Heat Treatment of Steel

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1 Heat Treatment of Steel Engineering 45 Materials Monday 8:00 AM Lab Troy Topping 10/31/11 Deanna Becker 11/21/11

2 Abstract This procedure compared properties of heat treated 4140 steels and accurately represented processes that material science engineers use in design. Two specimens of inch-diameter 4140 AR steel were heat treated into the quenched and tempered state and hardness and tensile tests were performed. From the stress-strain curves created, the elastic modulus, the 0.2% offset YS, the UTS, the % elongation, and the % reduction for samples on samples A and B were as follows respectively: 33,866 ksi and 29,162 ksi; 166 ksi and 168 ksi; 180,325 psi and 186,940 psi; 11.30% and 15.19%; and 36.03% and 37.46%. The hardnesses in the as quenched state on the Rockwell C scale for samples A and B were determined to be 47.7 and 47.3 respectively. The hardnesses in the quenched and tempered state on the Rockwell C scale for samples A and B were determined to be 31.9 and 33.4 respectively. Also, two specimens of inch-diameter 4140 AR steel were heat treated into the fully annealed state and hardness and tensile tests were performed. From the stress-strain curves created, the elastic modulus, the 0.2% offset YS, the UTS, the % elongation, and the % reduction for samples on samples A and B were as follows respectively: 27,740 ksi and 28,977 ksi; 54 ksi and 54 ksi; 62,776 psi and 78,772 psi; 28.48% and 27.00%; and 44.86% and 45.07%. The hardnesses on the Rockwell B scale for samples A and B were determined to be 86.0 and 86.6 respectively. Data was used from previous hardness and tensile tests done on inch-diamter 4140 AR specimens to compare properties. From the stress-strain curves created, the elastic modulus, the 0.2% offset YS, the UTS, the % elongation, and the % reduction for samples on Monday and Wednesday were as follows respectively: 23,917 ksi and 22,706 ksi; 74 ksi and 74 ksi; 103,230 psi and 103,315 psi; 15.46% and 17.80%; and 34.28% and 35.36%. The hardnesses on the Rockwell B scale were 96.5 and 96.3 for the Monday and Wednesday samples respectively.

3 Introduction Design engineers choose their materials with aims to accommodate for particular functions such as need for strength or ductility. Material science engineers heat-treat their materials to create a substance which holds the specific combination of properties needed for the functions. This lab s purpose was to acquaint the student with the Fe-Fe3C phase diagram and to instill them with the proper knowledge used in 4140 steel design. In this lab, 4140 AR steel is heat treated into the quenched and tempered state, and also into the fully annealed state. The properties of hardness, strength, and ductility are compared and carefully analyzed. Upon completion of this lab, the student will be able to properly discuss the purposes of quenching, tempering, and annealing 4140 steel. The student will also understand the microscopic transformations and will be able to relate them to the macroscopic outcomes. These skills will serve great purpose in future design and material choices.

4 Procedure 1. Put inch-diameter 4140 As Received (AR) steel specimen at room temperature in a carbon-rich substance. 2. Heat the specimen to about 1600 F into the Austenite γ region. 3. Using tongs, quench the Austenite in oil to room temperature to create Martensite. 4. Perform a Rockwell C hardness test on the As Quenched (AQ) specimen. 5. Temper the specimen at about 1000 F for 1 hour. 6. Perform a Rockwell C hardness test on the Quenched and Tempered (QT) specimen. 7. Perform a tensile test on the QT specimen. a. Be sure to remove the extensometer at 0.1 strain. 8. Create a Fully Annealed (FA) 4140 steel specimen by heating it to the γ region at about 1600 F then turning off the furnace and equilibrium cooling it overnight. 9. Perform a Rockwell B hardness test on the FA specimen. 10. Perform a tensile test on the FA specimen. a. Be sure to remove the extensometer at 0.2 strain. 11. Use excel to graph the data and determine mechanical properties for further analysis.

