tugas baja agung

8/19/2019 Tugas Baja Agung

1/71

.

BAB I

PENDAHULUAN

1.1 Deskripsi Struktur

1.1.1 Bentuk dan Dimensi Struktur

3.00 M3.00 M 3.00 M

4.50 M

3.00 M

3.00 M

4.00 M

4.00 M

1

Panjang masing masing elemen yaitu :

1.Kolom

H1 3:=

H2 3.:=

2. Balok

A 4:=

A1 4:=

A2 4:=

3. Kemi ingan Ata!


2/71

α 30 π180⋅:=

PanjangRafter

2 A

2⋅

cos α( ):=

PanjangRafter 4.619m=

1.1.2 Fungsi Struktur

"ungsi ge#ung a#ala$ as ama

1.1.3 Spesifkasi Material

%utu Baja & Bj'3(

%utu Baut & Bj'3)

Dari tabel 5.3 SNI 2002 didapat

fy 2100 g

cm2

:=

f! 3400 g

cm2

:=

*ekanan angin

Pang"n 1.93

g

m2:=

+enis Ata! & seng

#ata$ %c( ) 10 g

m2

:=

+a ak go #ing maksimum

s 0.6:=

%o#ulus Elastisitas

& 2 106

⋅ '

mm2

:=

%o#ulus ,ese

8 103

⋅ '

mm2

:=


3/71

Ni-a$ Poisson

µ 0.3:=

Koe sien !emuaian

a 12 10 6−

⋅:=BAB II

PE/EN0ANAAN , /DIN,

2.1 Analisa Pembebanan

b30o

4

cos30 0.866:=

s"n30 0.5:=

4

cos30:=

4.619=

+a ak %aksimum ,o #ing s

s 0.6:=

+umla$ ,o #ing n

n

s 1+:=

n 8.698=

Di-ulatkan se$ingga :

n 9:=

+a ak Anta ,o #ing 4

*

n 1−:=

* 0.5++=

, 4:=


4/71

a#ala$ -entang te !anjang sejaja go #ing

2.1.1 Beban Mati

Tentukan sendiri profl dari tabel profl baja light chane l

Be at P o l ,o #ing Lig$t 0$anel

%$ +.51:=

kg5m

(Baja Profl C Hal 50 dimensi 150 !5 20 3.2 tabel baja "

arena digunakan atap seng! maka "

Be at ata! 67

%c 10:=

kg5m2

(Bab 2 tabel 21 #al 12 at$ran

pembebanan "

Be at Penyam-ung

-$enyam !ng 5 0.05→:=

(bab 3 pasal 3.% at$ran

pembebanan"

arena nilai # tidak sama dengan 1 m! maka digunakan rumus "

%co %c⋅:=

%co 5.++4=

kg5m

Analisa " karena nilai $c untuk %m2 maka nilai $c terlebih dahulu dikalikandengan nilai #

*otal Be-an %ati 6#

Nilai 1.89 m & 188 ; 9 67 asumsi 188%/ %$ %co+( ) 1.05⋅:=

%/ 13.948=

kg5m

%omen #itenga$ -entang


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Kemi ingan ata! a

α 30:=

, 4:=

&'ilai bentang terpanjang pada bagian (ang mengalami tekuk)

M/ 1

8 %/⋅ cos30⋅ , 2⋅:=

M/ 24.158=

kg m

M/y 1

8 %/⋅ s"n30⋅ , 2⋅:=

M/y 13.948=

kg m

2.1.2 Beban *idup

Be-an o ang te !usat P

P 250:=

(bab 3 pasal 3.2 a&at 1 at$ran pembebanan"

Mo 1

4 P⋅ cos30⋅ , ⋅:=

Mo 216.5=

kg m

Moy 1

4 P⋅ s"n30⋅ , ⋅:=

Moy 125=

kg m

2.1.+ Beban Angin

(bab % pasal %.3 a&at 1b at$ran pembebanan

'tap se i)3 den an s$d$t *emirin an α

+30, jadi α -!5 se#in a r$m$sn&a (0.02 α)0.%""

c1 0.02 α⋅ 0.4−( ):=

c1 0.2=


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&Angin Tekan)

c2 0.4:=

&Angin *isap)

(bab % pasal %.3 a&at 1d at$ran pembebanan,

'tap se i)3 majem$* α-!5 se#in a r$m$sn&a 0.2 α ) 0.%"

Ke7e!atan Angin

a 5.56:=

m5s

&Dari Soal)

*ekanan Angin Pa

Pa a2

16:=

(bab % pasal %2 b$tir 3 at$ran pembebanan"

Pa 1.932=

kg5m2

Berdasarkan Peraturan Pembebanan ,ndonesia bab - pasal -.2 point 1 tekanan tiup minimum 2/ kg%m2

Pa 25:=

kg5m2

%a1 c1 Pa⋅:=

%a1 5=

kg5m2

%a2 c2 Pa⋅:=

%a2 10=

kg5m2

Tekanan ma0 $a $a2

%a 10:=

kg5m2

% %a *⋅:=


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% 5.++4=

kg5m

A a$ ' <

M 1

8

%⋅ , 2

⋅:=

M 11.54+=

kg m

A a$ ' y

M y 0:=

kg m

2.1.- Beban *ujan

% 1 40 0.8 α⋅−( ):=

(bab 3 pasal 3.2 a &at 2a at$ran pembebanan"

% 1 16=

kg5m2

Berdasarkan PP, bab + pasal +.2 point 2a P ma0 2 kg%m2

% 16:=

kg5m2

%r % *⋅ cos30⋅:=

%r 8=

kg5m

Mr 1

8 %r ⋅ cos30⋅ , 2⋅:=

Mr 13.856=

kg m

Mry 1

8 %r ⋅ s"n30⋅ , 2⋅:=

Mry 8=

kg m

2.1./ ombinasi Pembebanan


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(SNI 2002 #al 13, b$tir !.2.2"

Beban mati 3 &Beban 4rang atau *ujan) 3 Beban Angin

A a$ < :

M! 1 1.2 M/⋅ 1.6 Mo⋅+ 0.8 M⋅+:=

M! 1 384.62+=

kg m

M! 2 1.2 M/⋅ 0.5 Mo⋅+ 1.3 M⋅+:=

M! 2 152.251=

kg m

M! 3 1.2 M/⋅ 1.6 Mr⋅+ 0.8 M⋅+:=

M! 3 60.39+=

kg m

M! M! 1:=

M! 384.62+=

kg m

A a$ y :

