UNIVERSITI TUNKU ABDUL RAHMAN LEE KONG CHIAN FACULTY OF ENGINEERING AND SCIENCE UEME1263 SOLID MECHANICS I LABORATORY REPORT (II) Experiment Title: Tensile Test Date of Experiment: 11th March 2019 (Monday) Student’s Name Student ID no. Course Khoo Yi Feng 1700666 ME Mok Hou Yin 1706035 ME Ng Xing Yu 1800481 ME Teoh Kok Siang 1601997 ME Wan Shijun 1603870 ME Practical Group: P2 Lecturer in Charge: Ir Dr. Kwong Kok Zee Objective To perform tension test on steel specimen according to ASTM E8, to determine stress-strain relationship, yield strength, tensile strength, elongation and necking, modulus of elasticity and rupture strength. Introduction When an object was applied to an external tensile loading, the metal will undergo elastic and plastic deformation. Initially, the metal will elastically deform giving a linear relationship of stress and strain. By knowing the force applied, cross-sectional area and the 𝑃 change in length of that object, the stress, 𝜎 = 𝐴 and strain, 𝜀 = 𝐿𝑓 −𝐿𝑜 𝐿𝑜 can be determined. Figure 1: Stress-strain graph The yield strength is defined as the stress at which permanent deformation starts to occur. It is also a material constant that represents the limit of its elastic behaviour. Besides, yield point is the point where plastic deformation begins. Before the yield point the material deforms elastically and returns to its original shape when the applied stress is removed. Once the stress applied exceeds the yield point, some fraction of the deformation will be permanent and it is non-reversible. Tensile strength is the ratio of tensile force per unit cross sectional area. As the load is continuously and incrementally applied, the stress-strain curve will reach the maximum point, which is the ultimate tensile strength. At this point, the object can withstand the highest stress before necking takes place. Young’s modulus measures the ability of a material to withstand changes in length when undergo tension or compression. During elastic deformation, the engineering stress-strain relationship follows the Hooke’s 𝜎 Law and the slope of the curve indicates the Young’s modulus (E), 𝐸 = 𝜀 . After necking, plastic deformation is not uniform anymore and stress decreases accordingly until fracture. The fracture strength can be calculated from the load at fracture divided by the original crosssectional area, A, by using this formula, 𝜎𝑓𝑟𝑎𝑐𝑡𝑢𝑟𝑒 = 𝑃𝑓𝑟𝑎𝑐𝑡𝑢𝑟𝑒 𝐴𝑜 . Procedures 1. The diameter of the reinforced steel bar specimen was measured using a digital Vernier calliper. The length of the steel bar was measured using a ruler. Each specific dimension was measured three times and the average value of the three measurements was taken. 2. The gauge length on the steel bar specimen was measured by 5 cm each for both end and marked by a marker pen. 3. The specimen was then placed in the loading machine, which was the Instron 5582 Universal Testing Machine. 4. The extensometer was attached to the specimen. 5. The load reading was set to zero, then load was applied at a rate less than 690 kPa/min. Unless otherwise specified, any convenient speed of testing may be used up to half of the specified yield strength or yield point, or one quarter of the specified tensile strength, whichever was smaller. After the yield strength or yield point had been determined, the strain rate may be increased to a maximum of 15 mm/min of the gauge length per minute. 6. The tensile test was started. 7. The load was applied continuously to the specimen until it breaks. 8. The tested specimen was removed from the gripping head. 9. The stress versus strain curve and the results were recorded in the computer. 10. Steps 1 to 9 were repeated using the mild steel bar specimen. Apparatus 1. There was a testing machine capable of applying tensile load at a controlled rate of deformation or load. These types of machine can be classified into mechanical or closed loop electro hydraulic. A dial gauge could be equipped onto the machine to indicate the load or could be connected to a chart recorder or computer to measure and record the load and deformation. 