Data - data Konstruksi : Kuda - kuda type Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) Panjang Gudang ( L ) !arak antar ko"om Penutup atap %i'at dinding inggi ko"om ( )
= = = = = = = =
D 30 ˚ 20 m m #$ m %eng ge"om&ang ertutup m
P5 I P4
A4
P6
A5
K
G P3
P7
A6
A3 V4
M
E P2
A2
D3
V3
D4
P8
A7 V5
V2
D 1
P1 X 2
D5
D2
O
A1
1
V6
D1
A8
D6 V7
V1 A B1
C
F
B2
H
B3
B4
J
B5
L
N
B6
P
B7
B B8
2000
D*+,L KD+-KD+
0 0 2
G G G G G G KK G G G KK G G G
KK G G KK
G
N
0 0 0 2
G G KK
G
G
KK
G G
KK G G
KK
G G
KK
G G G
0 0 2
G G 2 00
P9 X 2
4000
200
.*/+/+ ++P
P*.1,/G+/ P+/!+/G B++/G : Batang &aa ( B ) = B : L :
B
=
20 4
→ = 4 & L = 20 m
= 2#$ m
Batang atas ( + ) = + : B : os 5
+
=
2#$ 0#466
Batang 8ertika" ( 9 ) = 9 = 97 = = = 92 = 96 = = = 93 = 9$ = = = 9 = = = Batang diagona" ( D ) = D = D6 = = = = D2 = D6 = = = = D3 = D = = = =
=
→ os 30 ˚ B
2#447
2#$ # 2#$ # #3 2#$ # 2#$ # 2#44$ 2#$ # 2#$ # #324 2#$ # 2#$ # $#770
= =
0#466 2#$ m
m
tg 3 0 0#$77 m tg 3 0 0#$77 m tg 3 0 0#$77 m tg 3 0 0#$77 m
˚
˚
#
2
#
3
#
˚
˚
( 2# 2#$ ² ; #3 ( 6.250 ; 2.081 ( 8.331 ) ½ 2.886 m ( 2# 2#$ ² ; 2#446 ( 6.250 ; 8.331 ( 14.581 ) ½ 3.818 m ( 2# 2#$ ² ; #324 ( 6.250 ; 18.727 ( 24.977 ) ½ 4.998 m
² )½
)½
² )½
)½
² )½
)½
P*.1,/G+/ P+/!+/G B++/G : Batang &aa ( B ) = B : L :
B
=
20 4
→ = 4 & L = 20 m
= 2#$ m
Batang atas ( + ) = + : B : os 5
+
=
2#$ 0#466
Batang 8ertika" ( 9 ) = 9 = 97 = = = 92 = 96 = = = 93 = 9$ = = = 9 = = = Batang diagona" ( D ) = D = D6 = = = = D2 = D6 = = = = D3 = D = = = =
=
→ os 30 ˚ B
2#447
2#$ # 2#$ # #3 2#$ # 2#$ # 2#44$ 2#$ # 2#$ # #324 2#$ # 2#$ # $#770
= =
0#466 2#$ m
m
tg 3 0 0#$77 m tg 3 0 0#$77 m tg 3 0 0#$77 m tg 3 0 0#$77 m
˚
˚
#
2
#
3
#
˚
˚
( 2# 2#$ ² ; #3 ( 6.250 ; 2.081 ( 8.331 ) ½ 2.886 m ( 2# 2#$ ² ; 2#446 ( 6.250 ; 8.331 ( 14.581 ) ½ 3.818 m ( 2# 2#$ ² ; #324 ( 6.250 ; 18.727 ( 24.977 ) ½ 4.998 m
² )½
)½
² )½
)½
² )½
)½
P*.1,/G+/ D,<*/%, G.D,/G Di>o Di>o&a &a mema memaka kaii gor gordding ing 2 2 1 = 20 mm B = $$ mm D = 7 mm ?@ = 60#7 mm ?y = # mm Berat pro'i" = 3# Kg#m¹ A
Bera Beratt gor gordi ding ng dan dan ata atapp ## ( K ) gording = 3# Kg#m¹ seng ge"om&ang = 0#0 Kg#m¹ ; C = 23# Kg#m² Baut 0 = 2#3 Kg#m² ; 2$#7 Kg#m² Cy
→ C tota" =
= C tota" # os 30 ˚ = 26 # 0#466 = 22#$6 Kg#m¹
⅛ <@₁ = # Cy = 0#2$ # 22#$6 = $6#EE36 Kg#m¹
# #
L ² #$ ²
⅛
A
Be&a Be&ann ang angin in pada pada atap atap # # ( K2 K2 ) Be&an angin = 0 Kg#m² kemiringan atap = 30 ˚ dipiak angin
= 0#02 = 0#02 = 0#6 di&e"akang angin 2 = -0#
# 5 # 30 0# = 0#2
0# 0#
Dari ta&e" pem&e&anan pengaru angin ditentukan : &e&an dipiak angin ? = # + # = 0#2 # 2#447 = 23#0E6E Kg#m¹ &e&an di&e"akang angin ?2 = 2 # + #
0 # 0 0
26 Kg#m²
= =
-0# # 2#447 -6#4E Kg#m¹
#
⅛ <@₂ = # ? # L ² = 0#2$ # 23#0E7 # #$ ² = $4#$4 Kg#m¹
<@₂ = = =
⅛ # ?2 # 0#2$ # -6#4E -6#E7 Kg#m¹
L ² # #$ ²
0
Be& Be&an idu idup # ( K3 ) untuk &e&an ini ditentukan P = 00
Kg terpusat di tenga &atang
<@₃ = = =
¼ # 0 00 # os 30˚ # 0#2$ # 00 # 0#466 E7#2$ Kg#m¹
¼ # 00 # %in 30 ˚ # ( # #$ 0#2$ # 00 # 0#$ # 24#2$ Kg#m¹
#$ #
#$
/ 2 ) 2#2$
Kom&inasi pem&e&anan ( P ditiadakan saat angin &ekerja )
