Hitungan Baja Xls

7/25/2019 Hitungan Baja Xls

1/133

Data - data Konstruksi :

Kuda - kuda type = D

= 30 Bentang Kuda - kuda ( B ) = 20 m

Panjang Gudang ( L ) = m

!arak antar ko"om = #$ m

Penutup atap = %eng ge"om&ang

%i'at dinding = ertutup

inggi ko"om ( ) = m

D*+,L KD+-KD+

.*/+/+ ++P

Kemiringan atap ( )

P1 X12

A

P2

D

P3

E

P4

G

P5

I

P6

K

P7

M

A1

V1D1

B1C F H J L N

B2 B3 B4 B5 B6 B7

V2

V3

V4

V5

V6

D2

D3 D4

D5

D6

A2

A3

A4 A5

A6

A7

2000

GGG

GGG

GGG

GGG

GGG

GGG

GGG

GGG

GGG

GGG

KK

KK

KK

KK

KK

KK

KK

KK

N

4000

2

000

200

200

200


2/133

P*.1,/G+/ P+/!+/G B++/G :

Batang &aa ( B ) =

B : L : = 4 &L = 20 m

B =20

= 2#$ m4

Batang atas ( + ) =

+ : B : os 5 os 30 = 0#466B = 2#$ m

+ =2#$

= 2#447 m0#466

Batang 8ertika" ( 9 ) =9 = 97 = 2#$ # tg 30

= 2#$ # 0#$77

= #3 m

92 = 96 = 2#$ # tg 30 = 2#$ # 0#$77 # 2

= 2#44$ m

93 = 9$ = 2#$ # tg 30 = 2#$ # 0#$77 # 3

= #324 m

9 = 2#$ # tg 30 = 2#$ # 0#$77 #

= $#770 m

Batang diagona" ( D ) =

D = D6 = ( 2#$ ; #3 = ( 6.250 ; 2.081

= ( 8.331

= 2.886 m

D2 = D6 = ( 2#$ ; 2#446 = ( 6.250 ; 8.331

= ( 14.581

= 3.818 m

D3 = D = ( 2#$ ; #324 = ( 6.250 ; 18.727

= ( 24.977

= 4.998 m

) )

)

) )

)

) )

)


3/133

P*.1,/G+/ D,o&a memakai gording 2

1 = 20 mm

B = $$ mm

D = 7 mm

?@ = 60#7 mm

?y = # mm

Berat pro'i" = 3#

A Berat gording dan atap # ( K )

gording = 3#

seng ge"om&ang = 0#0 ;C = 23#

Baut 0 = 2#3 ;

2$#7 C tota" = 26

Cy = C tota" # os 30 = 26 # 0#466

= 22#$6

= # Cy # L = 0#2$ # 22#$6 # #$ = $6#EE362$

= # C tota" # %in 5 # L = 0#2$ # 26 # # #$ = 0#2$ # 26 # 0#$ # 20#2$

= 32#E062$

A Be&an angin pada atap # ( K2 )

Be&an angin = 0kemiringan atap = 30

dipiak angin = 0#02 # 5 - 0#

= 0#02 # 30 - 0#

= 0#6 - 0# = 0#2

di&e"akang angin 2 = -0#

Dari ta&e" pem&e&anan pengaru angin ditentukan :

&e&an dipiak angin ? = # + # 0

= 0#2 # 2#447 # 0

= 23#0E6442

&e&an di&e"akang angin ?2 = 2 # + # 0

Kg#m

Kg#m

Kg#mKg#mKg#mKg#m Kg#m

Kg#m


4/133

= -0# # 2#447 # 0

= -6#4E376

= # ? # L = 0#2$ # 23#0E644 # #$ = $4#$43

= # ?2 # L = 0#2$ # -6#4E34 # #$ = -6#E7

= 0

A Be&an idup # ( K3 )

untuk &e&an ini ditentukan P = 00 Kg terpusat di tenga &atang

= # 00 # # #$= 0#2$ # 00 # 0#466 # #$

= E7#2$

= # 00 # # ( #$ / 2 )= 0#2$ # 00 # 0#$ # 2#2$

= 24#2$

Kom&inasi pem&e&anan ( P ditiadakan saat angin &ekerja )

Kombinasi K1 + K2


5/133

ota" m

JADI,


6/133

= ( 1.226 ; 0.052

= ( 1.278

= 1.130

iin = @ L =

@ $0 = #4

2$0 2$0

JADI, J iin

#30 J #4

OOOOKKKKKEEEEYYY

P*.1,/G+/ P*mH

Penggantung P"a'ond = 4 Kg#>mH

inggi ko"om ( ) = $ m

!arak gording = +=

2#447= 0#6 m

L #$

M Pem&e&anan

&e&an gording

= # 3#2 # #$ = 237#600 Kg#>mH

&e&an atap dan p"a'ond

= 24 # 2#447 # #$ = 363#7 Kg#>mH

&e&an kuda - kuda

= 27 # 2#447 # #$ = 3$0#7$ Kg#>mH

sam&ungan dan per"engkapan

= 30 3$0#7$ = 0$#22$ Kg#>mH

&e&an idup = 00#000 Kg#>mH ;

P = $7#32 Kg#>mH

muatan tak terduga 0 = P @ 0

= $7#37 @ 0

= $#732

P tota" = $#732 ; $7#37

= 273#0E

RA = R = !1 " !2 " !3 " !4 " !5 " !6 " !7 " !8 " !9

) )

