AERO214 Axial Loading

1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 0.0019 -0.0057 13.092 0.0762 21.819 0.1270 0.0095 30.547 32.729 0.1778 0.7620 34.911 3.0480 30.01 mm O F Rigid Plate Cylindrical Alloy -Steel casement Figure k Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.

1. Plot the engineering stress & strain diagram of an alloy having a tensile test result found in Table 1. The tensile test specimen has a diameter of 12.5mm and a gage length of 50.0mm. The given alloy is used to make a 30.0mm diameter cylinder, which is placed inside a hardened circular steel casement with a 30.01mm inner diameter. Table 1: Tensile test results of an alloy Change In Length (mm) Change In Diameter (mm) Load (kN) 0.000 0.0000 0.0000 4.364 0.0254 -0.0019 13.092 0.0762 -0.0057 21.819 0.1270 -0.0095 30.547 0.1778 32.729 0.7620 34.911 3.0480 30.01 mm Ø F Rigid Plate Cylindrical Alloy - Steel casement Figure 1: Section view of the steel casement encapsulating the cylindrical alloy 2. Determine the required minimum value of F such that the cylindrical alloy would touch the walls of the steel casement.

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 157 kN. - Length of the specimen is 23 mm. - The yield strength is 89 kN/mm2. - The percentage of elongation is 45 %. Determine the following Diameter of the specimen, Final length of the specimen, Stress under an elastic load of 18 kN, Young's Modulus if the elongation is 1.3 mm at 18 kN and Final diameter if the percentage of reduction in area is 25 %. Fine this ans 1-Initial Cross-sectional Area (in mm2) 2-The Diameter of the Specimen (in mm) 3-Final Length of the Specimen (in mm) 4-Stress at the elastic load (in N/mm2) 5-Young's Modulus of the Specimen (in N/mm2) 6-Final Area of the Specimen at Fracture (in mm) 7-Final Diameter of the Specimen after Fracture (in mm)

The following data were obtained from the tensile test of Aluminum alloy. The initial diameter of testspecimen was 0.505 inch and gauge length was 2.0 inch. Plot the stress strain diagram and determine(a) Proportional Limit (b) Modulus of Elasticity (c) Yield Stress at 0.2% offset (d) Ultimate Stress and(e) Nominal Rupture Stress.

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 143 kN. - Length of the specimen is 29 mm.  - The yield strength is 71 kN/mm2.  - The percentage of elongation is 48 %. Determine the following  Diameter of the specimen,  Final length of the specimen,  Stress under an elastic load of 18 kN,  Young's Modulus if the elongation is 1 mm at 18 kN  and  Final diameter if the percentage of reduction in area is 29 %.  Initial Cross-sectional Area 2.01 mm2. The Diameter of the Specimen 1.59 mm.  Final Length of the Specimen 42.92 mm.  Stress at the elastic load 8955.22 N/mm2. Find: Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 143 kN. - Length of the specimen is 29 mm.  - The yield strength is 71 kN/mm2.  - The percentage of elongation is 48 %. Determine the following  Diameter of the specimen,  Final length of the specimen,  Stress under an elastic load of 18 kN,  Young's Modulus if the elongation is 1 mm at 18 kN  and  Final diameter if the percentage of reduction in area is 29 %.  FIND: Young's Modulus of the Specimen (in N/mm2) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)

Define the given terms related to mechanical properties of materials. 1. Nanoindentation 2. Neckling 3. Newton’s material 4. Notch sensitivity test 5. Offset strain value

hings The following results were obtained during a tensile test to destruction on a mild steel test piece of diameter 15.96mm and 80mm gauge length Extension for a load of 40KN = 0.08mm Maximum load applied during test = 93kN Diameter at fracture = 12.85mm %3D Final length between gauge points 106mm From these results, determine: i) The modulus of elasticity for mild steel ii) The tensile strength iii) Percentage of elongation iv) Percentage of reduction in area

The following stress-strain curve was prepared based on a tensile test of a specimen that had a circular cross-section. The gage diameter of the specimen was 0.25 inches and the gage length was 4 inches. The stress scale of the stress-strain diagram is given with the factor a = 10 ksi. Estimate: (a) The modulus of elasticity. (b) The ultimate strength. (c) The yield strength (0.2% offset). (d) The percent elongation at fracture. 2013 Michael Swanbom STRESS VS. STRAIN BY NC SA 7a bat Sat 2at at 0.05 STRAIN 0.01 0.04 0.06 0.08 0.02 0.03 0.07 0.09 STRESS

In tensile test a plain carbon steel specimen has a (40mm) gauge length and the Final area (A final) of specimen after tensile test was 264.327. The load which caused fracture was (122.5 KN). After fracture, the final length was 47.516mm The true stress at fracture is less than engineering stress at fracture True False

The results of a tensile test are: Diameter of the specimen Gauge length Load at limit of Proportionality Extension at the limit of Proportionality Maximum Load : 10mm :40 mm :80kN : 0.06mm. :100 kN Calculate ultimate tensile stress and young's modulus.

