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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING

8BAB 2

PERENCANAAN GORDING

Gambar 2.1.Sketsa struktur atas

2.1Data Perencan aan

Jarak antar kuda kuda (Lb) = 5 m

Penutup Atap = Seng gelombang

Berat Penutup Atap = 10 kg/m2 (PBI 1983 Hal. 12)

em!r!ngan Atap 1 (1) = 25o

em!r!ngan Atap 2 (2) = "5o

Jarak #ord!ng 1 (B1) = 1$"% mJarak #ord!ng 2 (B2) = 1$5" m

&utu Ba'a = A" *u = +00 &Pa

*, = 250 &Pa

&utu -ulangan Ba'a = .2% dengan beugel.2+

eepatan Ang!n = 20 km/ 'am

= 5$555 m/ det!k


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Sifat Mekanis Baja Strktra! "an# $i#nakan %

&odulus last!s!tas () = 200000 &Pa (SNI 03-1729-2002 Point 5.1.3)

&odulus #eser (#) = %0000 &Pa (SNI 03-1729-2002 Point 5.1.3)

2.2Perkiraan Pr&fi! G&r$in#

Penentuan Pro!l #ord!ng berdasarkan kontrol bentang 3

20

bLh>

d!mana

==20

5

20

mLb

0$25 m = 250 mm

arena 4 250 mm d!paka! untuk bentang pan'ang$ maka d!gunakan pro!l

dengan 4 250 mm

Pro!l #ord!ng ,ang d!paka!Pro!lCNP 1' (Light Channl)

(!abl P"o#il $on%t"&'%i Baa I". *&+, &naan +ngan /t&n&' I". o"i%o)

6ata data pro!l CNP 1'

(!abl P"o#il $on%t"&'%i Baa I". *&+, &naan +ngan /t&n&' I". o"i%o)

A = 10 mm

b = 5 mm

t = 10$5 mm

Stion "a = +1$2 m2 = +120 mm2

ight = 1%$% kg/m

78 = 1$%+ m = 1%$+ mm

7, = 1%$+ m = 1%+ mm

98 = :25 m+ = :2510+ mm+

9, = %5$" m+

= %5$"10+

mm+

!8 = $21 m = 2$1 mm

!, = 1$%: m = 1%$: mm

S8 = 11 m" = 11000 mm"

S, = 1%$" m" = 1%"50 mm"

1()*

1'(

'*

Gambar 2.2. Pro!l ba'a 7;P 1


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2.+. Perencanaan G&r$in#

Gambar 2.,.uda kuda

Pan'ang 'ura! = :$0"2" m

Penutup atap = Seng gelombang

Spes!!kas! seng = 210 m 8 100 m (1 = 25o)

= 220 m 8 100 m (2 = "5o)

Pan'ang seng gelombang = 210 m dan 220 m

Gambar 2.+. Sketsa pro!l 7


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+


2.,Pembebanan

2.,.1 Pembebanan P&t&n#an I

em!r!ngan Atap (1) = 25o

Jarak #ord!ng (B1) = 1$"% m

a. Beban Mati -D1/

Beban send!r! gord!ng = 1%$% kg/m

Beban penutup atap = 1$"% m 8 10 kg/m = 1"$% kg/m

Berat la!n la!n = 20< 8 1%$% kg/m = "$ kg/m

>61 = "$" kg/m

Beban &at! Ara4 ? (>618) = >61os 1 = "$"os 25o = "2$:5"+ kg/m

Beban &at! Ara4 @ (>61,) = >61s!n 1 = "$"s!n 25o = 15$"+ kg/m

b. Beban 0i$ -P1/

Beban 4!dup d! tenga4 tenga4 gord!ng P = 100 kg (PBI 1983 Pa%al 3.2.(2).b)

Beban !dup Ara4 ? (PL18) = Pos 1 = 100os 25o = :0$"5% kg

Beban !dup Ara4 @ (PL1,) = Ps!n 1 = 100s!n25

o

= +2$21% kg

Gambar 2.*.6!str!bus! Pembebanan


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c. Beban Air 0jan -R1/

(PBI 1983 Pa%al 3.2.(2).a)

