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elasticity

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Just need to figure out how to draw the stress strain curve and calculate the modulus of elasticity
TEN Properties of Materials 30138 TABLE 10.2 Data for Problem 10.4 Load (k) 0 2.0 4.0 6.0 8.0 7.83 7.80 Deformation (in.) 0.0000 0.0006 0.0013 0.0019 0.0025 0.0043 0.0120 Load (k) 8.0 10.0 13.0 13.9 14.0 12.9 10.8 101 ms Deformation (in.) 0.0361 0.1100 0.2320 0.3200 0.3600 0.4000 0.4200 isua sd bloow.
Transcribed Image Text:TEN Properties of Materials 30138 TABLE 10.2 Data for Problem 10.4 Load (k) 0 2.0 4.0 6.0 8.0 7.83 7.80 Deformation (in.) 0.0000 0.0006 0.0013 0.0019 0.0025 0.0043 0.0120 Load (k) 8.0 10.0 13.0 13.9 14.0 12.9 10.8 101 ms Deformation (in.) 0.0361 0.1100 0.2320 0.3200 0.3600 0.4000 0.4200 isua sd bloow.
1 The data from the tension test of a steel specimen are given in Table 10.2. The gage length was 2.000 in. and the original diameter was 0.505 in. The final diameter was 0.397 in. Calculate stress and strain at each point and draw the stress-strain dia- and gram. Estimate the value of the modulus of elasticity, upper lower yield points, ultimate strength, rupture strength, and per- cent reduction in area. (Hint: Draw a second curve exaggerating the strain scale for the first eight points. See Figure 10.3.) Allowable Stresses and
Transcribed Image Text:1 The data from the tension test of a steel specimen are given in Table 10.2. The gage length was 2.000 in. and the original diameter was 0.505 in. The final diameter was 0.397 in. Calculate stress and strain at each point and draw the stress-strain dia- and gram. Estimate the value of the modulus of elasticity, upper lower yield points, ultimate strength, rupture strength, and per- cent reduction in area. (Hint: Draw a second curve exaggerating the strain scale for the first eight points. See Figure 10.3.) Allowable Stresses and
Expert Solution
Step 1

Original Diameter D1= 0.505 in

Hence

Original Area A0= π4D12= π4×0.5052= 0.2003 m2

Final Diameter D2= 0.397 in

Af= π4D22= π4×0.3972= 0.1238 m2

% Reduction in Area = 0.2003-0.12380.2003×100= 38.192

gauge Length= 2 in

Calculate stress= Load/ original area

Strain= deformation/ gauge length

Modulus of elasticity= stress/strain

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