Kemiringan atap ( )


7/133

= 636#$2 ; 273#0$ ; 273#0$ ; 273#0$ ; 273#0$ ; 27

2

273#0$ ; 636#$2

=04#3E0

2

= $0E2#E$ Kg

M Peritungan pem&e&anan angin

Be&an angin = 0

kemiringan atap = 30

dipiak angin = 0#02 # 5 - 0#= 0#02 # 30 - 0#

= 0#6 - 0# = 0#2



&e&an tekan angin ? = # 0

= 0#2 # 0

= 4

&e&an isap angin ?2 = 2 # 0

= -0# 0

= -6

Pengaru angin tekan ( ? ) = 4 (titik &uu")

+ =2#447

@ #$ @ 4 = $#E6304$ Kg2

D = 2#447 @ #$ @ 4 = 03#E260E7 Kg

* = 2#447 @ #$ @ 4 = 03#E260E7 Kg

G = 2#447 @ #$ @ 4 = 03#E260E7 Kg

, =2#447

@ #$ @ 4 = $#E6304$ Kg2

Pengaru angin isap ( ?2 ) = -6 (titik &uu")

, =2#447

@ #$ @ -6 = -03#E260E7 Kg2

K = 2#447 @ #$ @ -6 = -207#4$2E Kg

Kg#m

Kg#m

Kg#m


8/133

< = 2#447 @ #$ @ -6 = -207#4$2E Kg

= 2#447 @ #$ @ -6 = -207#4$2E Kg

B =2#447

@ #$ @ -6 = -03#E260E7 Kg2


9/133

.*+K%, G++ %,


10/133

D6 0 000 0

+/G,/ K+/+/ 1,%+P

B++/G %K+L+ B*%+. G++ B++

/+m ) G++ ( Kg) +.,K ( ; )

+ 0#62 000 0620

+2 0#62 000 0620

+3 0#62 000 0620

+ 0#62 000 0620

+$ E#2E 000 E2E0+6 #E$ 000 E$0

+7 #$E0 000 $E0

+4 7#27 000 7270

B 0 000 0

B2 0 000 0

B3 0 000 0

B 0 000 0

B$ #6 000 0

B6 E#E 000 0B7 3#4 000 0

B4 3#4 000 0

9 0 000 0

92 0 000 0

93 0 000 0

9 7#E6 000 7E60

9$ $#3 000 $3

96 2#6 000 260

97 0 000 0

D 0 000 0D2 0 000 0

D3 0 000 0

D E#2 000 E200


11/133

D$ 7#0 000 700

D6 $#32 000 $320

G++ .+/GK+ B++/G +K,B+ +/G,/ ( K


12/133

D2 0 3$06 0 0 0

D3 0 $E 0 0

D 0 0 E200 0 E200

D$ 0 0 700 0 700

D6 0 0 $320 0 $320

G++ .+/GK+ B++/G ( K


13/133

96 0 $$0 260 0 0E0

97 0 0 0 0 0

D 0 30 0 2634 0

D2 0 0 0 3$06 0

D3 0 $340 0 $E 0

D 0 $340 E200 0 3420

D$ 0 0 700 0 2E00

D6 0 30 $320 0 220

PERHITUNGAN PROFIL KUDA - KUDA

a!an" a!as A

/ama &atang = +O +2O +3O +O +$O +6O +7O +4

Gaya &atang m

Di>o&a menggunakan pro'i" 00 # 00 # 0 Baut = mm

,@ = 77#0

= E#2

i@ = 3#0 >m

s = 24#2 2#42 >m

e tota" = 2#42 ; 0#$ = 3#32 >m

iy = 2 ( ,@ ; )

= 2 ( 77#0 ; E#2 ( 3#32 )

= 777#2606 >mH

iy = ( 777#2606 Q 2 # ) = ( 777#2606 Q 2 # E#2 ) = ( 20#2$ ) = #EE >m

maka i@ = 3#0 >m ada"a i minimum ( i min )

e = 2# # 0

>m>m

mm

(e))


14/133

R =LK

=360#4

= 4#64i min 3#0

=3# # 2# # 0

=2070$60

= 4#64 04$#E2

# $%=

6E#E7= $2#27 Kg#>mHs 2#42

P=

E76= $#644 Kg#>mH

2 2 # E#2

I J I kt

$#644 Kg#>mH J $2#27 Kg#>mH

OOOOKKKKKEEEEYYY

Pro'i" 00 # 00 # 0 &isa dipakai#

a!an" ba#a$

/ama &atang = BO B2O B3O BO B$O B6O B7O B4Gaya &atang m

egangan iin = 600 Kg#>mH

Di>o&a menggunakan pro'i" E0 # E0 # E Baut = mm

,@ = 6#0

= $#$ 2 = 3

i@ = 2#7 >m

s = 2$# 2#$ >m

e'k = 3 - 2 ( #$ # 0#6 )

= 2E#2

R =LK

=32#$

= #0$i min 2#7

P=

30320= 034#3$6 Kg#>mH

e'k 2E#2

I J S

034#3$6 Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY

8kt=& . '

@

kt=

egangan yang tim&u" ( )

