LAB NO_02.pdf - University of Alberta Department of Civil...

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Unformatted text preview: University of Alberta Department of Civil and Environmental Engineering Civ E 2711 Laboratory 2 Name: 121).: Date: Mark: Lab Section: TENSION TEST 2.0 OBJECTIVES - To illustrate different types of tensile behaviour by performing tension tests on altunintun and steel bars. - To determine significant mechanical properties for aluminum and steel. - To demonstrate three different devices for measuring strain. 2.1 GENERAL BACKGROUND The tension test is a fundamental material test to determine the properties of materials needed for the design of structures. Typical stress vs. strain diagrams for aluminum and steel are shown in Figures 2.1 and 2.2. respectively. In addition to showing typical shapes for the diagrams. the figures define specific mechanical properties such as the modulus of elasticity. the proportional limit. the ultimate strength and the stress and strain at fi'acture. Figure 2.2 also shows the upper and lower yield points for low carbon steel. The most significant yield strength parameter for structural steel is the static yield stress. defined as the stress just subsequent to yielding that is maintained when the strain rate is zero. Of significance. as well. is the strain- hardening strain. which marks the end of the yield plateau. Because aluminum does not have a well-defined yield point. a yield strength is obtained by the offset method as shown in Figure 2.3. A line is drawn parallel to the initial portion of the stress vs. strain diagram and oHset by a strain of 0.2% (0.002). The yield strength is defined as the stress corresponding to the point of intersection of the offset line and the stress vs. strain curve. Typical tensile fractures are shown in Figure 2.4. Laboratory 2 Civ E 2 5’0 2 2.2 References - ASTM A3 20-09, Standard Methods and Definitions for Mechanical Testing of Steel Products5 American Society for Testing and Materials, Philadelphia, 2009. 2.3 Preparation of Test Specimens l. The aluminum specimen is 600 mm in length and the steel specimen is prepared based on the ASTM A3 20 standard. 2. A 200 mm gauge length is lightly marked with a center punch on each specimen. 3. A pair of strain gauges are mounted on opposite faces of each coupon. 2.4 Test Procedure 2.4.1 Aluminum 1. Measure and record the diameter of the specimen and the gauge length. 2. Place and align the specimen in the testing machine. 3. Connect the strain gauges. extensometer and MTS load and stroke outputs to the data acquisition system. 4. Apply an initial load of about ]_-"10th the estimated yield load. This ensures that the grips are properly biting into the specimen and that slip will not occur during the test. 5. Unload and zero instruments. (Take zero readings.) 6. Apply load continuously. 7. At approximately 5000 and T000 microstrain, measure the static stress level. 3. Continue loading to fi'acture, record the maximum load and the load at fracture. 9. After fracture= record the final length for the 200 mm gauge length and the diameter at the narrowest point. Examine and sketch the fracture surface, paying particular attention to the texture and shape. 2.4.2 Flat Steel Bar 1. Measure and record the width and thickness of the flat bar and the gauge length. 2. Place and align the specimen in the testing machine. 3. Connect the strain gauges, extensometer and MTS load and stroke outputs to data acquisition system. 4. Apply an initial load of about 1.:"10th the estimated yield load. Labor-stag: 2 Civ E 2 5'0 3 2.5 5. Unload and zero instruments. (Take zero readings.) 6. Apply load continuously. T. Take several static yield stress readings along the yield plateau. Observe the formation of Lilidersr lines on the specimen. 8. Continue loading until fracture. record the maximum load and the load at fiacture. 9. After fiacture, record the final length for the 200 mm gauge length and the Width and thickness at the narrowest point. Examine and sketch the fracture. paying particular attention to the texture and shape of the fracture surface. Record of Test Data and Analysis of Results 2. 