me a chapter 1

8/8/2019 Me a Chapter 1

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ME16A: INTRODUCTION TO

STRENGTH OF MATERIALS

COURSEINTRODUCTION


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Details of Lecturer

Course Lecturer: Dr. E.I. Ekwue

Room Number: 216 Main Block,Faculty of Engineering

Email: [email protected] ,

Tel. No. : 662 2002 Extension 3171

Office Hours: 9 a.m. to 12 Noon. (Tue,Wed and Friday)


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COURSE GOALS

This course has two specific goals:

(i) To introduce students to concepts of

stresses and strain; shearing force and

bending; as well as torsion and deflection of

different structural elements.

(ii) To develop theoretical and analyticalskills relevant to the areas mentioned in (i)

above.


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COURSE OUTLINECOURSE CONTENTS

1. General Concepts Stresses and strain, two and three-dimensional s

Generalized Hookes Law stress-strain relationships.

2. Properties of Materials Tension, Compression, Hardness and Impact tests.

3. Statically Determinate Stress Systems. St. Venants Principle. Stress Analysis o

loaded bars. Strains and deformations in axially loaded bars. Statically Indeterminat

systems

4. Shear Force and Bending Moment in Beams. Mathematical relationships betwe

intensity, shearing force and bending moment. Bending stresses in beams. B

two materials.

5. Analysis of Stresses in Two-Dimensions. Principal Stresses, Mohrs Circle

6. Deflection of Beams Simple cases. Direct integration and moment-area metho

7. Torsion of Circular Cross-Sections.


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CourseObjectivesUpon successful completion of this course,

students should be able to:

(i) Understand and solve simple problemsinvolving stresses and strain in two and three

dimensions.

(ii) Understand the difference betweenstatically determinate and indeterminate

problems.

(iii) Understand and carry out simple experimentsillustrating properties of materials in tension,

compression as well as hardness and impact tests.


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COURSE OBJECTIVES CONTD.

(iv) Analyze stresses in two dimensions andunderstand the concepts of principalstresses and the use of Mohr circles to solvetwo-dimensional stress problems.

(v) Draw shear force and bending momentdiagrams of simple beams and understandthe relationships between loading intensity,

shearing force and bending moment.

(vi) Compute the bending stresses in beamswith one or two materials.


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OBJECTIVES CONCLUDED

(vii) Calculate the deflection of beams

using the direct integration and moment-

area method.

(viii) Apply sound analytical techniques

and logical procedures in the solution of

engineering problems.


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Teaching Strategies

The course will be taug

ht viaLectures. Lectures will also

involve the solution of tutorialquestions. Tutorial questions are

designed to complement andenhance both the lectures and thestudents appreciation of the

subject. Course work assignments will be

reviewed with the students.


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Lecture Times

Wednesday: 2.00 to 2.50 p.m.

Thursday: 11.10 a.m. to 12.00 noon

Friday: 1.00 to 1.50 p.m. Lab Sessions: Two Labs per student on

Mondays (Details to be Announced Later)

Attendance at the Lectures and Labs is

Compulsory.


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Time-Table For Labs

M O N D A Y 1 :0 0 - 4 :0 0 P .M .

WeekG r o u p

1,5,9 2,6,10 3,7,11, 4 ,8,12

K - M E 1 3 A M E 1 6 A

(3,7)

M E 1 3 A

L M E 1 3A - M E 1 3A M E 1 6A

(4,8)

M M E 1 6 A(5,9)

M E 1 3 A - M E 1 3 A

N M E 1 3A M E 1 6A(6,10)

M E 1 3 A -


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More Course Details

BOOK Hearn, E.J. (1997), Mechanics of

Materials 1, Third Edition, Butterworth,

Heinemann

COURSE WORK

1. One Mid-SemesterTest (20%); 2. Practical report (15%) and

3. End of Semester 1 Examination (65%).


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ME16A: CHAPTER ONE

STRESS AND STRAINRELATIONS


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1.1 DIRECT OR NORMAL

STRESS When a force is transmitted through a body,

the body tends to change its shape or

deform. The body is said to be strained.

