MAAE 4102 – Strength and Fracture
MAAE 4102 - Strength and Fracture Problem Analysis Stress Life
1.
A method used to present mean stress fatigue data is to generate a family of curves on an S-N plot, with each curve representing representing a different stress stress ratio, R. Generate the curves for R values of -1, 0 and 0.5 for a steel with an ultimate strength of 100 ksi. For this example, use the Gerber relationship to to generate these curves. Use Eqn. 1 to estimate the 3 6 fully reversed (R = -1) -1) fatigue behaviour between between 10 and 10 cycles. S
=
1.62 S U N
0.085
−
(1)
2.
Another method used to present mean stress fatigue data is to generate a family of curves on an S-N plot, with each curve representing a different mean stress value, m. Generate the curves for mean stress values of 0, 20 and 40 ksi for a steel with an ultimate strength of 100 ksi. For this example, use the Goodman Goodman relationship to generate these curves. curves. Again use eqn. (1) to estimate the fully reversed (mean = 0) fatigue behaviour.
3.
Given a material with an ultimate strength of 70 ksi, an endurance limit of 33 ksi, and a true fracture strength of 115 ksi, determine the allo wable zero to maximum (R = 0) stress 3 4 5 6 which can be applied for 10 , 10 , 10 and 10 cycles. Make predictions using the Goodman, Gerber and Morrow relationships.
4.
A component undergoes a cyclic stress with a maximum value of 75 ksi and a minimum value of -5 ksi. Determine the mean stress, stress range, stress amplitude, stress stress ratio and amplitude ratio. If the component is made from a steel with an ultimate strength of 100 ksi, estimate its life using the Goodman relationship.
5.
A switching device consists of a rectangular cross-section metal cantilever 200 mm in length and 30 mm in width. The required operating displacement at the free end is 2.7 mm and the service life is to be 100 00 cycles. To allow for scatter in in life performance a factor of 5 is employed employed on endurance. Using the fatigue curves given in Fig. 1 determine the required thickness of the cantilever if made in (a) mild steel, (b) aluminum alloy. 2 2 Esteel = 208 GN/m , E Aluminum = 79 GN/m .
6.
A pressure vessel support bracket is to be designed so that it can withstand a tensile 2 loading cycle of 0-500 MN/m once every day for 25 years. years. Which of the following steels would have the greater tolerance to intrinsic defects in this application: (i) a -(3/2) -11 maraging steel (K IC , C = 0.15 x 10 , m = 4.1), or (ii) a mediumIC = 82 MN m -(3/2) -11 strength steel (K IC , C = 0.24 x 10 , m = 3.3)? For the loading situation situation a IC = 50 MN m geometry factor of 1.12 may be assumed.
Problem Set Page 1 of 8
MAAE 4102 – Strength and Fracture
x - Aluminium Alloy 24S-T3 reversed axial stress - Mild steel reversed axial stress Fig 1 7.
A series of crack growth tests on a moulding grade of polymethyl methacrylate gave the following results: da/dN (m/cycle) K (MN m-3/2)
-7
2.25 x 10 0.42
-7
4 x 10 0.53
-7
6.2 x 10 0.63
-7
17 x 10 0.94
-7
29 x 10 1.17
-3/2
If the material has a critical stress intensity factor of 1.8 MN m and it is known that the moulding process produces defects 40 m long, estimate the maximum repeated tensile 6 stress which could be applied to this material for at least 10 cycles without causing fatigue failure.
8.
A series of tensile fatigue tests on stainless steel strips containing a central through hole gave the following values for the fatigue endurance of the steel. If the steel strips were 100 mm wide, comment on the notch sensitivity of the steel. Stress concentration values are given in Fig 2. Hole diameter (mm) 2 Fatigue endurance (MN/m )
No hole 600
Problem Set Page 2 of 8
5 250
10 270
20 320
25 370
MAAE 4012 – Strength and Fracture
Figure 2.
9.
