tugas besar struktur baja 2 final ( ferdinand )
DESCRIPTION
Struktur BajaTRANSCRIPT
σy = 290 Mpa 2400 kg/cm2 < dilihat dari mutu baja BJ37
σy = ∂ ySF
=24001,5
=¿1600 kg/cm2
Perhitungan Sudut
Tan α= 16,524,25 = 0,6804
α = arch-1 0,6804 = 34,23°
Pembebanan Pada Gording
α = 34,23°
Coba profil Canal C = 200 . 75 . 8,5 berat sendiri = 29,4 kg/m (Try & Error )
Jarak antar Gording = 1 m
Beban Mati
Beban Atap (ᶐatap ) Atap yang dipilih Bluescope Lysaght TrimdekColorbound = 4,93 kg/mTebal = 0.48 mm
ᶐba = 4,93 x (0,5 x 1,65 x 2 ) x 1
= 8,15 kg/m
Berat Sendiri (ᶐbs ) = 25,3 kg/m Berat Sendiri (ᶐbm) = ᶐba + ᶐbs = 8,15 + 25,30 = 33,45 kg/m Beban Aksesoria (ᶐbA) = 25 % ᶐbm = 25 % x 33,45 = 8,36 kg/m Beban Mati total = ᶐbm + ᶐbA = 33,45 + 8,36 = 41,68 kg/m
ᶐbm Total = 41,68cos34,23 °= 48,68 kg/m
2
4
5
6789
101112
13
14
15
16
17
18
19
202122
23
24
25262728
29
30
31
32
33
34
Beban Hidup
Beban Orang = 100 kg Refernsi dari PBI ( Peratuan Pembebanan Indonesia)
Beban Hujan (W) = 20 kg
P = w x Jarak Kuda-kuda x Jarak antar gording= 20 x 6,8 x 2= 272 kg
Karena beban hujan ≥ beban orang maka, PBH = 272 kg
Px = Px . Sin α
= 272 Sin 34,23 = 153,004
Py = Py . Cos α
= 272 Cos 34,23 = 224
Pengaruh kemiringan Atap
ᶐbm x = ᶐbm x sin α
= 48,68 sin 34,23= 24,93 kg/m
ᶐbm y = ᶐbm x cos α
= 48,68 cos 34,23= 41,81 kg/m
Mbm x= ⅛ x ᶐbmx x (jarak kuda-kuda)2
= ⅛ x 24,93 x 6,82
= 144,09 kg/m
Mbm y= ⅛ x ᶐbmy x (jarak kuda-kuda)2 = ⅛ x 41,81 x 6,82
=271,43 kg/m
Mbh x =⅟4 x(Px) x (jarakkuda-kuda)
= ⅟4 x 153,004 x 6,8= 260,11 kg/m
Mbh y= ⅟4 x (P y) x (jarak kuda-kuda)
= ⅟4 x 224,89 x 6,8= 382,30 kg/m
Tegangan Lentur akibat Beban Atap
Mbt x = Mbmx + Mbhx
= 144,09 + 260,11
= 404,20 kg/m
Mbt y = Mbmy + Mbhy
= 271,43 + 382,30
= 653,73 kg/m
Tegangan Beban Atap Wx= 191 , Wy = 27
σbt x = Mbtx
Wy( table)= 404,20
27=1497
3
35
36373839404142
43
44
45
46
47
484950
51
5253
54
5556
57585960
6162
63
64
65
6667
68
6970
71
72
73
74
75
76
77
78
79
80
σbt y = Mbty
Wx(table)=653,73
191=342
σ Total Atap = √ (δbt x )2+( δbt y )2
= √ (1497 )2+(342 )2
= 1535 ≤ σ’= 1600 kg/m ok
jadi profil yang dipakai adalah “C” 200 . 75 . 8,5
Tegangan Geser Tetap
Lx tetap = ½ . ᶐbm x . Ɩ + ½Px
= (½ . 24,93) x 6,8 + (½ . 153,004)
= 84,76 + 76,502
= 161,26
Ly tetap = ½ . ᶐbm y . Ɩ + ½Py
= (½ . 41,81) x 6,8 + (½ . 224,89)
= 142,15 + 112,45
= 254,59
d= 8,5 r=6 Wx = 191
Ix = 1910 Lx = 161,26 Wy = 27
Iy = 148 Ly = 254,59
Ʈ x =Lx .Wxd
10× Iy
= 161,26× 191
8,510
×148
= 244,84
Ʈy =Lx .Wxd
10× Iy
= 254,59× 278,510
×1910
= 5,99
4
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
9697
98
99
100
101
102
103
104
105
106
107108
Ʈ Total = √ (δbs x )2+( δbs y )2
= √ (244,84 )2+(5,99 )2
= 244,91 ≤ Ʈ=0,6 . 1600
