ME16A: INTRODUCTION TO STRENGTH OF MATERIALS

COURSE INTRODUCTION

Details of Lecturer

Course Lecturer: Dr. E.I. Ekwue Room Number: 216 Main Block,

Faculty of Engineering

Email: ekwue@eng.uwi.tt , Tel. No. : 662 2002 Extension 3171 Office Hours: 9 a.m. to 12 Noon. (Tue,

Wed and Friday)

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 analytical skills relevant to the areas mentioned in (i) above.

COURSE OUTLINECOURSE CONTENTS 1. General Concepts – Stresses and strain, two and three-dimensional systems.

Generalized Hooke’s Law – stress-strain relationships.

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

3. Statically Determinate Stress Systems. St. Venant’s Principle. Stress Analysis of axially

loaded bars. Strains and deformations in axially loaded bars. Statically Indeterminate stress

systems

4. Shear Force and Bending Moment in Beams. Mathematical relationships between load

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

two materials.

5. Analysis of Stresses in Two-Dimensions. Principal Stresses, Mohr’s Circle

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

7. Torsion of Circular Cross-Sections.

         Course Objectives  Upon successful completion of this course,

students should be able to:

(i)  Understand and solve simple problems involving stresses and strain in two and three dimensions.

(ii) Understand the difference between statically determinate and indeterminate problems.

(iii) Understand and carry out simple experiments illustrating properties of materials in tension, compression as well as hardness and impact tests.

COURSE OBJECTIVES CONTD.

(iv) Analyze stresses in two dimensions and understand the concepts of principal stresses and the use of Mohr circles to solve two-dimensional stress problems.

(v) Draw shear force and bending moment diagrams of simple beams and understand the relationships between loading intensity, shearing force and bending moment.

  (vi) Compute the bending stresses in beams

with one or two materials.

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. 

Teaching Strategies The course will be taught via

Lectures. Lectures will also involve the solution of tutorial questions. Tutorial questions are designed to complement and enhance both the lectures and the students appreciation of the subject.

Course work assignments will be reviewed with the students.

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.

Time-Table For Labs

MONDAY 1:00 - 4:00 P.M.

Week Group

1,5,9

2,6,10

3,7,11,

4,8,12

K

-

ME13A

ME16A

(3,7)

ME13A

L

ME13A

-

ME13A

ME16A

(4,8)

M

ME16A (5,9)

ME13A

-

ME13A

N

ME13A

ME16A (6,10)

ME13A

-

More Course Details

BOOK – Hearn, E.J. (1997), Mechanics of Materials 1, Third Edition, Butterworth, Heinemann

 COURSE WORK 1. One Mid-Semester Test (20%); 2. Practical report (15%) and 3. End of Semester 1 Examination (65%). 

ME16A: CHAPTER ONE

STRESS AND STRAIN RELATIONS

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

Direct Stress Contd.

Direct stress may be tensile, t or

compressive, c and result from forces

acting perpendicular to the plane of the cross-section

Tension

Compression

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:

Direct or Normal Strain Contd.

Direct Strain ( ) = Change in Length

Original Length

i.e. = dl/L

dl

FF

L

Direct or Normal Strain Contd.

As strain is a ratio of lengths, it is dimensionless.

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.

1.3 Shear Stress and Shear Strain

Shear stresses are produced by equal and opposite parallel forces not in line.

The forces tend to make one part of the material slide over the other part.

Shear stress is tangential to the area over which it acts.

Shear Stress and Shear Strain Contd.

P Q

S R

FD D’

A B

C C’

L

x

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 = tan

Shear Stress and Shear Strain Concluded

For small ,

Shear strain then becomes the change in the right angle.

It is dimensionless and is measured in radians.

1.3 Complementary Shear Stress

a

1

1

2

2

P Q

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 1

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

. 2

Complimentary Shear Stress Contd.

Let t be the thickness of the material at right 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 = Force on QR (or PS) x b

1 2

1 2

x b t x a x a t x b

i e

. .

Complimentary Shear Stress Concluded

Thus: Whenever a shear stress occurs on a plane within a material, it is automatically accompanied by an equal shear stress on the perpendicular plane.

