mats 3 sol 3

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MatS 2001 - Fall 2012

Problem Set 3 Solutions: Assigned 09/14/12

3-1 A Ni-Cr steel beam is designed to support a load at 50% of the yield stress. The beam is bolted to a bridge truss using a hole drilled through the center. What is the maximum allowable hole diameter? Assume Y = 1.4. Note, given more than one set of material data, choose conservatively.

3-2 A poly(methyl methacrylate) (PMMA) window (1 m by 1 m by 2 mm thick), hangs freely

(vertically) from a frame (clamped at the top). It is pierced cleanly at the center by a 22 caliber steel jacketed bullet. Does the window fracture? Assume the stress at any vertical location in the window results from the mass hanging below that level. Also, the density of PMMA is 1 g/cm3. Give analysis to support your answer.

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3-3 Fatigue data for a brass alloy are given as follows:

Stress Amplitude (MPa) Cycles to failure 170 3.7x104 148 1.0x105

130 3.0x105

114 1.0x106

92 1.0x107

80 1.0x108 74 1.0x109

Make an S-N plot using these data. (A) Determine the fatigue strength at 4x106

cycles

(B) Determine the fatigue life in number of cycles for 120 MPa.

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3-4 The following fatigue data were acquired for a new metal alloy. K (MPam1/2) da/dN (m/cycle)

0.03 10-7

0.10 10-6 0.30 10-5

Using an appropriate plot determine the constants A and m.

m=1.99, A= 1.07x10-4 3-5 Consider a flat plate of some metal alloy that is to be exposed to repeated tensile-

compressive cycling in which the mean stress is 25 MPa. If the initial and critical surface crack lengths are 0.15 and 4.5 mm, respectively, and the values of m and A are 3.5 and 2

x 1014, respectively (for in MPa and a in m), estimate the maximum tensile stress to yield a fatigue life of 2.5 x107 cycles. Assume the parameter Y has a value of 1.4, which is independent of crack length.

Estimate the maximum tensile stress that will yield a fatigue life of 2.5 x 107 cycles, given values of ao, ac, m, A, and Y. Use the following equation:

Nf = 1

Aπm/2()mYm

ao

ac

a-m/2da

For m = 3.5

y = 1.9986x - 3.9709

-8

-7

-6

-5

-4

-3

-2

-1

0

-2 -1.5 -1 -0.5 0

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Nf = 1

Aπ1.75()3.5Y3.5

ao

ac

a-1.75da

= - 1.33

Aπ1.75()3.5Y3.5

1

ac0.75

- 1

ao0.75

Solving for yields:

1.33

NfAπ1.75Y3.5

1

ao0.75

- 1

ac0.75

1/3.5

1.33

(2.5 x 107)(2 x 10-14)(π)1.75(1.4)3.5

1

( )1.5 x 10-4 0.75 -

1

( )4.5 x 10-3 0.75

1/3.5

= 178 MPa

This 178 MPa will be the maximum tensile stress since we can show that the

minimum stress is a compressive one--when min is negative, is taken to be

max. If we take max = 178 MPa, and since m is stipulated in the problem to

have a value of 25 MPa, then from Equation (8.21)

min = 2m - max = 2(25 MPa) -178 MPa = -128 MPa

Therefore min is negative and we are justified in taking max to be 178 MPa.

3-6 Give the approximate temperature at which creep deformation becomes an important

consideration for each of the following metals: Creep becomes important at 0.4T

m, T

m being the absolute melting temperature of the

metal. For Ni, 0.4T

m = (0.4)(1455 + 273) = 691

K or 418°C

For Cu, 0.4Tm

= (0.4)(1085 + 273) = 543 K or 270°C

For Fe, 0.4Tm

= (0.4)(1538 + 273) = 725 K or 450°C

For W, 0.4Tm

= (0.4)(3410 + 273) = 1473 K or 1200°C

For Pb, 0.4Tm

= (0.4)(327 + 273) = 240 K or -33°C

For Al, 0.4Tm

= (0.4)(660 + 273) = 373 K or 100°C

3-7 The following creep data were taken on an aluminum alloy at 400°C and a constant stress

of 25 MPa. Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included.

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4030201000.0

0.1

0.2

0.3

0.4

T i m e ( m i n )

Strain

The steady-state creep rate (/t) is the slope of the linear region as

t =

0.230 - 0.0930 min - 10 min

= 7.0 x 10-3 min-1

3-8 Consider a thin metal plate 20 mm wide which contains a centrally positioned, through-

thickness crack. This plate is to be exposed to reversed tensile-compressive cycles of stress amplitude 125 MPa. If the initial and critical crack lengths are 0.20 and 8.0 mm,

respectively, and the values of m and A are 4 and 5x10-12, respectively (for in MPa and a in m), estimate the fatigue life of this plate.

