BAB IV PERHITUNGAN PENULANGAN STRUKTUR 4.1
BAB IV
PERHITUNGAN PENULANGAN STRUKTUR
4.1 Perhitungan Penulangan Balok 4.1.1 Perhitungan Penulangan Lentur Balok
Penulangan yang direncanakan adalah pada balok memanjang line 4 lantai 3 dengan balok no 107, 1909, 1911, 127.
● Data Perencanaan
b = mm
h = mm
f'c = MPa
fyulir = MPa
fypolos = MPa
selimut beton mm
dipakai tulangan pokok D mm
dipakai tulangan sengkang Ø mm
bentang balok L = mm
bentang bersih balok (Ln) = mm
d = h - selimut beton - diameter sengkang -½diameter tulangan rencana
= - - - ½
= mm
● Perencanaan Penulangan
Gambar 4.1 Lebar efektif balok(beff)
Lebar flens efektif (beff)
● beff = ¼L = ¼ x = mm
19 10
40 500
240
4700 40
19 10 5000 300
500 30 390
1250 5000
440,5
beff = 1250 mm
hf = 120 mm
b = 300 mm b = 300 mm b = 300 mm
Ln = 4200 mm Ln = 2700 mm
(2)
● beff = bw + 8 hfkr + 8 hfkn = 300+(8.120)+(8.120) = mm ● beff = bw +½Lnkr +½Lnkn = 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 = mm2 > mm2)
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 = mm2)
● Tulangan yang terpasang pada daerah tekan 3 D mm2)
● Tulangan bagi plat terpasang di sepanjang beff 6 Ø 10(Asplat = mm2)
Kontrol Momen Negatif
Tulangan tarik Asplat = 6 Ø = mm
2
Asbalok = 4 D = mm
2
Tulangan tekan As' = 3 D = mm2
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 1/2
10 19 19 1/2
mm2
As min = 1,4
fy
= 1,4 300 440,5
390
= 474,38
1133,54 (As' = 850,16
2220 3750
As min = 0,25
fy
bw d = 0,25 390
300 x 441 = 464,0
mm2
474,38
19 1250
120790000
471,00 '
(3)
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 . f'c . a . b + As' . fs' = As . fy
(0,85.f'c'.a.b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c
Substitusi nilai : a =β1.c
(0,85.f'c'.β1.c .b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c'.β1.b).c2 + 600As'.c - 600As'.d' = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c.β1.b)c2+ 600As'.c - 600As'.d' - Asplat.fypolos.c - Asbalok.fyulir.c = 0
(0,85.f'c.β1.b)c2+ (600As' - Asplat.fypolos - Asbalok.fyulir).c - 600As'.d' = 0
(0,85.30.0,85.300)c2 +(600.850,16`-471.240-1133,54.390).c-600.850,16.59,5 = 0
c2 - c - = 0
c = mm
a = β.c
= x = mm
-600 c
(0,85 . f'c . 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
= εc = 71,869 59,5 x 0,003 =
Asplat . fypolos + Asbalok . fyulir
1/2
(c - d')
Substitusi nilai : fs' = b
h
beff = 1250 mm
hf
d' y2
es
ec' = 0,003 es'
d
d -a
2 d - d' NT1 = As1 . fy NT2 = As2 . fy y1
ND1 = 0,85 fc' ab ND2 = As' . y
a As'
As
0,85 fc' c
(4)
-Karena εs > εy > εs'maka tulangan baja tarik telah leleh,baja tekan belum Dihitung tegangan pada tulangan baja tekan
f's = εs' x Es
= x
= < MPa
Menghitung gaya tekan dan tarik ND1 = 0,85 . f'c . a . b
= x x x
= N
ND2 = x f's
= x
= N
NT1 = x fypolos
= x
= N
NT2 = Asbalok x fyulir
= x
= N
ND1 + ND2 = NT1 + NT2
+ = +
= Z1 = d - (½ . a)
= - (1/2 . )
= 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
Asplat
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
(5)
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' = + = mm2
Tulangan tarik As = 3 D = mm2
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 . f'c . 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 1/2 10 25
40 10 1/2 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'
NT1 = As1 . fy NT2 = As2 . fy ND1 = 0,85 fc' ab