The direction of the complementary shear stress is such that their couple opposes that of the original shear stresses.

1.4 Volumetric Strain

Hydrostatic stress refers to tensile or compressive stress in all dimensions within or external to a body.

Hydrostatic stress results in change in volume of the material.

Consider a cube with sides x, y, z. Let dx, dy, and dz represent increase in length in all directions.

i.e. new volume = (x + dx) (y + dy) (z + dz)

Volumetric Strain Contd.

Neglecting products of small quantities: New volume = x y z + z y dx + x z dy + x y dz 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 v

v

v x y z

Strains Contd.

Note: By similar reasoning, on area x y

Also: (i) The strain on the diameter of a circle is equal to the strain on the circumference.

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

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

a x y

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

2

These can be proved using the theorem

of small errors

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,

c

D

Strains Contd.

Required: Prove the other two statements.

v D L 2

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

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

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

i.e. v D L 2

1.5 Elasticity and Hooke’s Law

All solid materials deform when they are stressed, 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, not all the induced strain is removed.

This is called the elastic limit.

Hooke’s Law

States that providing the limit of proportionality of a material is not exceeded, the stress is directly proportional to the strain produced.

If a graph of stress and strain is plotted as load 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, E or Young’s Modulus.

It is a measure of the stiffness of a material.

Hooke’s Law

Modulus of Elasticity, E = Directstress

Directstrain

Also: For Shear stress: Modulus of rigidity or shear modulus, G = Shearstress

Shearstrain

Also: Volumetric strain, is proportional to hydrostatic stress,

within the elastic range

i.e. : called bulk modulus.

v

/ v K

Stress-Strain Relations of Mild Steel

Equation For Extension

From the above equations:

EFA

dl L

FL

Adl

dlFL

AE

/

/

This equation for extension is very important

Extension For Bar of Varying Cross Section

F o r a b a r o f v a r y in g c r o s s s e c t io n :

P

A 1 A 2 A 3 P

L 1 L 2 L 3

d lF

E

L

A

L

A

L

A

LNM

OQP

1

1

2

2

3

3

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.

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.

Factor of safety = Ultimateoryieldstress

Designorworkingstress

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.

1.7 Practical Class Details

Each Student will have two practical classes: 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.

Practical Class Details Contd.

The test will be done up to fracture of the metals.

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.

Practical Class Details Contd.

(ii) The hardness test will be done using the same four metals and the Rockwell Hardness test.

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

1.8 MATERIALS TESTING

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

It is performed on a machine capable of applying a true axial load to the test specimen. 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.

Tensile Test Contd.

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

2. Up to maximum or ultimate stress, there is no visible reduction in diameter of specimen but after this stress, a local reduction in diameter called necking occurs and this is more well defined as the load falls off up to fracture point.

Original area of specimen is used for analysis.

Results From a Tensile Test

(a) Modulus of Elasticity, EStress up to it of proportionality

Strain

lim

(b) Yield Stress or Proof Stress (See below)

(c) Percentage elongation = Increase in gauge length

Original gauge lengthx 100

(d ) P ercentage reduction in a rea = Original area area at fracture

Original areax

100

(e ) Tensile S trength = Maximum load

Original cross tional areasec

The percentage o f e longation and percentage reduction in a rea g ive an ind ica tion o f th e

ductility o f the m ateria l i.e . its ab ility to w ithstand stra in w ithou t fracture occurring .

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%.

Determination of Proof Stress

PProof StressStress

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

The hardness of a material is determined by its ability to withstand indentation. There are four major hardness tests.

(i) Rockwell Hardness Test: This uses an indentor with a 120o conical diamond with a rounded apex for hard materials, or steel ball for 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 load F still acting. The two stages are shown as (b) and (c).

Rockwell Hardness Test

Hardness Test Contd.

Thus the permanent increase in the depth of penetration caused by the major load is d mm. The Rockwell hardness number, HR is:

HR = K - 500 d Where: K is a constant with value of

100 for the diamond indentor and 130 for the steel indentor.

1.8.3 Impact Testing

The toughness of a material is defined as its ability to withstand a shock loading without fracture. Two principal impact tests are the:

Izod and the Charpy tests. A test specimen is rigidly supported

and is impacted by a striker attached to a pendulum.