=

>m>m

mm

>m

=


15/133

Pro'i" E0 # E0 # E &isa dipakai#

a!an" %ia"ona& D

B++/G +.,K

/ama &atang = DO D$O D6

Gaya &atang mH

Di>o&a menggunakan pro'i" 40 # 40 # 0 Baut = T mm

,@ = 47#$

= $# 2 = 30#2

i@ = 2# >m

s = 23# 2#3 >m

R =LK

=$0#E

= 22#0i min 2#

R g = & ( e Q 0#7 e ) = 3# ( 2# # 0 Q 0#7 # 200 )= #06

R=

22#0= 2#026E336 U R g #06

untuk R U maka = 2#34 # R s

= 2#34 # 2#026E336 = #436$E

e'k = 30#2 - 2 ( #$ # #$ )

= 2$#7

e'k # V=

25.7 # 600= 4$2#37 Kg

#4

I M !4$2#37 Kg M 3420 Kg

OOOOKKKKKEEEEYYY


B++/G *K+/

/ama &atang = DO D2O D3Gaya &atang mH

>m>m >m

mm

Rs=

>m

=


16/133


,@ = 6#0

= $#$ 2 = 3

i@ = 2#7 >m

s = 2$# 2#$ >m

R =LK

=$0#E

= E7#0Ei min 2#7

e'k = 3 - 2 ( #$ # #$ )

= 26#$

P=

EE7= 32#72

e'k 3

I J S32#72 Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY


a!an" '()!i*a&

B++/G +.,K

/ama &atang = 9$O 96

Gaya &atang mH

Di>o&a menggunakan pro'i" 40 # 40 # 2 Baut = 20 mm,@ = 02#0

= 7#E 2 = 3$#4

i@ = 2#3E >m

s = 2# 2# >m

R =LK

=62#7

= 26#34i min 2#3E

e'k = 3$#4 - 2 ( #$ # #$ )

= 3#3

P 220

>m>m >m

mm

>m

=

>m>m >m

mm

>m


17/133

e'k 3#3#

I J S

70#3E Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY


B++/G *K+/

/ama &atang = 92O 93O 9

Gaya &atang mH


,@ = 02#0

= 7#E 2 = 3$#4i@ = 2#3E >m

s = 2# 2# >m

R =LK

=62#7

= 26#34i min 2#3E

R g = & ( e Q 0#7 e ) = 3# ( 2# # 0 Q 0#7 # 200 )= #06

R=

26#34= 2#3$77 U

R g #06


= 2#34 # 2#3$77 = $#60$E347

e'k = 3$#4 - 2 ( #$ # #$ )

= 3#3

e'k # V = 31.3 # 600 = 4E33#343 Kg $#606

I M !4E33#343 Kg M $30 Kg

OOOOKKKKKEEEEYYY


>m

>m >m

mm

Rs=

>m

=


18/133

PERHITUNGAN AUT PADA TIAP TITIK UHUL

Pro'i" se"uru kuda-kuda 00 # 00 # 0

te&a" p"at = 0 mm

diameter &aut = mmdiameter "u&ang = ; = $ mm

tegangan geser iin V = 600 Kg#>mH

/g = 0#2$ # & # d H # V= 0#2$ # 3# # #$ H # 600

= 2426 Kg

/tp = d # t # Vt # V /g M (= #$ # #0 # 0#E # 600 2426 M

= 260 OOOOKKKKKEEEE


19/133

!


20/133

n =gaya +

=6$E

= 2#246 3 &/g 2426

Batang +$

n =gaya +$

=4$3

= 2#EE 3 &/g 2426

Batang 9

n =

gaya 9

=

$30

= #4E0 3 &/g 2426

!


21/133

Batang B

n =gaya B

=203$0

= 7#20 $ &/g 2426

Batang B$

n =gaya B$

=3$0

= #7$E &/g 2426

Batang 9

n =gaya 9

=$30

= #4E0 3 &/g 2426

Batang D3

n =gaya D3

=EE7

= 3#$2E 3 &/g 2426

Batang D

n =gaya D

=3420

= #3$2 3 &/g 2426


22/133

/tp 604#

P*.1,/G+/ KLmH



!arak gording = + = 3#604 = 0#40 mL #$

M Pem&e&anan

&e&an gording

= # 3#2 # #$ = 237#600 Kg#>mH


= 24 # 3#604 # #$ = $#604 Kg#>mH

&e&an kuda - kuda

= 27 # 3#604 # #$ = 34#372 Kg#>mH


= 30 34#372 = 3#$2 Kg#>mH

&e&an idup = 00#000 Kg#>mH ;

P = 362#0E Kg#>mH

Peritungan reaksi ko"om : P

P P P

P P

P P

.+ .+

20

+,A = 0

Kemiringan atap ( )

.


