Bab III Gording Fix

7/26/2019 Bab III Gording Fix

1/11

BAB III

PERENCANAAN GORDING

3.1 Data yang Diketahui

Kemiringan atap : 22,5

Jarak antar kuda-kuda : 5 m

Bentang kuda-kuda : 30 mJarak miring antar gording : 1,624 m

Berat penutup atap (alumunium) : 10 kgm2

Be!an "idup (orang) : 100 kgBe!an angin (p#) : 25 kgm

2

$utu !a%a : BJ 3& (' 240 $pa, 'u 3&0 $pa)

*ro'il ang digunakan :Lipped Channel 100 + 50 + 20 + 2

3.2 Pembebanan Gording

3.2.1Beban mati !d"

Berat .endiri gording /p 4,& kgmBerat penutup atap !erat atap + %arak antar gording

10 kgm2 + 1,624 m

16,24 kgmBerat lat *enam!ung 5 + (!erat gording !erat penutup atap)

5 (4,& 16,24)

1,06 kgm

otal !e!an mati (/d) 22,1& kgm 0,221& mm

3.2.2 Beban #idu$

Be!an orang 100 kgBe!an "u%an (#") 40 - 0,

40 0, + 22,5

22 kgm2 karena 7 20 kgm2 maka !e!an air "u%an 20 kgm2

89 #" + Jarak miring antar gording

20 + 1,624

32,4 kgm 0,324 mm

3.2.3 Beban Angin

Be!an angin (p#) 25 kgm2

Koe' ngin tekan (1) 0,02 - 0,4

0,02 + 22,5 - 0,4 0,05

Be!an angin tekan (/#1) 1+ %arak miring gording + p#

0,05 + 1,624 + 25

2,03 kgm 0,0203 mm

Koe' ngin "i.ap (2) -0,40

Be!an angin "i.ap (/#2) 2+ %arak gording + p#


2/11

0,40 + 1,624 + 25

16,24 kgm

3.3 %ombina&i Pembebanan

Kom!ina.i pem!e!anan menurut ;< 2002, maka kom!inai pem!e!anan adala":1 = 1,4 >

2 = 1,2 > 1,6 ? 0,5 (?a atau 9)

3 = 1,2 > 1,6 (?a atau 9) (? ? atau 0 @)

4 = 1,2 > 1,3 @ ? ? 0,5 (?a atau 9)

5 = 1,2 > A 1,0 ? ?

6 = 0,C > A (1,3 @ atau 1,0 )

Keterangan :

> Be!an $ati

? Be!an 9idup ang ditim!ulkan ole" pengguna gedung

?a Be!an 9idup ang ditim!ulkan di .elama pera#atan ole" peker%a

9 Be!an 9u%an

@ Be!an ngin

Be!an Dempa

>engan E? 05 !ila ? F 5k*a, E?1 !ila ? 7 5 k*a

3.3.1Beban 'erbagi (erata8> 0,221& mm

8? 0 mm

8?a 0 mm

89 0,324 mm8@ 0,0203 mm

8 0 mm

GH$1 1,4 + 0,221& 0,31 mmGH$2 1,2 + 0,221& 1,6 + 0,0 0,5 + 0,324 0,43mm

GH$3 1,2 + 0,221& 1,6 + 0,0 0,5 + 0,0 0,2& mm

GH$4 1,2 + 0,221& 1,3 + 0,0203 0,5 + 0,0 0,5 + 0,324 0,45 mm

GH$5 1,2 + 0,221& 1,0 + 0 0,5 + 0,00 0,2& mmGH$6 0,C + 0,221& 1,3 + 0,0203 0,23 mm

