bab 6 - an tangga
TRANSCRIPT
200 cm
250 cm
200 cm
250 cm
BAB 6
PERENCANAAN TANGGA
6.1. Data Perencanaan Tangga
Type tangga = Tangga tetap Ruang yang tersedia = (400 x 400) cm Beda elevasi = 4,0 m = 157,480 inchi Panjang tangga = 2,5 m = 98,425 inchi Lebar bordes = 1,5 m = 59,055 inchi Elevasi bordes = 2,0 m = 78,740 inchi Panjang bordes = 4,0 m = 157,480 inchi Balok tangga 1 sampai bordes
Direncanakan optrede = 25 cm
n optrede =
20025 = 8 buah
n antrede = n optrede – 1
= 8 – 1 = 7 buah
Antrede =
2507 = 35,714 cm
= 40 cm
Tan α =
optredeantrede =
200250 =0,8
α = 38,660 o
Balok tangga 2 sampai lantai 2
Direncanakan optrede = 25 cm
n optrede =
20025 = 8 buah
n antrede = n optrede – 1
= 8 – 1 = 7 buah
Antrede =
2507 = 35,714 cm
= 40 cm
Tan α =
optredeantrede =
200250 =0,8
α = 38,660 o
6.2. Rencana Anak Tangga
Sin α =
dA
d1 = sin α . Antrede
= sin 38,660 o x 40
MC 10x25
2,0 m
MC 10x25
= 24,988 cm = 9,838 inch
d2 = d1 = 9,838 inch
Sehingga digunakan profil MC 10x25 dengan d = 10 inch ( LRFD I-52 dan 154 )
Anak tangga, digunakan :
Tebal baja = 3 cm = 1,181 inch
BJ Baja = 7850 kg/m3
Tangga terdiri 2 bagian
a. Bagian atas
- O = 8 @ 0,25 m
- A = 7 @ 0,40 m
b. Bagian bawah
- O = 8 @ 0,25 m
- A = 7 @ 0,40 m
Pembebanan Anak Tangga
1. Berat sendiri plat (qd)
qd = BJ baja x tebal x (L antrede + L optrede )
= 7850 x 0,03 x (0,40 + 0,25)
= 153,075 kg/m
= 0,103 kip/ft
2. Berat antrede (ql)
ql = 250 x Lantrede
= 250 x 0,40
= 100 kg/m
= 0,067 kip/ft
3. Berat terfaktor (qu)
qu = 1,2 qd + 1,6 ql
= 1,2 x 0,103 + 1,6 x 0,067 = 0,231 kip/ft
6.3. Rencana Balok Tangga
Profil yang digunakan MC 10x25 ( LRFD I-52)
Ix = 110 inch4
Zx = 25,8 inch3
Sx = 22 inch3
qx = 25 lb/ft = 0,025 kip/ft
Pembebanan balok tangga :
1. Berat sendiri plat (qd)
qd profil = 0,025 kip/ft
2. Berat antrede (ql)
ql = 0,5 . L antrede . 250
= 0,5 x 0,40 x 250
= 50 kg/m
= 0,034 kip/ft
3. Berat berfaktor (qu)
qu = 1,2 qd + 1.6 ql
= 1,2 x 0,025 + 1,6 x 0,034
= 0,084 kip/ft
6.4. Pembebanan Balok Bordes
1. Berat sendiri (qd)
qd = BJ baja x lebar bordes x tebal plat
= 7850 x 1,5 x 0,03
= 353,25 kg/m
= 0,237 kip/ft
2. Berat bordes (ql)
ql = lebar bordes x 250
= 1,5 x 250
= 375 kg/m
= 0,252 kip/ft
3. Berat terfaktor (qu)
qu = 1,2 qd + 1,6 ql
= 1,2 x 0,237 + 1,6 x 0,252
= 0,688 kip/ft
Maka :
qu1 = qu anak tangga + qu balok tangga
= 0,231 + 0,084
= 0,315 kip/ft
qu2 = qu balok bordes + qu balok tangga
= 0,688 + 0,084
= 0,772 kip/ft
Hasil Analisis SAP 2000 8 pada Tangga