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.

Example on Elongation

A flat plate of steel, 1 cm thick, and of trapezoidal form tapers from 5 cm width to 10 cm width in a length of 40 cm. Determine the elongation under an axial force of 50 kN. E = 2 x 107 N/cm2.

Diagram of a Trapezoidal Steel Plate

dxPP

x

L

B1

B2

t

Solution

Consider a length, dx at a distance, x from width, B1,

Width at that section

BB B

Lx B Kx

whereKB B

L

12 1

1

2 1

Area (Ax) of chosen c/section = ( B1 + K x ) t. If the length ‘dx’

elongates an amount du under load, its strain is:

du

dx

P

A E .

1

Solution Contd.

Total extension of bar, u

uP

AEdx

P

B Kx t Edx

uP

t E

dx

B kx

P

KtEB Kx

L

uP

Kt E

B KL

B

x

L L

L

z zz

01

0

10 1

1

1

0

( )

ln

ln

Solution Contd.

S u b s t i t u t i n g b a c k f o r K ,

uP

B B

Lt E

B B B

B

uP

B B

Lt E

B

B

( )l n

( )l n

2 1

1 2 1

1

2 1

2

1

I n p r o b l e m , t = 1 c m , B 1 = 5 c m , B 2 = 1 0 c m , L = 4 0 c m , P = 5 0 , 0 0 0 N , E = 2 x 1 0 7 N / c m 2

uN

x c m x xc m

5 0 0 0 01 0 5

4 01 2 1 0

1 0

50 0 1 3 8 6

7

,

( )l n .

Solution Concluded

S u b s t i t u t i n g b a c k f o r K ,

uP

B B

Lt E

B B B

B

uP

B B

Lt E

B

B

( )l n

( )l n

2 1

1 2 1

1

2 1

2

1

I n p r o b l e m , t = 1 c m , B 1 = 5 c m , B 2 = 1 0 c m , L = 4 0 c m ,

P = 5 0 , 0 0 0 N , E = 2 x 1 0 7 N / c m 2

uN

x c m x xc m

5 0 0 0 01 0 5

4 01 2 1 0

1 0

50 0 1 3 8 6

7

,

( )l n .

1.9 Lateral Strain and Poisson’s 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 a

compressive load.

Stress Effects

PP

Longitudinal Tensile Stress Effect

Longitudinal Compressive Stress Effect

PP

Poisson’s Ratio

Lateral strain is proportional to the longitudinal strain,

with the constant of proportionality called ‘Poisson’s ratio’ with symbol, .

Mathematically, Lateralstrain

Directorlongitudinalstrain

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

1.10 Thermal Strain

Most structural materials expand when heated,

in accordance to the law: T where is linear strain and is the coefficient of linear expansion; T is the rise in temperature.

That is for a rod of Length, L;

if its temperature increased by t, the extension,

dl = L T.

Thermal Strain Contd.

As in the case of lateral strains, thermal strains

do not induce stresses unless they are constrained.

The total strain in a body experiencing thermal stress

may be divided into two components:

Strain due to stress, and

That due to temperature, T.

Thus: = + T

= E

T

1.11. Principle of Superposition

It states that the effects of several actions taking place simultaneously can be reproduced exactly by adding the effect of each action separately.

The principle is general and has wide applications and holds true if:

(i) The structure is elastic (ii) The stress-strain relationship is linear (iii) The deformations are small.

1.12 General Stress-Strain Relationships

1.12 General Stress-Strain Relationships

F o r t h e e l e m e n t o f m a t e r i a l a s i n F i g u r e a b o v e

s u b j e c t e d t o u n i a x i a l s t r e s s , x , t h e e n s u i n g s t r a i n

i s a s s h o w n i n ( b ) .

S t r a i n i n x d i r e c t i o n ,

xx

E

S t r a i n s i n y a n d z d i r e c t i o n s a s a

r e s u l t o f s t r a i n i n x – d i r e c t i o n

=

x xxa n d

Ee a c h

N o t e : T h e n e g a t i v e s i g n i n d i c a t e s c o n t r a c t i o n .

