materi kuliah

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