Stress

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STRESS AND STRAIN

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Topics:

• Introduction• Main Principles of Statics

StressStress• Normal Stress• Shear Stress• Bearing Stress• Thermal Stre

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

Incompressible Compressible

41.1 Introduction

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.

Dynamics :1. Kinematics – concerned with the geometric aspects of the motion2. Kinetics – concerned with the forces causing the motion.

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External Loads

Surface Forces- caused by direct contact of one body withthe surface of another.

Body Force- developed when one body exerts a force on another body without direct physical contactbetween the bodies.- e.g earth’s gravitation (weight)

concentrated force

51.2 Main Principles of Statics

linear distributed load, w(s)

Axial LoadNormal StressShear StressBearing StressAllowable StressDeformation of Structural under Axial LoadStatically indeterminate problemh lThermal Stress

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Mechanics of material is a study of the relationship between the external loads applied to a deformable body and the intensity of i t l f ti ithi th b dinternal 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 inStrain describe the deformation by changes in length of line segments and the changes in the angles between them

• Normal Stress : stress which acts perpendicular, or normal to, the(σ) cross section of the load-carrying member.( ) y g

: 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.

81.1 Introduction

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Normal Stress, σthe intensity of force, or force per unit area, acting normal to ΔAnormal to ΔA

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

σ = P / A

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

(a)

(b)

•Unit: Nm -² •N/mm2 or MPa

Stress ( σ ) = Force (P)

Cross Section (A)

•N/mm2 or MPaN/m2 or Pa

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Assumptions :1. Uniform deformation: Bar1. 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

111.4 Axial Loading – Normal Stress

de o at o , o ce beapplied along centroidal axis of cross section C

AFFFAzRz =Σ=↑+ ∫ ∫ σ dd;

APAPA

=

=∫ ∫

σ

σ

σ = average normal stress at any point on cross sectional area

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on cross sectional areaP = internal resultant normal forceA = cross-sectional area of the bar

1.4 Axial Loading – Normal Stress

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• Use equation of σ = P/A for cross-sectional area of a member when section subjected to internal resultant force Psection 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

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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(b) rod BC.

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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)average normal stress in the mid section of (a) rod AB, (b) rod BC.

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Normal strain, ε is the elongation or contraction of a line segment per unit of length ε = ΔL / Lo

ΔL = elongationLo = length

δ strain normalL

=δ

=ε

* ΔL= δ

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Example 1.3:Determine the corresponding strain for a bar of length L=0.600m and uniform cross section which undergoes a deformation δ=150×10-6m.

66150 10 m 250 10 m m

L 0 600m/

.

−−δ ×

ε= = = ×

6250 10 250@−= × μ

1.4 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?

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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 lengthup with 4m length.

70kN 70kN

2.8m

Tensile test is an 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 mechanicalproperties of materials that can be deducedfrom the stress-strain diagram are illustratedin figure above.

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Point O-A = linear relationship between stressand strain

Point A = proportional limit (σPL)The ratio of stress to strain in this linear regionof stress-strain diagram is called Young Modulusor the Modulus of Elasticity given

σΔΔ

=Ε σ < σAt 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 increasePoint C-D = refer as the zone of strain hardeningPoint D = ultimate stress/strength ; specimen

b i t k d

εΔ σ < σPLUnit: MPa

begins to neck-downPoint E = fracture stress

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Point O to A

Point C to D

Point D to EPoint D to E

At point E

Normal or engineering stress can be determinedby dividing the applied load by the specimenoriginal cross sectional areaoriginal cross sectional area. True stress is calculated using the actual crosssectional area at the instant the load ismeasured.

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Some of the materials like aluminum (ductile), does not have clear yield point likesstructural steel. Therefore, stress valuecalled the offset yield stress, σYL 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 diagramportion of the stress strain diagramSecond line is then drawn parallel to the original line but offset by specified amount of strainThe intersection of this second line with

the stress-strain curve determine theoffset yield stress.

Commonly used offset value is 0.002/0.2%

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Brittle material such as ceramic and glass have low tensile stress value but high incompressive stress. Stress-strain diagram for brittle material.

