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# Mechanics of Material

## 1.

Mechanics of MaterialChapter II

Stress and Strain – Axial Loading

1

## 2.

Stress and StrainContents

Stress & Strain: Axial Loading

Normal Strain

Stress-Strain Test

Stress-Strain Diagram: Ductile Materials

Stress-Strain Diagram: Brittle Materials

Hooke’s Law: Modulus of Elasticity

Elastic vs. Plastic Behavior

Fatigue

Deformations Under Axial Loading

Example 2.01

Sample Problem 2.1

Static Indeterminacy

Example 2.04

Thermal Stresses

Poisson’s Ratio

2

Generalized Hooke’s Law

Dilatation: Bulk Modulus

Shearing Strain

Example 2.10

Relation Among E, n, and G

Sample Problem 2.5

Composite Materials

Saint-Venant’s Principle

Stress Concentration: Hole

Stress Concentration: Fillet

Example 2.12

Elastoplastic Materials

Plastic Deformations

Residual Stresses

Example 2.14, 2.15, 2.16

## 3.

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.

3

## 4.

DisplacementMovement 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 d

44

## 5.

DeformationIncludes changes in

both lengths and

angles.

55

## 6.

StrainA quantity used to measure the intensity of deformation.

Stress is used to measure the intensity of internal force.

Normal strain, e, used to measure change in size.

Shear strain, g, used to measure change in shape.

66

## 7.

Axial Strain at a Point77

dL dL

e

dL

d n

lim

L 0

L

## 8.

Axial Strain at a PointdL dL

e

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.

8

## 9.

Normal Strain/ Axial Strain at a PointdL dL

e

edL dL dL

dL

1 e dL dL

L

1 e dL L e dL

L

L

d L L e dL

L

d L L e L

L L d

e

L

L

9

## 10.

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

P

L0

d

Uniform

cross section

P

L1

Normal strain is essentially:

e

10

d

L0

with

normal strain

d L1 L0

Change in Length

Original Length

(Dimensionless)

## 11.

2.1 Normal Straineavg

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.

11

## 12.

2.1 Normal StrainExamples

L

d

2P P

2A A

e

P

d

L

d 0.2 mm

L0=0.5 m

12

d 0.2 *10 3

4

e

4 *10

L0

0.5

## 13.

2.1 Normal Strain: Examplesa. Determine the

expression for the

average

extensional strain

in rod BC as a

function of q for

0 q p/2

b. Determine the

approximation for

e q that gives

acceptable

accuracy for e

when q<<1 rad

13

When the “rigid” beam

AB is horizontal, the rod

BC is strain free.

## 14.

2.1 Normal Strain: ExamplesDeformation Diagram

L L

e

L BC 5a

L

L

3a c b

*

2

*

2

b* 4a cos q and

c* 4a sin q

4a

c*

θ

b*

14

## 15.

2.1 Normal Strain: ExamplesL L

e

L

3a 4a sin q 4a cos q 5a

2

e

e q

15

15

2

5a

3 4 sin q

2

4 cos q 5

2

5

e q

9 24 sin q 16 sin 2 q 16 cos 2 q 5

5

e q

25 24 sin q 5

24

1

sin q 1

5

25

## 16.

2.1 Normal Strain: Examplese q

24

1

sin q 1

25

q 1

Small angle approximation : sin q q

e q

24

1

q 1

25

By Binomial Theorem : 1 1

2

12

e q

q

25

16

16

The strain is dimensionless, as it should

be. At q = p/2, e(p /2) = 2/5. At this

point L* = 3a + 4a = 7a so this value is

correct.

## 17.

Mechanical Properties of MaterialsProperties are

determined by

mechanical tests

(Tension and

Compression.)

A typical test apparatus

is shown on the right.

17

## 18.

2.1 Stress Strain DiagramA variety of

testing

machine

types, and

sizes…

…and a variety of samples sizes.

18

## 19.

Gage LengthOriginal gage length is L0. This is not the

total length of the specimen.

Deformed Specimen

19

19

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.

## 20.

2.1 Stress Strain DiagramA plot of stress versus strain is called a stress strain

diagram. From this diagram we can find a number

of important mechanical properties.

