1. A test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5 in2. During the test the specimen yields under a load of 32,000 lb. The corresponding gage length = 2.0083 in. This is the 0.2 percent yield point. The maximum load of 60,000 lb is reached at a gage length = 2.60 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 2.92 in, determine the percent elongation. (e) If the specimen necked to an area = 0.25 in2, determine the percent reduction in area.

Answers

Answer 1
Final answer:

The yield strength is 64,000 psi, modulus of elasticity is approximately 15,422,000 psi, tensile strength is 120,000 psi, percent elongation is 46%, and percent reduction in area is 50% for the given tensile test specimen.

Explanation:

The question pertains to the calculation of various material properties from a tensile test. Specifically, we are looking to find the yield strength, modulus of elasticity, and tensile strength of a test specimen with given gage lengths and loads. Additionally, we will determine the percent elongation and percent reduction of area based on the provided gage lengths and areas before and after fracture.

Calculations:

Yield Strength: Yield strength is calculated as the load at which the material begins to deform permanently divided by the original cross-sectional area. Thus, yield strength = load / original area = 32,000 lb / 0.5 in2 = 64,000 psi.Modulus of Elasticity: Modulus of elasticity (Young's modulus) is a measure of the elasticity of a material and can be determined by the formula E = stress/strain. Here, E = (32,000 lb / 0.5 in2) / ((2.0083 in - 2.0 in) / 2.0 in) = 64,000 psi / 0.00415 = approximately 15,422,000 psi.Tensile Strength: Tensile strength is the maximum stress the material can withstand while being stretched or pulled before necking. For this specimen, tensile strength = maximum load / original area = 60,000 lb / 0.5 in2 = 120,000 psi.Percent Elongation: Percent elongation is an indication of ductility and is expressed as ((final gage length - initial gage length) / initial gage length) x 100%. Thus, percent elongation = ((2.92 in - 2.0 in) / 2.0 in) x 100% = 46%.Percent Reduction in Area: Percent reduction in area represents how much the cross-sectional area of a material decreases under tensile stress and is calculated as ((original area - reduced area) / original area) x 100%. Subsequently, the percent reduction in area = ((0.5 in2 - 0.25 in2) / 0.5 in2) x 100% = 50%.

Answer 2

a) The yield strength is 64,000 psi

b) The modulus of elasticity is 15,421,686.75 psi

c) The tensile strength is 120,000 psi.

d) The percent elongation at fracture is 46%

e) The percent reduction in area is 50%.

Given Data:

- Initial gage length, [tex]\( L_0 = 2.0 \)[/tex] in

- Initial area, [tex]\( A_0 = 0.5 \) in\(^2\)[/tex]

- Load at yield point, [tex]\( F_y = 32,000 \)[/tex] lb

- Gage length at yield point, [tex]\( L_y = 2.0083 \)[/tex] in

- Maximum load, [tex]\( F_{max} = 60,000 \)[/tex] lb

- Gage length at maximum load, [tex]\( L_{max} = 2.60 \)[/tex] in

- Gage length at fracture, [tex]\( L_f = 2.92 \)[/tex] in

- Final necked area, [tex]\( A_f = 0.25 \) in\(^2\)[/tex]

Calculations:

(a) Yield Strength [tex](\( \sigma_y \))[/tex]

[tex]\[ \sigma_y = \frac{F_y}{A_0} \][/tex]

[tex]\[ \sigma_y = \frac{32,000 \, \text{lb}}{0.5 \, \text{in}^2} \][/tex]

[tex]\[ \sigma_y = 64,000 \, \text{psi} \][/tex]

(b) Modulus of Elasticity [tex](\( E \))[/tex]

[tex]\[ \text{Strain at yield} = \frac{L_y - L_0}{L_0} = \frac{2.0083 - 2.0}{2.0} \][/tex]

[tex]\[ \text{Strain at yield} = 0.00415 \][/tex]

[tex]\[ E = \frac{\sigma_y}{\text{Strain at yield}} \][/tex]

[tex]\[ E = \frac{64,000 \, \text{psi}}{0.00415} \][/tex]

[tex]\[ E = 15,421,686.75 \, \text{psi} \][/tex]

(c) Tensile Strength [tex](\( \sigma_{ts} \))[/tex]

[tex]\[ \sigma_{ts} = \frac{F_{max}}{A_0} \][/tex]

[tex]\[ \sigma_{ts} = \frac{60,000 \, \text{lb}}{0.5 \, \text{in}^2} \][/tex]

[tex]\[ \sigma_{ts} = 120,000 \, \text{psi} \][/tex]

(d) Percent Elongation

[tex]\[ \text{Percent elongation} = \frac{L_f - L_0}{L_0} \times 100 \][/tex]

[tex]\[ \text{Percent elongation} = \frac{2.92 - 2.0}{2.0} \times 100 \][/tex]

[tex]\[ \text{Percent elongation} = \frac{0.92}{2.0} \times 100 \][/tex]

[tex]\[ \text{Percent elongation} = 46 \% \][/tex]

(e) Percent Reduction in Area

[tex]\[ \text{Percent reduction in area} = \frac{A_0 - A_f}{A_0} \times 100 \][/tex]

[tex]\[ \text{Percent reduction in area} = \frac{0.5 - 0.25}{0.5} \times 100 \][/tex]

[tex]\[ \text{Percent reduction in area} = \frac{0.25}{0.5} \times 100 \][/tex]

[tex]\[ \text{Percent reduction in area} = 50 \% \][/tex]


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A

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Answers

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