244 lab 2

Department of Building, Civil, & Environmental Engineering ENGR 244: Mechanics of Materials Experiment No.2: Tension Test on Metals Luis Alberto Alvarado Bravo 40193230 Group members: Erica Sciotto, Sofia Mendicino Section: CC-CM-X Summer 2022 July 11, 2022

1 OBJECTIVE The main purpose of this experiment is to test the tensile strength of an aluminum rod & a steel rod and see at what maximum load the rods reach their fracture points. INTRODUCTION Tensile strength is the ability of a material to resists and sustain uniaxial loads. The tensile strength of any material can be represented using a stress-strain curve. Figure 1- Stress-strain curve As shown in Figure 1, the stress-strain curve has three distinct points: yield strength, ultimate strength, and fracture point. The yield strength is the maximum stress a material can experience before permanent deformation is visible. When looking at figure, the yield strength can be identified as the point where the linear trend stops. This section of the curve is the elastic behavior of the material, meaning that no matter what the load was, the material will always return to its original state. The slope of the linear section of the curve is called the Young’s modulus , also called the modulus of elasticity, which means how easily a material can stretch. The second point on the curve, the ultimate strength, is the maximum

3 PROCEDURE To perform the experiment, the following materials were used: • Tension test machine • Deformation measurement device • Vernier caliper • Steel and aluminum rods of length 100 mm • V-groove tray The tension machine has two latch-like spaces where the material can be attached. The machine itself is connected to a hydraulic pump that increases the load with every stroke. The deformation measurement device has two ends with set screws that hold the metal rods in place. A thin rod sticks out of the device that determines how much elongation the material experiences. Each device is connected to a reader. The experiment follows these next steps: 1. Measure the diameters of the material rods at the center using the caliper. 2. Place the metal rod in the deformation measurement device, securing it with the set screws 3. Using a couple screw-on caps, place the rod in the tension test machine. The number of exposed threads on the rod should be equal on each end to make sure that the load is being applied evenly. 4. Apply the load steadily and record the load and deformation every 500 N. After reaching 5000 N, record data every 200 N. The load reader will begin fluctuating at some point, when that happens data is taken every 0.5 mm for aluminum, and every 0.25 mm for steel. 5. When the rod snaps, record the maximum load stored on the reader’s memory. 6. While still keeping the rod in the deformation device, using the v-groove tray, measure the diameter of the rod at the fracture point. 7. Repeat the process for the other test material.

4 RESULTS As stated earlier, measurements were taken of the diameter and initial lengths of the rods. These measurements were later retaken after the rods snapped. Table 1 - Rod measurements Table 2 - Load and deformation values Having taken all the values of the load and corresponding deformation, a stress-strain table can be created. Stress can be determined using the following formula, 𝜎 = 𝑃 𝐴 where P is the load in Newtons and A is the cross-sectional area of the material. Since the rod is cylindrical, the cross-sectional area formula is 𝐴 = 𝜋𝑟 2 Sample Steel Li = 100 mm di = 4.01 mm Lf = 103.2 mm df = 3.03 mm Aluminum Li = 100 mm di = 4.99 mm Lf = 111.4 mm df = 3.50 mm Before fracture After fracture Load (N) Deformation (δ) Load (N) Deformation (δ) Load (N) Deformation (δ) Load (N) Deformation (δ) 0 0.00 5775 3.91 0 0.00 6800 0.25 491 0.01 5857 4.45 449 0.00 7010 0.30 1001 0.01 5882 4.84 976 0.00 7210 0.30 1502 0.07 5906 5.33 1524 0.00 7399 0.31 2000 0.13 5932 5.89 2008 0.01 7606 0.53 2499 0.19 5961 6.47 2593 0.02 7696 0.77 3001 0.26 5992 6.78 3004 0.05 7744 1.05 3502 0.30 5998 7.26 3504 0.05 7770 1.29 4011 0.31 6007 7.89 4005 0.10 7780 1.54 4491 0.38 6012 8.35 4507 0.16 7789 1.76 4990 0.45 6034 8.87 4998 0.16 7750 1.98 5216 0.50 6026 9.32 5201 0.16 7494 2.40 5402 0.57 5959 9.99 5404 0.18 7199 2.93 5544 0.87 5931 10.60 5606 0.19 6892 3.65 5584 1.36 5502 11.17 5824 0.19 6629 3.75 5649 1.81 4838 11.71 6002 0.21 6425 3.94 5671 2.33 6189 0.21 5712 2.86 6389 0.24 5750 3.38 6619 0.25 Maximum Load, P max = 6062 N Maximum Load, P max = 7838 N ALUMINUM STEEL

