Strength of Materials PPB25403

Lecture 3: Mechanical Properties of Materials

Learning Outcomes;

Tension and Compression Tests

Stress-Strain Diagrams

Hooke’s Law

Poisson’s Ratio

The Tension and Compression Test

The strength of a material depends on its ability to

sustain a load.

This property is to perform under the tension or

compression test.

The following machine is designed to read the load

required to maintain specimen stretching.

The Stress–Strain Diagram

Conventional Stress–Strain Diagram

Nominal or engineering stress is obtained by dividing

the applied load P by the specimen’s original cross-

sectional area.

Nominal or engineering strain is obtained by dividing

the change in the specimen’s gauge length by the

specimen’s original gauge length.

0A

P

0L

The Stress–Strain Diagram

Conventional Stress–Strain Diagram

Stress-Strain Diagram

Elastic BehaviourStress is proportional to the strain.

Material is said to be

linearly elastic.

Yielding Increase in stress above

elastic limit will cause material

to deform permanently.

The Stress–Strain Diagram

Conventional Stress–Strain Diagram

Stress-Strain Diagram

Strain Hardening.

After yielding a further load will

reaches a ultimate stress.

Necking

At ultimate stress, cross-sectional

area begins to decrease in a

localized region of the specimen.

Specimen breaks at the

fracture stress.

The Stress–Strain Diagram

True Stress–Strain Diagram

The values of stress and strain computed from these

measurements are called true stress and true strain.

Use this diagram since most engineering design is done

within the elastic range.

Stress–Strain Behavior of Ductile and Brittle Materials

Ductile Materials

Material that can subjected to large strains before it

ruptures is called a ductile material.

Brittle Materials

Materials that exhibit little or no yielding before failure

are referred to as brittle materials.

Hooke’s Law

Hooke’s Law defines the linear relationship between

stress and strain within the elastic region.

E can be used only if a material has linear–elastic

behaviour.

Eσ = stress

E = modulus of elasticity or Young’s modulus

ε = strain

Hooke’s Law

Strain Hardening

When ductile material is loaded into the plastic region

and then unloaded, elastic strain is recovered.

The plastic strain remains and material is subjected to a

permanent set.

Strain Energy

When material is deformed by external loading, it will

store energy internally throughout its volume.

Energy is related to the strains called strain energy.

Modulus of Resilience

When stress reaches the proportional limit, the strain-

energy density is the modulus of resilience, ur.

Eu

pl

plplr

2

2

1

2

1

Strain Energy

Modulus of Toughness

Modulus of toughness, ut, represents the entire area

under the stress–strain diagram.

It indicates the strain-energy density of the material just

before it fractures.

Example 3.2The stress–strain diagram for an aluminum alloy that is used for making aircraft

parts is shown. When material is stressed to 600 MPa, find the permanent strain

that remains in the specimen when load is released. Also, compute the modulus of

resilience both before and after the load application.

Solution:When the specimen is subjected to the load,

the strain is approximately 0.023 mm/mm.

The slope of line OA is the modulus of elasticity,

From triangle CBD,

mm/mm 008.0100.7510600 9

6

CDCDCD

BDE

GPa 0.75006.0

450E

Solution:This strain represents the amount of recovered elastic strain.

The permanent strain is

(Ans) MJ/m 40.2008.06002

1

2

1

(Ans) MJ/m 35.1006.04502

1

2

1

3

3

plplfinalr

plplinitialr

u

u

(Ans) mm/mm 0150.0008.0023.0OC

Computing the modulus of resilience,

Note that the SI system of units is measured in joules, where 1 J = 1 N • m.

Poisson’s Ratio

Poisson’s ratio, v (nu), states that in the elastic range, the ratio of these strains is a constant since the

deformations are proportional.

Negative sign since longitudinal elongation (positive strain) causes lateral contraction (negative strain), and vice

versa.

long

latv Poisson’s ratio is dimensionless.

Typical values are 1/3 or 1/4.

Example 3.4A bar made of A-36 steel has the dimensions shown. If an axial force of is applied to

the bar, determine the change in its length and the change in the dimensions of its

cross section after applying the load. The material behaves elastically.

Solution:The normal stress in the bar is

mm/mm 108010200

100.16 6

6

6

st

zz

E

Pa 100.1605.01.0

1080 63

A

Pz

From the table for A-36 steel, Est = 200 GPa

The axial elongation of the bar is therefore

Solution:

The contraction strains in both the x and y directions are

m/m 6.25108032.0 6

zstyx v

(Ans) m1205.11080 6

z zz L

The changes in the dimensions of the cross section are

(Ans) m28.105.0106.25

(Ans) m56.21.0106.25

6

6

yyy

xxx

L

L

The Shear Stress–Strain Diagram

For pure shear, equilibrium

requires equal shear stresses

on each face of the element.

When material is

homogeneous and isotropic,

shear stress will distort the

element uniformly.

The Shear Stress–Strain Diagram

For most engineering materials the elastic behaviour is linear, so

Hooke’s Law for shear applies.

3 material constants, E, and G are actually related by the equation

G

G = shear modulus of elasticity

or the modulus of rigidity

v

EG

12

Example 3.5A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram

is shown. Find the shear modulus G, the proportional limit, and the ultimate shear

stress. Also, find the maximum distance d that the top of a block of this material

could be displaced horizontally if the material behaves elastically when acted upon

by a shear force V. What is the magnitude of V necessary to cause this

displacement?

Solution:The coordinates of point A are (0.008 rad, 360 MPa).

Thus, shear modulus is

(Ans) MPa 1045008.0

360 3G

Solution:

By inspection, the graph ceases to be linear

at point A. Thus, the proportional limit is

(Ans) MPa 504u

(Ans) MPa 360pl

This value represents the maximum shear stress,

point B. Thus the ultimate stress is

Since the angle is small, the top of the will be displaced horizontally by

mm 4.0mm 50

008.0rad 008.0tan dd

The shear force V needed to cause the displacement is

(Ans) kN 270010075

MPa 360 ; VV

A

Vavg

*Failure of Materials Due to Creep and

Fatigue

Creep

When material support a load for long period of time, it will deform

until a sudden fracture occurs.

This time-dependent permanent deformation is known as creep.

Both stress and/or temperature play a significant role in the rate of

creep.

Creep strength will decrease

for higher temperatures or

higher applied stresses.

*Failure of Materials Due to Creep and

Fatigue

Fatigue

When metal subjected to repeated cycles of

stress or strain, it will ultimately leads to fracture.

This behaviour is called fatigue.

Endurance or fatigue limit is a limit which no

failure can be detected after applying a load for

a specified number of cycles.

This limit can be

determined in S-N diagram.