Direct Stress = Applied Force (F)

Cross Sectional Area (A)

Units: Usually N/m2 (Pa), N/mm2, MN/m2,GN/m2 or N/cm2

Note: 1 N/mm2

= 1 MN/m2

= 1 MPa


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Direct Stress Contd.

Direct stress may be tensile, t or

compressive, c and result from forces

acting perpendicular to the plane of thecross-section

W

W

Tension

Compression


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1.2 Direct or Normal Strain

When loads are applied to a body,

some deformation will occur resulting to

a change in dimension.

Consider a bar, subjected to axial

tensile loading force, F. If the bar

extension is dl and its original length

(before loading) is L, then tensile strain

is:


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Direct or Normal Strain Contd.

Direct Strain ( ) = Change in LengthOriginal Length

i.e. = dl/L

dl

FF

L

I

I


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Direct or Normal Strain Contd.

As strain is a ratio of lengths, it isdimensionless.

Similarly, for compression by amount,

dl: Compressive strain = - dl/L Note: Strain is positive for an increase

in dimension and negative for a

reduction in dimension.


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1.3 Shear Stress and Shear Strain

Shear stresses are produced byequal and opposite parallel forces

not in line.

The forces tend to make one partof the material slide over the other

part.

Shear stress is tangential to thearea over which it acts.


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Shear Stress and Shear Strain

Contd.

P Q

S R

F

D D

A B

C C

L

x

J

Shear strain is the distortion produced by shear stress on

an element or rectangular block as above. The shear

strain, (gamma) is given as:

= x/L = tanK

K


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Shear Stress and Shear Strain

Concluded

For small ,

Shear strain then becomes the change

in th

e right angle.

It is dimensionless and is measured in

radians.

J K J!


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1.3 Complementary Shear Stress

a

X1

X1X2

X2

PQ

S R

Consider a small element, PQRS of the material in the

last diagram. Let the shear stress created on faces PQ

and RS be X1


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Complimentary Shear Stress

Contd.

The element is therefore subjected to a

couple and for equilibrium, a balancing

couple must be brought into action.

This will only arise from the shear stress on

faces QR and PS.

Let the shear stresses on these faces be

. X2


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Contd.

Let t be the thickness of the material atright angles to the paper and lengths of

sides of element be a and b as shown.

For equilibrium, clockwise couple =anticlockwise couple

i.e. Force on PQ (or RS) x a = Forceon QR (or PS) x b

X X

X X

1 2

1 2

x b t x a x a t x b

i e

!

!. .


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Concluded

Thus: Whenever a shear stress occurs on

a plane within a material, it is automatically

accompanied by an equal shear stress onthe perpendicular plane.

The direction of the complementary shear

stress is such

th

at th

eir couple opposes th

atof the original shear stresses.


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1.4 Volumetric Strain

Hydrostatic stress refers to tensile orcompressive stress in all dimensionswithin or external to a body.

Hydrostatic stress results in change involume of the material.

Consider a cube with sides x, y, z. Letdx, dy, and dz represent increase in

length in all directions. i.e. new volume = (x + dx) (y + dy) (z +

dz)


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Volumetric Strain Contd.

Neglecting products of small quantities:

N e w v o l u m e = x y z + z y d x + x z d y + x y d z

Original volume = x y z

= z y dx + x z dy + x y dz

Volumetric strain, = z y dx + x z dy + x y dz

x y z = dx/x + dy/y + dz/z

(V

Iv

Iv

I I I I v x y z!


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Strains Contd.

Note: By similar reasoning, on area x y

Also: (i) The strain on the diameter of acircle is equal to the strain on thecircumference.

(ii) The strain on the area of a circle, isequal to twice the strain on its diameter.

(iii) Strain on volume of a sphere, is equalto three times the strain on its diameter.