6 Once the calculations were completed, two stress-strain curves were plotted, one for aluminum and one for steel. Figure 2 - Steel stress-strain curve Figure 3 - Aluminum stress-strain curve

7 Table 4 - Additonal values of the rods With the curves, it is easier some of the values that are hard to identify, like the yield strength and the modulus of elasticity. To calculate the modulus of both materials, the offset method of 0.02% was used, and is represented by the orange and white lines. To calculate the percentage elongation and percentage area reduction the formula is the same, except for a small change. Since the elongation goes up, the formula is: % = 𝐹𝑖𝑛𝑎𝑙 𝑙?𝑛?𝑡ℎ (??) − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑙?𝑛?𝑡ℎ(?𝑖) 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑙?𝑛?𝑡ℎ(?𝑖) × 100 And since the area is decreasing the formula is: % = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑟?𝑎 − 𝐹𝑖𝑛𝑎𝑙 𝐴𝑟?𝑎 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑟?𝑎 × 100 As for the true fracture stress, it was found using the stress formula, however, instead of using the original area, in this case the area of the fracture point was used. The value of this area was calculated using the df value from Table 1. Aluminum Steel Yield strength (Mpa) 279.86 602.25 Ultimate strength (Mpa) 308.13 616.74 Percent elongation 11.40% 3.20% Elastic modulus (GPa) 52.52 175.8 Percent area reduction 50.80% 42.90% Proportional limit (Mpa) 266.71 555.06 True fracture stress (Mpa) 502.85 891.04

9 DISCUSSION & CONCLUSION One question that came up when gathering results was “ Why not sketch a load-elongation diagram instead of a stress-strain curve? ” . The reason, because the data is more reliable with a stress-strain curve. As implied by its name, the load-elongation diagram depends heavily on the length of the specimen. For example, if two specimens of same chemical compositions but different lengths were tested, the graphs would look different due the variations in length. However, if the same specimens were evaluated on a stress-strain curve, the values and curves would look alike, following a similar trend. As stated in the introduction, the modulus of elasticity indicates how easily can stretch, and how resistant it is to permanent deformation. This was quite noticeable, since at first the rods did not seem to stretch at first, it was after they had passed their yield strength that the elongation became more apparent. If the rods had instead being subjected to compression loads, the results would not have been completely different. They would have followed a similar trend for their elasticity modules and yield strength. Still, there would have been slight differences with their ultimate strength and fractures. According to the Merriam-Webster dictionary, ductility is the ability of a material to have its shape changed with failing. Meaning how much deformation a material can withstand before reaching its fracture point. Ductility can bi identified in the plastic region of stress-strain curve, between the yield strength and the ultimate strength. Although there is still some data before a material reaches fracture, anything after the ultimate strength point can be considered as failure.

10 Table 5 - Properties of aluminum & steel Lastly, comparing the values obtained from the experiment to those of Table 5, it can be seen that one rod experienced a lot of sources of error compared to the other. For the most part, the data obtained from the aluminum rod seems to be quite close to those of Table 5, aside from the modulus of elasticity. For steel however, the values are quite different. All the values, except for the modulus of elasticity, are not close enough to those of Table 5. There could have been many sources of error, and one of them can be noticed in Figure 4. Had the load been applied evenly, the rod should have broken at the middle, but on this case it broke close to the threads. This could have been the result of not leaving even number of threads on both sides of the rod. Another error might have occurred when applying the loads. Unlike aluminum where the load was applied at a steady rate, the load for steel was initially applied very fast, in about five seconds the load went from 0 to 3000 N.