Gambar 6 Axial Force Diagram(Dead) dari SAP 2000 8 Tangga
Gambar 6. Momen 3-3 Diagram(Dead) dari SAP 2000 8 Tangga
Gambar 6. Shear force 2-2 Diagram (Dead) dari SAP 2000 8 Tangga
6.5. Balok Tangga
Dipakai profil MC 10x25
Dari tabel LRFD 1-52 didapatkan :
A = 7,35 inch2
d = 10 inch
tw = 0,38 inch
bf = 3,405 inch
tf = 0,575 inch
eo = 1,036 inch
Ix = 110 inch4
J = 0,64 inch4
Sx = 22 inch
rx = 3,87 inch
Cw = 125 inch6
Zx = 25,8 inch
ry = 1,00 inch
Zy = 5,21 inch
Sy = 3,00 inch
Iy = 7,35 inch4
6.5.1. Kontrol Kekuatan Terhadap Momen
Mu = 3,4247 kip-ft ( hasil SAP )
Lb = 126,0455 inch
X1 =
πSx √ E .G .J . A
2
=
π22 √29000 .11200. 0 ,64 . 7 ,35
2 = 3946,88 inch
X2 =
4 .CwIy ( Sx
G . J )2
=
4 .1257 ,35 (22
11200.0 ,64 )2
= 6,4 x 10-4 inch
Lr =
ry . X1
Fy−Fr √1+√1+X2 .(Fy−Fr )2
=
1. 3946 ,87936−10 √1+√1+6,4 x10−4 (36−10 )2
= 225,015 inch
Lp =
300 . ry
√Fy =
300 .1
√36 = 50 inch
Lp = 50 inch
Lb = 126,0455 inch
Lr = 225,015 inch
Maka
Lp < Lb < Lr
M1 = 0,0516 kip-inch
M2 = 41,096 kip-inch
Cb = 1,75 + 1,05 (M 1
M 2)
+ 0,3 (M 1
M 2)2
= 1,75 + 1,05 ( 0 ,051641 ,096 )
+ 0,3 ( 0 ,051641 ,096 )
2
= 1,751 Cb < 2,3.......................................................................................LRFD 6-197
1,751 < 2,3 ………..( Aman )
Sehingga diambil Cb = 1,751
Mp = Fy . Zx = 36 x 25,8
= 928,80 kip-inch = 77,40 kip ft
Mr = (Fy-Fr) . Sx = (36-10) x 22
= 572 kip-inch = 47,667 kip ft
Mn = Cb . [Mp − (Mp − Mr ) . ( Lb− Lp
Lr − Lp )] = 1,751 x (928,80 – (928,80-572)) x
(126 ,0455−50225 ,015−50 )
= 355,1132 kip-inch = 112,926 kip ft
φ Mn = 0,9 . Mn = 0,9 x 112,9261
= 101,633 kip ftCheck :
φ Mn > Mu
101,633 kip ft > 3,4247 kip ft...................(Aman)
6.5.2. Kontrol Terhadap Geser
Vu = 1,616 kip ( hasil SAP )
tw = 0,38 inch
tf = 0,575 inch
h = d – 2.tf = 9 – 2.1,15 = 8,85 inch
htw ¿
418
√ fy7,1
0 ,38 ¿
418
√36
23,289 ¿ 69,667 ...........(Aman)
Vn = 0,6 . Aw. Fy = 0,6 . (8,85 . 0,38) . 36
= 72,641 kip
Ø Vn > Vu
0,9 . 72,641 > 1,616 kip 65,377 kip > 1,616 kip ...........(Aman)
6.5.3. Kontrol Terhadap Aksial
Pu = 10,333 kip
rx = 3,87 inch
Ag = 7,35 inch2
L = 126,0455 inch
Mencari nilai k
Dari LRFD hal 2 -18 diperoleh k = 0,65 (jepit – jepit)
Sehingga :
k . Lrx
=0 ,65 . 126 ,04553 ,87 = 21,171
Dari LRFD hal 6 – 147
k . Lrx
=21,171 → Øc.Fcr = 29,83 ksi
Øc.Pn = Øc.Fcr . Ag
= 29,83 . 7,35 = 219,2505 kip
Øc.Pn > Pu
219,2505 kip > 10,333 kip ……….(Aman)
6.5.4. Kontrol Terhadap Kompaksitas Bahan
bftf
≤65
√Fy3 ,4050 ,575
≤65
√36
5,922 ¿ 10,833 ( Kompak )
dtw
<640
√Fy100 ,38
