Pre-lab Assignment: Answer the following questions. Show your work. 1. A tensile
ID: 1766521 • Letter: P
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Pre-lab Assignment: Answer the following questions. Show your work. 1. A tensile test specimen is loaded in a machine while both load and elongation data are measured a. If an elon gation of 0.004 inch is measured over a 2.0 inch length, what is the engineering strain in the specimen? (p 154) If the specimen with circular cross-section has a diameter of 0.505 inch and is loaded with 7000 lbs, what is the engineering stress in the specimen? (p 154) b. The specimen is loaded to failure, which caused a measureable change in length of the gauge section. The original length was 2.00in and the final length was measured to be 2.26in. What is the percent elongation? (p 167) c. Using the stress-strain graph shown, determine the 0.2% offset yield strength. (p 165) 2. 140000 T 120000 100000 9 80000 60000 40000 20000 0 3. Using the stress-strain graph shown, determine the tensile strength. (p 165) 0 0.01 0.02 0.03 0.04 0.05 00 Strain (in/in)Explanation / Answer
1)
(a) Engineering Strain is the total strain for a specimen accounting the beginning and the end of a load deformation process. While True strain is measured at every instance during the course of deformation of the specimen.
Therefore, for a specimen, when the final deformation (elongation) is dL with the initial length L, then the engineering strain is given as
Engineering Strain = Change in length (dL) / Total length (L)
Therefore, Strain = 0.004/ 2
= 2 x 10-3
(b) Similarly, Engineering Stress is given as,
Engineering Stress = Force/ Initial Area
Given, diameter of circular cross-section, d = 0.505 inch = 12.827 mm
Therefore Area (A) = pie/4 * (diameter)^2
= 3.14/4 * (12.827^2) = 129.16 mm2
Force (F) = mass x acceleration = 3175.15 kg X 9.81 m/s2 ( 7000 lbs = 3175.15 kg)
= 31148 N
Therefore, Engg. Stress = 31148 / 129.16
= 241.16 N/mm2 or 241.16 kPa
(c) Original length = 2 " Final Length = 2.26"
Therefore, percentage elongation = (Final Length - Original length) / Original length * 100
= [(2.26 - 2) / 2]*100
= 13%
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