stress and strain renew

CHAPTER 1

STRESS AND STRAIN

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1. Today’s Objectives:

Students will be able to:

explain some of the important principles of statics.

use the principles to determine internal resultant loadings in a body.

3. explain the concepts of normal, shear, bearing and thermal stress.

Topics:

• Introduction

• Main Principles of Statics

Stress

• Normal Stress

• Shear Stress

• Bearing Stress

• Thermal Stre

4

Overview of Mechanics

1.1 Introduction

Mechanics : The study of how bodies react to forces acting on them

RIGID BODIES

(Things that do not change shape)

Statics : The study of bodies

in an equilibrium

DEFORMABLE BODIES

(Things that do change shape)FLUIDS

Mechanics of Materials :

The study of the relationships

between the external loads

applied to a deformable body and

the intensity of internal forces

acting within the body.

Incompressible Compressible

Dynamics :

1. Kinematics – concerned

with the geometric aspects

of the motion

2. Kinetics – concerned

with the forces causing the

motion.

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External Loads

1.2 Main Principles of Statics

External Loads

Surface Forces

- caused by direct contact of one body with

the surface of another.

Body Force

- developed when one body exerts a force on

another body without direct physical contact

between the bodies.

- e.g earth’s gravitation (weight)

concentrated force

linear distributed load, w(s)

STRESS AND STRAIN

• Axial Load• Normal Stress• Shear Stress• Bearing Stress• Allowable Stress• Deformation of Structural under Axial

Load• Statically indeterminate problem• Thermal Stress

Stress And Strain• Mechanics of material is a study of the

relationship between the external loads applied to a deformable body and the intensity of internal forces acting within the body.

• Stress = the intensity of the internal force on a specific plane (area) passing through a point.

• Strain = describe the deformation by changes in length of line segments and the changes in the angles between them

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Type of Stress

1.1 Introduction

• Normal Stress : stress which acts perpendicular, or normal to, the

(σ) cross section of the load-carrying member.

: can be either compressive or tensile.

• Shear Stress : stress which acts tangent to the cross section of

(τ) the load-carrying member.

: refers to a cutting-like action.

Normal Stress and Normal Strain

• Normal Stress, the intensity of force, or force per unit area, acting normal to A

A positive sign will be used to indicate a tensile stress (member in tension)

A negative sign will be used to indicate a compressive stress (member in compression)

= P / A

(a)

(b)

Stress ( ) = Force (P)

Cross Section (A)

•Unit: Nm -² •N/mm2 or MPa

N/m2 or Pa

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Examples of Axially

Loaded Bar

1.4 Axial Loading – Normal Stress

• Usually long and slender structural members

• Typical examples : truss members, hangers, bolts

Assumptions :

1. Uniform deformation: Bar

remains straight before and

after load is applied, and

cross section remains flat or

plane during deformation

2. In order for uniform

deformation, force P be

applied along centroidal axis

of cross section C

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Average Normal Stress

Distribution

A

P

AP

AFFFA

zRz

dd;

σ = average normal stress at any point

on cross sectional area

P = internal resultant normal force

A = x-sectional area of the bar

1.4 Axial Loading – Normal Stress

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Procedure of Analysis• Use equation of σ = P/A for x-sectional area of a member when

section subjected to internal resultant force P

Internal Loading

• Section member perpendicular to its longitudinal axis at pt

where normal stress is to be determined

• Draw free-body diagram

• Use equation of force equilibrium to obtain internal axial

force P at the section

• Determine member’s x-sectional area at the section

• Compute average normal stress σ = P/A

Average Normal Stress

1.4 Axial Loading – Normal Stress

Example 1.1:

Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1=30mm and d2=20mm, find average normal stress at the midsection of (a) rod AB, (b) rod BC.

Example 1.2Two solid cylindrical roads AB and BC are welded

together at B and loaded as shown. Knowing that d1 = 30 mm and d2 = 50 mm, find the average normal stress in the mid section of (a) rod AB, (b) rod BC.

• Normal strain, is the elongation or contraction of a line segment per unit of length

L = elongation

Lo = length

= L / Lo

strain normalL

* L=

Example 1.3:Determine the corresponding strain for a bar of length L=0.600m and uniform cross section which undergoes a deformation =15010-6m.

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6

150 10 m250 10 m m

L 0 600m

250 10 250

/.

@

Stress and Strain Example1.4 A cable and strut assembly ABC supports a vertical load

P=12kN. The cable has an effective cross sectional area of 160mm², and the strut has an area of 340mm².

