Question

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.

Answer 1

The mechanical properties of the tensile test specimen, such as yield strength, modulus of elasticity, tensile strength, percent elongation, and percent reduction in area, are determined by performing calculations based on the given dimensions, yield force, maximum load, and deformations observed during the test.

Step-by-step explanation:

To determine the mechanical properties of a tensile test specimen, the following calculations are performed:

• Yield strength (a) is the stress at which a material begins to deform plastically. Here, yield strength = Yield force / Area = 48,000 N / 100 mm² = 480 MPa (since 1 N/mm² = 1 MPa).
• Modulus of elasticity (b) (also known as Young's Modulus) which is a measure of the stiffness of a material, can be calculated using the formula Stress / Strain. The stress before yielding is 480 MPa, and strain is the change in length over the original length, which is (50.23 mm - 50 mm) / 50 mm = 0.0046. Thus, Modulus of elasticity = 480 MPa / 0.0046 = 104,347.83 MPa.
• Tensile strength (c) is the maximum stress that a material can withstand while being stretched or pulled before necking. Tensile strength = Maximum load / Original cross-sectional area = 87,000 N / 100 mm² = 870 MPa.
• To determine the percent elongation (d) at fracture, use the formula ((Final gage length - Initial gage length) / Initial gage length) * 100. Percent elongation = ((67.3 mm - 50 mm) / 50 mm) * 100 = 34.6%.
• The percent reduction in area (e) is calculated by ((Original area - Final area) / Original area) * 100. Percent reduction in area = ((100 mm² - 53 mm²) / 100 mm²) * 100 = 47%.

Answer 2

a) yield strength

`\sigma_y = (F_y)/(A) = =(48000)/(100) = 480 MPa`

b) modulus of elasticity

strain calculation

`\varepsilon_0=(L-L_0)/(L_0)=(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 = (\sigma_y)/(\varepsilon_(new))=(480)/(0.0026)`

= 184615.28 MPa = 184.615 GPa

c) tensile strength

`\sigma_u =(F_(max))/(A)=(87000)/(100)=870MPa`

d) percentage elongation

`\% Elongation = (L-L_0)/(L_0)* 100 = (67.3-50)/(50)* 100 = 34.6\%`

e) percentage of area reduction

`\% Area\ reduction = (A-A_f)/(A)* 100=(100-53)/(100)= 47 \%`