# MECHANICAL PROPERTIES OF MATERIALS

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?

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?

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

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?

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?

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, = (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.

Solution: Starting volume V = LoAo = 2.0(0.5) = 1.0 in3

(1) A = V/L = 1.0/2.5 = 0.4 in2

So, true stress = 24,000(0.5)/0.4 = 31,250 lb/in2 and = ln(2.5/2.0) = 0.223

(2) A = 1.0/3.2 = 0.3125 in2

So, true stress = 28,000(0.5)/0.3125 = 44,800 lb/in2 and = ln(3.2/2.0) = 0.470

These are two data points with which to determine the parameters of the flow curve equation.

(1) 31,250 = K(0.223)n and (2) 44,800 = K(0.470)n

44,800/31,250 = (0.470/0.223)n

1.4336 = (2.1076)n

ln(1.4336) = n ln(2.1076)

0.3602 = .7455 n n = 0.483

(1) K = 31,250/(0.223).483 = 64,513 lb/in2

(2) K = 44,800/(0.470).483 = 64,516 lb/in2 Use average K = 64,515 lb/in2

The flow curve equation is: = 64,515 0.483

3.12 A tensile specimen is elongated to twice its original length. Determine the engineering strain and true strain for this test. If the metal had been strained in compression, determine the final compressed length of the specimen such that (a) the engineering strain is equal to the same value as in tension (it will be negative value because of compression), and (b) the true strain would be equal to the same value as in tension (again, it will be negative value because of compression). Note that the answer to part (a) is an impossible result. True strain is therefore a better measure of strain during plastic deformation.

Solution: Engineering strain e = (2.0 – 1.0)/1.0 = 1.0

True strain = ln(2.0/1.0) = ln(2.0) = 0.693

(a) To be compressed to the same engineering strain (e = -1.0) the final height of the compression specimen would have to be zero, which is impossible.

(b) To be compressed to the same true strain value (e = -0.693) the final height of the compression specimen can be determined as follows:

= -.693 = ln(Lf/Lo)

Lf/Lo = exp.(-0.693) = 0.500 Therefore, Lf = 0.5 Lo

3.13 Derive an expression for true strain as a function of D and Do for a tensile test specimen of round cross section, where D = the instantaneous diameter of the specimen and Do is its original diameter.

Solution: Starting with the definition of true strain as = ln(L/Lo) and assuming constant volume, we have V = AoLo = AL

Therefore, L/Lo = Ao/A

A = D2 and Ao = Do2

Ao/A = Do2 /D2 = (Do/D)2

= ln(Do/D)2 = 2 ln(Do/D)

3.14 Show that true strain = ln(1 + e), where e = engineering strain.

Solution: Starting definitions: (1) = ln(L/Lo) and (2) e = (L – Lo)/Lo

Consider definition (2): e = L/Lo – Lo/Lo = L/Lo – 1

Rearranging, 1 + e = L/Lo

Substituting this into definition (1), = ln(1 + e)

3.15 Based on results of a tensile test, the flow curve strain-hardening exponent = 0.40 and strength coefficient = 551.6 MPa. Based on this information, calculate the (engineering) tensile strength for the metal.

Solution: Tensile strength occurs at maximum value of load. Necking begins immediately thereafter. At necking, n = . Therefore, = 551.6(0.4)0.4 = 382.3 MPa. This is a true stress.

TS is defined as an engineering stress. From Problem 3.15, we know that = 2 ln(Do/D). Therefore,

0.4 = 2 ln(Do/D)

ln(Do/D) = .4/2 = 0.2

Do/D = exp(0.2) = 1.221

Area ratio = (Do/D)2 = (1.221)2 = 1.4918

The ratio between true stress and engineering stress would be the same ratio.

Therefore, TS = 1.4918(382.3) = 570.3 MPa

3.16 A copper wire of diameter 0.80 mm fails at an engineering stress = 248.2 MPa. Its ductility is measured as 75% reduction of area. Determine the true stress and true strain at failure.

Solution: Area reduction AR = (Ao – Af)/Ao = 0.75

Ao – Af = 0.75 Ao

Ao – 0.75Ao = 0.25 Ao = Af

If engineering stress = 248.2 MPa, then true stress = 248.2/0.25 = 992.8 MPa

True strain = ln(Lf/Lo) = ln(Ao/Af) = ln(4) = 1.386. However, it should be noted that these values are associated with the necked portion of the test specimen.

3.17 A steel tensile specimen with starting gage length = 2.0 in and crosssectional area = 0.5 in2 reaches a maximum load of 37,000 lb. Its elongation at this point is 24%. Determine the true stress and true strain at this maximum load.

Solution: Elongation = (L – Lo)/Lo = 0.24

L – Lo = 0.24 Lo

L = 1.24 Lo

A = Ao/1.24 = 0.8065 Ao

True stress = 37,000/0.8065(0.5) = 91,754 lb/in2

True strain = ln(1.24) = 0.215

### Compression

3.18 A metal alloy has been tested in a tensile test with the following results for the flow curve parameters: strength coefficient = 620.5 MPa and strain-hardening exponent = 0.26. The same metal is now tested in a compression test in which the starting height of the specimen = 62.5 mm and its diameter = 25 mm. Assuming that the cross section increases uniformly, determine the load required to compress the specimen to a height of (a) 50 mm and (b) 37.5 mm.

