PERHITUNGAN PERHITUNGAN TULANGAN GESER BALOK ANAK :
PROYEK
RUSUNAMI PONDOK BAMBU
MATERIAL PROPERTIES Kuat leleh (f (f y )
=
400
MPa
400
MPa
Kuat leleh (f (f ys )
=
Kuat tekan (f (f c ' )
=
20,75 MPa
Faktor reduksi (Ø)
=
0,65
PENULANGAN GESER BALOK 1. Tumpuan Balok Penampang Balok : Lebar (b (b w ) =
300
mm
Tinggi balok (h (h ) Tinggi efektif (d (d ) Tulangan longitudinal Tulangan sengkang
500 475 16 10
mm mm mm mm
= = = =
β
→
= 0,85
Kuat geser ultimit (V (V u ) Vu
= ( 1.2DL + 1.6 LL) ( 1.4DL )
=
15900
=
1 6
= =
15900 159000
kg N
=
10 818 6, 4
N
=
10818,64
kg
=
1 364 28, 9 8
N
=
13642,9
kg
=
43 274 5, 6
N
=
43274,56
kg
=
21 637 2, 8
N
=
21637,28
kg
=
0,7180
mm /mm
=
0,2500
mm /mm
=
0,7180
mm /mm
Kuat geser balok (V (V c ) Vc
=
1 6
,
. f c .b w .d
Kuat geser perlu ( V s,perlu ) V u V C V s,perlu =
=
20,75^0.5 . 300 . 475
159000 0,65
-
108186,4
Kontrol kuat geser (Vs ) V s,max
Vs
=
=
2 3
1 3
'
. f c .b w .d
=
'
. f c .b w .d
=
2 3
1 3
20,75^0.5 . 300 . 475
20,75^0.5 . 300 . 475
jika V s,perlu
<
V s,max
→
penampang OK
V s,perlu
<
Vs
→
spasi sengkang d/2
V s,perlu
< < <
→
penampang OK
Cek 1 36428 , 98 N 13642,9 kg
V s,max 43 274 5, 6 N 43 2 74, 56 k g
Tulangan geser perlu ( A v perlu )
Av
=
Av , min
=
A v perlu
=
V S
=
f ys .d 1 Bw . 3 f ys
=
136428,98 400 . 475 1 3
300 400
=
2
2
2
Jarak antar sengkang (s) sengkang (s) A v pakai s
= =
2 Av , pakai Av , perlu
D
10
A v,pakai =
→
=
157 0,718
=
2
157
mm
2 18, 6 6
mm
Spasi maksimum sepanjang balok tidak boleh lebih besar dari SNI 03-2847-2002 pasal 23.10(4(2)) hal 231 : s max = = 475 = 23 7, 5 mm 2
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= 300 m m
=
Spasi tulangan geser s pakai =
219
300
=
30 0
mm
=
15900
= =
15900 1590
N kg
=
1 6
=
108 18 6
N
=
10819
kg
=
- 837 24, 4 6
N
=
-8372,45
kg
=
432 74 6
N
=
43274,6
kg
=
216 37 3
N
=
21637,3
kg
=
- 0, 440 7
mm /mm
=
0,2500
mm /mm
=
0,2500
mm /mm
=
157
mm
=
6 28, 0 0
mm
mm
2. Lapangan Balok Penampang Balok : Lebar (b (b w )
=
300
mm
Tinggi balok (h (h ) Tinggi efektif (d (d ) Tulangan longitudinal Tulangan sengkang
= = = =
500 475 16 10
mm mm mm mm
Kuat geser ultimit (V (V u ) Vu = ( 1.2DL + 1.6 LL) ( 1.4DL ) Kuat geser balok (V (V c ) 1 , Vc = . f c .bw .d 6
Kuat geser perlu ( V s,perlu ) V u V C V s,perlu =
20,75^0.5 . 300 . 475
15900 0,65
=
-
10 818 6
Kontrol kuat geser (Vs ) V s,max
Vs
=
=
2 3
1 3
'
. f c .b w .d
=
'
=
. f c .b w .d
2 3
20,75^0.5 . 300 . 475
1 3
20,75^0.5 . 300 . 475
jika V s,perlu
<
V s,max
→
penampang OK
V s,perlu
<
Vs
→
spasi sengkang d/2
V s,perlu
< < <
4 327 46 432 74, 6
→
penampang OK
Cek - 8 3724 , 46 N - 837 2, 45 k g
V s,max N kg
Tulangan geser perlu ( A v perlu )
Av
=
Av , min
=
A v perlu
=
V S
=
f ys .d 1 Bw . 3 f ys
=
-83724,46 400 . 475 1 3
300 400
=
2
2
2
Jarak antar sengkang (s) sengkang (s) A v pakai s
= =
2 Av , pakai Av , perlu
D
10
→
=
157 0,25
A v,pakai
2
Spasi maksimum sepanjang balok tidak boleh lebih besar dari SNI 03-2847-2002 pasal 23.10(4(2)) hal 231 : s max = = 475 = 23 7, 5 mm d 2
= 300 m m Spasi tulangan geser
=
2 300
=
30 0
mm