em!r!ngan Atap (1) = 25o

(25o

50o

Beban 4u'an d!anal!s!s ulang)Beban A!r u'an (>C perlu) = (+0 0$%1) = (+0 0$%25

o) = 20 kg/m2

Beban A!r &aks!mum (>C maks) = 20 kg/m2

d!paka! >C perlu = 20 kg/m2

Beban A!r u'an (>C1) =1$"% m 8 20 kg/m2 = 2$ kg/m

Beban A!r u'an Ara4 ? (>C18) = >C1os1 = 2$os25o = 25$01+1 kg/m

Beban A!r u'an Ara4 @ (>C1,) = >C1s!n 1 = 2$s!n 25

o

= 11$+" kg/m

$. Beban An#in -31/

(PBI 1983 Pa%al 4.2.(3) 4.2.(1))

eepatan Ang!n (V) = 5$555 m/det

-ekanan -!up (P rumus) =1

26

=1

555$5 2

= 1$:2: kg/m2

Beban Ang!n &!n!mum (P m!n) = 25 kg/m2

PrumusPm!n = 1$:2: kg/m2 25 kg/m2

d!paka! P m!n = 25 kg/m2

(PBI 1983 Pa%al 4.3)

Beban Ang!n (D) = 1$"% m 8 25 kg/m2 = "+$5 kg/m

Beban Ang!n -ekan (Dtekan) = koe!s!en ang!n tekan 8 D

= (0$021 0$+) 8 D ("&m&% &nt&' 5o)

= (0$0225o 0$+) 8 "+$5 = "$+5 kg/m

Beban Ang!n !sap (D4!sap) = koe!s!en ang!n 4!sap 8 D

= E0$+ 8 D

=E0$+ 8 "+$5 = E1"$% kg/m

Beban &erata Ang!n (>D1) = (Dtekan F D4!sap) os 1

= ("$+5F (E1"$%)) os 25o = E:$"%0% kg/m


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2.,.2 Pembebanan P&t&n#an II

em!r!ngan Atap (2) = "5o

Jarak #ord!ng (B2) = 1$5" m

a. Beban Mati -D2/

Beban send!r! gord!ng = 1%$% kg/m

Beban penutup atap alum!n!um = 1$5" m 8 10 kg/m = 15$" kg/m

Berat la!n la!n = 20< 8 1%$% kg/m = "$ kg/m

>62 = "$% kg/m

Beban &at! Ara4 ? (>628) = >62os 2 = "$%os "5o = "1$01"1 kg/m

Beban &at! Ara4 @ (>62,) = >62s!n 2 = "$%s!n "5o = 21$15 kg/m

b. Beban 0i$ -P2/

Beban 4!dup d! tenga4 tenga4 gord!ng

P = 100 kg(PBI 1983 Pa%al 3.2.(2).b)

Beban !dup Ara4 ? (PL28) = Pos 2 = 100os "5o = %1$:125 kg

Beban !dup Ara4 @ (PL2,) = Ps!n 2 = 100s!n "5o = 5$"5 kg

c. Beban Air 0jan -R2/

(PBI 1983 Pa%al 3.2.(2).a)

em!r!ngan Atap (1) = "5o("5o 50o Beban 4u'an d!anal!s!s ulang)

Beban A!r u'an (>C perlu) = (+0 0$%1) = (+0 0$% "5o) = 12 kg/m2

Beban A!r &aks!mum (>C maks) = 20 kg/m2

d!paka! >C perlu = 20 kg/m2

Beban A!r u'an (>C1) =1$5" m 8 12 kg/m2 = 1%$" kg/m

Beban A!r u'an Ara4 ? (>C18) = >C1os1 = 1%$"os "5o = 15$0": kg/m

Beban A!r u'an Ara4 @ (>C1,) = >C1s!n 1 = 1%$"s!n "5o = 10$5"0: kg/m

$. Beban An#in -31/


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(PBI 1983 Pa%al 4.2.(3) 4.2.(1))

eepatan Ang!n (V) = 5$555 m/det

-ekanan -!up (P rumus) =1

2

6

=1555$5

2

= 1$:2: kg/m2

Beban Ang!n &!n!mum (P m!n) = 25 kg/m2

PrumusPm!n = 1$:2: kg/m2 25 kg/m2

d!paka! P m!n = 25 kg/m2

(PBI 1983 Pa%al 4.3)

Beban Ang!n (D) = 1$5" m 8 25 kg/m

2

= "%$25 kg/mBeban Ang!n -ekan (Dtekan) = koe!s!en ang!n tekan 8 D

= (0$022 0$+) 8 D ("&m&% &nt&' 5o)

