bab2perencanaangording-140718225207-phpapp01
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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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
2.+. Perencanaan G&r$in#
Gambar 2.,.uda kuda
Pan'ang 'ura! = :$0"2" m
Penutup atap = Seng gelombang
Spes!!kas! seng = 210 m 8 100 m (1 = 25o)
= 220 m 8 100 m (2 = "5o)
Pan'ang seng gelombang = 210 m dan 220 m
Gambar 2.+. Sketsa pro!l 7
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
2.,Pembebanan
2.,.1 Pembebanan P&t&n#an I
em!r!ngan Atap (1) = 25o
Jarak #ord!ng (B1) = 1$"% 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$21% kg
Gambar 2.*.6!str!bus! Pembebanan
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
c. Beban Air 0jan -R1/
(PBI 1983 Pa%al 3.2.(2).a)
em!r!ngan Atap (1) = 25o
(25o
50o
Beban 4u'an d!anal!s!s ulang)Beban A!r u'an (>C perlu) = (+0 0$%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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
(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"$1""" kg/m
2.*6&mbinasi Pembebanan
B"+a%a"'an SNI 03-1729-2002 Pa%al .2.2
2.*.1 6&mbinasi Pembebanan P&t&n#an I (&nt&' 1 25o
: B1 138 m)
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
a. 6&mbinasi 1 -1), D/
Beban &erata
>u8 = 1$+ >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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
>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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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
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$22 kg
2.'Per5itn#an M&men2.'.1 M&men P&t&n#an I
a. M&men Ara5 Smb
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
Gambar 2.'. &omen ara4 sumbu ?
6ar! 4as!l anal!s!s menggunakan program SAP G1+$ d!dapat momen maks!mum 3
&tumpuan k!r! = 0 kgm
&tumpuan kanan = 250$+" kgm
&lapangan = 21%$ kgm
&u18 =2
)(
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
&tumpuan kanan = 2+:$2+ kgm
&lapangan = 22"$2 kgm
&u1, =2
)(
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TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
=2
)0"$220(
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5,0 m
TUGAS STRUKTUR BAJABAB 2 PERENCANAAN GORDING
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$22 kg
&omen dar! 4as!l per4!tungan seara manual berdasarkan konsep anal!sa struktur
(Asums! beban beker'a pada beam dengan pan'ang Lb H mem!l!k! 2 tumpuan
'en!s send!)
Gambar 2.1,.&omen ma8
&u8ma8 =
LbP&
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2.:6&ntr&! 6ekatan Pr&fi!2.:.1 6&ntr&! 6e!an#sin#an Penaman#
SNI 03-1729-2002 !abl 7.5-1
Asums! Penampang ompak
74ek 3
Flens< Sa"a Web< Ba$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