ND2 = As' . fs' a
(6)
(0,85 . f'c . a . b) . c + As' (c - d') x = As . fy.c
Substitusi nilai : a =β1.c
(0,85 . f'c .β1.c . b) . c + As' (c - d') 600 = As . fy.c
(0,85.f'c.β1.b) c2+ 600As'.c - 600As'.d' = As . fy.c
(0,85.f'c.β1.b) c2+ 600As'.c - 600As'.d' - As. fy . c = 0
(0,85.f'c.β1.b) c2+ (600As' - As . fy).c - 600As'.d' = 0
(0,85.30.0,85.300)c2+ (600.1604,54850,16.390).c -600.1604,54.49,373 = 0
c2 + 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 . f'c . a . beff + Asplat' . fs' = As1 . fs + As2 . fyulir
(0,85.f'c.a.beff).c + Asplat'.(c-y1).600 = As1 . fyulir.c + As2 . fyulir . c
Substitusi nilai : a =β1.c
49,780
Substitusi nilai : fs' = (c - y1) x 600
c dan fs =
As1 . fyulir + As2 . fyulir
fyulir
(c - y1)
600
c x
Substitusi nilai : fs' = x 600 c
(0,85 . f'c . a . b) + As' (c - d') x 600 = c
600
As . fy
6502,50
(0,85.f'c.a.beff) + Asplat' =
631163,6 47532378
b h
beff = 1350 mm
hf
y1
y2
d y = d'
As As'
d -a2
es ec' = 0,003
es'
NT1 = As1 . fy
NT2 = As2 . fy ND1 = 0,85 fc' ab
ND2 = As' . fs' 0,85 fc'
c a
(7)
(0,85.f'c.β1.c.beff).c + Asplat'.(c-y1).600 = As1 . fyulir.c + As2 . fyulir . c
(0,85.f'c.β1.beff).c2 + 600.Asplat'.c - 600.Asplat'.y1= As1 . fyulir.c +
As2 . fyulir . c
(0,85.f'c.β1.beff).c2 + (600.Asplat' - As1.fyulir - As2.fyulir).c
-600.Asplat'.y1 = 0
(0,85.30.0,85.1350).c2 + (600.471 1133,54.390 850,16.390).c -600.471.25 = 0
c2 - - = 0
c = mm
a = β.c
= x = mm
fs' = εs' . Es
-fs = fyulir = MPa
Menghitung gaya tekan dan tarik ND1 = 0,85 . f'c . a . beff
= x x x
= N
ND2 = x f's
= x
= N
NT1 = x
= x
= N
NT1 = x
= x
= N
ND1 + ND2 = NT1 + NT2
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 - y1
c
491041,050
0,85 26,050 22,142
24,181
390
(8)
+ = + =
Z1 = d - (1/2 . a)
= - (1/2 . )
= 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
Mu- = kNm
= Nmm
Mu+ = kNm
= Nmm
Dicoba pemasangan tulangan sebagai berikut :
● Tulangan yang terpasang pada daerah tarik 4 D (As = mm2)
● Tulangan yang terpasang pada daerah tekan 3 D mm2)
331560,450
197923200,570 230654569,708
197923200,570 115327284,854
197923200,570
0,9 197923200,570
178130880,513 120790000
442080,600 773641,050
705788,569 11389,396
717177,965
440,5 22,142
49,373 24,373
25
47463000
19
19 (As' = 850,16 1133,54
442080,600 429,429 331560,450 24,373
47,491 47491000 47,463
1,25
1,25 197923200,570
(9)
● Tulangan bagi plat terpasang di sepanjang beff 6 Ø 10(Asplat = mm2)
Kontrol Momen Negatif
Tulangan tarik Asplat = 6 Ø = mm
2
Asbalok = 3 D = mm
2
As = + = mm2
Tulangan tekan As' = 4 D = mm2
y1 = + = mm
y2 = + + = mm
x + x
d = - = mm
d' = + + = mm
Gambar 4.5 Penampang balok dan diagram tegangan momen negatif lapangan
Dimisalkan garis netral > d' maka perhitungan garis netral harus dicari menggunakan persamaan :
0,85 . f'c . a . b + As' . fs' = As . fy
(0,85.f'c'.a.b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c
Substitusi nilai : a =β1.c
850,16
500
40 10 1/2
10 471,00
19
mm 1321,16
(c - d')
x 600 =
c (0,85 . f'c . a . b) + As'
850,16 19
20
Substitusi nilai : fs' = (c - d') 600 c
y =
1133,54
59,5
= 47,201 471,00
1321,16
25
40 10 19
471
x 47,201 452,799
19 59,5
25
471,00
Asplat . fypolos + Asbalok . fyulir
1/2 10
1/2 59,5
850,16
b h
beff = 1250 mm
hf
d' y2
es
ec' = 0,003 es' d
d - a2 d - d'
NT1 = As1 . fy NT2 = As2 . fy