(a)Calculate the normal stresses in the cable and strut.

(b)If the cable elongates 1.1mm, what is the strain?

(c)If the strut shortens 0.37mm, what is the strain?

Answer• a)62.5Mpa

29.41 Mpa

b) 4.4x10-4 m

1.48x10-4 m

1.5 The bar shown has a square cross section (20mm x 40mm) and length, L=2.8m. If an axial force of 70kN is applied along the centroidal axis of the bar cross sectional area, determine the stress and strain if the bar end up with 4m length.

70kN 70kN

2.8m

The Stress-Strain Diagram

• Tensile test is a experiment to determine the load-deformation behavior of the material.

• Data from tensile test can be plot into stress and strain diagram.

• Example of test specimen

- note the dog-bone geometry

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• Universal Testing Machine - equipment used to subject a specimen to tension, compression, bending, etc. loads and measure its response

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Stress-Strain Diagrams

A number of important mechanical

properties of materials that be deduced

from stress-strain diagram are illustrated

in figure above.

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• Point O-A = linear relationship between stress

and strain

• Point A = proportional limit (PL)

The ratio of stress to strain in this linear region

of stress-strain diagram is called Young Modulus

or the Modulus of Elasticity given

At point A-B, specimen begins yielding.

• Point B = yield point

• Point B-C = specimen continues to elongate without any increase in stress. Its refer as perfectly plastic zone

• Point C = stress begins to increase

• Point C-D = refer as the zone of strain hardening

• Point D = ultimate stress/strength ; specimen

begins to neck-down

• Point E = fracture stress

< PL

Unit: MPa

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Point O to A

Point C to D

Point D to E

At point E

Normal or engineering stress can be determine

by dividing the applied load by the specimen

original cross sectional area.

True stress is calculated using the actual cross

sectional area at the instant the load is

measured.31

Some of the materials like aluminum (ductile), does not have clearly yield point such asstructural steel. Therefore, stress valuecalled the offset yield stress, YL is usedin line of a yield point stress.

As illustrated, the offset yield stress isdetermine by;• Drawing a straight line that best fits the data in initial (linear)

portion of the stress-strain diagram• Second line is then drawn parallel to the original line but offset

by specified amount of strain• The intersection of this second line with

the stress-strain curve determine theoffset yield stress.

• Commonly used offset value is 0.002/0.2% 32

Brittle material such as ceramic and glass

have low tensile stress value but high in

compressive stress. Stress-strain diagram for

brittle material.

Example 1.6

The 4 mm diameter cable BC is made of a steel

with E=200GPa. Knowing that the maximum

stress in the cable must not exceed 190MPa

and that the elongation of the cable must not

exceed 6mm, find the maximum load P that can

be applied as shown

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Elasticity and Plasticity• Elasticity refers to the property of a material such that

it returns to its original dimensions after unloading

• Any material which deforms when subjected to load and returns to its original dimensions when unloaded is said to

be elastic.

• If also the stress is proportional to the strain, the material is said to be linear elastic, otherwise it is non-

linear elastic.

• Beyond the elastic limit, some residual strain or permanent strains will remain in the material upon unloading

• The residual elongation corresponding to the permanent strain is called the permanent set

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• The amount of strain which is recovered upon unloading is called the elastic recovery.

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Poisson's Ratio, • When an elastic, homogenous and isotropic material is

subjected to uniform tension, it stretches axially but contracts laterally along its entire length.

• Similarly, if the material is subjected to axial compression, it shortens axially but bulges out laterally (sideways).

• The ratio of lateral strain to axial strain is a constant known as the Poisson's ratio,

where the strains are caused by uniaxial stress only

axial

lateral

v

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Example 2.8

A prismatic bar of circular cross-section

is loaded by tensile forces P = 85 kN. The

bar has length of 3 m and diameter of 30

mm. It is made from aluminum with modulus

of elasticity of 70 GPa and poisson's ratio

= 1/3. Calculate the elongation, l,and the

decrease in diameter d.

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Example 2.9

A 10 cm diameter steel rod is loaded with 862 kN axial force. Knowing that the E=207 GPa and = 0.29, determine the deformation of rod diameter after being loaded.

