ENGR 244 Lab 2

Lab Report 2 Tension Test On Metals ENGR 244 Lab Section CI - X Winter 2022 Professor Ahmed Soliman Concordia University Montreal, QC, Canada Tuesday, January 18 th , 2022

Objective: The purpose of this experiment is to gather more information about the strength and ductility of steel and aluminum. Introduction: The strength of a metal can be seen in many ways. Last lab, we tested the Brinell hardness of aluminum and steel which shows the metals resistance to permanent indentation. Following up the tests on aluminum and steel we have the tension test to measure the tensile strength, the metals’ ability to stretch. Much like the Brinell Hardness test, this test allows us to see which metal will better suit anything that we are designing based off the tensile loading. It will allow us to know if there would be permanent elongation, the thickness of the metal required for the load desired, the elasticity of the metal (how much stretching motion you have without permanent elongation), and the load the metal can take before failing. The only assumption made this lab was that both metals would reach their yield strengths at approximately 5000N which is why the scaling (see below in procedure) will change at that time to get more detailed results. Procedure: Measure the diameters of both your 100mm aluminum and 100mm steel rods using the vernier provided and record them. Place one of the rods into the deformation measuring device and make sure it is placed tightly so it cannot move. Reset the digital load reader and begin applying the load. Add 500N and record the corresponding length of deformation. Continue this step until you reach 5000N. Once you reach 5000N continue by adding only 200N of pressure until you reach a point where the deformation is much higher than before for a smaller load. After this point, take measurements based on the length of deformation (0.5mm and 0.25mm for aluminum and steel respectively). Continue this until the metal rod reaches failure and then record the maximum load recorded by the digital scale. Remove the two pieces of the fractured metal rod and record the final deformation by placing the two pieces together on the V groove device provided. Also record the diameter of the fracture section using the caliper provided.

Figure 1: Steel test results Figure 2: Aluminum test results

Sample Calculations Formulas used: σ = P(N)/A(mm 2 ) where σ is stress and A = (π/4)(d 2 ), ε = δ/L where ε is strain, δ is the change in length and L is the original length, %A reduc = (A final -A initial )/A initial , %Elongation = (L final -L initial )/L initial , E = σ/ε where E is the modulus of elasticity, ultimate strength = σ max /A initial , true fracture stress = FractureForce(N)/A final Legend: P(N) is the load in newtons D is the initial diameter of the gauge Fracture Force is the load, in newtons, when the gauge breaks ε(mm/mm) 0.0031 0.0035 0.00 39 0.00 43 0.0047 0.0049 0.0053 0.0056 0.00 6 0.00 68 0.00 74 0.00 99 0.019 0.02 52 0.0305 0.0337 0 100 200 300 400 500 600 Stress vs Strain of Steel σ(Mpa) Strain ε Stress σ Figure 3: Stress vs strain graph of steel Yield strength ~ 400MPa Proportional limit ~ 550MPa

Figure 6: Aluminum Fracture σ(Mpa) ε(mm/mm) 37.34 0.003 74.38 0.0031 112.86 0.0033 189.81 0.0035 229.96 0.0036 265.02 0.0039 303.5 0.0041 341.83 0.0043 379.09 0.0046 395.23 0.0047 409.54 0.0048 425.68 0.0049 441.05 0.0051 456.12 0.0053 472.56 0.0054 485.36 0.0056 501.49 0.0058 517.62 0.006 527.47 0.0065 537.47 0.0068 542.32 0.0071 553.91 0.0074 561.1 0.0082 568.75 0.0099 562.01 0.0162 562.39 0.019 548.83 0.0222 523.01 0.0252 493.24 0.028 467.71 0.0305 444.99 0.0322 422.34 0.0337 409.16 0.0342 σ(Mpa) ε(mm/mm) σ(Mpa) ε(mm/mm) 28.02 0.0016 282.51 0.063 71.16 0.0021 281.81 0.0659 81.86 0.0021 279.19 0.0694 114.84 0.0024 282.62 0.0719 140.61 0.0026 281.55 0.0744 158.58 0.0029 282.78 0.0772

184.35 0.0032 281.01 0.0801 209.58 0.0037 281.97 0.0828 240.7 0.0044 284.06 0.0856 255.24 0.0146 280.53 0.0886 268.18 0.028 273.74 0.0912 259.95 0.0309 271.39 0.0921 268.72 0.0338 265.83 0.0935 271.6 0.0366 256.63 0.0947 272.46 0.0392 253.42 0.0962 275.02 0.0418 247.01 0.0975 275.24 0.0502 233.91 0.0988 294.75 0.053 226.91 0.1008 276.68 0.0502 220.6 0.1019 276.95 0.053 213.06 0.103 278.61 0.0555 199.69 0.1048 278.18 0.0582 186.7 0.1065 279.94 0.0605 170.07 0.1083 Table 1: Calculated Stress and Strain of Aluminum Table 2: Calculated Stress and Strain of Steel Values for Steel: σ = P/A  σ = 493N/((π/4)(4.10mm) 2 ) = 37.34MPa = First stress value in table 2. ε = δ/L  ε = 0.3mm/100mm = 0.003 = First strain value in table 2. ultimate strength = σ max /A initial  ultimate strength = 7601N/((π/4)(4.10mm) 2 ) = 575.72MPa %Elongation = (L final -L initial )/L initial  %Elongation = (103.42mm-100mm)/100mm = 0.0342 = 3.42% E = σ/ε  E = (553.91MPa – 37.34MPa)/(0.0074-0.003) = 117.40 GPa

Figure 7: Expected values for steel and aluminum[ CITATION Con \l 4105 ] When comparing the results to the values in figure 7[ CITATION Con \l 4105 ], the values for aluminum are all similar. The same cannot be said for steel as the %elongation and modulus of elasticity values are far off by a large margin. The most likely cause for this would simply be a weakness or defect in the steel gauge we received. Any microscopic cracks or non-uniform diameters can change these values. Another possible reason for this could be that the steel used in the experiment had a different chemical composition as the steel used for figure 7[ CITATION Con \l 4105 ]. Each steel alloy should have its own unique modulus of elasticity and expected %elongation. References [1] C. U. E. a. C. Science, ENGR 244 Mechanics of Materials Lab Manual, Montreal: Concordia University. [2] T. Talk, "hevvypumps," 28 May 2019. [Online]. Available: https://hevvypumps.com/en-blog/tech- talk/what-is-the-difference-between-malleability-and-ductility/#:~:text=THE%20DIFFERENCE %20BETWEEN%20DUCTILITY%20AND%20MALLEABILITY&text=They%20are%20similar %20in%20that,malleability%20deals%20with%20compressive%20stres. [Accessed 30 01 2022]. Original Data: