Question

A plate of an aluminum alloy with a center crack is cycled between 48.1 kN and 92.2 kN with crack growth data as given in the table below. The plate is of a rectangular cross section with a width of 152.4 mm and a thickness of 2.29 mm

A plate of an aluminum alloy with a center crack is cycled between 48.1 kN and 92.2 kN with crack growth data as given in the table below. The plate is of a rectangular cross section with a width of 152.4 mm and a thickness of 2.29 mm A. Determine da/dN and ДК by adding two columns to the table below B- Plot the (da/dN) vs AK on log-log scale and determine the constants of Paris Equation mmcycles 5.08 7.62 9,500 10.16 14,300 12.70 17,100 15.24 19,100 17.78 20,500 20.32 21,500 22.86 22,300 25.40 22,900 30.4823,500 35.5624,000 0

Answer 1

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Solution:

Cross sectional area,

A-152.5 × 2.29-349mm"

Stresses coming on plate,

(92.2-48.1)×1000 349 ▲P 126.36M Pa

Assuming Y=1,

$\Delta K=\Delta \sigma \sqrt{\pi a}$

daa2a1

Using these two equations we plot following table,

7232 627 455 2333333333 87766555 2 552 636903 898612 13592 3 2 5 311 479 2 2 2 9 (-57

y4.7573x - 22.408 3.90 2.90 3.10 3.30 3.50 3.70 1.00 -2.00 3.00 -4.00 5.00 -6.00 7.00 -8.00 -9.00 In(AK)

The Paris law is given by,

da

da dN lolog(C)+mlog(AK)

Comparing with the equation we got,

da :) : dN 47573|og(ΔΚ)-22.4 ogl.

we get,

m=4.7575

and

log(C) -22.408

1.8549 * 10-10 C