kontrol kekuatan balok dan kolom - lrfd1

7/17/2019 Kontrol Kekuatan Balok Dan Kolom - Lrfd1

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Per tungan Ke utu an L t Da am 1 Zone

a. Data gedung

Luas lantai (a) = 28 m x 20 m =

Lantai 1 = Lobb ! Kantor

Lantai 2 " # = KantorLantai $ = %itness &enter

'otal umlah lantai (n) = $ lt

'inggi lantai e lantai (h) = * m

Luas lantai bersih +er orang (a,) = * m2-orang

Ke. /ata2 untu bangunan $ lantai (s) = *0 " 1$0 m-menit

= 2.$ m-deti

Ka+asitas lift (m) = a . n . . P

00 . a,

= $*0 x $ x00 x

= #200

$0

= $.* =

Keterangan

= 3atu menunggu minimum

P = Persentase em+iris terhada+ +enghuni bangunana ang harus terangat

dalam $ menit +ertama +ada am sibu (4)

P untu bangunan umum = $ 5 14

1 6eban +una lift

L = P (2a 5 m7) n

2a,

= $4 ( 2 x $*0 5 x * x 7 ) $

2 x *

= $4 x 1120 x $ 5 $4 x x * x 7 x $

12

= 2. 5 0 7

2 Daa angut lift dalam atu $ menit ()

= $ x *0 deti x m = 00 m7

' '

Dimana 9 = Daa angut ereta dalam $ menit

' = 3atu ang di+erluan oleh ereta dari dasar sam+ai e +una d



/ound 'ri+ 'ime (')

/umusna 9

' = (2 h : # s) (n 5 1) : s ( m : #)

s

' = ( 2 x * : # x 2.$ ) ( $ 5 1 ) : 2.$ x ( x

2.$

' = $.2 deti

Daa angut 7 Lift dalam $ menit

7 = 00 .m.7

'

7 = 00 x * x 7

$.2

= 1.#*; 7

Persamaan 9 L = 7

L = P (2a 5 m7) n = 00m72a, '

<ehingga 9 2. 5 0 7 = 1.#*; 7

aa 7 = 0.2 = 1 Lift * orang



$*0 m2

0 x $42.$

* orang

n embali edasar



* : #)



6>6 ?@. PA/A7&>7>>7 6>LBK KBPB<?'A

1. LA6>/ A%AK'?% 6>LBK >7>K KBPB<?' BA7 >K<?C PB<?'?%

bE bE

AK'A/?B/ ?7'A/?B/

b0 b0

6alo esterior bE L-8 : ara +usat balo e+usat +elat

bE 1-2 b0 : ara +usat balo e+usat +elat

6alo interior 9 bE L-#

bE b0

Kuat lentur nominal yang dihitung berdasarkan distribusi tegangan plastisdibagi menjadi 2 kasus:

bA 0.8$fE

0.8$fE

tsa

d-2 d1

titi berat

T

(>) (b) (

1. Sumbu netral Plastis Jatuh pada plat betonBerdasarka gambar diatas, maka gaya tekan :

C=0.85 f'c . A . bE

!rai keseimbangan gaya "# maka diperoleh:

a =As x fy

0.85 f'c bEmaka kual lentur nominal $ gambar b%

&n "atau&n " "

2.Sumbu netral plastis jatuh pada propil baja

gaya tekan ' , yang bekerja pada beton sebesar: $ lihat gambar '%

Cc = sehingga keseimbangan gaya :T' = Cc + Cs

s "

(s . )y * '

2

s " (s . )y * +,- /' . bE .ts

2

bf

sumber buku LRFD ( Agus setiawan) berdasarkan SNI 03-1729-2002

0aya tarik pada propil baja (T) =As . fy

F d1

#F d1 (sFy$d2ts * a2%Jika a > ts, maka asumsi dapat dirubah, hasil ini plat beton tidak 'ukup kuatuntuk menimbangi gaya tarik pada propil baja

(pabila kedalam blok tegangan beton , a ternyata melebihi tebal plat beton,

0.85•ƒ'c •bE • ts

Besarnya T lebih ke'il dari As•f'c

beton

baa



Kuat lentur nominal $&n%

&n " ' . d2 3 s . d2

" 2- Mutu baja !" #$

4#engangan putus baja 5u$minimum% " 61++4#engangan 7eleh baja 5y$minimum% " 2-+

4 8egangan $ minimum% " 1 4&odulus Elastisitas baja$Es% " 2++++

4&odulus 0eser $0% " ++++

48asio Paison " +.9

#ipe Balok Komposite $tanpa !e'k%

#abel Propil baja yang digunakan

)rame otasi!imensi

(&omen ;nersia

h b5 ro ;3

mm mm mm mm mm

Balok <)9++=9++ 9++ 9++ 1+ 1- 1 11>.+ 2+,6++

Balok <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++Kolom <)6++=6++ 6++ 6++ 19 21 22 21.@+ ??,?++

Balok ;nteriorBalok ;nterior Propil yang digunakan <)9++ = 9++ = 1+ = 1-

A

"7

"?++

" 1-+ 'm6 6

b+ " 9-+ 'm

&odulus Elastrisitas beton " 6@++ % 2- &PaE' " 29-++ &Pa

&odulus 8asio

n "Es

"2+++++ &Pa

" .-1+?9 G E' 29-++ &Pa

7ebar EiCalen Baja"

bE"

1-+ 'm" 1.@-

n

bE" 1.@ 1-+ 'mbEn

12 m

hs" 9++

&00 00 $0 $5

#.*. ata -r-caaa ( SNI 03-1729-2002 ) Mutu b-t/ ƒ'c

tw tw

'mH 'mH

a. --tua -bar -f-ktif (bE) balok interior

b

b



7etak garis etral (' " 1.@- 3 12 'm " 22-.++ m2

(s " <)9++ = 9++ = 1+ = 1- " 11>.+ m2

(s total " 966.+ m2

b)1tatis M/2- k-sisi atas 3at b-t/ (ya)

(s tot 3 ya " (' 3 h' (s 3 D2

966. 'm 3 ya " 22-.++ 312 'm

11>.+ 3 D2

96- 'm 3 ya " 19-+.+ 'm9 9296.? 'm9 "

ya "6-6.? 'm9

" 1.0 m966.+ 'm

C)1tatis M/2- k-sisi !a4a -s ba4a 3r/3i !aja (yb)

(s tot 3 yb " (' 3hs h'

2

966. 'm 3 yb " 22-.++ 39+ 12

2

96- 'm 3 yb " 1++ 1@>@.+ 'm9 "

yb ">>@.+ 'm9

" 2.@ 'm966.+ 'm

67T97:

ya yb " hs h'19.9 'm 2.@ 'm " 9+ 12 'm

62.+ 'm " 62 'm 76

ya= 13.30 cm

d1= .0 m

d2= 1. m yb= 28.7 cm

grs netral komposite

grs netral propil Wf

d% M/i2- i-rsia k/23/sit- (;) ,t-ra<a3 garis -tra

(' " 22-.++

"1

3 1.@- 'm 3 12 'm " 2@+12

d1 " ya *h'2

d1 " 19.9 'm *12 'm

" @.9+ 'm2

;' " ;o ( 3

2@++.++ 'm 22-.+ 'm 3 @.9 'm "

r/3i &00 00 $0 $5

(s " 11>.+ 'm;o " 2+6++ 'mF6

d2 " yb * hs2

" 2 @ 'm9+ 'm

" 19 @+ 'm

'mI

;o

d1I



;s " ;o (s 3

2+6++ 'm 11>. 'm 3 19.@ 'm "

(s " 7uas propil <);o " &omen inersia propil <)d2 " 7etak pusat berat propil <) terhadap garis netral komposite

;s " &omen inersia propil <) terhadap garis netral kompositef) M/2- ;-rsia -a23ag 6/23/sit-; " ;' ;s " [email protected] 'mF6 22+61.@ 'm; " 9?@2+.19 'mF6

&odulus Penampang elastis $S%*

S' ";

"9?@2+.19 'mF6

"ya 19.9+ 'm

*

Ssa ";

"9?@2+.19 'mF6

"ya *h' 1.2>? 'm

*

Ssb ";

"9?@2+.19 'mF6

"yb 2.@ 'm

M/2- asi ari 1A *000 ? $#CE6 97;: &00 00 $0 $5

ra2-M/2-(M) @*

M tu23u kiri M a3 M. tu23u kaa @u kirikg 2 kg 2 kg 2 kg 2

#00 *,*.5B $0,8$B.D5

!ipakai Gu tumpuan kanan " 11,6@1.@@ kgm " 116.@& lapangan " 29,>>2.-? kgm " 29>.>

T-gaga -tur 3a<a 3-a23ag k/23/sit- <a-ra a3aga

"& 3 ya

n 3 l

"@1?616++9 mm 3 192.>? mm

" 9+.612 &Pa.- 3 *2012;;mm

"& 3 $ya* 19+ mm%n 3 l

"@1?616++9 3 2.>? mm

" +.?@>6> &Pa.- 3 9?@2+12>>mm

"& 3 $ya* 19+ mm%n 3 l

"@1?616++9 3 2.>? mm

" -.@2@ &Pa9?@2+12>>mm

"& 3 ybn 3 l

"@1?616++9 3 2@.+6 mm

" -?+.+1+- &Pa9?@2+12>>mm

'a " tegangan lentur pada tepi (#(S plat beton'b " tegangan lentur pada tepi B(<(H plat betonsa " tegangan lentur pada tepi (#(S propil <)sb " tegangan lentur pada tepi baIah propil <)

Batas 7endutan "7

"-+++mm

" 19.> mm9?+ 9?+

#.*. EECE6A TE9AA M7ME :ETF9 97;:

-sai !a/k 6/23/sit-

!-ba 2ati (:)Berat sendiri pelat " +.12 3 26++ " 2 kgmSpesi $t " 2'm% " + +2 3 21++ " 62 kgm

d2I

Pada #epi atas 3-at b-t/

Pada #epi atas saya3 baja

pada tepi ba4a sayap baja

'a

'b

sa

sb



Berat plaond " 1- " 1- kgmBerat peralatan &E " @- " @- kgm

#otal !7 " 666 kgm!7 " 6.66 km

Beban hidup $77%Beban hidup " 2-+ " 2-+ kgm

ultimit " 1.2 3 !7 1.? 3 77

" 1.2 3 666 1.? 3 2-+ u " 92?6. kg m

7 " -.++ m

&ma3 "1

3 u 3

&ma3 "1

3 92?6.+ kg m 3 -.++ " 1+2+2

" $0*

Lperlu "&u

"116@1. mm

"J 3 y 0.;0 3 2-+

"

<i c/ba r/3i 3-a23ag yag <i guaka & 00 00 $0 $5M3 " bf 3 t $ hs 5 t % +.2- 3 tI" 00 3 1- $ 9++ 5 1- % +.2- 3 1+" 1#8$$*2.$ mm" 1#8$.$*2$ 'm9 +.-+>> 'm9 ok m

untuk ts " 12 'm dan a " 9.2- 'm $asumsi%

(s perlu "&u

N y Dhs

t *a

O2 2

"1+2+2-+++ mm

+.- 2-+ D9++

12+ *92.-

O2 2

" 1+2+2-+++ mm " 1+.- 2-+ D 1-+ 1+6 O

(s perlu " 1>.21 'm 11>.+ 'm memen

&enentukan letak sumbu netral plastis" 3

(sr " +.2- 3 ∏ 3 = +.2- 3 ∏ 3 12

yr " 26+ &pa

#sr " 119.+6 3 26+ = 2@19+ 0aya tekan nominal maksimum dari propil <) 9++=9++=1+=1-

(

7I

#sr (sr yr

!I



" 11>.+ 'm 3 2-+ " 2>,>-+

Keseimbangan gaya:#sr #s " ma3 * #s

2#s " ma3 * #sr

#s "D ma3 * #sr O

2

#s " D 2>,>-+ * 2@,19+ O " 1,61+ 2

M N " + O #sr 's = ma3 * #s 2@,19+ 1#10 " 2>,>-+ *

28$#0 " 2,-6+M N " + O ok

Sumbu netral plastis jatuh di lens,maka jarak sumbu netral pastis dari atas )lens#s

"161+.2+

" 2.-?6 y b! 2-+ 300

Lengan "(#$)

hs <) 1@9.>+ 1% plens *@.?> 29&'72

1&20'

y "29@.@

" 16.91 'm1??.2+

1-titik berat s

9++ mm2&%

2>.@2 mm

9++ $$ 1+ mm169.11@

=" +.- 3 2- 3 1.@- x 1$.0

" [email protected]

s " (s . )y * '

2" 1@9.>+ 3 2-+ * $$.81

= 216-.? 2

# " ' s " 22+1? 7

&n " ' d2 3 s : d

" --@. 1-? x 21#$8.* 221.>" [email protected]

&n 3 Q u

1-6?@19- 3 +.- = 11#.0* 4 $$#D$.8o

6/t/ t-ra<a3 -<uta7 -+++

ma3 4 #sr sehingga garis netral jatuh pada propil baja

0aris kerja gaya Cs yang diukur dari baIah lens

Luas A(#$I )

Cc 0.85•ƒ'c •bE • ts

Cc

Jb



9?+ 9?+

ek Penampang kompak

hs

)rame otasi!imensi

(&omen ;n

hs b5 t ;3

mm mm mm mm mmBalok 3%00R00 00 00 10 1$ 18 11;.80 2+,6++

I " >6.++ g-m

L3 " 1,9?+.++ m

Ly " 6-+.++ m

ys " 2-+ +a

us " 61++ +a

syarat "h/

1?+

"296

1?+

tI 1+ S 2-+

" 29.6 1+?.2-2-2>9

Sehingga kapasitas momen dianalisa se'ara plastis

M-tuka "u2a 1-ar k/-kt/r "-is A6F ( 16 1; 0$D**00*)

/' " 2- &pa Shear studE'" 29-++ &Pa u " #00Pa

! " 1> mm

t " 19+ mm

@ = C = 0.85 x fc x (' " 6@12-

! " 1> mm (s' " +.2- " +.2- T 1> " 29.9- mm2

" +.- 3 (sr 3 S /' 3 E' (s' 3" +.- 3 29.9- $2- 3 29-++ &Pa 29.9-

tw r o

'mI 'mH

S f

!I

Qn ! u



!u *egangan +utus +eng,ubung geser +aku (a)

As# Luas +ena$+ang satu +eng,ubung geser .enis +aku ($$2)

/n uat n$ina untuk +eng,ubung geser (N*4N)

!iameter maksimim shear konektor

2,- " 2.- 3 1- " [email protected] R 1> mm ....ok

Syarat minimum dam maksimun jarak *penghubung geser

jarak minimum longitudinal " ?d " 116 mm jarak minimum longitudinal " t " 1+6+ mmJarak tranCersal " 6d " @? mm

jumlah Shear konektor yang dibutuhkan

"Gh

"6@12-

Qn 1+?+-.26

" 6.6 ?d

&aka jumlah Shear kone'tor pakai jarak minimum

"?++

" $ batang11.6+

harus ditranser shear kone'tor paku

s-ar k/-kt/r $G* b-tag = *B.$B $$# 22

M-gitug kuat g-s-r 3-a23aghasil Sap 2+++ Gersi 16Gu " 116@1.@@ kg

J Gn " +.> $ +.? y % h " +.> $ +.? 2-+ % 296 1+" 91->++.+ kg

J Gn " 91->++.+ kg 4 Gu " 116@1.@@ kg

syarat :h

"296

" 29.6 11++

" ?>.-@+1+-2tI 1+ S y

Berarti penampang aman terhadap geser

Gu " gaya geser perluCn " gaya geser nominal

t!