y1
ND1 = 0,85 fc' ab ND2 = As' . fs'
y
a
As' As
0,85 fc'
(10)
(0,85.f'c'.β1.c .b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c'.β1.b).c2 + 600As'.c - 600As'.d' = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c.β1.b)c2+ 600As'.c - 600As'.d' - Asplat.fypolos.c - Asbalok.fyulir.c = 0
(0,85.f'c.β1.b)c2+ (600As' - Asplat.fypolos - Asbalok.fyulir).c - 600As.d' = 0
(0,85.30.0,85.300)c2 +(600.1133,54-471.240-850,16.390).c-600.1133,54.59,5 = 0
c2 + c - = 0
c = mm
a = β.c
= x = mm
-Karena εs > εy > εs'maka tulangan baja tarik telah leleh, baja tekan belum Dihitung tegangan pada tulangan baja tekan
f's = εs' x Es
= x
= < MPa
Menghitung gaya tekan dan tarik ND1 = 0,85 . f'c . a . b
= x x x
= N
ND2 = x f's
= x
= N
NT1 = x fypolos
= x
εy = Es = = 0,0020
200000
200000
31,801 390
0,85 0,00016
36047,687
Asplat
c
d - c
fy 390
62,830
εs = x εc = 452,799
471 240
0,00016
62,830
x 0,003 = 0,01862
c 62,830
c - d'
6502,50 235523,55
62,830
εc = 62,830
1133,54 31,801
59,5 x
30
0,003 =
53,406 300 408552,713
As'
εs'= x
40467378
(11)
= N
NT2 = Asbalok x fyulir
= x
= N
ND1 + ND2 = NT1 + NT2
+ = +
= Z1 = d - (1/2 . a)
= - (1/2 . )
= mm
Z2 = d - d'
=
-= mm
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 = 3 D = mm
2
As' = + = mm2
Tulangan tarik As = 4 D = mm2
y1 = + = mm
y2 = + + = mm
169434439,180 47491000
10 471,00
19 850,16
471,00 850,16 1321,16
19 1133,54
20 1/2 10 25
40 10 1/2 19 59,5
59,5 36047,687
452,799
850,16 390
331560,5
408552,713
393,299
188260487,978
0,9 188260487,978
452,799 113040
393,299 426,097
113040 331560,5
444600,4 444600,5
53,406
408552,713 426,097 36047,687
1,25
1,25 188260487,978
(1)
(0,85 . f'c . a . b) . c + As' (c - d') x = As . fy.c
Substitusi nilai : a =β1.c
(0,85 . f'c .β1.c . b) . c + As' (c - d') 600 = As . fy.c
(0,85.f'c.β1.b) c2+ 600As'.c - 600As'.d' = As . fy.c
(0,85.f'c.β1.b) c2+ 600As'.c - 600As'.d' - As. fy . c = 0
(0,85.f'c.β1.b) c2+ (600As' - As . fy).c - 600As'.d' = 0
(0,85.30.0,85.300)c2+ (600.1604,54850,16.390).c -600.1604,54.49,373 = 0
c2 + 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 . f'c . a . beff + Asplat' . fs' = As1 . fs + As2 . fyulir
(0,85.f'c.a.beff).c + Asplat'.(c-y1).600 = As1 . fyulir.c + As2 . fyulir . c Substitusi nilai : a =β1.c
49,780
Substitusi nilai : fs' = (c - y1) x 600
c dan fs =
As1 . fyulir + As2 . fyulir fyulir
(c - y1)
600
c x
Substitusi nilai : fs' = x 600 c
(0,85 . f'c . a . b) + As' (c - d') x 600 = c
600
As . fy
6502,50
(0,85.f'c.a.beff) + Asplat' =
631163,6 47532378
b h
beff = 1350 mm
hf
y1 y2
d y = d'
As As'
d -a2
es ec' = 0,003
es'
NT1 = As1 . fy
NT2 = As2 . fy ND1 = 0,85 fc' ab
ND2 = As' . fs' 0,85 fc'
c a
(2)
(0,85.f'c.β1.c.beff).c + Asplat'.(c-y1).600 = As1 . fyulir.c + As2 . fyulir . c
(0,85.f'c.β1.beff).c2 + 600.Asplat'.c - 600.Asplat'.y1= As1 . fyulir.c +
As2 . fyulir . c
(0,85.f'c.β1.beff).c2 + (600.Asplat' As1.fyulir As2.fyulir).c
-600.Asplat'.y1 = 0
(0,85.30.0,85.1350).c2 + (600.471 1133,54.390 850,16.390).c -600.471.25 = 0
c2 - - = 0
c = mm
a = β.c
= x = mm
fs' = εs' . Es
-fs = fyulir = MPa
Menghitung gaya tekan dan tarik ND1 = 0,85 . f'c . a . beff
= x x x
= N
ND2 = x f's
= x
= N
NT1 = x
= x
= N
NT1 = x
= x
= N
ND1 + ND2 = NT1 + NT2
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 - y1
c