Example 1.6The 4 mm diameter cable BC is made of a steelwith E=200GPa. Knowing that the maximumstress in the cable must not exceed 190MPaand that the elongation of the cable must notexceed 6mm, find the maximum load P that canbe applied as shown

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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 i id b l iis said to be elastic.If 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 unloadingunloading .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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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 outcompression, 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,

axial

lateralvεε

=

where the strains are caused by uniaxial stress only

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paksi x

sisi y

LLb db d

@

@

δε ε =

δ δε ε = − = −

Example 1.7A prismatic bar of circular cross-sectionis loaded by tensile forces P = 85 kN. Thebar has length of 3 m and diameter of 30mm. It is made from aluminum with modulusof elasticity of 70 GPa and poisson's ratio ν= 1/3. Calculate the elongation and thedecrease in diameter Δd.

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Example 1.8A 10 cm diameter steel rod is loaded with 862 kN by tensile forces. Knowing that the E=207 GPa and ν= 0.29, determine the deformation of rod diameter after being loaded. Solution

in rod σ MPaNxp 710910862 3

==σ in rod, σ =

Lateral strain,

MPamA

7.109)1.0(

41 22

==π

00053.010207

7.1093 ===

MPaxMPa

Eaσε

)000530(29)( ol −=−= ενε∴

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)00053.0(29.)( oal == ενε000154.0−=

)1.0)(000154.0()( −==Δ Dd lεcm00154.0−=

Exercises 11. A steel pipe of length L=1.2 m, outside diameter d2=150mm and

inside diameter d1=110mm is compressed by an axial force P= 620kN.The material has modulus of elasticity E= 200GPa and Poisson’s Ratio v = 0.30.Determine :

a) the shortening, δ ( ans :-0.455 mm)b) h l l i ( 113 9 10 6)b) the lateral strain,ε lateral (ans: 113.9x10-6)c) the increase ∆d2 in the outer diameter and the increase ∆d1 in the inner diameter

(ans: 0.0171 mm and 0.0125mm)d) the increase ∆t in the wall thickness

(ans: 0.00228 mm)

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2. A hollow circular post ABC as shown in Figure 2 supports a load P1=7.5 kN acting at the top. A second load P2 is uniformly distributed around the cap plate at B. The diameters and thicknesses of the upper and lower parts of the post are dAB=32 mm, tAB= 12mm, dBC 57 mm and tBC=9mm, respectively.a) Calculate the normal stress, σAB in the upper part

of the post (ans: 9 95 MPa)of the post. (ans: 9.95 MPa)b) If it is desired that the lower part of the post

have the same compressive stress as the upper part, what should be the magnitude of the load P2? (ans : P2=6kN)

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3. 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. D i h d l f l i h d l fDetermine the modulus of elasticy, the modulus of rigidity, and Poisson’s ratio of the material.

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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, τ, g ,

τ = shear stressV = shear forceA = 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

AV

=τ

Shear stresses can also arise indirectly as a result of tension, torsion or bending of a member.

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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

2dP

AV

πτ ==

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Rivet in Double Shear

Example 1.9For the 12 mm diameter bolt shown in the bolted joint below,

determine the average shearing stress in the bolt

222

)4

(2 dP

dP

AV

ππτ ===

determine the average shearing stress in the bolt.

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Single ShearDouble Shear

AF

AP==aveτ

AF

AP

2ave ==τ

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The effect of shear stress is to distort the shape of a body by inducing shear strains The shear strain, γ is a measure of the angular distortion of the body.

L

Vx

γ

Lx

=γ

(units: degrees, radians)

L

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Bearing stress is also known as a contact stressBearing stress in shaft key;

MMP 2

Bearing stress in rivet and plat;

rhLM

LhrM

AP

bb

2)2(

===σ

Pσtdb =σ

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Example 2.0A punch for making holes in steel plates is shown in the figure. Assume that a punch having diameter d=20 mm is used to punch a hole in an 8 mm plates, what is the average shear stress in the plate and the average compressive stress in the punch ifand the average compressive stress in the punch if the required force to create the hole is P = 110kN.

. P

20 mm

8 mm

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It also known as Shear Modulus of Elasticity or the Modulus of Rigidity.Value of shear modulus can be obtained from the linear region of shear stress-strain diagram.

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

γτ G= Unit : Pa

)1(2 ν+=

EG

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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, g p p p g ,b and c in the x, y and z directions, respectively.