5,gaya geser horiMontal total pada bidang kontak baja dan beton yang

tI



&

T

dE2

d,2



&pa

&pa&pa

T&pa&pa

Beratkgm

?,@-+ >6.++

19,?++ 19?.++22,6++ 1@2.++



hs h' O29+

12 O2

6-6.? 'm9

s 3 Dhs

O2

>.+ 3 D9+

O2

>>@.+ 'm9

'mF6

[email protected] 'mF6



22+61.@ 'mF6

2@?1.@ 'm9

2926.? 'm9

12@>.9 'm9

u kaa kg 2

$$,#D$.DD

kmkm

$#ekan%

#ekan

#ekan

tarik



9.-

9.-

kgm

0 22

-+.>> mm9

+.-+>> 'm9

hs * t %F29++ * 1- )U2

uhi

12a " 1+.6

mm

119.+6 mmI



1,61+

mm

( y $'m9%

2?+.-*22>.29@.@

&

T

fs

18.

.*

2*.#* 2*#.*

(@u asi 1a3 *000 @-rsi $#)

dE2

&<

d,2



;y

?,@-+

k . Penampang kompak

3 6++&Pa

'mH



2. >. PA/N?'C7V>7 KBPB<?' 6>LBK ?7DCK KBPB<?' BA7 >K<

LA6>/ A%AK'?% 6>LBK ?7DCK KBPB<?' BA7 >K<?C PB<?'?%

bE bE

AK'A/?B/ INTERIOR

b0 b0

6alo esterior bE L-8 : ara +usat balo e+usat +elat

bE 1-2 b0 : ara +usat balo e+usat +elat


bE b0


bA 0.8$fE

0.8$fE

tsa &

d-2 d1

titi berat

T

T

(>) (b) ()


C=0.85 f'c . A . bE


a =As x fy

0.85 f'c bEmaka kual lentur nominal $ gambar b%&n "atau&n " "


gaya tekan ' , yang bekerja pada beton sebesar: $ lihat gambar '%


s " (s . )y * '

2

s " (s . )y * +,- /' . bE .ts

2Kuat lentur nominal $&n%

& d2 d2

bf


&<

d,2


F d1

#F d1 (sFy$d2ts * a2%

Jika a > ts, maka asumsi dapat dirubah, hasil ini plat beton tidak 'ukup kuatuntuk menimbangi gaya tarik pada propil baja


0.85•ƒ'c •bE • ts


beton

baa



" 2- &pa Mutu baja !" #$ 2? &pa

4#engangan putus baja 5u$minimum% " 61++ &pa

4#engangan 7eleh baja 5y$minimum% " 2-+ &pa

4 8egangan $ minimum% " 1 T

4&odulus Elastisitas baja$Es% "2+++++ &pa

4&odulus 0eser $0% " ++++ &pa

48asio Paison " +.9



)rame otasi!imensi

(&omen ;nersia

h b5 ro ;3 ;y

mm mm mm mm mm

Balok <)9++=9++ 9++ 9++ 1+ 1- 1 11>.+ 2+,6++ ?,Balok <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++ 19,Kolom <)6++=6++ 6++ 6++ 19 21 22 21.@+ ??,?++ 22,

Balok ;nterior

Balok ;nterior Propil yang digunakan <) 9-+ = 9-+ = 12 = 1>A

"7

"?++

" 1-+ 'm6 6

b+ " 9-+ 'm

&odulus Elastisitas beton " 6@++ % 2- &PaE' " 29-++ &Pa

&odulus 8asio

n "Es

"2+++++ &Pa

" .-1+? G E' 29-++ &Pa

7ebar EiCalen Baja"

bE"

1-+ 'm" 1.@-n

bE" 1.@- 1-+ 'mbEn

12 m

hs" 9-+

& 50 50 $* $

7etak garis etral (' " 1.@- 3 12 'm " 22-.++ m2

(s " <) 9-+ = 9-+ = 12 = 1> " 1@9.>+ m2

(s total " 9> >+ m2


tw tw

'mH 'mH 'mH

a. --tua -bar -f-ktif (bE) balok interior

b

b



9>.> 'm 3 ya " 22-.++ 312 'm

1@9.>+ 3 D9-

2 2

9>> 'm 3 ya " 19-+.+ 'm9 -19+.1 'm9 " ?6

ya "?6+.1 'm9

" 1*.2# m9>.>+ 'm


(s tot 3 yb " (' 3hs h'

(s2

9>.> 'm 3 yb " 22-.++ 39- 12

1@9.>+2

9>> 'm 3 yb " >22- 9+69.9 'm9 " 122?

yb "122?.9 'm9

" 9+. 'm9>.>+ 'm

67T97:

ya yb " hs h'1?.2 'm 9+. 'm " 9- 12 'm

6@.+ 'm " 6@ 'm 76

ya= 1!.2" cm

d1= 10.2#


d2= 1. m yb= 30.8 cm


d% M/i2- i-rsia k/23/sit- (;) ,t-ra<a3 garis -tra

(' " 22-.++

"1

3 1.@- 'm 3 12 'm " 2@++.++ '12

d1 " ya *h'2

d1 " 1?.2 'm *12 'm

" 1+.26 'm2

;' " ;o ( 3

2@++.++ 'm 22-.+ 'm 3 1+.2 'm " 2?91

r/3i & 50 50 $* $

(s " 1@9.>+ 'm

;o " 2+6++ 'mF6

d2 " yb *hs2

" 9+. 'm *9- 'm

" 19.2? 'm2

;s " ;o (s 3

2+6++ 'm 1@9.> 'm 3 19.9 'm " 22@

(s " 7uas propil <);o " &omen inersia propil <)d2 " 7etak pusat berat propil <) terhadap garis netral komposite;s " &omen inersia propil <) terhadap garis netral komposite

f) M/2- ;-rsia -a23ag 6/23/sit-; " ;' ;s " 2?91-.+ 'mF6 22@+-.1 'mF6

'mI

;o

d1I

d2I



ya 1?.26 'm*

Ssa ";

"6>+2+.1- 'mF6

" 1ya *h' 6.26- 'm

*

Ssb ";

"6>+2+.1- 'mF6

" 1yb 9+. 'm

M/2- asi ari 1A *000 ? $#CE6 97;: & 50 50 $* $

ra2- M/2-(M) @*M tu23u kiri M a3 M. tu23u kaa @u kiri @u k

kg 2 kg 2 kg 2 kg 2 kg$0 $#BB#.B $0,5.8$ $$#$#.$0 ,$*.8

!ipakai &u tumpuan kanan " *11,616.1+ kgm " *116.16& lapangan " 1+,>>-.1 kgm " 1+>.>?


"& 3 yan 3 l

"1+>>-1++ mm 3 1?2.6- mm

" 6.21? &Pa.- 3 #;0201$#;mm

"& 3 $ya* 19+ mm%

n 3 l

"1+>>-1++ 3 92.6- mm

" +.--22 &Pa.- 3 6>+2+1-6>mm

"& 3 $ya* 19+ mm%n 3 l

"1+>>-1++ 3 92.6- mm

" @.2@6@ &Pa6>+2+1-6>mm

"& 3 ybn 3 l

"1+>>-1++ 3 [email protected] mm

" ?.>@? &Pa6>+2+1-6>mm

'a " tegangan lentur pada tepi (#(S plat beton

'b " tegangan lentur pada tepi B(<(H plat betonsa " tegangan lentur pada tepi (#(S propil <)sb " tegangan lentur pada tepi baIah propil <)

Batas 7endutan "7

"-+++mm

" 19.> mm9?+ 9?+


-sai !a/k 6/23/sit-!-ba 2ati (:)

Berat sendiri pelat " +.12 3 26++ " 2 kgmSpesi $t " 2'm% " +.+2 3 21++ " 62 kgmbin $keramik% " 26 " 26 kgmBerat plaond " 1- " 1- kgm

Berat peralatan &E " @- " @- kgm #otal !7 " 666 kgm

!7 " 6.66 kmBeban hidup $77%Beban hidup " 2-+ " 2-+ kgm

ultimit " 1.2 3 !7 1.? 3 77 " 1.2 3 666 1.? 3 2-+

u " 92?6. kg m

7 " -.++ m

&ma3 "1

3 u 3

&ma3 " 1 3 92?6.+ kg m 3 -.++ " 1+2+2.-" $0*0*5000



'a

'b

sa

sb

7I



<i c/ba r/3i 3-a23ag yag <i guaka & 50 50 $* $

M3 " bf 3 t $ hs 5 t % +.2- 3 tI $ h" $0 3 1> $ 9-+ 5 1> % +.2- 3 12 $ 9-" 2$2;8 mm

" 2$2;.8 'm9 *+.-+@9 'm9 ok memenuhi

untuk ts " 12 'm dan a " 9.2- 'm $asumsi% t*12a

(s perlu "&u

N y Dhs

t *a

O2 2

"1+2+2-+++ mm

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12+ *92.-

O2 2

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" [email protected]+

+.- 2-+ D 1@- 1+6 O

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(sr " +.2- 3 ∏ 3 = +.2- 3 ∏ 3 12 "

yr " 26+ &pa

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0aya tekan nominal maksimum dari propil <) 9++=9++=1+=1-" (s 3 y

" 1@9.>+ 'm 3 2-+ " 69,6@-


2#s " ma3 * #sr

#s "D ma3 * #sr O

2

#s "D 69,6@- * 2@,19+ O

" ,1@9 2

M N " + O #sr 's = ma3 * #s

2@,19+ 81 " 69,6@- *$02 " 9-,9+2

M N " + O ok

#sr (sr yr

!I

ma3

ma3 4 #sr sehingga garis netral jatuh pada propil baja



Lengan "(#$) hs <) 1@9.>+ 1'

plens *6@.?@ 36&319

12&22

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9-+ mm

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611.9 mm9-+ $$ 12 mm

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s " (s . )y * '

2" 1@9.>+ 3 2-+ * 0*.$*

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&n 3 Q u

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o6/t/ t-ra<a3 -<uta

Batas Lendtan =7

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ek Penampang kompak

hs

)rame otasi!imensi

(&omen ;nersia

hs b5 t ;3 ;

mm mm mm mm mmBalok 3%$0R$0 $0 $0 12 1; 20 1.;0 6+,9++

I " 19?.++ g-m

L3 " 2,?@+.++ mLy " >+>.++ m

ys " 2-+ +a

0aris kerja gaya Cs yang diukur dari baIah lensLuas A(#$I )


Cc

Jb

tw r o

'mI 'mH 'm



. . .




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t " 19+ mm

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! " 1> mm (s' " +.2- " +.2- T 1> " 29.9- mm2

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2,- " 2.- 3 1> " [email protected] R 1> mm ....ok




"Gh

"6@12-

Qn 1+?+-.26

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s-ar k/-kt/r $G* b-tag = $.5D $$# 22

M-gitug kuat g-s-r 3-a23aghasil Sap 2+++ Gersi 16Gu " +--.6- kg

J Gn " +.> $ +.? y % h " +.> $ +.? 2-+ % 2@2 12

" 66+?6+.+ kgJ Gn " 66+?6+.+ kg 4 &u " *11616.1+ kg

!I

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t!


tI



Berarti penampang aman terhadap geser




C PB<?'?%

dE2



Beratkgm

>6.++19?.++

1@2.++



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3 Dhs

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3 D9-

O2

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'mF6



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> 'm9

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kmkm

$#ekan%

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mm

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(sr " @ 3 $16 3 ∏ 3 = @.++ 3

yr " 26+ &pa

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0aya tekan nominal maksimum dari propil <) 9-+=9-+=12=1>

" (s 3 y

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2#s " ma3 * +

#s "D ma3 * #sr O

2

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22*80.*0

M N


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y 3 b! 26+ 3 3%0

Leng

lens ??.-+ 0&9%

Ieb 2+.2- 2'&%9

'&7%2

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Lengan "(#$

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'3&1%2

y1 "?62.6

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y2 "

2?>>.

" 92.6@ 'm9.1-2

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#s : &n2 " #s 3 $ y2 * y1 %

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&n " &n1 &n2 " 2+6>>>+9?.

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-@>.2?62.6

.@?- mm

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mmI

tuh pada propil baja



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@6.+- mm

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titi berat

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Berdasarka gambar diatas, maka gaya tekan :

C=0.85 f'c . A . bE


a =As x fy

0.85 f'c bEmaka kual lentur nominal $ gambar b%

&n "atau&n



F d1

#F d1 (sFy$d2ts a2%

beton

baa




gaya tekan ' , yang bekerja pada beton sebesar:

Cc = sehingga keseimbangan gayT' = Cc + Cs

s " (s . )y * '

2

s " (s . )y * +,- /' . bE .ts

2


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mm mm mm mm mm

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Jika a > ts, maka asumsi dapat dirubah, hasil ini plat beton tidak 'ukup kuatuntuk menimbangi gaya tarik pada propil baja


0.85•ƒ'c •bE • ts



tw tw

'mH



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"7

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b+ " 9-+ 'm

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&odulus 8asio

n "Es

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7ebar EiCalen Baja"

bE"

@- 'mn

bE" 1.@- @- 'mbEn

hs" 9++

&00 00 $0 $5

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(s total " 292.9+


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292 'm 3 ya " ?@-.+ 'm9

ya "9>+>.? 'm9

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(s tot 3 yb " (' 3hs

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b

b



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67T97:

ya yb " hs1?. 'm 2-.2 'm " 9+

62.+ 'm " 62 'm

ya=

d1= 10.8 m

d2= 10.2 m yb=



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Batas 7endutan " 7 " -+++mm "9?+ 9?+




'a

'b

sa

sb




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ultimit " 1.2 3 !7 1.?" 1.2 3 666 1.?

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t *2

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7I




(sr " +.2- 3 ∏ 3 = +.2-

yr " 26+ &pa#sr " 119.+6 3 26+ = 2@19+

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" (s 3 y

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2#s " ma3 * #sr

#s "D ma3 * #sr O

2

#s "D 2>,>-+ * 2@,19+ O

2M N " + O #sr 's = ma3

2@,19+ 1#10

28$#0

M N

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161+.2+y b! 2-+ 300

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hs <) 1@9.>+ 1%

plens *@.?> 29&'72

1&20'

y "29@.@

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Luas A(#$I )



1-9++ mm

9++ $$ 1+ mm

=" +.- 3 2- 3 1.@- x 1$.0

" [email protected]

s " (s . )y * '

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=2

# " ' s " 22+1? 7

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o6/t/ t-ra<a3 -<uta

Batas Lendtan =7

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Cc

Jb



ek Penampang kompak

hs

)rame otasi

!imensi

hs b5 t

mm mm mm mm mmBalok 3%00R00 00 00 10 1$ 18 11

I " >6.++ g-m

L3 " 1,9?+.++ m

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1*.8 m

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6

1?+

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beton

baa



maka kuat lentur nominal $ gambar b%&n "atau&n " "


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0.85•ƒ'c •bE • ts






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(h b5 ro

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12 m

hs" 9-+

& 50 50 $* $

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b

b




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2

2?.6 'm 3 ya " 112.-+ 312 'm

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yb " @?--. 'm9 "2?.6+ 'm

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"1+>>-1++ 3 @2.?> mm

".- 3 6?2?+-+?mm

"& 3 $ya* 19+ mm%n 3 l

"1+>>-1++ 3 @2.?> mm

"6?2?+-+?mm

"& 3 ybn 3 l

"1+>>-1++ 3 [email protected] mm

"6?2?+-+?mm

'a " tegangan lentur pada tepi (#(S plat beton'b " tegangan lentur pada tepi B(<(H plat betonsa " tegangan lentur pada tepi (#(S propil <)sb " tegangan lentur pada tepi baIah propil <)

Batas 7endutan "7

"-+++mm

" 19.9?+ 9?+

#.*. EECE6A TE9AA M7ME :ETF9 97;:-sai !a/k 6/23/sit-

!-ba 2ati (:)Berat sendiri pelat " +.12 3 26++ " 2Spesi $t " 2'm% " +.+2 3 21++ " 62bin $keramik% " 26 " 26Berat plaond " 1- " 1-Berat peralatan &E " @- " @-

#otal !7 " 666!7 " 6.66

Beban hidup $77%Beban hidup " 2-+ " 2-+

ultimit " 1.2 3 !7 1.? 3" 1.2 3 666 1.? 3

u " 92?6. kg m

7 " -.++ m

&ma3 "1

3 u 3

'a

'b

sa

sb

7I



&ma3 "1

3 92?6.+ kg m 3 -.++

Lperlu "&u

"*11616.1

J 3 y 0.;0 3

<i c/ba r/3i 3-a23ag yag <i guaka & 50 50 $* $M3 " bf 3 t $ hs 5 t %

" $0 3 1> $ 9-+ 5 1> % " 2$2;8 mm" 2$2;.8 'm9 *+.-+@9 '

untuk ts " 12 'm dan a " 9.2- 'm

(s perlu "&u

N y Dhs

t *a

2 2

" 1+2+2-+++ mm+.- 2-+ D

9-+ 12+ *

92.-2 2

"1+2+2-+++ mm

+.- 2-+ D 1@- 1+6

(s perlu " [email protected] 'm 1@9.>+ 'm




(sr " +.2- 3 ∏ 3 = +.2- 3

yr " 26+ &pa

#sr " 119.+6 3 26+ = 2@19+

0aya tekan nominal maksimum dari propil <) 9++=9++=1+=1-

" (s 3 y

" 1@9.>+ 'm 3 2-+ " 69,6@-


2#s " ma3 * #sr

#s "D ma3 * #sr O

2

#s "D 69,6@- * 2@,19+ O

"2

M N " + O #sr 's = ma3 * 2@,19+ 81 "

$02 "M N "


"1@2.@+

"y b! 2-+ 3%0

Lengan "(#$)

hs <) 1@9.>+ 1'

plens *6@.?@ 36&319

12&22

y "[email protected]

" 11.1- 'm12?.22?

