Concept explainers
The stress-strain curve.
Explanation of Solution
Given:
The initial area is
The final area is
The original length is
Formula used:
The true strain is given as,
Here,
The Actual area is given as,
The actual are for the initial load is given as,
The final length is given as,
Here,
The stress is given as,
Here, P is the applied load.
Calculation:
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,\
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
For
The final length can be calculated as,
The strain can be calculated as,
The actual area can be calculated as,
The stress can be calculated as,
The stress-strain curve from the above calculation is given as,
Figure 1.1
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Chapter 2 Solutions
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- 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)arrow_forwardThe 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)arrow_forwardThe 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 STRESSarrow_forward
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