5 Results Below experimental results for FA, AR, and QT 4140 steel specimens are listed and compared to literature values in Tables 1, 2, and 3 respectively. Additionally, Figure 1 shows a complete stress-strain curve for all six samples (two of each) with Tensile Strength (TS) and Breaking Strength (BS) labeled. The values of the Elastic Modulus (E) and Yield Strength (YS) are determined for FA, AR, and QT in Figures 2, 3, and 4 respectively. Table 1: Properties for 4140 FA specimen* Property Sample A Sample B Literature % Difference A Elastic Modulus 27,740 ksi 28,977 ksi 29,700 ksi 6.6% 2.4% Yield Strength 54 ksi 54 ksi 60.2 ksi 10.3% 10.3% Tensile Strength 62,776 psi 78,772 psi 95,000 psi 33.9% 17.1% Hardness (HRb) % 5.9% % Elongation 28.48% 27.00% 25.7% 10.8% 5.1% % Reduction in Area % Difference B 44.86% 45.07% 56.9% 21.8% 20.8% Table 2: Properties for 4140 AR specimen** Property Sample Sample % Difference Literature Mon Wed Mon Elastic Modulus 23,917 ksi 22,706 ksi 29,700 ksi 19.5% 23.6% Yield Strength 74 ksi 74 ksi 92.1 ksi 19.7% 19.7% Tensile Strength 103,230 psi 103,315 psi 141,000 psi 26.8% 26.7% Hardness (HRb) % 2.7% % Elongation 15.46% 17.80% 16.5% 6.3% 7.9% % Reduction in Area % Difference Wed 34.28% 35.36% 48.1% 28.7% 26.5% *Literature values found at 6b&ckck=1 **Literature values found at e9&ckck=1

6 Table 3: Properties for 4140 QT specimen*** % Difference B Property Sample A Sample B Literature % Difference A Elastic Modulus 33,866 ksi 29,162 ksi 29,700 ksi 14.0% 1.8% Yield Strength 166 ksi 168 ksi 140 ksi 18.6% 20.0% Tensile Strength 180,325 psi 186,940 psi 165,000 psi 9.3% 13.3% Hardness (HRc) AQ % 12.6% Hardness (HRc) QT % 4.6% % Elongation 11.30% 15.19% 18.0% 37.2% 15.6% % Reduction in Area 36.03% 37.46% 56.0% 35.7% 33.1% According to the literature values displayed in Tables 1, 2, and 3, strength and hardness seem to align themselves highest to lowest QT, AR, then FA respectively. Conversely, % elongation and % reduction in area seem to align themselves highest to lowest FA, QT, then AR respectively. High percent differences may be due to the furnace calibration needing some adjustment. ***Literature values for AQ found at 8b&ckck=1 Literature values for QT found at 18

7 Figure 1: Complete Stress-Strain Curves for 4140 Steels This graph shows the determination of the TS and BS of all samples of FA, AR, and QT 4140 steels.

8 Figure 2: FA YS and E (a) and (b) This graph shows the determination of the YS and E for the 4140 FA steel samples (a) and (b).

9 Figure 3: AR YS and E (mon) and (wed) This graph shows the determination of the YS and E for the 4140 AR steel samples (mon) and (wed).

10 Figure 4: QT YS and E (a) and (b) This graph shows the determination of the YS and E for the 4140 QT steel samples (a) and (b).

11 Discussion Heat treatment of steel is a common practice among material science engineers that helps control the properties of the steel such as hardness, strength, and ductility. For example, annealing steel can increase ductility, austenitizing and quenching steel significantly raises tensile strength, and quenching and tempering adds ductility but maintains a relatively high tensile strength. As demonstrated in Tables 1, 2, and 3, the elastic moduli for the heat-treated specimens were unaffected by the heat treatments and remained relatively the same as the AR specimens. This was expected because the modulus of elasticity is a linear material property that depends on inner atomic structure and does not change if the material is strengthened or made more ductile. As demonstrated in Tables 1, 2, and 3, the tensile strengths and breaking strengths for the QT specimens were much larger than the FA specimens. On the other hand, the tensile strengths for the AR specimens were higher than the FA specimens, but the breaking strengths were lower. These results were expected because quenching steel makes it stronger, while annealing sacrifices some strength in exchange for ductility. An annealed specimen consists of ferrite and pearlite. The pearlite consists of cementite and ferrite. This specimen is formed by raising the temperature of the steel to the austenite γ region, and then equilibrium cooling it to the α + Fe3C region. The as received specimen consists of pearlite that is made of ferrite and cementite. This is formed by maintaining the steel in the α + Fe3C region. An as quenched specimen consists of all metastable martensite. This is formed by heating the steel high enough to enter into the austenite γ region, and then quickly quenching it in oil to avoid equilibrium cooling. A quenched and tempered specimen consists of ferrite and martensite. This specimen is formed by heating the steel into the austenite γ region, quickly quenching it in oil to avoid equilibrium cooling, and finally heating it up to an intermediate temperature within the α + Fe3C region. As demonstrated in Tables 1, 2, and 3, an as received microstructure has medium hardness and strength, and medium ductility due to its equilibrium microstructure. Annealing a specimen reduces its hardness and tensile strength, but increases its ductility because the microstructure formed allows more slip. In contrast, quenching a specimen increases it hardness and strength significantly, but reduces its ductility to catastrophic levels due to too much microstructure slip impedance. To make this specimen safer, it can be tempered, which will maintain relatively high hardness and strength levels, but increase the ductility away from the brittle, unsafe, region.

