beban menurut cara kerjanya...
TRANSCRIPT
![Page 1: Beban menurut cara kerjanya dibedakanrahayukusuma.lecture.ub.ac.id/files/2014/09/3-Balok...2014/09/03 · Reaksi perletakan : 3,9625 ( ) 8 (1.25 8 ) ( 2.25 6) (1.5 4 ) ( 4 2) ( 2.9](https://reader036.vdocuments.site/reader036/viewer/2022081618/608933346634b75b2e4289af/html5/thumbnails/1.jpg)
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Beban menurut cara kerjanya dibedakan :
1. Muatan/beban langsung :
Suatu beban yang bekerja langsung pada suatu bagian konstruksi tanpa perantara konstruksi lain.
2. Muatan/beban tidak langsung
Suatu beban yang bekerja dengan perantara
konstruksi lain.
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Jembatan Sembayat, Gresik - Jatim
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Potongan melintang
Gelagar induk
Gel. melintang
aspal
arah muatan
Gel. memanjang
Skema gelagar tidak langsung dari suatu jembatan
Potongan Melintang
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Balok memanjang
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Balok melintang
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Penyederhanaan awal, gelagar tidak langsung
gel. melintang
gel. induk / utama
gel. memanjang
Penyederhanaan akhir untuk gelagar tidak langsung
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Muatan tak langsung akan mengurangi momen maksimum yang terjadi.
P
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P diuraikan menjadi P1 dan P2
Pa
caP
1
A B
PP1
c a-c
P2
a a a a
P2P1
a a 2aA B
Pa
cP 2
Lalu tinjau balok AB dengan gaya-
gaya P1 dan P2. diagram momen dan
reaksi dapat ditentukan.
Distribusi Beban
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q kg/m’ beban terbagi rata
gel. melintang
x x
beban terbagi rata diatas gel. memanjang
P P P
beban terpusat diatas gel. induk
Beban merata masuk ke gelagar induk menjadi beban terpusat,
dimana beban P bernilai sebesar
gelagar induk / utama
gelagar induk / utama
xqP .
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Lukis Diagram M, D akibat muatan tak langsung
A B
P1 P2
2 m 2 m 2 m 2 m
D E F C
P3 P5P4
2 m
0.75 m 1 m 1.4 m
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A B
P1 P2
2 m 2 m 2 m 2 m
D E F C
P3 P5P4
2 m
0.75 m 1 m 1.4 m
AB
D E F C
RA
2.25 t 1.5 t 4 t 2.1 t 2.9 t
tAtitik 25,12).75,02(2
1
ttitikD 25,23.2
12.75,0.
2
1
tEtitik 5.13.2
1
tFtitik 4
tBtitik 1,23.2
4,1
tctitik 9,23.2
6,02
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Reaksi perletakan :
)(9625,3
0)29.2()24()45.1()625.2()825.1(8
0
A
A
B
R
xxxxxxR
M
Gaya lintang :
tDAD 7125.225.19625.3
4625.025.225.19625.3 DED
0375.15.125.225.19625.3 EFD
0375.545.125.225.19625.3 FBD
9.2CBD
AB
D E F C
RA
2.25 t 1.5 t 4 t 2.1 t 2.9 t
![Page 15: Beban menurut cara kerjanya dibedakanrahayukusuma.lecture.ub.ac.id/files/2014/09/3-Balok...2014/09/03 · Reaksi perletakan : 3,9625 ( ) 8 (1.25 8 ) ( 2.25 6) (1.5 4 ) ( 4 2) ( 2.9](https://reader036.vdocuments.site/reader036/viewer/2022081618/608933346634b75b2e4289af/html5/thumbnails/15.jpg)
Momen :
0 AM
tmxM D 425.527125.2
tmxxM E 35.6)225.2()47125.2(
tmxxxM F 275.4)25.1()425.2()67125.2(
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AB
D E F C
RA
1.25 t 2.25 t 1.5 t 4 t 2.1 t 2.9 t
5.425tm
1.25 t
6.35tm
4.275tm
5.8tm
2.7125t
0.4625t
1.0375t
5.0375t
2.9t
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AB
q=20 t/m’
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
Lukis Diagram M, D akibat beban muatan tak langsung
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A. Mencari Besarnya Reaksi Per-letakan
MA=0 … q.12.6 – RG.12 = 0 RG = 120t
MG=0 … RA.12 – q.12.6 = 0 RA = 120t
V = 0 … RA+RG = q.12 120 + 120 = 20.12 … (ok)
AB
q=20 t/m’
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
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AB
q=20 t/m’
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
AB
20 t
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
40 t 40 t 40 t 40 t 40 t20 t
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Mencari Gaya – Gaya Dalam Balok Utama (Bidang D dan M)
Interval 0x2m
- Dx = RA – 20 = 120 – 20 =100t
- Mx = (RA-20).X = 100.X
X = 0m --- MA = 0
X = 2m --- MB = 200 t.m
AB
20 t
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
40 t 40 t 40 t 40 t 40 t20 t
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Interval 2x4m
- DX = RA – 20 – 40 = 120 – 60 = 60t
- MX = (RA-20).X –40(X-2)= 60X + 80
X = 2m --- MB = 200 t.m
X = 4m --- MC = 320 t.m
AB
20 t
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
40 t 40 t 40 t 40 t 40 t20 t
Interval 4x6m
- DX = RA –20 –40x2 = 20t
- MX=100X – 40(X-2) – 40(X-4)
X = 4m --- MC = 320 t.m
X = 6m --- MD = 360 t.m
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Interval 6x8m
- DX = RA–20–40x3 = –20t
- MX=100X–40(X-2)–40(X-4)–40(X-6)
X = 6m --- MD = 360 t.m
X = 8m --- ME = 320 t.m
Interval 8x10m
- DX = 100–40x4 = –60t
- MX=100X -40(X-2) -40(X-4) -40(X-6) -40(X-8) X = 8m --- ME = 320 t.m
X = 10m --- MF = 200 t.m
AB
20 t
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
40 t 40 t 40 t 40 t 40 t20 t
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Interval 10x12m
- DX = 100–40x5 = –100t
- MX=100X -40(X-2) -40(X-4) -40(X-6) -40(X-8) -40(X-10)
X = 10m --- MF = 200 t.m
X = 12m --- MG = 0
AB
20 t
2 m 2 m 2 m 2 m
D E FC
2 m 2 m
G
40 t 40 t 40 t 40 t 40 t20 t
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AB C D E F
q=20 t/m
6 x 2m
G
+
-
10
060
20
20
6010
0
Q
20
032
036
0
+M
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