23/133

- .B # 2$ ; P # 2#E ; P # E ; P # $#6 ; P

P # E# ; P # 6 ; P # 3# ; = 0

- .B # 2$ ; P ( 2$ ; 2#E ; E ; $#6 ; 3 ; E#

- .B # 2$ ; P ( 3 )

.B =P # 3

=3

P = #$ P2$ 2$

.+ = .B = #$ # P

= #$ # 362#0E

= 62E#2

Pendimensian ko"om

tinggi ko"om ( ) = $ m $00 mmV ko"om Bj 37 = 600 Kg#>mH

/ = 62E#2

Di>o&a menggunakan pro'i" ? 3$0 # 2$0

t = E mm

t 2 = mm

+ = 0#$

i @ = #6 >mH

i y = 6 >mH

?@ = 240

?y = 2E2

ara tegak "urus sum&u @

k =

R @ =L ky

=k L

= # $00

= 3#26$7$i @ i @ #6

R g = & ( e Q 0#7 e )

= 3# ( 2# # 0 Q 0#7 # 200 )= #06

R @=

3#27= 0#304

R g #06

untuk 0#43 J R s J maka

?@ =#

=#

= #0E4#$E3 - R s #$E3 - 0#304

T

n@ =& * +

=3# 2# # 0 0#$

=2#0*;00E

/ # 62E#2 # 3#27 74476

>m

>m->m-

Rs =

#

R @


24/133

n@=

2E2#32= #003

n@ - 2E2#32 -

B@ = 0#6 ; 0# #


25/133

0#043 Kg#>mH J 600 Kg#>mH

> ) /;

mH

0#$ 240

3$E#676 Kg#>mH J 600 Kg#>mH

Pro'i" ? 3$0 # 2$0 &isa dipakai#

Pe"at dasar ko"om pada pondasi

kip = #$36 # 0 Kg= 6#$2

Be&an aksia" ( P ) = 62E#2 = 3$#24

P W = 3 ksi

p = 0#3$ # P W

= 0#3$ # 3

= #0$ kgi

+ per"u = P Q P

= 3$#24 Q #0$

= 246#E3$

Pro'i" yang digunakan ada"a ? 3$0 # 2$0

+ = 0#$

d = 3 >m

& = 2$ >m

menentukan "e&ar ( B ) dan panjang ( / ) pe"at dimana arus memenui syarat :

B @ / U $$E0#46 >mH

untuk pemi"ian B dan / sedemikian ingga per&edaan ( se"isi ) diantara keduanya mendekati :

0#E2 d - 0#40 &

0#E2 # 3 - 0#4 # 2$ = #24 >m

di>o&a &er&agai gaya B dan / se&agai &erikut :

B @ / U $$E0#46 >mH

N as s(&isi$

7$ 76 $700

7$ 74 $6E 3

m >m

>m


26/133

7 40 $640 E

70 43 $70 3

gunakan B = 70 dan / = 43

tekanan aktua" ada"a

p =!

=62E#2

= #0$$ Kg#>mHB # / 70 # 43

itung < dan /

< =/ - 0#E$ d

=43 - 0#E$ # 3

= 2$#3$ >m2 2

/ =/ - 0#40 d

=70 - 0#40 # 2$

= 2$ >m2 2

itung te&a" pe"at ( tp )tp = ( ( P Q 0#2$ y )

= 2$ ( #0$$ Q 0#2$ # 200 )X

= #04 >m

70 # # 43


27/133

untuk mengetaui kuat atau tidaknya &aut menaan suatu &e&an arus memenui syarat &erikut :

V 0#4 @ 600 Kg#>mH

P

62E#2

2 # 0#2$ # 3# 2# . 3

22$#E3 Kg#>mH Y 0#4 @ 600 Kg#>mH

22$#E3 Kg#>mH Y 240 Kg#>mH

OOOOKKKKKEEEEYYY

2 # Q # & . . 3


28/133

P8

O

P9 X12

B

B8P

7

A8

KK


29/133


30/133


31/133


32/133

66 # $0 - # 36

00 # $0 - # 3#2

L-

L-


33/133


34/133

#0$ ; 273#0$ ;


35/133

G ( Kg )

*K+/ ( - )

2770

4660

$$$0

20

20

$$$0

4660

27700

0

0

0

0

0

0

0

0

$$0

30

E330

30

$$0

0

30

0

$340

$3400

30


36/133


37/133

0

G ( Kg )

*K+/ ( - )

0

0

0

0

00

0

0

0

0

0

0

600

EE03400

3400

0

0

0

0

0

0

0

00

0

0


38/133

0

0


39/133

3$06

$E

0

0

0

%, ( Kg )

*K+/

( - )

E76

$33$

04E7

6$E

4$3

4E03

E373

E403

0

0

0

0

0

0

0

0

0

264$7$0

$30

0


40/133

0

0

$74

766

EE7

0

0

0


41/133

6E#E7 Kg#>m


42/133


43/133


44/133


45/133

260

YY


46/133

67 3 &

70 3 &

70 3 &


47/133

00 3 &

$4 3 &

70 3 &


48/133


49/133


50/133

# 3 ;

; 6 ; 3# )=0

= 2E2#32


51/133

= E#373

>mH


52/133


53/133

= 2$0 ( 2$0 ; Q 2 ) M

= 33000

2


54/133


55/133


56/133


57/133


58/133


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60/133


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62/133


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68/133


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70/133


71/133


72/133


73/133


74/133


75/133

706#4

30#2


76/133


77/133


78/133

0#003206440E07$

66#247607$363

7E#$7262306E

2$#460230$432


79/133

$07#43

0#E66

4#4377$E

32#6447746733

06$#$3770$62


80/133

3#$

#$ diameter "u&ang


81/133


82/133

Data - data Konstruksi :

Kuda - kuda type = D

= 30 Bentang Kuda - kuda ( B ) = 2$ m

Panjang Gudang ( L ) = 0 m

!arak antar ko"om = #$ m

Penutup atap = +s&es Ge"om&ang

%i'at dinding = ertutup


D*+,L KD+-KD+

.*/+/+ ++P

P*.1,/G+/ P+/!+/G B++/G :