12 Figure 5: Property Trends of 4140 Steel**** This graph demonstrates the dependence of properties of 4140 steel upon tempering temperature. As seen in Figure 5 above, the tensile strength, yield strength, and hardness of 4140 steel decrease with increasing tempering temperature. Conversely, % elongation and % reduction in area increase with increasing temperature. This behavior is expected because at higher temperatures, dislocation motion occurs more due to inner atomic bonding. Comparing experimental data to literature values, one can observe that most values are representative of a tempering temperature around 800 F rather than the reported 1000 F. This might imply a calibration issue with the lab s furnace. ****Data from Materials & Processing Databook 81. Metal Progress. Vol No. 1. American Society for Metals (1981)

13 In order for a quenched specimen of 4140 steel to be put in the annealed condition, it would have to first be heated into the austenite γ region then equilibrium cooled to the α + Fe3C region. After annealing, its microstructure would consist of ferrite and pearlite. If a part required machining and hardness, it would be machined in the annealed state because it would be more ductile and easier to shape. Then it would be quenched and tempered to increase its strength. If a steel chisel was going to be made, it would require mostly strength and hardness but some ductility to avoid brittle fracture. The best heat treatment would be to anneal it for shaping, then quench and temper it for high strength and some ductility.

14 Conclusion The purpose of this lab was to get the student familiar with heat treatment of steel. From performing this lab, the student is now able to fully understand the Fe-Fe3C diagram and explain the microstructure of steel in any region of the phase-diagram. Also, the student is now able to make educated decisions on material choices and heat-treated steels. Due to this lab, the student now comprehends the microscopic and macroscopic effects of austenitizing, quenching, tempering, and annealing 4140 steel. Choices can now be made in designing materials to actuate specific combinations of properties in order to maximize usability and minimize catastrophic failure.

15 References *Literature for FA specimens: 6b&ckck=1 **Literature for AR specimens: e9&ckck=1 ***Literature for AQ specimens: 8b&ckck=1 Literature for QT specimens: 18 ****Discussion Data: Materials & Processing Databook 81. Metal Progress. Vol No. 1. American Society for Metals (1981)

16 Appendix Sample Calculations: Calculating Strain L L0 Strain = ε = L0 L0 = OriginalLength L = Length Calculating Total Strain ε TOTAL = ε PLASTIC +ε ELASTIC Calculating Stress Stress = σ = F = Force A 0 F A 0 = OriginalCrossSectionalArea Elastic Modulus: ElasticModulus = E = σ ε Calculating Percent Difference: Exp Lit %Diff = 100% Lit Exp = ExperimentalValue Lit = LiteratureValue Calculating the Ultimate Tensile Strength (UTS) F UTS = MAX A 0 F MAX = MaximumAppliedLoad A 0 = OriginalCrossSectionalArea Calculating the Breaking Strength (BS) BS = F BREAK F A BREAK 0 = ForceAppliedatRupture Calculating Elongation Percent: LBR L0 % EL = 100% L0 L = Length Between Gauge Marks on the Broken Specimen L BR 0 = Original Length Between Gauge Marks Calculating Percent Reduction in Area: %RA = A 0 A BR 100% A 0 A 0 = OriginalCrossSectionalArea A BR = CrossSectionalAreaatBreak

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