rolled surface finish.

Find: Determine the fatigue strength at 2 ✕ 105 cycles for reversed axial loading.


Schematic and Given Data:

Su = 950 MPa
25 mm
Sy = 600 MPa



Assumptions:
1. Actual fatigue data is not available for this material.
2. The estimated S-N curves constructed using Table 8.1 are adequate.
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,3. Fig. 8.13 can be used to estimate surface factor, Cs.




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, Analysis:
1. Endurance limit (106 cycle strength)
Sn = Sn CLCGCsCTCR
For axial,
Sn = 0.5Su = 0.5(950) = 475 MPa CL
= CT = CR = 1
CG = 0.8 (between 0.7 and 0.9)
Cs = 0.475
Sn = (475)(1)(0.8)(0.475)(1)(1) = 180.5 MPa ■
2. 103 cycle strength
For axial,
0.75Su = 0.75(950) = 712.5 MPa
3. S-N curves


712.5
248.7
Stress, MPa (log)




Axial


180.5
cycles
5
2 x 10




103 104 105 106 107
Cycles (log)



4. 2 ✕ 105 cycle strength
Axial: 248.7 MPa ■

Comments:
1. The surface factor, Cs is not used for correcting the 103-cycle 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) - log (180.5)]/(6 - 3) = [log (S) - log (180.5)]/(6 - log (200,000)).
SOLUTION (8.21)
Known: A steel bar having known Su and Sy has average machined surfaces.
Find: Plot on log-log 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
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