This problem asks that we estimate the fatigue life of a flat plate that has a centrally positioned through-thickness crack, given that W = 20 mm, 2ao = 0.20

mm, 2ac = 8.0 mm, m = 4.0, and A = 5 x 10-12. Furthermore, inasmuch as

reverse stress cycling is to be used = 125 MPa. For this plate and crack geometry, the parameter Y=1. Hence, the equation for Nf takes the form

Nf =

0

/2/2

1 ca

m mma

da

aA

=

0

4 212 2

1

5 10

ca

a

da

a

Which, upon integration, leads to the solution

1 1

412 2

11 2 0.004 0.0001

5 10 125

fN

= 8.09 x 105 cycles

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3-9 For an 18-8 Mo stainless steel (Figure 8.43), predict the time to rupture for a component

that is subjected to a stress of 80 MPa at 700°C.

Determine, for an 18-8 Mo stainless steel, the time to rupture for a component that is

subjected to a stress of 80 MPa at 700°C (973 K). From Figure 8.43, the value of the

Larson-Miller parameter at 80 MPa is about 23.5 x 103, for T in K and tr in h. Therefore,

23.5 x 103 = T(20 + log tr)

= 973(20 + log tr)

And, solving for tr

24.15 = 20 + log tr

which leads to tr = 1.42 x 104 h = 1.6 yr.

3-10 Consider an 18-8 Mo stainless steel component (Figure 8.43) that is exposed to a

temperature of 500°C. What is the maximum allowable stress level for a rupture lifetime of 5 years? 20 years?

Calculate the stress levels at which the rupture lifetime will be 5 years and 20 years

when an 18-8 Mo stainless steel component is subjected to a temperature of 500°C (773

K). It first becomes necessary, using the specified temperature and times, to calculate the values of the Larson-Miller parameter at each temperature. The values of tr

corresponding to 5 and 20 years are 4.38 x 104 h and 1.75 x 105 h, respectively.

Hence, for a lifetime of 5 years

T(20 + log tr) = 773[20 + log (4.38 x 104)] = 19.05 x 103

And for tr = 20 years

T(20 + log tr) = 773[20 + log (1.75 x 105)] = 19.51 x 103

Using the curve shown in Figure 8.43, the stress values corresponding to the five-

and twenty-year lifetimes are approximately 260 MPa and 225 MPa, respectively.

3-11 (a) Estimate the activation energy for creep (i.e., Qc in 2 expn CS

QK

RT

) for the

low carbon– nickel alloy having the steady-state creep behavior shown in Figure 8.39. Use data taken at a stress level of 55 MPa and temperatures of 427°C and 538°C.

Assume that the stress exponent n is independent of temperature. (b) Estimate S at

649°C. (a) Estimate the activation energy for creep for the low carbon-nickel alloy having the

steady-state creep behavior. Since is a constant, the equation takes the form

. s = K2

nexp

- QcRT

= K2' exp

- QcRT

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where K2' is now a constant. (Note: the exponent n has about the same value at

these two temperatures Problem 22.) Taking natural logarithms of the above expression

ln . s = ln K2' -

QcRT

For the case in which we have creep data at two temperatures (denoted as T1 and

T2) and their corresponding steady-state creep rates (. s1

and . s2

), it is possible to set

up two simultaneous equations of the form as above, with two unknowns, namely K2'

and Qc. Solving for Qc yields

Qc = -

R

ln .

s1 - ln

.

s2

1

T1 -

1T2

Let us choose T1 as 427°C (700 K) and T2 as 538°C (811 K); then from Figure

8.39, at = 55 MPa, . s1

= 0.01%/1000 h = 1 x 10-7 (h)-1 and . s2

= 0.8%/1000 h = 0.8

x 10-5 (h)-1. Substitution of these values into the above equation leads to

Qc = - (8.31 J/mol-K)[ ]ln( )10-7 - ln( )0.8 x 10-5

1

700 K -

1811 K

= 186,200 J/mol

(b) We are now asked to calculate . s at 649°C. It is first necessary to determine

the value of K2' , which is accomplished using the first expression above, the value of

Qc, and one value each of . s and T (say

. s1

and T1). Thus,

K2' = . s1

exp

Qc

RT1

= [ ]10-7 (h)-1 exp

186200 J/mol

(8.31 J/mol-K)(700 K) = 8.0 x 106 (h)-1

Now it is possible to calculate . s at 922 K as follows:

. s = K2' exp

- QcRT

= [ ]8.0 x 106 (h)-1 exp

-

186200 J/mol(8.31 J/mol-K)(922 K)

= 2.23 x 10-4 (h)-1 = 22.3 %/1000 h

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