<@₁ ; <@₂ = = <@₁ ; <@₂ = =
ota"
$6#EE36 ; $4#$4 $#$2 Kg#m¹ $6#EE36 ; -6#E7 -$E#E23 Kg#m¹ 32#E063 ; 0 32#E06 32#E063 ; 0 32#E06 $#$2 Kg#mF -$E#E23 Kg#mF ; $$#$ $$#$2E 2E Kg#m Kg#mFF 32#E06 Kg#mF 32#E06 Kg#mF ; 6$#4 6$#43 3 Kg#m Kg#mFF
<@₁ ; <@₃ = $6#EE36 = $#E
; E7#2$ Kg#m¹ ; 24#2$ Kg#m¹
0
ota" <@ =
$#E Kg#mF
ota"
6#03 Kg#mF
kom&inasi ter&esar yang menentukan ada"a kom&inasi
*G+/G+/ +/G *.!+D, <@
mH I Kg#>mH
40#227
J J
6#03
#
0 H
#
I Baja B! 37 Kg#>m² 600
K/.L L*/D+/ e = 2# # 0 ⁶
y
=
34
@
=
34
@
= =
@
=
34
@
=
34
@
= =
36E3 2E3$2E#6 0#03
= =
232 34#$ 0#06
C
# os α # * # ,@
26 # 2# #
74E 366E#2 0#2$
; ; C
0#466 0 ⁶
;
( @ ² ; ( #07 ²
0#$00 0 ⁶
$$6 3$$#$64 #06 y ;
;
4
⁴ $0 36
=
@
;
P
4
# os 5 * # ,@
@
00 # 2# #
#
L³
0#466 0 ⁶
# #
$0 ³ 36
# #
$0 ³ 3#2
0#224
# %in α # * # ,@
26 # 2# #
;
# #
L⁴
# #
L⁴
;
4
⁴ $0 3#2
=
² )½ 0#224 ² ) ½
#07
@
;
P
4
# %in 5 * # ,@
@
00 # 2# #
#
L³
0#$00 0 ⁶
= = = iMin =
0.052 ) ½
( 1.226 ; ( 1.278 ) ½ 1.130
2$0
@
L
2$0
=
#30
J J
@
$0
=
#4
iMin #4
P*.1,/G+/ P*
N
Data - data Konstruksi : Berat gording Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) !arak antar ko"om Penutup atap Penggantung P"a'ond inggi ko"om ( ) !arak gording
= = = = = = = =
3#2 Kg#mF 30 ˚ 2$ m #$ m %eng ge"om&ang : 4 Kg#>mH $m + 2#447 = L #$
0 Kg#>mH
=
0#6
Pem&e&anan &e&an gording = # 3#2
#
#$
=
237#600 Kg#>mH
&e&an atap dan p"a'ond = 24 # 2#447 #
#$
=
363#7 Kg#>mH
&e&an kuda - kuda = 27 # 2#447
#$
=
3$0#7$ Kg#>mH
=
0$#22$ Kg#>mH
= =
00#000 Kg#>mH ; $7#32 Kg#>mH
#
sam&ungan dan per"engkapan = 30 3$0#7$ &e&an idup P muatan tak terduga 0
= = =
P @ 0 $7#32 @ $#732
P tota" = =
$#732 ; 273#0E
0
$7#37
RA R! "1 # "2 # "3 # "4 # "5 # "6 # "7 # "8 # "9
m
= 636#$2 ; 273#0$ ;
273#0$
; 273#0$ ; 273#0$ ; 273#0$ ; 273#0$ ; 2
273#0$ ; 636#$2
N
=
04#3E0 2
=
$0E2#E$
Kg
Peritungan pem&e&anan angin Be&an angin = 0 Kg#m² kemiringan atap = 30 ˚ dipiak angin
di&e"akang angin 2
= 0#02 = 0#02 = 0#6 = -0#
# 5 # 30 0# = 0#2
Dari ta&e" pem&e&anan pengaru angin ditentukan : &e&an tekan angin ? = # = 0#2 # = 4 &e&an isap angin ?2 = 2 # = -0# = -6 Pengaru angin tekan ( ? ) 2#447 + = @ #$ @ 2
=
0# 0#
0 0 Kg#m¹ 0 0 Kg#m¹
4 (titik &uu")
4
= $#E630$ Kg
D
=
2#447
@ #$ @
4
= 03#E26 Kg
*
=
2#447
@ #$ @
4
= 03#E26 Kg
G
=
2#447
@ #$ @
4
= 03#E26 Kg
,
=
2#447 2
@ #$ @
4
= $#E630$ Kg
Pengaru angin isap ( ?2 ) = -6 (titik &uu") 2#447 , = @ #$ @ -6 = -03#E26 Kg 2 K
=
2#447
@ #$ @ -6 =
-207#4$2 Kg
<
=
2#447
@ #$ @ -6 =
-207#4$2 Kg
=
2#447
@ #$ @ -6 =
-207#4$2 Kg
B
=
2#447 2
@ #$ @ -6 =
-03#E26 Kg
<++/ *+P P*. B++/G B++/G /+<+ + +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$ D6
P+/!+/G ( >m ) 2#77 4#66 $#$$ 2# 2# $#$$ 4#66 2#77 4#4$ 4#4$ 6#6 3#6 3#6 6#6 4#4$ 4#4$ 0 #$$ 3# E#33 3# #$$ 0 3# # $#34 $#34 # 3#
%K+L+ G++ ( Kg) 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
B*%+. G++ B++/G ( Kg ) +.,K ( ; ) 0 0 0 0 0 0 0 0 44$0 44$0 660 360 360 660 44$0 44$0 0 0 0 0 0 0 0 0 0 0 0 0 0
*K+/ ( - ) 2770 4660 $$$0 20 20 $$$0 4660 2770 0 0 0 0 0 0 0 0 0 $$0 30 E330 30 $$0 0 30 0 $340 $340 0 30
.*+K%, G++ %,
G++ +/G,/
G++ +/G,/
+.,K
*K+/
+.,K
*K+/
(;)