= (0$02"5o 0$+) 8 "%$25 = 11$+5 kg/m

Beban Ang!n !sap (D4!sap) = koe!s!en ang!n 4!sap 8 D

= E0$+ 8 D

= E0$+ 8 "%$25 = E15$" kg/m

Beban &erata Ang!n (>D") = (Dtekan F D4!sap) os 2

= (11$+5 F (E15$")) os "5o = E"$1""" kg/m

4abe! 2.1 Cekap!tulas! Pembebanan ,ang beker'a

Pembebanan Ara5 Pembebanan

P&t&n#an I

Pembebanan

P&t&n#an II

Satan

Beban &at! (>6) 8 "2$:5"+ "1$01"1 kg/m

, 15$"+ 21$15 kg/m

Beban !dup (PL) 8 :0$"0% %1$:152 kg

, +2$21% 5$"5 kgBeban u'an (>C) 8 25$01+1 15$0": kg/m

, 11$+" 10$5"0: kg/m

Beban Ang!n (>D) 8 E:$"%0" E"$1""" kg/m

2.*6&mbinasi Pembebanan

B"+a%a"'an SNI 03-1729-2002 Pa%al .2.2

2.*.1 6&mbinasi Pembebanan P&t&n#an I (&nt&' 1 25o

: B1 138 m)


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a. 6&mbinasi 1 -1), D/

Beban &erata

>u8 = 1$+ >618 = +$1"+% kg/m>u, = 1$+ >61, = 21$51" kg/m

b. 6&mbinasi 2 -1)2 D 7 1)' 7 ()* R/

Beban &erata

>u8 = 1$2 >618 F 0$5 >C18 = 52$0511 kg/m

>u, = 1$2 >61, F 0$5 >C1, = 2+$21% kg/m

Beban -!t!k

Pu8 = 1$ PL18 = 1+5$00:" kg

Pu, = 1$ PL1, = $1%: kg

c. 6&mbinasi + -1)2 D 7 1)' R 7 ()83/

Beban &erata

>u8 = 1$2 >618 F 1$ >C18 F 0$%>D1= 2$02+ kg/m

>u, =1$2 >61, F 0$5 >C1, = "$102 kg/m

$. 6&mbinasi , -1)2 D 7 1)+ 3 7 ()* 7 ()* R/

Beban &erata

>u8 = 1$2 >618 F 1$">D1F 0$5 >C18 = ":$%5 kg/m

>u, =1$2 >61, F 1$ >C1, = 2+$21% kg/m

Beban -!t!k

Pu8 = 0$5 PL18 = +5$"15+ kg

Pu, = 0$5 PL1, = 21$1"0: kg

e. 6&mbinasi * -1)2 D 7 ()* /

Beban &erata

>u8 = 1$2 >618 = ":$5++1 kg/m


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>u, = 1$2 >61, = 1%$+": kg/m

Beban -!t!k

Pu8 = 0$5 PL18 = +5$"15+ kgPu, = 0$5 PL1, = 21$1"0: kg

2.*.2 6&mbinasi PembebananP&t&n#an II (&nt&' 2 35o: B2 153 m)

a. 6&mbinasi 1 -1), D/

Beban &erata

>u8 = 1$+ >628 = +"$+1%" kg/m

>u, = 1$+ >62, = "0$+01% kg/m

b. 6&mbinasi 2 -1)2 D 7 1)' 7 ()* R/

Beban &erata

>u8 = 1$2 >628 F 0$5 >C28 = ++$"55 kg/m

>u, = 1$2 >62, F 0$5 >C2, = "1$"2+2 kg/m

Beban -!t!k

Pu8 = 1$ PL28 = 1"1$0+" kg

Pu, = 1$ PL2, = :1$22 kg

c. 6&mbinasi + -1)2 D 7 1)' R 7 ()83/

Beban &erata

>u8 = 1$2 >628 F 1$ >C28 F 0$%>D2= 5%$2+ kg/m

>u, =1$2 >62, F 1$ >C2, = +2$:0%2 kg/m

$. 6&mbinasi , -1)2 D 7 1)+ 3 7 ()* 7 ()* R/

Beban &erata

>u8 = 1$2 >628 F 1$">D2F 0$5 >C28 = +0$22 kg/m

>u, =1$2 >62, F 0$5 >C2, = "1$"2+2 kg/m

Beban -!t!k

Pu8 = 0$5 PL28 = +0$:5 kg

Pu, = 0$5 PL2, = 2%$%% kg


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e. 6&mbinasi * -1)2 D 7 ()* /