Batang &aa ( B ) =

B : L : = 4 &L = 2$ m

B =2$

= 3#2$ m4

Batang atas ( + ) =

+ : B : os 5 os 30 = 0#466B = 3#2$ m

Kemiringan atap ( )

P1 X12

A

P2

D

P3

E

P4

G

P5

I

P6

K

P7

M

P8

O

P9 X12

B

A1

V1D1

B1 B8C F H J L N P

B2 B3 B4 B5 B6 B7

V2

V3

V4

V5

V6

V7

D2

D3 D4

D5

D6

A2

A3

A4 A5

A6

A7

A8

2000

GG

G

GGG

GGG

GGG

GGG

GGG

GGG

GGG

GGG

GGG

KK

KK

KK

KK

KK

KK

KK

KK

KK

N

4000

2000

200

200

200 200


83/133

Batang 8ertika" ( 9 ) =

9 = 97 = 3#2$ # tg 30 = 3#2$ # 0#$77

= #403 m

92 = 96 = 3#2$ # tg 30 = 3#2$ # 0#$77 # 2

= 3#606 m

93 = 9$ = 3#2$ # tg 30 = 3#2$ # 0#$77 # 3

= $#0E m9 = 3#2$ # tg 30

= 3#2$ # 0#$77 #

= 7#23 m

Batang diagona" ( D ) =

D = D6 = ( 3#2$ ; #403 = ( 9.766 ; 3.251

= ( 13.017

= 3.608 m

D2 = D$ = ( 3#2$ ; 3#604 = ( 9.766 ; 13.017

= ( 22.783= 4.773 m

D3 = D = ( 3#2$ ; $#0E = ( 9.766 ; 29.261

= ( 39.027

= 6.247 m

P*.1,/G+/ D,o&a memakai gording 2

1 = 20 mm

B = $$ mm

D = 7 mm

?@ = 60#7 mm

?y = # mm

Berat pro'i" = 3#

A Berat gording dan atap # ( K )

gording = 3#

as&es ge"om&ang = 0#3 ;

C = 23#7

Baut 0 = 2#37 ;

26# C tota" = 26

Cy = C tota" # os 30

= 26 # 0#466= 22#$6

= # Cy # L = 0#2$ # 22#$6 # #$ = $6#EE362$

= # C tota" # %in 5 # L = 0#2$ # 26 # # #$ = 0#2$ # 26 # 0#$ # 20#2$

= 32#E062$

A Be&an angin pada atap # ( K2 )

Be&an angin = 0


dipiak angin = 0#02 # 5 - 0#

= 0#02 # 30 - 0#

= 0#6 - 0# = 0#2

) )

)

) )

)

) )

)

Kg#m

Kg#mKg#mKg#mKg#mKg#m Kg#m

Kg#m


84/133



85/133


&e&an dipiak angin ? = # + # 0

= 0#2 # 3#60E # 0

= 24#46436024

&e&an di&e"akang angin ?2 = 2 # + # 0

= -0# # 3#60E # 0

= -$7#736720$$

= # ? # L = 0#2$ # 24#464 # #$

= 73#073037

= # ?2 # L = 0#2$ # -$7#736720$$3 # #$ = -6#6

= 0

A Be&an idup # ( K3 )

untuk &e&an ini ditentukan P = 00 Kg terpusat di tenga &atang

= # 00 # # #$

= 0#2$ # 00 # 0#466 # #$= E7#2$

= # 00 # # ( #$ / 2 )= 0#2$ # 00 # 0#$ # 2#2$

= 24#2$

Kom&inasi pem&e&anan ( P ditiadakan saat angin &ekerja )

Kombinasi K1 + K2


86/133

*G+/G+/ +/G *.!+D,

=mH J 600A.AN

K/.L L*/D+/

e = 2# # 0

y =

@C # #

;

@P # os 5 #

34 * # , @ 4 * # , @

= @ 26 # 0#466 # $0 ; @ 00 # 0#46634 2# # 0 # 36 4 2# # 0

=36E3

;74E

2E3$2E#6 366E#2

= 0#03 ; 0#2$ = 0#224

@ =

@C # #

;

@P # %in 5 #

34 * # , @ 4 * # , @

=

@26 # 0#$00 # $0

;

@00 # 0#$00

34 2# # 0 # 3#2 4 2# # 0

=232

;$$6

34#$ 3$$#$64

= 0#06 ; #06 = #07

= ( @ ; y = ( #07 ; 0#224 = ( 1.226 ; 0.052

= ( 1.278

= 1.130

iin = @ L =

@ $0 = #4

2$0 2$0

JADI, J iin

#30 J #4

OKEY

Kg#>m

JADI,


87/133

P*.1,/G+/ P*mH



!arak gording = +=

3#60E= 0#40 m

L #$

M Pem&e&anan

&e&an gording

= # 3# # #$ = 2#200 Kg#>mH


= 24 # 3#60E # #$ = $E#$4 Kg#>mH

&e&an kuda - kuda

= 27 # 3#60E # #$ = 34#34 Kg#>mH


= 30 34#34 = 3#$3 Kg#>mH

&e&an idup = 00#000 Kg#>mH ;