(-)
.+
0#62
7#02E
0620
.B
0#62
2#6$0
.+9
$#3
.+1
E#2
K
*K+/
(;)
(-)
702E
3$E
0
0620
26$0
7E70
0
$#30
$30
$30
0
0
#6
E200
600
600
0
:
000
:
000
+/G,/ K,., *K+/ B++/G /+<+ + +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$
P+/!+/G ( >m ) 4#6 7#2E$ $#E67 #63E $#303 $#303 $#303 $#303 #7 #7 E#E2 6#4E #$E #$E #$E #$E 0 #3 2#60 3#E70 0 0 0 2#634 3#$06 #$E 0 0
%K+L+ G++ ( Kg) 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
B*%+. G++ B++/G ( Kg ) +.,K ( ; ) 0 0 0 0 0 0 0 0 70 70 EE2 64E0 $E0 $E0 $E0 $E0 0 0 0 0 0 0 0 0 0 0 0 0
*K+/ ( - ) 46 72E$ $E67 63E $303 $303 $303 $303 0 0 0 0 0 0 0 0 0 3 260 3E70 0 0 0 2634 3$06 $E 0 0
D6
0
000
0
0
+/G,/ K+/+/ 1,%+P B++/G
%K+L+
B*%+. G++ B++/G ( Kg )
/+<+
P+/!+/G ( >m )
G++ ( Kg)
+.,K ( ; )
*K+/ ( - )
+ +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D
0#62 0#62 0#62 0#62 E#2E #E$ #$E0 7#27 0 0 0 0 #6 E#E 3#4 3#4 0 0 0 7#E6 $#3 2#6 0 0 0 0 E#2
000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
0620 0620 0620 0620 E2E0 E$0 $E0 7270 0 0 0 0 0 0 0 0 0 0 0 7E60 $3 260 0 0 0 0 E200
0 0 0 0 0 0 0 0 0 0 0 0 600 EE0 3400 3400 0 0 0 0 0 0 0 0 0 0 0
D$ D6
7#0 $#32
000 000
700 $320
0 0
G++ .+/GK+ B++/G +K,B+ +/G,/ ( K
+/G,/ 1,%+P ( Kg )
K
B++/G
+.,K
*K+/
+.,K
*K+/
+.,K
*K+/
+ +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D
(;) 0 0 0 0 0 0 0 0 70 70 EE2 64E0 $E0 $E0 $E0 $E0 0 0 0 0 0 0 0 0
(-) 46 72E$ $E67 63E $303 $303 $303 $303 0 0 0 0 0 0 0 0 0 3 260 3E70 0 0 0 2634
(;) 0620 0620 0620 0620 E2E0 E$0 $E0 7270 0 0 0 0 0 0 0 0 0 0 0 7E60 $3 260 0 0
(-) 0 0 0 0 0 0 0 0 0 0 0 0 600 EE0 3400 3400 0 0 0 0 0 0 0 0
(;) 2006 332$ 6$3 $E4 3E47 667 E247 E67 70 70 EE2 64E0 -0 0 0 0 0 0 0 3EE0 $3 260 0 0
(-) 0 0 0 0 0 0 0 0 0 0 0 0 0 600 E20 E20 0 3 260 0 0 0 0 2634
D2 D3 D D$ D6
0 0 0 0 0
3$06 $E 0 0 0
0 0 E200 700 $320
0 0 0 0 0
0
3$06 $E 0 0 0
E200 700 $320
G++ .+/GK+ B++/G ( K
B*B+/ +/G,/ ( Kg )
K
B++/G
+.,K
*K+/
+.,K
*K+/
+.,K
*K+/
+ +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$
(;) 0 0 0 0 0 0 0 0 44$0 44$0 660 360 360 660 44$0 44$0 0 0 0 0 0
(-) 2770 4660 $$$0 20 20 $$$0 4660 2770 0 0 0 0 0 0 0 0 0 $$0 30 E330 30
(;) 2006 332$ 6$3 $E4 3E47 667 E247 E67 70 70 EE2 64E0 -0 0 0 0 0 0 0 3EE0 $3
(-) 0 0 0 0 0 0 0 0 0 0 0 0 0 600 E20 E20 0 3 260 0 0
(;) 0 0 0 0 0 0 0 0 30320 30320 2$3$2 203$0 3$0 $60 E60 E60 0 0 0 0 220
(-) E76 $33$ 04E7 6$E 4$3 4E03 E373 E403 0 0 0 0 0 0 0 0 0 264 $7$0 $30 0
96 97 D D2 D3 D D$ D6
0 0 0 0 0 0 0 0
/ama &atang = Gaya &atang o&a menggunakan pro'i" ,@ = 77#0 >m⁴ = E#2 >m² i@ = 3#0 >m s = 24#2 mm → e tota" = iy = = = iy = = = =
2#42
$$0 0 30 0 $340 $340 0 30
260 0 0 0 0 E200 700 $320
+ +2 +3 + +$ +6 +7 +4 E76 Kg 3#604 m 360#4 >m 00 # 00 # 0
2#42
; 0#$ =
Baut = mm
>m 3#32
>m
2 ( ,@ ; (e)² ) 2 ( 77#0 ; E#2 ( 3#32 777#26 >mH ( 777#26 Q 2 # ) ½ ( 777#26 Q 2 # E#2 ) ½ ( 20#22 ) ½ #EE >m
maka i@ = e = 2#
0 0 2634 3$06 $E 0 0 0
3#0 >m ada"a i minimum ( i min ) # 0 ⁶
)²
)
0E0 0 0 0 0 3420 2E00 220
0 0 $74 766 EE7 0 0 0
LK i min
=
8 kt =
'² . ( @²
=
σ kt =
$ %& s
=
R=
360#4 3#0 3#
= 4#64
²
0 ⁶
# 2# # 4#64 ²
6E#E7 2#42
=
=
$2#27