Solution

in rod, =

< E axial strain

Lateral strain,

38

MPa

m

Nx

A

p7.109

)1.0(4

1

10862

22

3

00053.010207

7.1093

MPax

MPa

Ea

)00053.0(29.)( oal 000154.0

)1.0)(000154.0()( Dd l

cm00154.0

Latihan Tegasan dan Terikan

1. Satu rasuk tegar AD disokong oleh wayar keluli CF, sambungan tembaga BE dan penyangkut pada A. Rasuk berada dlm keadaan horizontal sebelum beban P dikenakan. Nilai modulus kekenyalan bagi keluli, Ek=210GPa dan bagi tembaga, Et=105GPa. Luas keratan rentas bagi keluli, Ak=0.00015m2 dan bagi tembaga At=0.0018m2. Carikan tegasan dlm wayar keluli dan sambungan tembaga.

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2. Satu silinder konkrit berisi keluli digunakan sebagai teras dalam bangunan. Daya mampatan yang disokong oleh teras ialah P=100kN. Tentukan tegasan paksi dalam silinder keluli dan konkrit dan juga pemendekan pada teras. Diberi Ekeluli = 200 GPa dan Ekonkrit= 24 GPa

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P

0.6m

keluli

konkrit

0.15m

0.18m

Shear Stress• A force acting parallel or tangential to a section taken through a material (i.e. in

the plane of the material) is called a shear force

• The shear force intensity, i.e. shear force divided by the area over which it acts, is called the average shear stress,

= shear stress

V = shear force

A = cross-sectional area

• Shear stress arises as a result of the direct action of forces trying to cut through a material, it is known as direct shear

force

•Shear stresses can also arise indirectly as a result of tension,

torsion or bending of a member.

A

V

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• Depending on the type of connection, a connecting element (bolt, rivet, pin) may be subjected to single shear or double shear as shown.

Rivet in Single Shear 4

2d

P

A

V

42

Rivet in Double Shear

Example 2.10

For the 12 mm diameter bolt shown in the

bolted joint below, determine the average

shearing stress in the bolt.

22

2

)4

(2d

P

d

P

A

V

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Shear Strain• The effect of shear stress is to distort the shape of a

body by inducing shear strains

• The shear strain, is a measure of the angular distortion of the body.

(units: degrees, radians)

L

Vx

L

x

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Bearing Stress

• Bearing stress is also known as a contact stress

Bearing stress in shaft key;

Bearing stress in rivet and plat;

rhL

M

Lh

rM

A

P

b

b

2

)2(

td

Pb

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Example 2.11

A circular punch, B can withstand the

maximum shear stress of 300MPa.

Determine the minimum load P that must

be place on the plate in order to make a

hole of 50mm diameter. Given the

thickness of plate is 10mm.

Die

P

50mm

10mm

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Example 2.12

Satu kunci segiempat tepat b x h x l

(15mm x 10mm x 75mm) digunakan untuk

menyambung gear dengan aci berdiameter,

d = 70mm. Momen yang dipindahkan oleh

kunci ialah 1000kNmm. Tentukan:

(a) Tegasan ricih dalam kunci

(b) Tegasan galas antara kunci dan shaft

r=35mm

P

b=15mm

h=10mm

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Shear Modulus

• It also known as Shear Modulus of Elasticity or the Modulus of Rigidity.

• Value of shear modulus can be occur from the linear region of shear stress-strain diagram.

• The modulus young (E), poisson’s ratio() and the modulus of rigidity (G) can be related as

GUnit : Pa

)1(2

EG

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Example 2.13

A standard tension test is used to determine the properties of an experimental plastic. The test specimen is a 15 mm diameter rod and it is subjected to a 3.5 kN tensile force. Knowing that an elongation of 11 mm and a decrease in diameter of 0.62 mm are observed in a 120 mm gage length. Determine the modulus of elasticy, the modulus of rigidity, and Poisson’s ratio of the material.

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P P

m

m

Stress on an inclined plane

Both normal stress and shear stress exist on an inclined plane (plane m-m).

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• The stresses are the same throughout the entire bar, the stresses on the inclined surface must be uniformly distributed.

• The bar is in equilibrium. Consequently, the resultant of the uniformly distributed stress must equal to P, even though the stress act on the inclined surface m-m.

P P

m

m

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• To investigate the stresses acting on the inclined plane, compute the components of the resultant force P acts on the perpendicular and parallel to the plane.

• The orientation of the inclined surface can be defined by the angle θ between the x axis and the normal n.

• When working with planes, the orientation of the plane is specified by the normal to the plane.