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The tensile force P causes an axial elongation of aεand lateral contractions of bνε and cνε in the x, y, andz directions respectively. Hence,

Initial b d

Initial volume of body V abc

body

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

= abc(1 + ε)(1 - νε)2

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

Final volume, Vf = abc(1 + ε - 2νε)

Change in volume, ΔV = Final Volume - Initial Volume

= abc(1 + ε - 2 νε) - abc = abc(1 + ε - 2 νε - 1) = abc(ε - 2 νε) = Vo ε(1 - 2 ν)

Hence,

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)21(

)21(

νσε

νε

−=

−=Δ

E

VV

o

I i i l i bj d l i i lIsotropic material is subjected to general triaxialstress σx, σy and σz.Since all strain satisfy ε << 1, so εv = εx + εy + εz

εx =

εy =

[ ])(1zyxE

σσνσ +−

[ ])(1zxyE

σσνσ +−

εz =

[ ])( zxyE

[ ])(1yxzE

σσνσ +−

)(21zyxv E

σσσνε ++−

=

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Example 2.1

A titanium alloy bar has the following original dimensions: x =10cm; y = 4cm; and z = 2cm. The bar is subjected to stresses σx= 14 N and σy = - 6 N, as indicated in figure below. Theremaining stresses (σz, τxy, τxz and τyz) are all zero. Let E = 16kN and ν = 0.33 for the titanium alloy.(a)Determine the changes in the length for

Δx, Δy and Δz.

(b) Determine the dilatation, εv.

y 6 N

z

x14 N14 N

6 N

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

One method of specifying the allowable load for the design oranalysis of a member is use a number called the Factor of Safety (FS).

Allowable-Stress Design

allow

fail

FF

FS =

FS > 1

or yieldyield ττ

σσ

FSor

FSy

allowy

allow τσ ==

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If a bar is fixed at both ends, as shown in fig. (a), two unknown axial reactions occurs, and the force equilibrium equation becomes; ;0FΣ↑

;

0PFF

;0F

AB

y

=−+

=Σ↑+

• In this case, the bar is called statically indeterminate, since the equilibrium equation are not sufficient to determine the reactions.

• the relative displacement of one end of the bar with respect to the other end is equal to zero

0B/A =δ

• the relationship between the forces acting on the bar and its changes in length are known as force-displacement relations

with respect to the other end is equal to zero since the ends supports are fixed. Hence;

AEPL,0B/A =δ=δ 0BA =δ+δ

• Realizing that the internal force in segment AC is +FA and in segment CB

BAAB FPF,0PFF −==−+

AC

CBBA

CBBACA

CBBACA

LFLAE

AELFF

AELF

AELF

0AELF

AELF

×=

=

=−

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

=−

1LLFP

FL

LFP

LLFFP

AC

CBB

BAC

CBB

AC

CBBB

Realizing that the internal force in segment AC is +FA, and in segment CB, the internal force is –FB. Therefore, the equation can be written as;

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

LLFP

LLLFP

LL

LLFP

B

AC

ACCBB

AC

AC

AC

CBB

AC

CBBA L

LFF =⎠⎝ LAC

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎠

⎜⎝

LL

PF

L

ACB

AC

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Example 2.2:Example 2.2:

3X A B

3B A

F 0 F F 20 10 N 0 1

F 20 10 F

, ( ) ................( )

( )

+ → Σ = − − + =

= −

( ) ( )

B A

A B

A AC B CB

A2 29 2 9 2

0 001m0 001m

F L F L0 001m

AE AEF 0 4m FB 0 8m 0 001m

0 0025m 200 10 Nm 0 0025m 200 10 Nm

/ ..

.

( . ) ( . ) .. .− −

δ =

δ − δ =

− =

− =⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤π × π ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

A B

orF 0 4m F 0 8m 3927 0N 2Substitute eq 1 oeq 2F

( . ) ( . ) . ................( )( )int ( )

− =

A A

A

0 4m 20 000N F 0 8m 3927 0NF 16 6kNFB 3 39kN

( . ) ( , )( . ) ..

.