>ipakai 8ma+ )*+, N-mm

3.3.2 Beban 'er$u&at

*> 0

*? 0

*?a 1000 *9 0

*@ 0


3/11

* 0

GH$1 1,4 + 0 0

GH$2 1,2 + 0 1,6 + 0 0,5 + 0 0

GH$3 1,2 + 0 1,6 + 1000 0,5 + 0 1600 GH$4 1,2 + 0 1,3 + 0 0,5 + 0 0,5 + 0 0

GH$5 1,2 + 0 1,0 + 0 0,5 + 0 0

GH$6 0,C + 0 1,3 + 0 0 >igunakan *ma+ 1600

Anai&a &truktur menggunakan /AP2))) 0.11* maka di$eroeh

(omen Nomina

$> &502C1CC mm

$? 0 mm

$?a 3C0&2405 mm

$9 5C6410 mm

$@ C&6101 mm

$ 0 mm

GH$1 1,4 + &502C1CC 105040& mm

GH$2 1,2 + &502C1CC 1,6 + 0 0,5 + 5C6410 C4333243 mm

GH$3 1,2 + &502C1CC 1,6 + 3C0&2405 0,5 + 0 152550& mm

GH$4 1,2 + &502C1CC 1,3 + C&6101 0,5 + 0 0,5 + 5C6410

10&031&&5mm

GH$5 1,2 + &502C1CC 1,0 + 0 0,5 + 0 C003503 mm

GH$6 0,C + &502C1CC 1,3 + C&6101 022410 mm

>igunakan $ma+ 152550& mm

Gaya ge&er Nomina

I> C3,4

I? 0

I?a C3&,&4

I9 206,31

I@ 234,43

I 0

GH$1 1,4 + C3,4 131,3

GH$2 1,2 + C3,4 1,6 + 0 0,5 + 206,31 215,&& GH$3 1,2 + C3,4 1,6 + C3&,&4 0,5 + 0 1612,CC

GH$4 1,2 + C3,4 1,3 + 234,43 0,5 + 0 0,5 + 206,31 520,53


4/11

GH$5 1,2 + C3,4 1,0 + 0 0,5 + 0 112,61

GH$6 0,C + C3,4 1,3 + 234,43 3C,22

>igunakan Ima+ 1612,CC

(omen nomina teraktor dan Gaya ge&er nomina teraktor

Kemiringan 22,5

$omen De.er

mm

ominal $n 152550&

"p .! $n+ $n + ;in 22,5 53&6C&

"p .! +$n $n + Go. 22,5 140C3642

ominal In 1612,CC

"p .! In In + ;in 22,5 61&26

"p .! + In+ In + Go. 22,5 14C021

3.+ De&ain Gording

3.+.1 Perhitungan %a$a&ita& Penam$ang

Data Bahan B 34"

' 240 $*a

'u 3&0 $*a

'r &0 $*a

20000

0$*a

Data Proi Ba5aLipped Channel 1)) 6 ,) 6 2) 6 2*7

"t 100 mm

! 50 mm

a 20 mm

t 2 mm

6205 mm2

22,5


5/11


6/11

2

5

-

21,16&410-

1C100

22000

10+5,1C64

=

0,0001C mm22

+ 14 + "t L t2 a + t + ("t - a) t + (! - 2 + t) ("t - t)

16&60 mm3

"t + t ( - t2) 2 + a + t (! - - t2) t + ( - t)2 t + (! - t - )2

100 + 2, (1,6 - 22) 2 + 20 + 2, (50 - 1,6 2,2) 2, + (1,6 - 2,)2

2, + (50 2, - 1,6)2

11165 mm3

Keterangan :

D modulu. ge.er + modulu. penampang pla.ti. t"d .! +

J Kon.tanta punitir tor.i modulu. penampang pla.ti. t"d .!


7/11

/yarat

8 9 8$ Mn=Mp MkompakN

8$ : 8 9 8r Mn=Mp (MpMr )(p)rp Mtak kompakN

8 ; 8rMn=Mr

r

2

Mlang.ingN

Jadi, pro'il gording terma.uk penampang tak kom$ak

Mnx=Mp(MpMr ) .(p)rp

Mnx=Mpx(MpxMrx )(p)rp

4022377(40223773247000 ) .

(17,85710,973)(28,37810,973)

3&15&02 mm

Mny=Mpy(MpyMry )

(p)

rp

2679667(26796673247000 )(17,85710,973)(28 ,37810,973)

20C&201 mm

(omen Nomina PengaruhLateral Buckling

/yarat

? O ?p Mn=Mp=fy Zx

?p O ? O ?r Mn=Cb[Mr+(MpMr) LrLLrLp ]

? 7 ?r Mn=Cb p

L [E Iy G J+(p E

L)2

Iy Iw ]

Mn=C b .p

L [E . Iy . G. J+(p . E

L)2

.Iy.Iw ]

Lp=1,76 ry

Efy

1,76x39,2x

200000240 C60 mm

Lr=ryx1fL

1+1+x 2 fL2

39,214994,67

170 1+1+0,00019x 1702


8/11

3145 mm

Cb= 12,5.Mux

(2,5Mux+3MA+4MB+3MC)

12,5x 583786.97

(2,5x583786.97+3x145947+4x 291893+3x 437840) 1,6&

Mpx=fy.Zx 240x 16760 40223&& mm

Mpy=fy.Zy 240x 11165 26&C666& mm

Mrx=Sx . ( fy fr )=19100 . (24070 ) 324&000 mm

Mry=Sy .(fyfr)=7100 .(24070) 120&000 mm

9a.il ? 7 ?r5000 7 3145PPP !entang pan%ang

nx=Cb

L [E Iy G J+(p E

L)2

Iy Iw ]

1,673,14

5000200000x220000x 80000x 1674,21

+(3,14

x200000

5000 )2

x 220000x 5,196.108

2C05564 mm F $p+, $n+ 2C05564 mm

ny=Cb

L [E Iy G J+(p E

L)2

Iy Iw ]200000

1,673,14

5000

+( 3,14x 2000005000 )2

x 220000x 5,196.108

2C05564 mm 7 $p, $n $p 26&C66& mm


9/11

'ahanan moment entur

/umbu 6Berda.arkanLocal Buckling$n+ 3&15&02 mm

Berda.arkanLateralBuckling$n+ 2C05564 mm

>ipili" nilai $n+ ang terkeil, aitu $n+ 2C05564 mm

a"anan momen lentur .! + ! + $n+ 0,C + 2C05564 261500& mm

/umbu y

Berda.arkanLocal Buckling$n 20C&201 mm

Berda.arkanLateralBuckling$n+ 26&C66& mm

>ipili" nilai $n+ ang terkeil, aitu $n 20C&201 mm

a"anan momen lentur .! + ! + $n+ 0,C + 20C&201 1&41 mm

$u+ 53&& mm

$u 140C36 mm

Mux . Mnx

583787

0,9.2905564

0,2232

Muy . Mny

14093860,9.2097201

0,&46&

;arat

Muxb.Mnx

+ Muyb.Mny O 10

0,C6CC O 10 PPPPPP $

'ahanan ge&erht

Bab III Gording Fix

Documents

Transcript of Bab III Gording Fix