(Model answers are posted below)

Question 8.4

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically-shaped surface crack of length 0.5 mm and having a tip radius of curvature of 5 x 10^-3 mm when a stress of 1035 MPa is applied.

Answer 8.4

This uses the formula

sigma_m= 2 sigma₀(a/rho_t)^0.5

where sigma₀, the applied stress, is 1035 MPa, a is 0.0005 m, and rho_t is 0.000005 m. Evaluating the formula then gives an effective stress of 20.7 GPa.

Question 8.14

A specimen of 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa m^0.5 is exposed to a stress of 1030 MPa. Will this specimen experience fracture if it is known that the largest surface crack is 0.5 mm long? Why, or why not? (Assume that the parameter Y has a value of 1.0.)

Answer 8.14

The critical stress, sigma, can be calculated from the equation

K_IC = Y sigma (pi a)^0.5

where K_IC=54.8 * 10⁶, Y=1 and a = 0.5 * 10^-3.

Using these figures, sigma evaluates to 1382 MPa. The applied stress, 1030 MPa, is less than this, so fracture will not occur.

Question 8.26

A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load amplitude is 66,700 N, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a safety factor of 2.0.

Answer 8.26

This question relies on the fact that steels have a fatigue limit, that is, a limiting stress below which fatigue failure will not occur.

From Figure 8.44 (Callister, 6th Edition), this limit is about 315 MPa (anything from 310 to 320 MPa is acceptable here).

So the stress amplitude must be less than 315/2 MPa, given the safety factor of 2.

So 315,000,000/2 = 66,700/(pi d²/4)

from which d = 23.2 mm. (Anything in the range 22-24 mm is acceptable.)

Question 8.49

Steady-state creep rate data are given here for some alloy taken at 200 C:

epsilons(h-1)	sigma (MPa)
2.5 x 10-3	55
2.4 x 10-2	69

If it is known that the activation energy for creep is 140,000 J/mol, compute the steady-state creep rate at a temperature of 250 C and a stress level of 48 MPa.

Answer 8.49

This relies on the equation for creep rate:

x = K₂epsilon ⁿ exp(-Q_c/(RT))

where x is the creep rate, often denoted by `epsilon_s with a dot on top'.

We will use the data supplied to find K₂ and n.

0.0025=K₂(55 * 10⁶)ⁿexp(-140,000/(8.31 * 473))

0.024=K₂(69 * 10⁶)ⁿexp(-140,000/(8.31 * 473))

To find n, divide the second equation by the first to obtain

24/2.5 = (69/55)ⁿ

Taking logs and solving, we get n = 9.974.

Finding K₂ is easier if we choose suitable units. If we choose h^-1MPa^-n it comes out to 3.22 * 10^-5 (whereas if we choose h^-1Pa^-n it comes out as a frighteningly small number.)

Plugging these numbers back into the creep rate equation gives us

x = 0.0194 hours^-1

(Don't omit the units.)

Question 9.6

For an alloy of composition 52 wt% zinc, 48 wt% copper, cite the phases present and their mass fractions at the following temperatures: 1000C, 800C, 500C and 300C.

Answer 9.6

At 1000 C, the only phase present is liquid, with the same composition (obviously) as the alloy.

At 800 C, the only phase present is beta, with the same composition (obviously) as the alloy.

At 500 C, the two phases present are beta, with composition about 65% Zn, and gamma, with composition about 35% Zn.

At 300 C, the two phases present are beta^/, with composition about 75% Zn, and gamma, with composition about 25% Zn.

Question 9.15

A magnesium-lead alloy of mass 7.5 kg consists of a solid alpha-phase that has a composition just slightly below the solubility limit at 300 C.

1. What mass of lead is in the alloy?
2. If the lead is heated to 400 C, how much more lead may be dissolved in the alpha phase without exceeding the solubility limit for this phase?

Answer 9.15

1. The composition just below the solubility limit at 300 C is about 18% Pb, so the mass of lead in the alloy is about 0.18 * 7.5 = 1.35 kg lead.
2. At 400 C, the solubility limit is about 32% lead. It's very tempting to think that if we add lead till we reach the solubility limit, we'll have a total of 0.32*7.5 = 2.4 kg of lead, implying 1.05 kg of lead have been added, but this is WRONG! Once we've added lead, the total mass is no longer 7.5 kg. Instead, we reason as follows:

   Suppose x kg of lead are added. Then the fraction of lead in the mixture is

   (x+1.35)/(x+7.5) = 0.32

   which can be solved to get x = 1.54 kg.

Question 9.28

For 5.7 kg of a magnesium-lead alloy of composition 50%wt Pb-50%wt Mg, is it possible, at equilibrium, to have alpha amd Mg₂Pb phases with respective masses of 5.13 and 0.57 kg? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why not.

Answer 9.28

It is not possible, as we can tell by applying the tie-line rule; we can't get a point sufficiently close to the alpha region.

Question 9.35

Consider the hypothetical eutectic phase diagram for metals A and B, which is similar to that for the lead-tin system (Figure 9.7 in Callister, 6th edition) Assume that:

1. alpha and beta phases exist at the A and B extremities of the phase diagram, respectively;
2. the eutectic composition is 36wt% A, 64wt% B and
3. the composition of the alpha phase at the eutectic temperature is 88 wt%A 12 wt% B.

Answer 9.35

It's probably helpful to sketch a phase diagram for this. Since the alloy is going to yield primary beta, that is, beta deposited before we get to the eutectic point, its composition must be to the `B' side of the eutectic composition. So the composition of the eutectic is 64% B, the composition of the alloy is to the B side of this, c₁, say, and the composition of the beta phase at the eutectic temperature is even further to the B side, c_B, say.

Now apply the lever rule to the system just above the eutectic temperature, when the two phases present are liquid of the eutectic composition and the primary beta-phase solid:

(c₁ - 0.64)/(c_B - 0.64) = 0.367

Next, apply the lever rule again, this time just below the eutectic temperature. Now the phases present are alpha and beta; we know that the alpha phase is 12% B, so we can apply the lever rule a second time to give the mass fraction of total beta:

(c₁ - 0.12)/(c_B - 0.12) = 0.768

Straightforward algebraic manipulation then gives us that c₁=0.75 and c_B = 0.9432. So we conclude that the alloy is 75% B, 25% A.

Question 9.63

Compute the mass fraction of eutectoid cementite in an iron-carbon alloy that contains 1.00 wt% C.

Answer 9.63

This is a hypereutectoid alloy, so as we cool, we will first deposit proeutectoid cementite, then, at the eutectoid temperature, the remaining austenite will convert to pearlite. This pearlite is made up of ferrite and eutectoid cementite.

The proeutectoid cementite mass fraction can be calculated from Equation 9.21:

Mass fraction = (1.00 - 0.76)/(6.7 - 0.76) = 4%

So the remaining 96% of the alloy will be pearlite. Since the pearlite has the eutectoid composition, 0.76% of it will be carbon. To find the weight fractions of alpha-ferrite and cementite in the pearlite, we apply the lever rule; the tie-line goes from the alpha-ferrite/pearlite boundary, which is at 0.022% C, through the eutectoid point, and ends at the end of the cementite zone at 6.7%C.

So mass fraction of cementite in pearlite = (0.76 - 0.022)/(6.7 - 0.022) = 11%.

So 96% of the alloy is pearlite, and 11% of the pearlite is eutectoid cementite, so the eutectoid cementite makes up 96% * 11% = 10.5% of the alloy.