P = 370#72 Kg#>mH

muatan tak terduga 0 = P @ 0

= 370#774E434 @ 0

= 37#072

P tota" = 37#072 ; 370#74

= $07#7E0

RA = R = !1 " !2 " !3 " !4 " !5 " !6 " !7 " !8 " !9= 7$3#4E40E ; $07#7E ; $07#7E ; $07#7E ; $07#7E ; $07#7E

2

$07#7E ; 7$3#4E40E

=2062#34

2

= 603#$E Kg

Kemiringan atap ( )


88/133

M Peritungan pem&e&anan angin

Be&an angin = 0


dipiak angin = 0#02 # 5 - 0#

= 0#02 # 30 - 0#

= 0#6 - 0# = 0#2



&e&an tekan angin ? = # 0= 0#2 # 0

= 4

&e&an isap angin ?2 = 2 # 0

= -0# 0

= -6

Pengaru angin tekan ( ? ) = 4 (titik &uu")

+ =3#60E

@ #$ @ 4 = 6#E$34062 Kg2

D = 3#60E @ #$ @ 4 = 2E#E07622$ Kg

* = 3#60E @ #$ @ 4 = 2E#E07622$ Kg

G = 3#60E @ #$ @ 4 = 2E#E07622$ Kg

, =3#60E

@ #$ @ 4 = 6#E$34062 Kg2

Pengaru angin isap ( ?2 ) = -6 (titik &uu")

, =3#60E

@ #$ @ -6 = -2E#E07622 Kg2

K = 3#60E @ #$ @ -6 = -2$E#4$22$ Kg

< = 3#60E @ #$ @ -6 = -2$E#4$22$ Kg

= 3#60E @ #$ @ -6 = -2$E#4$22$ Kg

B =3#60E

@ #$ @ -6 = -2E#E07622 Kg2

Kg#m

Kg#m

Kg#m


89/133


90/133

+/G,/ K,., *K+/

B++/G %K+L+ B*%+. G++ B++/G ( Kg

/+m ) G++ ( Kg) +.,K ( ; ) *

+ $#73 000 0

+2 3#0E6 000 0

+3 3#0E 000 0

+ 3#0E3 000 0

+$ 3#$3$ 000 0

+6 3#$3$ 000 0

+7 3#$3$ 000 0

+4 3#$3$ 000 0

B 7#6$2 000 76$#E

B2 7#6$2 000 76$#E

B3 6#24 000 624#

B #$E$ 000 $E$#

B$ 3#062 000 306#4

B6 3#062 000 306#4

B7 3#062 000 306#4

B4 3#062 000 306#4

9 0 000 0

92 0#440 000 0

93 #76$ 000 09 2#6$0 000 0

9$ 0 000 0

96 0 000 0

97 0 000 0

D #7$E 000 0

D2 3#0$7 000 0

D3 3#062 000 0

D 0 000 0

D$ 0 000 0

D6 0 000 0


91/133

+/G,/ K+/+/ 1,%+P

B++/G %K+L+ B*%+. G++ B++/G ( Kg

/+m ) G++ ( Kg) +.,K ( ; ) *

+ 7#042 000 7042#2

+2 7#042 000 7042#2

+3 7#042 000 7042#2

+ 7#042 000 7042#2

+$ 6#E6E 000 6E6#E

+6 #$03 000 $02#4

+7 E#732 000 E732#3

+4 #$ 000 $#3

B 0 000 0

B2 0 000 0

B3 0 000 0

B 0 000 0

B$ 3#06E 000 0

B6 6#24 000 0

B7 E#20$ 000 0

B4 E#20$ 000 0

9 0 000 092 0 000 0

93 0 000 0

9 $#36 000 $36#3

9$ 3#$ 000 3$#

96 #766 000 766#3

97 0 000 0

D 0 000 0

D2 0 000 0

D3 0 000 0

D 6#33 000 633#3

D$ 6#2 000 623#$

D6 3#$$2 000 3$$2#


92/133

G++ .+/GK+ B++/G +K,B+ +/G,/ ( K


93/133

G++ .+/GK+ B++/G ( K


94/133

PERHITUNGAN PROFIL KUDA - KUDA

a!an" a!as A

/ama &atang = +O +2O +3O +O +$O +6O +7O +4

Gaya &atang m

Di>o&a menggunakan pro'i" 00 # 00 # 0 Baut = mm,@ = 77#0

= E#2

i@ = 3#0 >m

s = 24#2 2#42 >m

e tota" = 2#42 ; 0#$ = 3#32 >m

iy = 2 ( ,@ ; )

= 2 ( 77#0 ; E#2 ( 3#32 )

= 777#2606 >mH

iy = ( 777#2606 Q 2 # )

= ( 777#2606 Q 2 # E#2 ) = ( 20#2$ ) = #EE >m

maka i@ = 3#0 >m ada"a i minimum ( i min )

e = 2# # 0

R =LK

=360#E

= 4#702i min 3#0

=3# # 2# # 0

=2070$60

= 6E#74#702 0E0#E4

# $%=

6E#73= $2#0E0 Kg#>mH

s 2#42

P=

27$#7= 32#444 Kg#>mH

2 2 # E#2

I J I kt

32#444 Kg#>mH J $2#0E0 Kg#>mH

OOOOKKKKKEEEEYYY


>m>m

mm

(e))

8kt=& . '

@

kt=

egangan yang tim&u" ( )