The results of a tensile test are: Diameter of the specimen Gauge length Load at limit of Proportionality Extension at the limit of Proportionality Maximum Load : 10mm :40 mm : 70kN :0.08mm. :100 kN Calculate stress at limit of proportionality and young's modulus.

C1. A mild steel tensile specimen of initial length 41 mm and initial diameter 6.2 mm is subjected to a tensile test and the following data are obtained. Yield Strength as 115 MPa Maximum Strength as 210 MPa Fracture Strength as 168 MPa Percentage of Elongation as 50 % Percentage of Reduction in area as 32% Determine the Final length, Final area, Yield load, Ultimate load & Fracture load. The yield load in N is The Ultimate load in N is The Fracture load in N is

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 158 kN. - Length of the specimen is 26 mm. - The yield strength is 75 kN/mm?. - The percentage of elongation is 40 %. Determine the following (v) Final diameter if the percentage of reduction in area is 21 %. Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm)

1. A tensile test was conducted on a metal "505" specimen and the following stress-strain curves were generated, both curves generated from the same set of data. Use the graphs to fill in the mechanical properties of the material tested in the box below. Don't forget units! Stress vs Strain Stress, psi Stress, psi 80000 70000 60000 50000 40000 30000 20000 10000 0 0.00 80000 70000 60000 50000 40000 30000 20000 10000 0.02 0 0.000 0.002 0.04 0.004 0.06 0.006 0.08 0.10 Strain Stress vs Strain 0.008 0.12 Elastic Modulus, E: 0.2% Offset Yield Strength, oo: Tensile Strength, ou: Breaking Strength, of: % Elongation: 0.14 0.010 0.012 0.014 Strain 0.16 0.18 0.016 0.018 0.20 0.020

I just can't find The percent elongation at fracture.

QUESTION ONE (a) Distinguish between physical and mechanical properties of materials. Give two examples of each. (b) Explain why in a stress versus strain curve, the plastic portion of the graph after necking tends to drop (ie the force drops) despite that the tension is increasing. (c) A tensile test uses a copper test specimen that has a gauge length of 80 mm and a di.ameter of 16 mm. During the test, the specimen yields under a load of 9,600 N. The corresponding gauge length is 80.24 mm. The maximum load reached is 148,000 N at a gauge length of 94.2 mm, while fracture happens at a load of 12,800 N and a gauge length of 102 6 mm Determine the following: (i) Modulus of elasticity E (ii) Yield strength Oy (iii) Fracture strength, ơt (iv) Tensile strength OTs. 1

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The data below are for a thin steel wire suitable for use as a guitar string.                         Ultimate tensile stress:            1.8 x 109 Pa                         Young Modulus:                     2.2 x 1011 Pa                         Cross-sectional area:               2.0 x 10-7 m2 In a tensile test, a specimen of the wire, of original length 1.5 m, is stretched until it breaks. Assuming the wire obeys Hooke’s law throughout, calculate the extension of the specimen immediately before breaking.

Draw a typical stress vs strain tensile test curve for the following material and label the axis. A typical brittle material subjected to a tensile stress that has been applied to the material till the sample breaks. 1- label the axis and draw the curve for a brittle material. 2- indicate the maximum strength of the material. 3- show on the portion of the curve where young's modulus can be calculated.

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 142 kN. - Length of the specimen is 23 mm.  - The yield strength is 82 kN/mm2.  - The percentage of elongation is 48 %. Determine the following  (i) Diameter of the specimen,  ii) Final length of the specimen,  iii) Stress under an elastic load of 15 kN,  iv) Young's Modulus if the elongation is 1.6 mm at 15 kN  and  (v) Final diameter if the percentage of reduction in area is 20 %.    Solution Initial Cross-sectional Area (in mm2) Answer for part 1 The Diameter of the Specimen (in mm) Answer for part 2 Final Length of the Specimen (in mm) Answer for part 3 Stress at the elastic load (in N/mm2) Answer for part 4 Young's Modulus of the Specimen (in N/mm2) Answer for part 5 Final Area of the Specimen at Fracture (in mm) Answer for part 6 Final Diameter of the Specimen after Fracture (in mm)

The following data are obtained from a tensile test of a copper specimen. - The load at the yield point is 147 kN. - Length of the specimen is 28 mm.  - The yield strength is 75 kN/mm2.  - The percentage of elongation is 49 %. Determine the following  (i) Diameter of the specimen,  ii) Final length of the specimen,  iii) Stress under an elastic load of 16 kN,  iv) Young's Modulus if the elongation is 1 mm at 16 kN  and  (v) Final diameter if the percentage of reduction in area is 22  Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm) Initial Cross-sectional Area (in mm2)