P

P

t n

x

θ

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• The perpendicular component (acting in the n direction) is termed the normal force N:

• The parallel component (acting in the t direction) is termed the shear force V:

P

P

t n

x

θN

V

N = P cos θ

V = - P sin θ

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• Since stress is defined as force per area, the magnitude of the normal and the shear stress on the inclined plane surface depends on the area exposed when the plane m-m cut the material.

• The area of the bar on the inclined plane depends on the angle θ

• As the angle θ of the inclined plane increase in magnitude, the exposed area An also increase.

• Both normal and shear stress magnitude will be directly affected by the orientation of the plane.

An = A / Cos θ

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• Normal stress on an inclined plane

• Average shear stress on an inclined plane

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)cos(

cos

A

P

A

N

n

n

22 CosCosA

P

)/(

CosA

PSin

A

V

n

22

1SinCosSin

A

P

Example 2.14

The bar shown has a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar’s cross section area, determine the average normal stress and average shear stress acting on the material along:

(a) Section plane a-a

(b) Section plane b-b

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800 N

a

a

b

b

20mm

20mm

60o

Example 2.15

The inclined member is subjected to a compressive force of 600 N. Determine the average compressive shear stress along the areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB.

57

600 N4

3

1cm

1.5 cm3 cm

2 cm

A

B

C

D

E

Volume Change

•Because of the change in the dimensions of a body as a result of tension or compression, the volume of the body also changes within the elastic limit.

•Consider a rectangular parallel piped having sides a, b and c in the x, y and z directions, respectively.

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• The tensile force P causes an axial elongation of a andlateral contractions of b and c in the x, y, and zdirections respectively. Hence,

Initial volume of body, Vo = abc Final volume, Vf = (a + a)(b - b)(c - c)

= abc(1 + )(1 - )2

Initial body

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Expanding and neglecting higher orders of (since is very small),

Final volume, Vf = abc(1 + - 2)

Change in volume,

V = Final Volume - Initial Volume = abc(1 + - 2 ) - abc = abc(1 + - 2 - 1) = abc( - 2 ) = Vo (1 - 2 )

Hence,

Dilatation,

V = Vo (1 - 2 )

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)21(

oV

V

)21(

E

e =

• Isotropic material is subjected to general triaxial stress x, y and z.

• Since all strain satisfy << 1, sov = x + y + z

x =

y =

z =

)(1

zyxE

)(1

zxyE

)(1

yxzE

)(21

zyxvE

61

Example 2.16

A titanium alloy bar has the following original dimensions: x =10cm; y = 4cm; and z = 2cm. The bar is subjected to stresses x

= 14 N and y = - 6 N, as indicated in figure below. Theremaining stresses (z, xy, xz and yz) are all zero. Let E = 16kN and = 0.33 for the titanium alloy.

(a)Determine the changes in the length for

x, y and z.

(b) Determine the dilatation, v.

z

x

y

14 N14 N

6 N

6 N

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Allowable Stress

• Applied load that is less than the load the member can fully support. (maximum load)

• One method of specifying the allowable load for thedesign or analysis of a member is use a number calledthe Factor of Safety (FS).

Allowable-Stress Design

allow

fail

F

FFS

FS > 1

FSor

FS

yield

allow

yield

allow

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Example 2.17

The suspended rod is supported at its end by a fixed-connected circular disk as shown in figure. If the rodpasses through a 40 mm diameter hole, determine theminimum required diameter of the rod and the minimumthickness of the disk needed to support the 20 kN load.The allowable normal stress for the disk is allow = 60MPa and the allowable shear stress for the disk is allow =35 MPa.

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40 mm

d

20kN

t

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.

0 RL

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Example 2.18

Determine the reactions at A and B for the steel barand loading shown, assuming a close fit at bothsupports before the loads are applied.

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Example 2.19

Two cylindrical rods, CD made of steel (E=29MPa) and

AC made of aluminum (E=10.4MPa), are joined at C and

restrained by rigid supports at A and D. Determine

(a) the reactions at A and D

(b) The deflection of point C

8 cm 10 cm 10 cm

18 N 14 N

cm cm

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Thermal Stresses• 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.

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.

TEA

P

TAEPPT

0

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Example 2.20

At room temperature (27oC) a 0.02 cm gap exists between the ends of the rods shown. At a later time when the temperature has reaches 2270C, determine

(a)The normal stress in the aluminum rod

(b)The change in length of the aluminum rod

12cm 10cm

0.02cm

cm2 cm2

Pa PaC

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