P

A

dn

e

L

20

e

## 21.

2.1 Stress Strain Diagram(Steel)Important Regions:

(pg 86-88 Hibbeler gives

detailed description)

• Elastic region() متمغّط

• Yielding() مرن

• Strain Hardening

() تصلب

• Necking() ِخنَاق

• Fracture() كسر

21

Recoverable()يسترد

deformation

Note: Very little difference

between engineering and

true values in elastic region

Permanent deformation() دائم

## 22.

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.

σ

E

ε

22

22

σ σ PL

## 23.

YieldingAt 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

23

23

## 24.

Perfectly Plastic ZoneFrom D to E the

specimen continues

to elongate without

any increase in

stress. The region

DE is referred to as

the perfectly plastic

zone.

24

24

## 25.

Strain HardeningThe 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.

25

25

## 26.

NeckingAt 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

26

26

## 27.

NeckingFracture

27

27

## 28.

True StressUse the current minimum area

rather than the original area :

σ true

28

28

P

A min

## 29.

True StrainUsing all of the successive متعاقبvalues 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)

L

e true e

L

L

dL

e true

ln 1 e

L

L0

or

29

29

e true

A0

ln

A min

## 30.

Design Properties1.

2.

3.

30

30

Strength

Stiffness

Ductility

## 31.

StrengthYield Strength: Highest stress that the material can

withstand يقاومwithout undergoing significant yielding and

permanent deformation.

Y Y P or Y Y S

Ultimate Strength: Highest value of stress (maximum value

of engineering stress) that the material can withstand.

σU

Fracture Stress: The value of stress at fracture.

σF

31

31

## 32.

StiffnessThe 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.

32

32

## 33.

Ductility تمدد33

33

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.

## 34.

Ductility Measures% Elongation

The final elongation exp ressed as a

percentage of the original gage length :

LF -L0

Percent Elongation=

L0

% Reduction in Area

×100%

A0 -AF

Percent Reduction in Area

A0

34

34

100%

## 35.

Ductile Materials1.

2.

3.

4.

5.

6.

35

35

Steel

Brass

Aluminum

Copper

Nickel

Nylon

## 36.

2.1 Stress Strain DiagramDuctile Materials() لَدْن

36

## 37.

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.

37

## 38.

2.1 Stress Strain DiagramElastic versus Plastic Behavior

• If the strain disappears when the

stress is removed, the material is

said to behave elastically.

• The largest stress for which this

occurs is called the elastic limit.

• When the strain does not return

to zero after the stress is

removed, the material is said to

behave plastically.

38

## 39.

Plastic Behavior39

39

## 40.

Mechanical propertiesdepend on the history of

the piece.

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

40

40

## 41.

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

Ee

E Youngs Modulus or

Modulus of Elasticity

• Strength is affected by() ُمتَأَثِّر

َ ), heat treating, and

alloying(خ ِّليط

manufacturing process but stiffness

(Modulus of Elasticity) is not.

(e ) a * e b

(e )e e 0 and (e )e e e

0

41

e

0 a * e 0 b and e a * e e b

e 0

a

tan ( ) E bei b 0

ee e0

## 42.

2.8 Deformations Under Axial Loading• From Hooke’s Law:

Ee

P

e

E AE

• From the definition of strain:

e

d

L

• Equating and solving for the deformation,

PL

d

AE

• With variations in loading, cross-section or

material properties,

Pi Li

d

i Ai Ei

42

## 43.

2.8 Deformation under Axial LoadingExample

E 29 10 6 psi

D 1.07 in. d 0.618 in.

Determine the deformation of

the steel rod shown under the

given loads.

• Apply a free-body analysis on each

component to determine the

internal force

d 75.9 10 in.

• Evaluate the total of the component

deflections.

3

43

SOLUTION:

• Divide the rod into components at

the load application points.

## 44.

2.8 Deformation under Axial LoadingExample

SOLUTION:

• Divide the rod into three

components:

• Apply free-body analysis to each

component to determine internal forces,

P1 60 103 lb

P2 15 103 lb

P3 30 103 lb

• Evaluate total deflection,

Pi Li 1 P1L1 P2 L2 P3 L3

E A1

A2

A3

i Ai Ei

d

60 103 12 15 103 12 30 103 16

6

0

.

9

0

.

9

0

.

3

29 10

1

75.9 10 3 in.

44

L1 L2 12 in.

L3 16 in.

A1 A2 0.9 in 2

A3 0.3 in 2

d 75.9 10 3 in.