=


95/133

a!an" ba#a$

/ama &atang = BO B2O B3O BO B$O B6O B7O B4

Gaya &atang m

egangan iin = 600 Kg#>mH


,@ = 6#0

= $#$ 2 = 3i@ = 2#7 >m

s = 2$# 2#$ >m

e'k = 3 - 2 ( #$ # 0#6 )

= 2E#2

R =LK

=32#$

= #0$i min 2#7

P=

26$0$#3= E07#76 Kg#>mH

e'k 2E#2

I J S

E07#76 Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY


a!an" %ia"ona& D

B++/G +.,K

/ama &atang = DO D$O D6

Gaya &atang mH

Di>o&a menggunakan pro'i" 40 # 40 # 0 Baut = T mm

,@ = 47#$

= $# 2 = 30#2

i@ = 2# >m

s = 23# 2#3 >m

R =LK

=62#7

= 2$E#24i min 2#

R g = & ( e Q 0#7 e ) = 3# ( 2# # 0 Q 0#7 # 200 )= #06

R=

2$E#24= 2#33E6662 U

R g #06

>m

>m

mm

>m

=

>m>m >m

mm

Rs=


96/133


= 2#34 # 2#33E6662 = $#$$E$$$$4

e'k = 30#2 - 2 ( #$ # #$ )

= 2$#7

e'k # V=

25.7 # 600= 73E6#27$ Kg

$#$60

I M !73E6#27$ Kg M 77 Kg

OOOOKKKKKEEEEYYY


B++/G *K+/

/ama &atang = DO D2O D3

Gaya &atang mH


,@ = 6#0 = $#$ 2 = 3

i@ = 2#7 >m

s = 2$# 2#$ >m

R =LK

=62#7

= 227#EE4i min 2#7

e'k = 3 - 2 ( #$ # #$ )

= 26#$

P=

44#$= 272#$32

e'k 3

I J S

272#$32 Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY


>m

=

>m>m >m

mm

>m

=


97/133

a!an" '()!i*a&

B++/G +.,K

/ama &atang = 9$O 96

Gaya &atang mH

Di>o&a menggunakan pro'i" 40 # 40 # 2 Baut = 20 mm

,@ = 02#0 = 7#E 2 = 3$#4

i@ = 2#3E >m

s = 2# 2# >m

R =LK

=$0#E

= 226#33i min 2#3E

e'k = 3$#4 - 2 ( #$ # #$ )

= 3#3

P=

3#= 3#773

e'k 3#3

I J S

3#773 Kg#>mH J 600 Kg#>mH

OOOOKKKKKEEEEYYY


B++/G *K+/

/ama &atang = 92O 93O 9

Gaya &atang mH


,@ = 02#0

= 7#E 2 = 3$#4

i@ = 2#3E >m

s = 2# 2# >m

R =LK

=72#3

= 30#774i min 2#3E

R g = & ( e Q 0#7 e ) = 3# ( 2# # 0 Q 0#7 # 200 )

= #06

R=

30#774= 2#74336434 U

R g #06


= 2#34 # 2#74336434 = 6#723600

e'k = 3$#4 - 2 ( #$ # #$ )

= 3#3

e'k # V=

31.3 # 600= 7737#$4 Kg

6#72

I M !7737#$4 Kg M 666 Kg

OOOOKKKKKEEEEYYY


>m>m >m

mm

>m

=

>m>m >m

mm

Rs=

>m

=


98/133


99/133

PERHITUNGAN AUT PADA TIAP TITIK UHUL

Pro'i" se"uru kuda-kuda 00 # 00 # 0

te&a" p"at = 0 mm

diameter &aut = mm

diameter "u&ang = ; = $ mm

tegangan geser iin V = 600 Kg#>mH

/g = 0#2$ # & # d H # V= 0#2$ # 3# # #$ H # 600

= 2426 Kg

/tp = d # t # Vt # V /g M (%= #$ # #0 # 0#E # 600 2426 M 260

= 260 OOOOKKKKKEEEEYYY

!


100/133

!


101/133

!


102/133

/g = 0#2$ # & # d H # V= 0#2$ # 3# # 2 H # 600

= $02 Kg

/tp = d # t # Vt # V

= #4 # 0#4 # 2 # 600

= 604

/g M /tp makaO /tp "a yang menentukan

n =D

=47#$47

= 0#0E 2 &/tp 604

P*.1,/G+/ KLmH



!arak gording = +=

3#60E= 0#402 m

L #$

M Pem&e&anan

&e&an gording

= # 3# # #$ = 2#200 Kg#>mH


= 24 # 3#60E # #$ = $E#$4 Kg#>mH

&e&an kuda - kuda

= 27 # 3#60E # #$ = 34#34 Kg#>mH


= 30 34#34 = 3#$3 Kg#>mH

&e&an idup = 00#000 Kg#>mH ;

P = 370#72 Kg#>mH

Kemiringan atap ( )


103/133

Peritungan reaksi ko"om : P

P P P

P P

P P

.+ .+

2$

+,A = 0

- .B # 2$ ; P # 2#E ; P # E ; P # $#6 ; P #

P # E# ; P # 6 ; P # 3# ; = 0

- .B # 2$ ; P ( 2$ ; 2#E ; E ; $#6 ; 3 ; E# ;

- .B # 2$ ; P ( 2#$0 )

.B =P # 2#$0

=2#$0

P = #$ P2$ 2$

.+ = .B = #$ # P

= #$ # 370#72= 664#23

Pendimensian ko"om

tinggi ko"om ( ) = $ m $00 mmV ko"om Bj 37 = 600 Kg#>mH

/ = 664#23

Di>o&a menggunakan pro'i" ? 3$0 # 2$0

t = E mm

t 2 = mm

+ = 0#$

i @ = #6 >mH

i y = 6 >mH

?@ = 240

?y = 2E2

ara tegak "urus sum&u @

k =

R @ =L ky

=k L

= # $00

= 3#26$7$i @ i @ #6

R g = & ( e Q 0#7 e ) = 3# ( 2# # 0 Q 0#7 # 200 )= #06

R @=

3#27= 0#304

R g #06

>m

>m->m-

Rs =

.


