perencanaan pelat injak
TRANSCRIPT
-
7/23/2019 PERENCANAAN PELAT INJAK
1/4
STUDIO PERANCANGAN 2
PERENCANAAN PELAT INJAK
200.00
Berat Isi
Beton Bertulang = 2500 Kg!"
As#al Beton = 22$0 Kg!"
Air %urni = &000 Kg!"
Data Pelat In'a(Te)al #elat in'a( = 20 *! = 0+2 !
Te)al la#isan as#al , o-erla. = &0 *! = 0+& !
Te)al genangan air /u'an = 5 *! = 0+05 !
e)ar #elat in'a( = 2 !
e)ar 'e!)atan total = 1 !
a. PEMBEBANAN
Analisis )e)an .ang ter'ai 3
&4 Be)an %ati 3
Berat seniri #elat )eton = 0+2 & 2500 = 500 Kg!
ABDULLAH GHIYATS D.U. AULIA NURUL ANNISA
-
7/23/2019 PERENCANAAN PELAT INJAK
2/4
STUDIO PERANCANGAN 2
Berat as#al = 0+2 & 22$0 = 22$ Kg!
Berat air /u'an = 0+05 & &000 = 50 Kg!
Total )e)an !ati = 66$ Kg!
24 Be)an 7iu# 3
Be)an T
8Pelat in'a( iasu!si(an satu ara/9
: = &$;25 Kg!
e)ar )iang (onta( roa tru( = $0 *!
200.00
Be)an Ko!)inasi 3
Be)an %ati = &+2 66$ = 12
-
7/23/2019 PERENCANAAN PELAT INJAK
3/4
STUDIO PERANCANGAN 2
= $;6;+$ Kg!
b. PENULANGAN
) = &00 *!
? = 200 > 50 = &50 !! = &5 *!
@?* = "0 %Pa
@. = "10 %#a
% = $;+6;$ (N!
(Berdasarkan BMS 1992 bagian 5 Perencanaan Komponen Beton halaman 5-13)
!in =
1,4
fy =1,4
390 = 0+00"51
=
Kc
KcR
x fy(R x fy)22,4x KcRx (
M
b x d2)x(
fy2
f'c)
1,2x KcR
x (fy2
f'c)
=
0,75x 390(0,75x 390)22,4x0,75x
46,764x106
1000x 1502
x (390
2
30)
1,2x 0,75x (390
2
30)
= 0+006;
!a = 0+65
0,85x f'c
fy
600
600+ fy = 0+650,85x 30
390 0,85
600
600+390 = 0+025"
ABDULLAH GHIYATS D.U. AULIA NURUL ANNISA
-
7/23/2019 PERENCANAAN PELAT INJAK
4/4
STUDIO PERANCANGAN 2
!in !a
Di#a(ai = 0+006;
Tulangan atas
As = ) = 0+006; &000 &50 = &&$0 !!2= &&+$0 *!2
Di#a(ai tulangan D13 100, As = 12,70 cm2
Tulangan )agi
As? = 0+5 As = 0+5 &&$0 = 560 !!2= 5+60 *!2
Di#a(ai tulangan D10 100, As = 7,13 cm2
ABDULLAH GHIYATS D.U. AULIA NURUL ANNISA