I I Ia x y!


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Strains Contd.

( )

,

iv Given and as strains on the diameter

and length of a cylinder

Strain on the volume is

D L

v D L

I I

I I I! 2

These can be proved using the theorem

of small errors


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Examples

(i) Diameter, D = 2 x radius, r i.e. D = 2 r

Taking logs: log D = log 2 + log r

Taking differentials: dD/D = dr/r

Also: Circumference, C = 2 r

i.e. log C = Log 2 + log r

dC/C = dr/r = dD/D

i.e. the strain on the circumference,

= strain on the diameter,

T

T

IcI


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Strains Contd.

Required: Prove the other two

statements.

I I Iv D L! 2

(iv) Volume of a cylinder, V = T r2 L where L is the length

Taking logs: log V = logT + 2 log r + log L

Taking differentials: dV/V = 2 dr/r + dL/L

i.e. I I Iv D L! 2


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1.5 Elasticity and Hookes Law

All solid materials deform when they arestressed, and as stress is increased,

deformation also increases.

If a material returns to its original size and

shape on removal of load causing

deformation, it is said to be elastic.

If the stress is steadily increased, a point is

reached when, after the removal of load, notall the induced strain is removed.

This is called the elastic limit.


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Hookes Law

States that providing the limit of proportionality of a material is not exceeded,

the stress is directly proportional to the strainproduced.

If a graph of stress and strain is plotted asload is gradually applied, the first portion of

the graph will be a straight line. The slope of this line is the constant of

proportionality called modulus of Elasticity, Eor Youngs Modulus.

It is a measure of the stiffness of a material.


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Hookes Law

Modulus of l sticit ,Direct stress

Direct strain!

W

I

Al or h r str ss: Modulus of ri idit or sh r odulus,Shear stress

Shear strain!

X

K

Also: Volumetric strain, is proportional to hydrostatic

stress, within the elastic range

i.e. : called bulk modulus.

Iv

I/v

K!


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Stress-Strain Relations of Mild

Steel


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Equation For Extension

From the above equations:

E

F A

dl L

F L

A dl

dlF L

A E

! ! !

!

I

/

/

This equation for extension isvery important


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Extension For Bar of Varying Cross

Section

Forabarofvaryingcrosssection:

P

A1 A2 A3 P

L1 L2 L3

dlF

E

L

A

L

A

L

A! LNM

OQP

1

1

2

2

3

3


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Factor of Safety

The load which any member of a machine

carries is called working load, and stress

produced by this load is the working stress.

Obviously, the working stress must be less

than the yield stress, tensile strength or the

ultimate stress.

This working stress is also called the

permissible stress or the allowable stress or

the design stress.


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Factor of Safety Contd.

Some reasons for factor of safety

include the inexactness or inaccuracies

in the estimation of stresses and the

non-uniformity of some materials.

actor of safetyUltimate or yield stress

Design or working stress

Note: Ultimate stress is used for materials e.g.

concrete which do not have a well-defined yield point,

or brittle materials which behave in a linear manner

up to failure. Yield stress is used for other materials

e.g. steel with well defined yield stress.


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1.7 Practical Class Details

Each Student will have two practicalclasses: one on :

Stress/strain characteristics and

Hardness and impact tests. (i) The stress/strain characteristics

practical will involve the measurement

of the characteristics for four metals,

copper, aluminium, steel and brass

using a tensometer.


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Practical Class Details Contd.

The test will be done up to fracture of themetals.

This test will also involve the accurate

measurement of the modulus of elasticity for

one metal. There is the incorporation of an

extensometer for accurate measurement of

very small extensions to produce an

accurate stress-strain graphs.

The test will be done up to elastic limit.


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Practical Class Details Contd.

(ii) The hardness test will bedone using the same four metalsand the Rockwell Hardness test.

The impact test with the four metals will be carried out using theIzod test.