= 244,91 ≤ Ʈ=960
Periksa tegangan ijin ( Efek Huber Hangky) dalam keadaan tetap
σi = √ (δ )²+σ (Ʈ )²
= √ (1535,57 ) ²+3(244,91) ²
= 1584,212 ≤ δ = 1600 Kg/cm² ok
Jadi profil yang digunakan adalah canal “C” 200 . 75 . 8,5
Beban Sementara
Ci = 0,02 (α)-0,4
= 0,02 (34,32)- 0,4 = 0,2846 --> jika (-) tidak perlu dihitung
Beban angin = w angina x C x Jarak gording= 45 x 0,2846 x 6,8= 12,81 kg/m
Beban angin terpusat
Pa = ᶐa x Jarak kuda−kudacosα
= 12,81 x 6,8cos34.23 = 101,39
Max = 0May = ⅟4 x Pa x jarak kuda-kuda
= ⅟4 x 101,39 x 6,8
= 172,37
5
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127128129130131132133134135136
137138139
140141142
Mbs x = Mbt y+ May= 404,20 + 0= 404,20
Mbs y = Mbt y+ May= 653,37 + 127,37= 826,10
Tegangan lentur akibat beban sementara
σbs x = Mbs x
Wy( table)= 404,20
27=14,97 x 100 = 1497
σbs y = Mbs y
Wy( table)=826,10
191=4,33 x 100= 433
σ Total sementara = √ (δbs x )2+( δbs y )2
= √ (1497 )2+(433 )2
= 1558 ≤ σ’ x 1,25
= 1558 ≤ 2000 ok
Tegangan geser sementara
Lx sementara= ½ . ᶐbm x . (jarak kuda-kuda) + ½Px= ½ . 24,93 . ( 6,8 ) + ½ 153,004= 161,264
Ly sementara= ½ . ᶐbm y . (jarak kuda-kuda) + Pa y . cosα= ½ . 41,81 . ( 6,8 ) +101,395 x 153,004 x cos34,23
= 229,24
Ʈ x =Lx .Wxd
10× Iy
= 161,26× 191
8,510
×148
= 244,84
Ʈy =Lx .Wxd
10× Iy
= 229,24 ×278,510
× 1910
= 5,40
6
143144145146147148149
150151152153154155156
157 158
159160161162163164
165
166
167
168
169170171172173174175176
177
178
179
180
181
182
183
184
Ʈ Total = √ (δbs x )2+( δbs y )2
= √ (244,84 )2+(5,40 )2
= 244,89 ≤ Ʈ=0,6 . 1600
= 244,89 ≤ Ʈ=960
Periksa tegangan ijin ( Efek Huber Hangky) dalam keadaan tetap
σi = √ (δ )²+σ (Ʈ )²
= √ (1696,20 ) ²+3(244,89) ²
= 1748 ≤ δ = 0,6 . 1600 Kg/cm²
= 1748 ≤ δ = 2000 Kg/cm² ok
Periksa Kekuatan lendutan
Lendutan izin (δ’) = jarak kuda−kudaFy
=6,8× 100240
=2,833 cm
Beban terbagi rata qx= qbm x+qa x
= 24,93+0=24,93
qy= qbm y+qa y
= 41,81+172,37=214,18
Beban terpusat Gording
Px=153,004 Py=224,885
Lendutan pada Gording
δx = 5. qx
100.l ⁴
384. E . Iy+ 1. Px . l ³
48. E . Iy
=5(24 .10 ¯ ²)(6,8 .10) ⁴384.(2,1.10 ¯ 6) .1910
+ 1.224,805 .(680) ³48.(2,1. 10 ¯ 6). 1910
= 5,45
7
185
186
187
188
189
190
191
192
193
194
195
196
197198199
200
201
202
203
204
205
206
207
208209
δy = 5. qy
100.l ⁴
384. E . Ix+ 1. Py . l ³
48. E . Ix
=5(214,18 .10 ¯ ²)(680) ⁴384.(2,1. 10 ¯ 6) .1910
+1. 224,805 .(680) ³
48.(2,1. 10¯ 6).1910
= 1,86
8
210
211
212
δ Total = √δx ²+δy ²
= √5 , 45²+1 ,86²
= 5,76 ≥ 2,83 Not ok
Karena Lendutan total lebih besar dari lendutan yang di izinkan, maka ada alternative lain untuk mengatasinya yaitu dengan menambahkan Track Stang. Ini lebih efisien dibandingkan dengan mengganti profil nya.
