123
5.6.1.3 Perhitungan Konstruksi Dinding Tegak
A. Perataan beban segitiga untuk pembebanan dinding Mmaks = q.L
2
9 √3 untuk x = L √3
Gambar 5.21
Perataan Beban Segitiga
Besarnya beban ekivalen : M
maks
= M
x
M
x
= 2
x L
x q
ek
−
3 9
2
L q
= 2
x L
x q
ek
− untuk x = L
√3 q
ek
= 0,5246 q
Gambar 5.22
Diagram Tegangan Tanah Tiap Segmen Dinding Gerbang A
X q
L qek
L
A
4,00
0,47 1,50
0,40 1,50
sa
7
sa
8
sa
9
sa
10
sa
w
1,00
1,12
2,00 0,35
sa
1
sa
2
segmen 1
segmen 2
segmen 3
sa
3
sa
4
sa
5
sa
6
124
Segmen 1
a1 = Ka
1
x q = 0,729 x 1
= 0,729 tm
2
a2 =
1
x h x Ka
1
– 2 x C
1
x √Ka
1
= 1,6453 x 1 x 0,729 – 0,853 = 0,346 tm
2
q
ek
= a1 + 0,5246 x a2 = 0,729 + 0,5246 x 0,346
= 0,911 tm
2
Segmen 2
a3 = a1 + a2 = 0,729 + 0,346
= 1,075 tm
2
a4 =
1
x h x Ka
1
– 2 x C
1
x √Ka
1
= 1,6453 x 1 x 0,729 – 0,853 = 0,346 tm
2
a5 = a3 + a4 = 1,075 + 0,346
= 1,421 tm
2
a6 =
2
x h x Ka
2
- 2 x C
1
x √Ka
1
= 1,7099 x 0,12 x 0,679 – 0,118 = 0,021 tm
2
q
ek
= a3 + 0,5246 x a4 + a5 + 0,5246 x a6 = 1,075 + 0,5246 x 0,346 + 1,421 + 0,5246 x 0,021
= 2,689 tm
2
Segmen 3
a7 = a5 + a6 = 1,421+ 0,021
= 1,442 tm
2
a8 =
2
x h x Ka
2
= 1,7099 x 1,88 x 0,679
125 = 2,183 tm
2
a9 = a5 + a6 = 1,442 + 2,183
= 3,625 tm
2
a10 =
sub
x h x Ka
2
= 0,6738 x 0,12 x 0,679 = 0,055 tm
2
aw =
w
x h x Kw = 1 x 0,12 x 1
= 0,12 tm
2
q
ek
= a7 + 0,5246 x a8 + a9 + 0,5246 x a10 + 0,5246 x aw = 1,442 + 0,5246 x 2,183 + 3,625+ 0,5246 x 0,055 + 0,5246 x 0,12
= 6,304 tm
2
B. Penulangan Dinding
Segmen 1
q = 0,911 tm
2
Ix = h1 = 1,00 m Iy = jarak antar counterfort = 2,675 m
IyIx = 2,6751,00 = 2,675 Diasumsikan pelat terjepit di kedua sisinya
M
Ix
= 0,001.q
u
.Ix
2
.x M
Iy
= 0,001.q
u
.Ix
2
.x M
ty
= -0,001.q
u
.Ix
2
.x M
tiy
= ½.M
Ix
Iy Ix
126 Dimana nilai x berturut-turut berdasarkan perbandingan lylx = 2,675
adalah 107; 20,4; dan 112 Tabel Hal 26 Grafik dan Tabel Perhitungan
Beton Bertulang, Seri Beton-4, W.C. Vis, Gideon Kusuma M
Ix
= 0,001 x 0,911 x 1,0
2
x 107 = 0,097 tm M
Iy
= 0,001 x 0,911 x 1,0
2
x 20,4 = 0,019 tm M
ty
= -0,001 x 0,911 x 1,0
2
x 112 = -0,102 tm M
tiy
= ½ x 0,097 = 0,049 tm Diambil momen yang maksimum
Tulangan Arah X Tumpuan dan Lapangan Mu = 0,097 tm = 9,7 x 10
5
Nmm Mn = Mu
φ = 9,7 x 10
5
0,8 = 1,213 x 10
6
Nmm b1 = 300 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b1 – d’ – Ø2 = 300 – 50 – 202 = 240 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 1,213 x 10
6
1000 x 240
2
x 19,125 = 1,101 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
1,101 2
1
-3
− = 1,102 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 1,102 x 10
-3
x 1000 x 240 x 19,125240 = 21,071 mm
2
ρ = Asb.d = 21,0711000 x 240 = 8,779 x 10
-5
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton-1, W.C. Vis, Gideon Kusuma atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
127 As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 240 = 1399,2 mm
2
Tulangan pokok terpasang = Ø 20 – 200 As = 1571 mm
2
Cek : ρ = As
terpasang
b.d = 15711000x240 = 6,546 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Arah Y Tumpuan dan Lapangan
