METALS AND METAL ALLOYS

Question 1

four most common materials widely used today

Answer 1

Metals, ceramics, polymers and composites

Question 2

produces an elongation and positive linear strain

Answer 2

tensile load

Question 3

produces contraction and a negative linear strain

Answer 3

compressive load

Question 4

To compare specimens of different sizes, the __ is calculated per unit area

Answer 4

load

Question 5

Engineering stress

Answer 5

σ = F / Ao, F is load applied perpendicular to specimen’s cross section; Ao is cross-sectional
area (perpendicular to the force) before application of the load.

Question 6

Engineering strain

Answer 6

ε = Δl / lo (× 100 %), Δl is change in length, lo is the original length

Question 7

Stress and strain are – for tensile loads, – for compressive loads

Answer 7

positive, negative

Question 8

shear stress

Answer 8

Shear stress: τ = F / Ao, F is load applied parallel to the upper and lower faces each of which has an
area A0.

Question 9

shear strain

Answer 9

Shear strain: γ = tgθ (× 100 %)
θ is strain angle

Question 10

variation of pure shear, change in shape not volume

Answer 10

torsion

Question 11

twisting or rotating force applied to a structural member, causing it to resist twisting

Answer 11

torsion

Question 12

in torsion, The shear stress in this case is a function of applied —, shear strain is related to —.

Answer 12

torque T, angle of twist φ

Question 13

measure of the twisting force applied to a shaft or beam. It is calculated as the product of force and the perpendicular distance from the axis of rotation.
Units: Newton-meters (Nm) or pound-feet (lb-ft).

Answer 13

torque

Question 14

angular deformation of a shaft or beam due to the applied torque. It is measured in radians or degrees

Answer 14

angle of twist

Question 15

— is a measure of a material’s ability to resist a pulling force before it breaks or fractures. It is expressed in units of force per unit area, such as pascals (Pa) or pounds per square inch (psi).

Answer 15

tensile strength

Question 16

In tensile tests, if the deformation is elastic, the stress-strain relationship is called –

Answer 16

Hooke’s Law:
σ=Eε, E is the Young’s Modulus or modulus of elasticity and has the same units as 𝛔, N/m2 or Pa.

Question 17

Deformation in which stress and strain are proportional is called –

Answer 17

elastic deformation

Question 18

A plot of stress (ordinate) versus strain (abscissa) results in a linear relationship. The slope of this linear segment corresponds to the –

Answer 18

modulus of elasticity, E

Question 19

stiffness, or a material’s resistance to elastic deformation

Answer 19

modulus of elasticity, E

Question 20

greater the modulus, the – the material, or the – the elastic strain that results from the application of a given stress

Answer 20

stiffer, smaller

Question 21

– is the deformation of a material that is reversible, meaning the material will return to its original shape and size once the stress causing the deformation is removed

Answer 21

elastic strain

Question 22

– is nonpermanent, which means that when the applied load is released, the piece returns to its original shape

Answer 22

elastic deformation

Question 23

There are some materials (i.e., gray cast iron, concrete, and many polymers) for which this elastic portion of the stress–strain curve is –; hence, it is not possible to determine a modulus of elasticity as described previously

Answer 23

not linear

Question 24

nonlinear behavior, either the – or – modulus is normally used

Answer 24

tangent, secant

Question 25

– is taken as the slope of the stress–strain curve at some specified level of stress

Answer 25

tangent modulus

Question 26

– represents the slope of a secant drawn from the origin to some given point of the curve.

Answer 26

secant modulus

Question 27

– is manifested as small changes in the interatomic spacing and the stretching of interatomic bonds and corresponding slight atomic displacements.

Answer 27

macroscopic elastic strain

Question 28

Values of the modulus of elasticity —; caused by –

Answer 28

ceramic and metals are the same, polymers have lower; atomic bonding

Question 29

increasing temperature, the modulus of elasticity –.

Answer 29

decreases

Question 30

— occurs when a material is subjected to a force that causes it to twist or slide relative to its neighboring layers. This type of deformation is characterized by the relative displacement of particles within the material, without any change in volume

Answer 30

shear elastic deformation

Question 31

For shear elastic deformations, shear stress and strain are proportional to each other through the expression

Answer 31

τ = Gγ
where G is the shear modulus, the slope of the linear elastic region of the shear stress–
strain curve

Question 32

elastic deformation is – ; meaning that the strain produced in a material is directly proportional to the applied stress at any given instant, regardless of how long the stress has been applied

Answer 32

time independent

Question 33

in reality elastic deformation takes time (finite rate of
atomic/molecular deformation processes) - continues after initial loading, and after load release

Answer 33

time-dependent, anelasticity

Question 34

anelasticity effect is normally small for metals but can be significant for polymers

Answer 34

visco-elastic behavior

Question 35

– is a combination of elastic and viscous properties exhibited by certain materials. These materials exhibit both solid-like and liquid-like characteristics, depending on the rate of deformation.

Answer 35

Viscoelastic behavior

Question 36

Materials subject to tension –

Answer 36

shrink laterally

Question 37

Those subject to compression –

Answer 37

bulge

Question 38

ratio of lateral and axial strains is called

Answer 38

Poisson’s ratio υ

Question 39

Sign in the above equations shows that lateral strain is in opposite sense to longitudinal strain υ is dimensionless

Answer 39

T

Question 40

Axial (z) elongation (positive strain) and lateral (x and y) contractions (negative strains) in response to an imposed –

Answer 40

tensile stress

Question 41

Poisson’s ratio – Theoretical value for isotropic material:

Answer 41

isotropic material: 0.25
➢ Maximum value: 0.50, Typical value: 0.24 - 0.30

Question 42

For isotropic materials, shear and elastic moduli are related to each other and
to Poisson’s ratio according to
E = 2G (1 + υ)

Answer 42

E = 2G (1 + υ), G is about 0.4E; thus, if the value of one modulus is known, the other may be approximated.

Question 43

• stress and strain are not proportional to each other
• the deformation is not reversible
• deformation occurs by breaking and re-arrangement of atomic bonds

Answer 43

Plastic deformation