Mec_Ch_2_2011

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Mechanics of MaterialChapter II

Stress and Strain – Axial Loading

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Stress and StrainContentsStress & Strain: Axial LoadingNormal StrainStress-Strain TestStress-Strain Diagram: Ductile MaterialsStress-Strain Diagram: Brittle Materials Hooke’s Law: Modulus of ElasticityElastic vs. Plastic BehaviorFatigueDeformations Under Axial LoadingExample 2.01Sample Problem 2.1Static IndeterminacyExample 2.04Thermal StressesPoisson’s Ratio

Generalized Hooke’s LawDilatation: Bulk ModulusShearing StrainExample 2.10Relation Among E, , and GSample Problem 2.5Composite MaterialsSaint-Venant’s PrincipleStress Concentration: HoleStress Concentration: FilletExample 2.12Elastoplastic MaterialsPlastic DeformationsResidual StressesExample 2.14, 2.15, 2.16

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Stress and StrainAxial loading

• Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced under loading. Statics analyses alone are not sufficient.

• Considering structures as deformable allows determination of member forces and reactions which are statically indeterminate.

• Determination of the stress distribution within a member also requires consideration of deformations in the member.

44

Displacement

Movement of a point w.r.t. a reference system. Maybe caused by translation and or rotation of object (rigid body). Change in shape or size related to displacements are called deformations. Change in linear dimension causes deformation

55

Deformation

Includes changes in both lengths and angles.

66

Strain

A quantity used to measure the intensity of deformation. Stress is used to measure the intensity of internal force.

Normal strain, , used to measure change in size.Shear strain, , used to measure change in shape.

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Axial Strain at a Point

n

L 0

dL dL

dL

limL

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Axial Strain at a Point

dL dL

dL

If the bar stretches (dL’>dL), the strain is positive and called a tensile strain.

If the bar contracts (dL’<dL), the strain is negative and called a compressive strain.

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Normal Strain/ Axial Strain at a Point

L L

L

dL dLdL dL dL

dL

1 dL dL

L 1 dL L dL

L L dL

L L L

L L

L L

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Normal StrainNormal Strain: is the deformation of the Member per unit length.

01

0

strain normal

LLwith

L

(Dimensionless)

Uniform cross sectionL0

P

L1

P

Change in LengthOriginal LengthNormal strain is essentially:

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2.1 Normal Strain

avg

L L L

L L

If the bar stretches (L’>L), the strain is positive and called a tensile strain.

If the bar contracts (L’<L), the strain is negative and called a compressive strain.

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2.1 Normal Strain Examples

L

A

P

A

P

2

2

L

L0=0.5 m

P

mm2.0

43

0

10*45.0

10*2.0

L

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2.1 Normal Strain: Examples

a. Determine the expression for the average extensional strain in rod BC as a function of for

b. Determine the

approximation for that gives acceptable accuracy for when <<1 rad

When the “rigid” beam AB is horizontal, the rod BC is strain free.

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Deformation Diagram

2 2* *

*

*

L LL BC 5a

L

L 3a c b

b 4acos and

c 4asin

θ

*b

*c

a4

1515

2 2

2 2

2 2

L L

L

3a 4asin 4acos 5a

5a

3 4sin 4cos 5

5

9 24sin 16sin 16cos 5

5

25 24sin 5 241 sin 1

5 25


1616

Small angle approximation

By Binomial Theorem

241 sin 1

25

1

: sin

241 1

25

: 1 12

12

25


The strain is dimensionless, as it should be. At = /2, ( /2) = 2/5. At this point L* = 3a + 4a = 7a so this value is correct.

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Mechanical Properties of Materials

Properties are determined by mechanical tests (Tension and Compression.)

A typical test apparatus is shown on the right.

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2.1 Stress Strain Diagram

A variety of testing machine types, and sizes…

…and a variety of samples sizes.

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Gage Length

Original gage length is L0. This is not the total length of the specimen.

Deformed Specimen

Original gage length is deformed to L*. The load and the elongation are carefully measured. The load is slowly applied. This is a static tension test.