Jarak Track Stang
Jarak kuda-kuda = (JK . 100 )+103
=(6,8 .100 )+10
3=236,67
Lendutan Izin δ '=236,67240
=0,99
Lendutan pada Gording
δx = 5. qx
100. l4
384. E . Iy+ 1 . Px . l3
48. E . Iy
= 5(24 /100)(236,67) ⁴384.(2,1 .10 ¯ 6) .148
+ 1 .224,805.(236,67)³48.(2,1 .10 ¯ 6) .148
= 0,168
δy = 5. qx
100. l4
384. E . Ix+ 1. Py . l3
48. E . Ix
= 5(24/100)(236,67) ⁴384.(2,1. 10 ¯ 6).1910
+1 . 224,805 .(236,67) ³48.(2,1 . 10¯ 6) . 1910
= 0,037
δ Total = √δx ²+δy ²
= √0 , 168²+0 , 037²
= 0,172 ≤ δ’= 0,99 Ok
10
213
214
215
216
217218219
220
221
222223224225226227
228
229
230231
232
233
234235
236
237
238239
240
241
PEMBEBANAN PADA KUDA – KUDA RANGKA BAJA
BEBAN ATAP
q atap = Berat atap x jarak kuda-kuda
q atap= 8.154 x1.25cos(34.23)
x 6.8=83.91
Pa 1 = Pa 11= ½ bentang x q atap= ½ x 3.25 x 83.91 = 136.35 kg
Pa 2 = Pa 10= {(½ bentang) + (½ bentang)} x q atap= {(½ x 3.255)+ (½ x 6)} x 83.91 = 388.08 kg
Pa 3 = Pa 4 = Pa 8 = Pa 9= {(½ bentang) x 2} x q atap= {(½ x 6) x 2} x 83.91 = 503.46 kg
Pa 5 = Pa 7= {(½ bentang + ½ bentang)} x q atap= {(½ x 6 + ½ x 3)} x 83.91 = 377.59 kg
Pa 6 = {(½ atap) x 2} x q atap= {(½ x 3) x 2} x 83.91 = 251.73 kg
11
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256257258259260261262263264265266267268269270271272273
BEBAN GORDING
G = qbs gording (berat canal) x jarak kuda-kuda
G = 25.3 x 6.8
G = 172.04 kg
PG 1 = PG 11= (2 x ½ ) x G = 172.04 kg
PG 2 = PG 10= {(2 x ½) + (3 x ½)} x G = 430.1 kg
PG 3 = PG 9= {(3 x ½) + (4 x ½)} x G = 602.14 kg
PG 4 = PG 8={(4 x ½) + (4 x ½)} x G = 688.16 kg
PG 5 = PG 7= {(4 x ½) + (2 x ½)} x G = 516.12 kg
PG 6 = {(2 x ½) + (2 x ½)} x G = 344.08 kg
BEBAN HIDUP / BEBAN HUJAN
qh = W hujan x Jarak kuda – kuda
= 20 x 6.8
qh = 136 kg
Ph 1 = Ph 11= qh x (3.25 x ½) = 221 kg
Ph 2 = Ph 10= {(3.25 x ½) + (6 x ½)} x qh = 629 kg
Ph 3 = Ph 4 = Ph 9 = Ph 8= {(6 x ½) + (6 x ½)} x qh = 816 kg
Ph 5 = Ph 7= {(6 x ½) + (3 x ½)} x qh = 612 kg
Ph 6 = {(3 x ½) + (3 x ½)} x qh = 408 kg
BEBAN TOTAL / BEBAN TETAP TOTAL
12
274
275
276
277
278279280281282283284285286287288289
290
291
292
293
294295296297298299300301302303304
305
306
307
PBT = Pa + Pg + Ph
- PBT 1 = PBT 11 = Pa + Pg + Ph = 136.3538 + 172.04 + 221 = 529.3938 kg
- PBT 2 = PBT 10 = Pa + Pg + Ph = 388.0838 + 430.1 + 629 = 1447.1838 kg
- PBT 3 = PBT 9 = Pa + Pg + Ph = 503.46 + 602.14 + 816 = 1921.6 kg
- PBT 4 = PBT 8 = Pa + Pg + Ph = 503.46 + 688.16 + 816 = 2007.62 kg
- PBT 5 = PBT 7 = Pa + Pg + Ph = 377.595 + 516.12 + 612 = 1505.715 kg
- PBT 6 = Pa + Pg + Ph = 251.73 + 34.08 + 408 = 1003.81 kg
GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN TETAP KUDA-KUDA
13
308