2. There was a gripping device. It was used to transmit the load from the testing machine to the test specimen. Moreover, it was used to ensure axial stress within the gauge length of the specimen. 3. An extensometer with an LVDT or dial gauge was used to measure the deformation of the specimen during the experiment. 4. Callipers were used to measure the dimensions of the steel specimen. Results: Mild Steel: Before loading: 1st readings 162 9.98 Length(mm) Diameter(mm) 2nd readings 163 9.95 3rd readings 163 9.97 Average 163 9.97 Original Area, 𝐴0 = 𝜋𝑟 2 = 𝜋( 9.97 2 ) 2 = 78.07 mm2 After loading: Length(mm) Diameter(mm) 1st readings 176 5.35 2nd readings 175 5.37 3rd readings 175 5.39 Average 175 5.37 Final Area, 𝐴 = 𝜋𝑟 2 = 𝜋( 5.37 2 ) 2 = 22.65 mm2 Reinforced Steel: Before loading: Length(mm) Diameter(mm) 1st readings 202 8.39 Original Area, 𝐴0 = 𝜋𝑟 2 = 𝜋( 8.39 2 ) 2 = 55.29 mm2 2nd readings 203 8.37 3rd readings 203 8.40 Average 203 8.39 After Loading: 1st readings 231 5.22 Length(mm) Diameter(mm) 2nd readings 230 5.21 3rd readings 230 5.22 Final Area, 𝐴 = 𝜋𝑟 2 = 𝜋( 5.22 2 ) 2 = 21.40 mm2 Calculations: Calculate the stress and strain for each load increment until failure Stress, 𝜎 = Strain, 𝜀 = 𝑃 𝐴𝑜 𝐿𝑓 −𝐿𝑜 𝐿0 = ∆𝐿 𝐿𝑜 Young Modulus, 𝐸 = 𝜎 𝜀 Average 230 5.22 Mild Steel: σ = ε= 2200 N 78.07 mm2 175−163 163 = 28.18 MPa = 0.0736 𝐸𝑒𝑥𝑝 = 1.882 × 1011 Pa 𝐸𝑡ℎ𝑒𝑜 = 2.05 × 1011 Pa Force 2.2 pulled (kN) Tensile 28.18 Stress (MPa) Tensile 0.015 Strain (%) Force pulled (kN) Tensile Stress (MPa) Tensile Strain (%) 4.4 6.6 8.8 11 13.2 15.4 17.6 19.8 56.36 84.54 112.7 140.9 169.1 197.3 225.4 253.6 0.030 0.045 0.060 0.075 0.090 0.105 0.120 0.135 22 24.2 26.4 28.6 30.8 33 35.2 281.8 310.0 338.2 366.3 394.5 422.7 Plastic deformation 0.150 0.165 0.180 0.195 0.210 0.225 Failure begun Reinforced Steel: σ = ε= 2200 N 55.29 mm2 230−203 203 = 39.79 MPa = 0.133 𝐸𝑒𝑥𝑝 = 1.551 × 1011 Pa 𝐸𝑡ℎ𝑒𝑜 = 2.10 × 1011 Pa Force pulled (kN) 2.2 Tensile Stress (MPa) Tensile Strain (%) 4.4 6.6 8.8 11 13.2 15.4 17.6 19.8 43.6 39.79 79.5 8 119. 4 159. 2 199. 0 238. 7 278. 5 318. 3 358. 1 Plastic deformation 0.026 0.05 1 0.07 7 0.10 3 0.12 8 1.53 9 0.18 0 0.20 5 0.23 1 Failure begun Stress versus strain curve Mild steel: Reinforced steel: Load at yield point Mild steel, Py = 30225 N Reinforced steel, Py = 23538 N Yield Strength Mild Steel: Yield Strength = 385 MPa 𝜎𝑌𝑆 (experimental result from graph) = 385 MPa 𝜎𝑌𝑆 (from formula) = 𝑃𝑦 𝐴𝑜 = 30225 78.07 𝜎𝑌𝑆 (theoretical) = 370 MPa = 387 MPa Reinforced Steel: Yield Strength = 310 MPa 𝜎𝑌𝑆 (experimental result from graph) = 310 MPa 𝜎𝑌𝑆 (from formula) = 𝑃𝑦 𝐴𝑜 = 23538 55.29 𝜎𝑌𝑆 (theoretical) = 460 MPa = 426 MPa Maximum Load Mild steel, Pmax = 33.354 kN Reinforced steel, Pmax = 27.470 kN Ultimate Tensile Strength (UTS) For mild steel 𝜎UTS (experimental from graph) = 425.53 MPa 𝜎UTS (from formula) = 𝑃𝑚𝑎𝑥 𝐴0 33.354 kN = 78.07 mm2 = 427.23 MPa 𝜎UTS (theoretical) = 440 MPa For reinforced steel 𝜎UTS (experimental from graph) = 361.96 MPa 𝜎UTS (from formula) = 𝑃𝑚𝑎𝑥 𝐴0 27.470 kN = 55.29 mm2 = 496.83 MPa 𝜎UTS (theoretical) = 595 MPa Elongation For mild steel: Original Length, Lo = 163 mm Final Length, Lf = 175 mm Therefore, the Elongation = [(Lf – Lo) / Lo] × 100 = [(175 – 163) / 163] × 100 = 7.36% For reinforced steel: Original Length, Lo = 203 mm Final Length, Lf = 230 mm Therefore, the Elongation = [(Lf – Lo) / Lo] × 100 = [(230 – 203) / 203] × 100 = 13.3% Modulus of Elasticity For mild steel: E (experimental from graph) = 188171 MPa E (theoretical) = 205 GPa = 205000 MPa For reinforced steel: E (experimental from graph) = 155148 MPa E (theoretical) = 210000 MPa Rupture Strength For mild steel Rupture Strength 1 MPa 𝜎𝑟𝑢𝑝𝑡𝑢𝑟𝑒 = 1 MPa For reinforced steel Rupture Strength 49 MPa 𝜎𝑟𝑢𝑝𝑡𝑢𝑟𝑒 = 49 MPa Reduction of cross-sectional area For mild steel: Original Area, A0 = 78.07 mm2 Final Area, As = 22.65 mm2 Percent reduction in cross-sectional area = [ (𝐴𝑜 −𝐴𝑠 ) 𝐴0 =[ ] × 100 (78.07−22.65) ] × 100 78.07 = 70.99 % For reinforced steel: