(Model answers will be posted shortly after the due date.)
Question 6.3
A specimen of aluminium having a rectangular cross-section of 10 mm by 12.7 mm is pulled in tension with 35,500 N force, producing only elastic deformation. Calculate the resulting strain.
Answer 6.3
The stress on the aluminium is 35,500/(0.01 * 0.0127) Pa. Young's modulus for aluminium is 69 GPa, so the strain is 35,500/(0.01 * 0.0127 * 69,000,000,000) = 0.00405.
Question 6.8
A cylindrical rod of copper (E=110 GPa) having a yield strength of 240 Mpa, will be subjected to a load of 6660 N. If the rod is 380 mm long, what must its diameter be to allow an elongation of 0.5 mm?
Answer 6.8
Let the diameter be d. Then the stress on the copper is 4 * 6660/(pi * d2). The strain is going to be (0.5/380), and Young's modulus is 110 GPa.
So, stress = Young's modulus * strain
4*6660/(pi * d2)=110,000,000,000 * 0.5 /380 = 145,000,000
So d = (4 * 6660 * 380/(pi * 110,000,000,000 * 0.5))0.5 = 7.654 mm
Question 6.23
A cylindrical rod 120 mm long, diameter 15 mm, is to be deformed using a tensile load of 35 kN. It must experience neither plastic deformation nor a diameter reduction of greater than 0.012 mm. Which of the materials listed below, if any, are possible candidates?
| Material | E (GPa) | Yield Strength (MPa) | Poisson's Ratio |
|---|---|---|---|
| Aluminium | 70 | 250 | 0.33 |
| Titanium | 105 | 850 | 0.360 |
| Steel | 205 | 550 | 0.27 |
| Magnesium | 45 | 170 | 0.2 |
Answer 6.23
First, consider the constraint that the rod must not experience plastic deformation, i.e., it must not yield. The imposed stress is 35000/(pi * (0.015/2)2) = 198 MPa. So we eliminate any candidate with a yield stress lower than this, which gets rid of magnesium.
Next we calculate the diameter reduction for each candidate. This will be given by the expression
Diameter reduction = - nu sigma d0 / E
Plugging in the numbers for the remaining candidates, this evaluates to:
1.4 * 10-5 m for aluminium
1.02 * 10-5 m for titanium
0.391* 10-5 m for steel
The maximum allowable diameter reduction is 1.2 * 10-5, so this constraint eliminates aluminium, leaving titanium and steel as the remaining candidates.
Question 6.42
For a brass alloy, the following engineering stresses produce the corresponding plastic engineering strains prior to necking:
| Engineering Stress (MPa) | Engineering Strain |
|---|---|
| 315 | 0.105 |
| 340 | 0.220 |
On the basis of this information, calculate the engineering stress necessary to produce an engineering strain of 0.28.
Answer 6.42
You can't solve this by assuming that the stress-strain relationship is linear in the plastic region, because it isn't. The only way to proceed is to use Equation 6.19, which specifies a non-linear relationship between true stress and true strain. So we first convert the engineering stress and strain to their true counterparts:
True stress 1 = stress1(1+ strain1)=315(1.105) = 348 MPa
True stress 2 = stress2(1+ strain2)=340(1.220) = 415 MPa
True strain 1 = loge(1+ strain1)=log(1.105) = 0.0998
True strain 2 = loge(1+ strain2)=log(1.220) = 0.199
Substituting these into Equation 6.19 and solving, we get
n = 0.255
K = 626 MPa
Now, to produce an engineering strain of 0.28, we must produce a true strain of log(1.28) = 0.247. So the true stress = 626 * (0.247)0.255 = 438 MPa
But we need to know the engineering stress, which is the true stress/(1.28) = 342 MPa.
Question 7.11
Consider a metal single crystal oriented with the normal to the slip plane and the slip direction at angles of 60 degrees and 35 degrees respectively to the tensile axis.
If the critical resolved shear stress is 5.7 MPa [or 6.2 MPa if you're working from the 5th edition], will an applied stress of 12 MPa cause the crystal to yield? If not, what will?
Answer 7.11
This calls for Schmid's Law, which is Equation 7.1 in Callister:
resolved shear stress = applied stress * cos(phi) * cos(lambda)
Plugging in numbers, we see that it would take an applied stress of 14 MPa to produce a resolved shear stress of 5.7, and an applied stress of 15.1 MPa to produce a resolved shear stress of 6.2 MPa. So the crystal won't fail, no matter which edition you're working from.
Question 7.30
(i) What is the approximate ductility (%EL) of a brass that has a yield strength of 275 Mpa?
(ii)What is the approximate Brinell hardness of a 1040 steel having a yield strength of 690 Mpa?
Answer 7.30
(i) This can be answered by referring to Figure 7.17: from part (a) of the figure, a yield strength of 275 MPa corresponds to 10% cold work; and from part (c), 10% cold work corresponds to 41-46 %EL ductility.
(ii) From Fig. 7.17(a), 1040 steel has a yield strength of 6980 MPa at 11% cold work. From 7.17 (b), 11% cold work corresponds to 780 MPa tensile strength. We can then use Equation 6.20a to estimate the Brinell hardness:
HB = TS/3.45 = 226
Question 7.41
An uncold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 Mpa. Estimate the yield strength of this alloy after it has been heated to 500 degrees Celsius for 1000 seconds, if it is known that the value of sigma0 is 25 Mpa.
Answer 7.41
We know that heating is going to make the grains bigger, and we know that bigger grains mean a softer material. So we expect that the answer is going to be somewhat less than 150 MPa.
We have to estimate the size of the grains after heating. For this we consult Figure 7.23, noting that the x-axis is in minutes, not seconds. Interpolating by eye on a small log-log graph is difficult; the final grain size is somewhere in the range 0.015 to 0.02 mm. (If you ever need to calculate a more accurate figure, you will be provided with a more detailed graph.)
To find the new strength, we use the Hall-Petch equation:
strength = sigma0+kyd-1/2
We first substitute in the initial strength in order to find ky. The value we get for ky depends on the units we use for d. It doesn't matter what units we use, as long as we use the same units for the grain diameter before and after heating. (I used millimeters, and got k=12.5.)
Putting the new grain size into the equation, we get the final yield strength to be about 121 MPa; which is less than 150 MPa, as we expected.