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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