```MSE 227
Homework 6 Chapter 9: Failures
Fall, 2010
1.
The value of KIc for a glass optical fiber is 0.75 MPa√m. The strength of the glass fiber is
69 MPa. What is the largest surface flaw that can safely exist in such a fiber? Assume Y = 1.0
1  2
= √ →  = ( )

a = 3.7 x 10-5 m
2.
A structural component fabricated from a wide plate of a steel alloy that has a plane strain
fracture toughness KIc of 98.9 MPa√m and a tensile strength of 860 MPa is to be inspected
rdiographically using x-rays to assess internal quality. If the x-ray system is capable of detecting
a flaw larger than 3.0 mm and the design stress does not exceed half the alloy’s yield strength,
would a critical flaw for the plate be detectable? Prove why or why not. Assume Y=1.0.
design max = ½  = 860/2 = 430 MPa
KIc = 98.9 MPa√m
amin = 3 mm
a = 1/(98.9/430)2 = 0.0168 m or 16.8 mm >> 3 mm
Thus the flaw would be detectable.
3.
The fracture strength of glass can be increased by etching a thin layer form the surface
using hydrofluoric acid. The assumption is that etching alters the surface by reducing the length
and increasing the tip radius of any inherent surface flaws. How much would the crack length
have to be shortened to increase the fracture strength of the glass fourfold? How much would the
tip radius have to be increased to increase the fracture strength twofold?
From the stress intensity factor, we know:  ≈ o√(a/)
Thus stress scales directly with √a and inversely as 1/√

If fracture strength increases fourfold, the stress intensity in case 2 is ¼ that in case 1:
2 = ¼ 1 or 4√a2 = √a1
therefore
a2 = 1/16 a1
Similarly, 1/2√(1/2) = √(1/1)
or
2 = 41`
4.
Estimate the magnitude of the maximum stress that exists at the tip of an internal crack
having a tip radius of 2.5 x 10-4 mm and a crack length of 2.5 x 10-2 mm. if the applied tensile
stress is 170 MPa.
Using:
 =o[2√(a/)+1]
where a = 1.25 x 10-2 mm,
 = 2.5 x 10-4 mm
And o = 170 MPa
 = 15.14 x o = 2574 MPa
If we neglect the additional 1,
 = 14.14 x o = 2400 MPa
In most cases, 2√(a/) >> 1, and it is safe to disregard the 1.
5.
The fatigue data for a steel alloy are given as follows:
Stress Amplitude
Cycles to
MPa
Failure
470
104
440
3 x 104
390
105
350
3 x 105
310
106
290
3 x 106
290
107
290
108
a.
Make an S-N plot (stress amplitude versus logarithm cycles to failure) using the
given data.
What is the fatigue limit for this alloy?
Determine the fatigue lifetimes at stress amplitudes of 415 MPa and 275 MPa.
Estimate fatigue strength at 2 x 104 and 6 x 105cycles.
b.
c.
d.
SN Curve
500
450
Stress. MPs
400
350
300
250
200
1.00E+03
1.00E+04
1.00E+05
1.00E+06
Cycles
(a)
Using the curve,
(b)
fatigue limit = 290 MPa
(c)
5 x 104, unlimited
(d)
450 MPa, 320 MPa
1.00E+07
1.00E+08
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