Tutorial work - log scale example with fatigue problems PDF

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Example for Log Scales

Data for S-N Curve # Cycles Log # Cycles 10,000 4 100,000 5 1.00E+06 1.00E+07

S (MPa) 410 350

6 7

290 290

Stress Amplitude, MPa

450

400

350

300

250 3

4

5

6

7

8

Number of Cycles to Failure

1) What is stress amplitude level for 50,000 cycle life? Answer: Find log (50,000) = 4.69. At point 4.69 on log scale draw a vertical line and see where it intersects the S-N curve. Answer: 369 MPa.

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2) What is the number of cycles to failure for a stress cycle with a stress amplitude of 330 MPa? Answer: Method 1: Find the intersection of the line at 330 MPa with the curve and drop a vertical line down to the log scale. (Get 5.32). Take the antilog of this number to find the number of cycles (Answer 210,000 cycles)

Stress Amplitude, MPa

450

400

350

300

A 250 1.0E+3

1.0E+4

B

1.0E+5

1.0E+6

1.0E+7

1.0E+8

Number of Cycles to Failure

= (+5)+(A/B) =5+0.32=5.32 Number of cycles to failure =105.32 = 210,000 cycles Method 2: Estimate from the scale if minor ticks are shown

Stress Amplitude, MPa

450

400

350

300

2

250 1.0E+3

3 4 5 6 789

1.0E+4

2

3 4 5 6 789

1.0E+5

2

3 4 5 6 789

1.0E+6

2

3 4 5 6 789

1.0E+7

2

3 4 5 6 789

1.0E+8

Number of Cycles to Failure

2

Sample Problems Three identical fatigue specimens (denoted A, B, and C) are fabricated from an Aluminum alloy. Each is subjected to one of the maximum-minimum stress cycles listed below; the frequency is the same for all three tests.

• •

Specimen

max (MPa)

min (MPa)

A B C

+450 +300 +500

-150 -300 -200

Rank the fatigue lifetimes of these three specimens from the longest to the shortest. Now justify this ranking using a schematic S-N plot.

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Sample Problem A rotational shaft is made of a medium carbon steel and is serviced under cyclic loading condition. Rank the following situations in terms of expected fatigue life:

• • • •

max = 350 MPa, min = 150 MPa max = 450 MPa, min = 250 MPa max = 350 MPa, min = 150 MPa, and the surface of shaft was “case hardened” by carburization. max = 450 MPa, min = 250 MPa, and the shaft had been in contact with corrosive environments.

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Review of some math A=Bxnexp(-C/RT) Take ln of both sides…and rearrange ln A = nlnx+ln(B exp(-C/RT)) Looks a lot like… y=mx+b where m = n and b = ln(B exp(-C/RT)) Curve fit ln A vs ln x and determine slope and intercept See the correlation? strain rate

stress exponent (material parameter)

 Q c  . n s  K 2  exp   activation energy for creep  RT  (material parameter)

material const.

applied stress

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