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A plot of stress versus strain is called a stress strain diagram. From this diagram we can find a number of important mechanical properties.

nP

A L

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2.1 Stress Strain Diagram(Steel)

Recoverable(يسترد) deformation Permanent

deformation(دائم )

Important Regions:

• Elastic region(متمّغ-ط )• Yielding(مرن )• Strain Hardening

( تصلب)• Necking(َن8اق ( ِخ:

(pg 86-88 Hibbeler gives detailed description)

Note: Very little difference between engineering and true values in elastic region.

• Fracture(كسر )

2222

2.1 Stress Strain Diagram(Steel)

In the figure above the region from A to B has a linear relationship between stress and strain. The stress at point B is called the proportional limit, PL. The ratio of stress to strain in the linear region is called E, the Young’s modulus or the modulus of elasticity.

PLE σσε

σ

2323

Yielding

At point B, the specimen begins yielding. Smaller load increments are required to to produce a given increment of elongation. The stress at C is called the upper yield point, (YP)u The stress at D is called the lower yield point, (YP)l The upper yield point is seldom used and the lower yield point is often referred to simply as the yield point, YP

2424

Perfectly Plastic Zone

From D to E the specimen continues to elongate without any increase in stress. The region DE is referred to as the perfectly plastic zone.

2525

Strain Hardening

The stress begins to increase at E. The region from E to F is known as the strain hardening zone. The stress at F is the ultimate stress.

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Necking

At F the stress begins to drop as the specimen begins to “neck down.” This behavior continues until fracture occurs at point G, at the fracture stress, F

2727

Necking

Fracture

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True Stress

min

σA

Ptrue

:area originalthe than rather

area minimum currentthe Use

2929

True Strain

0

0

min

ln 1

ln

true

L

true

L

true

L

L

dL

L

or

A

A

Using all of the successiveمتعاقب values of L that they have recorded. Dividing the increment dL of the distance between the gage marks, by the corresponding value of L. (sum of the incremental elongations divided by the current gauge length)

3030

Design Properties

1. Strength

2. Stiffness

3. Ductility

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Strength

Ultimate Strength: Highest value of stress (maximum value of engineering stress) that the material can withstand.

Fracture Stress: The value of stress at fracture.

Uσ

Yield Strength: Highest stress that the material canwithstand يقاوم without undergoing significant

yielding andpermanent deformation.

orY Y P Y Y S

Fσ

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Stiffness

The ratio of stress to strain (or load to displacement.) Generally of interest in the linear elastic range. The Young’s modulus or modulus of elasticity, E, is used to represent a material’s stiffness.

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Ductilityتمدد

1. Materials that can undergo a large strain before fracture are classified as ductile materials.

2. Materials that fail at small values of strain are classified as brittle materials.

3. Really referring to modes of fracture.

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Ductility Measures

% Elongation

% Reduction in Area

exp

:

F 0

0

L -LPercent Elongation= ×100%

L

The final elongation ressed as a

percentage of the original gage length

100%0 F

0

A -APercent R eduction in A rea

A

3535

Ductile Materials

1. Steel

2. Brass

3. Aluminum

4. Copper

5. Nickel

6. Nylon

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Ductile Materials(نAَل8د )

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2.1 Stress Strain DiagramBrittle Materials(هش )

Typical stress-strain diagram for a brittle material showing the proportional limit (point A) and fracture stress (point B)No yielding, or necking is evident. For brittle materials that fail the pieces still fit together e.g. glass or ceramics.

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2.1 Stress Strain Diagram Elastic versus Plastic Behavior

• If the strain disappears when the stress is removed, the material is said to behave elastically.

• When the strain does not return to zero after the stress is removed, the material is said to behave plastically.

• The largest stress for which this occurs is called the elastic limit.

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Plastic Behavior

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After reloading of a piece the elastic and proportional limit can be increased.

Mechanical properties depend on the history of the piece.

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2.2 Hooke’s Low: Modulus of elasticity• Below the yield stress

E

• Strength is affected by(رN Pِّث Pَأ (ُمSَت

alloying(يطWِلPَخ ), heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.