General Stress-Strain RelationshipsContd.

F o r a n e l e m e n t s u b j e c t e d t o t r i a x i a l s t r e s s e s ,

x y za n d, , t h e t o t a l s t r a i n i n x d i r e c t i o n w i l l b e

d u e t o x a n d l a t e r a l s t r a i n s d u e t o y za n d .

U s i n g t h e p r i n c i p l e o f s u p e r p o s i t i o n , t h e r e s u l t a n t s t r a i n i n x - d i r e c t i o n i s :

xx y z

x x y z

E E E

i eE

. . { ( ) }1

y y x zE

1{ ( ) } G e n e r a l i s e d H o o k e ’ s L a w i n t h r e e d i m e n s i o n s

z z x yE

1{ ( ) }

General Stress-Strain RelationshipsContd.

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 complex strain systems.

Plain Stress and Plain Strain

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

Using strain in Z direction as zero in this case does not mean that stress in the z direction is zero.

Strain Caused by Stress and Temperature

I n a d d i t i o n t o s t r a i n c a u s e d b y s t r e s s , t h e r e m a y a l s o b e t h e r m a l s t r a i n

d u e t o c h a n g e i n t e m p e r a t u r e . T h e g e n e r a l f o r m o f t h e s t r e s s / s t r a i n

r e l a t i o n s i s :

x x y zEt

1{ ( ) }

y y x zEt

1{ ( ) }

z z x yEt

1{ ( ) }

x yx y

y zy z

z xz x

G G G ; ;

Try On Your Own

Show that : v x y zE

1 2( )

ExampleE x a m p l e : A p l a t e o f u n i f o r m t h i c k n e s s 1 c m a n d d im e n s io n 3 x 2 c m i s a c t e d u p o n b y

t h e l o a d s s h o w n . T a k in g E = 2 x 1 0 7 N / c m 2 , d e t e r m in e x ya n d . P o i s s o n ’ s r a t i o i s

0 . 3 . 4 2 k N

y

1 8 k N 2 c m 1 8 k N

x

4 2 k N

3 c m

Solution x

N

c m x c mN c m

1 8 0 0 0

2 19 0 0 0 2/

y

N

c m x c mN c m

4 2 0 0 0

3 11 4 0 0 0 2/

H o o k e ’ s l a w i n t w o d i m e n s i o n s s t a t e s t h a t :

x x yE xx 1 1

2 1 09 0 0 0 0 3 1 4 0 0 0 2 4 0 1 07

6[ ] [ . ( ]

a n d y y xE xx 1 1

2 1 01 4 0 0 0 0 3 9 0 0 0 5 6 5 1 0

76[ ] [ . ( ]

1.13 Relationship between Elastic Modulus (E) and Bulk Modulus, K

It has been shown that : v x y z

x x y z

x y z

x

y z

v x y z

v

v

v

EFor hydrostatic stress

i eE E

Similarly and are eachE

Volumetric strain

E

E

Bulk Modulus KVolumetric or hydrostatic stress

Volumetric strain

i e E K and KE

1

12 1 2

1 2

31 2

31 2

3 1 23 1 2

( )

,

. .

,

,

. .

Maximum Value For Poisson’s Ratio

F r o m t h e e q u a t i o n , i f v = 0 . 5 , t h e v a l u e o f K b e c o m e s i n f i n i t e l y l a r g e .

H e n c e t h e b o d y i s i n c o m p r e s s i b l e . I f v > 0 . 5 , K b e c o m e s n e g a t i v e

i . e . t h e b o d y w i l l e x p a n d u n d e r h y d r o s t a t i c p r e s s u r e w h i c h i s

i n c o n c e i v a b l e . I t m a y b e c o n c l u d e d t h a t t h e u p p e r l i m i t o f P o i s s o n ’ s r a t i o

i s 0 . 5 .

N o t e : KG

a n d E G

2 1

3 1 22 1

W h e r e : G i s S h e a r M o d u l u s

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 or

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 in Compound Bars

I f a c o m p o u n d b a r i s l o a d e d i n c o m p r e s s i o n b y a f o r c e , F ,

S i n c e t h e r o d a n d t u b e a r e o f t h e s a m e l e n g t h a n d m u s t r e m a i n

t o g e t h e r , t h e t w o m a t e r i a l s m u s t h a v e t h e s a m e s t r a i n i . e .