## 45.

4545

## 46.

2.9 Static Indeterminacy• Structures for which internal forces and reactions

cannot be determined from statics alone are said

to be statically indeterminate.

• A structure will be statically indeterminate

whenever it is held by more supports than are

required to maintain its equilibrium.

• Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

• Deformations due to actual loads and redundant

reactions are determined separately and then

added or superposed.

d dL dR 0

46

## 47.

2.9 Static IndeterminacySOLUTION:

• Solve for the displacement at B due to the applied

loads with the redundant constraint released,

P1 0 P2 P3 600 103 N

A1 A2 400 10 6 m 2

P4 900 103 N

A3 A4 250 10 6 m 2

L1 L2 L3 L4 0.150 m

Pi Li 1.125 109

dL

A

E

E

i i i

• Solve for the displacement at B due to the redundant

constraint,

P1 P2 RB

A1 400 10 6 m 2

L1 L2 0.300 m

47

A2 250 10 6 m 2

Pi Li

1.95 103 RB

δR

E

i Ai Ei

## 48.

2.9 Static Indeterminacy• 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

R A 323kN

RB 577 kN

48

## 49.

2.10 Thermal StressesA

B

L

A

B

dT

dP

A

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.

• Treat the additional support as redundant and apply

the principle of superposition.

PL

d T T L

dP

AE

thermal expansion coef.

• The thermal deformation and the deformation from

the redundant support must be compatible.

P

d dT d P 0

T L

49

PL

0

AE

d dT d P 0

P AE T

P

E T

A

## 50.

2.10 Poisson’s Ratio• For a slender bar subjected to axial loading:

ex

with

x

E

y z 0

Li li 0

ei

li 0

• 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),

e y ez 0

• Poisson’s ratio is defined as

ey

e

lateral strain

n

z

axial strain

ex

ex

50

## 51.

2.10 Poisson’s RatioSiméon Poisson

“Life is good for only two

things, discovering

mathematics and teaching

mathematics.”

etransverse n e longitudinal

n (Greek letter nu) is called

51

the Poisson’s ratio. Typical values are

in the 0.2 – 0.35 range.

## 52.

5252

## 53.

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 stress

2) deformations are small

• With these restrictions:

x

ex

E

ey

53

ez

n x

E

n x

E

n y

E

n z

n z

y

E

n y

E

E

E

z

E

## 54.

5454

## 55.

2.11 Generalized Hooke’s LawA 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 n = 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.

55

## 56.

2.11 Relation Among E, n, and GSOLUTION:

• Apply the generalized Hooke’s Law to • Evaluate the deformation components.

find the three components of normal

d B A e x d 0.533 10 3 in./in. 9 in.

strain.

ex

x n y n z

E

E

E

1

12

ksi

0

20

ksi

3

10 106 psi

1

0.533 10 3 in./in.

ey

n x y n z

E

E

E

1.067 10 3 in./in.

ez

n x n y

E

z

E

E

1.600 10 3 in./in.

56

dC

D

dB

A

4.8 10 3 in.

e z d 1.600 10 3 in./in. 9 in.

dC

D

14.4 10 3 in.

d t e y t 1.067 10 3 in./in. 0.75 in.

d t 0.800 10 3 in.

• Find the change in volume

e e x e y e z 1.067 10 3 in 3/in 3

V eV 1.067 10 3 15 15 0.75 in 3

V 0.187 in 3

## 57.

َ ِست2.11 Dilatation(طالَة

ْ ِ) ا: Bulk( ) حجمModulus

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

e 1 e x 1 e y 1 e z 1 1 e x e y e z 1

ex e y ez

1 2n

x y z

E

dilatation (change in volume per unit volume)

• For element subjected to uniform hydrostatic pressure,

3 1 2n

p

e p

k

E

k

E

bulk modulus

3 1 2n

• Subjected to uniform pressure, dilatation must be

negative, therefore

0 n 12

57

## 58.

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

rhomboid() شبيه المعين. The corresponding shear strain is quantified in

terms of the change in angle between the sides,

58

58

xy f g xy

## 59.

Shear Strainπ

*

γ θ

2

Small angle approximation

g sin g tan g

ds

p

*

g q tan q

2

2

Ls

p

59

59

*

## 60.

Hooke’s Law for ShearGg

G is the shear modulus or

the shear modulus of elasticity

or the modulus of rigidity.

xy G g xy

yz G g yz

zx G g zx

60

60

## 61.

2.11 Shearing StrainSOLUTION:

• Determine the average angular

deformation or shearing strain of

the block.

