Perencanaan plat untuk rumah 2 lantai (one ways, two ways)

2. PEMBEBANAN
PEMBEBANAN PADA PLAT
LANTAI 2
PEMBEBANAN PADA
ATAP
Keramik granit = 0,02 x 2,6kN/m3 = 0,5kN/𝑚2
Spesi = 0,02 x 21 kn/m3 = 0,42kN/𝑚2
Plat = 0,1 x 24 kn/m3 = 2,4k𝑁/𝑚2
Plafon menggantung = 0,11 + 0,07kn/m2 = 0,18kN/𝑚2
Instalasi = 0,25kN/𝑚2
= 3,77kN/𝑚2
Rumah Tinggal / hunian = 1,92 kN/𝑚2
Wu = 1,2 DL + 1,6 LL
= 1,2 (3,77) + 1,6 (1,92)
= 7,60 kN/𝑚2
Spesi = 0,03 X 21kN/m3 = 0,63 kN/m2
Plat = 0,09 x 24 kn/m3 = 2,16 kN/𝑚2
Instalasi = 0,25 kN/𝑚2
Plafon menggantung = 0,18kN/𝑚2
= 3,22 kN/𝑚2
Atap Datar = 0,96 kN/𝑚2
Wu = 1,2 DL + 1,6 LL + 0,5 H
= 1,2 (3,22) + 1,6 (0,96) + 0,5 (0,5)
= 5,65 kN/𝑚2
Beban air hujan 5cm = 0,05 x 10 kn/m2 = 0,5
kN/𝑚2
(DEAD LOAD)
(LIVE LOAD)
(DEAD LOAD)
(LIVE LOAD)

3. BALOK = 25
KOLOM = 18
TINJAU ONE WAY
&
TWO WAY

4. BALOK = 25
KOLOM = 18
TINJAU ONE WAY
&
TWO WAY

5. PERENCANAAN PLAT
ONE WAY SLAB PLAT
LANTAI 2
𝑙 𝑛 =
1
2
𝐿 𝑥 + 2.
1
2
1
4
𝐿 𝑥 = 2250 𝑚𝑚
ℎ 𝑝𝑙𝑎𝑡 = 100 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
1
24
𝑙 𝑛 = 93,45 𝑚𝑚
Beban: 𝑊𝑢 = 7,60 𝑘𝑁/𝑚2
Momen:
𝑀𝐴 = 𝑀 𝐶 =
1
24
𝑊𝑢. 𝑙 𝑛
2
= 1,60 𝑘𝑁. 𝑚
𝑀𝑙𝑎𝑝 𝐴𝐵 = 𝑀𝑙𝑎𝑝 𝐵𝐶 =
1
11
𝑊𝑢. 𝑙 𝑛
2
= 3,50 𝑘𝑁. 𝑚
𝑀 𝐵 =
1
9
𝑊𝑢. 𝑙 𝑛
2
= 4,28 𝑘𝑁. 𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = ℎ − 𝑝 −
1
2
∅ 𝑑 = 100 − 20 −
1
2
10
= 75 𝑚𝑚
TINJAU I & J

6. PERENCANAAN PLAT
Tulangan tak terduga; 𝑀𝐴 = 𝑀 𝐶 = 1,60 𝑘𝑁.
𝑚
• 𝑅 𝑛 =
𝑀𝑢
𝑏.𝑑2 = 0,28 𝑀𝑃𝑎
• 𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
• 𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚.𝑅 𝑛
𝑓𝑦
= 0,00088
Tidak perlu mengecek 𝜌 𝑚𝑖𝑛 < 𝜌 < 𝜌 𝑚𝑎𝑥
• 𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 66 𝑚𝑚2
As perlu < As
OK Tapi boros
Tulangan lapangan; 𝑀𝑙𝑎𝑝 𝐴𝐵 = 𝑀𝑙𝑎𝑝 𝐵𝐶 = 3,
50 𝑘𝑁. 𝑚
• 𝑅 𝑛 =
𝑀𝑢
𝑏.𝑑2 = 0, 62𝑀𝑃𝑎
• 𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
• 𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚.𝑅 𝑛
𝑓𝑦
= 0,0019
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044 yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
• 𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 142,5 𝑚𝑚2
As perlu < As
OK Tapi boros

