The document describes the design of a two-way slab using the direct design method, with given parameters such as building data, loads, panel and column sizes. It details the calculations of slab thickness, moments, and reinforcement requirements. The slab thickness is determined to be 170mm based on stiffness requirements. Distribution of bending moments are calculated along the x- and y-directions. Reinforcement amounts are designed for slab sections at column strips and center strips based on the bending moments.

1. STRUKTUR BETON II

2. 1. Rencanakan plat slab dengan metode desain
langsung (DDM) dengan data sebagai berikut :
- Bangunan kantor :
- Live load : 2,4 kN/m2  (SNI 1727-2020)
- Dead load : 1,2 kN/m2
- Ukuran panel : 7,8*7,2 m. 7,8 m
- Ukuran kolom : 50*50 cm
- Ukuran balok : 30*50m. 7,8 m
- Mutu beton : fc’ = 30 MPa.
- Mutu baja : fy = 300MPa. 7,8 m
7,2 m 7,2 m 7,2 m

3. Penyelesaian :
Cara DDM (Direct Design Method) SNI 2847-2019  8.10.2 :
- Sedikitnya tiga bentang masing-masing arah  8.10.2.1 ok
- Panjang bentang berturutan tidak beda dari ⅓ bentang yang panjang  8.10.2.2
- Rasio bentang :  = Ly = 7,8 = 1,08 < 2 (two way slab) :  8.10.2.3
Lx 7,2
- Pergeseran kolom maksimum 10 %  8.10.2.4
- Semua beban untuk beban gravitasi dan merata  8.10.2.5
- Beban hidup terbagi rata, LL < 2 DL  8.10.2.6
Asumsi  > 2, Asumsi h min terbesar h = 90 mm  tabel 8.3.1.2 7,8
Cek :
DLt = 1,2 + 0,09*24 = 3,36 kN/m2 7,8
2DL = 2*3,36 = 6,72 kN/m2 > LL = 2,4 kN/m2  ok
Ln1 = 780 – 50 = 730 cm 7,8
Ln2 = 720 – 50 = 670 cm
7,2 7,2 7,2

4. untuk αtm > 2,0, ketebalan pelat minimum terbesar dari :  Tabel : 8.3.1.2
- h = Ln(0,8 + fy/1.400)  (d)
36 + 9β
- h = 90 mm  (e)
h = 7.300(0,8 + 300/1.400) = 161,63 mm > 90 mm  ok bw + 2hb ≤ bw + 8hf
36 + 9*1,09
h = 162,63 ≈ 170 mm h = 17 cm
Balok T :  table 6.3.2.1/ R8.4.1.8
Ukuran Balok 30/50. hb = 33 cm
hb = 500 – 170 = 330 mm
bf = bw + 2.hb = 300 + 2*330 = 960 mm (lebar balok T) bw= 30
bw + 8hf = 300 + 8*170 = 1.660 mm > 960 mm  ok

5. Titik berat :
ya = 960*170*85 + 300*330*335 = 47,037*106 = 179,39 mm
960*170 + 300*330 262.200
yb = 500 – 179,39 = 320,61 mm
Ib = 1/12*96*173 + 96*17*9,4392 + 1/12*30*333 + 30*33*15,5612
Ib= 39.304+145.402,5847+89.842,5+239.723,2738= 514.272,3585 cm4
Is1 = 1/12*780*173 = 319.345 cm4  arah sb y
Is2 = 1/12*720*173 = 294.780 cm4  arah sb x
dengan E sama, (8.10.2.7b) : bf = 960 mm
αf1 = Ib = 514.272,3585 = 1,61 0
Is1 319.345 h = 170 ya
αf2 = Ib = 514.272,3585 = 1,74
Is2 294.780 hb = 330
αfm = αf1 + αf2 = 1,61 + 1,74 = 1,675 yb
2 2
αfm = rasio kekakuan rata-rata bw = 300 mm