0)(tan

**

)()(

*)(

0

0

00

00

bbeiEa

baandba

and

ba

e

e

ee

ee

Elasticity of Modulus

or Modulus YoungsE

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2.8 Deformations Under Axial Loading

AE

P

EE

• From Hooke’s Law:

• From the definition of strain:

L

• Equating and solving for the deformation,

AE

PL

• With variations in loading, cross-section or material properties,

i ii

iiEA

LP

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2.8 Deformation under Axial LoadingExample

Determine the deformation of the steel rod shown under the given loads.

in. 618.0 in. 07.1

psi1029 6

dD

E SOLUTION:

• Divide the rod into components at the load application points.

• Apply a free-body analysis on each component to determine the internal force

• Evaluate the total of the component deflections. in. 109.75 3

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2.8 Deformation under Axial LoadingExampleSOLUTION:

• Divide the rod into three components:

221

21

in 9.0

in. 12

AA

LL

23

3

in 3.0

in. 16

A

L

• Apply free-body analysis to each component to determine internal forces,

lb1030

lb1015

lb1060

33

32

31

P

P

P

• Evaluate total deflection,

in.109.75

3.0

161030

9.0

121015

9.0

121060

1029

1

1

3

333

6

3

33

2

22

1

11

A

LP

A

LP

A

LP

EEA

LP

i ii

ii

in. 109.75 3

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2.9 Static Indeterminacy

• Structures for which internal forces and reactions cannot be determined from statics alone are said to be statically indeterminate.

0 RL

• Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.

• Redundant reactions are replaced with unknown loads which along with the other loads must produce compatible deformations.

• A structure will be statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium.

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SOLUTION:

• Solve for the displacement at B due to the applied loads with the redundant constraint released,

EEA

LP

LLLL

AAAA

PPPP

i ii

ii9

L

4321

2643

2621

34

3321

10125.1

m 150.0

m10250m10400

N10900N106000

• Solve for the displacement at B due to the redundant constraint,

i

B

ii

iiR

B

E

R

EA

LPδ

LL

AA

RPP

3

21

262

261

21

1095.1

m 300.0

m10250m10400

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• Require that the displacements due to the loads and due to the redundant reaction be compatible,

• Find the reaction at A due to the loads and the reaction at B

kN577

kN323

BRAR

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2.10 Thermal Stresses

A

L

B • A temperature change results in a change in length or thermal strain. There is no stress associated with the thermal strain unless the elongation is restrained by the supports.

T

A B

A BP

P

coef.expansion thermal

AE

PLLT PT

• Treat the additional support as redundant and apply the principle of superposition.

0

0

AE

PLLT

PT

• The thermal deformation and the deformation from the redundant support must be compatible.

TE

A

PTAEP

PT

0

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2.10 Poisson’s Ratio

• For a slender bar subjected to axial loading:

0

0

0

ililiL

iwith

zyEx

x

• The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic and homogeneous (no direction and position independence),

0 zy • Poisson’s ratio is defined as

x

z

x

y

strain axial

strain lateral

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2.10 Poisson’s Ratio

“Life is good for only two things, discovering mathematics and teaching mathematics.”

Siméon Poisson

transverse longitudinal

(Greek letter nu) is called the Poisson’s ratio. Typical values

are in the 0.2 – 0.35 range.

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2.11 Generalized Hooke’s Law

• For an element subjected to multi-axial loading, the normal strain components resulting from the stress components may be determined from the principle of superposition. This requires:

1) strain is linearly related to stress2) deformations are small

EEE

EEE

EEE

zyxz

zyxy

zyxx

• With these restrictions:

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A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi.

For E = 10x106 psi and = 1/3, determine the change in:

a) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.

