2005/2006 RETM YILI GZ YARIYILI MUKAVEMET 1 DERS

FNAL SORU VE CEVAPLARI

SORU 1 ekildeki gerilme durumunda; a) Asal gerilmeleri ve dzlemlerini bularak eleman zerinde gsteriniz. b) Maksimum kayma gerilmesi ve dzlemini bularak eleman zerinde gsteriniz. c) C dzlemindeki gerilmeleri bulunuz.

300

C

80 MPa

16 MPa

24 MPa

Mohr emberi yntemi ile zm:

maks.

min

ort

y

x

2B

YR

CO A

X

ED

a) X(16,-24), Y(80,24)

16 80 48 482 2

x yort ortOC MPa MPa

+ += = = = =

80 48 32 32CD OD OC MPa CD MPa= = = = ,

2 232 24 40 40R MPa R MPa= + = =

max max48 40 88 88OC R MPa MPa = + = + = =

min min48 40 8 8OC R MPa MPa = = = =

( )12 cos 48 18 40 36.88 18, 44 = = = D b) Maksimum kayma gerilmesi Mohr emberinde F noktasndadr ve min n bulunduu B noktasndan itibaren saat ynnde 90o gelinerek ulalabilir.

o90

maxF

ABO C

max 40 = =R MPa Maksimum kayma gerilmesinin bulunduu dzlem;

2 90 45 = =

a80 MPa

24 MPa

=18.44o p

y

x16 MPa =88 MPamax

min =8 MPa=40 MPamax

=8 MPamax

=45os

(a)

(b)

(a) Asal gerilmelerin ve (b)Maksimum kayma gerilmesinin eleman zerinde gsterilmesi.

TN

60o

'

'

Y

X

X

Y

ABO CR

c) , , = +x OC CT

' 60 36.88 90 6.88CX T = + =

, 48 sin 6.87 52.78x MPa = + =

, 40cos6.87 39.71xy MPa = = Formlasyon yntemi ile zm:

a) 2 2

2 2.

16 80 16 80 24 882 2 2 2

x y x ymaks xy MPa

+ + = + + == + + = ,

2 22 2

min.16 80 16 80 24 8

2 2 2 2x y x y

xy MPa + + = + == + =

2 2 24tan 2 0.75 18.44

16 80xy

x y

x = = = = D

b) 2 2

2 2max

16 80 24 402 2

= + = + = x y

xy MPa

16 80tan 2 1.33 26.56

2 2 24x y

xy x

= = = = D

c) '16 80 16 80cos 2 sin 2 cos 60 24sin 60 52.78

2 2 2 2x y x y

x xy MPa + + = + + = + + =

' 52.78x MPa =

' 16 80sin 2 cos 2 sin 60 24cos 60 39.712 2

x yxy xy MPa

= + = + = ' 39.71xy MPa =

a

80 MPa

=18.44o p

y

xmax

min

16 MPa =88 MPa

=8 MPa

(a)

24 MPa(b)

=40 MPamax

s=26.56o

(a) Asal gerilmelerin ve (b) Maksimum kayma gerilmesinin eleman zerinde gsterilmesi.

SORU 2 ekildeki ykleme durumu iin n-n kesitinin; a) a noktasndaki, b) b noktasndaki kayma gerilmelerini bulunuz. 0.75m 1.2 m

200 kN 200 kN

0.75m

A B

n

n

50 mm

150 mm

75 mm

a

b

75 mm 75 mm

50 mm

50 mm

DC

1.2 m 0.75 m0.75 m

200 kN200 kNKirie etkiyen kuvvetlerin simetrik olmasndan,

BR

BA

200AR kN= , 200BR kN=

R AKesme kuvveti diyagram aada grld gibidir.