Kg#>mH
=
$#644
Kg#>mH
J J
I kt $2#27
Kg#>mH
2070$60 04$#E2
egangan yang tim&u" ( σ ) P 2
σ =
=
2
E76 # E#2
$#644
Pro'i"
I Kg#>mH
00 # 00 # 0
/ama &atang Gaya &atang
= = = =
&isa dipakai#
B B2 B3 B B$ B6 B7 B4 30320 Kg 3#3 m 33 >m 600 Kg#>mH
Di>o&a menggunakan pro'i" E0 # E0 # E ,@ = 6#0 >m⁴ = $#$ >m² 2 = 3 i@ = 2#7 >m s = 2$# mm → 2#$ >m e'k = =
3 - 2 ( 2E#2 >m²
R=
LK i min
=
σ =
P e'k
=
32#$ 2#7
#$
#
)
= #0$
30320 2E#2
034#3$6
0#6
Baut = mm
I Kg#>mH
=
034#3$6
Kg#>mH
J J
S 600
Kg#>mH
=
6E#E7
Kg#>m²
Pro'i"
E0 # E0 # E
&isa dipakai#
B++/G +.,K /ama &atang Gaya &atang
= = = =
D D$ D6 3420 Kg $#0E m 600 Kg#>mH
Di>o&a menggunakan pro'i" 40 # 40 # 0 ,@ = 47#$ >m⁴ = $# >m² 2 = 30#2 i@ = 2# >m s = 23# mm → 2#3 >m
R=
LK i min
( R g = ' = 3# ( = #06
R s =
$0#E 2#
=
R
e Q 2# #
Rg
untuk R U maka
= =
I
Pro'i"
40 # 40 # 0
B++/G *K+/ /ama &atang Gaya &atang
200
= 2#026E3 U
2#34 2#34
Kg
>m²
0#7 e ) ½ 0 ⁶ Q 0#7 #
# #
e'k = 30#2 - 2 ( #$ # = 2$#7 >m² 25.7 e'k # V # 600 σ = = #4
4$2#37
Baut = T mm
= 22#0
22#0 #06
=
$0#E >m
Rs 2#026E3 #$
N N
)
=
#436$
)
=
4$2#37
" 3420
Kg
Kg
&isa dipakai#
= = = =
D D2 D3 EE7 Kg $#0E m 600 Kg#>mH
$0#E >m
Di>o&a menggunakan pro'i" E0 # E0 # E ,@ = 6#0 >m⁴ = $#$ >m² 2 = 3 i@ = 2#7 >m s = 2$# mm → 2#$ >m LK i min
R= e'k = =
$0#E 2#7
3 - 2 ( 26#$ >m²
P e'k
σ =
=
#
I Kg#>mH
E0 # E0 # E
>m²
= E7#0E #$
EE7 3
=
32#72
Pro'i"
#$
Baut = mm
)
=
32#72
J J
S 600
&isa dipakai#
B++/G +.,K /ama &atang Gaya &atang
= = = =
9$ 96 220 6#27 600
Di>o&a menggunakan pro'i" 40 # 40 # 2 ,@ = 02#0 >m⁴ = 7#E >m² 2 = 3$#4 i@ = 2#3E >m s = 2# mm → 2# >m LK i min
R= e'k = =
3$#4 3#3 P
=
62#7 2#3E
- 2 >m²
Kg#>mH
(
220
= 26#34 #$
#
#$
)
Kg m Kg#>mH
62#7 >m
Baut = 20 mm >m²
e'k
#
3#3 I Kg#>mH
70#3E
Pro'i"
40 # 40 # 2
J J
S 600
&isa dipakai#
B++/G *K+/ /ama &atang Gaya &atang
= = = =
92 93 9 $30 Kg 6#27 m 600 Kg#>mH
Di>o&a menggunakan pro'i" 40 # 40 # 2 ,@ = 02#0 >m⁴ = 7#E >m² 2 = 3$#4 i@ = 2#3E >m s = 2# mm → 2# >m
R=
LK i min
( R g = ' = 3# ( = #06
R s =
62#7 2#3E
=
R
e Q 2# #
Rg
untuk R U maka
= =
2#34 2#34
I
Pro'i"
40 # 40 # 2
Kg
Baut = mm >m²
0#7 e ) ½ 0 ⁶ Q 0#7 #
200
= 2#3$7 U # #
e'k = 3$#4 - 2 ( #$ # = 3#3 >m² 31.3 e'k # V # 600 σ = = $#606
4E33#343
62#7 >m
= 26#34
26#34 #06
=
Kg#>mH
Rs 2#3$7 #$
N N
&isa dipakai#
)
=
$#60$E
)
=
4E33#343
" $30
Kg
Kg
Pro'i" se"uru kuda-kuda te&a" p"at diameter &aut diameter "u&ang tegangan geser iMin V /g
= = =
/tp = = =
00 # 00 # 0 0 mm mm ; = 600 Kg#>mH
= = = =
0#2$ # ' # 0#2$ # 3# 2426 Kg d
# #$ 260
t #
#
Vt #0
$ mm
d H # V # #$ H #
# #
V 0#E
600
/g #
600
2426
N N
&* 260
!
=
gaya + = /g
E76 2426
=
6#EE363 →
$
&
=
gaya B = /g
30320 2426
=
0#72E →
7
&
Batang B n
!
E76
+
$33$
2426 =
2#03
→
3
&
=
0
→
3
&
=
#$67
→
3
&
=
#$70
→
3
&
=
#$70
→
3
&
!
$33$
+
04E7
2426 =
2#6E$
→
3
&
=
0#EE
→
3
&
!
04E7
+
6$E
2426 =
3#$2E
→
3
&
=
2#03$
→
3
&
n
=
gaya + = /g
6$E 2426
=
2#246
→
3
&
=
gaya +$ = /g
4$3 2426
=
2#EE
→
3
&
=
gaya 9 = /g
$30 2426
=
#4E0
→
3
&
Batang +$ n Batang 9 n
!
30320
+
30320
2426 =
0#000
→
3
=
0#000
→
3
&
=
#7$4
→
3
&
=
#770
→
3
&
&
!
30320
+
2$3$2
2426 =
2#03
→
3
&
=
0#EE
→
3
&
!