Original Area, A0 = 55.29 mm2 Final Area, As = 21.40mm2 Percent reduction in cross-sectional area = [ (𝐴𝑜 −𝐴𝑠 ) =[ 𝐴0 ] × 100 (55.29−21.40) 55.29 = 61.29 % ] × 100 Fracture strain Mild steel Fracture strain = 0.43 Reinforced steel Fracture strain = 0.40 Work hardening exponent (n) 𝜎 = 𝐾𝜀 𝑛 where σ represents the applied stress on the material, ε is the strain, and K is the strength coefficient (Kmild steel = 530 MPa, Kreinforced steel = 640 MPa) 𝜎 𝐾 = 𝜀𝑛 𝜎 log 𝐾 = log(𝜀 𝑛 ) 𝜎 log 𝐾 = n log 𝜀 𝑛= 𝜎 𝐾 log log 𝜀 Mild steel 𝑛= 385.601 530 log log 0.39286 = 0.3404 ; the yield stress and yield strain were used in this calculation Reinforced steel 𝑛= 310.145 640 log log 0.39306 = 0.7758 ; the yield stress and yield strain were used in this calculation Fracture mode and fracture surface Mild steel Fracture mode: Mode I – Opening Fracture surface: Nearly to cup and cone surface Reinforced steel Fracture mode: Mode I – Opening Fracture surface: Nearly to cup and cone surface Percentage error 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%) = 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 − 𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 × 100 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 Percentage error of maximum tensile strength 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 = 440 − 425.53 × 100 = 3.289 % 440 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 = 595 − 361.96 × 100 = 39.17 % 595 Percentage error of Young’s Modulus 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 = 205 − 188.171 × 100 = 8.209 % 205 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 = 210 − 155.148 × 100 = 26.12 % 210 Percentage error of yield’s stress 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑚𝑖𝑙𝑑 𝑠𝑡𝑒𝑒𝑙 = |370 − 385| × 100 = 4.054 % 370 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑟𝑟𝑜𝑟 (%)𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑠𝑡𝑒𝑒𝑙 = 460 − 310 × 100 = 32.609 % 460 Discussion In this experiment, tensile test experiment had been performed on a mild steel bar and a reinforced steel bar. After conducting the experiment, the data for mild steel bar had been tabulated and some of the key properties were calculated. The mild steel bar used experienced a percentage of elongation about 7.36% and percentage of reduction in cross-sectional area of 70.99%. Besides, yield stress strength, 𝜎𝑦 , of 385 MPa, Young’s Modulus, E, of 1.882 × 1011 Pa and ultimate tensile strength, 𝜎𝑈𝑇𝑆 , of 425.53 MPa were obtained from the experiment table. However, there were some percentage errors occurred between experimental values and theoretical. Lastly, the mild steel bar had rupture strength of 1MPa. In the reinforced steel bar, the reinforced steel bar experienced a percentage of elongation about 13.3% and percentage of reduction in cross-sectional area at 61.29%. From the tabulated data and graph, the yield stress, 𝜎𝑦 , Young’s Modulus E, and ultimate tensile strength, 𝜎𝑈𝑇𝑆 obtained from the experiment were 310 MPa, 1.551× 1011 Pa and 361.96 MPa respectively. However, some percentage errors were obtained between experimental values and theoretical. The reinforced steel bar had rupture strength at 49 MPa. Due to the specimen was a ductile material, the group were able to observe every phase that the specimen undergone under tensile load which were the elastic deformation, plastic deformation, necking and fracture. However, the experimental results suggested that uncertainties existed in the experiment because there was deviation of experimental result from the theoretical one. These deviations were caused by the imperfections of the experimental set up. One of the main imperfections was the diameter of the specimen at breaking end (necking region) at one end was bigger than the other one. This indicates that the initial diameter throughout the specimen itself was not uniform, which results in different stress concentration throughout the entire specimen. Moreover, due to the reinforced steel specimen had spiral extruded pattern, this factor affected the stress concentration within the specimen as well. Not only that, the clamping of the specimen might not be rigid and strong enough and this caused the specimen displaces when tensile load was applied. Consequently, the stress on the specimen was scattered and inconcentrated. Lastly, the strength of the specimen that used might be affected due to the specimen has been put for a long period and the surface of the specimen was rusted. 