M!y1 1.2 M/y⋅ 1.6 Moy⋅+ 0.8 M y⋅+:=

M!y1 216.+3+=

kg m

M!y2 1.2 M/y⋅ 0.5 Moy⋅+ 1.3 M y⋅+:=

M!y2 +9.23+=

kg m

M!y3 1.2 M/⋅ 1.6 Mry⋅+ 0.8 M⋅+:=

M!y3 51.02+=

kg m

M!y M!y1:=

M!y 216.+3+=

kg m

2.2 Desain 5ording

2.2.1 Perhitungan apasitas Penampang


9/71

P o!e ties !enam!ang go #ing Light Channel

(Baja Profl C Hal 50

dimensi 150 !5 20 3.2 "

H 15:=

c

44.3:=

7m3

6.5:=

c

y 12.2:=

7m3

ε

/ 2:=

c

332:=

7m(

t 0.32:=

c

y 53.8:=

7m(

tf 0.32:=

c

r 5.89:=

c

A 9.56+:=

7m2

ry 2.3+:=

c

%ate ial Baja

(SNI 2002 #al , b$tir 5.1.3"


10/71

(SNI 2002 #al 11 , tabel 5.3"

& 2000000:=

kg57m2

+enis : B+ 3(

fy 2100:=

kg57m2

( 800000:=

kg57m2

fr 0.3 fy⋅:=

= & tegangan sisa

fr 630=

kg57m2

6ek Terhadap Tekuk 7ateral

"akto Pengali %omen 0-

1.136:=

&8ntuk balok sederhana)

Nilai 7- #i!e ole$ -e #asa kan tu unan umus :


2,3

3.M4.M3.MM2,5M12,5

CCBAmax

maxb ≤+++

=

1.136:=

.....$nt$* balo* seder#ana (SNI 2002 #lm3/ point .3.1"

&Sumber " S', Baja *al.+9 :.+.1 point d)

12.5 Mma⋅

2.5Mma 3MA+ 4 M-⋅+ 3M+ 2.30≤:= Mma

Dimana :%A & %omen #i 15( -entang & 3532.6.l2

%B & %ma< & %omen #i 152 -entang & 15>.6.l2

%0 & %omen #i 35( -entang & 3532.6.l2


11/71

12.5 1

8 ,

2⋅

⋅

2.5 1

8⋅ ,

2⋅ ⋅ 3 3

32⋅ ,

2⋅ ⋅+ 4 1

8⋅ ,

2⋅ ⋅+ 3 3

32⋅ ,

2⋅ ⋅+

1.1363636363636363636→:=

1.136 2.30≤

.......... 4 ;;;

?e$ingga #i!e ole$ nilai 0- & 1.13@

1.136=

7 400:=

c

&'ilai bentang terpanjang tanpa pengekang lateral)

(SNI 2002 #al 3 ,tabel .3)2"

7$ 1.+6 ry⋅ &

fy⋅:=

7$ 128.+26=

c

Dimana + yaitu konstanta !unte to si #engan umus :

3ii .hb3

1J ∑=

&Sumber " *andout Dosen)?e$ingga umus + menja#i :

1

3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=

0.35=

7m(

1 π & ⋅ ⋅ A⋅

2⋅:=

1 1.16 105×=

c

tf 3⋅

H2

12⋅

3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=


12/71

1.51 103×=

7m@

2 4

y

⋅

⋅

2⋅:=

2 2.819 10 6−×=

c

7r ry 1

fy fr − ⋅ 1 1 2 fy fr −( ) 2⋅++⋅:=

7r 35+.899=

c

7 400:=c

L L-

(SNI 2002 #al 3 , b$tir .3.5"

35+.899 400<

....4 ;;;

M "sa, !nt! $en am$an g seg" 4

1 ,5hb61hb41

SZ 2

2

===ξ

:n t ! $rof", # ; ξ ∼ 1.09 < 1=18= )"asanya /"am )",ξ > 1=12

&Sumber " *andout Dosen)

M$ 1.5 ⋅ fy⋅:=

M$ 1.395 105×=

kg 7m

Mr fy fr −( ) ⋅:=

Mr 6.512 104×=

kg 7m

(SNI 2002 #al 3 , tabel .3)2.b"


13/71

Mn Mr M$ Mr −( ) 7r 7−( )7r 7$−( )⋅+⋅:=

Mn 5.846 104×=

kg 7m

uat 7entur


14/71

kg 7m

M!y 1.063 104

⋅:=

kg 7m

M!y 1.063 10 4×=kg 7m

M!y φMny≤

(SNI 2002 #al 3%, b$tir .1.2"

1.063 104× 2.+6+ 10

4×≤

....... 4 ;;;

2.2.2 Pemeriksaan ekuatan dan 7endutan

Pemeriksaan ekuatan

M! 2 104×=

kg 7m

φMn 5.262 104×=

kg 7m

M!y 1.063 104×=

kg 7m

φMny 2.+6+ 104×=

kg 7m

Mn M$:=

(SNI 2002 #al 3!, b$tir .2.3"

φM$ φMn:=

φM$y φMny:=

M!

φM$ ζ M!y

φM$y ζ+ 1.0≤

#imana

M!

φM$M!y

φM$y+ 0.+64=


15/71

0.+64 1.0<

...... 4 ;;;

Pemeriksaan 7endutan

Len#utan Aki-at Be-an %ati Be-an ang Be-an Hujan #an Be-an Angin

Arah Sumbu x

7 400:=

c

Be-an %ati

∆ /0.05 %/⋅ cos α π

180⋅

⋅ 74⋅

384 &⋅ ⋅:=

∆ / 0.061=c

Be-an ang

∆ oP cos α π

180⋅

⋅ 73⋅

192 &⋅ ⋅:=

∆ o 0.109=

c

Be-an Hujan

∆ r 0.05 %r ⋅ cos α π

180⋅

⋅ 74⋅

384 &⋅ ⋅:=

∆ r 0.035=

c

Be-an Angin

∆ 0.05 %⋅ 74

⋅384 &⋅ ⋅:=

∆ 0.029=

c

%aka :