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1.8 MATERIALS TESTING

1.8.1. Tensile Test: This is the mostcommon test carried out on a material.

It is performed on a machine capable of applying a true axial load to the testspecimen. The machine must have:

(i) A means of measuring the applied

load and (ii) An extensometer is attached to the test

specimen to determine its extension.


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Tensile Test Contd.

Notes: 1. For iron or steel, the limit of proportionality and the elastic limit arevirtually same but for other materials likenon-ferrous materials, they are different.

2. Up to maximum or ultimate stress, thereis no visible reduction in diameter of specimen but after this stress, a localreduction in diameter called necking occurs

and this is more well defined as the load fallsoff up to fracture point.

Original area of specimen is used for analysis.


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Results From a Tensile Test

(a) Modulusof Elasti ity, EStress up to it of proportionality

Strain!

lim

(b) Yi ld StressorProofStress (See below)

(c) Percentageelongation =Increase in gauge length

Original gauge lengthx 100

(d) Percentagereductionin area =Original area area at fracture

Original area x

100

(e) TensileStrength =Maximum load

Original cross tional areasec

The ercentageofelongationand ercentage reduction inareagi ean indicationof the

ductilityof the aterial i.e. its ability towithstand strainwithout fractureoccurring.


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Proof Stress

High carbon steels, cast iron and most of the

non-ferrous alloys do not exhibit a well

defined yield as is the case with mild steel. For these materials, a limiting stress called

proof stress is specified, corresponding to a

non-proportional extension.

The non-proportional extension is a specified

percentage of the original length e.g. 0.05,

0.10, 0.20 or 0.50%.


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Determination of Proof Stress

PProof Stress

Stress

The proof stress is obtained by drawing AP parallel to the initial

slope of the stress/strain graph, the distance, OA being the strain

corresponding to the required non-proportional extension e.g. for

0.05% proof stress, the strain is 0.0005.

A

Strain

1 8 2 Hardness Test


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1.8.2 Hardness Test

The hardness of a material is determined byits ability to withstand indentation. There arefour majorhardness tests.

(i) Rockwell Hardness Test: This uses an

indentor with a 120o conical diamond with arounded apex forhard materials, or steel ballfor softer materials.

A minor load, F is applied to cause a small

indentation as indicated in Fig. (a) below. The major load, Fm is then applied and

removed after a specified time to leave loadF still acting. The two stages are shown as

(b) and (c).

R k ll H d T


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Rockwell Hardness Test


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Hardness Test Contd.

Thus the permanent increase in thedepth of penetration caused by the

major load is d mm. The Rockwellhardness number, HR is:

HR = K - 500 d

Where: K is a constant with value of100 for the diamond indentor and 130for the steel indentor.


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1.8.3 Impact Testing

The toughness of a material is definedas its ability to withstand a shockloading without fracture. Two principal

impact tests are the: Izod and the

Charpy tests.

A test specimen is rigidly supportedand is impacted by a striker attached toa pendulum.


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Impact Test Concluded

The difference in height from which a

pendulum is released and the height to

which it rises after impact gives a

measure of the energy absorbed by the

specimen and this is recorded on a dial

mounted on a tester.


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Example on Elongation

A flat plate of steel, 1 cm thick,

and of trapezoidal form tapersfrom 5 cm width to 10 cm widthin a length of 40 cm.

Determine th

e elongation underan axial force of 50 kN. E = 2 x107 N/cm2.


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Diagram of a Trapezoidal Steel Plate

dx

PP

x

L

B1

B2

t

S l ti


55/92

Solution

Considera length,dx at a distance,xfromwidth, B1,

Width atthatsection

!

!

!

BB B

Lx B Kx

where KB B

L

12 1

1

2 1

Area (Ax) of chosen c/section = ( B1 + K x ) t. Ifthelength dx

elongates an amountduunderload,itsstrainis:

du

dx

P

A E! .

1

S l ti C td


56/92

Solution Contd.

