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2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 10 mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 10° psi 300 MPa 2000 200 200 1000E 100- 100 0.000 0.005 0.010 0.015 Strain 0.000 0.020 0.040 0.060 0.080 Strain Figure 5.10 Tensile stress-strain behavior for an alloy steel Stress (MPa) Stress Stress (10 psi)

2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 10 mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 10° psi 300 MPa 2000 200 200 1000E 100- 100 0.000 0.005 0.010 0.015 Strain 0.000 0.020 0.040 0.060 0.080 Strain Figure 5.10 Tensile stress-strain behavior for an alloy steel Stress (MPa) Stress Stress (10 psi)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior
shown in Figure 5.10 that has a cross sectional diameter of 10 mm.
a. Will the specimen experience elastic or plastic deformation? Why?
b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied?

2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 1o mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 10 psi 300 MPa 2000 200 200 1000 1000E 100 100 0.000 0.005 0.010 0.015 Strain 0.040 Strain 0.000 0.020 0.060 0.080 Figure 5.10 Tensile stress-strain behavior for an alloy steel Stress (MPa) Stress Stress (10 psi)
Transcribed Image Text:2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 1o mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 10 psi 300 MPa 2000 200 200 1000 1000E 100 100 0.000 0.005 0.010 0.015 Strain 0.040 Strain 0.000 0.020 0.060 0.080 Figure 5.10 Tensile stress-strain behavior for an alloy steel Stress (MPa) Stress Stress (10 psi)
a. modulus of elasticity b. yield strength c. maximum load (F) if original diameter is 12.8 mm d. Al equivalent to a stress of 345 MPa Required: Solution: a. Solve for the modulus of elasticity The modulus of elasticity is the slope in the linear portion of the stress-strain curve. E = slope = A6 /Aɛ Using Figure 5.7, select 2 values of stress in the linear portion of the stress-strain curve and read the strain corresponding to the value of stress. E= 8,-8, / E2 – E, E = (150-0)/ (0.0016-0) E = 93.8 GPa b. Solve for the yield strength Construct a parallel line to the linear portion of the stress-strain curve with an offset of o.002. The intersection between the parallel line and the stress-strain curve is the yield strength. The yield strength is around 250 MPa. c. Solve for maximum load if original diameter is 12.8 mm In calculating the maximum load (F) use engineering stress: 8 = F/ A, Rearrange the formula of engineering stress to get F: F= SA. The maximum stress gives the maximum load. The maximum stress is the tensile strength, read in the graph the tensile strength and this is equal to 450 MPa. 1 F= 8 A, = 8 (d./2)² F= (450 MPa) * (10° N/m²/1MPA) * (12.8 x 103 m/2)² n F= 57,900 N d. Solve for change in length (Al) equivalent to a stress of 345 MPa Use the formula of engineering strain: ɛ = Al / lo Rearrange the formula to get change in length: Al = ɛl, Locate in Figure 5.7 the stress equivalent to 345 MPa and read the strain corresponding to this stress. The strain corresponding to a stress of 345 MPa is o.06. Al = e lo Al = (0.06) (250 mm) Al = 15 mm Clicker Exercise 2 1 A bar of steel alloy that exhibits the stress-strain behavior shown in Figure 5.10 is subjected to a tensile load; the specimen is 375 mm long and square cross section 5.5 mm on a side. a. Compute the magnitude of the load necessary to produce an elongation of 2.25 mm. b. What will be the deformation after the load has been released? 2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 1o mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 103 psi 300 MPa 2000 200 200 1000 1000E 100E 100 0.000 0.005 0.010 0.015 Strain 0.000 0.020 0.040 0.060 0.080 Strain Figure 5.10 Tensile stress-strain behavior for an alloy steel 4.3 Ductility Ductility is a measure of the degree of plastic deformation that has been sustained at fracture. A material that experiences very little or no plastic deformation upon fracture is termed as brittle. The material that experiences plastic deformation is termed ductile. The tensile stress-strain behavior of both brittle and ductile materials is shown in Figure 5.8. Stress (MPa) Stress Stress (103 psi)
Transcribed Image Text:a. modulus of elasticity b. yield strength c. maximum load (F) if original diameter is 12.8 mm d. Al equivalent to a stress of 345 MPa Required: Solution: a. Solve for the modulus of elasticity The modulus of elasticity is the slope in the linear portion of the stress-strain curve. E = slope = A6 /Aɛ Using Figure 5.7, select 2 values of stress in the linear portion of the stress-strain curve and read the strain corresponding to the value of stress. E= 8,-8, / E2 – E, E = (150-0)/ (0.0016-0) E = 93.8 GPa b. Solve for the yield strength Construct a parallel line to the linear portion of the stress-strain curve with an offset of o.002. The intersection between the parallel line and the stress-strain curve is the yield strength. The yield strength is around 250 MPa. c. Solve for maximum load if original diameter is 12.8 mm In calculating the maximum load (F) use engineering stress: 8 = F/ A, Rearrange the formula of engineering stress to get F: F= SA. The maximum stress gives the maximum load. The maximum stress is the tensile strength, read in the graph the tensile strength and this is equal to 450 MPa. 1 F= 8 A, = 8 (d./2)² F= (450 MPa) * (10° N/m²/1MPA) * (12.8 x 103 m/2)² n F= 57,900 N d. Solve for change in length (Al) equivalent to a stress of 345 MPa Use the formula of engineering strain: ɛ = Al / lo Rearrange the formula to get change in length: Al = ɛl, Locate in Figure 5.7 the stress equivalent to 345 MPa and read the strain corresponding to this stress. The strain corresponding to a stress of 345 MPa is o.06. Al = e lo Al = (0.06) (250 mm) Al = 15 mm Clicker Exercise 2 1 A bar of steel alloy that exhibits the stress-strain behavior shown in Figure 5.10 is subjected to a tensile load; the specimen is 375 mm long and square cross section 5.5 mm on a side. a. Compute the magnitude of the load necessary to produce an elongation of 2.25 mm. b. What will be the deformation after the load has been released? 2. A load of 140,000N is applied to a cylindrical specimen of a steel alloy displaying a stress-strain behavior shown in Figure 5.10 that has a cross sectional diameter of 1o mm. a. Will the specimen experience elastic or plastic deformation? Why? b. If the original specimen length is 500 mm, how much will it increase in length when this load is applied? 300 2000 103 psi 300 MPa 2000 200 200 1000 1000E 100E 100 0.000 0.005 0.010 0.015 Strain 0.000 0.020 0.040 0.060 0.080 Strain Figure 5.10 Tensile stress-strain behavior for an alloy steel 4.3 Ductility Ductility is a measure of the degree of plastic deformation that has been sustained at fracture. A material that experiences very little or no plastic deformation upon fracture is termed as brittle. The material that experiences plastic deformation is termed ductile. The tensile stress-strain behavior of both brittle and ductile materials is shown in Figure 5.8. Stress (MPa) Stress Stress (103 psi)
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