309310311312313314315316317318319320321322323324325
326327328329330331332333334
MENCARI BEBAN SENDIRI
Menentukan Profil Awalo Periksa Angka Kelangsingan (λ)
Kestabilan = x < 200
X = Limin ; imin = L
λ ( Batang terpanjangketentuan kestabilan
)
imin = 1002.63200
=5.01315
Karena yang dipakai menggunakan profil ganda semua, maka imin = ix = iy dan dicoba menggunakan profil siku 180.180.16
Ix = Iy = 5.51 ≤ imin OK
Panjang Batang Luar + Panjang Batang Dalamo Panjang Batang Luar = Panjang (∑A + ∑B)
= a1+a2+a3+a4+a5+a6+a7+a8+a9+a10+b1+b2+b3+b4+b5+b6
= 393.10 + (725.72 x 3) + (361.86 x 2) + (725.72 x 3) + 393.10 + (647.06 x 2) + (621.18 x 6)
= 10887.44100 = 108.874 m
Panjang Batang Luar = 108.874 x 2 = 217.748 m
o Panjang Batang Dalam = d1+d2+d3+d4+d5+d6+d7+d8+d9+d10+d11+d12+d13+d14
= 304.75 + 550.74 + 424.99 + 770.11 + 625.14 + 1002.63 + 850.54 + 850.54 + 1002.63 + 625.14 + 770.11 + 424.99 + 550.74 + 304.75 + 1000
= 10057.8100 = 100.578 m
Panjang Batang Dalam = 100.578 x 2 = 201.156 m
Panjang Total = 217.748 + 201.156 = 418.904 m
qbs = Panjang total xberat profil x berat aksesorispanjang batang
14
335
336337338
339
340
341342
343
344
345346347348349350
351
352
353354355356
357
358
359
360
361
362
363
= 418.904 x 43.5 x1.2548.5 = 469.647 kg/m
PBS = qbs x L (Panjang batang x ½)
- PBS 1 = PBS 9 = qbs x L= 469.647 x (6.25 x ½ ) = 1467.6468 kg
- PBS 2 = PBS 8 = qbs x L= 469.647 x {(6.25 x ½ )+(6 x ½ )} = 2876.5878 kg
- PBS 3 = PBS 7 = qbs x L= 469.647 x {(6 x ½ )+(6 x ½ )} = 2817.882 kg
- PBS 4 = PBS 6 = qbs x L= 469.647 x {(6 x ½ )+(6 x ½ )} = 2817.882 kg
- PBS 5 = qbs x L= 469.647 x x {(6 x ½ )+(6 x ½ )} = 2817.882 kg
GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN SEMENTARA KUDA-KUDA
Berat Sendiri Kuda-Kuda
15
364
365
366367368369370371372373374375
376
377378
379380
381
382
qbs = Panjang total xberat profil x berat aksesorispanjang batang
= 418.904 x 43.5 x1.2548.5 = 469.647 kg/m
Pemilihan Profil
Batang Terpanjang (lk) = 10.02 m = 1002.63i min
≤ 200
= 1002.63200
=5.01315≤ i
(i yang di gunakan = 5.51) = 1002.63240
=4.177 ≤ i
BEBAN AKIBAT ANGIN KIRI
MUKA
q angin = W angin x c x Jarak kuda−kudacos α ; α = 34.23O ; C = 0.2846
= 45 x o.2846 x 6.8cos 34.23°
q angin = 105.3326 kg/m
o P 1 = q angin x ½ L= 105.3326 x (3.931 x ½ ) = 207.031 kg
o P 2 = q angin x (½ L + ½ L)= 105.3326 x {(3.931 x ½) + (7.257 x ½)} = 589.230 kg
o P4 & P3 = q angin x 2 x ½L = 105.3326 x 2 x (7.252 x ½) = 764.3986 kg
o P5 = q angin x (½ L + ½ L)= 105.3326 x {(7.257 x ½) + (3.628 x ½)} = 573.5726 kg
o P6 = q angin x ½ L= 105.3326 x (3.628 x ½) = 191.0733 kg
16
383
384
385
386
387
388
389
390
391