2$3$2
+
203$0
2426 =
2#6E$
→
3
&
=
2#03$
→
3
&
Batang B n
=
gaya B = /g
203$0 2426
=
7#20
→
$
&
=
gaya B$ = /g
3$0 2426
=
#7$E
→
&
=
gaya 9 = /g
$30 2426
=
#4E0
→
3
&
=
gaya D3 = /g
EE7 2426
=
3#$2E
→
3
&
=
gaya D = /g
3420 2426
=
#3$2
→
3
&
Batang B$ n Batang 9 n Batang D3 n Batang D n
<*/*/K+/ KP*L jum"a pe"at koppe" = te&a" pe"at ( ; ) = gaya ma@ = "e&ar pe"at =
4 & 4 mm E403 Kg 4 >m
gaya &atang yang dipiku" pe"at koppe" ( D ) : D = 0#0$ # P = 0#0$ # E403 = 7#0$ Kg
/g
= = =
/tp = = = /g
0#2$ # ' # 0#2$ # 3# $02 Kg d
# #4 604
t #
#
Vt 0#4
d H # V # 2 H #
600
# #
600
V 2
#
N /tp maka /tp "a yang menentukan D
7#0$
/tp
#
604
P*.1,/G+/ KL< N
N
Data - data Konstruksi : Berat gording Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) !arak &agin Kuda - kuda !arak antar ko"om Penutup atap Penggantung P"a'ond inggi ko"om ( ) !arak gording
= = = = = = = = =
3#2 30 ˚ 2$ m 3# m #$ m %eng ge"om&ang : 4 Kg#>mH $m + 3#604 = L #$
0 Kg#>mH
=
0#40
m
Pem&e&anan &e&an gording = # 3#2
#
#$
=
237#600 Kg#>mH
&e&an atap dan p"a'ond = 24 # 3#604 #
#$
=
$#604 Kg#>mH
&e&an kuda - kuda = 27 # 3#604
#$
=
34#372 Kg#>mH
=
3#$2 Kg#>mH
#
sam&ungan dan per"engkapan = 30 34#372 &e&an idup P
Peritungan reaksi ko"om :
= =
00#000 Kg#>mH ; 362#0E Kg#>mH
P P
P
P
P
P
P
.+
.+ 20 .
,-A 0
P
- .B
#
2$
;
P
#
#
E# ;
P
#
6
;
- .B
#
2$
;
P
(
2$
;
- .B
#
2$
;
P
( 3 )
P
# 3 3 = P 2$ 2$
P
.B =
2#E
;
P
#
E
;
P
#
P
#
3#
;
=
0
2#E
;
E
;
$#6
= #$
P
$#6
;
3
;
P
#
3
;
;
E# ;
6
; 3# )=0
.+ = .B = #$ # P = #$ # 362#0E = 62E#2 Pendimensian ko"om tinggi ko"om ( ) = V ko"om Bj 37 = / =
$m → $00 600 Kg#>mH 62E#2
Di>o&a menggunakan pro'i" t = E mm t2 = mm + = 0#$ >m² i@ = #6 >mH iy = 6 >mH ?@ = 240 >m³ ?y = 2E2 >m³ ara tegak "urus sum&u @ k = L ky R@ = = i@ ( R g = ' = 3# ( = #06 R@ Rg
Rs= untuk
0#43
?@ =
n@
=
? 3$0 # 2$0
k
# #$E3 '²
*
/
# R @²
+
L
e Q 2# #
=
i@
J
Rs T =
# $00 #6
0#7 e ) 0 ⁶ Q
3#27 #06
= J Rs
mm
=
= 3#266
½
0#7 #
200
)
0#304
maka
=
#$E3
3#
² #
# -
2# #
62E#2
=
0#304
#
0 ⁶
#0E4
0#$ ² 3#27
=
2*;0E 7*;06
= 2E2#32
n@ n@ - B@
= = =
2E2#32 2E2#32 -
=
=
ara tegak "urus sum&u @ k = L ky Ry = = iy ( R g = ' = 3# ( = #06 Ry Rg
Rs= 0#43
?y =
ny
=
ny ny -
# #$E3 '²
*
/
# R @²
+
L
=
iy e Q 2# #
J
Rs T
#
2$#460
?@
/ +
0#7 e ) 0 ⁶ Q
=
=
#67
= 43#3333
½
0#7 #
200
)
0#7$
# -
#$E3
3#
² #
=
#02
→
600
=
0#7$
2# #
62E#2
Kg#>mH J
J
$00
maka
=
E#373 E#373 -
$00
6
43#333 #06
= J Rs
=
k
#
ek persamaan interaksi / ny <@ a) ?@ ; B@ @ + ny - ?@ 62E#2 #0E4 ; 0#6 @ 0#$
&)
0#6 ; 0# # <@Q<@2 0#6 ; 0# # 0 0#6
<@ = 0#2$ # / # L = 0#2$ # 62E#2 = 343044#262$
untuk
→
#003
0 ⁶
0#$ ² 43#333
#
J
600
=
2*;0E *;07
=
J
V 343044#262$ J 240
Kg#>mH
V
62E#2 0#$
#67
Kg#>mH
600
Kg#>mH
E#373
0#043 >)
Kg#>mH
600
/ <@ ; J V + ?@ 62E#2 343044#262$ ; 0#$ 240 3$E#676
Pro'i"
J
Kg#>mH
? 3$0 # 2$0
J
p = = =
0#3$ 0#3$ #0$
+ per"u = = =
J
600
600
Kg#>mH
Kg#>mH
&isa dipakai#
Pe"at dasar ko"om pada pondasi kip = #$36 # 0 ² m² = 6#$2 >m² Be&an aksia" ( P ) PW
Kg#>mH
= =
Kg
62E#2 3 ksi
=
3$#24
# PW # 3 kgi
P Q P 3$#24 246#E3$
Q
#0$
Pro'i" yang digunakan ada"a + = 0#$ >m² d = 3 >m & = 2$ >m
? 3$0 # 2$0
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
7$ 7$
$$E0#46
76 74
>mH
$700 $6E
3
7 70
40 43
$640 $70
gunakan B = 70 dan / = tekanan aktua" ada"a " p = = B # /
E 3
43 62E#2 = 70 # 43
#0$$
43
-
0#E$ 2
#
3
70
-
0#40 2
#
2$
Kg#>mH
itung < dan /
<
=
/ - 0#E$ d 2
=
/
=
/ - 0#40 d 2
=
