آموزش استاتیک - بخش دوم

: 1-2( ) 2-2( ) 3-2 4-2 5-2

1

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1-1

2

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3

.

.

.

. ( )

:

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4

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5

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( ) 1-2

6

:

( ) ( 1

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:

ft 6ft 12lb 1000ft 24lb 2000

0

E

MC

lb 000,10E

xx CF 0 0xC

yy CF lb 10,000 lb 1000 - lb 20000 lb 7000yC

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Joint A is subjected to only two unknown member forces. Determine these from the

joint equilibrium requirements.

534

lb 2000 ADAB FF CF

TF

AD

AB

lb 2500

lb 1500

There are now only two unknown member

forces at joint D.

DADEDADB

FF

FF

532

CF

TF

DE

DB

lb 3000

lb 2500

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lb 3750

25001000054

54

BE

BEy

F

FF

CFBE lb 3750

lb 5250

375025001500053

53

BC

BCx

F

FF

TFBC lb 5250

lb 8750

37503000053

53

EC

ECx

F

FF

CFEC lb 8750

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checks 087507000

checks 087505250

54

53

y

x

F

F

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: ( 2

.

.

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: FH,GH,GI

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kN 5.12

kN 200

kN 5.7

m 25kN 1m 25kN 1m 20

kN 6m 15kN 6m 10kN 6m 50

A

ALF

L

L

M

y

A

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kN 13.13

0m 33.5m 5kN 1m 10kN 7.50

0

GI

GI

H

F

F

M

TFGI kN 13.13

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kN 82.13

0m 8cos

m 5kN 1m 10kN 1m 15kN 7.5

0

07.285333.0m 15

m 8tan

FH

FH

G

F

F

M

GL

FG

CFFH kN 82.13

kN 371.1

0m 10cosm 5kN 1m 10kN 1

0

15.439375.0m 8

m 5tan

32

GH

GH

L

F

F

M

HI

GI

CFGH kN 371.1

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( ) 3-2

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. 3 . .

= 0 = F

. 2 3 . .

= 0

( 180 ) 2 .

= 0

0

= 0

0

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:

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100N

7

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Pb

aQbQaPM A 0

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4-2

. 15 :

:

( ) ( = ( + ) )-( 2 )( ) ( = ( + ) )-( 2 )

.

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5-2

: CD -1

1)p65

2 )p65

3 )p56

4 )p65

83

= 0 , 4 = 0

sin =9

15=3

5 = 8 sin =

24

5

24

5 4 = 0 =

5

6

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-2

1)CD

2 )BH

3 )HG

4 )BG

89

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: B -3

(1N400

(2N480/7

(3N333/3

(4N427/2

85

tan =8

6=4

3, sin =

4

5, cos =

3

5, =

1

2 = 4

= 0 ( sin ) 4 50 4 4 = 0 = 250

= 0 , 4 50 4 8 = 0 = 400

= 0 , 250 cos = 0 = 250 3

5= 150 = 4002 + 1502 = 427/2

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.

.

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http://faradars.org/fvmec94053