Total extension of bar, u

uP

A E

dxP

B Kx t E

dx

uP

t E

dx

B kx

P

KtE B Kx

L

uP

K tE

B K L

B

x

L L

L

! !

!

!

!

z zz0 10

10

1

1

1

0

( )

ln

ln


57/92

Solution Contd.

Substitutingback forK,

uP

B B

L

t E

B B B

B

uP

B B

Lt E

B

B

!

!

( )

ln

( )

ln

2 1

1 2 1

1

2 1

2

1

In roble , t =1 c , B1= c , B2=10 c , L= 0 c , P= 0,000N, E=2x107N/c2

uN

x cm x x

cm!

!50 000

10 5

401 2 10

10

5001386

7

,

( )

ln .


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Solution Concluded

Substitutingback forK,

uP

B B

Lt E

B B B

B

u PB B

Lt E

B

B

!

!

( )

ln

( )

ln

2 1

1 2 1

1

2 1

2

1

In roble , t =1 c , B1= c , B2=10 c , L= 0 c ,

P= 0,000N, E=2x107N/c2

uN

x cm x x

cm!

!50 000

10 5

401 2 10

10

50 01386

7

,

( )

ln .

1 9 L t l St i d P i R ti


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1.9 Lateral Strain and Poissons Ratio

Under the action of a longitudinal

stress, a body will extend in the

direction of the stress and contract in

the transverse or lateral direction

(see Fig. below).

The reverse occurs under acompressive load.


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Stress Effects

PP

Longitudinal Tensile Stress Effect

Longitudinal Compressive Stress Effect

PP


61/92

Poissons Ratio

Lateral strain is proportional to the longitudinal strain,

with the constant of proportionality called Poissons ratio withsymbol,R .

Mathematically, R!Lateralstrain

Director longitudinalstrain

For most metals, the range ofR is 0.28 to 0.33.

1 10 Thermal Strain


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1.10 Thermal Strain

Most structural aterials expandwhen heated,

inaccordance to the law: I E! T

where I is linear strainand

E is the coefficient of linearexpansion;

T is the rise in te perature.

That is fora rodofLength, L;if its te perature increasedby t, theextension,

dl =E LT.

h l S i C d


63/92

Thermal Strain Contd.

As i t se f l ter l str i s, t er l str i s

t i estresses less t ey re str i ed.

T e t t l str i i dyexperiencing t er l stress

y edi ided int t components:

tr indue tostress, IW nd

T t due to temper ture, IT

.

T us: I IW IT

I EE

T

1 11


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1.11. Principle of Superposition

It states that the effects of several actionstaking place simultaneously can be

reproduced exactly by adding the effect of each action separately.

The principle is general and has wideapplications and holds true if:

(i) The structure is elastic (ii) The stress-strain relationship is linear

(iii) The deformations are small.

1 12 G l St St i


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1.12 General Stress-Strain

Relationships


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1.12 General Stress-Strain

RelationshipsFor theelement ofmaterial as inFigureabove

subjected touniaxial stress, Wx , theensuing strain

is as shown in (b).

Strain inxdirection, IW

xx

E!

Strains inyand z directions as a

result of strain inx direction

= ! R I R I RW

x xxand

Eeach

Not Thenegative sign indicates contraction.


67/92

General Stress-Strain Relationships

Contd.

For an element subjected to triaxial stresses,

W W W x y zand, , the total strain in x direction will be

due to Wx and lateral strains due to W Wy zand .

Using the principle of superposition, the resultant strain in x-direction is:

IW R W R W

I W R W W

xx y z

x x y z

E E E

i eE

!

! . . { ( )}1

I W R W W y y x zE

! 1

{ ( )} Generalised Hookes Law in threedimensions

I W R W W z z x y

E

! 1

{ ( )}


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General Stress-Strain Relationships

Contd.

Note: In the case of shear

strain, there is no lateral strain,hence the shear stress/shear

strain relationship is the same

for both uniaxial and complexstrain systems.