∆ tota, ∆ / ∆ o+ ∆r +:=


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∆ tota, 0.204=

c

Arah Sumbu

7 400:=

c

Be-an %ati

∆y/0.05 %/⋅ s"n α π

180⋅

⋅ 74⋅

384 &⋅ y⋅:=

∆y/ 0.216=

c

Be-an ang

∆yoP s"n α π

180⋅

⋅ 73⋅

192 &⋅ y⋅:=

∆yo 0.38+=

c

Be-an Hujan

∆yr 0.05 %r ⋅ s"n α π

180⋅

⋅ 74⋅

384 &⋅ y⋅:=

∆yr 0.124=

c

Be-an Angin :

∆y 0:=

%aka :

∆ytota, ∆y/ ∆yo+ ∆yr +:=

∆ytota, 0.+2+=

c

∆ ∆ tota, 2 ∆ytota, 2+:=

∆ 0.+55=


17/71

c

end$tan I in

7 400:=

c

∆" 7240

:=

(SNI 2002 #al 15, b$tir !.%.3"

∆" 1.66+=

c

∆ ∆"≤

0.+55 1.66+≤

...... 4 ;;;

BAB III

ANALI?A PE%BEBANAN

+.1 Analisa Pembebanan

+.1.1 Beban Mati

Beban pada


18/71

A2 4:=

m

A 4:=

m

$da)*$da 1

Rafter1( )

-gor/"ng1 %$ 0.5⋅ A1⋅:=

-gor/"ng1 15.02=

kg

$da)*$da 2

Rafter2( )

-gor/"ng2 %$ 0.5⋅ A1 A2+( )⋅:=

-gor/"ng2 30.04=

kg

$da)*$da 3

Rafter3( )

-gor/"ng3 %$ 0.5⋅ A2 A2+( )⋅:=

-gor/"ng3 30.04=

kg

$da)*$da %

Rafter4( )

-gor/"ng4 %$ 0.5⋅ A2( )⋅:=

-gor/"ng4 15.02=

kg

Beban Atap " Beban atap dijadikan beban terpusat disepanjang

ra=ter pada titik gording

$da)*$da 1

Rafter1( )

-ata$1 %co 0.5⋅ A1⋅ *⋅:=

-ata$1 6.66+=


19/71

kg

$da)*$da 2

Rafter2( )

-ata$2 %co 0.5⋅ A1 A2+( )⋅ ⋅:=

-ata$2 13.334=

kg

$da)*$da 3

Rafter3( )

-ata$3 %co 0.5⋅ A2 A2+( )⋅ ⋅:=

-ata$3 13.334=

kg

$da)*$da %

Rafter4( )

-ata$4 %co 0.5⋅ A2( )⋅ *⋅:=

-ata$4 6.66+=

kg

Beban Mati pada Balok

A

B

A A A

A1

1 2 3 4

C

D

A2

A2

Be-an #ija#ikan -e-an t a!e4oi# atau -e-an t a!esium #an5atau -e-an segitiga

aj$r 1 dan %


20/71

Asumsi te-al !lat lantai 18

7m

?$,at 0.10:=

(Bab 2 tabel 2.1 #al 11 4 12 at$ran pembebanan "

- eton 2400:=

kg5m3

- $,afon 40:=

kg5m2

- s$es" 21:=

kg5m3

- ! "n 24:=

kg5m2

a. Be-an Plat

-$,at1 - eton ?$,at⋅ A

2⋅:=

-$,at1 480=

kg5m

-. Be-an Pla=on ; /angka

-$r1 - $,afon A

2⋅:=

-$r1 80=

kg5m

7. Be-an ?!esi 2 7m

?s$es" 0.02:=

-s1 - s$es" ?s$es"⋅ A

2⋅:=

-s1 0.84=

kg5m

#. Be-an U-in 2 7m


21/71

?! "n 0.02:=

-!1 - ! "n ?! "n⋅ A

2⋅:=

-!1 0.96=

kg5m

?ota,M1 -$,at1 -$r1+ -s1+ -!1+:=

?ota,M1 561.8=

kg5m

aj$r 2 dan 3

a. Be-an Plat

-$,at2 - eton ?$,at⋅ A

2

A

2+ ⋅:=

-$,at2 960=

kg5m


-$r2 - $,afon A

2

A

2+ ⋅:=

-$r2 160=

kg5m

7. Be-an ?!esi 2 7m

-s2 - s$es" ?s$es"⋅ A

2

A

2+ ⋅:=

-s2 1.68=

kg5m

#. Be-an U-in 2 7m

-!2 - ! "n ?! "n⋅ A

2

A

2+ ⋅:=

-!2 1.92=

kg5m

?ota,M2 -$,at2 -$r2+ -s2+ -!2+:=


22/71

?ota,M2 1.124 103×=

kg5m

aj$r '