Beban &erata>u8 = 1$2 >628 = "$215 kg/m

>u, = 1$2 >62, = 2$05% kg/m

Beban -!t!k

Pu8 = 0$5 PL28 = +0$:5 kg

Pu, = 0$5 PL2, =2%$%% kg

2.*.+ Pembebanan Ak5ir

;a"i ha%il &i %tia/ 'ombina%i mngg&na'an /"og"am SP 614 ma'a +i+a/at

omn !"#a'to" a'%im&m inim&m /a+a

a. Ga"a 4erfakt&r P&t&n#an I (&nt&' 1 25o: B1 138 m)

>u8 = 52$0511 kg/m

>u, = 2+$21% kg/m

Pu8 = 1+5$00:" kg

Pu, = $1%: kg

b. Ga"a 4erfakt&r P&t&n#an II (&nt&' 1 35o: B2 153 m)

>u8 = ++$"55 kg/m

>u, = "1$"2+2 kg/m

Pu8 = 1"1$0+" kg

Pu, = :1$22 kg

2.'Per5itn#an M&men2.'.1 M&men P&t&n#an I

a. M&men Ara5 Smb


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Gambar 2.'. &omen ara4 sumbu ?

6ar! 4as!l anal!s!s menggunakan program SAP G1+$ d!dapat momen maks!mum 3

&tumpuan k!r! = 0 kgm

&tumpuan kanan = 250$+" kgm

&lapangan = 21%$ kgm

&u18 =2

)(


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&tumpuan kanan = 2+:$2+ kgm

&lapangan = 22"$2 kgm

&u1, =2

)(


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=2

)0"$220(


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5,0 m


6engan ga,a ga,a ,ang beker'a pada gord!ng 3

>u8 = 50$0511 kg/m

>u, = "1$"2+2 kg/mPu8 = 1+5$00:" kg

Pu, = :1$22 kg

&omen dar! 4as!l per4!tungan seara manual berdasarkan konsep anal!sa struktur

(Asums! beban beker'a pada beam dengan pan'ang Lb H mem!l!k! 2 tumpuan

'en!s send!)

Gambar 2.1,.&omen ma8

&u8ma8 =

LbP&


15/23


16/23

2.:6&ntr&! 6ekatan Pr&fi!2.:.1 6&ntr&! 6e!an#sin#an Penaman#

SNI 03-1729-2002 !abl 7.5-1

Asums! Penampang ompak

74ek 3

Flens< Sa"a Web< Ba$an

I p I t

#ln%t

b

I

#,

250

.bt

h

I

#,

5

t

b

I

#,

250

t

2

I

#,

5

5$10

5

I250

250

5$10

10

I250

5

$1:05 I 15$%11+ 15$2"%1 I +2$05%"

Penaman# 6&mak Penaman# 6&mak

&aka Asums! Pro!l adala4 Penampang ompak adala4 Benar

2.:.2 6&ntr&! en$tan

SNI 03-1729-2002 !abl .4.3

(Asums! beban beker'a pada beam dengan pan'ang Lb H mem!l!k! 2 tumpuan

'en!s send!)

a. Displacement Ara5 Smb

K8 =

I,>

LbP&

Lb=&Mu=3267430N . mm

Jad! pro!l 7;P 16?A4mena4an &u

2.:.* 6&ntr&! Geser

SNI 03-1729-2002 !abl 8.8

6etentan 1

h

tw1,10

knE

fy

kn=5+5

a

h

2=5,0051

h

t1,10

knE

fy

A

t 1,10

knE

fy

15,238169,6056 A$

On = 0$ , A

= 252000 ;

6etentan 2


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1,10knE

fyh

t1,37

knE

fy

69,6056 I 15$2"%1 I %$:0

On = 0$ *,A 1,10knE

fy

1

h

tw

= 1151102 ;

Gambar 2.1*.#a,a geser maks!mum

Berdasarakan 4as!l anal!s! menggunakan program SAPG1+$ d!dapat3

Ou = 202$51 kg = 2025$1 ;

Check$

Ou I On

2025$1 ; I 22%00 ; O6) AMAN 4ER0ADAP GESER

Berdasarkan 4as!l per4!tungan seara manual$ asums! beban beker'a pada beam

dengan pan'ang Lb H 2 tumpuan 'en!s send!

Ou =1

2.qu.l+

1

2. Pu


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= 202$"2+ kg

= 202$"2+ ;

Check$

Ou I On

202$"2+; I 10"5::2 ; O6) AMAN 4ER0ADAP GESER

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