1 2

1

1

2

22

1 2

1

1

S t r a i nS t r e s s

Ei e

E E

E

E. , . . . . . ( )

W h e r e E 1 a n d E 2 a r e t h e e l a s t i c m o d u l i o f m a t e r i a l s 1 a n d 2 r e s p e c t i v e l y .

A l s o : T h e t o t a l l o a d , F m u s t b e s h a r e d b y t h e t w o m a t e r i a l s , i . e . F = F 1 + F 2

W h e r e : F 1 a n d F 2 a r e t h e l o a d s i n t h e i n d i v i d u a l e l e m e n t s .

Compound Bars Contd.

N o w : a s f o rc e = s t re s s x a re a : T h e n : F = 1 1 2 2A A . . . . . . . . . . . . . . . (2 )

W h e re A 1 a n d A 2 a re th e a re a s o f m a te r ia ls 1 a n d 2 re s p e c t iv e ly .

S u b s t itu t in g f o r 2 f ro m E q n . 1 in to E q n 2 :

F AE A

EA

E A

E

F E

E A E Aa n d

F E

E A E A

LNM

OQP

1 11 2 2

11 1

2 2

1

11

1 1 2 22

2

1 1 2 2

1.14.2 Temperature stresses in compound bars

1 1

2 2

L

(a) L1T

1

L2T

2 {b} FL

AE1 1

F 1 F

F 2 F

(c) FL

AE2 2

Temperature stresses in compound bar Contd.

Consider a compound bar, see (a) above of length, L consisting of 2

different materials (1) and (2) having coefficients of expansion

1 and 2 respectively with 1>2. If the bar is subjected to a

1 uniform temperature rise, T and the right hand fixing released,

1the bar (1) will expand more than (2) as shown in diagram (b).

However, because of the end fixing, free expansion cannot occur.

Diagram (c) shows that the end fixing must supply a force which

decreases the length of bar (1) and increases the length of bar (2)

until equilibrium is achieved at a common length.

As no external forces are involved, a self equilibrating

(balancing force system is created).

Temperature Stresses Contd.

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

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

FL

AE1 1

and the increase in length of (2) is FL

AE2 2

.

At Equilibrium:

L TFL

AEL T

FL

AE

i e FAE AE

T

i e AAE AE

EE AAT

T AEE

AE AE

T AEE

AE AE

11 1

22 2

1 1 2 21 2

1 12 2 1 1

1 2 1 21 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 1

2 2

L

(a) L1T

1

L2T

2 {b} FL

AE1 1

F 1 F

F 2 F

(c) FL

AE2 2

Example

A steel tube having an external diameter of 36 mm and an internal diameter of 30 mm has a brass rod of 20 mm diameter inside it, the two materials being joined rigidly at their ends when the ambient temperature is 18 0C. Determine the stresses in the two materials: (a) when the temperature is raised to 68 0C (b) when a compressive load of 20 kN is applied at the increased temperature.

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

3 0 B r a s s r o d 2 0 3 6

S t e e l t u b e

A r e a o f b r a s s r o d ( A b ) = x

m m2 0

43 1 4 1 6

22 .

A r e a o f s t e e l t u b e ( A s ) = x

m m( )

.3 6 3 0

43 1 1 0 2

2 22

A E x m x x N m x Ns s 3 1 1 0 2 1 0 2 1 0 1 0 0 6 5 3 1 4 2 1 06 2 9 2 8. / .

11 5 3 1 0 6 1 0 8

A Ex

s s

.

Solution Contd.

A E x m x x N m x Nb b 3 1 4 1 6 1 0 8 0 1 0 0 2 5 1 3 2 7 1 06 2 9 2 8. / .

13 9 7 8 8 7 3 6 1 0 8

A Ex

b b

.