C D

200 kN

-200 kN

200 kN

x

V

200 kN

A B

150 mm

50 mm

1

2

225 mm

y2y

1y

75 mm

y

x

Tarafsz Eksend1

2d

1 1 2 2

1 2

. . (75.150).125 (225.50).25 7575.150 225.50

75

A y A yy mA A

y mm

+ += = =+ + =

m

1 1 75 125 50d y y m= = = m

2 2 75 25 50d y y mm= = =

3 32 21 1

1 1 1

41

. 75.150. (75.150).( 50)12 12

49218750

b hI A d

I mm

= + = + =

3 3

2 22 22 2 2 2

. 225.50. (225.50).(50) 3046875012 12

b h 4I A d I mm= + = + =

4 4 41 2 49218750 30468750 79687500 79687500

4I I I mm mm mm I mm= + = + = = n-n kesitindeki kesme kuvveti: V=200 kN

T.E

50 mma

x

y

75 mm

y=100 mm

a) 4 4. (75.50).100 375000 375000a aQ A y mm Q mm= = = =

75t mm=

3. 200.10 .375000 12,55 12,55

. 79687500.75a

a aV Q MPa MPa

I t = = = =

y=75mm

75 mm

y

x

b

100 mm

b) 4 4. (75.100).75 562500 562500b bQ A y mm Q mm= = = =

75t mm=

T.E

3. 200.10 .562500 18,82 18,82. 79687500.75

bb b

V Q MPa MPaI t

= = = =

SORU 3

B

D

A

C

10 kN

200 mm

150 mm

6 mm 44 mm

z x

y

H K

AB silindirinin ii bo olup, d ap 88 mm ve et kalnl 6 mmdir. CD rijit kolunun C ucuna etkiyen 10 kN luk kuvvetin etkisiyle, a) H noktasnda meydana gelen asal gerilmeleri ve maksimum kayma gerilmesini, b) K noktasnda meydana gelen asal gerilmeleri ve maksimum kayma gerilmesini bulunuz.

T

KH

y

xz

150 mm

10.200 2000T k= = Nmm 10 kN

A

B

H ve K noktalarndaki eilme momenti; 10.150 1500M kNmm= =

( )4 4 444 38 2612172 26121722

J mm J 4mm= = =

42612172 130608962 2JI I m= = = m

M10 kN

T.E

Ta) H Noktas : H noktas tarafsz eksen zerinde olduundan eilme

H

K

Z

X

nedeniyle normal gerilme meydana gelmez, yani 0 = dr. Burulmadan dolay:

3

. .. 2000.10 .44 33,69 33,69

2612172bur burT c MPa MPaJ

= = = = Kesme kuvvetinden dolay:

( )2 2 3 32 2 1 1 2 14 4 2. 2 3 2 3 3c c c cQ A y c c = = = C21CH T.E ( )3 3 3210.10 . 44 383 12,89 12,89

12.1306086kesme kesmeVQ MPa MPaIt

= = = =

. 33,69 12,89 46,8 46,8bur kesme MPa MPa = + = + = =

Y

X

H

. 46,8maks MPa = = . 46,8maks MPa = min. 46,8MPa =

b) K Noktas : K noktasnda V= 10 kNluk kesme kuvveti olmasna ramen Q=0 olduundan dolay 0 = dr.

Burulmadan dolay:

3

. .. 2000.10 .44 33,69 33,69

2612172bur burT c MPa MPaJ

= = = = Eilme momentinden dolay:

3. 1500.10 .44 50,53 50,531306086y y

M c MPa MPaI

= = = = , 0z =

2 22 2

.50,53 50,53 33,69 16,375

2 2 2 2z y z y

maks zy MPa + = + + = + + =

2 2

2 2min.

50,53 50,53 33,69 67,2652 2 2 2

z y z yzy MPa

+ = + = + =

2 22 250,53 33,69 42

2 2z y

zy MPa = + = + =

KZ

Y

. 42maks MPa = = . 16,375maks MPa = min. 67,265MPa =

SORU 4

A

D

B

060

15 kN

10 kN/m

700 mm400 mm

C

500 mm

A Balants

8 mm8 mm

ekildeki sistemde; a) 700Y MPa = ve emniyet katsays n=2 olduuna gre BD kablosunun apn, b) 400Y MPa = , n=2 ve A piminin ap 10 mm olduuna gre A pimini kayma gerilmesine gre kontrol ediniz, c) Emniyetli yatak gerilmesi 300 MPa olduuna gre A noktasndaki yatak kontroln yapnz.

ADC elemannn serbest cisim diyagramn izelim,

0AM = 7 kNBF

AF y

xFA

500 mm

C

400 mm 700 mm

15 kN

600

D

7.750 15.1100 .sin 60.400 .cos 60.500 0B BF F + + =

A 36,46BF = kN

0 0 36, 46cos 60 0

18, 23x

x

x A

A

F F

F kN

= = = =

0 sin 60 7 15 0 36,46sin 60 22 0 9,57

y yy A B A AF F F F F+ = + = + = = y kN

a) 700 350 350

2Y

em emMPa MPan = = = =

3 3

22 2

4.36,46.10 4.36,46.10350 11,51.350

4

Bem DB DB

DB DB

F MPa d d mmd d

= = = = =

b) .400 2002

Yem MPan

= = =

2 2 2 2

. 2 2 2

.

4. 4. (18,23) ( 9,57)2 131,01 2002 2 10

4131,01 200

x yA

A Aort

A A

ort

F F FMPa MPa

d d

MPa MPa

+ + = = = =