itung te&a" pe"at ( tp ) tp = ( P Q 0#2$ y ) ½ = 2$ ( #0$$ Q # 0#2$ → = #04 >m
2$#3$
>m
=
2$
>m
200
70 #
pe"at "andasan yang dipakai ada"a
=
)X
#
43
2$0 # ) =
E
#
/g
= = = = = = =
/tp = = = /g
n
2 2 3 3
d # d #
23
;
2
=
4
2
=
72
0#2$ # ' # 0#2$ # 3# 723#$6 Kg d
# 2# 3$6
t #
#
=
# 2
23
2 mm
d H # V # 2# H #
600
V 0#E
#
600
N /tp maka /tp "a yang menentukan
=
P /tp
ek kekuatan &aut :
=
62E#2 3$6
=
#77
→
2
&
mm
untuk mengetaui kuat atau tidaknya &aut menaan suatu &e&an arus memenui syarat &erikut : V 0#4 @ 600 Kg#>mH P 2 # Q # ' . ² . 3
0#2$
62E#2 # 3#
22$#E3 Kg#>mH
Y 0#4 @
22$#E3 Kg#>mH
Y
2
#
240
2# ² . 600 Kg#>mH
Kg#>mH
3
706#4 30#2
0#003206440E07$
66#247607$363 7E#$7262306E 2$#460230$432
$07#43 0#E66 4#4377$E 32#6447746733
06$#$3770$62
3#$ l
2
= =
2$0 33000
(
2$0
;
Q
2
)
N
#$
diameter "u&ang
Data - data Konstruksi : Kuda - kuda type Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) Panjang Gudang ( L ) !arak antar ko"om Penutup atap %i'at dinding inggi ko"om ( )
= = = = = = = =
D 30 ˚ 2$ m 0 m #$ m +s&es Ge"om&ang ertutup $m
P5 I P4
A4
P6
A5
G P3
K P7
A6
A3 V4
M
E P2
A2
D3
V3
D4
P8
A7 V5 D5
D2
V2
D 1 P1 X 2
O
A1
1
V6
D1
A8
D6
A B1
P9 X 2
V7
V1
C
F
B2
H
B3
J
B4
B5
L
N
B6
P
B7
B B8
2000
D*+,L KD+-KD+
0 0 2
G G G G G G KK G G G KK G G G
KK G G KK
G
N
0 0 0 2
G G KK
G
G
KK
G KK
G G G
KK
G G
KK
G G G
0 0 2
G G 200
4000
200
.*/+/+ ++P
P*.1,/G+/ P+/!+/G B++/G : Batang &aa ( B ) = B : L :
B
=
2$ 4
=
Batang atas ( + ) = + : B :
+
=
3#2$ 0#466
→
3#2$
os
=
= L=
4 & 2$ m
m
5
→
3#60E
os B m
30 ˚
= =
0#466 3#2$
m
Batang 8ertika" ( 9 ) = 9 = 97 = = = 92 = 96 = = = 93 = 9$ = = = 9 = = =
3#2$ 3#2$ #403 3#2$ 3#2$ 3#606 3#2$ 3#2$ $#0E 3#2$ 3#2$ 7#23
Batang diagona" ( D ) = D = D6 = = = = D2 = D$ = = = = D3 = D = = = =
tg 30 ˚ 0#$77
# # m # #
tg 30 ˚ 0#$77 #
2
tg 30 ˚ 0#$77 #
3
tg 30 ˚ 0#$77 #
m # # m # # m
( 3#2$ ² 9.766 ( 13.017 (
; ; )½ m ; ; )½ m ; ; ½ ) m
3.608
( 3#2$ ² 9.766 ( 22.783 ( 4.773
( 3#2$ ² 9.766 ( 39.027 ( 6.247
#403 3.251
3#604 13.017
$#0E 29.261
² )½
)½
² )½
)½
² )½
)½
P*.1,/G+/ D,<*/%, G.D,/G Di>o&a memakai gording 1 = 20 mm B = $$ mm D = 7 mm ?@ = 60#7 mm ?y = # mm Berat pro'i" = A
Kg#m¹
3#
Berat gording dan atap # ( K ) gording = as&es ge"om&ang = C = Baut 0 =
Cy
A
2
3# 0#3 23#7 2#37 26#
Kg#m¹ Kg#m¹ Kg#m² Kg#m² Kg#m²
; ; →
30 ˚
= C tota" # os = 26 # 0#466 = 22#$6 Kg#m¹
<@₁ = ⅛ # = 0#2$ # ¹ Kg#m = $6#EE36
Cy 22#$6
⅛
C tota" 26 # 26 #
# #
L #$
# %in 5 %in 30˚ 0#$ #
² ²
# L # #$ ² 20#2$
Be&an angin pada atap # ( K2 ) Be&an angin = 0 Kg#m² kemiringan atap = 30 ˚ dipiak angin
= = =
C tota" =
0#02 0#02 0#6
-
# # 0#
5 30 = 0#2
0# 0#
²
26 Kg#m²
di&e"akang angin 2
=
-0#
Dari ta&e" pem&e&anan pengaru angin ditentukan : &e&an dipiak angin ?
&e&an di&e"akang angin
<@₂ = = =
⅛ # 0#2$ # 73#073 Kg#m¹
<@₂ = = =
⅛ # 0#2$ # -6#6
= = = = = =
?2
?
# 24#464
?2 # -$7#736720$$3 Kg#m¹
# + # 0 0#2 # 3#60E # 24#46436 Kg#m¹ 2 # + # 0 -0# # 3#60E -$7#73672 Kg#m¹
L ² # #$
²
L ² # #$
²
00
Kg
0
#
0
Be&an idup # ( K3 ) untuk &e&an ini ditentukan P = <@₃ = = =
¼ # 00 0#2$ # E7#2$ Kg#m¹
os 30˚ # 00 #
# 0#466
#$ #
¼ # 00 0#2$ # 24#2$ Kg#m¹
%in 30˚ # 00 #
#(
#$ #
0#$
Kom&inasi pem&e&anan ( P ditiadakan saat angin &ekerja )
<@₁ ; <@₂ = = ₁ ₂ <@ ; <@ = =
ota"
<@₁ ; <@₃ = =
$6#EE362$ ; Kg#m¹ 30#067 $6#EE362$ ; Kg#m¹ -4E#$2 32#E062$ ; 32#E06 32#E062$ ; 32#E06 30#067 Kg#mF -4E#$2 Kg#mF 0#E Kg#mF 32#E06 Kg#mF 32#E06 Kg#mF 6$#43 Kg#mF
$6#EE362$ $#E 32#E062$ 6#03