1) What does the area under the stress-strain curve represent? The area under the stress-strain curve shows the toughness of a material. The definition of toughness is the ability of a material to absorb energy and plastically deform without fracturing. Another definition of material toughness is the amount of energy per unit volume that a material can absorb before rupturing. It is also defined as a material's resistance to fracture the object is under stress. The toughness in the elastic region is called the modulus of resilience. Strength and toughness are different in describing a material. The larger the elongation and stress applied the higher the toughness requires. Thus, the area under the graph is larger. 2) How would you distinguish between “brittle” and “ductile” modes of failure of a specimen under tensile test? There are three ways to differentiate between “brittle” and “ductile” modes of failure of a specimen under tensile test. Firstly, it can be distinguished by observing the stress-strain graph. For ductile materials, they undergo extensive plastic deformation and energy absorption before fracture occurs. In contrast, for brittle materials, they undergo little plastic deformation and they have much lower energy absorption ability before fracture occurs. Hence, the stressstrain curve for ductile material must be extended longer than the one for brittle materials. The area under the stress-strain curve as known as toughness of the material for ductile material also must be larger than the one for brittle material. (People.virginia.edu, n.d.) Figure 2: The stress-strain graph for brittle and ductile materials In this experiment, by observing the stress-strain graphs plotted, the lengths of the stress-strain curve for both materials were almost the same and it was uncertain to judge which one was longer. However, the area under the stress-strain curve for reinforced steel was larger than the one for mild steel. This indicates that reinforced steel is more ductile and can absorb more energy than mild steel before fracture occurs in this way of analysis. Next, the second way to differentiate the modes of failure is by observing the fracture surface. If the fracture surface has cup and cone characteristics (Figure 3), this indicates that the metal has high ductility. When the stress-strain curve has passed the ultimate tensile strength (maximum point of the curve), necking starts to occur due to the uneven plastic deformation. The formation of micro voids occurs at the necking region. The micro voids become larger and consequently merge together with each other when the tensile load increases. Crack with a plane normal to the external tensile load occurs. Before the specimen breaks, a shear plane is formed along the outer surface of the specimen with a slanted angle of 45° to the tensile axis. Then, this shear plane merges with the cracks and the cup and cone surface is generated. Figure 3: Cup and cone surface for ductile fracture In contrary, for brittle fracture, since the plastic deformation for this material is little, the crack propagates rapidly and the direction of the propagation is nearly perpendicular to the tensile axis. Metallic materials normally form crystalline solids and they have crystallographic planes. In brittle material, the cracking mostly propagates through these specific crystallographic planes and therefore, the breaking of the atomic bonds is more uniform. Hence, the fracture surface of brittle materials is flat and smooth. (People.virginia.edu, n.d.) Figure 4: Flat surface for brittle fracture