491041,050
0,85 26,050 22,142
24,181
390
(3)
+ = + =
Z1 = d - (1/2 . a)
= - (1/2 . )
= 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
Mu- = kNm
= Nmm
Mu+ = kNm
= Nmm
Dicoba pemasangan tulangan sebagai berikut :
● Tulangan yang terpasang pada daerah tarik 4 D (As = mm2)
● Tulangan yang terpasang pada daerah tekan 3 D mm2)
331560,450
197923200,570 230654569,708
197923200,570 115327284,854
197923200,570
0,9 197923200,570
178130880,513 120790000
442080,600 773641,050
705788,569 11389,396
717177,965
440,5 22,142
49,373 24,373
25
47463000
19
19 (As' = 850,16 1133,54
442080,600 429,429 331560,450 24,373
47,491 47491000 47,463
1,25
1,25 197923200,570
(4)
● Tulangan bagi plat terpasang di sepanjang beff 6 Ø 10(Asplat = mm2)
Kontrol Momen Negatif
Tulangan tarik Asplat = 6 Ø = mm2
Asbalok = 3 D = mm2
As = + = mm2
Tulangan tekan As' = 4 D = mm2
y1 = + = mm
y2 = + + = mm
x + x
d = - = mm
d' = + + = mm
Gambar 4.5 Penampang balok dan diagram tegangan momen negatif lapangan
Dimisalkan garis netral > d' maka perhitungan garis netral harus dicari menggunakan persamaan :
0,85 . f'c . a . b + As' . fs' = As . fy
(0,85.f'c'.a.b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c Substitusi nilai : a =β1.c
850,16
500
40 10 1/2
10 471,00
19
mm 1321,16
(c - d')
x 600 =
c (0,85 . f'c . a . b) + As'
850,16 19
20
Substitusi nilai : fs' = (c - d') 600 c
y =
1133,54
59,5
= 47,201 471,00
1321,16
25
40 10 19
471
x
47,201 452,799
19 59,5
25
471,00
Asplat . fypolos + Asbalok . fyulir 1/2 10
1/2 59,5
850,16
b h
beff = 1250 mm
hf
d' y2
es
ec' = 0,003 es' d
d - a2 d - d' NT1 = As1 . fy NT2 = As2 . fy
y1
ND1 = 0,85 fc' ab ND2 = As' . fs' y
a As'
As
0,85 fc' c
(5)
(0,85.f'c'.β1.c .b).c + As'(c-d').600 = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c'.β1.b).c2 + 600As'.c - 600As'.d' = Asplat.fypolos.c + Asbalok.fyulir.c
(0,85.f'c.β1.b)c2+ 600As'.c - 600As'.d' - Asplat.fypolos.c - Asbalok.fyulir.c = 0
(0,85.f'c.β1.b)c2+ (600As' - Asplat.fypolos - Asbalok.fyulir).c - 600As.d' = 0
(0,85.30.0,85.300)c2 +(600.1133,54-471.240-850,16.390).c-600.1133,54.59,5 = 0
c2 + c - = 0
c = mm
a = β.c
= x = mm
-Karena εs > εy > εs'maka tulangan baja tarik telah leleh, baja tekan belum Dihitung tegangan pada tulangan baja tekan
f's = εs' x Es
= x
= < MPa
Menghitung gaya tekan dan tarik ND1 = 0,85 . f'c . a . b
= x x x
= N
ND2 = x f's
= x
= N
NT1 = x fypolos
= x
εy = Es = = 0,0020
200000
200000
31,801 390
0,85 0,00016
36047,687
Asplat c
d - c
fy 390
62,830
εs = x εc = 452,799
471 240
0,00016
62,830
x 0,003 = 0,01862
c 62,830
c - d'
6502,50 235523,55
62,830
εc = 62,830
1133,54 31,801
59,5 x
30
0,003 =
53,406 300
408552,713 As'
εs'= x
40467378
(6)
= N
NT2 = Asbalok x fyulir
= x
= N
ND1 + ND2 = NT1 + NT2
+ = +
= Z1 = d - (1/2 . a)
= - (1/2 . )
= mm
Z2 = d - d'
=
-= mm
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 = mm2
As'balok = 3 D = mm2
As' = + = mm2
Tulangan tarik As = 4 D = mm2
y1 = + = mm
y2 = + + = mm
169434439,180 47491000
10 471,00
19 850,16
471,00 850,16 1321,16
19 1133,54
20 1/2 10 25
40 10 1/2 19 59,5
59,5 36047,687
452,799
850,16 390
331560,5
408552,713
393,299
188260487,978
0,9 188260487,978
452,799 113040
393,299 426,097
113040 331560,5
444600,4 444600,5
53,406
408552,713 426,097 36047,687
1,25
1,25 188260487,978