KITES TEAM

STRUCTURE

LECTURE 5

1 2

1

0

0

_

0

X

A B

F

P R R

Super position

PStress

AP

StrainEA

Pa

EA

Thermal Expension coefficient

1

2

1 1 2

1 2

( )

0

T

T

T

L T L L

11

22

1 22

1 2

B

B

B B

R

EA

R

EA

R L R L

EA EA

1 21 2

1 2

1 21 2

1 2

( ) 0

(

B

B

R L LT L L

E A A

R L LT L L

E A A

2

1 2 0 0

( 1)

( )0

B B

B

B

aB a B

B

B

A B

R R

A EAR L

EAR LPa

EA EAP

P P R LL

R P

aR

LPa a

R R P P PL L

P a La L

L

A250mm bar of 15x30mm rectangular cross-section of two aluminum layers .5mm thick ,brazed to a center brass layer of the same thickness . If it is subjected to centric forces of magnetude P=30KN. And Knowing that E of a =70GPa &E of B=105GPa . Determine the normal stress (a ) in the aluminum layers (b) in the brass layer

3

9 6

8

3

9 6

: Is the force in the aluminium layers

F : Is the force in the steel layers

30 (1)

(250)(10 )

70(10 )(10)(30)(10 )

1.19 10

(250)(10 )

105(10 )(10)(5)(10 )

a

S

a S

a a aa

a a

a a

S S SS

S S

S

F

F F KN

F L F

E A

x F mm

F L F

E A

81.587 10

_

1.334

12.85

17.15

85.67

57.17

S

S a

a S

S

a

S

a

x F mm

Combatibility conditon

F F

F KN

F KN

MPa

MPa

Compressive centric forces of 160KN are applied at both ends of the assembly shown by means of rigid plates Knowing that E of S=200GPa and E of a=70GPa determine (a) the normal stresses in the steel core and the aluminum shell,(b) the deformation of the assembly

3

9 2 2 6

9

3

9 2

: Is the force in the aluminium layers

F : Is the force in the steel layers

160 (1)

(250)(10 )

70(10 )( )(62 25 )(10 )4

1.4126 10

(250)(10 )

200(10 )( )(254

a

S

a S

a a aa

a a

a a

S S SS

S S

F

F F KN

F L F

E A

x F mm

F L F

E A

6

9

4

)(10 )

2.5465 10

_

1.8027

57.1

102.9

116.32

40.7

1.454(10 )

S S

S a

a S

S

a

S

a

S a

x F mm

Combatibility conditon

F F

F KN

F KN

MPa

MPa

m

Three steel rods (E=200GPa) support a 36-KN load P Each of the rods AB and CD has a 200mm2 cross-sectional area .Neglecting the deformation of rod BED . Determine (a) the change in length of rod EF (b) the stress in each rod

3

9 6

6

3

36

(2 ) ( ) 36( ) 0

0.5(36 )

_

0.5(36 )(1000)(500)(10 )

200(10 )(200)(10 )

6.25(36 )(10 )

(1000)(400)(10 )

20

Y CD BA EF

B CD EF

CD EF

AB CD

D D EFB D

D D

D EF

E E EFE

E E

F F F F KN

M F a F a a

F F

from symmetry F F

F L F

E A

F m

F L F

E A

9 6

6

0(10 )(625)(10 )

3.2 10

sin _ _ _ _

_

23.81

6.095

38.096

30.47

0.0762

E EF

E D

EF

CD AB

EF

AB

E

F x m

ce the BED is rigid

Combatibility conditon

F KN

F F KN

MPa

MPa

mm

Two cylindrical rods , one of steel and the other of brass ,are joined at c and restrained by rigid support at Aand E . For the loading shown and Knowing that E of s=200GPa and E of b=105GPa , determine (a) the reactions atAand E(b) the deflection of point C

3

9 2 6

5

3

9

1-Eliminating the support reaction_at_E

0, 40

40 , 100

(100)(1000)(180)(10 )