7. PERENCANAAN PLAT
Tulangan tumpuan; 𝑀 𝐵 = 4,28 𝑘𝑁. 𝑚
• 𝑅 𝑛 =
𝑀𝑢
𝑏.𝑑2 = 0, 76𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
• 𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚.𝑅 𝑛
𝑓𝑦
= 0,0024
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 330 𝑚𝑚2
As perlu < As
OK Tapi boros
• Tulangan pembagi:
𝐴 𝑠 =
0,18 𝑏. ℎ
100
= 180 𝑚𝑚2
As perlu < As
OK Tapi boros

8. TWO WAYS
PLAT
LANTAI 2
TINJAU D TIPE
𝑙 𝑥𝑛 =
1
2
𝐿 𝑥 + 2.
1
2
1
4
𝐿 𝑥 = 1125 𝑚𝑚
𝑙 𝑦𝑛 =
1
2
𝐿 𝑦 + 2.
1
2
1
4
𝐿 𝑦 = 2250 𝑚𝑚
ℎ 𝑝𝑙𝑎𝑡 = 100 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
Beban: 𝑊𝑢 = 7,60 𝑘𝑁/𝑚2
Momen:
𝑀𝑙𝑥 = 0,001 𝑊𝑢 . 𝑙 𝑥𝑛
2
. 72 = 0,693 𝑘𝑁. 𝑚
𝑀𝑙𝑦 = 0,001 𝑊𝑢 . 𝑙 𝑦𝑛
2
. 19 = 0, 731 𝑘𝑁. 𝑚
𝑀𝑡𝑥 = −0,001 𝑊𝑢 . 𝑙 𝑥𝑛
2
. 117 = −1,125 𝑘𝑁. 𝑚
𝑀𝑡𝑦 = −0,001 𝑊𝑢 . 𝑙 𝑦𝑛
2
. 76 = −2,924 𝑘𝑁. 𝑚
𝑀𝑡𝑖𝑥 =
1
2
𝑀𝑙𝑥 = 0,347 𝑘𝑁. 𝑚
𝑀𝑡𝑖𝑦 =
1
2
𝑀𝑙𝑦 = 0,366 𝑘𝑁. 𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑥 = ℎ − 𝑝 −
1
2
∅ 𝑑𝑥
= 100 − 20 −
1
2
10 = 75 𝑚𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑦 = ℎ − 𝑝 −
1
2
∅ 𝑑𝑦 − ∅ 𝑑𝑥
= 100 − 20 −
1
2
10 − 10 = 65 𝑚𝑚

9. PERENCANAAN PLAT LANTAI 2
• 𝑀𝑙𝑥 = 0,693 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑥
2 = 0,12 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00037
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 330 𝑚𝑚2
As perlu < As
OK Tapi boros
• 𝑀𝑙𝑦 = 0,731 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 0,17 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00053
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 286 𝑚𝑚2
As perlu < As
OK Tapi boros

10. PERENCANAAN PLAT LANTAI 2
• 𝑀𝑡𝑥 = 1,125 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑥
2 = 0,2 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00062
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 330 𝑚𝑚2
As perlu < As
OK Tapi boros
• 𝑀𝑡𝑦 = 2,924 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 0,69 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,0022
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 286 𝑚𝑚2
As perlu < As
OK Tapi boros

11. PERENCANAAN PLAT LANTAI 2
• 𝑀𝑡𝑖𝑥 = 0,347 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑥
2 = 0,062 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00019
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 1,425 𝑚𝑚2
As perlu < As
OK Tapi boros
• 𝑀𝑡𝑖𝑦 = 0,366 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 0,15 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00047
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 30,55 𝑚𝑚2
As perlu < As
OK Tapi boros

12. ONE WAY
SLAB
PLAT ATAP
𝑙 𝑛 =
1
2
𝐿 𝑥 + 2.
1
2
1
4
𝐿 𝑥 = 2250 𝑚𝑚
ℎ 𝑝𝑙𝑎𝑡 = 90 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
1
24
𝑙 𝑛 = 93,45 𝑚𝑚
Beban: 𝑊𝑢 = 5,65 𝑘𝑁/𝑚2
Momen:
𝑀𝐴 = 𝑀 𝐶 =
1
24
𝑊𝑢. 𝑙 𝑛
2
= 1,19 𝑘𝑁. 𝑚
𝑀𝑙𝑎𝑝 𝐴𝐵 = 𝑀𝑙𝑎𝑝 𝐵𝐶 =
1
11
𝑊𝑢. 𝑙 𝑛
2
= 2,6 𝑘𝑁. 𝑚
𝑀 𝐵 =
1
9
𝑊𝑢. 𝑙 𝑛
2
= 3,18 𝑘𝑁. 𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = ℎ − 𝑝 −
1
2
∅ 𝑑 = 93,45 − 20 −
1
2
10 = 68,45 𝑚𝑚
TINJAU I & J