The fatigue endurances from the S-N curve for a certain steel are: Stress 2 (MN/m )
Fatigue endurance (cycles)
350 380 410
2,000,000 500,000 125,000
2
If a component manufactured from this steel is subjected to 600,000 cycles at 350 MN/m 2 and 150,000 cycles at 380 MN/m , how many cycles can the material be expected to 2 withstand at 410 MN/m before fatigue failure occurs, assuming that Miner’s cumulative damage theory applies?
Problem Set Page 3 of 8
MAAE 4102 – Strength and Fracture
Strain Life
1.
3
It has been determined that a certain steel (E = 30 x 10 ksi) follows the following true stress, , true plastic strain, p, relation: σ =
. ( 350ksi)(ε P ) 0 11
The true plastic strain at fracture was found to be 0.48. Determine: a) True fracture strength, f b) Total true strain at fracture c) Strength coefficient, K d) Strain hardening exponent, n e) Strength at 0.2% offset, Sy f) Percent reduction in area, % RA g) True fracture ductility, f 2.
The following stress-strain and strain-life properties are given for a steel: E = 30 x 10 ksi σ ′ =
120ksi
b = -0.11
ε ′ =
0 95 .
c = -0.64
f
f
a) b)
c) d) e) f)
K = 137 ksi n = 0.22
Draw on log-log coordinates the elastic strain-life, and total strain-life curves. Determine the transition life (2Nt). Draw the hysteresis loops corresponding to strain amplitude (/2) values of 0.05, 0.00125 and 0.0007. Determine the fatigue life in reversals at these three strain levels. 6 Determine the elastic, plastic and total strain amplitude for a life (2Nf ) of 2 x 10 reversals. Determine the elastic, plastic, and total strain amplitude for a life (2Nf ) of 500 reversals. Determine the cyclic stress amplitude corresponding to fatigue lives of 500 and 2 6 x 10 reversals. 4 A component made from this material is required to have a life of no less than 10 reversals. The loading on the component causes a total strain amplitude of 0.008. Determine if the component will meet the life requirements.
Problem Set Page 4 of 8
MAAE 4012 – Strength and Fracture
3.
Smooth aluminum specimens are subjected to two series of cyclic load-controlled tests. The first test (level A) varies between a maximum stress value, max, of 21.3 ksi and a minimum value, min, of -30.1 ksi. The second test (level B) varies between 61.5 and 10.1 ksi. Predict the life to failure, in reversals, at the two levels. Use the Morrow, Manson-Halford and Smith-Watson-Topper relationships for the predictions. Assume that there is no mean stress relaxation. The material properties for the aluminum are 3
E = 10.6 x 10 ksi
K = 95 ksi
f = 160 ksi
b = -0.124
f = 0.22
c = -0.59
n = 0.065
Listed below are actual test results at the two levels. Three tests were run at each of the levels. Compare the predictions to these values. Level
Test Results: Lives in Reversals, 2Nf
A
5.4 x 10
B
5.6 x 10
5
5.5 x 10
5
4
6.4 X 10
4
Problem Set Page 5 of 8
5
7.2 x 10
4
6.9 X 10
MAAE 4102 – Strength and Fracture
Fracture Mechanics
1.
A large plate made of AISI 4340 steel contains an edge crack and is subjected to a tensile stress of 40 ksi. The material has an ultimate strength of 260 ksi and a K c value of 45 ksi in. Assume that the crack is much smaller than the width of the plate. Determine the
critical crack size. 2.
A large cylindrical bar made of 4140 steel (y = 90 ksi) contains an embedded circular (penny shaped) crack with a 0.1 in. diameter. Assume that the crack radius, is much smaller than the radius of the bar, R, so that the bar may be considered infinitely large compared to the crack. The bar is subjected to a tensile stress of 50 ksi. Determine the plastic zone size at the crack tip. Are the basic LEFM assumptions violated?
3.