; Kg#m¹ ; Kg#m¹
ota" <@ =
$#E Kg#mF
ota"
6#03 Kg#mF
73#073037 -6#6 0 0
;
;
E7#2$ 24#2$
kom&inasi ter&esar yang menentukan ada"a kom&inasi
terpusat di tenga &atang
#$
/
2 2#2$
)
0
*G+/G+/ +/G *.!+D, <@ mH
40#227
0
²
6#03
;
#
0 H
#
$E#43
I
J J
C
#
os α * # ,@
26 2#
# #
Kg#>mH
I Baja B! 37 Kg#>m² 600
K/.L L*/D+/ e = 2# # 0 ⁶ y
=
34
@
=
34
@
= = @
=
34
@
=
34
@
= =
36E3 2E3$2E#6 0#03
= = = = =
iMin =
;
232 34#$ 0#06 ( @ ² ( #07 1.226 ( 1.278 (
0#466 0 ⁶ 74E 366E#2 0#2$
;
C
#
26 2#
# #
;
y
²
;
; )½
0.052
² 0#224 )½
;
4
⁴ $0 36
=
@
;
L⁴
;
4
⁴ $0 3#2
# #
=
@
;
#07
)½ ²
)½
1.130
2$0
@
L
#30
2$0
= J J
@ iMin #4
P
4
# os 5 * # ,@ @
00 # 2# #
#
L³
0#466 0 ⁶
# #
$0 ³ 36
# #
$0 ³ 3#2
0#224 #
0#$00 0 ⁶
;
L⁴
# #
%in α * # ,@
$$6 3$$#$64 #06
;
#
$0
=
#4
P
4
# %in 5 * # ,@ @
00 # 2# #
#
L³
0#$00 0 ⁶
P*.1,/G+/ P*
N
Data - data Konstruksi : Berat gording Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) !arak antar ko"om Penutup atap Penggantung P"a'ond inggi ko"om ( ) !arak gording
= = = = = = = =
3# Kg#mF 30 ˚ 2$ m #$ m as&es ge"om&ang : 4 Kg#>mH $m + 3#60E = L #$
0#3 Kg#>mH
=
0#40
m
Pem&e&anan &e&an gording = #
3#
#
#$
=
2#200 Kg#>mH
&e&an atap dan p"a'ond = 24 # 3#60E
#
#$
=
$E#$4 Kg#>mH
&e&an kuda - kuda = 27 # 3#60E
#
#$
=
34#34 Kg#>mH
=
3#$3 Kg#>mH
= =
00#000 Kg#>mH 370#72 Kg#>mH
sam&ungan dan per"engkapan = 30 34#34 &e&an idup P muatan tak terduga 0
= = =
P @ 0 370#77E @ 37#072
P tota" = =
37#072 ; $07#7E0
0
;
370#74
RA R! "1 # "2 # "3 # "4 # "5 # "6 # "7 # "8 # "9
=
7$3#4E4
;
$07#7E
$07#7E
;
7$3#4E4
=
2062#34 2
=
603#$E
Kg
;
$07#7E
; 2
$07#7E
;
$07#7E
; $07#7E ; $07#7E ;
N
Peritungan pem&e&anan angin Be&an angin = 0 Kg#m² kemiringan atap = 30 ˚ dipiak angin
di&e"akang angin 2
= = = =
0#02 0#02 0#6 -0#
-
Dari ta&e" pem&e&anan pengaru angin ditentukan : &e&an tekan angin ?
&e&an isap angin
?2
Pengaru angin tekan ( ? ) 3#60E + = @ 2
# # 0#
= = = = = =
5 30 = 0#2
0#2
# # 4
2 # -0# -6
=
0# 0#
0 0 Kg#m¹ 0 0 Kg#m¹
4 (titik &uu")
#$
@
4
=
6#E$34 Kg
D
=
3#60E
@
#$
@
4
=
2E#E0762 Kg
*
=
3#60E
@
#$
@
4
=
2E#E0762 Kg
G
=
3#60E
@
#$
@
4
=
2E#E0762 Kg
,
=
3#60E 2
@
#$
@
4
=
6#E$34 Kg
=
-6 (titik &uu")
Pengaru angin isap ( ?2 ) 3#60E , = @ 2
#$
@
-6
=
-2E#E076 Kg
K
=
3#60E
@
#$
@
-6
=
-2$E#4$2 Kg
<
=
3#60E
@
#$
@
-6
=
-2$E#4$2 Kg
=
3#60E
@
#$
@
-6
=
-2$E#4$2 Kg
B
=
3#60E 2
@
#$
@
-6
=
-2E#E076 Kg
<++/ *+P P*. B++/G /+<+ + +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$ D6
B++/G P+/!+/G ( >m ) 3#4$ E#330 6#220 3#0 3#0 6#220 E#330 3#4$ 4#4$3 4#4$3 6#60 3#67 3#67 6#60 4#4$3 4#4$3 0 #$$$ 3#0 E#330 3#0 #$$$ 0 3#0 # $#347 $#347 # 3#0
%K+L+ G++ ( Kg) 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
B*%+. G++ B++/G ( Kg ) +.,K ( ; ) *K+/ ( - ) 0 34$ 0 E330 0 6220 0 30 0 30 0 6220 0 E330 0 34$ 44$3# 0 44$3# 0 660 0 366#7 0 366#7 0 660 0 44$3# 0 44$3# 0 0 0 0 $$$ 0 30 0 E330 0 30 0 $$$ 0 0 0 30 0 # 0 $346#7 0 $346#7 0 # 0 30
.*+K%, G++ %,
+.,K (;)
*K+/ (-)
G++ +/G,/ +.,K : 000
G++ +/G,/ *K+/ : 000
K
.+
7#042
#647
7042#2
646#6
23E$#6
0
.B
7#04
#764
7043#6
767#7
$3$#E
0
.+9
3#$
3#$4$
3$#
3$4#4
0
3#7
.+1
6#E6
3#067
6E$#4
3066#7
32E#
0
+/G,/ K,., *K+/ /+<+ + +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$ D6
B++/G P+/!+/G ( >m ) $#73 3#0E6 3#0E 3#0E3 3#$3$ 3#$3$ 3#$3$ 3#$3$ 7#6$2 7#6$2 6#24 #$E$ 3#062 3#062 3#062 3#062 0 0#440 #76$ 2#6$0 0 0 0 #7$E 3#0$7 3#062 0 0 0
%K+L+ G++ ( Kg) 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
