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barxatty [35]
3 years ago
13

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 t

he 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.
Physics
1 answer:
Shalnov [3]3 years ago
7 0

Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

 \sigma_u =\dfrac{F_{max}}{A}=\dfrac{87000}{100}=870MPa

d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

e) percentage of area reduction

\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

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