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BAB 1
PERENCANAAN RANGKA KUDA KUDA
1.1. Ketentuan umum
1. Lokasi Konstruktur = Daaerah Pedalaman
2. Jenis atap = Genteng
3. Dukungan = Sendi rol
4. Langit-langit ada = da
!. Jenis ka"u = Kruing
#. lat sam$ung = %aut
&. Jarat antara Kuda-kuda = 3'(( m
). *kuran L +m, = ('( m
. Sudut = 4(/
1.2. Ukuran Panjang Batang
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%atang 0=0=G=G=K=K%=KJ =l
cos=
0,9
cos 40=1,1748m
%atang D=D==J=JL=L% = (' m
%atang 0D=KL = D .tg
= (' . tg 4(/ = ('&!!1 m
%atang = J = +2.L,.tg
= +2 5 (',.tg 4(/ = 1'!1(3 m
%atang G = +35L6,.t6
= +35(',.tg 4(/ = 2'2#!! m
%atang = = (EF)2+(I)
= (1,5103)2+(1,4)2=2,0593
1.3.Tabel Panjang Batang
7o %atang Pan8ang+9m, 7o %atang Pan8ang+9m,
1 0 1'!1(3 12 G 1'!1(3
2 D (' 13 2'(!3
3 0D ('&!!1 14 J ('
4 0 1'!1(3 1! J 1'!1(3
! 0 1'!1(3 1# K 1'!1(3
# D (' 1& JK 1'!1(3
& 1'!1(3 1) JL (') G 1'!1(3 1 KL ('&!!1
2'(!3 2( K% 1'!1(3
1( (' 21 L% ('
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1.4. Hitungan Sudut
K%L = CAD=40
D0 = %LK = 0D = = J = KLJ = (/
0D = D0 = LKJ = %KL = !(/
0 = KJL= !(/
0 = KJ = )(/
= J = ar9 sin1,51032,593 = !3/3&:23.#&::
0 = KJ = !(/
= J = (/ - = 3#/22:3#'33::
G = JG = (/ - = 3#/22:3#'33::
G = G = 1)(/ -+0;, = 1(3/3&:23.#&::
G = G = !(/
D0 = K%L = 4(/
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1.5. Tabel Sudut
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BAB II
PEE!"A!AA! #$%I!#
Jarak kuda kuda = 3'(( m
Jarak Gording = 1'!1(3 m
*kuran Gording = 1(>12
%erat tap = 4( Kg
Kemiringan atap = 4(/
%erat 8enis ka"u +Kruing, = (( Kg/m3
lt = 100 Kg /cm2
= 1(('((( Kg/cm2
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2.1.1 Beban &ati '&(
%erat sendiri gording +1(>12, = ('1.('12.((.1
= 1(') Kg>m
%erat sendiri atap = 0..
= 4(.1'!1(3.1
= #('412 Kg>m
?total = 1(') ; #('412 = &1'212 Kg>m
?5 = 6.9os
6 = &1'212.9os 4(/ = !4'!!1! Kg /cm2
?" = 6 sin @ 6 = &1'212. Sin 4(/ = 4!'&&41 Kg/cm2
A5i = +1>),.+6.9os ,. (L)2
@ +1>),.+!4'!!1!,. (3,0)2
= #1'3&(4 Kg>9m
A"i = +1>),.+6.sin ,. (L)2
@ +1>),.+4!'&&41,. (3,0)2
= !1'4!) Kg>9m
2.12) Beban Hidu* 'H(
%erdasarkan peraturan pem$e$anan ndonesia +PP 1)3, pasal B3.2.2 C$e$an hidup atap
"ang tidak dapat di9apai dan di $e$ani oleh orang harus diam$il "ang paling menentukandiantara dua ma9am $e$a$n $erikut
. %e$an air hu8an
%e$an ter$agi rata per m2
$idang datar B
6 = + 4( - (')., 2( L Kg /cm2
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6 = + 4( - (')., L 2( L
= + 4( (').4(,.1'(( 2(.1'((
= 1# kg>m 2( kg>m
65 = 6.9os @ 1#.9os 4(/ = 12'2!#& Kg/cm2
65 = 6.sin @ 1#.sin 4(/ = 1('2)4# Kg /cm2
A5i= +1>),.+6.9os ,. (L)2
@ +1>),.+12'2!#&,. (3,0)2
= 13'&))&Kg>9m
A"2= +1>),.+6.sin ,. (L)2
@ +1>),.+1('2)4#,. (3,0)2
= 11'!&(1Kg>9m
B. Beban TerpusatBeban terpusat dari seorang pekerja atau seorang pemadam
kebakaran dengan peralatan sebesar P = 100 kg
P5 = P 9os
@1((.9os 4(= '#(44 kg>mP" = P sin @ 1((.sin 4(= #4'2&)&
A52 = +1>4,.+P.9os ,.+L, @ +1>4,.+'#(44,.+3'(, = !&'4!33 kg>m
A"2 = +1>4,.+P.sin ,.+L, @ +1>4,.+#4'2&)&,.+3.(, = 4)'2(( kg>m
Pada peran9angan di pakai $e$an hidup "ang ter$esar antara $e$an air hu8an dan terpusat.
Pada peran9angan ini digunakan $e$an hidup "ang $erasal dari $e$an terpusat dengan B
A52 = !&'4!33 kg>mA"2 = 4)'2(( kg>m
2.1.3 %e$an
%erdasarkan PP 1)3 B@
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= (0,02.!00,!".2$.1,$10%= 1$,10% kg&m
'% = 1&).q tekan. L2
= 1&).1$,10%. 32
= 1*,++0) kg&m
'% = 0
2.1.!. -ombinasi pembebanan
a.Pembebanan tetap