To determine the nominal or engineering stress and strain experienced by a specimen of a material while it is subjected to a tension test, and to be able to read important values from a conventional stress-strain diagram obtained from the test. A tension test is being conducted on a steel-rod specimen with a gauge length of L0=50 mm and initial diameter of d0=13 mm. Data were collected to form the conventional stress-strain diagram as shown. From the diagram, f = 506 MPa , e = 689 MPa , g = 585 MPa , and h = 0.146 mm/mm . A) Assuming that the strain remains constant throughout the region between the gauge points, determine the nominal strain ε experienced by the rod if it is elongated to L = 53.0 mm . B) Assuming that the stress is constant over the cross-sectional area and if the tension force used is P = 16.0 kN , find the nominal stress experienced by the rod. C)Determine the force P needed to reach the ultimate stress in the steel-rod specimen.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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To determine the nominal or engineering stress and strain experienced by a specimen of a material while it is subjected to a tension test, and to be able to read important values from a conventional stress-strain diagram obtained from the test.

A tension test is being conducted on a steel-rod specimen with a gauge length of L0=50

mm and initial diameter of d0=13 mm. Data were collected to form the conventional stress-strain diagram as shown. From the diagram, f = 506 MPa , e = 689 MPa , g = 585 MPa , and h = 0.146 mm/mm .

A) Assuming that the strain remains constant throughout the region between the gauge points, determine the nominal strain ε experienced by the rod if it is elongated to L = 53.0 mm .

B) Assuming that the stress is constant over the cross-sectional area and if the tension force used is P = 16.0 kN , find the nominal stress experienced by the rod.

C)Determine the force P needed to reach the ultimate stress in the steel-rod specimen.

σ (MPa)
750
500+-o=f
250-
σ=e
do
-Lo
σ=g
02.04 06 08 10 12 14 16 18 20 22 24
E = h
& (cm/cm)
Transcribed Image Text:σ (MPa) 750 500+-o=f 250- σ=e do -Lo σ=g 02.04 06 08 10 12 14 16 18 20 22 24 E = h & (cm/cm)
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