a. Be-an Plat

-$,at3 - eton ?$,at⋅ A1

4

⋅:=

-$,at3 240=

kg5m


-$r3 - $,afon A1

4

⋅:=

-$r3 40=

kg5m

7. Be-an ?!esi 2 7m

-s3 - s$es" ?s$es"⋅ A1

4 ⋅:=

-s3 0.42=

kg5m

#. Be-an U-in 2 7m

-!3 - ! "n ?! "n⋅ A1

4 ⋅:=

-!3 0.48=

kg5m

?ota,M3 -$,at3 -$r3+ -s3+ -!3+:=

?ota,M3 280.9=

kg5m

aj$r B

a. Be-an Plat

-$,at4 - eton ?$,at⋅ A2

2

A1

4+ ⋅:=

-$,at4 +20=


23/71

kg5m


-$r4 - $,afon A2

2

A1

4+ ⋅:=

-$r4 120=

kg5m

7. Be-an ?!esi 2 7m

-s4 - s$es" ?s$es"⋅ A2

2

A1

4+ ⋅:=

-s4 1.26=

kg5m

#. Be-an U-in 2 7m

-!4 - ! "n ?! "n⋅ A2

2

A1

4+ ⋅:=

-!4 1.44=

kg5m

?ota,M4 -$r3 -s3+ -!3+ -$,at4+:=

?ota,M4 +60.9=

kg5m

aj$r C

a. Be-an Plat

-$,at5 - eton ?$,at⋅ A2

2

A2

2+ ⋅:=

-$,at5 960=

kg5m


-$r5 - $,afon A2

2

A2

2+ ⋅:=

-$r5 160=

kg5m

7. Be-an ?!esi 2 7m


24/71

-s5 - s$es" ?s$es"⋅ A2

2

A2

2+ ⋅:=

-s5 1.68=

kg5m

#. Be-an U-in 2 7m

-!5 - ! "n ?! "n⋅ A2

2

A2

2+ ⋅:=

-!5 1.92=

kg5m

?ota,M5 -$r5 -s5+ -!5+ -$,at5+:=

?ota,M5 1.124 103×=

kg5m

aj$r D

a. Be-an Plat

-$,at6 - eton ?$,at⋅ A2

2 ⋅:=

-$,at6 480=

kg5m


-$r6 - $,afon A2

2 ⋅:=

-$r6 80=

kg5m

7. Be-an ?!esi 2 7m

-s6 - s$es" ?s$es"⋅ A2

2 ⋅:=

-s6 0.84=

kg5m

#. Be-an U-in 2 7m

-!6 - ! "n ?! "n⋅ A2

2 ⋅:=


25/71

-!6 0.96=

kg5m

?ota,M6 -$,at6 -$r6+ -s6+ -!6+:=

?ota,M6 561.8=

kg5m

-e)ean MerataA ")at -e)an@"n/"ng

Beban Dinding

a. Lantai 1

Be-an #in#ing #ija#ikan -e-an me ata #ise!anjang -alok lantai

Be-an #in#ing -ata

(Bab 2 tabel 2.1 #al 12 at$ran pembebanan "

- ata 250:=

H7 3:=

-/ - ata H⋅:=

kg5m2

-/ +50=

kg5m

+.1.2 Beban *idup

Beban 4rang Terpusat &P)


26/71

P 250:=

kg5m2

(Bab 3 pasal 3.2 a&at 1 #al 13 at$ran pembebanan "

Be-an #iletakkan !a#a titik go #ing !a#a a=te

Angin *ekan Angin Hisa!

Pada ra ter

-ot P:=

-ot 250=

g

Beban *idup pada Balok

Be-an $i#u! -e u!a -e-an segitiga & Be-an %ati Pa#a Balok

A

B

A A A

A1

1 2 3 4

C

D

A2

A2

- "/!$ - ata:=

- "/!$ 250=


27/71

kg5m2

aj$r 1 dan %

- 1 - "/!$ A

2⋅:=

- 1 500=

kg5m

aj$r 2 dan 3

- 2 - "/!$ A

2

A

2+ ⋅:=

- 2 1 103×=

kg5m

aj$r '

- 3 - "/!$ A1

4 ⋅:=

- 3 250=

kg5m

aj$r B

- 4 - "/!$ A2

2

A1

4+ ⋅:=

- 4 +50=

kg5m

aj$r C

- 5 - "/!$ A2

2

A2

2+ ⋅:=

- 5 1 103×=

kg5m

aj$r D

- 6 - "/!$ A2

2 ⋅:=

- 6 500=

kg5m


28/71

+.1.+ Beban Angin

Pa#a a=te yang #ite!i -e-an angin & C52

Pa#a a=te yang #itenga$ -e-an angin & C

Angin Tekan

Kemi ingan a=te a & 38

Pa 25=

kg5m2

c1 0.2:=

%a1 5=

kg5m2

*ekanan Angin P

%a1 c1 Pa⋅:=

Bi#ang Ke ja D & +a ak anta ku#a'ku#a < ja ak go #ing

go #ing

a=te

-A2

+2 A2

2A A11

!a"ter 1

@1 0.5 A1⋅ ⋅:=

@1 1.155=

m2

!a"ter 2

@2 0.5 A1 A2+( )⋅ ⋅:=

@2 2.309=

m2


29/71

!a"ter 3

@3 0.5 A2 A2+( )⋅ ⋅:=

@3 2.309=

m2

!a"ter #

@4 0.5 A2( )⋅ ⋅:=

@4 1.155=

m2

Beban Angin Tekan

!a"ter 1

#1 0.866 %a1⋅ @1⋅:=

#1 5=

g

!a"ter 2

#2 0.866 %a1⋅ @2⋅:=

#2 10=

g

!a"ter 3

#3 0.866 %a1⋅ @3⋅:=

#3 10=

g

!a"ter #

#4 0.866 %a1⋅ @4⋅:=

#4 5=

g

Angin *isap

*ekanan Angin P

Pa 25=

kg5m2

c2 0.4:=


30/71

%a2 c2 Pa⋅:=

%a2 10=

kg5m2

Ang"n ?e an Ang"n H"sa$

Be-an Angin Hisa!

!a"ter 1

#5 0.866 %a2⋅ @1⋅:=

#5 10=

g

!a"ter 2

#6 0.866 %a2⋅ @2⋅:=

#6 20=

g

!a"ter 3

#+ 0.866 %a2⋅ @3⋅:=

#+ 20=

g

!a"ter #

#8 0.866 %a2⋅ @4⋅:=

#8 10=

g

+.2 Analisa Struktur

Analisa st uktu #engan ?AP 2888 3D

Kom-inasi !em-e-anan yang #i!akai ?K?NI'2882


31/71

1 1.( D

2 1.2 D ; 1.@ L

3 1.2 D ; 1.@ L ; 8.>

( 1.2 D ; 1.@ L ' 8.>

S8M> B>BA'

