Mechanical Properties of Materials

Question 1

What are the three types of static stresses to which materials are subjected?

Answer 1

tensile, compressive, and shear.

Question 2

State Hooke’s law.

Answer 2

Hooke’s Law defines the stress-strain relationship for an elastic material: sigma = Ee, where E = a constant of proportionality called the modulus of elasticity.

Question 3

What is the difference between engineering stress and true stress in a tensile test?

Answer 3

Engineering stress divides the load (force) on the test specimen by the original area; whereas true stress divides the load by the instantaneous area which decreases as the specimen stretches

Question 4

Define tensile strength of a material.

Answer 4

The tensile strength is the maximum load experienced during the tensile test divided by the original area.

Question 5

Define yield strength of a material.

Answer 5

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 stress-strain 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%.

Question 6

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

Answer 6

Because of necking that occurs in the test specimen.

Question 7

What is work hardening?

Answer 7

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

Question 8

Under what circumstances does the strength coefficient have the same value as the yield
strength?

Answer 8

When the material is perfectly plastic and does not strain harden

Question 9

How does the change in cross-sectional area of a test specimen in a compression test differ from its counterpart in a tensile test specimen?

Answer 9

In a compression test, the specimen cross-sectional area increases as the test progresses; while in a tensile test, the cross-sectional area decreases.

Question 10

What is the complicating factor that occurs in a compression test that might be considered analogous to necking in a tensile test?

Answer 10

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

Question 11

Tensile testing is not appropriate for hard brittle materials such as ceramics. What is the test commonly used to determine the strength properties of such materials?

Answer 11

A three-point bending test is commonly used to test the strength of brittle materials. The test provides a measure called the transverse rupture strength for these materials.

Question 12

How is the shear modulus of elasticity G related to the tensile modulus of elasticity E, on average?

Answer 12

G = 0.4 E, on average

Question 13

How is shear strength S related to tensile strength TS, on average?

Answer 13

S = 0.7 TS, on average.

Question 14

What is hardness, and how is it generally tested?

Answer 14

Hardness is defined as the resistance to indentation of a material. It is tested by pressing a hard object (sphere, diamond point) into the test material and measuring the size (depth, area) of the indentation

Question 15

Why are different hardness tests and scales required?

Answer 15

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.

Question 16

Define the recrystallization temperature for a metal

Answer 16

The recrystallization temperature is the temperature at which a metal recrystallizes (forms new grains) rather than work hardens when deformed.

Question 17

Define viscosity of a fluid

Answer 17

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

Question 18

What is the defining characteristic of a Newtonian fluid?

Answer 18

A Newtonian fluid is one for which viscosity is a constant property at a given temperature. Most liquids (water, oils) are Newtonian fluids.

Question 19

What is viscoelasticity, as a material property?

Answer 19

Viscoelasticity refers to the property most commonly exhibited by polymers that defines the strain of the material as a function of stress and temperature over time. It is a combination of viscosity and elasticity.

Question 20

A tensile test specimen has a gage length = 2.0 in and diameter = 0.798 in.
Yielding occurs at a load of 31,000 lb. The corresponding gage length = 2.0083 in, which is
the 0.2 percent yield point. The maximum load of 58,000 lb is reached at a gage length =
2.47 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If
fracture occurs at a gage length of 2.62 in, determine the percent elongation. (e) If the
specimen necked to an area = 0.36 in2
, determine the percent reduction in area.

Answer 20

(a) Area A = π(0.798)2
/4 = 0.50 in2
.
Yield strength Y = 31,000/0.50 = 62,000 lb/in2
(b) s = E e. Subtracting the 0.2% offset, e = (2.0083  2.0)/2.0  0.002 = 0.00215. This is
the elastic strain to calculate modulus of elasticity.
E = s/e = 62,000/0.00215 = 28.84  106
lb/in2
(c) TS = 58,000/0.5 = 116,000 lb/in2
(d) EL = (2.62  2.0)/2.0 = 0.62/2.0 = 0.31 = 31%
(e) AR = (0.5  0.36)/0.5 = 0.28 = 28%

Question 21

In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa.
When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and the
strain-hardening exponent in the flow curve equation.

Answer 21

Two equations: (1) 250 = K(0.12)
n
 and (2) 350 = K(0.26)
n
350/250 = (0.26/0.12)
n 1.4 = (2.1667)
n
n ln(2.1667) = ln(1.4) 0.7732 n = 0.3365 n = 0.435
Substituting this value with the data back into the flow curve equation to obtain the value
of the strength coefficient K:
(1) K = 250/(0.12)
.435 = 629 MPa
Check: (2) K = 350/(0.26)
.435 = 629 MPa
The flow curve equation is:  = 629
0.435

Question 22

) The flow curve for a certain metal has a strain-hardening exponent = 0.20
and its strength coefficient = 26,000 lb/in2
. Determine (a) flow stress at a true strain = 0.15
and (b) true strain at a flow stress = 20,000 lb/in2
. (c) What is the likely metal in this
problem?

Answer 22

a) Yf = 26,000(0.15).20 = 17,791 lb/in2
(b)  = (20,000/26,000)1/.20 = (0.7692)
5.0 = 0.269
(c) The flow curve parameters indicate that the metal is probably annealed pure aluminum,
according to Table 3.4

Question 23

(USCS Units) In a compression test, a steel test specimen (E = 30 * 10^6
lb/in2
) 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 cross-sectional area increases uniformly during the test.

Answer 23

(a) Starting volume of test specimen V = hD
2
/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

Question 24

One of the inspectors in the quality control department has frequently used the Brinell and
Rockwell hardness tests, for which equipment is available in the company. He claims that
the Rockwell hardness test is based on the same principle as the Brinell test, which is that
hardness is measured as the applied load divided by the area of the impression made by an
indentor. Is he correct? If not, how is the Rockwell test different?

Answer 24

No, the inspector’s claim is not correct. Not all hardness tests are based on the
applied load divided by area. The Rockwell hardness test measures the depth of
indentation of a cone or small-diameter ball resulting from an applied load.