Plain Stress and Plain Strain


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Plain Stress and Plain Strain

A plain stress condition is said to exist whenstress in the z direction is zero.

The above equations may be applied for but strain in the z direction is not zero.

Also plain strain condition exists when thestrain in z direction is zero.

Using strain in Z direction as zero in thiscase does not mean that stress in the zdirection is zero.

Strain Caused by Stress and


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Strain Caused by Stress and

Temperature

Inaddition to strain causedby stress, theremayalsobe thermal strain

due to change in temperature. Thegeneral formof the stress/strain

relations is:

I W R W W x x y zE

t! 1

{ ( )}

I W R W W y y x zE

t! 1

{ ( )}

I W R W W z z x yE

t! 1

{ ( )}

KX

KX

KX

xy

xy

yz

yz

zxzx

G G G! ! !; ;


71/92

Try On Your Own

Show that : IR

W W Wv x y z

E

!

1 2

( )


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ExampleExample: A plate of uniform thickness 1 cm and dimension 3 x 2 cm is acted upon b

the loads shown. Taking E = 2 x 107 N/cm2, determine I Ix yand . Poissons ratio is

0.3. 42 kN

y

18 kN 2 cm 18 kN

x

42 kN

3cm


73/92

SolutionWx

cm x cmcm! !

18000

2 19000 2/

Wycm x cm

cm! !42000

3 114000 2/

Hookes law in two dimensions states that:

I W R W x x yE x

x! ! ! 1 1

2 109000 0 3 14000 240 10

7

6[ ] [ . ( ]

and I W R W y y xE x

x! ! ! 1 1

2 1014000 0 3 9000 565 10

7

6[ ] [ . ( ]


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1.13 Relationship between Elastic

Modulus (E) and Bulk Modulus, K

It has been shown that : I I I I v x y z!

I W Y W W

W W W W

I W W YW

Y

I IW

Y

I I I I

I

W

Y

W

IY

W

I

Y

Y

x x y z

x y z

x

y z

v x y z

v

v

v

E

For hydrostatic stress

i eE E

Similarly and are eachE

Volumetric strain

E

E

Bulk Modulus KVolumetric or hydrostatic stress

Volumetric strain

i eE

K and K

E

!

! ! !

! !

! !

!

!

! !

! !

1

12 1 2

1 2

3

1 2

31 2

3 1 2 3 1 2

( )

,

. .

,

,

. .

Maximum Value For Poissons


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Maximum Value For Poisson s

Ratio

From theequation, ifv=0. , thevalueofKbecomes infinitely large.

ence thebody is incompressible. Ifv > 0. , Kbecomes negative

i.e. thebodywill expandunder hydrostatic pressurewhich is

inconceivable. It maybe concluded that theupper limit ofPoissons ratio

is 0. .

Not KG

and E G!

! 2 1

3 1 2 2 1Y

YY

Where: G is ShearModulus

1 14 Compound Bars


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1.14 Compound Bars

A compound bar is one comprising two or more parallel elements, of different materials,

which are fixed together at their end. The compound bar may be loaded in tension o

compression.

1 2

F F

2

Section through a typical compound bar consisting of a circular bar (1) surrounded by a

tube (2)

1 14 1 Stresses Due to Applied Loads


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1.14.1 Stresses Due to Applied Loads

in Compound Bars

Ifa compoundbar is loaded in compressionbya force, F,

Since the rodand tubeareof the same length andmust remain

together, the twomaterialsmust have the same strain i.e.

I I

W WW

W

1 2

1

1

2

2

21 2

1

1

!

! ! !StrainStress

Ei e

E E

E

E. , ..... ( )

WhereE1andE2are theelasticmoduli ofmaterials 1and2 respectively.

Al o: The total load, Fmust be sharedby the twomaterials, i.e. F=F1+F2

Where: F1andF2are the loads in the individual elements.


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Compound Bars Contd.

