6.52 A cylindrical specimen of a brass alloy 10.0 mm (0.39 in.) in diameter and 120.0 mm (4.72 in.) long is pulled in tension with a force of 11,750 N (2640 lbf); the force is subsequently released.
(a) Compute the final length of the specimen at this time. The tensile stress–strain behavior for this alloy is shown in Figure 6.12.
(b) Compute the final specimen length when the load is increased to 23,500 N (5280 lbf) and then released.
Solution
(a) In order to determine the final length of the brass specimen when the load is released, it first becomes necessary to compute the applied stress using Equation 6.1; thus
= F
Upon locating this point on the stress-strain curve (Figure 6.12), we note that it is in the linear, elastic region;
therefore, when the load is released the specimen will return to its original length of 120 mm (4.72 in.).
(b) In this portion of the problem we are asked to calculate the final length, after load release, when the load is increased to 23,500 N (5280 lbf). Again, computing the stress
= 23,500 N
The point on the stress-strain curve corresponding to this stress is in the plastic region. We are able to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic region; this line intersects the strain axis at a strain of about 0.012 which is the amount of plastic strain. The final specimen length li may be determined from a rearranged form of Equation 6.2 as
li = l0(1 + ) = (120 mm)(1 + 0.012) = 121.44 mm (4.78 in.)
6.53 A steel alloy specimen having a rectangular cross section of dimensions 19 mm × 3.2 mm (34 in. × 18 in.) has the stress–strain behavior shown in Figure 6.22. This specimen is subjected to a tensile force of 110,000 N (25,000 lbf).
(a) Determine the elastic and plastic strain values.
(b) If its original length is 610 mm (24.0 in.), what will be its final length after the load in part (a) is applied and then released?
Solution
(a) We are asked to determine both the elastic and plastic strain values when a tensile force of 110,000 N (25,000 lbf) is applied to the steel specimen and then released. First it becomes necessary to determine the applied stress using Equation 6.1; thus
From Figure 6.22, this point is in the plastic region so the specimen will be both elastic and plastic strains. The total strain at this point, t, is about 0.020. From this point on the stress-strain curve we are able to estimate the amount of strain recovery e from Hooke's law, Equation 6.5 as
e = E
And, since E = 207 GPa for steel (Table 6.1)
e = 1810 MPa
207 103MPa = 0.0087
The value of the plastic strain, p is just the difference between the total and elastic strains; that is
p = t – e = 0.020 – 0.0087 = 0.0113
(b) If the initial length is 610 mm (24.0 in.) then the final specimen length li may be determined from a rearranged form of Equation 6.2 using the plastic strain value as
li = l0
(
1 + p)
= (610 mm)(1 + 0.0113) = 616.9 mm (24.26 in.)Hardness
6.54 (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this material.
(b) What will be the diameter of an indentation to yield a hardness of 300 HB when a 500-kg load is used?
Solution
(a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the equation in Table 6.5 for HB, where P = 1000 kg, d = 2.50 mm, and D = 10 mm. Thus, the Brinell hardness is
(b) This part of the problem calls for us to determine the indentation diameter d that will yield a 300 HB when P = 500 kg. Solving for d from the equation in Table 6.5 gives
d = D2 D 2P
6.55 (a) Calculate the Knoop hardness when a 500-g load yields an indentation diagonal length of 100 m.
(b) The measured HK of some material is 200. Compute the applied load if the indentation diagonal length is 0.25 mm.
Solution
(a) Knoop microhardness is related to the applied load (P in kg) and indentation length (l in mm) according to the equation given in Table 6.5—viz.
HK14.2P l2
The applied load is 500 g = 0.500 kg, whereas the indentation length is 100 m = 0.100 mm. Incorporating these values into the equation above leads to
HK(14.2)(0.500 kg) (0.100 mm)2 710
(b) This portion of the problem asks for the applied load for HK = 200 and l = 0.25 mm. Assigning P in the above equation to be the dependent variable yields the following expression:
P(HK)l2 14.2
When we insert the above values for HK and l into this expression, the value of the P is computed as follows:
P(200)(0.25 mm)2
14.2 0.880 kg
6.56 (a) What is the indentation diagonal length when a load of 0.60 kg produces a Vickers HV of 400?
