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7/31/2019 Garis Netral
1/2
PORTAL TENGAH MEMANJANG L.1 PROFIL INP
fc' = 24 Mpa L = 560 cm
E. Baja = Mpa bo= 670 cm
E. Beton = 4700 fc' = Mpa
be = 140 cm
Data profil INP -
Ix = cm4 ts = 12 cm ts = 12 cm
h = cm (tebal plat)
As = cm2 h = 17.3 cm
bf = cm
garis netral
h = 42.5 cm 27.3 cm
1/2h cm
bf = 16.3 cm
- Lebar effektif (be)
be= 1/4.L
= 1/4 . 560
= 140 cm
be = bo(jarak antar balok) = 670 cm
Be = bf + 16.ts = cm
diambil beff = 140 cm
n =
=
= 8.69 ------
Ac = be . Ts Ac
= 140 x 12 n
= cm2 = cm2
At = As + be
= 132 + 140= cm2
dt = 1/2. h + 1/2.ts Ic = 1/12 . (beff/n) . 10^3
= 1/2 x 42.5 + 1/2x 12 = 1 140 10 ^3
= cm 12 9
dc = = cm4
= It = Ic + (Ac . dc^2) + Ix + (As . ds^2)
= cm4
= cm
ds = (beff/At) . Dt
=
= x 10^-6
= cm
=E
7/10/2012 3:35
200000
23025
42 1/2
E. Baja
E. Beton
200000
23025
1680
208.3
42 1/2
36970
42.5
132
16.3
Ac(eq) =
21.3
272.0
27.25
9.0
x
=1680
9
186.7
272.0x 27.25
14.03
x
1296.2963
358033.88
358033.88
560
272.0x 27.25
13.22
140
132
(As/At) . dt
10^-4
6.393
=K = 639.34621 cm3
639.34621
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7/31/2019 Garis Netral
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PORTAL TEPI B &c MEMANJANG L.2 (ATAP DAG) PROFIL INP
fc' = 24 Mpa L = 560 cm
E. Baja = Mpa bo= 540 cm
E. Beton = 4700 fc' = Mpa
be = 46.67 cm
Data profil INP -
Ix = cm4 ts = 10 cm ts = 10 cm
h = cm (tebal plat)
As = cm2 h = 21.2 cm
bf = cm
garis netral
h = 38.0 cm 24.0 cm
1/2h cm
bf = 14.9 cm
- Lebar effektif (be)
be= 1/12.L
= 1/12 . 560
= 46.7 cm
be = bo(jarak antar balok) = 540 cm
Be = bf + 16.ts = cm
diambil beff = 47 cm
n =
=
= 8.69 ------
Ac = be . Ts Ac
= 46.7 x 10 n
= cm2 = cm2
At = As + be
= 107 + 46.7
= cm2
dt = 1/2. h + 1/2.ts Ic = 1/12 . (beff/n) . 10^3
= 1/2 x 38.0 + 1/2x 10 = 1 46.7 10 ^3
= cm 12 9
dc = = cm4
= It = Ic + (Ac . dc^2) + Ix + (As . ds^2)
= cm4
= cm
ds = (beff/At) . dt
=
= x 10^-6
= cm= E
38
200000
23025
7/10/2012 3:3538
38
24010
107
14.9
19.0
174.9
E. Baja
200000
23025
Ac(eq) ==
466.666667
9
466.666667 51.9
x x
2.865
9.0
16.71
K =160454.11
= 286.5251946.6666667
x 24560
(As/At) . dt 432.09877
107x 24
153.7 286.52519
7.29 10^-4
cm3
153.7 160454.11
153.7
24
E. Beton