B>BA' MAT,

Beban mati pada ra ter

-gor/"ng1 15.02=

kg

-ata$1 6.66+=

kg

-gor/"ng2 30.04=

kg

-ata$2 13.334=

kg

-gor/"ng3 30.04=

kg

-ata$3 13.334=

kg

-gor/"ng4 15.02=

kg

-ata$4 6.66+=

kg

beban mati atap pada pin ir ra ter

Rafter1 -ata$1 0.5⋅ 3.3335289003621545+96+→:=

Rafter2 -ata$2 0.5⋅ 6.66+05+800+2430915935→:=

Rafter3 -ata$3 0.5⋅ 6.66+05+800+2430915935→:=

Rafter4 -ata$4 0.5⋅ 3.3335289003621545+96+→:=

6otal beban mati pada ten a# ra ter

-?1 -gor/"ng1 -ata$1+ 21.68+05+800+243091593→:=


32/71

-?2 -gor/"ng2 -ata$2+ 43.3+4115601448618318+→:=

-?3 -gor/"ng3 -ata$3+ 43.3+4115601448618318+→:=

-?4 -gor/"ng4 -ata$4+ 21.68+05+800+243091593→:=

6otal beban mati pada pin ir ra ter

-P1 -gor/"ng1 Rafter1+ 18.3535289003621545+9+→:=

-P2 -gor/"ng2 Rafter2+ 36.+0+05+800+243091593→:=

-P3 -gor/"ng3 Rafter3+ 36.+0+05+800+243091593→:=

-P4 -gor/"ng4 Rafter4+ 18.3535289003621545+9+→:=

Beban mati pada balo*

Laju 1 #an (

?ota,M1 561.8=

kg5m

Laju 2 #an 3

#1 s "n π

3

⋅ 4.33=

#6 s "n π

3 ⋅ 1+.321=

?ota,M2 1.124 103×=

kg5m

Laju A #an D

#1 cos π

3

⋅ 2.5=

?ota,M3 280.9=

kg5m

#6 cos π

3 ⋅ 10=

Laju B #an 0

?ota,M4 +60.9=

kg5m

Beban Dindin

-/ +50=


33/71

kg5m

B>BA' *,D8P

#2 s "n π

3 ⋅ 8.66=

#+ s "n π

3 ⋅ 1+.321=

Beban #id$p pada ra ter

-ot 250=

g

#2 cos π

3

⋅ 5=

#+ cos π

3

⋅ 10=

Beban #id$p pada balo*

Laju 1 #an (

- 1 500=

kg5m

Laju 2 #an 3

Laju A #an D

- 2 1 103×=

kg5m

- 3 250=

kg5m

Laju B #an 0

- 4 +50=

kg5m

B>BA' A'5,'

'n in te*an pada ra ter

/a=te 1

#1 5=

kg


34/71

/a=te 2

#2 10=

kg

/a=te 3

#3 10=

kg

/a=te (

#4 5=

kg

'n in #isap pada ra ter

/a=te 1

#5 10=

kg

/a=te 2

#6 20=

kg

/a=te 3

6atatan " 8ntuk ra=ter tepi beban angin ?%2

8ntuk ra=ter tengah beban angin ?

#+ 20=

kg

/a=te (

#8 10=

kg

AP,T87AS, 5A@A '4


35/71

< 1 &*,SAP T> A')

Tarik Tekan) 2113. ( '

> 2113. ( '

1)23.8 '18 1@>2.3@ '

11 2832.) '

12 2832.) '

13 18@ .)( '

19 32).2> '

1) 8 '

1 33>.@2 '

21 8 '

1( ' '@@3.22

1@ ' '( .>

1> ' '@>@.13

28 ' '91).13

22 ' '233 .8(

23 ' '1 29.82

2( ' '1913.8@

29 ' '1(>>.>@

2@ ' '1 2(.9(

31 ' '[email protected]

ATAS

Posisi Batang5a(a Batang

*4 A' *,SAP)

Tarik Tekan) 2893. ) '

> 2893. ) ' 1@)@.>( '

18 1@[email protected] '

11 282).91 '

12 282).91 '

13 18@ .)9 '

19 33 .1) '

1) 8 '

1 32@.)( '

21 9.@>(E'19 '

1( ' '@>).3(

1@ ' '91). )

1> ' '@@2.8)

28 ' '( >.

22 ' '239@.>

23 ' '1 2(. )

2( ' '1(> .31

29 ' '1912.@(

2@ ' '1 2(.99

31 ' '233>.31

ATAS


*4


36/71

kg

kg

kg

< 2&T> A' *,SAP)

Tarik Tekan) 2(91.@2 '> 2(91.@2 '

1 .3 '18 1 2).>@ '11 2(8(.)) '12 2(8(.)) '13 12)1.>9 '19 (13.29 '1) 8 '

1 3 1.>1 '21 2.9()E'12 '1( ' '>3).(91@ ' '@31.81> ' ') 3. 128 ' '9 >.3922 ' '2>8(.2923 ' '22>>.@12( ' '1)@@.2229 ' '1>8).22@ ' '22>>.1331 ' '2))2.2>

ATAS


*4


37/71

< 2&*,SAP T> A')

Tarik Tekan) [email protected]) '

> [email protected]) '

28>8.39 '18 288>.>( '

11 2(13.>3 '

12 2(13.>3 '

13 12)1.>3 '

19 3 2.(3 '

1) 8 '

1 (12.9 '

21 2.9()E'12 '

1( ' ') 9.29

1@ ' '9 .3

1> ' '>[email protected]

28 ' 8

22 ' '1.13)E'1(

23 ' 8

2( ' 8

29 ' '1.)89E'1(

2@ ' 8

31 ' '3.(11E'1(

ATAS


*4 A' *,SAP)

Tarik Tekan) 231 .8@ '

> 231 .8@ '

1> 1.>@ '

18 1>38.9 '

11 22) .81 '

12 22) .81 '

13 128(.(> '

19 3>>.99 '

1) 8 '

1 3)8.12 '

21 8 '

1( ' ')>).(1

1@ ' '9 3.3>

1> ' ')( . @28 ' '[email protected]

22 ' '[email protected]

23 ' '21@).3

2( ' '1@)3. 1

29 ' '1)8 .8@

2@ ' '21@@. 3

31 ' '2@2).@1

Posisi 5a(a Batang

*4


38/71

< +&*,SAP T> A')

Tarik Tekan) 2(8 .81 '

> 2(8 .81 '

1 @1.2@ '

18 1 88.81 '11 22>@.)> '

12 22>@.)> '

13 128(.(@ '

19 3)8.)1 '

1) 8 '

1 3>). 3 '

21 8 '

1( ' ')91.2(

1@ ' '9@@.13

1> ' ')>@.8@

28 ' '9 2.((

22 ' '2@2>.((

23 ' '21@).(@

2( ' '1)8 .93

29 ' '1@)3.(

2@ ' '21@@. 1

31 ' '2@9(.2>

ATAS


*4).(( [email protected]( 19@9.( *ekan'Hisa! 193.83 1>3 . 2 1 1).((Hisa!'*ekan 193.82 1 1).>3 1>3 .9(