<640
√36
26,316 < 106,667 ( Kompak )
6.5.5. Kontrol Terhadap Defleksi
qu = 0,315 kip/ft
L = 78,740 inch
Δ= 1384
×q .L4
E . Ix
=
1384 x
0 ,0262. 78 ,7404
29000 .110 = 8,231 x 10-4 inch
Δmax= L360
=78 ,740360 = 0,219 inch
Δ<Δmax 8,231 x 10-4 inch < 0,219 inch ( Aman )
6.6. Balok Borde s
Dipakai profil MC 9x25,4
Dari tabel LRFD 1-54 didapatkan :
A = 7,47 inch Ix = 88 inch4
Iy = 7,65 inch4
d = 9,00 inch Zx = 23,2 inch Zy = 5,23 inch
tw = 0,45 inch Sx = 19,6 inch Sy = 3,02 inch
bf = 3,5 inch rx = 3,43 inch ry = 1,01 inch
tf = 0,55 inch Cw = 104 inch6
eo= 0,986 inch J = 0,69 inch4
6.6.1. Kontrol Kekuatan Terhadap Momen
Mu = 16,5467 kip-ft ( hasil SAP )
Lb = 59,055 inch
X1 =
πSx √ E .G .J . A
2 =
π19 ,6 √29000 .11200. 0 ,69 .7 ,47
2 = 4637,370 inch
X2 =
4 .CwIy ( Sx
G . J )2
=
4 .1047 ,65 (19 ,6
11200.0 ,69 )2
= 3,5 x 10-4 inch
Lr =
ry . X1
Fy−Fr √1+√1+X2 .(Fy−Fr )2
=
1,01 . 4637 ,36936−10 √1+√1+3,5 x 10−4 (36−10)2
= 261,796 inch
Lp =
300 . ry
√Fy =
300 .1 ,01
√36 = 50,5 inch
Lp = 50,5 inch
Lb = 59,055 inch
Lr = 261,796 inch
Maka
Lp < Lb < Lr
M1 = 0,071 kip ft = 0,852 kip-inch
M2 = 16,5467 kip ft = 198,560 kip-inch
Cb = 1,75 + 1,05 (M 1
M 2)
+ 0,3 (M 1
M 2)2
= 1,75 + 1,05 ( 0 ,852198 ,560 )
+ 0,3 ( 0 ,852198 ,560 )
2
= 1,755
Cb < 2,3.......................................................................................LRFD 6-197
1,755 < 2,3 ………..(OK!)
Sehingga diambil Cb = 1,755
Mp = Fy . Zx = 36 x 23,2 = 835,20 kip-inch
Mr = (Fy-Fr) . Sx = (36-10) x 19,6 = 509,6 kip-inch
Mn = Cb . [Mp − (Mp − Mr ) . ( Lb− Lp
Lr − Lp )]= 1,755 x (835,20 – (835,20 -509,6)) x
(59 ,055−50 ,5261 ,796−50 ,5 )
= 1442,238 kip-inch
= 120,186 kip ft
φ Mn = 0,9 . Mn = 0,9 x 120,186 = 108,168 kip-ft
Check :
φ Mn > Mu 108,168 kip-ft > 16,5467 kip-ft...................( Aman)
6.6.2. Kontrol Terhadap Geser
Vu = 5,065 kip ( hasil SAP )
h = d . 2.tf = 8 x 2 x 0,55
= 7,9 in
htw ¿
418
√ fy7,9
0 ,45 ¿
418
√36
17,556 ¿ 69,667 ...........( Aman )
Vn = 0,9 . Aw. Fy = 0,9 . (7,9 . 0,45) . 36
= 76,788 kip
Ø Vn > Vu
0,6 . 76,788 > 3,799 kip
69,109 kip > 5,065 kip ...........( Aman )
6.6.3. Kontrol Terhadap Kompaksitas Bahan
bftf
≤65
√Fy3,5
0 ,55≤65
36
6,264 ¿ 10,833 ( Kompak )
dtw
<640
√Fy
90 ,45
<640
√36
20 < 106,667 ( Kompak )
6.6.4. Kontrol Terhadap Defleksi
qu = 0,772 kip/inch
L = lebar bordes = 59,055 inchi
Δ= 5384
×q .L4
E . Ix
=
5384 x
0 ,772 .59 ,0554
29000 . 88 = 7,984 x 10-4 inch
Δmax= L360
=59 ,055360
=0 ,1640inch
Δ<Δmax 7,984 x 10-4 < 0,164 ( Aman )