13. PERENCANAAN PLAT ATAP
• Tulangan tak terduga; 𝑀𝐴 = 𝑀 𝐶 = 1,19
𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑2
= 0,25 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00079
Tidak perlu mengecek 𝜌 𝑚𝑖𝑛 < 𝜌 < 𝜌 𝑚𝑎𝑥
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 54,0755 𝑚𝑚2
Tulangan yang digunakan D10-175mm
• Tulangan lapangan; 𝑀𝑙𝑎𝑝 𝐴𝐵 =
𝑀𝑙𝑎𝑝 𝐵𝐶 = 2,6 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑2
= 0,55 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,0017
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 301,18 𝑚𝑚2
Tulangan yang digunakan D10-175mm

14. PERENCANAAN PLAT ATAP
• Tulangan tumpuan; 𝑀 𝐵 = 3,18 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑2
= 0,68 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,0022
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒 = 301,18 𝑚𝑚2
Tulangan yang digunakan D10-175mm
• Tulangan pembagi:
𝐴 𝑠 =
0,18 𝑏. ℎ
100
= 123,21 𝑚𝑚2
Tulangan yang digunakan
D10-175mm

15. TWO
WAYS
SLAB
PLAT
ATAP
TINJAU D
TIPE
𝑙 𝑥𝑛 =
1
2
𝐿 𝑥 + 2.
1
2
1
4
𝐿 𝑥 = 1125 𝑚𝑚
𝑙 𝑦𝑛 =
1
2
𝐿 𝑦 + 2.
1
2
1
4
𝐿 𝑦 = 2250 𝑚𝑚
ℎ 𝑝𝑙𝑎𝑡 = 125 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
Beban: 𝑊𝑢 = 5,65 𝑘𝑁/𝑚2
Momen:
𝑀𝑙𝑥 = 0,001 𝑊𝑢 . 𝑙 𝑥𝑛
2
. 72 = 0,515 𝑘𝑁. 𝑚
𝑀𝑙𝑦 = 0,001 𝑊𝑢 . 𝑙 𝑦𝑛
2
. 19 = 0, 544 𝑘𝑁. 𝑚
𝑀𝑡𝑥 = −0,001 𝑊𝑢 . 𝑙 𝑥𝑛
2
. 117 = −0,837 𝑘𝑁. 𝑚
𝑀𝑡𝑦 = −0,001 𝑊𝑢 . 𝑙 𝑦𝑛
2
. 76 = −2,174 𝑘𝑁. 𝑚
𝑀𝑡𝑖𝑥 =
1
2
𝑀𝑙𝑥 = 0,257 𝑘𝑁. 𝑚
𝑀𝑡𝑖𝑦 =
1
2
𝑀𝑙𝑦 = 0,272 𝑘𝑁. 𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑥 = ℎ − 𝑝 −
1
2
∅ 𝑑𝑥
= 90 − 40 −
1
2
10 = 45 𝑚𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑦 = ℎ − 𝑝 −
1
2
∅ 𝑑𝑦 − ∅ 𝑑𝑥
= 90 − 40 −
1
2
10 − 10
= 35 𝑚𝑚

16. PERENCANAAN PELAT ATAP
• 𝑀𝑙𝑥 = 0,515 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏.𝑑 𝑒𝑥
2 = 0,25
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,00078
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 198 𝑚𝑚2
• 𝑀𝑙𝑦 = 0,544 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 0,44 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,001
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 154 𝑚𝑚2

17. PERENCANAAN PELAT ATAP
• 𝑀𝑡𝑥 = 0,837 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑥
2 = 0,4 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,001
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 198 𝑚𝑚2
• 𝑀𝑡𝑦 = 2,174 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 1,78 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,006
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 210 𝑚𝑚2

18. PERENCANAAN PLAT ATAP
𝑀𝑡𝑖𝑥 = 0,257 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑥
2 = 0,13 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,0004
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑥 = 18 𝑚𝑚2
𝑀𝑡𝑖𝑦 = 0,272 𝑘𝑁. 𝑚
𝑅 𝑛 =
𝑀𝑢
𝑏. 𝑑 𝑒𝑦
2 = 0,02 𝑀𝑃𝑎
𝑚 =
𝑓𝑦
0,85𝑓𝑐′
=17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 =
1
𝑚
1 − 1 −
2𝑚. 𝑅 𝑛
𝑓𝑦
= 0,0006
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 𝜌. 𝑏. 𝑑 𝑒𝑦 = 35 𝑚𝑚2