9-+ mm

#sr (sr yr

!I

ma3

ma3 4 #sr sehingga garis netral jatuh pa




1>13&2

611

9-+ $$ 12 mm

=

" +.- 3 2- 3 1.@- x 1;.0" @+?.-?

s " (s . )y * '

2" 1@9.>+ 3 2-+ * 0*.$*

=2

# " ' s " 22+>1 7

&n " ' d2 3 s : d

" @+?.? 2@ x 218#.2 " 2199211.9

&n 3 Q u

2199211 3 +.- = 1812.$ 4

o6/t/ t-ra<a3 -<uta

Batas Lendtan =7

"-+++mm

"9?+ 9?+

ek Penampang kompak

hs


Cc

Jb



)rame otasi!imensi

(hs b5 t

mm mm mm mm mmBalok 3%$0R$0 $0 $0 12 1; 20 1.;0

I " 19?.++ g-m

L3 " 2,?@+.++ mLy " >+>.++ m

ys " 2-+ +a

us " 61++ +a

syarat "h/

1?+

"2@2

tI 12

" 22.???????@




! " 1> mm

t " 19+ mm

@ = C = 0.85 x fc x (' "

! " 1> mm (s' " +.2- " +.2- T 1> "

" +.- 3 (sr 3 S /' 3 E'

" +.- 3 29.9- $2- 3 29-++ &

"1+?+-.26

1199-6 &Pa!u *egangan +utus +eng,ubung geser +aku (a)




2,- " 2.- 3 1> " [email protected] R 1>


tw r o

'mI

f

!I

Qn

t!





"Gh

"29>+?2.-

Qn 1+?+-.26 " 2.2 ?d


"?++

" * batang>.-+


s-ar k/-kt/r $G* b-tag = $.5D $$# 22

M-gitug kuat g-s-r 3-a23aghasil Sap 2+++ Gersi 16

Gu " +--.6- kg

J Gn " +.> $ +.? y % h " +.> $ +.? 2-+ % 2@2 " 66+?6+.+ kg

J Gn " 66+?6+.+ kg 4 &u "

syarat :h

"2@2

" 22.??@ 11++

" ?>.-@+

tI 12 S yBerarti penampang aman terhadap geser





BPB<?' BA7 >K<?C PB<?'?%

PB<?'?%

t

usat +elat

0.8$fE

&

T

()

bf

&<

dE2

d,2



ambar '%

" 2- &pa2? &pa

" 61++ &pa

" 2-+ &pa" 1 T" 2+++++ &pa" ++++ &pa" +.9



&omen ;nersia

Berat;3 ;y

kgm2+,6++ ?,@-+ >6.++6+,9++ 19,?++ 19?.++??,?++ 22,6++ 1@2.++

1>

A

Pa

>.9

'mH 'mH



(s 3 Dhs

h' O2

1@9.>+ 3 D9-

12 O2

'm9 " -+-.1 'm920.2 m

h' (s 3 D

hsO

212

1@9.>+ 3 D9-

2

'm9 " @?--. 'm9

2?.@ 'm

h'12 'm

6

cm

te

cm

f

" 19-+.++ 'mF6

O



6.9 'm " 262--.- 'mF6

.2 'm " 22++-.9 'mF6

22++-.9 'mF6

mF6" 222.9 'm9

mF6" -->6.- 'm9

mF6" 1@9+.? 'm9

@*@u kiri @u kaa kg 2 kg 2

,$*.8 8,055.#5

" *116.16 km" 1+>.>? km

d1I

d2I



" -.??+> &Pa $#ekan%

2.+9+1- &Pa #ekan

[email protected]@@? &Pa #ekan

?9.-9@2 &Pa tarik

> mm

kgmkgmkgmkgmkgm kgmkm

kgm

77 9.-2-+ 9.-



" 1+2+2.- kgm

" $0*0*5000 22 mm

" *-+.@9 mm92-+

" *+.-+@9 'm9

+.2- 3 tI $ hs * t +.2- 3 12 $ 9-+ * 1> )U2

9 ok memenuhi

asumsi% t*12a " 1+.6

O

O

" [email protected]+ mmO

memenuhi

F



∏ 3 12 " 119.+6

,1@9

#s 69,6@- * ,1@9

9-,9+2+ O ok

19.?21 1> mm

( y $'m9%9+69.9*[email protected]

&

mmI

a propil baja



titik berat s

.9 mm

T

199.@@2 mm

fs

196.2

21.2

$$#$#.$ kg2 (Mu asi 1a3 *000 @-rsi $#)

16 mm

&< dE2

d,2



&omen ;nersia

;3 ;y

6+,9++ 19,?++

1?+

2-+

1+?.2-2-2> ok . Penampang kompak

29>+?2.-

9.9- mm2

(s' 3 a 29 3 6++&Pa

ok

mm ....ok

'mH 'mH

! u



12

*11616.1+ kg

+-26

tI



18.

.*

2*.#* 2*#.*



2. 6. PA/N?'C7V>7 KBPB<?' 6>LBK ?7DCK&enentukan letak sumbu netral plastis

" 3

(sr " @ 3 $1 3 ∏ 3 = @.++ 3

yr " 26+ &pa

#sr " 9>-.?6 3 26+ = >6>-6

0aya tekan nominal maksimum dari propil <) 9-+=9-+=12=1>

" (s 3 y

" 1@9>+.++ 3 2-+ " 696@

Keseimbangan gaya:

#sr #s " ma3 * #s2#s " ma3 * +

#s "D ma3 * #sr O

2

#s "D 696@-++.++ * >6>-9.?+ O

2M N " + O #sr 's = ma3

>6>-9.?+ 212*2.20

222122*.80

M N


"[email protected]+

y 3 b! 26+ 3 3%0

lens ??.-+ 0&9

Ieb 22.1- 2'&%

''&%1

Karena letaknya tidak jatuh di leng, maka sumbu netal jatuh di <eb Proil#s * $b . t . y%

6?9@@9.2+

" 1y 3 tw 2-+ 3 12

Lengan "(#

lens ??.-+ 36&0%

Ieb 1.-- 2&1%

'%&0%1

?>? @

#sr (sr yr

!I

ma3

ma3 4 #sr sehingga garis netral jatu


0aris kerja gaya Ts yang diukur dari baIah lensLuas A(#$I )



.?-1.

y2 "2@6>.6

" 92.99 'm-.+-1

Hitung &omen terhadap garis kerja s :#sr : &n1 " #sr 3 $h tb * y * -+%

" >6>-6 3 $9-+12+*@6,+-*-+%" 926>1>@.> mm

#s : &n2 " #s 3 $ y2 * y1 %

" -2+2?99+-.>

" 19>9++?61.+ mm

&n " &n1 &n2 " 1@216>9.>

Wb.n " +.- 3 1@216>9.>

" 16?92@9?9.1 mm 1#.* t



BPB<?' BA7 >K<?C 7AV>'?%

1 3 ∏ 3 12 " 9>-.?6

-++.++

" [email protected]+

* #s " 696@-++.++ * [email protected]+ " 222122?.+

" + O ok

" 2-.919 1>

engan "(#$) ( y $'m9%?9.2

9 ?99.-?>?.@

6.->1 mm

) ( y $'m9%22?6.96-.1

2@6>.6

mmI

pada propil baja

mm



.

929.2@ mm

on.m 4 11#1# tm (&(



6>6. @. PA/N?'C7V>7 KBLB1 PA/N?'C7V>7 KBLB 'A7V>N D? L>7'>? 1 <C6C R

%/>A #*

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx

= 1; = ;$0*1*1 I r = 5 = 5 Y Z

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 188.08 7

ux = 8$. 7m ux5base = #.$$ 7m

u = ;.;1 7m u5base = .$ 7m

6. >K<? KBLB >/>N R

= 1.00 (AP?')

V6 = [(?-L) KBLB ( *$*1$81 - $000 ) : (2.00 R ;$0*1*

= 1.#$ (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emu

+ = = *001 x 2$0 = ;0002$0 7mm ;00.0

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm $88.2$

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingga 9Jb x nx = 0.; x ;00.0 = 810.0 7m

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\



&. PA/6A<>/>7 BA7

_x.Lx=

1.#$ R $000=

#2.2$

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.$$ = 0

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

2*;*.7

#22$U2

= &m = 0.= 0.#1 1

157c-7l15

188.08

2*;*.

>mbil = 1.00

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momen harus berhubungan dengan sumbu lenturna dalaadalah sumbu x sehingga 9

(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 1 <C6C R

%/>A 1

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx =

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z =

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx =

= 1; = ;$0*1*1 I r = 5 = 5 Y Z =

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1*.*# 7

ux = 12.;$ 7m ux5base = *.10 7m

u = 20.; 7m u5base = #.$1 7m

6. >K<? KBLB >/>N R

= 1.00 (AP?')

V6 = [(?-L) KBLB ( *$*1$81. - $000 ) :

2.00 R ;$0*1*

= 1.#$ (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emu

+ = = *001 x 2$0 = ;0002$0 7mm ;00

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm $88

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\



n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +


D. PA/6A<>/>7 BA7

_x.Lx=

1.#$ R $000=

#2.2$

/x 1#.$$&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.#1 = 0

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

2*;*. 7#22$U2

= &m = 0.#12= 0.#2 1

157c-7l 15 1*.*#

2*;*.

>mbil = 1.00


(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 1 <C6C R

%/>A $*

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx =

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z =

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx =

= 1; = ;$0*1*1 I r = 5 = 5 Y Z =

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1211.$$ 7

ux = 12.0 7m ux5base = 18.$# 7m

u = 2.$1 7m u5base = 22.* 7m

6. >K<? KBLB >/>N R

= 1.00 (AP?')

V6 = [(?-L) KBLB ( *$*1$81 - $000 ) :

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>



2.00 R ;$0*1

= 1.#$ (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada em

+ = = *001 x 2$0 = ;0002$0 7mm ;0

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm $8

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12$*.* 7m +


&. PA/6A<>/>7 BA7

_x.Lx=

1.#$ R $000=

#2.2$

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.** = 0

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

2*;*. 7#22$U2

= &m = 0.= 0.$ 1

157c-7l 15 1211.$$

2*;*.

>mbil = 1.00

Kx

Zx.\


(_.L-r)I



1 PA/N?'C7V>7 KBLB 'A7V>N L>7'>? 1 <C6C X

%/>A #*

P/B%?L KBLB 9 #00 R #00 R 1 R 21

asti odulus

*001 Y = 1 >rea (>) = 21#$#.0 I

1*;$12* Y = 21 = *$*1$81 I

/ = 22 = 22#0*$$## I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

5 Y = 12 >rea (>) = 2180.00 I

5 Y = 1; = ;$0*1*1 I

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 188.08 7

ux = 8$. 7m

u = ;.;1 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

$*1$81. - #000 ) = 2.*1 V6 = [(?-L) KBLB (000

K = 1.01 L?N>' 7BBV/>

an n sama dengan + = .L=

1.01 R $000

r 102.20

= 1R

.LR S

` r

02 12

= 1.#=

1

1.*5(0.* ) 1.*5(0.*

']

'^ ?x

?

']

'^ ?x

?

7u



7n = = >g (\-) =

7u=

188.08= 0.$1$

J 7n $1.#$

&. >K<? 6>LBK

b\ =

$0=

;.21

2 x t\ 2 x 1;

7u=

188.08 x=

Jb x t\ 0.; x 2$0 x 2180

+ =$00 ( 2. 5 7u )

S\ Jbn

+ =$00

x2. 5

S2$0

=h

=$0 5 2 (1;)

= 2*t 12

L+ =;0 x /

=;0.00 R 1

S\ S 2$0.00

= 1-j(2b x tfY) : (h 5 2 x tf) tYk

= 21$. mmU

R] = ` R SA.V..> = .1#

<x 2 2*80

= ;.*0808

= 20#00.$

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 - & =?

us ini sumbu lentur

>g x \r

10H



Lr = /(R]-\ 5 \r) x S1:S1:R (̂\ 5 \r

= 11$82.$ x 1.$8

= 181.8#1;* mm

L+ = $.10*2 m L =

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) na

+ = Z.\ = 1*;$12* x

r = < (\5\r) = 112028 x (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= #20.12 7m + BK

Jb x n = 0.; x

D. PA/6A<>/>7 BA7

_.L = 1.01 R $000

/ 102.20

&m = 0.* 5 0.#

7el = `IA.>g = .1

= &m = 0.28

157c-7l 15 188.10##1

>mbil = 1.00

A. Perisa terhada+ +ersamaan

ux = .ux =u = .u =7c

:1 ux

27n 1 Jb x nx

0.2$:

1.00x

8$.

1.00 810.0

0.2$ : 0.10$

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momesehingga 9

(_.L-r)I



0.8;

adi Prol 3% $0 x $0 x 1 x 21 menuu+i untu memiul beban5beb

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 1 <C6C X

%/>A 1

P/B%?L KBLB 9 #00 R #00 R 1 R 21

asti odulus

*001 Y = 1 >rea (>) = 21#$#.0 I

1*;$12* Y = 21 = *$*1$81 I

/ = 22 = 22#0*$$## I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

Y = 12 >rea (>) = 2180.00 I

Y = 1; = ;$0*1*1 I

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1*.*# 7

ux = 12.;$ 7m

u = 20.; 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

*$*1$81. - #000 )= 2.*1

V6 = [(?-L) KBLB (

000

K = 1.#2 L?N>' 7BBV/>


1.#2 R $000

r 102.20

7m

7m = 1 .L

']

'^ ?x

?

']

'^ ?x

?

7u



` r

02 12

= 1.#=

1.*5(0.* ) 1.*5(0

7n = = >g (\-)

7u=

1*.*#= 0.20;

J 7n #28.02

&. >K<? 6>LBK

b\ =

$0= ;.21

2 x t\ 2 x 1;

7u=

1*.*# x=

Jb x t\ 0.; x 2$0 x 2180

+ =$00 ( 2. 5 7u )

S\ Jbn

+= $00

x 2. 5 0.20S2$0

= h

=$0 5 2 (1;)

= 2*t 12

L+ =;0 x /

=;0.00 R 1

S\ S 2$0.00

= 1-j(2b x tfY) : (h 5 2 x tf) tYk

= 21$. mmU

R] = ` R SA.V..> = .1#

<x 2 2*80

= ;.*081A

>g x \r

us ini sumbu lentur

10H



= 20#00.$

& = ?\ . NI-2 = 20#00

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 - & =?