104/133

untuk 0#43 J R s J maka

?@ =#

=#

= #0E4#$E3 - R s #$E3 - 0#304

T

n@ =& * +

=3# 2# # 0 0#$

=2#*;00E

=/ # 664#23 # 3#27 723273

n@=

2E0#$02= #003

n@ - 2E0#$02 -

B@ = 0#6 ; 0# #


105/133

& )?@

/J V

+

#67664#23

J 600 Kg#>mH0#$

0#723 Kg#>mH J 600 Kg#>mH

> ) /;

mH0#$ 240

36#E$ Kg#>mH J 600 Kg#>mH

Pro'i" ? 3$0 # 2$0 &isa dipakai#

Pe"at dasar ko"om pada pondasi

kip = #$36 # 0 Kg= 6#$2

Be&an aksia" ( P ) = 664#23 = 3$E#43E

P W = 3 ksi

p = 0#3$ # P W

= 0#3$ # 3

= #0$ kgi

+ per"u = P Q P

= 3$E#43E Q #0$

= 2E$#04$

Pro'i" yang digunakan ada"a ? 3$0 # 2$0

+ = 0#$

d = 3 >m

& = 2$ >m

menentukan "e&ar ( B ) dan panjang ( / ) pe"at dimana arus memenui syarat :

B @ / U $$E0#46 >mH

m >m

>m


106/133

untuk pemi"ian B dan / sedemikian ingga per&edaan ( se"isi ) diantara keduanya mendekati :

0#E2 d - 0#40 &

0#E2 # 3 - 0#4 # 2$ = #24 >m

di>o&a &er&agai gaya B dan / se&agai &erikut :

B @ / U $$E0#46 >mH

N as s(&isi$

7$ 76 $700

7$ 74 $6E 3

7 40 $640 E

70 43 $70 3

gunakan B = 70 dan / = 43

tekanan aktua" ada"a

p =!

=664#23

= #062 Kg#>mHB # / 70 # 43

itung < dan /

< =/ - 0#E$ d

=43 - 0#E$ # 3

= 2$#3$ >m2 2

/ =/ - 0#40 d

=70 - 0#40 # 2$

= 2$ >m2 2

itung te&a" pe"at ( tp )

tp = ( ( P Q 0#2$ y ) = 2$ ( #062 Q 0#2$ # 200 )X

= #0$2 >m

70 # # 43


107/133

/tp = d # t # V

= 2# # 0#E # 600

= 3$6

/g M /tp makaO /tp "a yang menentukan

n =P

=664#23

= #74$ 2 &/tp 3$6

ek kekuatan &aut :

untuk mengetaui kuat atau tidaknya &aut menaan suatu &e&an arus memenui syarat &erikut :V 0#4 @ 600 Kg#>mH

P

664#23

2 # 0#2$ # 3# 2# . 3

227#362 Kg#>mH Y 0#4 @ 600 Kg#>mH

227#362 Kg#>mH Y 240 Kg#>mH

OOOOKKKKKEEEEYYY

2 # Q # & . . 3


108/133


109/133


110/133


111/133


112/133

# $0 -# 36

# $0 -# 3#2


113/133

; $07#7E ;


114/133


115/133

)

K+/ ( - )

34$

E330

6220

30

30

6220

E330

34$

0

0

0

0

0

0

0

0

0$$$

30

E330

30

$$$

0

30

#

346#7

346#7

#

30

,/+%,

*K+/

( - )

0

0

3#7

0


116/133

)

K+/ ( - )

72#E

0E$#6

0E#2

0E2#4

$3$#

$3$#

$3$#

$3$#

0

0

0

0

0

0

0

0

0

47E#6

76#E6$0#

0

0

0

7$E#2

30$7

06#4

0

0

0


117/133

)

K+/ ( - )

0

0

0

0

0

0

0

0

0

0

0

0

06E#

24#

20#E

20#E

00

0

0

0

0

0

0

0

0

0

0

0


118/133

%, ( Kg )

*K+/

( - )

0

0

0

00

0

0

0

0

0

0

0

0

3066#6

63#

63#

0

47E#6

76#E

0

0

0

0

7$E#2

30$7

306#4

0

0

0


119/133

)

*K+/

( - )

27$#7

33#

2232

-47E#

4#$

77#

33#

43E#

0

0

0

0

0

00

0

0

3#6

47#E

6$#4

0

0

0

46E#2

7#

4#$0

0

0


120/133

3 Kg#>m


121/133


122/133


123/133


124/133


125/133

3 &

2 &


126/133

3 &

2 &

2 &


127/133

2 &


128/133


129/133

3 ;

6 ; 3# )=0


130/133

2E0#$02

E#062


131/133


132/133


133/133

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