A very wide late made from Al 7075-T651 (K Ic = 27 ksi in., y = 80 ksi) contains an edge crack. Plot the allowable nominal stress (ksi) as a function of crack size, (in inches), if the design requirements specify a factor of safety o f 2 on the critical stress intensity factor. If the plate specifications were changed so that Al 7050-T73651 was used (K Ic = 35 ksi in., y = 70 ksi), re-plot the curve. For a nominal stress of one-half the yield stress, determine the increase in allowable flaw size by changing from the Al 7075 alloy to the Al 7050 alloy.
4.
Design a pressure vessel that is capable of withstanding a static pressure of 1000 psi and that will “leak-before-burst.” The required material has a fracture toughness of 60 ksi in. and a yield strength of 85 ksi. The diameter of the vessel is specified to be 4 ft. A
crack with surface length of 1 in. can reliably be detected. Since the cost of the vessel is related directly to the amount of material used, optimize the design so that the cost is minimized. 5.
A component made from 7005-T53 aluminum contains a semi-circular surface crack (a/c = 1) and is subjected to R = 0.1 loading with a stress range, , of 250 MPa. (Refer to Example 1 for an expression for the stress intensity range, K .) the following crack growth data were obtained in laboratory air environment. Using these data: a) b) c)
Plot crack length, (mm), versus cycles, N Plot da/dN versus K . Identify the three regions of crack growth. Determine the Paris law constants, C and m, for the linear region of crack growth.
Problem Set Page 6 of 8
MAAE 4012 – Strength and Fracture
________________________________________________________ N (cycles) a (mm) da/dN (mm) _________________________________________________________
95,000 0.244 -7 100,000 0.246 7.00 x 10 -6 105,000 0.251 3.920 x 10 -6 110,000 0.285 9.665 x 10 -5 115,000 0.347 1.053 x 10 -5 125,000 0.414 1.230 x 10 -5 130,000 0.490 2.063 x 10 -5 135,000 0.621 4.661 x 10 -5 140,000 0.956 9.565 x 10 -4 145,000 1.577 3.964 x 10 -3 147,000 2.588 1.105 x 10 -3 147,400 3.078 1.554 x 10 -3 147,500 3.241 8.758 x 10 147,500 3.445 __________________________________________________________
K I
=
Problem Set Page 7 of 8
1 Q
1.12σ
a
π
MAAE 4102 – Strength and Fracture
6.
Calculate the critical defect size for each of the following steels assuming they are each subjected to a stress of 0.5y. Comment on the results obtained. Steel
Yield Strength 2 (MN/m )
y
Fracture Toughness -3/2 (MN )
Mild Steel
207
200
Low-alloy Steel
500
160
Medium Carbon Steel
1000
280
High-carbon Steel
1450
70
18% Ni (Maraging) Steel
1900
75
Tool Steel
1750
30
7.
A sheet of glass 0.5 m wide and 18 mm thick is found to contain a number of surface cracks 3 mm deep and 10 mm long. If the glass is placed horizontally on two supports, calculate the maximum spacing of the supports to avoid the fracture of the glass due to its -3/2 3 own weight. For glass K IC = 0.3 (MN ) and density = 2600 kg/m .
8.
The accident report on a steel pressure vessel which fractured in a brittle manner when 2 the internal pressure of 19 MN/m had been applied to it shows that the vessel had a longitudinal crack 8 mm long and 3.2 mm deep. A subsequent fracture mechanics test on -3/2 a sample of the steel showed that it had a K IC value of 75 MN . If the vessel diameter was 1 m and the thickness was 10 mm, determine whether the data reported are consistent with the observed failure.
9.
An aluminum alloy plate with a yield stress of 450 MN/m fails in service at a stress of 2 110 MN/m . The conditions are plane stress and there is some indication of ductility at the fracture. If a surface crack of 20 mm long is observed at the fracture plane calculate the size of the plastic zone at the crack tip. Calculate also the percentage error likely if LEFM was used to obtain the fracture toughness of this material.
2
Problem Set Page 8 of 8