struktur kayu eka wijaya i i gelagar melintang
TRANSCRIPT
Eka wijaya03071001061
PERHITUNGAN GELAGAR MELINTANG
Diketahui : Jarak gelagar memanjang = 0,5 mJumlah gelagar memanjang = 11 buahJarak gelagar melintang = 0,75 mLebar Jembatan = 5 m
Pembebanan1. Akibat Lantai Kendaraan
Beban P1
Berat gelagar memanjang (15/25) : 0,15 x 0,25 x 500 = 18,75 kg/mBerat papan lantai kend. (5/25 & 15/25) : 0,20 x 0,25 x 500 = 25 kg/mBerat air hujan : 0,05 x 0,25 x 1000 = 12,5 kg/m Berat aspal : 0,065 x 0,25 x 2500 = 40,625 kg/m +
= 96,875 kg/m
Toeslag 10% = 9,6875 kg/m +
q = 106,56 kg/m
Maka,
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Eka wijaya03071001061
P1 = q x L = 106,56 kg/m x 0,75 m = 79,92 kg
Beban P2
Berat gelagar memanjang (15/25) : 0,15 x 0,25 x 500 = 18,75 kg/mBerat papan lantai kend. (5/25 & 15/25) : 0,20 x 0,50 x 500 = 50 kg/mBerat air hujan : 0,05 x 0,50 x 1000 = 25 kg/m Berat aspal : 0,065 x 0,50 x 2500 = 81,25 kg/m +
= 175 kg/m
Toeslag 10% = 17,5 kg/m +
q = 192,5 kg/mMaka,
P2 = q x L = 192,5 kg/m x 0,75 m = 144,375 kg
Beban P3
Berat gelagar memanjang trotoar (10/15): 0,10 x 0,15 x 500 = 7,5 kg/mBerat papan lantai trotoar : 0,07 x 0,25 x 500 = 8,75 kg/mBerat balok sandaran trotoar : 3 x 0,07 x 0,10 x 500= 10,75 kg/m+
= 27 kg/mToeslag 10% = 2,7
kg/m+ q = 29,7kg/m
Maka,P3 = q x L = 29,7 kg/m x 0,75 m = 22,275 kg
Beban P4
Berat gelagar memanjang trotoar (10/15): 0,10 x 0,15 x 500 = 7,5 kg/mBerat papan lantai trotoar : 0,07 x 0,25 x 500 = 8,85 kg/m +
=16,25 kg/m
Toeslag 10% = 1,625 kg/m +
q = 17,875 kg/m
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Eka wijaya03071001061
Maka,P4 = q x L = 17,875 kg/m x 0,75 m = 13,41 kg
Beban P5
Berat gelagar memanjang (15/25) : 0,15 x 0,25 x 500 = 18,75 kg/mBerat papan lantai trotoar (7/20) : 0,07 x 0,50 x 500 = 17,5 kg/m +
= 36,25 kg/m
Toeslag 10% = 3,625 kg/m +
q = 39,875 kg/m
Maka,P5 = q x L = 39,875 kg/m x 0,75 m = 29,9 kg
Reaksi Perletakan
RA = RB = P1 +P3 + P4 + P5 + 2,5 P2
= 79,92kg + 22,275 kg + 13,41 kg + 29,9 + 2,5 (144,375 kg )
= 506,44 kg
Momen Maksimum yang terjadi ditengah bentang
M max1 = RA (2,5 m) – P3 (2,5 m) – P5 (2 m) – P1 (1,5 m) – P4 (1,5
m) – P2
(1 + 0,5 m)
= 506,44 kg (2,5 m) – 22,275 kg(2,5 m) – 29,9 kg (2 m)
– 79,92 kg (1,5 m) – 13,41 kg (1,5 m) – 144,375 kg (1,5
m)
= 794,055 kgm
Q max1 = RA = RB = 506,44 kg
2. Akibat Muatan Loading
