●
beff = bw + 8 hf
kr
+ 8 hf
kn
= 300+8.120+8.120 =
mm
●
beff = bw + ½Ln
kr
+ ½Ln
kn
= 300+½.4200+½.2700 =
mm dipakai nilai beff terkecil yaitu
= mm
Tulangan minimal sedikitnya harus dihitung menurut SNI 2847-2013 Pasal 10.5.1 :
x dan
bw d x
x Maka dipakai tulangan minimal
2 D As =
mm
2
mm
2
A. Perhitungan penulangan tumpuan kiri joint 63
Mu
-
= kNm
= Nmm
Mu
+
= kNm
= Nmm
Dicoba pemasangan tulangan sebagai berikut :
●
Tulangan yang terpasang pada daerah tarik 4 D
As = mm
2
●
Tulangan yang terpasang pada daerah tekan 3 D
mm
2
●
Tulangan bagi plat terpasang di sepanjang beff
6 Ø 10
As
plat
= mm
2
Kontrol Momen Negatif
Tulangan tarik As
plat
= 6 Ø =
mm
2
As
balok
= 4 D =
mm
2
Tulangan tekan As = 3 D
= mm
2
y1 = +
= mm
y2 = +
+ =
mm 471,00
1133,54 850,16
20 19
59,5 19
566,77
78,103 78103000
120,790
19
10 25
40 10
12 10
19 19
12 mm
2
A
s min
= 1,4
fy =
1,4 300
440,5 390
= 474,38
1133,54 As =
850,16 2220
3750
A
s min
= 0,25
fy bw d =
0,25 390
300 x 441 = 464,0
mm
2
474,38
19 1250
120790000
471,00
fc
30
85
x +
x d
= -
= mm
d = +
+ =
mm
Gambar 4.2 Penampang balok dan diagram tegangan momen negatif tumpuan kiri
Dimisalkan garis netral d maka perhitungan garis netral harus dicari menggunakan persamaan :
0,85 . fc . a . b + As . fs = As . fy
0,85.fc.a.b.c + Asc-d.600 = As
plat
.fy
polos
.c + As
balok
.fy
ulir
.c Substitusi nilai : a = β 1.c
0,85.fc.β 1.c .b.c + Asc-d.600 = As
plat
.fy
polos
.c + As
balok
.fy
ulir
.c 0,85.fc.β 1.b.c
2
+ 600As.c - 600As.d = As
plat
.fy
polos
.c + As
balok
.fy
ulir
.c 0,85.fc.β 1.bc
2
+ 600As.c - 600As.d - As
plat
.fy
polos
.c - As
balok
.fy
ulir
.c = 0 0,85.fc.β 1.bc
2
+ 600As - As
plat
.fy
polos
- As
balok
.fyulir.c - 600As.d = 0 0,85.30.0,85.300c
2
+600.850,16`-471.240-1133,54.390.c- 600.850,16.59,5 = 0
c
2
- c
- = 0
c =
mm a
= β. c =
x =
mm -
600 c
0,85 . fc . a . b + As 19
59,5
=
0,85 71,869
61,089 71,869
1604,54 = 49,373 mm
71,869 6502,50
45027,6 30350534
y = 471
25 1133,54
59,5 500
49,373 450,627
10 40
c x
600 c - d
x
0,00052 εs
c - d x
c =
71,869 59,5
εc = x
0,003 =
As
plat
. fy
polos
+ As
balok
. fy
ulir
12
c - d Substitusi nilai : fs
=
b h
beff = 1250 mm hf
d y2
es
ec = 0,003 es
d d - a
2 d - d
N
T1
= A
s1
. fy N
T2
= A
s2
. fy
y1
N
D1
= 0,85 f
c
ab N
D2
= A
s
. y
a As
As
0,85 f
c
c
86
-
Karena εs εy
εs maka tulangan baja tarik telah leleh,baja tekan belum Dihitung tegangan pada tulangan baja tekan
fs = εs x
Es =
x =
MPa Menghitung gaya tekan dan tarik
ND
1
= 0,85 . fc . a . b =
x x
x =
N ND
2
= x
fs =
x =
N NT
1
= x
fy
polos
= x
= N
NT
2
= As
balok
x fy
ulir
= x
= N
ND
1
+ ND
2
= NT
1
+ NT
2
+ =
+ =
Z1 =
d - ½ . a
= - 12 .
= mm
Z2 =
d - d
= -
= mm
εy = εs =
d - c x
εc = 450,627
x c
71,869 71,869
0,003 = 0,01581
61,089 300
467329,738 As
850,16 103,265
87790,869 As
plat
0,00052 200000
103,265 390
0,85 30
390 200000
= 0,0020
471 240
113040 1133,54
390 442080,6
467329,738 87790,869
113040 442080,6
555120,6 555120,6
450,627 61,089
420,083 450,627
59,5 391,127
fy Es
=
87
Mn = ND1 . Z1 + ND2 . Z2
= x
+ x
= Nmm
Mr =
Mn =
. =
Nmm Mu =
Nmm Aman
Mpr =
. Mn
= .
