sifat penampang datar

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purbolaras.wordpress.com SIFAT PENAMPANG DATAR REFERENSI: Mekanika Bahan Edisi Keempat, J.M. Gere & S.P. Timoshenko (Sebagian rumus) Persegi / Persegi Panjang x c 1 2 h = y c 1 2 b = A bh = I x 3 1 12 bh = I y 3 1 12 hb = S x 2 1 6 bh = S y 2 1 6 hb = I p ( ) 2 2 1 12 bh b h = + Segitiga Sama Sisi x c 1 2 a = y c 3 6 a = A 2 3 4 a = I x 4 3 96 a = I y 4 3 96 a = S x 3 3 48 a = S y 3 1 16 a = I p 4 3 48 a = Lingkaran x c 1 2 d = y c 1 2 d = A 2 4 d π = I x 4 64 d π = I y 4 64 d π = S x 3 32 d π = S y 3 32 d π = I p 4 32 d π = h b X Y y x c c a X Y y x c c a X Y d x y c c

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Page 1: Sifat Penampang Datar

purbolaras.wordpress.com

SIFAT PENAMPANG DATAR REFERENSI: Mekanika Bahan Edisi Keempat, J.M. Gere & S.P. Timoshenko

(Sebagian rumus)

Persegi / Persegi Panjang

xc 1

2h=

yc 1

2b=

A bh=

Ix 31

12bh=

Iy 31

12hb=

Sx 21

6bh=

Sy 21

6hb=

Ip ( )2 21

12bh b h= +

Segitiga Sama Sisi

xc 1

2a=

yc 3

6a=

A 23

4a=

Ix 43

96a=

Iy 43

96a=

Sx 33

48a=

Sy 31

16a=

Ip 43

48a=

Lingkaran

xc 1

2d=

yc 1

2d=

A 2

4dπ=

Ix 4

64dπ=

Iy 4

64dπ=

Sx 3

32dπ=

Sy 3

32dπ=

Ip 4

32dπ=

h

b

X

Y

y

x

c

c

a

X

Y

y

x

c

c

a

X

Y

dx

y

c

c

Page 2: Sifat Penampang Datar

purbolaras.wordpress.com

Tabung Pipa Dinding Tebal

xc 1

2d=

yc 1

2d=

A ( )t d tπ= −

Ix,y ( )( )44

642d d tπ= − −

( )3 2 2 3 4

83 4 2d t d t dt tπ= − + −

Sx,y ( )( )33

322d d tπ= − −

( )3 2 2 3 4

43 4 2

dd t d t dt tπ= − + −

Ip ( )3 2 2 3 4

43 4 2d t d t dt tπ= − + −

Tabung Pipa Dinding Tipis

xc 1

2d=

yc 1

2d=

A dtπ=

Ix 3

8d tπ=

Iy 3

8d tπ=

Sx 2

4d tπ=

Sy 2

4d tπ=

Ip 3

4d tπ=

Tabung Kotak

xc 1

2h=

yc 1

2b=

A ( ) 22 4t b h t= + −

Ix ( ) ( )331

122 2bh b t h t = − − −

Iy ( ) ( )331

122 2hb h t b t = − − −

Sx ( )( )331

62 2

hbh b t h t = − − −

Sy ( )( )331

62 2

bhb h t b t = − − −

Ip ( )( ) ( )( )3 33 31

122 2 2 2bh hb b t h t h t b t = + − − − − − −

X

Y

d

t

x

y

c

c

X

Y

d

t

x

y

c

c

X

Y

b

h

t

ty

x

c

c

Page 3: Sifat Penampang Datar

purbolaras.wordpress.com

Penampang T

xc 1

2b=

yc ( )( )

( )( )

221 1

2 2bh b w h t

bh b w h t

− − −=

− − −

A ( )bt w h t= + −

Ix 231 1

12 2 cbh bh h y= + −

( )( ) ( )( ) ( )23

1 1

12 2 cb w h t b w h t h t y− − − − − − − −

Iy ( )( )3 31

12h t w tb= − +

Sx x

c

I

y=

Sy y

c

I

x=

Ip x yI I= +

Penampang I / H

xc 1

2b=

yc 1

2h=

A ( ) ( )2bh b w h t= − − −

Ix ( ) ( )( )331

122bh b w h t= + − −

Iy ( )( )3 31

122 2h t w tb= − +

Sx ( )( )( )331

62

hbh b w h t= + − −

Sy ( )( )3 31

62 2

bh t w tb= − +

Ip ( ) ( )( ) ( ) ( )( )3 33 31

122 2bh hb b w h t h t b w = + − − − + − −

Keterangan : xc = jarak titik berat (arah sumbu-X) yc = jarak titik berat (arah sumbu-Y) A = luas penampang Sx = momen statis (tinjauan sumbu-X) Sy = momen statis (tinjauan sumbu-Y) Ix = momen inersia (tinjauan sumbu-X) Iy = momen inersia (tinjauan sumbu-Y) Ip = momen inersia polar terhadap pusat sumbu

X

Yxc

yc

b

h

t

w

X

Yxc

yc

b

h

t

w