In this experiment, based on the observation (Figure 5 and 6), both of the steel specimens tended to appear as cup and cone surface after fracture. This indicates that the fracture was moderately ductile and both of the materials were ductile. Comparing the flatness of the specimens, the fracture surface of the mild steel was flatter than the one of reinforced steel. This indicates that reinforced steel is more ductile than mild steel in this analysis. This method of analysis agrees with the previous method used. Figure 5: The fracture surface of mild steel Figure 6: The fracture surface of reinforced steel Thirdly, the results of the elongation of length and reduction of cross sectional area are helpful to determine the mode of fracture. The summary of the calculations for these two parameters is shown below: Elongation of length Mild steel Reinforced steel 7.36% 13.3% Reduction of cross sectional 70.99% 61.29% area Table 1: The summary of the calculations for the elongation of length and reduction of cross sectional area In term of elongation of length, the elongation of reinforced steel is larger than the one of mild steel. This indicates that reinforced steel was easier to be deformed permanently and it had better ductility than mild steel. In term of reduction of cross sectional area, the reduction of cross sectional area of mild steel is larger than the one of reinforced steel. This indicates that mild steel was easier to be deformed permanently and it had better ductility than reinforced steel. In short, both of these parameters contradicted to each other. In fact, the ductility of the mild steel should be greater than reinforced steel because mild steel has lower content of carbon (0.05% - 0.25%) (Quora, 2018) compared to reinforced steel (0.2% - 0.4%) (Concretecorrosion.net, n.d.). Most of the mild steel can be cold formed easily which means mild steel can be machined and fabricated under room temperature due to its great ductility. However, reinforced steel has to be hot formed in the manufacturing processes which means it is less ductile. In this experiment, the results were not expected with the desired result due to the errors and uncertainties discussed previously. 3) What are the reasoning to prepare the gauge length region of a specimen (region away from the gripped end-regions) with reduced cross-sectional area? The cross-sectional area of the gauge length region is relative smaller than the remaining region of the specimen is because deformation and failure are expected to be occurred at the gauge length region. The gauge length region is where meausrements are made and it is centered within the reduced region. The ends of the gauge length region and the shoulders of the specimen have to be further enough with the purpose to prevent constrain deformation of the larger ends within the gauge length region (Asminternational.org, 2004). Moreover, the gauge length should be larger than the diameter of the specimen or else a more complex stress mode other than simple tension will be occurred. Conclusion In conclusion, based on the experimental result, it showed that the mild steel bar can resist higher stress compared to the reinforced steel. However, theoretically, reinforced steel should have higher resistance to the high stress. Hence, the results of this experiment were not reliable. This was caused by a few of uncertainties such as the uneven stress concentration on the specimen, the condition of the specimen and also the status of clamping. Lastly, the objective was achieved as the stress-strain relationship, yield strength, tensile strength, elongation and necking, modulus of elasticity and rupture strength were obtained in this experiment. References Asminternational.org. (2004). Introduction to Tensile Testing. [online] Available at: https://www.asminternational.org/documents/10192/3465262/05105G_Chapter_1.pdf/e13396 e8-a327-490a-a414-9bd1d2bc2bb8 [Accessed 17 Mar. 2019]. Bestech.com.au. (2019). 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