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 )