200(10 )( )(40 )(10 )4

7.162(10 )

(40)(1000)(120)(10 )

200(10 )( )(4

ED CB

CD AB

BA BABA

BA BA

BA

CB CBCB

CB CB

P P KN

P KN P KN

F L

E A

m

F L

E A

2 6

5

3

9 2 6

5

41

40 )(10 )

1.91 10

(40)(1000)(100)(10 )

105(10 )( )(30 )(10 )4

5.3894 10

01.446(10 )

CB

DC DCDC

DC DC

DC

E BA CB DC

x m

F L

E A

F x m

m

2

62

2 2

92

2 1

2-Deflection_at_E due_to the support reaction

300 200(10 )

200( )(40 ) 105( )(30 )4 4

3.888 10

0

37.19

100 37.19 62.8

AC EDE E

AC AC ED ED

E E

E E

E E

E

A

EC BA BC

L LR

E A E A

R

x R

R KN

R KN

R

3 35 5

9 2 6

5

(37.19)(10 )(300)(10 )7.162(10 ) 1.91(10 )

200(10 )( )(40 )(10 )4

4.633(10 )

AC

AC AC

C

C

L

E A

m

A rod consiststing of two cylindrical portions AB and BC is restrained at both ends potion AB is made of steel and portion BC is made of brass Knowing that the rod is initially unstressed determine the comperssive force induced in ABC when there is a temperature rise of 50C

6 3

4

6 3

4

1

1-Eliminate_the_ reaction_force_at_A

11.7(10 )(50)(250)(10 )

1.4625(10 )

20.9(10 )(50)(300)(10 )

3.135(10 )

4

_ _

_ _

.597

A AB BC

AB S AB

AB

BC B BC

BC

A

USING THE PRINICIPLE

OF SUPPER

T

PO

L

m

TL

SI N

m

TIO

45(10 )m

6200 , 11.7 10s sE GPa x C 6105 , 20.9 10b bE GPa x C

2

3 3

29 2 6 9 2 6

92

1 2

2-Deflection_at_A due_to the support reaction

250(10 ) 300(10 )

200(10 )( )(30 )(10 ) 105(10 )( )(50 )(10 )4 4

3.224 (10 )

142.6

BCABA A

AB AB BC BC

A A

A A

A A A

A

LLR

E A E A

R

R m

R KN

Knowing that a 0.5mm gap exists when the temperature is24C, determine (a) the temperature at which the normal stress in the aluminum bar will be equal to 75MPa (b) the corresponding exact length of the aluminum bar

2

6

1500

105

21.6 10

Bronze

A mm

E GPa

x C

2

6

min

1800

73

23.2 10

Alu um

A mm

E GPa

x C

6 3

5

t:Is the bar total displacement due to temperature rise

T:Is the change in the temperature

1-The elong

23

atio

.2(1

n of the

0 )(4

bar due to

50)(10 )

temperature r

1

is

.044 10

21.

e

a a a

a

b b b

TL

Tx m

TL

6 3

6

5

6(10 )(350)(10 )( )

7.56 (10 )

1.8 (10 ) (1

b

t a b

T

T m

Tx m

6

9

2

2-Deflection of the due to the reaction

450 350(10 )

73(1800) 105(1500)

5.647 (10 )

555.56 /

Forin ducing -75 MPa of in the alumium bar

,the reaction

a bR

a a b b

R

R

aa

L LR

E A E A

R

R m

RRN m

A

force must be

135R KN

3

3

_

_ Re

_

Due to the 0.5mm GPa ,the compatibility equation will be :

0.5 10

1.2623 10

70

24 94

3-The delection in the aluminum rod

0.7308

0.4623

t R

t

Surrounding

a thermal

a action

a a ther

x m

x m

T C

T T C

mm

mm

_ Re 0.2685

450.2685

mal a action

a

mm

L mm