392
393
394
395
396397398399400401402403404405
406
407
408
BELAKANG
q angin = W angin x c x Jarak kuda−kudacos α ; α = 34.23O ; C = 0.4
= 45 x o.4 x6.8cos34.23°
q angin = 148.0431 kg/m
o P6 = q angin x ½ L= 148.0431 x (3.628 x ½) = 268.550 kg
o P7 = q angin x (½ L + ½ L)
= 148.0431 x {(7.257 x ½) + (3.628 x ½)} = 805.724 kgo P8 & P9 = q angin x (½ L + ½ L)
= 148.0431 x {(7.257 x ½) + (7.257 x ½)} = 1074.3488 kgo P10 = q angin x (½ L + ½ L)
= 148.0431 x {(7.257 x ½) + (3.931 x ½)} = 788.1815 kgo P11 = q angin x ½ L
= 148.0431 x (3.931 x ½) = 290.9787 kg
GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN ANGIN KIRI
17
409
410411
412
413414415416417418419420421422423424425426427428
430431432
433
434
BEBAN AKIBAT ANGIN KANAN
MUKA
q angin = W angin x c x Jarak kuda−kudacos α ; α = 34.23O ; C = 0.2846
= 45 x o.2846 x 6.8cos 34.23°
q angin = 105.3326 kg/m
o P 11 = q angin x ½ L= 105.3326 x (3.931 x ½ ) = 207.031 kg
o P 10 = q angin x (½ L + ½ L)= 105.3326 x {(3.931 x ½) + (7.257 x ½)} = 589.230 kg
o P8 & P9 = q angin x 2 x ½L = 105.3326 x 2 x (7.252 x ½) = 764.3986 kg
o P7 = q angin x (½ L + ½ L)= 105.3326 x {(7.257 x ½) + (3.628 x ½)} = 573.5726 kg
o P6 = q angin x ½ L= 105.3326 x (3.628 x ½) = 191.0733 kg
BELAKANG
q angin = W angin x c x Jarak kuda−kudacos α ; α = 34.23O ; C = 0.4
= 45 x o.4 x6.8cos34.23°
q angin = 148.0431 kg/m
o P6 = q angin x ½ L= 148.0431 x (3.628 x ½) = 268.550 kg
o P5 = q angin x (½ L + ½ L)= 148.0431 x {(7.257 x ½) + (3.628 x ½)} = 805.724 kg
o P3 & P4 = q angin x (½ L + ½ L)= 148.0431 x {(7.257 x ½) + (7.257 x ½)} = 1074.3488 kg
o P2 = q angin x (½ L + ½ L)= 148.0431 x {(7.257 x ½) + (3.931 x ½)} = 788.1815 kg
o P1 = q angin x ½ L
18
435
436
437
438
439
440
441442443444445446447448449450451452453454455
456
457458459460461462463464465466467468
= 148.0431 x (3.931 x ½) = 290.9787 kg
GAMBAR PEMBEBANAN UNTUK PEMBEBANAN BEBAN ANGIN KANAN
KONTROL MOMEN
1. Beban Total / Beban Tetap TotalKontrol ∑M = 0 , terhadap beban tetap (PBT)
o ∑MA = 0-Vb x 48.50 + P1 x 0 + P2 x 3.25 + P3 x 9.25 + P4 x 15.25 + P5 x 21.25 + P6 x 24.25 + P7 x 27.25 + P8 x 33.25 + P9 x 39.25 + P10 x 45.25 + P11 x 48.5 = 0
-Vb x 48.50 = -383809.6
Vb = 383821.648.5
Vb = 7913.8 kg
19
469
470
471472473
474
476477478479
480
481482483484485486487488
489
490491
o ∑MB = 0Va x 48.50 - P11 x 0 - P10 x 3.25 - P9 x 9.25 - P8 x 15.25 - P7 x 21.25 - P6 x 24.25 - P5 x 27.25 - P4 x 33.25 - P3 x 39.25 - P2 x 45.25 - P1 x 48.5 = 0
Va x 48.50 = 383765.9
Va = 383765.948.5
Va = 7912.7 kg
Va + Vb = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9 + P10 + P1115826.5 = 15826.8 = 0.3 ≈ 0 !OK!