Lr = /(R]-\ 5 \r) x S1:S1:

= 11$82.$ x 1.$8

= 181.8#1;* mm

L+ = $.10*2 m L =


+ = Z.\ = 1*;$12* x 2$0

r = < (\5\r) = 112028 x (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

=

= #20.12 7m + BK

Jb x n = 0.; x #2

D. PA/6A<>/>7 BA7

_.L = 1.#2 R $000/ 102.20

&m = 0.* 5 0.#

7el = `IA.>g = .1

= &m =157c-7l 15

>mbil = 1.00



(_.L-r)I



ux = .ux = 1

u = .u = 1

7c:

8 ux

27n ; Jb x nx

0.10$:

8.00x

12.;$

;.00 810.0

0.10$ : 0.01#

0.1*8

adi Prol 3% $0 x $0 x 1 x 21 menuu+i untu memiul beban5beb

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 1 <C6C X

%/>A $*

P/B%?L KBLB 9 #00 R #00 R 1 R 21

asti odulus

*001 Y = 1 >rea (>) = 21#$#.0 I

1*;$12* Y = 21 = *$*1$81 I

/ = 22 = 22#0*$$## I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

Y = 12 >rea (>) = 2180.00 I

Y = 1; = ;$0*1*1 I

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1211.$$ 7

ux = 12.0 7m

u = 2.$1 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

*$*1$81. - #000 ) V6 = [(?-L) KBLB (

']

'^ ?x

?

']

'^ ?x

?

7u



000.

K = 1.#2 L?N>' 7BBV/>


1.#2 R $000

r 102.20

7m

7m = 1R

.LR S

` r

02 12

= 1.#=1.*5(0.* ) 1.*5(0

7n = = >g (\-)

7u = 1211.$$ = 0.$ 0.20

J 7n #28.02

&. >K<? 6>LBK

b\ =

$0= ;.21

2 x t\ 2 x 1;

7u=

1211.$$ x=

Jb x t\ 0.; x 2$0 x 2180

+ =$00 ( 2. 5 7u )

S\ Jbn

+= $00

x2. 5

S2$0

= h

=$0 5 2 (1;)

= 2*t 12

L+ =;0 x /

=;0.00 R 1

S\ S 2$0.00

>g x \r

us ini sumbu lentur

10H



= 1-j(2b x tfY) : (h 5 2 x tf) tYk

= 21$. mmU#

R] = ` R SA.V..> = .1#

<x 2 2*80

= ;.*081A

= 20#00.$

& = ?\ . NI-2 = 20#00

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 - & =?

Lr = /(R]-\ 5 \r) x S1:S1:= 11$82.$ x 1.$8

= 181.8#1;* mm

L+ = #.#1*8 m L =


+ = Z.\ = 1*;$12* x 2$0

r = < (\5\r) = 112028 x (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

=

= #20.12 7m + BK

Jb x n = 0.; x #2

D. PA/6A<>/>7 BA7

_.L = 1.#2 R $000

/ 102.20

&m = 0.* 5 0.#

7el = `IA.>g = .1


(_.L-r)I



= &m =157c-7l 15

>mbil = 1.00


ux = .ux =u = .u =7c

:1 ux

27n 1 Jb x nx

0.1:

1.00x

12.0

1.00 810.0

0.1 : 0.01$

0.28

adi Prol 3% $0 x $0 x 1 x 21 menuu+i untu memiul beban5b



6>6. @. PA/N?'C7V>7 KBLB

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7 Plasti odulus

rx = 1#.$$ <x = 2*80; Y Zx = *001 Y

r = 102.20 < = 112028 Y Z = 1*;$12* Y

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

rx = 5 = 5 Y Zx = 5 Y

r = 5 = 5 Y Z = 5 Y

ase = #.$$ 7m

ase = .$ 7m

22#0*$$## - $ ) : ( 22#0*$$## - # ) = 2.22.00 R 1$81$*1 - *

= *.*$

\=

1R *.*$ R S

2$0= 0.1*

A .1#28* 200000

= 1.2*$0.1* )

Zx

Z



21#$#.0 x (2$0-) = #201.1 7

0.200

+ S10

= 10.$2$0

.$ 0.12$

Q **$

S\

= .1 **$

= #2.0*S2$0

+ (PA7>P>7V KBP>K)

= $10*.2= $.10*2

1000

R S2R10 x *;208 x 1801#$ x 21#$#

2

111;A500 x 2122*812.*0#

+a = 20#00.$* +a

.8*;A500$



$.00 m Lr = 18.2 m K><C<

alam asus ini &b = 218$ sehingga ada emunginan n sama dengan +

= #281$ 7mm #2.8 7m

= 201*$8;8;.#2 7mm 201.** 7m

#20.12 = 8.11 7m

= *.*$

1-2 = 0.* 5 0.# x 0.8 = 0.28

200000 x 21#$# = 10##1.*; 7**$I

= 0.$ 1.00

x 8$. = 8$. nm

x ;.;1 = ;.;1 nm

:u

Jb x n

:;.;1

8.11

: 0.02*

us berhubungan dengan sumbu lenturna dalam asus ini sumbu lentur adalah sumbu



1.00 $%

ersebut sesuai desain L/%D

6>6. @. PA/N?'C7V>7 KBLB


rx = 1#.$$ <x = 2*80; Y Zx = *001 Y

r = 102.20 < = 112028 Y Z = 1*;$12* Y

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

rx = 5 = 5 Y Zx = 5 Y

r = 5 = 5 Y Z = 5 Y

ase = *.1 7m

ase = #.$1 7m

22#0*$$## - $ ) : ( 22#0*$$## - # )= 2.2

2.00 R 1$81$*1 - *

= *;.*2

\ 1 2$0

Zx

Z



A .1#28*.

200000.

= 1.2;;0.8 )

= 21#$#.0 (2$0-) = #02.; 7

0.200

+ S10

= 10.$2$0

1.#* 0.12$

Q **$

S\

= .# **$

= #2.0*S2$0

+ (PA7>P>7V KBP>K)

=$10*.2

= $.10*21000

R S2R10 x *;208 x 1801#$ x 21#$#

2.00

x 2122*812.*0#



+a

+a

.8A50$

5 \r)I

$.00 m Lr = 18.2 m K><C<


= #281$ 7mm #2.8 7m

= 201*$8;8;.#2 7mm 201.** 7m

= 8.11 7m

= *;.*2

1-2 = 0.* 5 0.# x 0.21 = 0.$1

200000 x 21#$# = 828.02 7*;.*2I

= 0.$**# 1.00

.02




x 12.;$ = 12.;$ nm

x 20.; = 20.; nm

:u

Jb x n

:20.;

8.11

: 0.0#;

1.00 $%


6>6. @. PA/N?'C7V>7 KBLB


rx = 1#.$$ <x = 2*80; Y Zx = *001 Y

r = 102.20 < = 112028 Y Z = 1*;$12* Y

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

rx = 5 = 5 Y Zx = 5 Y

r = 5 = 5 Y Z = 5 Y

ase = 18.$# 7m

ase = 22.* 7m

22#0*$$## - $ ) : ( 22#0*$$## - # )

Zx

Z



2.00 R 1$81$*1 - *.

= *;.*2

\=

1R *;.*2 R S

2$0= 0.8

A .1#28* 200000

= 1.2;;0.8 )

= 21#$#.0 (2$0-) = #02.; 7

+ S10

= 10.$2$0

2.#* 0.12$

Q **$

S\

= . **$

= #2.0*S2$0

+ (PA7>P>7V KBP>K)

=##1.*8

= #.#1*8100



R S2R10 x *;208 x 1801#$ x 21#$#

2.00x 2122*812.*0#

+a

+a

.8*;A500$

5 \r)I

$.00 m Lr = 18.2 m K><C<


= #281$ 7mm #2.8 7m

= 201*$8;8;.#2 7mm 201.** 7m

= 8.11 7m

= *;.*2

1-2 = 0.* 5 0.# x 0.*88 = 0.2$

200000 x 21#$# = 828.02 7

**$I




$

= 0.8.$$ 1.00

.02

x 12.0 = 12.0 nm

x 2.$1 = 2.$1 nm

:u

Jb x n

:2.$1

8.11

: 0.08*

1.00 $%




6>6. @. P1 PA/N?'C7V>7 KBLB 'A7V>N D? L>7'>? 2 <C6C R

%/>A 82

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$

= 21 = *$*1$81. I r = 102.20

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1#;.#* 7

ux = 2#.* 7m ux5base

u = 1;.1$ 7m u5base

6. >K<? KBLB >/>N R

= 1.00 (AP?')

V6 = [(?-L) KBLB ( *$

= 1.* (D>/? '>6AL 7BBV/>)Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) na

+ = = *001 x 2$0

r = <x (\5\r) = <x . (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingJb x nx = 0.; x ;00.0 =

&. PA/6A<>/>7 BA7

_x.Lx=

1.* R $000=

#8.0#

/x 1#.$$

&m = 0.* 5 0.# 1-2 =

7el = `IA.>g=

.1#U2 x 200000 x 21#$

#22$U2

= &m = 0.2*8=

1 7 -7 l 1#; #*

']

'^ ?x

?

']

'^ ?x

?

7u

V>

Kx

Zx.\

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momesumbu x sehingga 9

(_.L-r)I



181.8$

>mbil = 1.00



6>6. @. P

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 2 <C6C R%/>A 8

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$

= 21 = *$*1$81. I r = 102.20

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ;#8.1 7

ux = 8.11 7m ux5base

u = $0.0; 7m u5base

6. >K<? KBLB >/>N R

= 1.00 (AP?')

V6 = [(?-L) KBLB (

= 1.* (D>/? '>6AL 7BBV/>)


+ = = *001 x 2

r = <x (\5\r) = <x . (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

']

'^ ?x

?

']

'^ ?x

?

7u

V>

Kx

Zx.\



Karena n tida boleh melebihi + maa dalam asus ini n = + sehing

Jb x nx = 0.; x ;00.0 = 81

D. PA/6A<>/>7 BA7

_x.Lx = 1.* R $000 = #8.0#/x 1#.$$

&m = 0.* 5 0.# 1-2 =

7el = `IA.>g=

.1#U2 x 200000

#22$U

= &m = 0.1

157c-7l 15 ;#8.1

181.8$

>mbil = 1.00


(_.L-r)I



6>6. @. P

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 2 <C6C R

%/>A *2

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$

= 21 = *$*1$81. I r = 102.20

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 1; = ;$0*1*1 I r = 5

/ 20 1$81$*1 I

']

'^ ?x

?

'] '^ ?x

?



>. A7VN?'C7V 6A6>7 'A/%>K'B/

= #2;.;0 7

ux = 50. 7m ux5base

u = $.0# 7m u5base

6. >K<? KBLB >/>N R

= 1.00 (AP?')V6 = [(?-L) KBLB (

= 1.* (D>/? '>6AL 7BBV/>)


+ = = *001 x 2

r = <x (\5\r) = <x . (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12$*.* 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehing

Jb x nx = 0.; x ;00.0 = 81

&. PA/6A<>/>7 BA7

_x.Lx=

1.* R $000=

#8.0#

/x 1#.$$

&m = 0.* 5 0.# 1-2 =

7el = `IA.>g=

.1#U2 x 20000

#22$U

= &m = 0.80

157c-7l 15 #2;.;0

181.8$

>mbil = 1.00

7u

V>

Kx

Zx.\


(_.L-r)I



A/N?'C7V>7 KBLB

BDCLC< B% <A&'?B7 Plasti odulus

<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

= 20.#* 7m

= 1. 7m

*1$81 - $000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000

un dalam asus ini &b = 218$ sehingga ada emunginan n sama dengan +

= ;0002$0 7mm ;00.0 7m

= $882$#220 7mm $88.2$ 7m

ga 9810.0 7m

0.* 5 0.# x 0.81 = 0.2*8

=181.8$

7

0.2; 1

Zx

Z

harus berhubungan dengan sumbu lenturna dalam asus ini sumbu lentur adalah



A/N?'C7V>7 KBLB


<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

= 11.#* 7m

= #;.1; 7m

*$*1$81. - $000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000


0 = ;0002$0 7mm ;00.0 7m

= $882$#220 7mm $88.2$ 7m

Zx

Z



ga 9

.0 7m

0.* 5 0.# x 0.0 = 0.1

21#$#=

181.8$ 7

= 0. 1




A/N?'C7V>7 KBLB


<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y= 5 Y Z = 5 Y

ZxZ



= 0.* 7m

= 2.## 7m

*$*1$81 - $000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000


0 = ;0002$0 7mm ;00.0 7m

= $882$#220 7mm $88.2$ 7m

ga 9

.0 7m

0.* 5 0.# x 50.#; = 0.;*

x 21#$#=

181.8$ 7

= 0.82 1




1 PA/N?'C7V>7 KBLB 'A7V>N L>7'>? 2 <C

%/>A 82

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1#;.#* 7

ux = 2#.* 7m

u = 1;.1$ 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

= 2.*1V6 = [(?-L)

K = 1.00 L?N>' 7BBV/> = .L

=1.00

r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u=

1#;.#*=

J 7n 88.2;

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u 1#; #* x

']

'^ ?x

?

']

'^ ?x

?

7u

>g x \r



Jb x t\ 0.; x 2$0 x 2

+ =$00 ( 2. 5

S\

+ =$00

xS2$0

= h = $0 5 2 (1;)t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

= ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m

Karena L+ L Lr maa n terleta antar

+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.00

/

&m = 0.*

7el = `IA.>g

= &m157c-7l

/asio elangsingan _L-r ang di +erhitunsumbu sehingga 9

(_.L-r)I



>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.1#:

1.00x

1.00

0.1# : 0.00

0.2$$

adi Prol 3% $0 x $0 x 1 x 21 menu

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 2%/>A 8

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ;#8.1 7

ux = 8.11 7m

u = $0.0; 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

= 2.*1V6 = [(?-L)

K = 1.$*0 L?N>' 7BBV/>

= .L=

1.$*

r

= 1R

.L

` r

02 12

']

'^ ?x

?

']

'^ ?x

?

7u



= 1.#=

1.*5(0.* )

7n = = >g

7u=

;#8.1=

J 7n *.*

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

;#8.1 x

Jb x t\ 0.; x 2$0 x 2

+ = $00 ( 2. 5S\ J

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S = 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

= ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m

+ = Z.\ = 1*;$12*

>g x \r

Karena L+ L Lr maa n terleta antar

+



r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.$*0

/

&m = 0.*

7el = `IA.>g

= &m157c-7l

>mbil =


ux = .ux

u = .u

7c:

8

27n ;

0.10 : 8.00 x;.00

0.10 : 0.00;

0.2$


PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? 2

%/>A *2

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) == 1; =

/ 20


(_.L-r)I

']

'^ ?x

?

'] '^ ?x

?



>. A7VN?'C7V 6A6>7 'A/%>K'B/

= #2;.;0 7

ux = 50. 7m

u = $.0# 7m

6. >K<? KBLB >/>N X

V> = 1.00 (AP?')

= 2.*1V6 = [(?-L)

K = 1.$*0 L?N>' 7BBV/>

= .L=

1.$*0

r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u = #2;.;0 = 0.118

J 7n *.*

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

#2;.;0 x

Jb x t\ 0.; x 2$0 x 2

+ =$00 ( 2. 5

S\ J

+ = $00 xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

7u

>g x \r



= ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = #.#1*8 m

+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.$*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.0$;:

1.00x

1.00

0.0$; : 0.000

0.1$1


Karena L+ L Lr maa n terleta antar+


(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB6C X

/>D?C< B% VX/>'?B7

21#$#.0 I rx = 1#.$$

*$*1$81 I r = 102.20

22#0*$$## I

/>D?C< B% VX/>'?B7

2180.00 I rx = 5

;$0*1*1 I r = 5

1$81$*1 I

ux5base = 20.#* 7m

u5base = 1. 7m

KBLB ( 22#0*$$## - $ ) : (

2.00 R 1$

R #000= $0.88

102.20

R S \

=1

R $0.88A .1#28*

1.#= 1.1$$

1.*5(0.* R 0.$2 )

(\-) = 21#$#.0 x (2$0-) = #$*2.0

0.#8 0.200

;.21 + S

10= 10.$

2$0

10H



180. .