Mu = 0,102 tm = 1,02 x 10
6
Nmm Mn = Mu
φ = 1,02 x 10
6
0,8 = 1,275 x 10
6
Nmm b1 = 300 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b1 – d’ – Ø tulangan arah X - Ø2 = 300 – 50 – 20 - 202 = 220 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 1,275 x 10
6
1000 x 220
2
x 19,125 = 1,377 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
1,377 2
1
-3
− = 1,378 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 1,378x10
-3
x 1000 x 220 x 19,125240 = 24,158 mm
2
ρ = Asb.d = 24,1581000 x 220 = 1,098 x 10
-4
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton-1, W.C. Vis, Gideon Kusuma atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 220 = 1282,6 mm
2
Tulangan pokok terpasang = Ø 20 – 225 As = 1396 mm
2
128 Cek :
ρ = As
terpasang
b.d = 13961000x220 = 6,345 x 10
-3
ρ
min
ρ ρ
maks
ok
Segmen 2
q = 2,689 tm
2
Ix = h2 = 1,12 m Iy = jarak antar counterfort = 2,675 m
IyIx = 2,6751,12 = 2,388 Diasumsikan pelat terjepit di kedua sisinya
M
Ix
= 0,001.q
u
.Ix
2
.x M
Iy
= 0,001.q
u
.Ix
2
.x M
ty
= -0,001.q
u
.Ix
2
.x M
tiy
= ½.M
Ix
Dimana nilai x berturut-turut berdasarkan perbandingan lylx = 2,388 adalah 99,4; 21,6; dan 112
Tabel Hal 26 Grafik dan Tabel Perhitungan Beton Bertulang, Seri Beton-4, W.C. Vis, Gideon Kusuma
M
Ix
= 0,001 x 2,689 x 1,12
2
x 99,4 = 0,335 tm M
Iy
= 0,001 x 2,689 x 1,12
2
x 21,6 = 0,073 tm M
ty
= -0,001 x 2,689 x 1,12
2
x 112 = -0,378 tm M
tiy
= ½ x 0,335 = 0,168 tm Diambil momen yang maksimum
Tulangan Arah X Tumpuan dan Lapangan Mu = 0,335 tm = 3,35 x 10
6
Nmm Mn = Mu
φ = 3,35 x 10
6
0,8 = 4,188 x 10
6
Nmm b2 = 350 mm
Iy Ix
129 f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b2 – d’ – Ø2 = 350 – 50 – 202 = 290 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 4,188 x 10
6
1000 x 290
2
x 19,125 = 2,604 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
604 ,
2 2
1
-3
− = 2,607 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 2,607 x 10
-3
x 1000 x 290 x 19,125240 = 60,251 mm
2
ρ = Asb.d = 60,2511000 x 290 = 2,078 x 10
-4
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton-1, W.C. Vis, Gideon Kusuma atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 290 = 1690,7 mm
2
Tulangan pokok terpasang = Ø 20 – 175 As = 1795 mm
2
Cek : ρ = As
terpasang
b.d = 17951000x290 = 6,190 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Arah Y Tumpuan dan Lapangan
Mu = 0,378 tm = 3,78 x 10
6
Nmm Mn = Mu
φ = 3,78 x 10
6
0,8 = 4,725 x 10
6
Nmm b2 = 350 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm
130 d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b2 – d’ – Ø tulangan arah X - Ø2 = 350 – 50 – 20 - 202 = 270 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 4,725 x 10
6
1000 x 270
2
x 19,125 = 3,389 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
389 ,
3 2
1
-3
− = 3,394 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 3,394 x 10
-3
x 1000 x 270 x 19,125240 = 73,041 mm
2
ρ = Asb.d = 73,0411000 x 270 = 2,705 x 10
-4
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton-1, W.C. Vis, Gideon Kusuma atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 270 = 1574,1 mm