2. Beban Sendirio ∑MA = 0-Vb x 48.50 + P1 x 0 + P2 x 6.25 + P3 x 12.25 + P4 x 18.25 + P5 x 24.25 + P6 x 30.25 + P7 x 36.25 + P8 x 42.25 + P9 x 48.25 = 0
-Vb x 48.50 = -552363.57
Vb = 552363.5748.5
Vb = 11388.94 kg
o ∑MB = 0Va x 48.50 - P9 x 0 - P8 x 6.25 - P7 x 12.25 - P6 x 18.25 - P5 x 24.25 - P4 x 30.25 - P3 x 36.25 - P2 x 42.25 - P1 x 48.25 = 0
Vb x 48.50 = 552363.57
Va = 552363.5748.5
Va = 11388.94 kg
Va + Vb = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9 + P10 + P11
22777.88 = 22777.87 = 0.01 ≈ 0 !OK!
20
492493494495496
497
498499500501502503504505506507
508
509510511512513514515
516
517518519520
521
522
523
524
525
PERHITUNGAN GAYA BATANG RANGKA KUDA-KUDA
1. Gambar Rangka Kuda Kuda Baja
2. Pendimensian Rangka batang dalam Perhitungan SAP 2000
21
526
527
528
529
530
531
532
533534
535
536537
538
539
3. Pembebanan pada Rangka Kuda Kuda (Pada Joint)
3.1 Pembebanan PBT (Pembebanan Beban Tetap)
3.2 Pembebanan PBS (Pembebanan Beban Sementara)
3.3 Pembebanan Angin Kiri
22
540
541
542
543
544
545546
547548549
550551552553
3.4 Pembebanan Angin Kanan
4. Reaksi Batang yang ditimbulkan yang terinput di SAP 2000
23
554555556557558
559560
561
562
4.1 Akibat Load Dead
4.2 Akibat Load Live
4.3 Akibat Left Load Wind
4.4 Akibat Right Load Wind
24
563
564565
566567568569
570571
Kombinasi 2
Kombinasi 3
Dari tabel combo d SAP dicari batang tekan & tarik, yg terbesar, profil ganda
26
578579
580581582
583
584
DESAIN BATANG TARIK
Dari tabel kombinasi pembebanan didapat
27
No Batang Station (m) OutputCase Gaya Tekan (-) Gaya Tarik
(+)A1 3.93585 Comb 2 -15672.71A2 7.22578 Comb 1 -19096.07A3 7.22578 Comb 1 -16561.68A4 7.25578 COMB1 -13928.74A5 3.62789 COMB1 -11085.66A6 3.62789 COMB1 -11085.66A7 7.25578 COMB1 -13928.74A8 7.25578 COMB1 -16561.68A9 7.25578 COMB1 -19096.07A10 3.93585 COMB3 -21163.73B1 6.47185 COMB1 18029.59B2 6.20455 COMB1 15670.37B3 6.2201 COMB1 13346.11B4 6.16117 COMB1 10677.28B5 6.16117 COMB1 10677.28B6 6.2201 COMB1 13346.11B7 6.20455 COMB1 15670.37B8 6.47185 COMB1 18029.59D1 3.04821 COMB1 -1646.54D2 5.50858 COMB2 1239.01D3 4.27102 COMB2 -2194.59D4 7.72622 COMB2 2253.65D5 6.24743 COMB1 -2823.45D6 10.01966 COMB2 3893.66D7 8.694 COMB2 -3734.23D8 8.694 COMB3 -5066.18D9 10.01966 COMB3 5160.34D10 6.24743 COMB3 -4197.07D11 7.72622 COMB3 3224.57D12 4.27102 COMB3 -3060.73D13 5.50858 COMB3 1696.63D14 3.04821 COMB3 -2489.86V1 10.2 COMB1 11548.76