7u ) Q **$

Jbn S\

2. 5 0.#8= .

**$=

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20 = $10*.2= $.10*2

2$0.00 1000

) tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 2

2*80; 2= ;.*0808;81;111;A500 x 2122*812.*0#

= 20#00.$$*$ +a = 20#00.$* +a

& = .8*;A500$?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =

+ dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emung

x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mm

L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $0.88

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 1*#0.2$ 7

**$I

= 0.2= 0.2* 1.0015 1#;.#*

1*#0 2$

an dalam +erbesaran momen harus berhubungan dengan sumbu lenturna dalam a



1.00

= 1.00 x 2#.* = 2#.* nm

= 1.00 x 1;.1$ = 1;.1$ nm

ux:

u

Jb x nx Jb x n

2#.*:

1;.1$

810.0 8.11

: 0.0$1

1.00 $%

u+i untu memiul beban5beban tersebut sesuai desain L/%D

6>6. @. PA/N?'C7V>7 KBLB

C6C X

/>D?C< B% VX/>'?B7 B

21#$#.0 I rx = 1#.$$ <x

*$*1$81 I r = 102.20 <

22#0*$$## I

/>D?C< B% VX/>'?B72180.00 I rx = 5

;$0*1*1 I r = 5

1$81$*1 I

ux5base = 11.#* 7m

u5base = #;.1; 7m

KBLB ( 22#0*$$## - $ ) : (

2.00 R

R #000= *1.0*

102.20

R

S

\=

1R *1.0*

A .1#28*

Zx

Z



1.#= 1.2$#*

1.*5(0.* R 0.*8 )

(\-) = 21#$#.0 (2$0-

0.2*1 0.200

;.21 + S10

= 10.$2$0

= 1.; 0.12$180

7u ) Q **$n S\

2. 5 0.2*1= .#2

**$=

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

) tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 2

2*80;.* 2.00= ;.*080;A500 x 2122*812.*

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8A50$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =

x 2$0 = #281$ 7mm

10H




x (2$050) = 201*$8;8;.#2 7mm

L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = *1.0*

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 11#.#0 7*;.*2I

= 0.20= 0.215 ;#8.1 1.00

11#.#0

1.00

= 1.00 x 8.11 = 8.11 nm

= 1.00 x $0.0; = $0.0; nm

ux:

u

Jb x nx Jb x n

8.11 : $0.0;810.0 8.11

: 0.118

1.00 $%


6>6. @. PA/N?'C7V>7 KBLB

C6C X

/>D?C< B% VX/>'?B7 B

21#$#.0 I rx = 1#.$$ <x

*$*1$81 I r = 102.20 <

22#0*$$## I

/>D?C< B% VX/>'?B7

2180.00 I rx = 5 ;$0*1*1 I r = 5

1$81$*1 I


ZxZ



ux5base = 0.* 7m

u5base = 2.## 7m

KBLB ( 22#0*$$## - $ ) : (

2.00 R

R #000= *1.0*

102.20

R S \

=1

R *1.0*A .1#28*

1.#= 1.2$#*

1.*5(0.* R 0.*8 )

(\-) = 21#$#.0 (2$0-

0.200

;.21 + S10

= 10.$2$0

= 0.8 0.12$180

7u ) Q **$

n S\

2. 5 0.118 = .$$ **$ =S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

##1.*8= #.#1*8

2$0.00 100

) tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 2

2*80; * 2 00

10H



= ;.*080;A500 x 2122*812.*

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =

x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mmL+ ) +

+ BK

x #20.12 = 8.11 7m

R $000 = *.2

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 2*2. 7**$I

= 0.28

= 0.015 #2;.;0 1.00

2*2.

1.00

= 1.00 x 50. = 50. nm

= 1.00 x $.0# = $.0# nm

ux:

u

Jb x nx Jb x n

50.:

$.0#

810.0 8.11

: 0.0;

1.00 $%







<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

22#0*$$## - # )= 2.2

81$*1 - *

R S2$0

= 0.$2200000

7

Zx

Z



#2.0*

#$#

18.2 m K><C<

inan n sama dengan +

#2.8 7m

201.** 7m

x 0.;0 = 0.2

sus ini sumbu lentur adalah



CLC< B% <A&'?B7 lasti odulu

= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

22#0*$$## - # )= 2.2

1$81$*1 - *

R

S

2$0= 0.*8

200000



) = #2$.01 7

#2.0*

#$#

18.2 m K><C<

#2.8 7m

inan n sama dengan



201.** 7m

x 0.;82 = 0.20

CLC< B% <A&'?B7 Plasti odulus

= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y= 5 Y Z = 5 Y




22#0*$$## - # )= 2.2

1$81$*1 - *

R S2$0

= 0.*8200000

) = #2$.01 7

#2.0*

#$#



18.2 m K><C<

#2.8 7m

201.** 7m

x 0.8 = 0.28

inan n sama dengan




6>6. @. PA/N?'C7V>7 KBLB1 PA/N?'C7V>7 KBLB 'A7V>N D? L>7'>? <C6C R

%/>A 11

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx =

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z =

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx =

= 1; = ;$0*1*1 I r = 5 = 5 Y Z =

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 102*.$# 7

ux = 2#.; 7m ux5base = 22.1* 7m

u = 18.0 7m u5base = 20.*2 7m

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1$81 - #000 ) :

2.00 R ;$

V6 = [(?-L) KBLB ( *$*1$81 - #000 ) :2.00 R ;$

= 1.0 (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada

+ = = *001 x 2$0 = ;0002$0 7mm ;

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm $

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +


']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\



&. PA/6A<>/>7 BA7

_x.Lx=

1.0 R #000=

;.12

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.88 =

7el = `IA.>g=

.1#U2 x 200000 x 21#$=

2*#$.$*7

.80U2

= &m = 0.2#$= 0.2$ 1

157c-7l15

102*.$#

2*#$.$*

>mbil = 1.00

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momen harus berhubungan dengan sumbu lenturnadalah sumbu x sehingga 9

(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? <C6C R

%/>A 118

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx

= 1; = ;$0*1*1 I r = 5 = 5 Y Z

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= *82. 7

ux = *.$0 7m ux5base = 2.* 7m

u = #;.01 7m u5base = ##.81 7m

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1$81.0 - #000 )

2.00 R ;$

V6 = [(?-L) KBLB ( *$*1$81.0 - #000 )

2.00 R ;$

= 1.0 (D>/? '>6AL 7BBV/>)


+ = = *001 x 2$0 = ;0002$0 7mm

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\



n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +


&. %>K'B/ PA/6A<>/>7 BA7

_x.Lx=

1.0 R #000=

;.12

/x 1#.$$&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.#11# =

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

2*#$.$* 7.80U2

= &m = 0.##= 0.#$ 1

157c-7l 15 *82.

2*#$.$*

>mbil = 1.00


(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? <C6C R

%/>A ;

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx

= 21 = *$*1$81. I r = 102.20 < = 112028 Y Z

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

= 12 >rea (>) = 2180.00 I rx = 5 = 5 Y Zx

= 1; = ;$0*1*1 I r = 5 = 5 Y Z

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1.88 7

ux = .2 7m ux5base = 1.;2 7m

u = #;.$0 7m u5base = 2*.#* 7m

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1$81. - #000 )

2.00 R ;$

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>



V6 = [(?-L) KBLB ( *$*1$81 - #000 )

2.00 R ;

= 1.$; (D>/? '>6AL 7BBV/>)


+ = = *001 x 2$0 = ;0002$0 7mm

r = <x (\5\r) = <x . (2$050) = $882$#220 7mm

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +


D. PA/6A<>/>7 BA7

_x.Lx=

1.$; R #000=

$.2

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5 0.# x 0.$$ =

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

#010.* 7.80U2

= &m = 0.*;

= 0.1 1157c-7l 15 1.88

#010.*

>mbil = 1.00

Kx

Zx.\


(_.L-r)I



1 PA/N?'C7V>7 KBLB L>7'>? <C6C X

%/>A 11

P/B%?L KBLB 9 #00 R #00 R 1 R 21

Plasti odulus

*001 Y = 1 >rea (>) =

1*;$12* Y = 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

5 Y = 12 >rea (>) =

5 Y = 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 102*.$# 7

ux = 2#.; 7m

u = 18.0 7m

6. >K<? KBLB >/>N X

( *$*1$81. - #000 )= 2.;0

V> = [(?-L)

0*1*1 - 000

( *$*1$81. - #000 ) = 2.;0 V6 = [(?-L)0*1*1 - 000

K = 1.1$ L?N>' 7BBV/>

emunginan n sama dengan + = .L=

1.1$

r

00.0 7m

88.2$ 7m = 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

']

'^ ?x

?

']

'^ ?x

?

7u



7n = = >g

7u=

102*.$#=

J 7n 8*#.18

0.2#$

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

102*.$# x

Jb x t\ 0.; x 2$0 x 21

+ =$00 ( 2. 5

S\

+ =$00

xS2$0

=h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf)

= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

a dalam asus ini sumbu lentur

>g x \r



Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8#2 mm

L+ = $.10*2 m

Karena L+ L Lr maa n terleta antara +

+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

= #20.12 7m

Jb x n = 0.;

D. %>K'B/ PA/6A<>/>7 BA7

_.L = 1.01

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =A. Perisa terhada+ +ersamaan

ux = .ux

u = .u

7c:

8

27n ;

0.1:

8.00x

;.00

0.1 : 0.02

/asio elangsingan _L-r ang di +erhitungasumbu sehingga 9

(_.L-r)I



0.20#

adi Prol 3% $0 x $0 x 1 x 21 menuu

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? <C

%/>A 118

P/B%?L KBLB 9 #00 R #00 R 1 R 21

Plasti odulus

= *001 Y = 1 >rea (>) =

= 1*;$12* Y = 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 5 Y = 12 >rea (>) =

= 5 Y = 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= *82. 7

ux = *.$0 7m

u = #;.01 7m

6. >K<? KBLB >/>N X

: ( *$*1$81.0 - #000 )= 2.;0

V> = [(?-L)

0*1*1 - 000

: ( *$*1$81.0 - #000 )= 2.;0

V6 = [(?-L)

0*1*1 - 000

K = 1.$8* L?N>' 7BBV/>


1.$8*

r

;00.0 7m

$88.2$ 7m = 1 .L

']

'^ ?x

?

']

'^ ?x

?

7u



` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u=

*82.=

J 7n *0;.1

0.#$

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

*82. x

Jb x t\ 0.; x 2$0 x 21

+ =$00 ( 2. 5

S\ Jb

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf)

= 21$.

R] = ` R SA.V..>

<x 2

>g x \r




& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8#2 mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.$8*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =



(_.L-r)I



ux = .ux

u = .u

7c:

1

27n 1

0.0;$:

1.00x

1.00

0.0;$ : 0.008

0.22


PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? <C

%/>A ;

P/B%?L KBLB 9 #00 R #00 R 1 R 21

Plasti odulus

= *001 Y = 1 >rea (>) =

= 1*;$12* Y = 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 5 Y = 12 >rea (>) =

= 5 Y = 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 1.88 7

ux = .2 7m

u = #;.$0 7m

6. >K<? KBLB >/>N X

: ( *$*1$81. - #000 )= 2.;0

V> = [(?-L)

0*1*1 - 000

']

'^ ?x

?

']

'^ ?x

?

7u



: ( *$*1$81. - #000 )= 2.;0

V6 = [(?-L)

0*1*1 - 000

K = 1.##* L?N>' 7BBV/>


1.##*

r

;00.0 7m

$88.2$ 7m = 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u = 1.88 = 0.0

J 7n #0.;8

0.*;

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u

=

1.88 x

Jb x t\ 0.; x 2$0 x 21

+ =$00 ( 2. 5

S\ Jb

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

>g x \r




= 1-j(2b x tfY) : (h 5 2 x tf)

= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8#2 mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

== #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.##*

/

&m = 0.*

7el = `IA.>g




= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.018:

1.00x

1.00

0.018 : 0.00#

0.1$


(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7 Plasti od

21#$#.0 I rx = 1#.$$ <x = 2*80;.$ Y Zx = *00

*$*1$81 I r = 102.20 < = 11202.1; Y Z = 1*;$

22#0*$$## I

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

2180.00 I rx = 5 = 5 Y Zx = 5

;$0*1*1 I r = 5 = 5 Y Z = 5

1$81$*1 I

ux5base = 22.1* 7m

u5base = 20.*2 7m

KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # ) =2.#

2.00 R 1$81$*1 - *

KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # ) = 2.#2.00 R 1$81$*1 - *

R #000= $1.#

102.20

R S \

=1

R $1.# R S2$0

= 0.$;A .1#2; 200000

1.#= 1.1;8

1.*5(0.* R 0.$; )

Zx

Z



(\-) = 21#$#.0 x (2$0-) = #$#*.10 7

0.2** 0.200

;.21 +

S

10= 10.$

2$0

= 2.08* 0.12$80

7u ) Q **$

Jbn S\

2. 5 0.2**=

.#2 **$= #2.0*

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20 = $10*.2= $.10*2

2$0.00 1000

Yk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80; 2= ;.*0808;81;111;A500 x 2122*812.*0#

= 20#00.$$*$ +a = 20#00.$* +a

& = .8*8802*A500$?

10H



S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2 m K><C<

dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emunginan n sama dengan +

x 2$0 = #281$ 7mm #2.8 7m

x (2$050) = 201*$8;8;.#2 7mm 201.** 7m

+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $0.;2

102.20

5 0.# 1-2 = 0.* 5 0.# x 0.;0 = 0.2

= .1#U2 x 200000 x 21#$# = 1*1$.1# 7$0.;2I

= 0.2 = 0.2$ 1.00

15 102*.$#1*1$.1#

1.00

= 1.00 x 2#.; = 2#.; nm

= 1.00 x 18.0 = 18.0 nm

ux:

u

Jb x nx Jb x n

2#.;:

18.0

810.0 8.11

: 0.0##

dalam +erbesaran momen harus berhubungan dengan sumbu lenturna dalam asus ini sumbu lentur adalah



1.00 $%

i untu memiul beban5beban tersebut sesuai desain L/%D

6>6. @. PA/N?'C7V>7 KBLB

C X

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7 Plasti

21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx =

*$*1$81 I r = 102.20 < = 112028 Y Z =

22#0*$$## I

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

2180.00 I rx = 5 = 5 Y Zx = 5

;$0*1*1 I r = 5 = 5 Y Z = 5

1$81$*1 I

ux5base = 2.* 7m

u5base = ##.81 7m

KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # )=

2.00 R 1$81$*1 - *

KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # )=

2.00 R 1$81$*1 - *

R #000= *2.08

102.20

\ 1 2$0

Zx

Z



A .1#2;.