2
Tulangan pokok terpasang = Ø 20 – 175 As = 1795 mm
2
Cek : ρ = As
terpasang
b.d = 17951000x270 = 6,648 x 10
-3
ρ
min
ρ ρ
maks
ok
Segmen 3
q = 6,304 tm
2
Ix = h3 = 2,00 m Iy = jarak antar counterfort = 2,675 m
IyIx = 2,6752 = 1,338 Diasumsikan pelat terjepit di ketiga sisinya :
131 M
Ix
= 0,001.q
u
.Ix
2
.x M
Iy
= 0,001.q
u
.Ix
2
.x M
tx
= -0,001.q
u
.Ix
2
.x M
ty
= -0,001.q
u
.Ix
2
.x M
tiy
= ½.M
Ix
Dimana nilai x berturut-turut berdasarkan perbandingan lylx = 1,338 adalah 41, 20, 73, dan 55
Tabel Hal 26 Grafik dan Tabel Perhitungan Beton Bertulang, Seri Beton-4, W.C. Vis, Gideon Kusuma
M
Ix
= 0,001 x 6,304 x 2
2
x 41 = 1,034 tm M
Iy
= 0,001 x 6,304 x 2
2
x 20 = 0,504 tm M
tx
= -0,001 x 6,304 x 2
2
x 73 = -1,841 tm M
ty
= -0,001 x 6,304 x 2
2
x 55 = -1,387 tm M
tiy
= ½ x 1,034 = 0,517 tm Diambil momen yang maksimum
Tulangan Arah X Tumpuan dan Lapangan Mu = 1,841 tm = 1,841 x 10
7
Nmm Mn = Mu
φ = 1,841 x 10
7
0,8 = 2,301 x 10
7
Nmm b3 = 400 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b3 – d’ – Ø2 = 400 – 50 – 202 = 340 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 2,301 x 10
7
1000 x 340
2
x 19,125 = 0,011
Iy Ix
132 F = 1 -
k 2
1 −
= 1 -
0,011 2
1 −
= 0,011 Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced As = F.b.d.R
l
fy = 0,011 x 1000 x 340 x 19,125240 = 298,031 mm
2
ρ = Asb.d = 298,0311000 x 340 = 8,765 x 10
-4
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton-1, W.C. Vis, Gideon Kusuma atau
ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 340 = 1982,2 mm
2
Tulangan pokok terpasang = Ø 20 – 150 As = 2094 mm
2
Cek : ρ = As
terpasang
b.d = 20941000x340 = 6,159 x 10
-3
ρ
min
ρ ρ
maks
ok Tulangan Arah Y Tumpuan dan Lapangan
Mu = 1,387 tm = 1,387 x 10
7
Nmm Mn = Mu
φ = 1,387 x 10
7
0,8 = 1,734 x 10
7
Nmm b3 = 400 mm
f’c = 225 kgcm
2
= 22,5 Nmm
2
fy = 2400 kgcm
2
= 240 Nmm
2
Dicoba tulangan pokok Ø 20 mm d’ = tebal selimut beton = 50 mm Tabel 3. Dasar-Dasar Perencanaan
Beton Bertulang, Seri Beton –1,W. C. Vis, Gideon Kusuma d = b3 – d’ – Ø tulangan arah X - Ø2 = 400 – 50 – 20 - 202 = 320 mm
k = Mnb.d
2
.R
1
R
l
=
1
.f’c = 0,85 x 22,5 = 19,125 Nmm
2
k = 1,734 x 10
7
1000 x 320
2
x 19,125 = 8,854 x 10
-3
F = 1 - k
2 1
− = 1 -
10 x
854 ,
8 2
1
-3
− = 8,894 x 10
-3
Fmax =
1
.450600 + fy = 0,85 x 450600 + 240 = 0,455 F Fmax
tulangan tunggal under reinforced
133 As = F.b.d.R
l
fy = 8,894x10
-3
x 1000 x 320 x 19,125240 = 226,789 mm
2
ρ = Asb.d = 226,7891000 x 320 = 7,087 x 10
-4
ρ
min
= 1,4 fy = 1,4240 = 5,83 x 10
-3
ρ
maks
= 0,03635 Tabel-8 Dasar-Dasar Perencanaan Beton Bertulang,
Seri Beton- 1, W.C. Vis, Gideon Kusuma
atau ρ
maks
=
1
.450600 + fy.R
l
fy = 0,85x450600+240x 19,125240 = 0,03629
ρ ρ
min
dipakai ρ
min
As
min
= ρ
min
.b.d = 5,83 x 10
-3
x 1000 x 320 = 1865,6 mm
2
Tulangan pokok terpasang = Ø 20 – 150 As = 2094 mm
2
Cek : ρ = As
terpasang
b.d = 20941000x320 = 6,544 x 10
-3
ρ
min
ρ ρ
maks
ok
5.6.1.4 Perhitungan Bagian Perkuatan Dinding Counterfort
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