Dari soal diketahui jembatan jalan kelas II, maka diambil 70
%
Muatan garis P = 5 ton x 70 % = 3,5 ton
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Muatan merata q = 1 ton/m x 70 % = 0,7 ton/m
Jarak gelagar melintang = 0,75 m
Faktor kejut = 1 + 2050 + 0 ,75 = 1,394
Jarak gelagar memanjang = 0,5 m maka beban yang bekerja
pada tiap gelagar adalah :
P = 3,5 ton0 ,75m x 0,5 m x 1,394 = 3,253 t = 3253 kg
q = 0,7 t /m0 ,75m x 0,5 m x 1,394 = 0,651 t/m = 651 kg/m
Q = ½ qL
= ½ x 651 kg/m x 0,75 m = 244,125 kg
Maka besarnya gaya :
P2 = P + Q
= 3253 kg + 244,125kg
= 3497,125 kg
P1 = ½ . P2
= ½ . 3497,125 kg
= 1748,56 kg
Reaksi Perletakan
RA = RB = P1 + 2,5 P2
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= 1748,56 kg + 2,5 (3497,125 kg)
= 10491,37 kg
M max2 = RA (2,5 m) – P1 (1,5 m) – P2 (1 + 0,5 + 0)m
= 10491,37 kg (2,5 m) – 1748,56 kg (1,5 m) –
3497,125 kg (1,5 m)
= 18359,896 kgm
Q max2 = RA
= 10491,37 kg
3. Akibat berat sendiri gelagar melintang
Gelagar melintang direncanakan 30/80
q = 0,3 x 0,8 x 500
= 120kg/m
M max3 = 1/8 . q . L2
= 1/8 . 120 kg/m . (5m)2
= 375 kgm
Q max3 = ½ . q . L
= ½ . 120kg/m . 5 m
= 300 kg
Momen Maksimum Total
Mmax total = M max1 + M max2 + M max3
= 794,055 kgm + 18359,896 kgm + 375 kgm
= 19528,95 kgm
Qmax total = Q max1 + Q max2 + Q max3
= 506,44 kg + 10491,37kg + 300 kg
= 11297,81 kg
Kontrol Tekuk
w = 1/6 . b . h2
= 1/6 . 30 cm . (80 cm)2
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= 32000 cm3
σ tk =Mmax total
w =
1952895 kgcm
32000cm3 = 41,03 kg/cm2
Syarat Aman : σ tk ll <
σ tk ll
41,03 kg/cm2 < 50 kg/cm2 .................... AMAN!!!
Kontrol Tegangan Geser
τ =
3.Qtotal
2.b .h
=
3 x11297,81 kg
2x 30 x80 cm2 = 6,06 kg/cm2
Syarat Aman
τ < τ ll
6,06 kg/cm2 < 6,67 kg/cm2.....................AMAN!!!
Kontrol Lendutan
q = 120 kg/m
E = 80.000 kg/cm2 ( kayu kelas I )
I =
112
bh3
= 1
12 . 30 . 803
= 1280000 cm4
Lendutan maksimum terjadi di tengah bentang
Akibat P5 di titik 1
P = 29,9 kg
a = 0,5 meter
b= 5 m – 0,5 m = 4,5 meter
δ1 =P .a2 .b2
3 EI . L =
29 ,9 kg . (50 )2 . 4502
3. 80 x 103 . 128.104 . 500
= 0,000099 cm
Civil Engineering of Sriwijaya University
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Akibat P1 dan P4 di titik 2
P = 79,92 kg + 13,41 kg + 1748,56 kg = 1841,89 kg
a = 1 meter
b= 5 m – 1 m = 4 meter
δ2 =P .a2.b2
3 EI . L =