6. Untuk 0,2 < αtm = 1,675 < 2 ketebalan pelat minimum terbesar dari : 8.3.1.2
- h = Ln(0,8 + fy/1.400)  (b) tabel 8.3.1.2
36 + 5β(αfm – 0,2)
h = 7.300(0,8 + 300/1.400) = 168,13 mm < 170 mm  ok
36 + 5*1,09(1,675 – 0,2)
- h = 125 mm < 168,13 mm
Ambil h = 168,13 m ≈ 170 mm, dicek :
DLt = 1,2 + 0,17*24 = 5,28 kN/m2
2DL = 2*5,28 = 10,56 kN/m2 > LL = 2,4 kN/m2  ok
Kombinasi pembebanan :  Tabel 5.3.1
U = 1,4DL = 1,4*5,28 = 7,392 kN/m2.
U = 1,2DL + 1,6LL = 1,2*5,28 + 1,6*2,4 = 10,176 kN/m2 (kPa)
Maka diambil U terbesar yaitu : wu = 10,176 kN/m2.
Ukuran kolom : 500*500 mm.
Ln1 = 780 – 50 = 730 cm  arah y yang digunakan
Ln2 = 720 – 50 = 670 cm  arah x yang digunakan
ln1 = 730 cm > 0,65l1 = 0,65*780 = 507 cm . → 8.10.3.2.1
ln2 == 670 cm > 0,65l2 = 0,65*720 = 468 cm .

7. Arah x :
M0 = wu.l2.ln1
2 = 10,176*7,2*7,32 = 488,05 kNm → (8.10.3.2)
8 8
Momen distribusi untuk interior → 8.10.4.1
Momen negatip interior : Mu
- = 0,65*M0 = 0,65*488,05 = 317,23 kNm
Momen positip interior : Mu
+ = 0,35*M0 = 0,35*488.05 = 170,82 kNm
Momen distribusi di lajur kolom → (8.10.2.7b)
αf = Ecb.Ib2 = 514.272,3585 = 1,74 → modulus elastisitas : Ebalok = Eslab
Ecs.Is2 294.780
Momen terfaktor di lajur kolom untuk Momen negatif interior Tabel 8.10.5.1
L2/L1 = 7,2/7,8 = 0,923 dan  antara 0,5 dan 1
αL2/L1 = 1,74*0,923 = 1,606 > 1  antara 0,90 dan 0,75
Interpolasi :
koefisien distribusi M- = 0,9 - (0,5 – 0,923)(0,9 – 0,75) = 0,77 Tabel 8.10.5.1
(0,5 – 1,0)
koefisien distribusi M+ = 0,9 - (0,5 – 0,923)(0,9 – 0,75) = 0,77 Tabel 8.10.5.5
(0,5 – 1,0)

8. Momen terfaktor negatip interior lajur kolom : 8.10.5.1
Tabel 8.10.5.1 : Momen negatip interior Mu lajur kolom
Tabel 8.10.5.5 : Momen positif Mu lajur kolom
Tabel 8.10.5.7.1 : Momen positif Mu lajur kolom

9. Tulangan geser :
Anggap diameter tulangan pelat menggunakan D10 dan concrete cover terlindung
setebal 20 mm  (20.6.1.3.1 : pelindung beton untuk tulangan).
d = tebal pelat – concrete cover – 1,5 diameter besi.
d = 170 – 20 – 1,5*10 = 135 mm = 13,5 cm.
Gaya geser :
Vu = 1,15*½wu.ln2 = 1,15*½(10,176 *6,7) = 39,203 kN. → tabel 6.5.4
Vc = 0,17fc’.bw.d = 0,1730*1.000*135*10-3 = 125,702 kN. → (22.5.5.1)
Φ.Vc = 0,75*125,702 = 94,2765kN > Vu = 39,203 kN → (Φ = 0,75 → tabel 21.2.1)
Ketebalan cukup untuk menahan geser
 = 1 untuk beton normal  Tabel 19.2.4.2
Tulangan susut  Tabel : 24.4.3.2
As-min  0,0020.bw.d = 0,0020*1.000*135 = 270 mm2.  fy = 300 MPa < 420 MPa
As-D10 = ¼..D2 = ¼**102 = 78,5 mm2 .
sperlu = As-D10.b = 78,5*1.000 = 290,74 mm ≈ 290 mm.  yang menentukan, terkecil
As
+
perlu 270
Dipakai D10-290 mm = (1.000/290)*78,5 = 270,6897 mm2.
Syarat : s  5h = 5*170 = 850 mm dan s  450 mm. ok. → 24.4.3.3