*ekan'Hisa! 131.1) 1)(>.91 1>1(. 3Hisa!'*ekan 131.1@ 1>19.2 1)(>.1@

*ekan'Hisa! 193.83 1>3 . 2 1 1).((Hisa!'*ekan 193.82 1 1).>3 1>3 .9(

*ekan'Hisa! >).(@ 19@9.)2 1@8 . 1Hisa!'*ekan >).(( [email protected]( 19@9.(

/A"*E/ 1

/A"*E/ 2

/A"*E/ 3

/A"*E/ (

/A"*E/ 9

'S9;SI B87'6 S8NDI7I D'7I 9D') 9D'Berat sendiri ra=ter 1

-P1 2⋅( ) 14 -?1⋅( )+ 16 -ot⋅( )+ #1 8⋅( )+ #6 8⋅( )+ B 0.1⋅ 454.033=

kg

Berat sendiri ra=ter 2


39/71

-P2 2⋅( ) 14 -?2⋅( )+ 16 -ot⋅( )+ #2 8⋅( )+ #+ 8⋅( )+ B 0.1⋅ 492.065=

kg

Berat sendiri ra=ter +

-P3 2⋅( ) 14 -?3⋅( )+ 16 -ot⋅( )+ #3 8⋅( )+ #8 8⋅( )+ B 0.1⋅ 484.065=

kg

Berat sendiri ra=ter -

-P4 2⋅( ) 14 -?4⋅( )+ 16 -ot⋅( )+ #4 8⋅( )+ #9 8⋅( )+ B 0.1⋅ =#9

kg

BAB ,C

P>'6A'AA' P>'AMPA'5

-.1 Perencanaan an"

(SNI 2002 #al 11 , tabel 5.3"

P o!e ties Penam!ang

+enis : B+ 3(

14.8:=

7m

ry 2.3+:=

7m

fy 2100:=

kg57m2

f 10:=

7m

138:=

7m3

fr 0.3 fy⋅:=

= & tegangan sisa

tf 0.9:=

7m


40/71

A 26.84:=

7m2

fr 630=

kg57m2

t 0.6:=

7m

f! 3400:=

kg57m2

f 2tf −:=

1020:=

7m(

(SNI 2002 Hal Point 5.1 ? 3"

8.2=

c

y 151:=

7m(

& 2000000:=

kg57m2

r 6.1+:=

7m

( 800000:=

kg57m2

-.1.1 Perencanaan Batang Tekan

, 400:=

7m

7 450:=

7m


Bagian ' -agian !enam!ang $a us mem!unyai nilai kelangsingan yang le-i$ ke7il#a i


41/71

(SNI 2002 #al 30 ) 31 , tabel /.5.1"

C#e@* *elan sin an pofl

lens

λf f

tf :=

λf 11.111=

fy 250:=

M$a

λrf 250fy

:=

λrf 15.811=

λf λrf <

11.111 15.811<

...................CD

?eb

λ t

:=

λ 13.66+=

λr 665

fy:=

λr 42.058=

λ λr<

13.66+ 42.058<

.................CD

C#e@* *elan sin an elemen str$*t$r

λ 7 ry

:=

λ 189.8+3=

(SNI 2002 #al 2 , b$tir /.!.%"

λ 200<

..........CD


42/71


43/71

0ek kon#isi -atas te $a#a! :

'! 1218.08:=

g

(Diperole# Nilai Dari 'nalisa Beban Den an S'P"

Lele$

φ 'n 0.9A fy⋅:=

(SNI 2002 #al /0, b$tir 10.1"

φ 'n 5.0+3 10 4×=

2. " aktu

7 3.465:=

c: 1:=

(SNI 2002 #al /0, b$tir 10.2"

#i en7anakan sam-ungan anta ku#a ' ku#a #engan menggunakan -aut

/"ameter

/ 2.2:=

c

jumla$ -autn 4:=

te-al !enam!ang

t 0.8:=

c

An A n /⋅ t⋅−:=

An 19.8=

(SNI 2002 #al /1, b$tir 10.2.1"

Ae An :⋅:=

Ae 19.8=

φ 'n 0.+5 Ae⋅ f!⋅:=

φ 'n 5.049 104×=


44/71

φ 'n '!≥

(SNI 2002 #al /0, b$tir 10.1.1)1"

9.185 104× 2556.5+≥

.................CD

3. Ke untu$an -lok ujung

a. kon#isi gese mu ni

Bidang geser (v)

Bidang geser (v)

Agv

Anv

s s s

#

s1

t

e 1:=

/).t33s(s1AA

s).tss(s1A

21

nns

g

−+==+++=

s 4:=

c

s1 2:=

c

Ans s1 3s+ 3 /2⋅− t⋅:=

Ans 8.56=

φ 'n 0.+5 0.6 f!⋅( )⋅ Ans⋅:=

φ 'n 1.31 10 4×=

φ 'n '!≥

(SNI 2002 #al /0, b$tir 10.1.1)1"

1.425 104× 2556.5+≥

.................CD

-. Kom-inasi gese ' ta ik

Ant 150 e− 0.5 /⋅−( )t:=


45/71

Ant 118.32=

7m2

An s1 3 s⋅+( )t B 4 /⋅ t⋅( )−:=

Ag s1 3 s⋅+( )t B:=

An 4.16=

7m2

Ag 11.2=

7m2

f! Ant⋅ 4.023 105×=

0.6 f!⋅ An⋅ 8.486 103×=

f! Ant⋅ 0.6 f!⋅ An⋅≥ ma a,φ 'n 0.+5 0.6 fy⋅ Ag⋅ f! Ant⋅+( )⋅:=

φ 'n 3.123 10 5×=

φ 'n '!≥

(SNI 2002 #al /0, b$tir 10.1.1)1"

3.409 105× 2556.5+≥

.................CD A

-.2 Perencanaan Balok

P o l IC" 1)9. 8.@.

(6abel Baja Profl I


46/71

7m3

+enis : B+ 3(

tf 0.8:=

7m

A 23.04:=

7m2

fy 2100:=

kg57m2

t 0.5:=

7m

fr 0.3 fy⋅:=

= & tegangan sisa

(SNI 2002 Hal Point 5.1 ? 3"

1210:=

7m(

fr 630=

kg57m2

& 2000000:=

kg57m2

y 9+.5:=

7m(

800000:=

kg57m2

r +.26:=

7m

-.2.1 6ek Terhadap 7entur


1.136:=




47/71

2,3

3.M4.M3.MM2,5M12,5

CCBAmax

maxb ≤+++

=


Dimana :

%A & %omen #i 15( -entang & 3532.6.l2


%0 & %omen #i 35( -entang & 3532.6.l2


7 400:=

c


(SNI 2002 #al 3 ,tabel .3)2"

7$ 1.+6 ry⋅ &

fy⋅:=

7$ 111.888=

c


3ii .hb3

1J ∑=


?e$ingga umus + menja#i ::

1

3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=

1.558=

7m(

1 π & ⋅ ⋅ A⋅

2⋅:=

1 1.211 105×=

c

tf 3⋅

H2

12⋅

3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=


48/71


49/71

kg 7m


M! 4383.98:=

kg 7m

6ek 7entur "

M! φMn≤

(SNI 2002 #al 3%, b$tir .1.1"

4383.98 1.4+5 109×≤

.......4 ;;;

-.2.2 Perencanaan Batang Tekan

7 450:=c


λc7

fy

&⋅

π ry⋅:=

λc 2.253=

(SNI 2002 #al 5!, b$tir .2"

ω 1.25 λc( ) 2⋅:=

ω 6.346=

(SNI 2002 #al 5 , b$tir .3)!"

