Engineering strain formula

3 MECHANICAL PROPERTIES OF MATERIALS

Review Questions

3.1 What is the dilemma between design and manufacturing in terms of mechanical properties?

Answer. To achieve design function and quality, the material must be strong; for ease of manufacturing, the material should not be strong, in general.

3.2 State Hooke's law.

Answer. Hooke's Law defines the stressstrain relationship for an elastic material: = E, where E = a constant of proportionality called the modulus of elasticity.

3.3 Define yield strength of a material.

Answer. The yield strength is the stress at which the material begins to plastically deform. It is usually measured as the 0.2% offset value, which is the point where the stressstrain curve for the material intersects a line that is parallel to the straight-line portion of the curve but offset from it by 0.2%.

3.4 Why cannot a direct conversion be made between the ductility measures of elongation and reduction in area using the assumption of constant volume?

Answer. Because of necking that occurs in the test specimen.

3.5 What is work hardening?

Answer. Work hardening, also called strain hardening, is the increase in strength that occurs in metals when they are strained.

3.6 How does the change in crosssectional area of a test specimen in a compression test differ from its counterpart in a tensile test specimen?

Answer. In a compression test, the specimen crosssectional area increases as the test progresses; while in a tensile test, the crosssectional area decreases.

3.7 What is the complicating factor that occurs in a compression test?

Answer. Barreling of the test specimen due to friction at the interfaces with the testing machine platens.

3.8 Why are different hardness tests and scales required?

Answer. Different hardness tests and scales are required because different materials possess widely differing hardnesses. A test whose measuring range is suited to very hard materials is not sensitive for testing very soft materials.

3.9 Define viscosity of a fluid.

Answer. Viscosity is the resistance to flow of a fluid material; the thicker the fluid, the greater the viscosity.

Multiple Choices

3.1 Which of the following are the three basic types of static stresses to which a material can be subjected (three correct answers): (a) compression, (b) hardness, (c) reduction in area, (d) shear, (e) tensile, (f) true stress, and (f) yield?

answer. (a), (d), and (e).

3.2 The plastic region of the stressstrain curve for a metal is characterized by a proportional relationship between stress and strain: (a) true or (b) false?

Answer. (b). It is the elastic region that is characterized by a proportional relationship between stress and strain. The plastic region is characterized by a power function the flow curve.

3.3 Which one of the following is the correct definition of ultimate tensile strength, as derived from the results of a tensile test on a metal specimen: (a) the stress encountered when the stressstrain curve transforms from elastic to plastic behavior, (b) the maximum load divided by the final area of the specimen, (c) the maximum load divided by the original area of the specimen, or (d) the stress observed when the specimen finally fails?

Answer. (c).

3.4 Which one of the following materials has the highest modulus of elasticity: (a) aluminum, (b) diamond, (c) steel, (d) titanium, or (e) tungsten?

Answer. (b).

3.5 The shear strength of a metal is usually (a) greater than or (b) less than its tensile strength?

Answer. (b).

3.6 Most hardness tests involve pressing a hard object into the surface of a test specimen and measuring the indentation (or its effect) that results: (a) true or (b) false?

Answer. (a).

3.7 Which one of the following materials has the highest hardness: (a) alumina ceramic, (b) gray cast iron, (c) hardened tool steel, (d) high carbon steel, or (e) polystyrene?

Answer. (a).

3.8 Viscosity can be defined as the ease with which a fluid flows: (a) true or (b) false?

Answer. (b). Viscosity is the resistance to flow.

Problems

Strength and Ductility in Tension

3.1 A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 168,000 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 = 92 mm2, determine the percent reduction in area.

Solution:

(a) Y = 98,000/200 = 490 MPa.

(b) s = E e

Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026

E = s/e = 490/0.0026 = 188.5 x 103 MPa.

(c) TS = 168,000/200 = 840 MPa.

(d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6%

(e) AR = (200 - 92)/200 = 0.54 = 54%

3.2 During a tensile test in which the starting gage length = 125.0 mm and the crosssectional area = 62.5 mm2, the following force and gage length data are collected (1) 17,793 N at 125.23 mm, (2) 23,042 N at 131.25 mm, (3) 27,579 N at 140.05 mm, (4) 28, 913 N at 147.01 mm, (5) 27,578 N at 153.00 mm, and (6) 20,462 N at 160.10 mm. The maximum load is 28,913 N and the final data point occurred immediately prior to failure. (a) Plot the engineering stress strain curve. Determine (b) yield strength, (c) modulus of elasticity, and (d) tensile strength.

Solution:

(a) Student exercise.

(b) From the plot, Y = 310.27 MPa.

(c) First data point is prior to yielding.

Strain e = (125.23 - 125)/125 = 0.00184, E = 310.27/0.00184 = 168,625 MPa.

(d) From the plot, TS = 462.6 MPa. Also, TS = 28,913/62.5 = 462.6 MPa.

Flow Curve

3.6 During a tensile test, a metal has a true strain = 0.10 at a true stress = 37,000 lb/in2. Later, at a true stress = 55,000 lb/in2, true strain = 0.25. Determine the strength coefficient and strain-hardening exponent in the flow curve equation.

Solution: (1) 37,000 = K(0.10)n and (2) 55,000 = K(0.25)n

55,000/37,000 = (0.25/0.10)n 1.4865 = (2.5)n

n ln(2.5) = ln(1.4865) 0.9163 n = 0.3964 n = 0.4326

Substituting this value with the data back into the flow curve equation, we obtain the value of the strength coefficient K:

(1) K = 37,000/(0.10)0.4326 = 100,191 lb/in2

(2) K = 55,000/(0.25)0.4326 = 100,191 lb/in2

The flow curve equation is: = 100,191 0.4326

3.7 In a tensile test a metal begins to neck at a true strain = 0.28 with a corresponding true stress = 345.0 MPa. Without knowing any more about the test, can you estimate the strength coefficient and the strain-hardening exponent in the flow curve equation?

Solution: If we assume that n = when necking starts, then n = 0.28.

Using this value in the flow curve equation, we have K = 345/(0.28).28 = 492.7 MPa

The flow curve equation is: = 492.7 0.28

3.9 The flow curve for a certain metal has a strain-hardening exponent of 0.22 and strength coefficient of 54,000 lb/in2. Determine (a) the flow stress at a true strain = 0.45 and (b) the true strain at a flow stress = 40,000 lb/in2.

Solution: (a) Yf = 54,000(0.45).22 = 45,300 lb/in2

(b) = (40,000/54,000)1/.22 = (0.7407)4.545 = 0.256

3.10 A metal is deformed in a tension test into its plastic region. The starting specimen had a gage length = 2.0 in and an area = 0.50 in2. At one point in the tensile test, the gage length = 2.5 in, and the corresponding engineering stress = 24,000 lb/in2; at another point in the test prior to necking, the gage length = 3.2 in, and the corresponding engineering stress = 28,000 lb/in2. Determine the strength coefficient and the strain-hardening exponent for this metal.