20. DESAIN ULANG PLAT LANTAI 2 (ONE-WAY SLAB): plat I dan J
𝑙 𝑛 = 2250 𝑚𝑚
ℎ 𝑝𝑙𝑎𝑡 = 135 𝑚𝑚; ℎ 𝑚𝑖𝑛 = 93,45 𝑚𝑚
Beban: 𝑊𝑢 = 7,60 𝑘𝑁/𝑚2
Momen:
𝑀𝐴 = 𝑀 𝐶 = 1,60 𝑘𝑁. 𝑚
𝑀𝑙𝑎𝑝 𝐴𝐵 = 𝑀𝑙𝑎𝑝 𝐵𝐶 = 3,50 𝑘𝑁. 𝑚
𝑀 𝐵 = 4,28 𝑘𝑁. 𝑚
𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = 110 𝑚𝑚

21. • Tulangan tak terduga
; 𝑀𝐴 = 𝑀 𝐶 = 1,60 𝑘𝑁. 𝑚
 𝑅 𝑛 = 0,1322 𝑀𝑃𝑎
 𝑚 = 17,11
 𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0004
Tidak perlu mengecek 𝜌 𝑚𝑖𝑛 < 𝜌 < 𝜌 𝑚𝑎𝑥
 𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 44 𝑚𝑚2
Tulangan yang digunakan D10-150mm
• Tulangan lapangan; 𝑀𝑙𝑎𝑝 𝐴𝐵 =
𝑀𝑙𝑎𝑝 𝐵𝐶 = 3,50 𝑘𝑁. 𝑚
 𝑅 𝑛 = 0,29 𝑀𝑃𝑎
 𝑚 = 17,11
 𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0009
 𝜌 𝑚𝑖𝑛 = 0,0044
 𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
 𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 484 𝑚𝑚2
Tulangan yang digunakan D10-150mm

22. • Tulangan tumpuan; 𝑀 𝐵 =
4,28 𝑘𝑁. 𝑚
 𝑅 𝑛 = 0,3537 𝑀𝑃𝑎
 𝑚 = 17,11
 𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,001
 𝜌 𝑚𝑖𝑛 = 0,0044
 𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 484 𝑚𝑚2
Tulangan yang digunakan D10-
150mm
• Tulangan pembagi:
𝐴 𝑠 = 243 𝑚𝑚2
Tulangan yang digunakan D10-
150mm

23. • 𝑙 𝑥𝑛 = 1125 𝑚𝑚
• 𝑙 𝑦𝑛 = 2250 𝑚𝑚
• ℎ 𝑝𝑙𝑎𝑡 = 135 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
• Beban: 𝑊𝑢 = 7,60 𝑘𝑁/𝑚2
• Momen:
 𝑀𝑙𝑥 = 0,693 𝑘𝑁. 𝑚
 𝑀𝑙𝑦 = 0, 731 𝑘𝑁. 𝑚
 𝑀𝑡𝑥 = −1,125 𝑘𝑁. 𝑚
 𝑀𝑡𝑦 = −2,924 𝑘𝑁. 𝑚
 𝑀𝑡𝑖𝑥 = 0,347 𝑘𝑁. 𝑚
 𝑀𝑡𝑖𝑦 = 0,366 𝑘𝑁. 𝑚
• 𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑥 = 110 𝑚𝑚
• 𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑦 = 100 𝑚𝑚

24. • 𝑀𝑙𝑥 = 0,693 𝑘𝑁. 𝑚
 𝑅 𝑛 = 0,057 𝑀𝑃𝑎
 𝑚 = 17,11
 𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0002
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 484 𝑚𝑚2
Tulangan yang digunakan D10-150mm
• 𝑀𝑙𝑦 = 0,731 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,073 𝑀𝑃𝑎
𝑚 = 17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0002
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 440 𝑚𝑚2
Tulangan yang digunakan D10-150mm

25. • 𝑀𝑡𝑥 = 1,125 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,093 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0003
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 484 𝑚𝑚2
Tulangan yang digunakan D10-150mm
• 𝑀𝑡𝑦 = 2,924 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,029 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0003
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 440 𝑚𝑚2
Tulangan yang digunakan D10-150mm

26. • 𝑀𝑡𝑖𝑥 = 0,347 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,029 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,00009
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 9,9 𝑚𝑚2
Tulangan yang digunakan D10-150mm
• 𝑀𝑡𝑖𝑦 = 0,366 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,037 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0001
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 10 𝑚𝑚2
Tulangan yang digunakan D10-150mm