1841,89 kg . (100 )2. 4002
3. 80 x 103 . 128.104 . 500
= 0,019 cm
Akibat P2 di titik 3
P = 144,375 kg + 3497,125 kg = 3641,5 kg
a = 1,5 meter
b= 5 m – 1,5 m = 3,5 meter
δ3 =P .a2 .b2
3 EI . L =
3641,5 kg . (150 )2 . 3502
3. 80 x 103 . 128.104 . 500
= 0,065 cm
Akibat P2 di titik 4
P = 144,375 kg + 3497,125kg = 3641,5 kg
a = 2 meter
b= 5 m – 2 m = 3 meter
δ4 = P .a2 .b2
3 EI .L =
3641,5 kg . (200 )2 .3002
3. 80 x 103 . 128.104 . 500
= 0,085 cm
Akibat P2 di titik 5
P = 144,375 kg + 3497,125 kg = 3641,5 kg
a = 2,5 meter
b= 5 m – 2,5 m = 2,5 meter
δ5 =P .a2 .b2
3 EI . L =
3641,5 kg . (250 )2 . 2502
3. 80 x 103 . 128.104 . 500
= 0,093 cm
Akibat q
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δq = 5.q . L4
384 EI =
5.1 ,20 kg/cm . 5004
384 . 80 x103 . 128 .104
= 0,00952 cm
max = 21 + 22 + 23 +2 4 + 5 + q
= (2(0,000099) + 2(0,019) + 2(0,065) +
2(0,085)+0,093+ 0,0095) cm
= 0,44 cm
δ = L400 =
500400 = 1,25 cm
Syarat aman :
δ < δ
0,44 cm < 1,25 cm......................AMAN!!!
PERHITUNGAN PASAK GELAGAR MELINTANG
Untuk gelagar melintang digunakan balok 30/80 disusun tersusun 3
lapis :
Bidang Lintang gelagar tersebut :
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25
30
25
30
Eka wijaya03071001061
Bidang Lintang Diketahui :
P1 = 79,92 + 1748,56 = 1828,48 kg
P2 = 144,375 + 3497,125 = 3641,5kg
P3 = 22,275 kg
P4 = 13,41 kg
q = 120 kg/m
RA = RB
= ½ x q x L + P1 + P2 + P3 + 2,5 P4
= ½ x 120 kg/m x 5 m + 1828,48 kg + 3641,5 kg
22,275 kg
+ 2,5 (13,41 kg)
= 5825,78 kg
Karena batang simetris maka ditinjau setengah saja
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3m
Eka wijaya03071001061
QA = 5825,78 kg
QA kanan = 5850,83 kg – 1828,48 kg = 3997,3 kg
QB kiri = QA kanan - (0,5 x 120 ) = 3937,3 kg
QB kanan = QB kiri – 3641,5 kg = 295,8 kg
QC kiri = QB kanan - ( 0,5 x 120) = 235,8 kg
QC kanan = QC kiri – 22,275 kg = 213,525 kg
QD kiri = QC kanan - ( 0,5 x 120) = 153,525 kg
QD kanan = QD kiri – 13,41 kg = 140,115 kg
QE kiri = QD kanan - ( 0,5 x 120) = 80,115 kg
QE kanan = QE kiri – 13,41 kg = 66,705 kg
QF kiri = QE kanan - ( 0,5 x 120) = 6,705 kg
QF kanan = QF kiri – ½. 13,41 kg = 0 kg
GAYA GESER PER SATUAN PANJANG
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τ b = τ = D .SI
Dengan catatan I dapat diambil tanpa reduksi ( felix yap hal 71 ). t max dapat dicari dengan menggunakan sifat-sifat titik parabola.