T x xb s( ) ( ) 5 0 1 7 1 1 1 0 3 1 06 4

W i t h i n c r e a s e i n t e m p e r a t u r e , b r a s s w i l l b e i n c o m p r e s s i o n w h i l e

s t e e l w i l l b e i n t e n s i o n . T h i s i s b e c a u s e e x p a n d s m o r e t h a n s t e e l .

i e FA E A E

Ts s b b

b s. . [ ] ( )1 1

i . e . F [ 1 . 5 3 1 0 6 + 3 . 9 7 8 8 7 3 6 ] x 1 0 - 8 = 3 x 1 0 - 4

F = 5 4 4 4 . 7 1 N

Solution Concluded

S t r e s s i n s t e e l t u b e = 5 4 4 4 7 1

3 1 1 0 21 7 5 1 1 7 5 1

22 2.

.. / . / ( )

N

m mN m m M N m T e n s i o n

S t r e s s i n b r a s s r o d = 5 4 4 4 7 1

3 1 4 1 61 7 3 3 1 7 3 32

2 2.

.. / . / ( )

N

m mN m m M N m C o m p r e s s i o n

( b ) S t r e s s e s d u e t o c o m p r e s s i o n f o r c e , F ’ o f 2 0 k N

ss

s s b b

F E

E A E A

x N x x N m

xM N m C o m p r e s s i o n

' /

. .. / ( )

2 0 1 0 2 1 0 1 0

0 6 5 3 1 4 2 0 2 5 1 3 2 7 1 04 6 4 4

3 9 2

82

bb

s s b b

F E

E A E A

x N x x N m

xM N m C o m p r e s s i o n

' /

. .. / ( )

2 0 1 0 8 0 1 0

0 6 5 3 1 4 2 0 2 5 1 3 2 7 1 01 7 6 9

3 9 2

82

R e s u l t a n t s t r e s s i n s t e e l t u b e = - 4 6 . 4 4 + 1 7 . 5 1 = 2 8 . 9 3 M N / m 2 ( C o m p r e s s i o n )

R e s u l t a n t s t r e s s i n b r a s s r o d = - 1 7 . 6 9 - 1 7 . 3 3 = 3 5 . 0 2 M N / m 2 ( C o m p r e s s i o n )

Example

A composite bar, 0.6 m long comprises a steel 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 the copper bar in order that the extension of each material shall be the same when the composite bar is subjected to an axial load.

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

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 LEL

FL

AE

c s

Thus: F L

A E

F L

A Ec c

c c

s s

s s

Solution Concluded

A l s o : T o t a l f o r c e , F i s t r a n s m i t t e d b y b o t h c o p p e r a n d s t e e l

i . e . F c = F s = F

i eL

A E

L

A Ec

c c

s

s s

. .

S u b s t i t u t e v a l u e s g i v e n i n p r o b l e m :

0 4

4 1 1 0 1 0

0 2

4 0 0 4 0 2 1 0 1 02 2 9 2 2 9 2

.

/ /

.

/ . /

m

d m x N m

m

x x x N m

dx x

m d m m m22

22 2 1 0 0 0 4 0

1 1 00 0 7 8 1 6 7 8 1 6

.; . . .

T h u s f o r a l o a d i n g o f 3 0 k N

S t r e s s i n s t e e l , s

x N

x xM N m

3 0 1 0

4 0 0 4 0 1 02 3 8 7

3

2 62

/ .. /

S t r e s s i n c o p p e r , c

x N

x xM N m

3 0 1 0

4 0 0 7 8 1 6 1 09

3

2 62

/ ./

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.

Elastic Strain Energy Contd.

Figure below shows the load-extension graph of a uniform bar.

The extension dl is associated with a gradually applied load, P

which is within the elastic range. The shaded area represents

the work done in increasing the load from zero to its value

Load

P

Extension

dl

Work done = strain energy of bar = shaded area

Elastic Strain Energy Concluded

W = U = 1/2 P dl (1)

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

Strain = Stress/E

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

Substituting for P and dl in Eqn (1) gives:

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

A L is the volume of the bar.

i.e U = 2/2E x Volume

The units of strain energy are same as those of work i.e. Joules. Strain energy

per unit volume, 2/2E is known as resilience. The greatest amount of energy that can

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

elastic limit stress.