Now: s force stressx rea: T en: W W1 1 2 2A A ...............( )

ereA1andA2are t eareasofmaterials1and2 respecti ely.

ubstituting for W2 from n. 1 into n2:

F A E A

EA E A

E

F E

E A E Aand

F E

E A E A

! ! LNM OQP

!

!

W W W

W W

1 1

1 2 2

1

1 1

2 2

1

1

1

1 1 2 2

2

2

1 1 2 2

1 14 2 T i


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1.14.2 Temperature stresses in

compound bars

1 E 1

2 E 2

L

(a) L E 1 T

1LE 2 T

2 {b}FL

A E1 1

F 1 F

F 2 F

(c )FL

A E2 2

T t t i d


80/92

Temperature stresses in compound

bar Contd.

Considera compoundbar, see (a) aboveof length, L consistingof2

different materials (1) and (2) having coefficients ofexpansion

E 1 andE 2 respectivelywith E 1 >E 2 . If thebar is subjected toa

E 1 uniform temperature rise, Tand the right hand fixing released,

E 1 thebar (1) will expandmore than (2) as shown indiagram (b).

owever, becauseof theend fixing, freeexpansion cannot occur.

iagram (c) shows that theend fixingmust supplya forcewhich

decreases the length ofbar (1) and increases the length ofbar (2)

until equilibrium is achievedat a common length.

As noexternal forces are involved, a selfequilibrating

(balancing force system is created).

T t St C td


81/92

Temperature Stresses Contd.

Free expansions in bars (1) and (2) are L T and L TE E1 2 respectively.

Due to end fixing force, F: the decrease in length of bar (1) is

FL

A E1 1and the increase in length of (2) is

FL

A E2 2.

At Equilibrium:

L TFL

A EL T

FL

A E

i e F A E A E

T

i e AA E A E

E E A AT

T A E E

A E A E

T A E E

A E A E

E E

E E

W E E

WE E

W

E E

1

1 1

2

2 2

1 1 2 2

1 2

1 12 2 1 1

1 2 1 2

1 2

11 2 2 1 2

1 1 2 2

21 2 1 1 2

1 1 2 2

1 1

!

!

LNM

OQP

!

!

!

. . [ ] ( )

. . ( )

( )

( )

Note: As a result of Force, F, bar (1) will be in compression while (2) will be in tension.

1 E1

2 E2

L

(a) LE1T

1

LE2 T

2 {b}FL

A E1 1

F 1 F

F 2 F

(c)FL

A E2 2

Example


82/92

Example

A steel tube having an external diameter of36 mm and an internal diameter of 30 mmhas a brass rod of 20 mm diameter inside it,the two materials being joined rigidly at their

ends when the ambient temperature is 180C. Determine the stresses in the twomaterials: (a) when the temperature is

raised to 68 0C (b) when a compressiveload of 20 kN is applied at the increasedtemperature.

Example Contd


83/92

Example Contd.

For brass: Modulus of elasticity = 80

GN/m2; Coefficient of expansion = 17 x

10 -6 /0C

For steel: Modulus of elasticity = 210

GN/m2; Coefficient of expansion = 11 x

10 -6 /0C

Solution


84/92

Solution

30 Brass rod 20 36

Steel tube

Areaofbrass rod (Ab) =T x

mm20

4 31416

22

! .

Areaof steel tube (As) =T x

mm( )

.36 30

431102

2 22

!

A E x m x x m xs s ! !31102 10 210 10 0 653142 10

6 2 9 2 8. / .

1153106 10 8

A E

xs s

! .

Solution Contd


85/92

Solution Contd.

A E x m x x N m x Nb b ! !31416 10 80 10 0251327 106 2 9 2 8. / .

139788736 10 8

A Ex

b b

! .