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608609610611612613614615616617
Desain batang tarik (profil ganda)Beban terbesar Batang B1 = 18.029,6 kg
Panjang batang = 6,50 m
Data-data profil → L 18 0. 18 0.1 6 A = 55,4 IX = IY = 1680 cm4
imin = iη = 3,5 cm ix = iy = 5,51 cm ex = ey = 50,2 mm = 5,02 cm = 0,0502 m
a = 2ey + t
= (2 x 5,02) + 1.9
= 11.94 cm
Sumbu bahan → sb.x= 5,51 → ix
Sumbu bebas bahan → sb.y = iy = √ ( iy )2+(12
a)2
= √ (5,51 )2+ (5,97 )2
= 8,12 cm
Diambil → ix = 5,51
Periksa Kestabilan
λx = LKimin ≤ 240 ; λy =
LKimin ≤ 240
λx = 6505,51 ≤ 240 ; λy =
6508.12 ≤ 240
= 117,9 ≤ 240 ….. ok ; = 80.05 ≤ 240 …. ok
28
618619
620621622623624625626627628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
Periksa Kekuatan
tr = P
2x An ≤ (dimana = 1600 kg/cm2)
Dimana → An = 80% x 55,4 = 44,32
tr = 18.029,6
88,64 ≤
= 203.4 ≤ 1600 ….. ok
BATANG TEKAN
Dari tabel kombinasi pembebanan didapat
Desain batang tekan (profil ganda)Beban terbesar Batang A10 = -21085.9 kg
Panjang batang = 3.93 m
Data-data profil → L 18 0. 18 0.1 6 A = 55,4 IX = IY = 1680 cm4
imin = iη = 3,50 cm ix = iy = 5,51 cm ex = ey = 502 mm = 5,02 cm = 0,0502 m Iη = 679 cm2
a = 2ey + t
= (2 x 5,02) + 1.9
= 11.94 cm
Periksa Kekuatan
P . ω
A ≤
→ = = 1,21
Karena = 1.21 maka
29
…Ok.
647
648
649
650
651
652
653
654
655656
657658659660661662663664665666
667
668
669
670
671
672
673
674
= 1,25 (1,21)2
= 1,83
Periksa Kestabilan
λx = LKimin ≤ 200
λx = 3933.5 ≤ 200
= 112.3 ≤ 200 ….. ok
Sumbu bebas bahan → sb.y = iy = √ ( iy )2+(12
a)2
= √ (5,51 )2+ (5,97 )2
= 8,12
→ y = lkiy ≤ 200
= 3935,51 ≤ 200
= 71.32 ≤ 200 ….. ok
Desain Plat koppel
w² = x²
→ x² = I² +y²
I = √ (112.3)2−(71.32)²
I = 86.7
ii = 3,5 cm
i1 = I . i1
= 86.7 x 3,5 = 303.45 cm
30
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
Jumlah medan
n 393
303.45
n = 1,29 pembulatan menjadi 3
lI = 3933 = 131 cm
Periksa angka kelangsingan terhadap sumbu bebas bahan
1 = l 1imin
= 1313,5 = 37,43
w = iy = √ ( y )2+1²
= √ (71.32 )2+ (37,43 )2
= 80,54
Hitung lebar plat Koppel ; (dicoba t = 1/4 “ = 0.6 cm)
112 . t . h3
10 . I .2al i
112 . t . b3
10 .679 .(2 ×5,97)131
b3 618,874 x12
0.6
b 23, 131
H = (2 x 18) + 1.6 = 37.6
ukuran plat Koppel
H x b x t = 37,6 x 23, 131 x 0,6
31
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
Periksa kekuatan plat Koppel
ukuran plat Koppel : 37,5 x 20.04 x 0,6
lI = 131 cm
w = 80,54
Mencari besarnya Q
Q = 2% x 21085.9 kg = 421.7 kg
Atau → Q = ww131 x 21085.9
Ww …..?