maka : τ max = τ = 3 Q
2 h
Gaya geser yang terjadi untuk satu balok :
τ A = 3 x 5825 ,78 kg
2 x 80 cm = 109,23 kg/cm
τ A25 = τ A−(15
40 )2
×τ A =93 ,87kg /cm
τ A kanan =
3 x 3997 ,3 kg2 x 80 cm = 74,95kg/cm
τ 25 Akanan = τ A kanan −(15
40 )2
×τ A kanan =64 ,41kg/cm
τ B kiri =
3 x 3937 ,3 kg2 x 80 cm = 73,82kg/cm
τ 25Bkiri = τ Bkiri−(15
40 )2
×τ Bkiri=63 ,44 kg/cm
τ B kanan =
3 x 295,8 kg2 x 80 cm = 5,55 kg/cm
τ 25Bkanan = τ Bkanan−(15
40 )2
×τ Bkanan=4 ,77kg /cm
τC kiri =
3 x 235 ,8 kg2 x 80 cm = 4,42 kg/cm
τ 25Ckiri = τ Ckiri−(15
40 )2
×τCkiri=3 ,80kg/cm
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τC kanan =
3 x 213 ,525 kg2 x 80 cm = 4 kg/cm
τ 25Ckanan = τ Ckanan−(15
40 )2
×τ Ckanan=3 ,44 kg /cm
τ D kiri =
3 x 153,525 kg2 x 80 cm = 2,88 kg/cm
τ 25Dkiri = τ Dkiri−(15
40 )2
×τ Dkiri=2, 475 kg/cm
τ D kanan =
3 x 140 ,115 kg2 x 80 cm = 2,63 kg/cm
τ 25Dkanan = τ Dkanan−(15
40 )2
×τ Dkanan=2 ,26kg /cm
τ E kiri =
3 x 80 ,115 kg2 x 80 cm = 1,50 kg/cm
τ 25Dkiri = τ Dkiri−(15
40 )2
×τ Dkiri=1 ,29kg /cm
τ E kanan =
3 x 66 ,705 kg2 x 80 cm = 1,25 kg/cm
τ 25Ekanan = τ Ekanan−(15
40 )2
×τ Ekanan=1 ,07 kg/cm
τ F kiri =
3 x 6,705 kg2 x 80 cm = 0,13 kg/cm
τ 25Fkiri= τ Fkiri−(15
40 )2
×τ Fkiri=0 ,11kg /cm
τ Fkanan = 0 kg/cm
τ 30Fkanan= 0kg/cm
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ut
Eka wijaya03071001061
Kekuatan Pasak
τ ll kayu kelas III mutu A = 6,67 kg/cm2
dipakai pasak kayu dengan syarat :
u > 5 t
t > 1,5 cm
u > 15 cm
Dipakai pasak :
- u = 15 cm
- t = 3 cm
- b = 40 cm
Maka kekuatan pasak :
S1 = u . b . τ ll pasak = 15 cm x 40 cm x 6,67 kg/cm2
= 4002 kg
S2 = b . t . τ tk balok = 40 cm x 3 cm x 12,5 kg/cm2 =
1500 kg
Diambil nilai S minimum yaitu S = 1500 kg
Gaya Geser yang Harus Dipikul pasak
LAB =
τ25 A kanan + τ25B kiri
2 x 50 cm
=
64 .41 kg /cm + 63 . 44 kg /cm2 x 50 cm = 3196.25 kg
LBC =
τ25B kanan + τ25C kiri
2 x 50 cm
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diambil t > 3 cm
sehingga u = 5t =5 x 3 =
15 cm
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=
4 .77 kg /cm + 3 .80 kg/cm2 x 50 cm = 214.25 kg
LCD =
τ25C kanan + τ25 D kiri
2 x 50 cm
=
3.44 kg /cm + 2 .475kg /cm2 x 50 cm = 147.875 kg
LDE =
τ25 D kanan + τ25 E kiri
2 x 50 cm
=
2.26 kg /cm + 1.29 kg /cm2 x 50 cm = 88.75 kg
LEF =
τ25 E kanan + τ25 F kiri
2 x 50 cm
=
1.07kg /cm + 0 ,11 kg /cm2 x 50 cm = 29.5 kg
Jumlah Pasak yang Diperlukan Per Satu Baris
hAB =
LAB
S =
3169 .25 kg1500 kg = 2.13 @ 2 pasak
hBC =
LBC
S =
214 . 25 kg1500 kg = 0,14 @ 1 pasak
hCD =
LCD
S =
147 . 875 kg1500 kg = 0,098 @ 0 pasak
hDE =
LDE
S =
88 . 75 kg1500 kg = 0,059 @ 0 pasak
hEF =
LEF
S =
29 .5 kg1500 kg = 0,020 @ 0 pasak
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