• Apply Hooke’s law for shearing stress

and strain to find the corresponding

shearing stress.

A rectangular block of material with

• Use the definition of shearing stress to

modulus of rigidity G = 90 ksi is

find the force P.

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.

61

## 62.

2.11 Shearing Strain• Determine the average angular deformation

or shearing strain of the block.

g xy tan g xy

0.04 in.

2 in.

g xy 0.020 rad

• Apply Hooke’s law for shearing stress and

strain to find the corresponding shearing

stress.

xy Gg xy 90 103 psi 0.020 rad 1800 psi

• Use the definition of shearing stress to find

the force P.

P xy A 1800 psi 8 in. 2.5 in. 36 103 lb

P 36.0 kips

62

## 63.

2.11 Relation Among E, n, and G• An axially loaded slender bar will

elongate in the axial direction and

contract in the transverse directions.

• An initially cubic element oriented as in

top figure will deform into a rectangular

parallelepiped. The axial load produces a

normal strain.

• 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.

• Components of normal and shear strain are

related,

E

1 n

2G

63

## 64.

Generalized Hooke’s Lawg xy

64

64

1

e x x n y z

E

1

e y y n x z

E

1

e z z n x y

E

1

1

1

xy

g yz yz

g xz xz

G

G

G

## 65.

Generalized Hooke’s LawE

1 n e x n e y e z

x

1 n 1 2n

E

1 n e y n e x e z

y

1 n 1 2n

E

1 n e z n e x e y

z

1 n 1 2n

xy Gg xy

65

65

yz Gg yz

xz Gg xz

## 66.

Plane StressA 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.

66

66

## 67.

Generalized Hooke’s Law1

1

e x x n y z e y y n x z

E

E

1

e z z n x y

E

1

1

1

g xy xy

g yz yz

g xz xz

G

G

G

1

e x E x n y

1

n

z yz xz 0 e y

y n x e z

x y

E

E

1

g xy G xy

67

67

## 68.

Hooke’s Law for Plane Straine z g yz g xz 0

1

e x x n y z

E

1

e y y n x z

E

1

0 z n x y

E

z n x y

g xy

68

68

1

xy

G

## 69.

2.12 Composite Materials• Fiber-reinforced composite materials are formed

َ ) of

from lamina( َ ) َر َّققَ ال َمعْدِّنof fibers(خيْط

graphite, glass, or polymers embedded( ) محشوin a

resin matrix.

• Normal stresses and strains are related by Hooke’s

Law but with directionally dependent moduli of

elasticity,

y

x

Ex

Ey

Ez z

ex

ey

ez

• Transverse contractions are related by directionally

dependent values of Poisson’s ratio, e.g.,

n xy

ey

e

n xz z

ex

ex

• Materials with directionally dependent mechanical

properties are anisotropic() متباين الخواص.

69

## 70.

2.12 Composite Materialsx n yx y n z

ex

zx

Ex

ey

ez

70

70

n xy x

Ex

Ey

Ez

y n zy z

Ey

n xz x n yz y

Ex

Ey

Ez

z

Ez

## 71.

2.12 Composite Materials71

71

## 72.

2.12 Stress Concentration: HoleDiscontinuities of cross section may result in high localized or concentrated

stresses.

max

K

ave

72

## 73.

2.12 Stress Concentration: HoleDiscontinuities of cross section may result in high localized or concentrated

stresses.

73

## 74.

2.12 Stress Concentration: HoleExample: 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.

74

## 75.

2.12 Stress Concentration: Hole• Determine the geometric ratios and

find the stress concentration factor

from Fig. 2.64b.

D 60 mm

1.50

d 40 mm

r

8 mm

0.20

d 40 mm

K 1.82

• Find the allowable average normal

stress using the material allowable

normal stress and the stress

concentration factor.

ave

max

K

165 MPa

90.7 MPa

1.82

• Apply the definition of normal stres

to find the allowable load.

P A ave 40 mm 10 mm 90.7 MPa

36.3 103 N

P 36.3 kN

75

## 76.

2.12 Stress Concentration: Hole76