10. Tabel Distribusi momen arah x : L2/L1 = 7,2/7,8 = 0,923 dan
α1(L2/L1) = 1,74*0,923 = 1,606
Lajur kolom M- (kNm) M+ (kNm)
Mu = ΦM0 (8.10.4.1) 0,65*488,05 = 317,23 0,35*488,05= 170,82
Koefisien distribusi 0,77 0,77
Momen terfaktor lajur klm: Muc 0,77Mu = 244,267 0,77Mu = 131,531
Momen balok, 85%(8.10.5.7.1) 0,85Mu= 269,646 0,85Mu = 145,197
Momen slab lajur kolom 0,15Mu = 47,585 0,15Mu = 25,623
Momen slab lajur tengah 0,23Mu = 72,963 0,23Mu = 39,289

11. Lajur kolom :
blk T
lajur kolom lajur tengah
L ½L
¼L ¼L

12. Perhitungan momen :
bf = bw + 2.hb = 300 + 2*330 = 960 mm  lebar balok T R8.4.1.8
Lebar lajur kolom = 2(¼.L2) = 2(¼*7,2) = 3,6 m  8.4.1.5
Lebar total slab lajur kolom = ½.L2 – bf = ½*7,2 – 0,96 = 2,64 m
Momen slab lajur kolom :
Muc
- = 47,585 = 20,027 kNm/m Φ = 0,9 → Tabel 21.2.1
0,9*2,64
Muc
+ = 25,623 = 10,784 kNm/m
0,9*2,64
Lebar lajur tengah : ½.L2 = ½*7,2 = 3,6 m
Momen slab lajur tengah :
Mul
- = 72,963 = 22,519 kNm/m
0,9*3,6
Mul
+ = 39,289 = 12,126 kNm/m
0,9*3,6

13. Penulangan slab lajur kolom :
Mu = As.fy(d – ½.a)
As = Mu
fy(d – ½.a) a
As
- = Mu = 20,027*106 = 549,438 mm2
0,9.d.fy. 0,9*135*300
a = As
-.fy = 549,438*300 = 6,5
0,85.fc’.b 0,85*30*1.000
As
- = 20,027*106 = 506,692 mm2
300*(135 – ½*6,5)
As-D10 = ¼..D2 = ¼**102 = 78,5 mm2 .
sperlu = As-D10.b = 78,5*1.000 = 154,926 mm ≈ 150 mm. (Dipakai D10-150 mm = 523,33 mm2)
As
-
perlu 506,692 (1.000 : 150) * 78,5 = 523,33
Syarat : s  3h = 3*170 = 510 mm dan s  450 mm. ok. → 8.7.2.2 (tidak kritis)
As
+ = 10,784*106 = 295,857 mm2
0,9*135*300
a = As
+.fy = 295,857*300 = 3,5
0,85.fc’.b 0,85*30*1.000
As
+ = 10,784*106 = 269,769 mm2
300*(135 – ½*3,5)
As-D10 = ¼..D2 = ¼**102 = 78,5 mm2 .
sperlu = As-D10.b = 78,5*1.000 = 290,99 mm ≈ 275 mm (Dipakai D10-275 mm = 285,45 mm2)
As
+
perlu 269,769
Syarat : s  3h = 3*170 = 510 mm dan s  450 mm. ok. → 8.7.2.2 (tidak kritis)

14. Penulangan slab lajur tengah :
As
- = 22,519*106 = 617,805 mm2
0,9*135*300
a = As
-.fy = 617,805*300 = 7,3
0,85.fc’.b 0,85*30*1.000
As
- = 22,519*106 = 571,476 mm2
300*(135 – ½*7,3)
As-D10 = ¼..D2 = ¼**102 = 78,5 mm2 .
sperlu = As-D10.b = 78,5*1.000 = 137,36 mm ≈ 130 mm. (Dipakai D10-130 mm = 603,846 mm2)
As
-
perlu 571,476
Syarat : s  3h = 3*170 = 510 mm dan s  450 mm. ok. → 8.7.2.2 (tidak kritis)
As
+ = 12,126*106 = 332,675 mm2
0,9*135*300
a = As
+.fy = 332,675*300 = 3,9
0,85.fc’.b 0,85*30*1.000
As
+ = 12,126*106 = 303,796 mm2
300*(135 – ½*3,9)
As-D10 = ¼..D2 = ¼**102 = 78,5 mm2 .
sperlu = As-D10.b = 78,5*1.000 = 258,4 mm ≈ 250 mm. (Dipakai D10-250 mm = 314 mm2)
As
+
perlu 303,796
Syarat : s  3h = 3*170 = 510 mm dan s  450 mm. ok. → 8.7.2.2 (tidak kritis)

15. Hitung penulangan terhadap arah y