200000

1.#= 1.2*1

1.*5(0.* R 0.*;8 )

(\-) = 21#$#.0 (2$0-) = #2#*.2# 7

0.18; 0.200

;.21 + S10

= 10.$2$0

= 1.8 0.12$80

7u ) Q **$

n S\

2. 5 0.18;= .#;

**$= #2.0*

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

Yk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80;.* 2.00= ;.*081A500 x 2122*8128

10H



= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2 m K><C<


x 2$0 = #281$ 7mm #2.8 7m

x (2$050) = 201*$8;8;.#2 7mm 201.** 7m

+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = *2.08102.20

5 0.# 1-2 = 0.* 5 0.# x 0.;1# = 0.2#

= .1#U2 x 200000 x 21#$# = 10;8.#0 7$0.;2I

= 0.2#

= 0.2$15 *82. 1.00

10;8.#01.00




= 1.00 x *.$0 = *.$0 nm

= 1.00 x #;.01 = #;.01 nm

ux:

u

Jb x nx Jb x n

*.$0:

#;.01

810.0 8.11

: 0.10

1.00 $%

i untu memiul beban5beban tersebut sesuai desain L/%D

6>6. @. PA/N?'C7V>7 KBLB

C X


21#$#.0 I rx = 1#.$$ <x = 2*80; Y Zx =

*$*1$81 I r = 102.20 < = 112028 Y Z =

22#0*$$## I

/>D?C< B% VX/>'?B7 BDCLC< B% <A&'?B7

2180.00 I rx = 5 = 5 Y Zx = 5

;$0*1*1 I r = 5 = 5 Y Z = 5

1$81$*1 I

ux5base = 1.;2 7m

u5base = 2*.#* 7m

KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # )=

2.00 R 1$81$*1 - *

Zx

Z



KBLB ( 22#0*$$## - # ) : ( 22#0*$$## - # )

2.00 R 1$81$*1 - *

R #000= $*.*0

102.20

R S \

=1

R $*.*0 R S2$0

= 0.*A .1#2; 200000

1.#= 1.218

1.*5(0.* R 0.* )

(\-) = 21#$#.0 (2$0-) = ##01.1$ 7

0.200

;.21 + S10

= 10.$2$0

= 0.280 0.12$80

7u ) Q **$

n S\

2. 5 0.0=

.*#

**$=

#2.0*

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

10H



Yk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80;.* 2.00

= ;.*081A500 x 2122*8128

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2 m K><C<


x 2$0 = #281$ 7mm #2.8 7m

x (2$050) = 201*$8;8;.#2 7mm 201.** 7m

+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $*.*0102.20

5 0.# 1-2 = 0.* 5 0.# x 0.$# = 0.8*

= .1#U2 x 200000 x 21#$# = 120.1# 7




$0.;2I

= 0.8*

= 0.;15 1.88 1.00

120.1#1.00

= 1.00 x .2 = .2 nm

= 1.00 x #;.$0 = #;.$0 nm

ux:

u

Jb x nx Jb x n

.2:

#;.$0

810.0 8.11

: 0.11

1.00 $%

+i untu memiul beban5beban tersebut sesuai desain L/%D



Y

Y

Y

Y



lus

*001 Y

1*;$12* Y

Y

Y

2.#

2.#



.



*001 Y

1*;$12* Y

Y

Y

2.#



2.#



6>6. @. PA/1 PA/N?'C7V>7 KBLB 'A7V>N D? L>7'>? # <C6C R

%/>A 11#

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x == 21 = *$*1$81. I r = 102.20 < =

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5 =

= 1; = ;$0*1*1 I r = 5 =

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= **.0 7

ux = 21.;$ 7m ux5base =

u = 1.00 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1$81

2.0

V6 = [(?-L) KBLB ( *$*1$81

2.0

= 1.0 (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun dal

+ = = *001 x 2$0 = ;

r = <x (\5\r) = <x . (2$050) = $

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingga 9

Jb x nx = 0.; x ;00.0 = 81

&. PA/6A<>/>7 BA7

_x.Lx=

1.0 R #000=

;.12

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

2.80U2

'] '^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momen harussehingga 9

(_.L-r)I



= &m = 0.2*= 0.2

157c-7l15

**.0

2*#$.$*

>mbil = 1.00



6>6. @. PA/

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? # <C6C R

%/>A 11;

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x

= 21 = *$*1$81. I r = 102.20 <

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ##.; 7

ux = 0.* 7m ux5base =

u = ;.$* 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>



V6 = [(?-L) KBLB ( *$*1

= 1.0 (D>/? '>6AL 7BBV/>)


+ = = *001 x 2$0 =

r = <x (\5\r) = <x . (2$050) =

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingga 9Jb x nx = 0.; x ;00.0 = 810.0 7

&. %>K'B/ PA/6A<>/>7 BA7

_x.Lx=

1.0 R #000=

;.12

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5

7el = `IA.>g=

.1#U2 x 200000 x 21#$#

.80U2

= &m = 0.$2=

157c-7l 15 ##.;

2*#$.$*

>mbil = 1.00

Kx

Zx.\


(_.L-r)I



6>6. @. PA/

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? # <C6C R

%/>A 10#

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x

= 21 = *$*1$81. I r = 102.20 <

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I>. A7VN?'C7V 6A6>7 'A/%>K'B/

= #.$ 7

ux = 8.; 7m ux5base =

u = 2.8 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*

V6 = [(?-L) KBLB ( *$*

= 1.0 (D>/? '>6AL 7BBV/>)


+ = = *001 x 2$0 =

r = <x (\5\r) = <x . (2$050) =

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +


Jb x nx = 0.; x ;00.0 = 810.0 7D. PA/6A<>/>7 BA7

_x.Lx=

1.0 R #000=

;.12

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.* 5

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\




7el = `IA.>g=

.1#U2 x 200000 x 21#$#

.80U2

= &m = 0.2*;=

157c-7l 15 #.$

2*#$.$*

>mbil = 1.00

(_.L-r)I



?'C7V>7 KBLB


2*80; Y Zx = *001 Y112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

5 Y Zx = 5 Y

5 Y Z = 5 Y

2*.02 7m

12.;# 7m

- #000 ) : ( *$*1$81. - #000 )=

0 R ;$0*1*1 - 000

- #000 ) : ( *$*1$81. - #000 )=

0 R ;$0*1*1 - 000

lam asus ini &b = 218$ sehingga ada emunginan n sama dengan +

002$0 7mm ;00.0 7m

82$#220 7mm $88.2$ 7m

.0 7m

0.# x 0.8# = 0.2*

*#$.$*7

erhubungan dengan sumbu lenturna dalam asus ini sumbu lentur adalah sumbu x



1



?'C7V>7 KBLB


= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

.#0 7m

8.*1 7m

81.0 - #000 ) : ( *$*1$81.0 - #000 )=

2.00 R ;$0*1*1 - 000



81.0 - #000 ) : ( *$*1$81.0 - #000 )=

2.00 R ;$0*1*1 - 000


;0002$0 7mm ;00.0 7m

$882$#220 7mm $88.2$ 7m

0.# x 0.1;8 = 0.$21

=2*#$.$* 7

0.$ 1




?'C7V>7 KBLB


= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

*.;# 7m

1;.*# 7m

$81. - #000 ) : ( *$*1$81. - #000 )=

2.00 R ;$0*1*1 - 000

1$81 - #000 ) : ( *$*1$81. - #000 )=

2.00 R ;$0*1*1 - 000


;0002$0 7mm ;00.0 7m

$882$#220 7mm $88.2$ 7m

0.# x 0.82* = 0.2*;




=2*#$.$* 7

0.2 1



1 PA/N?'C7V>7 KBLB L>7'>? # <C6C X

%/>A 11#

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) == 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= **.0 7

ux = 21.;$ 7m

u = 1.00 7m

6. >K<? KBLB >/>N X

2.;0V> = [(?-L)

2.;0V6 = [(?-L)

K = 1.1$ L?N>' 7BBV/>

= .L=

1.1$r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u=

**.0=

J 7n 8*#.18

&. >K<? 6>LBK

b\ $0

'] '^ ?x

?

']

'^ ?x

?

7u

>g x \r



2 x t\ 2 x 1;

7u=

**.0 x

Jb x t\ 0.; x 2$0 x 2

+ =$00 ( 2. 5

S\

+ =$00

xS2$0

=h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf)

= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m

Karena L+ L Lr maa n terleta antara

+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

= #20.12 7m

Jb x n = 0.;

D. %>K'B/ PA/6A<>/>7 BA7

_.L = 1.01

/

&m = 0.*




7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.08:

1.00x

1.00

0.08 : 0.02

0.1*0


2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? # <

%/>A 11;

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ##.; 7

ux = 0.* 7m

u = ;.$* 7m

6. >K<? KBLB >/>N X

2.;0V> = [(?-L)

(_.L-r)I

']

'^ ?x

?

']

'^ ?x

?

7u



2.;0V6 = [(?-L)

K = 1.$8* L?N>' 7BBV/>

= .L=

1.$8*

r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u=

##.;=

J 7n *0;.1

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

##.; x

Jb x t\ 0.; x 2$0 x 2

+ =

$00 ( 2. 5

S\ Jb

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf)

= 21$.

R] = ` R SA.V..>

<x 2

>g x \r



& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.$8*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c : 127n 1

0.0*0:

1.00x

1.00

0.0*0 : 0.001

0.1**



(_.L-r)I



PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? # <

%/>A 10#

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =>. A7VN?'C7V 6A6>7 'A/%>K'B/

= #.$ 7

ux = 8.; 7m

u = 2.8 7m

6. >K<? KBLB >/>N X

2.;0V> = [(?-L)

2.;0V6 = [(?-L)

K = 1.##* L?N>' 7BBV/>

= .L=

1.##*

r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )7n = = >g

7u = #.$ = 0.0;

J 7n #0.;8

&. >K<? 6>LBK

']

'^ ?x

?

']

'^ ?x

?

7u

>g x \r



b\ =

$0=

2 x t\ 2 x 1;

7u=

#.$ x

Jb x t\ 0.; x 2$0 x 2

+ =

$00 ( 2. 5

S\ Jb

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf)= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7




_.L = 1.##*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.0#*:

1.00x

1.00

0.0#* : 0.010

0.12;


(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

/>D?C< B% VX/>'?B7 BDCL

21#$#.0 I rx = 1#.$$ <x =*$*1$81 I r = 102.20 < =

22#0*$$## I


2180.00 I rx = 5 =

;$0*1*1 I r = 5 =

1$81$*1 I

ux5base = 2*.02 7m

u5base = 12.;# 7m

KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1$81

KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1$81

R #000= $1.#

102.20

R S \

=1

R $1.# RA .1#28$

1.#= 1.1;8

1.*5(0.* R 0.$; )

(\-) = 21#$#.0 x (2$0-) = #$#*.10

0.1$ 0.200

;.21 10

Zx

Z



2$0.

= 1.# 0.12$80

7u ) Q **$

Jbn S\

2. 5 0.1$ = .$1 **$

= #2.0*S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20 = $10*.2= $.10*2

2$0.00 1000

tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80; 2= ;.*0808;81;111;A500 x 2122*812.*0#

= 20#00.$$*$ +a = 20#00.$* +a

& = .8*8802*A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2

+ dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emungina

x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mm

L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $0.;2

102.20

5 0.# 1-2 = 0.* 5 0.# x

10H

gan dalam +erbesaran momen harus berhubungan dengan sumbu lenturna dalam asu



= .1#U2 x 200000 x 21#$# = 1*1$.1# 7$0.;2I

= 0.2;$ = 0.1 1.00

15 **.0

1*1$.1#

1.00

= 1.00 x 21.;$ = 21.;$ nm

= 1.00 x 1.00 = 1.00 nm

ux:

u

Jb x nx Jb x n

21.;$:

1.00

810.0 8.11

: 0.0#$

1.00 $%


6>6. @. PA/N?'C7V>7 KBLB

6C X


21#$#.0 I rx = 1#.$$ <x =

*$*1$81 I r = 102.20 < =

22#0*$$## I

/>D?C< B% VX/>'?B7 B

2180.00 I rx = 5 =

;$0*1*1 I r = 5 =

1$81$*1 I

ux5base = .#0 7m

u5base = 8.*1 7m

KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1

Zx

Z



KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1

R #000= *2.08

102.20

R S \

=1

R *2.08 RA .1#28$

1.#= 1.2*1

1.*5(0.* R 0.*;8 )

(\-) = 21#$#.0 (2$0-)

0.121 0.200

;.21 + S10

= 10.$2$0

= 0.88# 0.12$80

7u ) Q **$

n S\

2. 5 0.121= .$*

**$= #2.0*

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80;.* 2.00= ;.*08A500 x 2122*812.*0#

= 20#00.$$*$ +a

10H



= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2


x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mm

L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = *2.08

102.20

5 0.# 1-2 = 0.* 5 0.# x

= .1#U2 x 200000 x 21#$# = 10;8.#0 7$0.;2I

= 0.210

= 0.2215 ##.; 1.00

10;8.#01.00

= 1.00 x 0.* = 0.* nm

= 1.00 x ;.$* = ;.$* nm

ux : uJb x nx Jb x n

0.*:

;.$*

810.0 8.11

: 0.10$

1.00 $%





6>6. @. PA/N?'C7V>7 KBLB

6C X


21#$#.0 I rx = 1#.$$ <x =

*$*1$81 I r = 102.20 < =

22#0*$$## I

/>D?C< B% VX/>'?B7 B

2180.00 I rx = 5 =

;$0*1*1 I r = 5 =

1$81$*1 I

ux5base = *.;# 7m

u5base = #.#* 7m

KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1

KBLB ( 22#0*$$## - # ) : ( 22

2.00 R 1

R #000= $*.*0

102.20

R S \

=1

R $*.*0 RA .1#28$

1.#= 1.218

1.*5(0.* R 0.* )(\-) = 21#$#.0 (2$0-)

0.200

Zx

Z



;.21 + S10

= 10.$2$0

= 0.0* 0.12$80

7u ) Q **$

n S\

2. 5 0.0;=

.$;

**$=

#2.0*

S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

tYkmmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#$#

2*80;.* 2.00= ;.*08A500 x 2122*812.*0#

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr = 18.2


x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mmL+ ) +

+ BK

x #20.12 = 8.11 7m

10H




R #000 = $*.*0

102.20

5 0.# 1-2 = 0.* 5 0.# x

= .1#U2 x 200000 x 21#$# = 120.1# 7

$0.;2I

= 0.$$

= 0.$$15 #.$ 1.00

120.1#

1.00

= 1.00 x 8.; = 8.; nm

= 1.00 x 2.8 = 2.8 nm

ux:

u

Jb x nx Jb x n

8.;:

2.8

810.0 8.11

: 0.02

1.00 $%




< B% <A&'?B7 Plasti odulus

2*80;.$ Y Zx = *001 Y11202.1; Y Z = 1*;$12* Y

< B% <A&'?B7

5 Y Zx = 5 Y

5 Y Z = 5 Y

#0*$$## - # ) =2.#

*1 - *

#0*$$## - # ) =2.#

*1 - *

S2$0

= 0.$;200000

7



m K><C<

n sama dengan +

2.8 7m

01.** 7m

0.*1 = 0.2;$

ini sumbu lentur adalah




2*80; Y Zx = *001 Y

112028 Y Z = 1*;$12* Y

CLC< B% <A&'?B7

5 Y Zx = 5 Y

5 Y Z = 5 Y

#0*$$## - # )= 2.#

$81$*1 - *



#0*$$## - # )= 2.#

$81$*1 - *

S2$0

= 0.*;8200000

= #2#*.2# 7



m K><C<

n sama dengan +

2.8 7m

01.** 7m

0.;* = 0.210





2*80; Y Zx = *001 Y

112028 Y Z = 1*;$12* Y

CLC< B% <A&'?B7

5 Y Zx = 5 Y

5 Y Z = 5 Y

#0*$$## - # )= 2.#

$81$*1 - *

#0*$$## - # )= 2.#

$81$*1 - *

S2$0

= 0.*200000

= ##01.1$ 7



m K><C<

n sama dengan +

2.8 7m

01.** 7m




0.1* = 0.$$



6>6. @. PA1 PA/N?'C7V>7 KBLB 'A7V>N D? L>7'>? $ <C6C R

%/>A 1#0

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$ <x

= 21 = *$*1$81. I r = 102.20 <

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 188.$8 7

ux = 108.1* 7m ux5base =

u = *.8# 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$*1$8

V6 = [(?-L) KBLB ( *$*1$8

= 1.;8 (D>/? '>6AL 7BBV/>)Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun

+ = = *001 x 2$0 =

r = <x (\5\r) = <x . (2$050) =

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingga 9Jb x nx = 0.; x ;00.0 =

&. PA/6A<>/>7 BA7

_x.Lx=

1.;8 R #000=

##.#1

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.*

7el = `IA.>g=

.1#U2 x 200000 x 21#$#=

.80U2

= &m = 0.2= 0.2

1 7 -7 l 188 $8

']

'^ ?x

?