;cr fy

ω:=

;cr 330.925=

kg57m2

φ 'n 0.85 A⋅ ;cr ⋅:=

φ 'n 6.481 103×=

g


'! 568.44:=


50/71

g

φ 'n '!≥

(SNI 2002 #al ! , b$tir . .2"

8.202 103

× 568.44≥......4 ;;;

ombinasi Momen 7entur Dan 'ormal Tekan

'!

φ 'n0.088=

(SNI 2002 #al 0, b$tir 12.5"

:nt!

'!

φ 'n 0.2≥ma a

'!

φ 'n8

9

M!

φMnM!y

φMny+ ⋅+ 0.+6+=

(SNI 2002 #al 0, b$tir 12.5)2"

0.+6 1≤

......4 ;;;-.2.+ Perencanaan Batang Tarik & 'u Tidak ada batang tarik )

(SNI 2002 #al /0, b$tir 10.1"

'n A fy⋅:=

'n 4.838 104×=

g

"akto /e#uksi

7e,e

φ 'n1 0.9 'n⋅:=

φ 'n1 4.355 104×=

g

(SNI 2002 #al /0, b$tir 10.1.1)2a"


51/71

;ract!re

φ 'n2 0.+5 'n⋅:=

φ 'n2 3.629 10 4×=

g

(SNI 2002 #al /0, b$tir 10.1.1)2b"

Diambil nilai φ 'n (ang terbesar! maka "

φ 'n φ 'n1:=

φ 'n 4.355 104×=

g


'! 0:=

g

'! φ 'n≤

(SNI 2002 #al /0, b$tir 10.1.1)1"

0 4.355 104×≤

....4 ;;;

-.2.- Pemeriksaan Terhadap 5eser

2 tf −:=

2 16.+=

c

a 400:=

c

A 2 t−:=

A 16.2=

7m2

2

t33.4=

a

223.952=

Terdiri dari + s(arat! di antaran(a (aitu "


52/71

Dn 5 5

a

2 2

+:=

(SNI 2002 #al %5, b$tir . )2a"

Dn 5.009=

1.1 Dn &⋅

fy⋅ +5.9+3=

ma a

(SNI 2002 #al %5, b$tir . )2a"

2

t1.1

Dn &⋅

;y⋅≤

33.4 +5.9+3≤

....4 ;;;

En 0.6 fy⋅ A⋅:=

En 2.041 104×=

g

(SNI 2002 #al %!, b$tir . )3a"

oefsien . aktor


53/71

E! 6.51 103×=

φMn 1.4+5 10 9×=

φEn 1.83+ 10 4×=

(SNI 2002 #al % , b$tir . )2"

M!

φMn0.625

E!

φEn⋅+ 0.221=

0.221 1≤

....4 ;;;

-.+ Perencanaan olom

P o l IC" 288.198.@.

(6abel Baja Profl I


54/71

7m

fr 0.3 fy⋅:=

= & tegangan sisa

(SNI 2002 Hal Point 5.1 ? 3"

2690:=

7m(

fr 630=

kg57m2

& 2000000:=

kg57m2

y 50+:=

7m(

( 800000:=

kg57m2

r 8.30:=

7m

-.+.1 6ek Terhadap 7entur


) 1.136:=



2,33.M4.M3.MM2,5

M12,5C

CBAmax

maxb ≤+++

=


Dimana :

%A & %omen #i 15( -entang & 3532.6.l2


%0 & %omen #i 35( -entang & 3532.6.l2


7) 400:=


55/71

c


(SNI 2002 #al 3 ,tabel .3)2"

7$ 1.+6 ry⋅ &

fy⋅:=

7$ 196.0+6=

c


3ii .hb3

1J ∑=


?e$ingga umus + menja#i :: 1

3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=

3.55+=

7m(

1 π & ⋅ ⋅ A⋅

2⋅:=

1 1.195 105×=

c

tf 3⋅

H2

12⋅

3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=

3.+02 104×=

7m@

2 4

y ⋅

⋅

2⋅:=

2 2.+68 10 6−×=

c

7r ry 1

fy fr − ⋅ 1 1 2 fy fr −( ) 2⋅++⋅:=

7r 560.019=


56/71

c

7 400:=

c

7$ 7< 7r <

(SNI 2002 #al 3 , b$tir .3.5"

196.0+6 400< 560.019<

6ermas$* Bentan ;enen a#

...... 4

(SNI 2002 #al 3/, tabel .3)1"

Mcr & y⋅ ⋅ ⋅ π &⋅

7 2 y⋅ ⋅+⋅:=

Mcr 8.925 109×=

kg 7m

uat 7entur


57/71

-.+.2 Perencanaan Batang Tekan

7 450:=

c


λc7

fy

&⋅

π ry⋅:=

λc 1.286=

(SNI 2002 #al 5!, b$tir .2"

ω 1.25 λc( ) 2⋅:=

ω 2.066=

(SNI 2002 #al 5 , b$tir .3)!"

;cr fy

ω:=

;cr 1.016 103×=

kg57m2

φ 'n 0.85 A⋅ ;cr ⋅:=

φ 'n 3.3+ 10 4×=

g


'! 12269.33:=

g

φ 'n '!≥

(SNI 2002 #al ! , b$tir . .2"

4.+35 104× 12269.33≥

......4 ;;;

ombinasi Momen 7entur Dan 'ormal Tekan

'!

φ 'n0.364=

(SNI 2002 #al 0, b$tir 12.5"


58/71

:nt!

'!

φ 'n0.2≥

ma a

'!

φ 'n8

9

M!