Solution: Starting volume of test specimen V = hDo2/4 = 62.5(25)2/4 = 30679.6 mm3.

(a) At h = 50 mm, = ln(62.5/50) = ln(1.25) = 0.223

Yf = 620.5(.223).26 = 420.1 MPa

A = V/L = 30679.6/50 = 613.6 mm2

F = 420.1(613.6) = 257,770 N

(b) At h = 37.5 mm, = ln(62.5/37.5) = ln(1.667) = 0.511

Yf = 620.5(0.511).26 = 521.1 MPa

A = V/L = 30679.6 /37.5 = 818.1 mm2

F = 521.1(818.1) = 426,312 N

3.19 The flow curve parameters for a certain stainless steel are strength coefficient = 1100 MPa and strain-hardening exponent = 0.35. A cylindrical specimen of starting crosssectional area = 1000 mm2 and height = 75 mm is compressed to a height of 58 mm. Determine the force required to achieve this compression, assuming that the cross section increases uniformly.

Solution: For h = 58 mm, = ln(75/58) = ln(1.293) = 0.257

Yf = 1100(.257).35 = 683.7 MPa

Starting volume V = 75(1000) = 75,000 mm3

At h = 58 mm, A = V/L = 75,000/58 = 1293.1 mm2

F = 683.7(1293.1) = 884,095 N.

3.20 A steel test specimen (modulus of elasticity = 30 x 106 lb/in2) in a compression test has a starting height = 2.0 in and diameter = 1.5 in. The metal yields (0.2% offset) at a load = 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a) yield strength and (b) flow curve parameters (strength coefficient and strain-hardening exponent). Assume that the crosssectional area increases uniformly during the test.

Solution: (a) Starting volume of test specimen V = h D2/4 = 2(1.5)2/4 = 3.534 in3.

Ao = Do/4 = (1.5)2/4 = 1.767 in2

Y = 140,000/1.767 = 79,224 lb/in2

(b) Elastic strain at Y = 79,224 lb/in2 is e = Y/E = 79,224/30,000,000 = 0.00264

Strain including offset = 0.00264 + 0.002 = 0.00464

Height h at strain = 0.00464 is h = 2.0(1 – 0.00464) = 1.9907 in.

Area A = 3.534/1.9907 = 1.775 in2.

True strain = 140,000/1.775 = 78,862 lb/in2.

At F = 260,000 lb, A = 3.534/1.6 = 2.209 in2.

True stress = 260,000/2.209 = 117,714 lb/in2.

True strain = ln(2.0/1.6) = 0.223

Given the two points: (1) = 78,862 lb/in2 at = 0.00464, and (2) = 117,714 lb/in2 at = 0.223.

117,714/78,862 = (0.223/0.00464)n

1.493 = (48.06)n

ln(1.493) = n ln(48.06)

0.4006 = 3.872 n n = 0.103

K = 117,714/(0.223)0.103 = 137,389 lb/in2.

The flow curve equation is: = 137,389 .103

Bending and Shear

3.21 A bend test is used for a certain hard material. If the transverse rupture strength of the material is known to be 1000 MPa, what is the anticipated load at which the specimen is likely to fail, given that its width = 15 mm, thickness = 10 mm, and length = 60 mm?

SolutionF = (TRS)(bt2)/1.5L = 1000(15 x 102)/(1.5 x 60) = 16,667 N.

3.22 A special ceramic specimen is tested in a bend test. Its width = 0.50 in and thickness = 0.25 in. The length of the specimen between supports = 2.0 in. Determine the transverse rupture strength if failure occurs at a load = 1700 lb.

SolutionTRS = 1.5FL/bt2 = 1.5(1700)(2.0)/(0.5 x 0.252) = 163,200 lb/in2.

3.23 A torsion test specimen has a radius = 25 mm, wall thickness = 3 mm, and gage length = 50 mm. In testing, a torque of 900 Nm results in an angular deflection = 0.3. Determine (a) the shear stress, (b) shear strain, and (c) shear modulus, assuming the specimen had not yet yielded. (d) If failure of the specimen occurs at a torque = 1200 Nm and a corresponding angular deflection = 10, what is the shear strength of the metal?

Solution: (a) = T/(2R2t) = (900 x 1000)/(2(25)2(3)) = 76.39 MPa.

(b) = R/L, = 0.3(2/360) = 0.005236 radians

= 25(0.005236)/50 = 0.002618

(c) = G, G = / = 76.39/0.002618 = 29,179 MPa.

(d) S = (1200(103))/(2(25)2(3)) = 101.86 MPa.

3.24 In a torsion test, a torque of 5000 ftlb is applied which causes an angular deflection = 1 on a thinwalled tubular specimen whose radius = 1.5 in, wall thickness = 0.10 in, and gage length = 2.0 in. Determine (a) the shear stress, (b) shear strain, and (c) shear modulus, assuming the specimen had not yet yielded. (d) If the specimen fails at a torque = 8000 ftlb and an angular deflection = 23, calculate the shear strength of the metal.

Solution: (a) = T/(2R2t) = (5000 x 12)/(2(1.5)2(0.1)) = 42,441 lb/in2.

(b) = R/L, = 1(2/360) = 0.01745 rad., = 1.5(0.01745)/2.0 = 0.01309

(c) = G, G = / = 42,441/0.01309 = 3.24 x 106 lb/in2.

(d) S = (8000 x 12)/(2(1.5)2(0.1)) = 67,906 lb/in2.