28. • 𝑙 𝑛 = 2250 𝑚𝑚
• ℎ 𝑝𝑙𝑎𝑡 = 125 𝑚𝑚; ℎ 𝑚𝑖𝑛 = 93,45 𝑚𝑚
• Beban: 𝑊𝑢 = 5,65 𝑘𝑁/𝑚2
• Momen:
 𝑀𝐴 = 𝑀 𝐶 = 1,19 𝑘𝑁. 𝑚
 𝑀𝑙𝑎𝑝 𝐴𝐵 = 𝑀𝑙𝑎𝑝 𝐵𝐶 = 2,6 𝑘𝑁. 𝑚
 𝑀 𝐵 = 3,18 𝑘𝑁. 𝑚
• 𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 = 80 𝑚𝑚

29. • Tulangan tak terduga; 𝑀𝐴 = 𝑀 𝐶 =
1,19 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,19 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0006
Tidak perlu mengecek 𝜌 𝑚𝑖𝑛 < 𝜌 < 𝜌 𝑚𝑎𝑥
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 48 𝑚𝑚2
Tulangan yang digunakan D10-175mm
• Tulangan lapangan; 𝑀𝑙𝑎𝑝 𝐴𝐵 =
𝑀𝑙𝑎𝑝 𝐵𝐶 = 2,6 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,41 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,001
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 352 𝑚𝑚2
Tulangan yang digunakan D10-175mm

30. • Tulangan tumpuan; 𝑀 𝐵 =
3,18 𝑘𝑁. 𝑚
𝑅 𝑛 = 0, 5𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,002
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 352 𝑚𝑚2
Tulangan yang digunakan D10-175mm
• Tulangan pembagi:
𝐴 𝑠 = 225 𝑚𝑚2
Tulangan yang digunakan D10-
175mm

31. • 𝑙 𝑥𝑛 = 1125 𝑚𝑚
• 𝑙 𝑦𝑛 = 2250 𝑚𝑚
• ℎ 𝑝𝑙𝑎𝑡 = 125 𝑚𝑚; ℎ 𝑚𝑖𝑛 =
• Beban: 𝑊𝑢 = 5,65 𝑘𝑁/𝑚2
• Momen:
 𝑀𝑙𝑥 = 0,515 𝑘𝑁. 𝑚
 𝑀𝑙𝑦 = 0, 544 𝑘𝑁. 𝑚
 𝑀𝑡𝑥 = −0,837 𝑘𝑁. 𝑚
 𝑀𝑡𝑦 = −2,174 𝑘𝑁. 𝑚
 𝑀𝑡𝑖𝑥 = 0,257 𝑘𝑁. 𝑚
 𝑀𝑡𝑖𝑦 = 0,272 𝑘𝑁. 𝑚
• 𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑥 = 80 𝑚𝑚
• 𝑑 𝑒𝑓𝑒𝑘𝑡𝑖𝑓 𝑦 = 70 𝑚𝑚

32. • 𝑀𝑙𝑥 = 0,515 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,081 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0003
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 352 𝑚𝑚2
Tulangan yang digunakan D10-175mm
• 𝑀𝑙𝑦 = 0,544 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,111 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0004
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 308 𝑚𝑚2
Tulangan yang digunakan D10-175mm

33. • 𝑀𝑡𝑥 = 0,837 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,131 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0004
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan adalah 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 352 𝑚𝑚2
Tulangan yang digunakan D10-
175mm
• 𝑀𝑡𝑦 = 2,174 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,444 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,001
𝜌 𝑚𝑖𝑛 = 0,0044
𝜌 𝑚𝑖𝑛 = 0,0044
yang digunakan ada𝑙𝑎ℎ 𝜌 𝑚𝑖𝑛
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 308 𝑚𝑚2
Tulangan yang digunakan D10-175mm

34. • 𝑀𝑡𝑖𝑥 = 0,257 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,029 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,00009
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 8 𝑚𝑚2
Tulangan yang digunakan D10-
175mm
• 𝑀𝑡𝑖𝑦 = 0,272 𝑘𝑁. 𝑚
𝑅 𝑛 = 0,037 𝑀𝑃𝑎
𝑚 =17,11
𝜌 𝑝𝑒𝑟𝑙𝑢 = 0,0001
𝐴 𝑠 𝑝𝑒𝑟𝑙𝑢 = 14 𝑚𝑚2
Tulangan yang digunakan D10-175mm

35. GAMBAR PENULANGAN