c = w❑ √ fy
E
= 80 ,54❑ √ 2400
2,1 x10⁶ = 0.867
Jadi : 0,25 c 1,2
ww= 1,43
1,6−0,6 c
ww= 1,43
1,6−0,6 ( 0.91 ) = 1.3243
jadi →Q = ww131 x 21085.9
Q = 1.3243
131 x 21085.9 = 213.163 kg
Tentukan gaya geser (T) atau (L)
T = Q2a
= 145.9 x 131
2 x5,97 = 2463 kg
32
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
= Lsxbix
= 32 x
Tb xh (1600 x 0,6 = 960)
= 32 x
246320.04 x37,5 ( 960)
= 4.82 ….. ok
Periksa kekuatan batang tekan
w x
90.76 112.3….. ok
Periksa kritis (c)
w = 90.76 → c = w❑ √ FY
E
= 90.76❑ √ 2100
2,1 .10⁶ = 0.91
Jadi : 0,25 c 1,2
Sehingga → ω = 1,43
1,6−0,67 c
ω = 1,43
1,6−0,67 (0.91) = 1,44
cek= ρ . w2x A (1600)
= 21085.9 x 1,44
2× 55,4 1600
= 274.04 1600….. ok
Jadi profil 10.200.18 bisa dipakai
33
744
745
746
747748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
Detail SambunganPada sambungan desain kuda-kuda ini menggunakan jenis sambungan Las yang digunakan sebagai pengikat antara profil kuda-kuda dan plat kopel yang terdapat di desain kuda kuda ini.
P = P4 ; P = gaya batang
A=Px √sin α+¿¿¿¿ ; α = sudut
a= t√2
; t = 1.6 ; a = 1.13
Ln = Aa
Lbr = Ln + 3a
Lbr total = Lbr x 4
DETAIL BATANG A1
P = 15672.71
4=3918.18
A = 3918.18 x√sin 34.23+¿¿¿¿ = 3.77
ln = 3.771.13
=3.33
lbr = 3.33 + (3 x 1.13) = 6.72
Lbr total Batang A1 = 6.72 x 4 = 26.88 ~ 27
DETAIL BATANG A2
34
766
767
768
769
770
771772
773
774775776
777
778
779
780
781782783
784
785
786
787
788789790
791
P = 19096.07
4=4774.02
A = 4774.02 x √sin 34.23+¿¿¿¿ = 4.59
ln = 4.591.13
=4.06
lbr = 4.06+ (3 x 1.13) = 7.45
Lbr total Batang A2 = 7.45 x 4 = 29.8 ~ 30
DETAIL BATANG B1
P = 18029.59
4=4507.40
A = 4507.40 x √sin 19.23+¿¿¿¿ = 4.70
ln = 4.701.13
=4.15
lbr = 4.15+ (3 x 1.13) = 7.55
Lbr total Batang B1 = 7.55 x 4 = 30.2 ~ 31
DETAIL BATANG B2
P = 15670.37
4=3917.59
A = 3917.59 x√sin 19.23+¿¿¿¿ = 4.08
ln = 4.081.13
=3.61
lbr = 3.61+ (3 x 1.13) = 7.00
Lbr total Batang B2 = 7.00 x 4 = 28
35
792
793
794
795796
797798
799
800
801
802803
804
805
806
807
808
809810
811812
α 56.99no. Batang gaya batang P bar Á ln lbr
D2 1239.01 309.75 0.33 0.29 3.69 4.00D13 1696.63 424.16 0.45 0.40 3.80 4.00
α 45.10no. Batang gaya batang P bar Á ln lbr
D3 2194.59 548.65 0.59 0.52 3.91 4.00D12 3060.73 765.18 0.82 0.72 4.12 5.00
α 67.07no. Batang gaya batang P bar Á ln lbr
D4 2194.59 548.65 0.59 0.52 3.91 4.00D11 3060.73 765.18 0.82 0.72 4.12 5.00
α 61.32no. Batang gaya batang P bar Á ln lbr
D5 2823.45 705.86 0.76 0.67 4.06 5.00D10 2823.45 705.86 0.76 0.67 4.06 5.00
α 72.59no. Batang gaya batang P bar Á ln lbr
D6 3893.66 973.42 1.04 0.92 4.32 5.00D9 5160.34 1290.09 1.38 1.22 4.62 5.00
α 69.35no. Batang gaya batang P bar Á ln lbr
D7 3734.23 933.56 1.00 0.88 4.28 5.00D8 3567.29 891.82 0.96 0.84 4.24 5.00
α 90.00no. Batang gaya batang P bar Á ln lbr
V1 11548.76 2887.19 3.09 2.73 6.13 7.00
37
820
822
GAMBAR KUDA – KUDA BAJA TYPE SCISSORS BENTANG 48,5 M
38
35
823824825826827828829830831832833834835836837838839840841842843844845846847