5.] Aluminum (a) Initial diameter, Di = m Initial area. Ai = m2 (b) Gauge length, L = m (c) Fracture Maximum load = kN Load at fracture = kN Final gauge length = m Final diameter = m ((1) Description and sketch of fi‘acture surface: (e) Stress-strain data for the aluminum specimen is presented in two graphs to be handed out in the lab period. The first shows the elastic and initial plastic behaviour at an expanded scale and the second shows the complete stress-strain behaviour. Laboratory 2 Cfv E 270 4 Using the curves based on extensometer data. report the following material properties and show them on the graphs. Modulus of elasticity. E = MPa Proportional limit= Up = MPa 0.2% offset static yield stress Guys = MPa Ultimate strength. on = MPa Ultimate strain, on = mnlmm Percent elongation at fracture = 9’2: (f) Using the modulus of elasticity. E. as obtained by students in your class and recorded in the table shown, compute the mean= standard deviation (maintain at least five significant figures) and coefficient of variation. (The differences in the values of E that are reported arise from differences in plotting and slope calculations by different persons using identical test data). Mean.— 11 E Standard deviation.s= m HZ{X1-)—(Z:Xi2) Coefficient offvariation= V: (S! i} x100% where xi = observation I] = total number of observations 2.5.2 Flat Steel Bar (a) Initial width. w = _ mm= initial thickness= t = _ _ mm Initial area. Ai = mm2 (1)) Gauge length. L = m (c) Fracture Maximum load = W Fracture load = kN Final gauge length = m Final width = m. final thickness = mm Labor-drag.- 2 Cir: E 2 70 5 (d) Description and sketch of fi‘aeture surface: (e) Stress-strain data for the steel specimen is presented in two graphs to be handed out in the lab period. The first shows the elastic and initial plastic behaviour at an expanded scale and the second shows the complete stress-strain behaviour. Using the curves based on extensometer data, report the following material properties and show them on the graphs. Modulus of elasticity, E = MPa Upper yield point, “3’11 = MPa Lower yield point, (33,] = MPa Static yield stress, 073,5 = MPa Yield strain, a? = min-"mm Strain hardening strain, as, = min-'rum Ratio of strain hardening strain-"'yield strain, smile)f = Ultimate strength, on = MPa Ultimate strain, a“ = min-"mm Percent elongation at fracture = ‘34: True stress at fi‘acture = MPa 2.6 Comment on the difierenees, if any, between strain readings obtained from the extensometer and fi‘om the strain gauges. What advantages and disadvantages can you see to these two methods of strain measurement? Laboratory 2 Civ E 270 6 Ultimate Strength Fracture Strain fl Proportional Limit i E """"""" .’ E a r g .E E 1|' 2 2 i . E “3 i it Elastic unloading 1% i i of the two parts g i : after fracture 3 i I: E i l' l r " 1 i Permanent Strain i Strain (mmrmm) Figure 2.1 Stress vs. Strain Diagram for Almujmtm Ultimate Strength strain Permanent Strain ,‘ \Upperyiald stress i Fracture Strain A _ ———— ."""" Lower yield stress (dynamic) 'f E i i t s. i i .' ! i i E Static yield stress E 5 : F5 i E i i' Elastic unloading Proportional Limit % 5 f of the two parts i g i ; after fracture i t i i i 3 i .' E i i i' i Strain—hardening i f 1 i i ' i ! Strain (mmimm) Figure 2.2 Stress vs. Strain Diagram for Steel Laboratory 2 Civ E 270 T Static yield strees level Yield strength Stress {M Pa) |<—>| Strain (mmi‘mm) \_ 0.2% Offset Figure 2.3 Offset Yield Strength for Altunjmun {c} (d) {e} (f) (9] . Partial 'Star Irregular. Cup—eerie. Shear granular. silky Cu p-cona. fracture" fibrous silky failure cleavage silky (flat specimen} Figure 2.4 Typical Tensile Fractures for Metals (Ref. Davis, Troxell and Hauek= "The Testing of Engineering Materials", Figure 8.20, p. 140.} Labor-atria! 2 Civ E 25'0 3 3.0 Design Problem Member AB in the stmeture shown consists of a rod made of the same grade of aluminum as the grade tested in this laboratory. The link EF is made of steel of the same grade as steel specimen tested. Using a factor of safetyr of 2.5 on the ultimate stress, determine: (a) the required diameter of rod AB= (b) the required cross-sectional area of link EF. Assume that the tensile and compressive strengths are the same. ...
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