T x xb s( ) ( )E E ! ! 50 17 11 10 3 106 4

With increase in temperature, brass will be in compressionwhile

steel will be in tension. This is becauseexpands more than steel.

i e FA E A E

Ts s b b

b s. . [ ] ( )1 1

! E E

i.e. F[1. 106+ . 788736]x10-8= 3x10

-

F = 5444.71N


86/92

Solution Concluded

Stress insteel tube544471

311021751 1751

2

2 2.

.. / . / ( )

N

mmN mm MN m Tension! !

Stress inbrass rod544471

314161733 1733

2

2 2.

.. / . / ( )

N

mmN mm MN m Compression! !

(b) Stressesdue tocompression force, of20

Wss

s s b b

F E

E A E A

x x x m

xMN m Compression!

!

!

' /

. .. / ( )

20 10 210 10

0653142 0251327 104644

3 9 2

8

2

Wbb

s s b b

F E

E A E A

x N x x N m

x MN m Compression! ! !' /

. . . / ( )20 10 80 10

0653142 0251327 10 1769

3 9 2

8

2

esultant stress insteel tube - . 17. 1 28. /m2 ( ompression)

esultant stress inbrass rod -17. - 17. .02 /m2 ( ompression)


87/92

Example

A composite bar, 0.6 m long comprises asteel bar 0.2 m long and 40 mm diameter which is fixed at one end to a copper bar having a length of 0.4 m.

Determine the necessary diameter of thecopper bar in order that the extension of each material shall be the same when thecomposite bar is subjected to an axial load.

What will be the stresses in the steel andcopper when the bar is subjected to an axialtensile loading of 30 kN? (For steel, E = 210GN/m2; for copper, E = 110 GN/m2)


88/92

Solution

0.2 mm

0.4 mm

F 40 mm dia d F

Let the diameter of the copper bar be d mm

Specified condition: Extensions in the two bars are equal

dl dl

dl LE

L FL

AE

c s!

! ! !I W

Thus:F L

A E

F L

A E

c c

c c

s s

s s

!


89/92

Solution Concluded

Also: Totalforce,Fistransmitted by both copperandsteel

i.e. Fc = Fs = F

i eL

A E

L

A E

c

c c

s

s s

. . !

Substitutevalues giveninproblem:

0 4

4 110 10

0 2

4 0 040 210 102 2 9 2 2 9 2

.

/ /

.

/ . /

m

d m x N m

m

x x x N mT T!

dx x

m d m mm2

222 210 0 040

1100 07816 7816! ! !

.; . . .

Thusfora loading of30 kN

Stressinsteel, WT

sx N

x xMN m! !

30 10

4 0 040 102387

3

2 6

2

/ .. /

Stressin copper, WT

c

x N

x xMN m! !

30 10

4 0 07816 109

3

2 6

2

/ ./


90/92

1.15 Elastic Strain Energy

If a material is strained by a gradually

applied load, then work is done on the

material by the applied load. The work is stored in the material in the form

of strain energy.

If the strain is within the elastic range of the

material, this energy is not retained by the

material upon the removal of load.


91/92

Elastic Strain Energy Contd.

Figurebelow shows the load-extensiongraph ofauniformbar.

Theextensiondl is associatedwith agraduallyapplied load, P

which is within theelastic range. The shadedarea represents

thework done in increasing the load from zero to its value

Load

P

Extension

dl

Work done= strainenergyofbar= shadedarea

Elastic Strain Energy Concluded


92/92

Elastic Strain Energy Concluded

W= U =1/2Pdl (1)

Stress, W =P/A i.e P=W A

Strain = Stress/E

i.edl/L = W /E , dl = (W L)/E L= original length

Substituting forPanddl inEqn (1) gives:

W= U =1/2W A . (W L)/E = W 2/2ExAL

AL is thevolumeof thebar.

i.e U =W2/2Ex olume

Theunits of strainenergyare sameas thoseofwork i.e. Joules. Strainenergy

perunit volume, W2/2E is knownas resilience. Thegreatest amount ofenergy that can

stored in a material without permanent set occurring will be whenW is equal to the

me a chapter 1

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