= Nmm
Kontrol Momen Positif
Tulangan tekan As
plat
= 6 Ø =
mm
2
As
balok
= 4 D =
mm
2
As =
+ =
mm
2
Tulangan tarik As = 3 D
= mm
2
y1 = +
= mm
y2 = +
+ =
mm x
+ x
d =
- =
mm
Gambar 4.3 Penampang balok dan diagram tegangan momen positif tumpuan kiri
Dimisalkan garis netral y2 maka perhitungan garis netral harus dicari menggunakan persamaan :
0,85 . fc . a . b + As . fs = As . fy =
mm
Substitusi nilai : fs =
c - d x
600 y =
59,5 440,5
d =
471 25
1133,54 59,5
1604,54 500
471,00 1133,54
20 12
10 25
40 10
12 19 59,5
850,16 230654569,708
0,9 230654569,708
207589112,737 78103000
1604,54 467329,738
420,083 87790,869
391,127
1,25 1,25
230654569,708 288318212,134
10 471,00
19 1133,54
19
49,373
b h
beff = 1250 mm hf
y1
y2 d
y = d
As As
d - a
2 d - d
es c
ec = 0,003 es
N
T1
= A
s1
. fy N
T2
= A
s2
. fy N
D1
= 0,85 f
c
ab N
D2
= A
s
. f
s
a 0,85 f
c
88
0,85 . fc . a . b . c + As c - d
x = As . fy
.
c Substitusi nilai : a = β 1.c
0,85 . fc . β 1.c . b . c + As c - d 600
= As . fy
.
c 0,85.fc.β 1.b c
2
+ 600As.c - 600As.d = As . fy
.
c 0,85.fc.β 1.b c
2
+ 600As.c - 600As.d - As. fy . c = 0
0,85.fc.β 1.b c
2
+ 600As - As . fy.c - 600As.d = 0
0,85.30.0,85.300c
2
+ 600.1604,54-850,16.390.c - 600.1604,54.49,373 = 0
c
2
+ c
- = 0
c =
mm Karena c y2, tulangan tekan sebagian mengalami gaya tarik maka nilai c
harus dihitung ulang.
Gambar 4.4 Penampang balok dan diagram tegangan momen positif tumpuan kiri yang sudah dihitung ulang
Dimisalkan garis netral diantara y1 dan y2 maka perhitungan garis netral dicari dengan menggunakan persamaan :
0,85 . fc . a . beff + As
plat
. fs = As1 . fs + As2 . fy
ulir
0,85.fc.a.beff.c + As
plat
.c-y
1
.600 = As1 . fy
ulir
.c + As2 . fy
ulir
. c Substitusi nilai : a = β 1.c
49,780
Substitusi nilai : fs =
c - y
1
x 600
c dan
fs = As1 . fy
ulir
+ As2 . fy
ulir
fy
ulir
c - y
1
600 c
x Substitusi nilai : fs
= x
600 c
0,85 . fc . a . b + As c - d
x 600
= c
600 As . fy
6502,50
0,85.fc.a.beff + As
plat
= 631163,6
47532378
b h
beff = 1350 mm hf
y1
y2 d
y = d
As As
d -
a 2
es ec = 0,003
es
N
T1
= A
s1
. fy N
T2
= A
s2
. fy N
D1
= 0,85 f
c
ab N
D2
= A
s
. f
s
0,85 f
c
c a
d - y1
89
0,85.fc.β 1.c.beff.c + As
plat
.c-y
1
.600 = As1 . fy
ulir
.c + As2 . fy
ulir
. c 0,85.fc.β 1.beff.c
2
+ 600.As
plat
.c - 600.As
plat
.y
1
= As1 . fy
ulir
.c + As2 . fy
ulir
. c 0,85.fc.β 1.beff.c
2
+ 600.As
plat
- As1.fy
ulir
- As2.fy
ulir
.c - 600.As
plat
.y1 = 0 0,85.30.0,85.1350.c
2
+ 600.471 - 1133,54.390 - 850,16.390.c - 600.471.25 = 0
c
2
- -
= 0 c
= mm
a = β. c
= x
= mm
fs = εs . Es
- fs = fy
ulir
= MPa
Menghitung gaya tekan dan tarik ND
1
= 0,85 . fc . a . beff =
x x
x =
N ND
2
= x
fs =
x =
N NT
1
= x
= x
= N
NT
1
= x
= x
= N
ND
1
+ ND
2
= NT
1
+ NT
2
MPa
11389,396 As1
390 331560,450
0,85 30
22,142 1250
705788,569 As
471,00 24,181
200000 27093,75
850,16 1133,5
390 fy
442080,600 As1
fy 7065000
26,050
. εc . Es x
0,003 x
26,050 =
= c - y
1
c 491041,050
0,85 26,050
22,142
24,181 390
= 26,050
25
90
+ =
+ =
Z1 =
d - 12 . a
= - 12 .
= 429,43 mm
Z2 =
d - y1
= -
= mm
Mn = NT1 . Z1 + NT2 . Z2
= x
+ x
= Nmm
Mr =
Mn =
. =
Nmm Mu =
Nmm Aman
Mpr =
. Mn
= .
= Nmm
Syarat kuat momen yang terpasag menurut SNI 2847-2013 pasal 21.5.2.2 : Mn
+
≥ ½ Mn
-
Nmm ≥ ½ .
Nmm Nmm ≥
Nmm
B. Perhitungan penulangan lapangan