2.11 Generalized Hooke’s Law

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2.11 Relation Among E, and GSOLUTION:

• Apply the generalized Hooke’s Law to find the three components of normal strain.

in./in.10600.1

in./in.10067.1

in./in.10533.0

ksi203

10ksi12

psi1010

1

3

3

3

6

EEE

EEE

EEE

zyxz

zyxy

zyxx

• Evaluate the deformation components.

in.9in./in.10533.0 3 dxAB

in.9in./in.10600.1 3 dzDC

in.75.0in./in.10067.1 3 tyt

in.108.4 3AB

in.104.14 3DC

in.10800.0 3t

• Find the change in volume

33

333

in75.0151510067.1

/inin10067.1

eVV

e zyx

3in187.0V

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2.11 Dilatation(ت:َط8اَل8ة Aا:ْس ): Bulk(حجم ) Modulus

• Relative to the unstressed state, the change in volume is

1 1 1 1 1 1

1 2

dilatation (change in volume per unit volume)

x y z x y z

x y z

x y z

e

E

• For element subjected to uniform hydrostatic pressure,

modulusbulk 213

213

Ek

k

p

Epe

• Subjected to uniform pressure, dilatation must be negative, therefore

210

5858

Shear Strain

• A cubic element subjected to a shear stress will deform into a

rhomboid( المعين The corresponding shear strain is .( شبيهquantified in terms of the change in angle between the sides,

xyxy f

5959

Shear Strain

*θ2

πγ

* *

sin tan

tan2 2

s

s

Small angle approximation

L

6060

Hooke’s Law for Shear

G is the shear modulus or

the shear modulus of elasticity

or the modulus of rigidity.

xy xy

yz yz

zx zx

G

G

G

G

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2.11 Shearing Strain

A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 0.04 in. under the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

SOLUTION:

• Determine the average angular deformation or shearing strain of the block.

• Use the definition of shearing stress to find the force P.

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.

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2.11 Shearing Strain

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.

psi1800rad020.0psi1090 3 xyxy G

• Use the definition of shearing stress to find the force P.

lb1036in.5.2in.8psi1800 3 AP xy

kips0.36P

• Determine the average angular deformation or shearing strain of the block.

rad020.0in.2

in.04.0tan xyxyxy

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2.11 Relation Among E, and G

• An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions.

12G

E

• Components of normal and shear strain are related,

• If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain.

• An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.

6464

Generalized Hooke’s Law

x x y z

y y x z

z z x y

xy xy yz yz xz xz

1

E1

E1

E1 1 1

G G G

6565


x x y z

y y x z

z z x y

xy xy yz yz xz xz

E1

1 1 2

E1

1 1 2

E1

1 1 2

G G G

6666

Plane Stress

A body that is in a two-dimensional state of stress with z = xz = yz = 0 is said to be in a state of plane stress.

6767


x x y z y y x z

z z x y

xy xy yz yz xz xz

x x y

z yz xz y y x z x y

xy xy

1 1

E E1

E1 1 1

G G G1

E1

0E E1

G

6868

Hooke’s Law for Plane Strain

z yz xz 0

x x y z

y y x z

z x y

z x y

xy xy

1

E1

E1

0E

1

G

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2.12 Composite Materials

• Fiber-reinforced composite materials are formed

from lamina( �ْع�ِد�َن� الَم َق� �� of ( َخPيbط)of fibers ( َر�َّق

graphite, glass, or polymers embedded(ُمحشو ) in a resin matrix.

z

zz

y

yy

x

xx EEE

• Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity,

x

zxz

x

yxy

• Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g.,

• Materials with directionally dependent mechanical

properties are anisotropic( الخواص .( متباين

7070

z

z

y

yyz

x

xxzz

z

zzy

y

y

x

xxyy

z

z

y

yyx

x

xx

EEE

EEE

EEEzx


7171


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2.12 Stress Concentration: Hole

ave

max

K

Discontinuities of cross section may result in high localized or concentrated stresses.

73


Discontinuities of cross section may result in high localized or concentrated stresses.

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Example: Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8 mm. Assume an allowable normal stress of 165 MPa.

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• Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.

MPa7.9082.1

MPa165maxave

K

• Apply the definition of normal stress to find the allowable load.

N103.36

MPa7.90mm10mm40

3

aveAP

kN3.36P

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b.

82.1

20.0mm40

mm850.1

mm40

mm60

K

d

r

d

D

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