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xσ

yσ

z σ

yxτ

yz τ

xyτ

xz τ zx

τ zy

τ

Chapter.3 Axially Loaded Bars

E-mail: modish33@naver.com

HP: 010-9249-5551

2014 2 ( )

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1. :

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1. :

0,0

A

zdA z

A

ydA y A A

y, z

y, z

0

0

A A

A A

zdAdA z

ydAdA y

, y z 0 A P

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1. :

• P

•

• ,

A P

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2. Example 3-1• A 2in , B 4in . (a) A , (b) B

.

• Solve)

A

A

kip4 P

psi1273)1(

40002

A P

B

kip8,0412 P P

psi637)2(

80002

A P

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3. Example 3-2• A=400mm 2 . m=3400kg .

?

• Solve

B ( )

6.26)8/4arctan(8/4tan

mg P

mg mg mg P

mg P F

P P F

BC

BD

BD y

BD BC x

2

24.26.26sinsin

0sin

0cos

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3. Example 3-2• ,

• m=3400kg, g=9.81m/s 2 A=400X10 -6m2 ,

BC BD CD CE DE

2mg -2.24mg -1.33mg 2.40mg -2mg

BC BD CD CE DE

(MPa) 167 -186 -111 200 -167

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4. : •

• P . θ P .

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2cos

0cos)cos( A A

.

θ ,

A

->

->

cossin

0sin)cos( A A

4. :

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θ =0~180°

.

,

45°. 2/

Prove)

0 ,

)cos(sin 22

d d

cossin

2/135

2/45

4. :

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5. Example 3-3• . 200N .

? ( ?)

• Solve)

Pa000,400)05.0)(01.0(

200 A P

: 40°

Pa000,23540cos)000,4000(cos 22

Pa000,19740cos40sin)000,400(cossin

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6. :

L

L L L

'

• ( )

: (modulus of elasticity) (Young’s modulus)

- ( / )

-

• :

E E

(GPa)

Al 70

Cast iron 80~170

Concrete ( ) 18~30Mg 40

Rubber 0.0001~0.004

Steel 190~220 EA PL

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6. : • .

• ( , )

dLdLdL '

lat L L L

L L

latlat

lat

'

)1('

Ex.)

D D )1(' lat

(Poisson’s ratio)

-

lat

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7. Example 3-4• L=200mm , D=10mm P=16 dmf .

, L’=200.60mm, D’=9.99mm . ?

•

Solve)

MPa7.203)005.0(

160002

A P :

:

:

003.0200

20060.200' L

L L

001.010

1099.9'lat

D

D D

:

GPa9.67003.0

107.203 6

E

333.0003.0001.0lat

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8. Example 3-5

• A=0.4in 2. , E=12X10 6psi . 10kip B B?

•

Solve)

1.53)12/16arctan(

kip5.12

0sin10

P

P F y

( )

in.0521.0)4.0)(1012()20)(500,12(

6

EA PL

※ Psi: pound per square in.

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9.

• :

• : statically indeterminate

• , , ( )

• :

• Ex.) A B . .

0=+ B A P P F -

A

A A A EA

L P δ =

B

B B B EA

L P δ =

,

0=+ B A δδ : 0

( : compatibility condition)

0=+ B

B B

A

A A

EA L P

EA L P

B A A B B

A B B A A A L A L

F P

A L A L F

P /+1

=,/+1

=-

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9.

• , .

• , .

1)

2)

3)

: /

A

A A A EA

L P δ =

B

B B B EA

L P δ = -

0=+ B A δδ

0=+ B A P P F -

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9.

• (or )

1) - .

2) .

3) .

- .

• (or )

1) - .

2) .

3) . -

.

A A A A

A A P f P EA

Lδ == B B B

B

B B P f P EA

Lδ ==

B B B A A A EA L f EA L f /=,/= (flexibilities)

0=+ B B A A P f P f

A P B P

A A A A

A A δk δ L

EA P == B B B

B

B B δk δ L

EA P ==

B B B A A A L EAk L EAk /=,/= (stiffness)

0=+ B B A A δk δk F -

Aδ

Bδ

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10. Example 3-6

• E=12X10 6psi . A L A=10in. , 2in. . B LB=8in. , 4in. . A , B 0.02in. . 160kip A B , A B

?

• Solve)

lb000,160

0000,160

A

A x

P

P F

1) 2) A

.in0424.0)1)(1012(

)10)(000,160(26

A

A A A EA

L P

, , .b A

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B

10. Example 3-6

3) 0000,160 B A x P P F

4) - B

B B B

A

A A A EA

L P EA

L P ,

5) b B A

02.0)2)(1012(

)8(

)1)(1012(

)10(2626

B A

B

B B

A

A A

P P

b EA

L P EA

L P

lb500,70lb,500,89 B A P P

6)

psi610,5)2(

500,70

psi490,28)1(

500,89

2

2

B

B

B

A

A A

A

P

A P

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11. Example 3-7

• (E Al=10.0X106psi) , . (E Fe=28.5X106psi), b=0.02in. .

A=0.5in2. , L=10in. . , ?

• Solve) . .

. b .

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11. Example 3-71) 02 FeAl P P F x

2) - A E

b L P A E L P

Fe

FeFe

Al

AlAl

)(,

3) b AlFe

4) - ,

b A E L P

A E b L P

Al

Al

Fe

Fe )(

5) ,

lb760,11lb,880,5 FeAl P P

6) ,

psi520,235.0

760,11 psi760,115.0

5880

FeFe

AlAl

A P A

P

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12.

:

( ) ,

)( x A

dx xdA /)(

)( x A P

)(

' x EA

Pdxdxdx L x EA

Pdx L L0 )(

'

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12.

qdx(dx) :

q : /

x

2

30

2

0

3

0

L x

Lq

P

dx

L

xq P

L

x

(dx)

dx L x

L EAq

EA Pdx

2

30

element 3

L

EA Lqdx

L x L

EAq

0

302

30

43

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13. Example 3-8

• E=120GPa . A(x)=0.03+0.008x 2m2

. P=200MN , (a) x=1m (b) .

•

Solve)x=1m : 22 m038.0)1(008.003.0)1( A

x=1m : MN526038.0

1020)1(

6

A P

: 2

0 29

6

0mm62.8

)008.003.0)(10120()1020(

)( xdx

x EA Pdx L

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14. Example 3-9

• 1ft E=2X10 6psi . F=480kip.

. . (a) , (b) .

.in12ft1

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14. Example 3-9

• Solve)

lb/in.000,4120/

120

0120

0

F q

q F

F qdx

x

lb000,400

xqx P P qdx

x

: lb000,480 P

: psi240,4)6(

000,4802

A

P

x=0 x=120in. ,

X=120in. x=192in. .

lb000,4 x P

lb000,480 P

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14. Example 3-9

.in153.0)6)(102()72)(000,480(

26above

EA PL

(dx)

EA xdx

EA Pdx 000,4

element

.in127.0)120(21

)6)(102(4000

214000000,4

226

120

0

2120

0 below

x EA

dx EA

x

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14. Example 3-10

• . L, A, ,E , ?

• Solve )

x

x

A P

Adx P

00

0,

W AL

(dx) : EA

dx A EA Pdx x element

: EA

WL EA

AL EA

dx A L x22

2

0

: EA

WL L L

2