'] Zx

'^ ?x Z

?

7u

V>

Kx

Zx.\

/asio elangsingan _L-r ang di +erhitungan dalam +erbesaran momen hax sehingga 9

(_.L-r)I



21##.;0

>mbil = 1.00



6>6. @. PA

2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? $ <C6C R%/>A 1$0

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$

= 21 = *$*1$81. I r = 102.20

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5

/ = 20 = 1$81$*1 I

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ;1.00 7

ux = $.; 7m ux5base =

u = #;.# 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *$

V6 = [(?-L) KBLB ( *$

= 1.;8 (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun

+ = = *001 x 2$0

r = <x (\5\r) = <x . (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0 82 7m +

']

'^ ?x

?

']

'^ ?x

?

7u

V>

Kx

Zx.\




Jb x nx = 0.; x ;00.0 = 810.0

&. %>K'B/ PA/6A<>/>7 BA7

_x.Lx = 1.;8 R #000 = ##.#1/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.*

7el = `IA.>g=

.1#U2 x 200000 x 21#$#

.80U2

= &m = 0.2

157c-7l 15 ;1.00

21##.;0

>mbil = 1.00


(_.L-r)I



6>6. @. PA

PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? $ <C6C R

%/>A 1$

P/B%?L KBLB 9 #00 R #00 R 1 R 21

/>D?C< B% VX/>'?B7

= 1 >rea (>) = 21#$#.0 I rx = 1#.$$

= 21 = *$*1$81. I r = 102.20

/ = 22 = 22#0*$$#.8 I

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

/>D?C< B% VX/>'?B7

= 12 >rea (>) = 2180.00 I rx = 5

= 1; = ;$0*1*1 I r = 5 / = 20 = 1$81$*1 I

']

'^ ?x

?

']

'^ ?x?



= 10.2; 7

ux = *2.01 7m ux5base =

u = #.22 7m u5base =

6. >K<? KBLB >/>N R

= [(?-L) KBLB ( *

V6 = [(?-L) KBLB (

= 1.;8 (D>/? '>6AL 7BBV/>)

Karena L+ L Lr maa n terleta antara + dan r (untu &b = 1) namun

+ = = *001 x 2$0

r = <x (\5\r) = <x . (2$050)

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr 5 L+ ) +

= 12;0.82 7m +

Karena n tida boleh melebihi + maa dalam asus ini n = + sehingga 9Jb x nx = 0.; x ;00.0 = 810.0

D. PA/6A<>/>7 BA7

_x.Lx=

1.;8 R #000=

##.#1

/x 1#.$$

&m = 0.* 5 0.# 1-2 = 0.*

7el = `IA.>g=

.1#U2 x 200000 x 21#$

.80U2

= &m = 0.0

157c-7l 15 10.2;

21##.;0

>mbil = 1.00

7u

V>

Kx

Zx.\


(_.L-r)I



/N?'C7V>7 KBLB


= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

88.0 7m

*.11 7m

1 - #000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000

1 - #000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000

dalam asus ini &b = 218$ sehingga ada emunginan n sama dengan +

;0002$0 7mm ;00.0 7m

$882$#220 7mm $88.2$ 7m

810.0 7m

5 0.# x 0.81* = 0.2

21##.;07

8 1

rus berhubungan dengan sumbu lenturna dalam asus ini sumbu lentur adalah sumbu



/N?'C7V>7 KBLB


<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

$0.18 7m

##.82 7m

*1$81.0 - #000 ) : ( *$*1$81.0 - #000 )

2.00 R ;$0*1*1 - 000

*1$81.0 - #000 ) : ( *$*1$81.0 - #000 )

2.00 R ;$0*1*1 - 000


= ;0002$0 7mm ;00.0 7m

= $882$#220 7mm $88.2$ 7m

Zx

Z



7m

5 0.# x 0.; = 0.228

=21##.;0 7

= 0.2 1




/N?'C7V>7 KBLB


<x = 2*80; Y Zx = *001 Y

< = 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

Zx

Z



#$.;8 7m

#.2# 7m

*1$81. - #000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000$*1$81 - #000 ) : ( *$*1$81. - #000 )

2.00 R ;$0*1*1 - 000


= ;0002$0 7mm ;00.0 7m

= $882$#220 7mm $88.2$ 7m

7m

5 0.# x 0.#2 = 0.0

#=

21##.;0 7

= 0.0$ 1




1 PA/N?'C7V>7 KBLB L>7'>? $ <C6C X

%/>A 1#0

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= 188.$8 7

ux = 108.1* 7m

u = *.8# 7m

6. >K<? KBLB >/>N X

= 2.;0V> = [(?-L)

= 2.;0V6 = [(?-L)

K = 1.28 L?N>' 7BBV/> = .L

=1.28

r

= 1R

.L

` r

02 12

= 1.#=

1.*5(0.* )

7n = = >g

7u=

188.$8=

J 7n ;*.*0

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u 188 $8 x

']

'^ ?x

?

']

'^ ?x

?

7u

>g x \r



Jb x t\ 0.; x 2$0 x 2

+ =$00 ( 2. 5

S\

+ =$00

xS2$0

= h = $0 5 2 (1;)t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028

n = &b x ( r : ( + 5 r ) L+ 5 L - Lr

= #20.12 7m

Jb x n = 0.;

D. %>K'B/ PA/6A<>/>7 BA7

_.L = 1.01

/

&m = 0.*

7el = `IA.>g

= &m157c-7l

/asio elangsingan _L-r ang di +erhitungsehingga 9

(_.L-r)I



>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.02#:

1.00x

1.00

0.02# : 0.1#

0.1*


2 PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? $ <%/>A 1$0

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =

/ = 20 =

>. A7VN?'C7V 6A6>7 'A/%>K'B/

= ;1.00 7

ux = $.; 7m

u = #;.# 7m

6. >K<? KBLB >/>N X

= 2.;0V> = [(?-L)

= 2.;0V6 = [(?-L)

K = 1.##* L?N>' 7BBV/>

= .L=

1.##*

r

= 1

R

.L

` r

0 2 1 2

']

'^ ?x

?

']

'^ ?x

?

7u



= 1.#=

1.*5(0.* )

7n = = >g

7u=

;1.00=

J 7n #0.;8

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

;1.00 x

Jb x t\ 0.; x 2$0 x 2

+ = $00 ( 2. 5S\ J

+= $00

xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S = 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

>g x \r



n = &b x ( r : ( + 5 r ) L+ 5 L - Lr

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.##*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.012:

1.00x

1.000.012 : 0.0*

0.210


PA/N?'C7V>7 KBLB P?7VV?/ D? L>7'>? $ <

%/>A 1$

P/B%?L KBLB 9 #00 R #00 R 1 R 21

= 1 >rea (>) =

= 21 =

/ = 22 =

P/B%?L 6>LBK ?7DCK 9 $0 R $0 R 12 R 1;

= 12 >rea (>) =

= 1; =/ = 20 =


(_.L-r)I

']

'^ ?x

?

']

'^ ?x?



= 10.2; 7

ux = *2.01 7m

u = #.22 7m

6. >K<? KBLB >/>N X

= 2.;0V> = [(?-L)

= 2.;0V6 = [(?-L)

K = 1.##* L?N>' 7BBV/>

= .L=

1.##*

r

= 1R

.L

` r

02 12 = 1.#

=1.*5(0.* )

7n = = >g

7u = 10.2; = 0.028

J 7n #0.;8

&. >K<? 6>LBK

b\ =

$0=

2 x t\ 2 x 1;

7u=

10.2; x

Jb x t\ 0.; x 2$0 x 2

+ =$00 ( 2. 5

S\ J

+ = $00 xS2$0

= h

=$0 5 2 (1;)

t 12

L+ =;0 x /

=;0.00

S\ S

= 1-j(2b x tfY) : (h 5 2 x tf

= 21$.

R] = ` R SA.V..>

<x 2

7u

>g x \r



& = ?\ . NI-2

= 8.;*A:12 mmU*

R^ = # (<x-V.)U2 -

Lr = /(R]-\ 5 \r) x

= 11$82.$ x 1.$8

= 181.8# mm

L+ = $.10*2 m


+ = Z.\ = 1*;$12*

r = < (\5\r) = 112028n = &b x ( r : ( + 5 r ) L+ 5 L - Lr

=

= #20.12 7m

Jb x n = 0.;

D. PA/6A<>/>7 BA7

_.L = 1.##*

/

&m = 0.*

7el = `IA.>g

= &m

157c-7l

>mbil =


ux = .ux

u = .u

7c:

1

27n 1

0.01#:

1.00x

1.00

0.01# : 0.0

0.20$



(_.L-r)I



6>6. @. PA/N?'C7V>7 KBLB

/>D?C< B% VX/>'?B7

21#$#.0 I rx = 1#.$$ <x

*$*1$81 I r = 102.20 <

22#0*$$## I

/>D?C< B% VX/>'?B7

2180.00 I rx = 5

;$0*1*1 I r = 5

1$81$*1 I

ux5base = 88.0 7m

u5base = *.11 7m

KBLB ( 22#0*$$## - # ) : (

2.00 R

KBLB ( 22#0*$$## - # ) : (

2.00 R

R #000= #8.#*

102.20

R S \

=1

R #8.#*A .1#28*

1.#= 1.1$81

1.*5(0.* R 0.$#$ )

(\-) = 21#$#.0 x (2$0-) = #*1.0

0.0#8 0.200

;.21 + S

10= 10.$

2$0

Zx

Z

10H



80. .

7u ) Q **$

Jbn S\

2.5

0.0#8 = .* **$

=S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20 = $10*.2= $.10*2

2$0.00 1000

tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 21#

2*80; 2

= ;.*0808;81;111;A500 x 2122*812.*0#

= 20#00.$$*$ +a = 20#00.$* +a

& = .8*8802$;A500$?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =

+ dan r (untu &b = 1) namun dalam asus ini &b = 218$ sehingga ada emun

x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mm

L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $0.;2

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 1*1$.1# 7$0.;2I

= 0.2#2 = 0.2# 1.0015 188.$8

1*1$ 1#

an dalam +erbesaran momen harus berhubungan dengan sumbu lenturna dalam



1.00

= 1.00 x 108.1* = 108.1* nm

= 1.00 x *.8# = *.8# nm

ux:

u

Jb x nx Jb x n

108.1*:

*.8#

810.0 8.11

: 0.018

1.00 $%


6>6. @. PA/N?'C7V>7 KBLB

6C X

/>D?C< B% VX/>'?B7

21#$#.0 I rx = 1#.$$ <x

*$*1$81 I r = 102.20 <

22#0*$$## I

/>D?C< B% VX/>'?B72180.00 I rx = 5

;$0*1*1 I r = 5

1$81$*1 I

ux5base = $0.18 7m

u5base = ##.82 7m

KBLB ( 22#0*$$## - # ) : (

2.00 R

KBLB ( 22#0*$$## - # ) : (

2.00 R

R #000= $*.*0

102.20

R S \

=

1

R $*.*0A .1#28*

Zx

Z



1.#= 1.218

1.*5(0.* R 0.* )

(\-) = 21#$#.0 (2$0

0.02# 0.200

;.21 + S10

= 10.$2$0

= 0.18$ 0.12$80

7u ) Q **$n S\

2. 5 0.02# = .** **$

=S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 2

2*80;.* 2.00

= ;.*081A500 x 2122*812.*

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =


x 2$0 = #281$ 7mm

10H



L+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $*.*0

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 120.1# 7$0.;2I

= 0.2

= 0.2#15 ;1.00 1.00120.1#

1.00

= 1.00 x $.; = $.; nm

= 1.00 x #;.# = #;.# nm

ux:

u

Jb x nx Jb x n

$.;:

#;.#

810.0 8.11: 0.11

1.00 $%


6>6. @. PA/N?'C7V>7 KBLB

6C X

/>D?C< B% VX/>'?B7

21#$#.0 I rx = 1#.$$ <x

*$*1$81 I r = 102.20 <

22#0*$$## I

/>D?C< B% VX/>'?B7

2180.00 I rx = 5

;$0*1*1 I r = 5 1$81$*1 I


Zx

Z



ux5base = #$.;8 7m

u5base = #.2# 7m

KBLB ( 22#0*$$## - # ) : (

2.00 RKBLB ( 22#0*$$## - # ) : (

2.00 R

R #000= $*.*0

102.20

R S \

=1

R $*.*0A .1#28*

1.#= 1.218

1.*5(0.* R 0.* )

(\-) = 21#$#.0 (2$0

0.200

;.21 + S10

= 10.$2$0

= 0.210 0.12$80

7u ) Q **$

n S\

2. 5 0.028 = .*$ **$ =S2$0

= 2* + (PA7>P>7V KBP>K)

R 102.20=

$10*.2= $.10*2

2$0.00 1000

tYk

mmU#

= .1# R S2R10 x *;208 x 1801#$ x 2

2*80;.* 2.00

10H



= ;.*081A500 x 2122*812.*

= 20#00.$$*$ +a

= 20#00.$* +a

& = .8*;A500$

?

S1:S1:R^(\ 5 \r)I

L = $.00 m Lr =


x 2$0 = #281$ 7mm

x (2$050) = 201*$8;8;.#2 7mmL+ ) +

+ BK

x #20.12 = 8.11 7m

R #000 = $*.*0

102.20

5 0.# 1-2 = 0.* 5 0.#

= .1#U2 x 200000 x 21#$# = 120.1# 7$0.;2I

= 0.200

= 0.2015 10.2; 1.00

120.1#

1.00

= 1.00 x *2.01 = *2.01 nm

= 1.00 x #.22 = #.22 nm

ux:

u

Jb x nx Jb x n

*2.01:

#.22

810.0 8.11

: 0.11#

1.00 $%






= 2*80;.$ Y Zx = *001 Y

= 11202.1; Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

22#0*$$## - # ) =2.#

1$81$*1 - *

22#0*$$## - # ) =2.#

1$81$*1 - *

R S2$0

= 0.$#$200000

7



#2.0*

$#

18.2 m K><C<


#2.8 7m

201.** 7m

x 0.8;# = 0.2#2

asus ini sumbu lentur adalah sumbu




= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y

22#0*$$## - # )= 2.#

1$81$*1 - *

22#0*$$## - # )= 2.#

1$81$*1 - *

R S2$0

= 0.*200000



) = ##01.1$ 7

#2.0*

1#$#

18.2 m K><C<


#2.8 7m



x 0.;0 = 0.2


= 2*80; Y Zx = *001 Y

= 112028 Y Z = 1*;$12* Y

BDCLC< B% <A&'?B7

= 5 Y Zx = 5 Y

= 5 Y Z = 5 Y




22#0*$$## - # )= 2.#

1$81$*1 - *22#0*$$## - # )

= 2.#1$81$*1 - *

R S2$0

= 0.*200000

) = ##01.1$ 7

#2.0*

1#$#



18.2 m K><C<


#2.8 7m

201.** 7m

x 1.000 = 0.200




@?. <>6C7V>7 P>D> KB7

1. <>6C7V>7 6>LBK >7>K DA7V>7 6>LBK ?7DCK

#(BE7 P8U);7 B(J( V(0 !;0

U#(S; P8U);7

!;&ES;

(

&U&E

h b5 t ro ;3mm mm mm mm mm

Balok (nak <)9++=9++ 9++ 9++ 1+ 1- 1 11>.+ 2+,6++Balok ;nduk <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++

#ahanan #umpu pada bagian <eb dari balok.