φMnM!y

φMny+ ⋅+ 1.043=

(SNI 2002 #al 0, b$tir 12.5)2"

0.849 1≤

......4 ;;;

-.+.+ Perencanaan Batang Tarik

(SNI 2002 #al /0, b$tir 10.1" 'n A fy⋅:=

'n 8.192 104×=

g

"akto /e#uksi

7e,e

φ 'n1 0.9 'n⋅:=

φ 'n1 +.3+3 10 4×=

g

(SNI 2002 #al /0, b$tir 10.1.1)2a"

;ract!re

φ 'n2 0.+5 'n⋅:=

φ 'n2 6.144 104×=

g

(SNI 2002 #al /0, b$tir 10.1.1)2b"

Diambil nilai φ 'n (ang terbesar! maka "

φ 'n φ 'n2:=

φ 'n 6.144 10 4×=


59/71

g


'! 286.+2:=

g

'! φ 'n≤

(SNI 2002 #al /0, b$tir 10.1.1)1"

286.+2 6.144 104×≤

....4 ;;;

-.+.- Pemeriksaan Terhadap 5eser

2 tf −:=

2 18.5=c

a 400:=

c

A 2 t−:=

A 1+.9=

7m2

2

t 30.833=

a

221.622=

Terdiri dari + s(arat! di antaran(a (aitu "

Dn 5 5

a

2 2

+:=

(SNI 2002 #al %5, b$tir . )2a"

Dn 5.011=

1.1 Dn &⋅

fy⋅ +5.988=

ma a

(SNI 2002 #al %5, b$tir . )2a"


60/71

2

t1.1

Dn &⋅

;y⋅≤

20 69.613≤

....4 ;;;

En 0.6 fy⋅ A⋅:=

En 2.255 104×=

g

(SNI 2002 #al %!, b$tir . )3a"

oefsien . aktor


61/71

BAB C

P>'6A'AA' SAMB8'5A'

?am-ungan Balok #an Kolom

#;

-a,o

Do,om

%u &dari nilai momen balok (ang di chek)

M! 4383.98:=

kg5mm2

Fu &dari nilai geser balok (ang di check)

E! 62+8.03:=

g

P o l Balok yang #i!akai

IC" 1)9G 8

) 1+5:=

m

90:=

m

Baut yang #igunakan


62/71

-a,o ( #;)

Pe,at

- 34−

f! 340:=

MPa

fy 210:=

MPa

(SNI 2002 #al 11, tabel 5.3"

Diamete Baut =-

φ 58 2.54⋅ 10⋅:=

φ 15.8+5=

m

Luas Baut A-

A 1

4 π⋅ φ 2⋅:=

A 19+.933=

mm2

+umla$ -aut n

n 4:=

6heck 5eser

(SNI 2002 #al 100, b$tir 13.2)%"

+umla$ -i#ang gese m

m 1:=


63/71


64/71

Asumsi : Aki-at momen semua -aut mengalami ta ik

/1/2/3/4/

/1/2/3/4/

R$

a

Rn

-a!t

/! & /n

/! & a < - < =y

Rn n ft⋅ 0.+5⋅ A⋅:=

Rn 3.68+ 105×=

a Rn

fy⋅:=

a 10.034=F " a aF /1<

Asumsi semua -aut mengalami ta ik Bena

apasitas Momen

s 4 φ ⋅( )−

5:=


65/71

s 5.3=

/1 s 1

2 φ ⋅+:=

/1 13.238=

/2 2 s⋅ 3

2 φ ⋅+:=

/2 34.413=

/3 3 s⋅ 5

2 φ ⋅+:=

/3 55.58+=

/4 4 s⋅ +

2 φ ⋅+:=

/4 +6.+63=

/ :=

/ 90=

M/ φf 2⋅ ft⋅ 0.+5⋅ A⋅ /1 /2+ /3+ /4+( )⋅ B φy a⋅ ⋅ fy⋅ / a2

− ⋅+:=

M/ 5.309 10+×=

6heck

M/ M!>

5.309 10+× 4563.31>

........4

BAB C,,

>S,MP87A'

DALA% PE/EN0ANAAN BA+A DI,UNAKAN P/ "IL

1. , /DIN,

P o l 0 Lig$t 0$annel : 198 G @9 G 28 G 3.2

H 15:=

c

44.3:=

7m3


66/71

) 6.5:=

c

y 12.2:=

7m3

/ 2:=

c

332:=

7m(

t 0.32:=

c

y 53.8:=

7m(

tf 0.32:=

c

r 5.89:=

c

A 9.56+:=

7m2

ry 2.3+:=

c

%ate ial Baja

& 2000000:=

kg57m2

fy 2100:=

kg57m2

fr 0.3 fy⋅:=

( 800000:=

kg57m2

fr 630=

kg57m2


67/71

2. /A"*E/

P o l IC" 198.188.>.12


14.8:=

7m

ry 2.3+:=

7m

10:=

7m

138:=

7m3

tf 0.9:=

7m

A 26.84:=

7m2

t 0.6:=

7m

& 2000000:=

kg57m2

1020:=

7m(

800000:=

kg57m2

y 151:=

7m(

fy 2100:=

kg57m2

r 6.1+:=

7m

fr 630:=


68/71

kg57m2

3.-A7CD

P o l IC" 1)9. 8.11.1)


1+.5:=

7m

ry 2.06:=

7m

) 9:=

7m

139:=

7m3

tf 0.8:=

7m

A 23.04:=

7m2

t 0.5:=

7m

& 2000000:=

kg57m2

1215:=

7m(

( 800000:=

kg57m2

y 9+.5:=

7m(

fy 2100:=

kg57m2

r +.26:=

7m


69/71

fr 630:=

kg57m2

(. K L %

P o l IC" P o l IC" 288.198.11.1)


19.4:=

7m

ry 3.60:=

7m

15:=

7m

2++:=

7m3

tf 0.9:=

7m

A 39.01:=

7m2

t 0.6:=

7m

& 2000000:=

kg57m2

2690:=

7m(

800000:=

kg57m2

y 509:=

7m(

fy 2100:=

kg57m2

r 8.30:=


70/71

7m

fr 630:=

kg57m2

cos 2 π6

:=

y s"n 2 π6

:=

B>BA' A'5,'

0.5=

y 0.866=

'n in te*an pada ra ter

/a=te 1 #an (/a=te 1 #an (

#1 ⋅ 2.5=

kg

#1 y⋅ 4.33=

kg

/a=te 2 #an 3

/a=te 2 #an 3#2 ⋅ 5=

kg

#2 y⋅ 8.66=

kg

'n in #isap pada ra ter

/a=te 1 #an (

/a=te 1 #an (#6 ⋅ 10=

kg

#6 y⋅ 1+.321=

/a=te 2 #an 3


71/71

/a=te 2 #an 3

#+ ⋅ 10=

kg

#+ y⋅ 1+.321=

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