− − =

=

=

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Example 2.3:Example 2.3:

Solution:Solution: 3y A C E

C3

A E

F 0 F F F 1 5 1 0 N 0 1

C C W M 0

F 0 4 1 5 1 0 0 2 F 0 4 0 2

, ( ) . . . . . . . . . . . . . . . . ( )

( . ) ( ) ( . ) ( . ) . . . . . . . . . . . ( )

+ ↑ Σ = + + − =

+ =∑

− + + =

The applied load will cause the horizontal line ACE move to inclined line A’C’E’

δ δδ δ C EA E

C E A E

A EC E

A EC E

C A E

C CD A AB E EF

0 8 0 4

0 4 0 8

0 40 8

0 4 0 40 8

0 5 0 5

F L F L F L0 5 0 5

. .

. .

..

. ..

. .

δ − δδ − δ=

δ − δ δ − δ=

δ − δδ − δ = ×

δ − δδ = + δ

δ = δ + δ

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥C CD A AB E EF

5 5 5st st st

C A5 5

st st

0 5 0 51 5 10 E 2 5 10 E 2 5 10 E

F 0 5 F 0 50 51 5 10 E 2 5 10 E

. .. . .

( . ) ( . ).. .

− − −

− −

= +⎢ ⎥ ⎢ ⎥× × ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡=

× ×⎣

E5

st3 3 3

C A E3 3

A EC 3

C A E

F 0 50 52 5 10 E

33 33 10 F 10 10 F 10 10 F

10 10 F 10 10 FF33 33 10

F 0 3F 0 3F eq 3

( . )..

.

.. . ................. ( )

−

⎤ ⎡ ⎤+⎢ ⎥ ⎢ ⎥

×⎢ ⎥ ⎢ ⎥⎦ ⎣ ⎦

× = × + ×

× + ×=

×= +

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3y A C E

C3

A E

C A E

3

F 0 F F F 15 10 N 0 1

CCW M 0

F 0 4 15 10 0 2 F 0 4 0 2F 0 3F 0 3F eq 3

Substituteeq 3 o eq 1

, ( ) ................( )

( . ) ( )( . ) ( . ) ...........( ). . ................. ( )

( ) int ( )

+ ↑ Σ = + + − =

+ =∑

− + + =

= +

3A E

3 3A A

Substituteeq 4 o eq 2

F 0 4 15 10 0 2 F 0 4 0

F 0 4 3 10 0 4 11 538 10 F 0

( ) int ( )

( . ) ( )( . ) ( . )

( . ) ( ) ( . ) . ( )

− + + =

⎡ ⎤− + + − =⎣ ⎦3A C E

A

F F F 15 10 N 0 1

F

( ) ................( )+ + − =3

A E E3

A E3

AE

3E A

0 3F 0 3F F 15 10 0

1 3F 1 3F 15 10

15 10 1 3FF1 3

F 11 538 10 F eq 4

( . . ) ( )

. . ( )

( ) ..

. ( ) ....................... ( )

+ + − =

+ =

−=

= −

A A

3 3A A

3

A

3

F 0 4 3 10 4 615 10 0 4F 0

7 61510F0 8

9 519 109 52kN

( ) ( ) ( ) ( )

( . ) ( ) . ( ) .

..

. ( ).

⎣ ⎦

− + + − =

−=

−

==

Aplace F 9 52kN o eq 49 52kN

Re . int ( ).=

=3

E A3 3

E

C A E3 3

F 11 538 10 F

11 538 10 9 52 102 02kN

place F 2 02kN o eq 3F 0 3F 0 3F

0 3 9 519 10 0 3 2 02 103 462 kN

. ( )

. ( ) . ( ).

Re . int ( ). .

. ( . ( ) . ( . ).

= −

= −=

=

= +

= + ×=

A change in temperature can cause material to change its dimensions.If the temperature increases, generally a material expands, whereas if the temperature decreases, the material will contract.If this is the case, and the material is homogenous and isotropic, it has been found from experiment that the deformation of a member having a length L can be calculated using the formula;

δT=αΔTLWhere

α=linear coefficient of thermal expansion (unit: 1/C°)

ΔT=change in temperatureL=original length of the memberδT=change in length of the member

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Example 2.4:Example 2.4:

Given: α=12x10-6/C°

FFF0F

BA

Y===Σ↑+

Solution:Solution:

