A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and the corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 87,000 N is reached at a gage length = 64.2 mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to an area = 53 mm2, determine the percent reduction in area.

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wagonbelleville wagonbelleville · Answer 1 · 2019-09-13T08:17:42+02:00

Answer:

a) yield strength

[tex]\sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa[/tex]

b) modulus of elasticity

strain calculation

[tex]\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046[/tex]

strain for offset yield point

[tex]\varepsilon_{new} = \varepsilon_0 -0.002[/tex]

=0.0046-0.002 = 0.0026

now, modulus of elasticity

[tex]E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}[/tex]

= 184615.28 MPa = 184.615 GPa

c) tensile strength

[tex]\sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa[/tex]

d) percentage elongation

[tex]\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%[/tex]

e) percentage of area reduction

[tex]\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%[/tex]