Bidang Terendam - Djoko Luknanto

Mekanika Fluida
Kuliah Mekanika Fluida
Gaya pada bidang terendam
Ir. Djoko Luknanto, M.Sc., Ph.D.
3/14/2005
Jack la Motta
1
Bidang Datar
muka air
O
„
„
h0
h
„
p
F = Ap0
„
y
„
Gaya pada pintu: dF = p dA
p harus terbagi merata
p tekanan hidrostatika
p = hγ
p = (y sin α) γ
dy
dA
pintu air/benda
terendam air
G
α
y
„
„
„
„
„
3/14/2005
Djoko Luknanto
y0
P
yP
x
dF = γ sin α × ydA
F = ∫ dF = γ sin α ∫ ydA
= γ sin α × Ay0 = ( y0 sin α )γ × A
= h0γ × A
y0 = jarak pusat berat pintu (G
(G)
p0 = tekanan hidrostatika di G
A = luas pintu
∫ ydA = momen statis bidang A
yP = jarak pusat tekanan pintu (P
(P)
F = p0 × A
y0 =
∫ ydA = ∫ ydA
A
∫ dA
2
1
Mekanika Fluida
Besar dan Pusat Gaya
„
Besar Gaya
„
dF = γ sin α × ydA
Pusat Gaya
Fy P = ∫ ydF = ∫ ypdA
F = ∫ dF = γ sin α ∫ ydA
= ∫ yhγdA = γ ∫ y ( y sin α )dA
= γ sin α × Ay0 = ( y0 sin α )γ × A
= γ sin α ∫ y 2 dA
= h0γ × A = p0 × A
F = p0 × A
y0 =
„
„
„
„
„
„
yP =
∫ ydA = ∫ ydA
A
∫ dA
γ sin α ∫ y 2 dA
F
γ sin α ∫ y dA
=
γ sin α ∫ y 2 dA
=
∫ y dA = I
2
=
F = besar gaya hidrostatika
h0 = jarak vertikal pusat berat pintu (G
(G)
y0 = jarak pusat berat pintu (G
(G)
yP = jarak pusat tekanan pintu (P
(P)
∫ ydA = momen statis bidang A
∫ y2dA = momen inersia bidang A
3/14/2005
A( y0 sin α )γ
Ap0
2
Ay0
S
2
=
I 0 + Ay0
I
= y0 + 0
Ay0
S
y P = y0 +
I0
S
Jack la Motta
3
Segi 4: Luas dan Momen Statis
„
y
Luas:
y =+ h / 2
y =+ h / 2
y =− h / 2
y =− h / 2
A=
dy
dA = bdy
x
h
„
∫ dA = ∫ bdy = b[ y ]
+h / 2
−h / 2
= bh
Pusat Berat:
Ay0 =
y =+ h / 2
y =+ h / 2
y =− h / 2
y =− h / 2
∫ ydA = ∫ bydy
+h / 2
b
3/14/2005
Djoko Luknanto
1 ⎤
⎡1 ⎤
⎡1
= b⎢ y 2 ⎥
= b⎢ h2 − h2 ⎥ = 0
8 ⎦
⎣ 2 ⎦ −h / 2
⎣8
y0 = 0
Jack la Motta
4
2
Mekanika Fluida
Segi 4: Momen Inersia di Pusat Berat
Momen Inersia thd sb-x:
„
y
I0 =
y =+ h / 2
∫ y dA = ∫ by dy
2
y =− h / 2
dy
dA = bdy
x
h
y =+ h / 2
2
b
3/14/2005
y =− h / 2
+h / 2
⎡1 ⎤
= b⎢ y 3 ⎥
⎣ 3 ⎦ −h / 2
1 ⎤
⎡1
= b⎢ h3 + h3 ⎥
24 ⎦
⎣ 24
1
I 0 = bh 3
12
Jack la Motta
5
Segi 4: Momen Inersia Ix
G pusat berat segi 4
Momen Inersia Ix
„
y
„
Ix =
dy
x
Djoko Luknanto
⎤
⎥
⎦ y0 − h / 2
b ⎡⎛
h⎞ ⎛
h⎞
⎢⎜ y0 − ⎟ − ⎜ y0 − ⎟
3 ⎢⎣⎝
2⎠ ⎝
2⎠
=
2
2
b ⎡⎛ 3 3hy0 3h 2 y0 h 3 ⎞ ⎛ 3 3hy0 3h 2 y0 h 3 ⎞⎤
+
+ ⎟⎟ − ⎜⎜ y0 −
+
− ⎟⎟⎥
⎢⎜⎜ y0 +
3 ⎣⎢⎝
2
4
8⎠ ⎝
2
4
8 ⎠⎦⎥
3
3
⎤
⎥
⎥⎦
2
2
b ⎡ 3hy0
h 3 3hy0
h3 ⎤
+ +
+ ⎥
⎢
3⎣ 2
8
2
8⎦
1
2
I x = bh 3 + bhy0
12
2
I x = I 0 + Ay0
=
b
3/14/2005
y0 + h / 2
3
=
G
y0
⎡1
∫ y dA = b⎢⎣ 3 y
y = y0 − h / 2
dA = bdy
h
y = y0 + h / 2
2
Jack la Motta
6
3
Mekanika Fluida
Segitiga
1
A = bh
2
„
Luas:
„
Pusat Berat:
„
Momen Inersia, I0
h
y0 =
1
h
3
G
h/3
b/2
b/2
3/14/2005
I0 =
1 3
bh
36
Jack la Motta
7
Lingkaran
1
πD 2
4
„
Luas: A =
„
Pusat Berat: y 0
„
Momen Inersia, I0
G
=
D
1
D
2
r
1
I0 =
πD 4
64
3/14/2005
Djoko Luknanto
Jack la Motta
8
4
Mekanika Fluida
Setengah Lingkaran
Luas: A =
„
Pusat Berat:
„
Momen Inersia, I0
r
G
y0
1 2
πr
2
„
y0 =
4r
3π
I 0 = 0,1102r 4
3/14/2005
Jack la Motta
9
Bidang Lengkung
muka air
D
C
„
„
h
dF
α
dFx
„
dFy
α
B
dA
„
A
3/14/2005
Djoko Luknanto
Gaya dF selalu tegak lurus
bidang kontak dA
Nilai dF = hγ dA
Komponen x-y bid. kontak
dAx = dA cos α
dAy = dA sin α
Komponen x-y gaya dF
dFy = hγ dA cos α
dFx = hγ dA sin α
Jack la Motta
10
5
Mekanika Fluida
Gaya pada bidang lengkung
muka air
Ax
D
C
„
Fy
h
h0
dF
Ay
dFy
α
dFx
α
B
dA
„
G
Fx
ϕ
A
2
F = Fx + Fy
ϕ = arctan
3/14/2005
Djoko Luknanto
Fy
Fx
2
F
Komponen x gaya dF
dFx= dF sin α
= hγ dA sin α
= γ hdAy
Fx = γ ∫hdAy = γ h0Ay
Komponen y gaya dF
dFy= dF cos α
= hγ dA cos α
= γ hdAx
Fy = γ ∫hdAx = γ V
Jack la Motta
11
6