δAB 0=δ The change in length of the bar is zero (because the supports do not move)

AB T F( )+ ↑ δ = δ −δ

To determine the change in length, remove the upper support of the bar and obtain a bar is fixed at the base and free to displace at the upper end.So the bar will elongate by an AB T F( ) So the bar will elongate by an amount δT when only temperature change is actingAnd the bar shortens by an amount δF when only the reaction is acting

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AB T F

T F 0FLTL 0AE

F 1

( )

( )

+ ↑ δ = δ −δ

δ −δ =

αΔ − =

62 9

42 9

4 2 9

F 112 10 60 30 1 00 01 200 10

F 13 6 100 01 200 10

F 3 6 10 0 01 200 107 2kN

( )( )( ). ( )

( ).. ( )

. . ( ).

−

−

−

× ° − ° − =×

× =×

= × × ×=

Average normal thermal stress:

2F 7 2kN 72MPaA 0 01

.;.

σ= =

Average normal thermal stress:

Example 2.5Example 2.5

Given:

6st 12 10 C/−α = × °st

6al

9st

9al

12 10 C

23 10 C

E 200 10 Pa

E 73 1 10 Pa

/

/

.

−=

α ×

α × °

= ×

= ×

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3y st alF 0 2F F 90 10 N 0 eq 1, ( ) ......... ( )+ ↑ Σ = + − =

st al

st st T st F

al al T al F

eq 2............................... ( )

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( )

δ = δ

+↑ δ = δ − δ

δ = δ − δ

δ δ δ δst T st F al T al F

st al

st al

6 st2 9

6

F L F LTL TLA E A E

F 0 2512 10 80 20 0 25

0 02 200 10

23 10 80 20

( ) ( ) ( ) ( )

( . )( )( . )( . ) ( )

(

−

−

δ − δ = δ − δ

αΔ − =αΔ −

× °− ° − =π ×

× °− ° al2 9

4 4

F 0 250 250 03 73 1 10

F 0 25 F 0 25

( . ))( . )( . ) ( . )

( ) ( )

−π ×

4 4st al6 6

4 10 4 9st al

10 4 9 4st al

st

F 0 25 F 0 251 8 10 3 45 10

251 327 10 206 685 10

1 8 10 9 947 10 F 3 45 10 1 21 10 F

9 947 10 F 3 45 10 1 21 10 F 1 8 10

1 65F

( . ) ( . ). .. ) ( . )

. . . .

. . . .

.

− −

− − − −

− − − −

× − = × −× ×

× − × = × − ×

− × = × − × − ×

×=

4 9al

10

3al

10 1 21 10 F9 947 10

165 88 10 1 216F eq 3

..

. . ............... ( )

− −

−− ×

− ×

= − × +

3st al

3 3al al

3 3al al

Substituteeq 3 o eq 1

2F F 90 10 N 0

2 165 88 10 1 216F F 90 10 N 0

331 76 10 2 432F F 90 10 N 0

( )int ( )

( )

( . . ) ( )

. . ( )

+ − =

− × + + − =

− × + + − =3

al

al

al3

st

3 432F 421 76 10F 122 89 kN

Substitute F 122 89kN o eq 3

F 165 88 10 1 216F

. ..

. int ( )

. .

= ×

=

=

= − × + al3 3165 88 10 1 216 122 89 10

16 445kN. . ( . )= − × + ×

= −16 445kN.=

The negative value for F steel indicates that the force acts opposite to arrow shown.THE STEEL POSTS ARE IN TENSION and ALUMINIUM POSTS IS IN COMPRESSION

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TUTORIAL 1

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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Answer, RA= 323 kN, Rb= 577kN

TUTORIAL 2Two cylindrical rods, CD made of steel (E=200 GPa) and AC made of aluminum (E=72 GPa), are joined at C andrestrained by rigid supports at A and D. Determine (a) the reactions at A and D (RA=52.9kN, RD= 87.1 kN)(b) The deflection of point C (0 086 mm)(b) The deflection of point C (0.086 mm)

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TUTORIAL 3At room temperature (21oC) a 0.5 mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 1600C, determine (a)The normal stress in the aluminum rod (σa =-150.6 ( ) ( aMPa)(b)The change in length of the aluminum rod (δa= 0.369 mm)

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