8n = 0.$ (2.# x \c) x d x tp

= 0.$ x ( 2.# ) x ( #10 ) x ( 1;

'ahanan Veser 6aut dengan dua bidang geser 9

= 0.$ (0.$ x \cq) x m x >

= 0.$ x ( 0.$ ) x ( 82$ ) x ( 2

Perhitungan umlah 6aut

3% 00 n = 2.;;= 1.1

1#.02

Perisa Veser 6alo +ada 3% 00 9= 180 x ( 10 ) = 1800

= ( 180 5 2.$ ( 1; : 2 ) )

= #0 x ( 10 )

= ( #0 5 0.$ ( 1; : 2 ) )

= #10 ( $0.$0 )

= 0.* x ( #10 ) x 12$.00 )

=

= 1*$0 : ( 2#0 ) x ( #00 ) =

= #0;*$0= #

10000

#n = 0.$ x ( #0.; ) = 0.28 ton

2. <>6C7V>7 6>LBK ?7DCK DA7V>7 KBLB

>. <>6C7V>7 6>LBK ?7DCK DA7V>7 KBLB D? 6>V?>7 'A7V>N L>7'>? 1

tw

'mH 'mH

Sambungan antara Balok (nak $<) 9++ 3 9++% dengan Balok ;nduk $<) 9-+ 3 9-+% dengan menggunak9++% adalah sebesar 29.>> ton dan &utu Baja Proil BJ 61.

R n

Ag mmI

An

Agt

Ant

\c.>nt

0.*\c.>n

Karena \c.>nt 0.*\c.>n maa 9

&n 0.*\c.>n : \ .>gt



#(BE7 P8U);7 B(J( V(0 !;0

U#(S; P8U);7

!;&ES;

(&U&E

h b5 t ro ;3mm mm mm mm mm

Balok ;nduk <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++Kolom <)6++=6++ 6++ 6++ 19 21 22 21.@+ ??,?++

= 1.2 D : 1.? L = 1.2 x ( $.#$

= 1.2 D : 1.? L = 1.2 x ( 8.

en(itn( taanan nominal bat *

+eser *

1. = 0.$ x ( 0.#0

2. = 2 x 0.1# =

Tu23u 1. " 0.$ x ( 2.#0

2. " 0.$ x ( 2.#0

&arik *

= +.@- (0.$ x \cq) x m x >

= 0.$ x ( 0.$ ) : ( 82$ ) x 28.; ) =

,eritn(an -ik ,enyambn( tas dan Ba/a *

Dioba dua buah baut +ada masing5masing +rol siu sehingga

d =

=1.** x 10

=1**.*0

=2' 2 x 11.$1 2*.02

ara baut terhada+ ens atas balo = v ( $00

Vunaan +rol siu 80 x 120 x 10 sehingga 9w $ 5 5 = $ 5 10 5

dengan d = $00 mm maa gaa ang beera +ada +rol siu adal

' = =

1.** x 10=

1**

d $00 $0

Vaa ini menimbulan momen +ada +rol siu sebesar 9

= 0.$

Sambungan sebuah Balok ;nduk dengan Kolom, antara Balok ;nduk $<) 9-+ 3 9-+% dengan Kolom $<)baut (92- N1> mm dengan ulir pada bidang geser, diketahui momen ujung sebesar @-.6- km $!% dan 1dan >@.-1 k $7%.

tw 'mH 'mH

u

u

6idang geser Rn

6idang geser Rn

3eb 6alo Rn

%lens 6alo Rn

R n

siu rsiu

. '.w



= 0.$ x ( *2.2 ) x ( $# ) = 10;8#

Ka+asitas nominal +enam+ang +ersegi adalah 9

n = 0.;

#

sehingga di+eroleh 9 b = # x 10;8#

=

*8;

0.; x 2#0 x 10I 21*

Vunaan siu 80 x 120 x 10 dengan +anang $0 mm +ada ens olom

,eritn(an -ambn(an ada lens Balok *

Vaa geser +ada ens balo adalah = 1.** x 10=

1**#

$0 $0

6aut +enambung adalah baut dengan satu bidang geser sehingga 9

n = ;0.#= 1.2; 2 buah baut

0.1#

,eritn(an -ambn(an eb Balok den(an sik 80 120 10 *

'ahanan dua bidang geser ( 1#0.28 ) 7 lebih besar dari+ada tahanan tum+u

sehingga tahanan baut di tentuan oleh tahanan tum+u.

n = 2$0.#*= 1.#; 2 buah baut

1*8.2*

,eritn(an -ambn(an eb Balok den(an lens %olom *

n = 2$0.#*

= .$ # buah baut0.1#

6. <>6C7V>7 6>LBK ?7DCK DA7V>7 KBLB D? 6>V?>7 P?7VV?/ K?/? L>7'>? 1

#(BE7 P8U);7 B(J( V(0 !;0

U#(S; P8U);7

!;&ES;

(&U&E

h b5 t ro ;3

mm mm mm mm mm

Balok ;nduk <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++Kolom <)6++=6++ 6++ 6++ 19 21 22 21.@+ ??,?++

= 1.2 D : 1.? L = 1.2 x ( $.#8

= 1.2 D : 1.? L = 1.2 x ( 8.0

(b.dI) . y

6aut ang menghubungan balo dengan ens olom adalah sambungan dengan sa

Sambungan sebuah Balok ;nduk dengan Kolom, antara Balok ;nduk $<) 9-+ 3 9-+% dengan Kolom $<)baut (92- N1> mm dengan ulir pada bidang geser, diketahui momen ujung sebesar @-.6 km $!% dan 1dan >@.-1 k $7%.

tw

'mH 'mH

u

u




+eser *

1. = 0.$ x ( 0.#0

2. = 2 x 0.1# =

Tu23u 1. " 0.$ x ( 2.#0

2. " 0.$ x ( 2.#0

&arik *

= 0.$ (0.$ x \cq) x m x >

= 0.$ x ( 0.$ ) : ( 82$ ) x 28.; ) =

,eritn(an -ik ,enyambn( tas dan Ba/a *Dioba dua buah baut +ada masing5masing +rol siu sehingga

d =

=1.* x 10

=1**.20

=2' 2 x 11.$1 2*.02

ara baut terhada+ ens atas balo = v ( $00

Vunaan +rol siu 80 x 120 x 10 sehingga 9

w $ 5 5 = $ 5 10 5


' =

=

1.* x 10

=

1**

d $00 $0


= 0.$

= 0.$ x ( *#.# ) x ( $# ) = 11002;


n = 0.;

#

sehingga di+eroleh 9 b = # x 11002;=

*8#0

0.; x 2#0 x 10I 21*



Vaa geser +ada ens balo adalah = 1.* x 10=

1**

$0 $0


6idang geser Rn

6idang geser Rn

3eb 6alo Rn

%lens 6alo Rn

R n

siu rsiu

. '.w

(b.dI) . y



n = ;0.#8= 1.2; 2 buah baut

0.1#




n = 2$0.#* = 1.#; 2 buah baut1*8.2*


n = 2$0.#*= .$ # buah baut

0.1#

&. <>6C7V>7 6>LBK ?7DCK DA7V>7 KBLB D? 6>V?>7 P?7VV?/ K>7>7 L>7'>? 1

#(BE7 P8U);7 B(J( V(0 !;0

U#(S; P8U);7

!;&ES;

(&U&E

h b5 t ro ;3

mm mm mm mm mm

Balok ;nduk <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++Kolom <)6++=6++ 6++ 6++ 19 21 22 21.@+ ??,?++

= 1.2 D : 1.? L = 1.2 x ( ;.2

= 1.2 D : 1.? L = 1.2 x ( 1.21


+eser *

1. = 0.$ x ( 0.#0

2. = 2 x 0.1# =

Tu23u 1. " 0.$ x ( 2.#0

2. " 0.$ x ( 2.#0

&arik *

= 0.$ (0.$ x \cq) x m x >

= 0.$ x ( 0.$ ) : ( 82$ ) x 28.; ) =


Sambungan sebuah Balok ;nduk dengan Kolom, antara Balok ;nduk $<) 9-+ 3 9-+% dengan Kolom $<)baut (92- N1> mm dengan ulir pada bidang geser, diketahui momen ujung sebesar >9.92 km $!% dan dan [email protected] k $7%.

tw

'mH 'mH

u

u

6idang geser Rn

6idang geser Rn

3eb 6alo Rn

%lens 6alo Rn

R n



,eritn(an -ik ,enyambn( tas dan Ba/a *

Dioba dua buah baut +ada masing5masing +rol siu sehingga

d =

=2#.0* x 10

=2#0*0.80

=2' 2 x 11.$1 2*.02

ara baut terhada+ ens atas balo = v ( $00Vunaan +rol siu 80 x 120 x 10 sehingga 9

w $ 5 5 = $ 5 10 5


' = =

2#.0* x 10=

2#0*0

d $00 $0


= 0.$

= 0.$ x ( #8121.* ) x ( $# ) = 12;;28


n = 0.;

#

sehingga di+eroleh 9 b = # x 12;;28=

$1;

0.; x 2#0 x 10I 21*



Vaa geser +ada ens balo adalah = 2#.0* x 10 = 2#0*1$0 $0


n = *8.$= 0.;8 2 buah baut

0.1#




n = 1;.#8= 1.1$ 2 buah baut1*8.2*


n = 1;.#8= 2.* # buah baut

0.1#

siu rsiu

. '.w

(b.dI) . y




. <>6C7V>7 6>LBK >7>K DA7V>7 6>LBK >7>K

#(BE7 P8U);7 B(J( V(0 !;0

U#(S; P8U);7!;&ES;

(&U&E

h b5 t ro ;3

mm mm mm mm mm

Balok (nak <)9++=9++ 9++ 9++ 1+ 1- 1 11>.+ 2+,6++Balok ;nduk <)9-+=9-+ 9-+ 9-+ 12 1> 2+ 1@9.>+ 6+,9++

#ahanan #umpu pada bagian <eb dari balok.

8n = 0.$ (2.# x \c) x d x tp

= 0.$ x ( 2.# ) x ( #10 ) x ( 1;

'ahanan Veser 6aut dengan dua bidang geser 9

= 0.$ (0.$ x \cq) x m x >

= 0.$ x ( 0.$ ) x ( 82$ ) x ( 2

Perhitungan umlah 6aut

3% 00 n = 22.0#= 1.$

1#.02

Perisa Veser 6alo +ada 3% 00 9

= 180 x ( 10 ) = 1800

= ( 180 5 2.$ x ( 1; : 2 ) )= #0 x ( 10 )

= ( #0 5 0.$ ( 1; : 2 ) )

= #10 ( $0.$0 )

= 0.* x ( #10 ) x 12$.00 )

=

= 1*$0 : ( 2#0 ) x ( #00 ) =

= #0;*$0= #

10000

#n = 0.$ x ( #0.; ) = 0.28 ton

#. <>6C7V>7 KBLB DA7V>7 PB7D><?

tw

'mH 'mH

Sambungan antara Balok (nak $<) 9++ 3 9++% dengan Balok (nak $<) 9++ 3 9++% dengan menggunak9++% adalah sebesar 22.6+ ton dan &utu Baja Proil BJ 61.

R n

Ag mmI

An

Agt

Ant

\c.>nt

0.*\c.>n

Karena \c.>nt 0.*\c.>n maa 9

&n 0.*\c.>n : \ .>gt



'/CK<? 6>7VC7>7 VADC7V

(K(

;E8S;(

BE8(# 8E(KS; #E8)(K#U8;ykgm ton.mm

?,@-+ >6.++ 6++ 29.>>

19,?++ 19?.++

) x ( 10 ) = 1#0220.00= 1#.02 ton-baut (3% 00)

10000

) x ( 28.; ) = 1$##.#= 1.$ ton-baut (3% 00)

10000

2 buah baut

x ( 10 ) = 12$.00

= #00

x ( 10 ) = $0.$0

= 200$ 7

= 1*$0 7

0;*$0 7

0.; ton

2.;; ton BK

)8(&E

B(#(0'mH

an baut (92- N1> mm, reaksi teraktor Balok (nak $<) 9++ 3

mmI

mmI

mmI

Pul



(K(

;E8S;(BE8(#

;ykgm

19,?++ 19?.++ 91+

22,6++ 1@2.++

) : 1.* x ( 1#1.1 ) = 1*.*#= 1.** 7m

10

) : 1.* x ( ;.$1 ) = 2$0.#* 7

) x ( 82$ ) x ( 28.; ) = 01.8$ = 0.1# k1000

1#0.28 7

) x ( #10 ) x ( 1;.00 ) x ( 12 ) = 1*82*#=

1*8.2*

1000

) x ( #10 ) x ( 1;.00 ) x ( 1; ) = 2**#18=

2**.#2

1000

11$08= 11.$1 k

1000

120.; mm $00 mm

5 $0 ) = $ mm

11.00 = $#.00 mm

ah 9

.*0= * k

++ 3 6++% dengan &utu Baja Proil BJ 61 dan menggunakan1.91 $7%, gaya geser ujung yang bekerja sebesar @.@+ k $!%

)8(&E

B(#(0'mH



7mm

8

= 1*.*# mm0

= ;0 k

( 1*8.2* 7 )

0.1# 7) sehingga 9

(K(

;E8S;(BE8(#

;y

kgm

19,?++ 19?.++ 2>?

22,6++ 1@2.++

) : 1.* x ( 1#1.1 ) = 1*.*= 1.* 7m

10

) : 1.* x ( ;.$1 ) = 2$0.#* 7

u bidang geser (Rn =

++ 3 6++% dengan &utu Baja Proil BJ 61 dan menggunakan1.91 $7%, gaya geser ujung yang bekerja sebesar @.@+ k $!%

)8(&EB(#(0

'mH



) x ( 82$ ) x ( 28.; ) = 01.8$= 0.1# k

1000

1#0.28 7

) x ( #10 ) x ( 1;.00 ) x ( 12 ) = 1*82*#=

1*8.2*

1000

) x ( #10 ) x ( 1;.00 ) x ( 1; ) = 2**#18=

2**.#2

1000

11$08= 11.$1 k

1000

120.#0 mm $00 mm

5 $0 ) = $ mm

11.00 = $#.00 mm

ah 9

.20

= * k

7mm

1$= 1*.* mm

0

= ;0 k



( 1*8.2* 7 )

0.1# 7) sehingga 9

(K(

;E8S;(BE8(#

;y

kgm

19,?++ 19?.++ 2>?

22,6++ 1@2.++

) : 1.* x ( 80.; ) = 2#0.*1= 2#.0* 7m

10

) : 1.* x ( *.$2 ) = 1;.#8 7

) x ( 82$ ) x ( 28.; ) = 01.8$= 0.1# k

1000

1#0.28 7

) x ( #10 ) x ( 1;.00 ) x ( 12 ) = 1*82*#=

1*8.2*

1000

) x ( #10 ) x ( 1;.00 ) x ( 1; ) = 2**#18=

2**.#2

1000

11$08

u bidang geser (Rn =

++ 3 6++% dengan &utu Baja Proil BJ 61 dan menggunakan+.9> $7%, gaya geser ujung yang bekerja sebesar @1.21 k $!%

)8(&EB(#(0

'mH



1000.

;1.#8 mm $00 mm

5 $0 ) = $ mm

11.00 = $#.00 mm

ah 9

.80= #8 k

7mm

= 2#0.*1 mm

0

= *; k

( 1*8.2* 7 )

0.1# 7) sehingga 9u bidang geser (Rn =



(K(

;E8S;(BE8(# 8E(KS; #E8)(K#U8

;y

kgm ton.mm

?,@-+ >6.++ 9> 22.+6 19,?++ 19?.++

) x ( 10 ) = 1#0220.00= 1#.02 ton-baut (3% 00)

10000

) x ( 28.; ) = 1$##.#= 1.$ ton-baut (3% 00)

10000

2 buah baut

x ( 10 ) = 12$.00= #00

x ( 10 ) = $0.$0

= 200$ 7

= 1*$0 7

0;*$0 7

0.; ton

22.0# ton BK

)8(&EB(#(0

'mH

n baut (92- N1> mm, reaksi teraktor Balok (nak $<) 9++ 3

mmImmI

mmI

Pul



k

k

kontrol kekuatan balok dan kolom - lrfd1

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