(b) Calculate the Vickers hardness when a 700-g load yields an indentation diagonal length of 0.050 mm.
Solution
(a) The equation given in Table 6.5 for Vickers microhardness is as follows:
HV1.854P d12
Here P is the applied load (in kg) and d1 is the diagonal length (in mm). This portion of the problem asks that we compute the indentation diagonal length. Rearranging the above equation to make d1 the dependent variable gives
d1 1.854P HV
Using values for P and HV given in the problem statement yields the following value for d1:
d1 (1.854)(0.60 kg)
400 0.0527 mm
(b) For P = 700 g (0.700 kg) and d1 = 0.050 mm, the Vickers hardness value is
HV1.854P d12
(1.854)(0.700 kg) (0.050 mm)2 519
6.57 Estimate the Brinell and Rockwell hardnesses for the following:
(a) The naval brass for which the stress–strain behavior is shown in Figure 6.12.
(b) The steel alloy for which the stress–strain behavior is shown in Figure 6.22.
Solution
This problem calls for estimations of Brinell and Rockwell hardnesses.
(a) For the brass specimen, the stress-strain behavior for which is shown in Figure 6.12, the tensile strength is 450 MPa (65,000 psi). From Figure 6.19, the hardness for brass corresponding to this tensile strength is about 125 HB or 85 HRB.
(b) The steel alloy (Figure 6.22) has a tensile strength of about 1950 MPa (283,000 psi) [Problem 6.26(d)].
This corresponds to a hardness of about 560 HB or ~55 HRC from the line (extended) for steels in Figure 6.19.
6.58 Using the data represented in Figure 6.19, specify equations relating tensile strength and Brinell hardness for brass and nodular cast iron, similar to Equations 6.20a and 6.20b for steels.
Solution
This problem calls for us to specify expressions similar to Equations 6.20a and 6.20b for nodular cast iron and brass. These equations, for a straight line, are of the form
TS = C + (E)(HB)
where TS is the tensile strength, HB is the Brinell hardness, and C and E are constants, which need to be determined.
One way to solve for C and E is analytically--establishing two equations using TS and HB data points on the plot, as
(TS)1 = C + (E)(HB)1 (TS)2 = C + (E)(HB)2 Solving for E from these two expressions yields
E = (TS)1(TS)2 (HB)2(HB)1
For nodular cast iron, if we make the arbitrary choice of (HB)1 and (HB)2 as 200 and 300, respectively, then, from Figure 6.19, (TS)1 and (TS)2 take on values of 600 MPa (87,000 psi) and 1100 MPa (160,000 psi), respectively.
Substituting these values into the above expression and solving for E gives
E =600 MPa1100 MPa
200 HB300 HB = 5.0 MPa/HB (730 psi/HB)
We may determine the value for the constant C by inserting this value of E into either of the (TS)1 or (TS)2 equations cited above. Using the (TS)1 expression, we solve for C as follows:
C = (TS)1 – (E)(BH)1
= 600 MPa - (5.0 MPa/HB)(200 HB) = – 400 MPa (– 59,000 psi) Thus, for nodular cast iron, these two equations take the form
TS(MPa) = – 400 + 5.0 HB TS(psi) = – 59,000 + 730 HB
Now for brass, we take (HB)1 and (HB)2 as 100 and 200, respectively, then, from Figure 6.19, (TS)1 and (TS)2 take on values of 370 MPa (54,000 psi) and 660 MPa (95,000 psi), respectively. Substituting these values into the above expression for E gives
E =370 MPa660 MPa
100 HB200 HB = 2.9 MPa/HB (410 psi/HB)
We determine the value of C in using the same procedure outlined above:
C = (TS)1 – (E)(BH)1
= 370 MPa – (2.9 MPa/HB)(100 HB) = 80 MPa (13,000 psi) Thus, for brass these two equations that relate tensile strength and Brinell hardness take the form
TS(MPa) = 80 + 2.9 HB TS(psi) = 13,000 + 410 HB