B*%+. G++ B++/G ( Kg ) +.,K ( ; ) *K+/ ( - ) 0 $72#E 0 30E$#6 0 30E#2 0 30E2#4 0 3$3$# 0 3$3$# 0 3$3$# 0 3$3$# 76$#E 0 76$#E 0 624# 0 $E$# 0 306#4 0 306#4 0 306#4 0 306#4 0 0 0 0 47E#6 0 76#E 0 26$0# 0 0 0 0 0 0 0 7$E#2 0 30$7 0 306#4 0 0 0 0 0 0
+/G,/ K+/+/ 1,%+P /+<+ + +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$ D6
B++/G P+/!+/G ( >m ) 7#042 7#042 7#042 7#042 6#E6E #$03 E#732 #$ 0 0 0 0 3#06E 6#24 E#20$ E#20$ 0 0 0 $#36 3#$ #766 0 0 0 0 6#33 6#2 3#$$2
%K+L+ G++ ( Kg) 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000
B*%+. G++ B++/G ( Kg ) +.,K ( ; ) *K+/ ( - ) 7042#2 0 7042#2 0 7042#2 0 7042#2 0 6E6#E 0 $02#4 0 E732#3 0 $#3 0 0 0 0 0 0 0 0 0 0 306E# 0 624# 0 E20#E 0 E20#E 0 0 0 0 0 0 $36#3 0 3$# 0 766#3 0 0 0 0 0 0 0 0 0 633#3 0 623#$ 0 3$$2# 0
G++ .+/GK+ B++/G +K,B+ +/G,/ ( K
+/G,/ *K+/ ( Kg ) +.,K *K+/ (;) (-) 0 $72#E 0 30E$#6 0 30E#2 0 30E2#4 0 3$3$# 0 3$3$# 0 3$3$# 0 3$3$# 76$#E 0 76$#E 0 624# 0 $E$# 0 306#4 0 306#4 0 306#4 0 306#4 0 0 0 0 47E#6 0 76#E 0 26$0# 0 0 0 0 0 0 0 7$E#2 0 30$7 0 306#4 0 0 0 0 0 0
+/G,/ 1,%+P ( Kg ) +.,K *K+/ (;) (-) 7042#2 0 7042#2 0 7042#2 0 7042#2 0 6E6#E 0 $02#4 0 E732#3 0 $#3 0 0 0 0 0 0 0 0 0 0 306E# 0 624# 0 E20#E 0 E20#E 0 0 0 0 0 0 $36#3 0 3$# 0 766#3 0 0 0 0 0 0 0 0 0 633#3 0 623#$ 0 3$$2# 0
K
G++ .+/GK+ B++/G ( K
B*B+/ +/G,/ ( Kg )
K
B++/G
+.,K
*K+/
+.,K
*K+/
+.,K
*K+/
+ +2 +3 + +$ +6 +7 +4 B B2 B3 B B$ B6 B7 B4 9 92 93 9 9$ 96 97 D D2 D3 D D$ D6
(;) 0 0 0 0 0 0 0 0 44$3# 44$3# 660 366#7 366#7 660 44$3# 44$3# 0 0 0 0 0 0 0 0 0 0 0 0 0
(-) 34$ E330 6220 30 30 6220 E330 34$ 0 0 0 0 0 0 0 0 0 $$$ 30 E330 30 $$$ 0 30 # $346#7 $346#7 # 30
(;) 33E#3 3E46#6 3E44 3E4E# 266#$ 7E67# 6E6#E 7E7$#E 76$#E 76$#E 624# $E$# -7#3 0 0 0 0 0 0 27#2 3$# 766#3 0 0 0 0 633#3 623#$ 3$$2#
(-) 0 0 0 0 0 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
(;) 0 0 0 0 0 0 0 0 26$0$#3 26$0$#3 22244# 406#4 3$E# 30E3# 270#3 270#3 0 0 0 0 3# 2#3 0 0 0 0 76#6 200E# 2#
(-) 27$#7 $33# 2232 -47E# 4#$ -77# 333# $43E# 0 0 0 0 0 0 0 0 0 23#6 47#E 66$#4 0 0 0 46E#2 77# 44#$ 0 0 0
/ama &atang Gaya &atang
= = =
+ +2 +3 + +$ +6 +7 +4 27$#7 Kg 3#60E m
Di>o&a menggunakan pro'i" ,@ = 77#0 >m⁴ = E#2 >m² i@ = 3#0 >m s = 24#2 mm → e tota" = iy = = = iy = = = =
2#42
00 # 00 # 0
2#42
;
0#$
2 ( ,@ ; 2 ( 77#0 777#26 >mH ( 777#26 Q ( 777#26 Q ( 20#22 ) #EE >m
maka i@ = e = 2#
#
3#0 0
LK i min
=
8 kt =
'² . ( @²
=
σ kt =
$ %& s
=
R=
360#E
3#32
(e)² E#2
2
mm
>m
) (
# #
2
Baut =
>m
=
;
>m
)
3#32
)²
)
½
E#2
)
½
½
>m ⁶
ada"a i minimum ( i min )
360#E 3#0
= ²
3#
4#702
# 2# # ² 4#702
6E#73 2#42
0 ⁶
=
=
$2#0E0
Kg#>mH
=
32#444
Kg#>mH
J J
I kt $2#0E0
egangan yang tim&u" ( σ ) σ =
P 2
=
2
27$#7 #
E#2 I
32#444
Pro'i"
00 # 00 # 0
Kg#>mH
&isa dipakai#
Kg#>mH
2070$60 0E0#E4
=
6E#73
Kg#>m²
/ama &atang Gaya &atang
= = = =
B B2 B3 B B$ B6 B7 B4 26$0$#3 Kg 3#3 m 600 Kg#>mH
Di>o&a menggunakan pro'i" ,@ = 6#0 >m⁴ = $#$ >m² i@ = 2#7 >m s = 2$# mm → e'k = =
3 2 2E#2 >m²
R=
LK i min
=
σ =
P e'k
=
E0 # E0 # E
2 = 2#$
(
#$
32#$ 2#7
=
#
Di>o&a menggunakan pro'i" ,@ = 47#$ >m⁴ = $# >m² i@ = 2# >m s = 23# mm →
LK i min
62#7 2#
( R g = ' = 3# ( = #06
R s =
R Rg
=
E07#76
J J
S 600
= = = =
=
=
)
Kg#>mH
Kg#>mH
&isa dipakai#
B++/G +.,K /ama &atang Gaya &atang
R=
mm
#0$
Kg#>mH
E0 # E0 # E
Baut = 3
0#6
I E07#76
>m
>m
26$0$#3 2E#2
Pro'i"
33
D D$ D6 76#6 Kg 6#27 m 600 Kg#>mH
40 # 40 # 0 2 = 2#3
e 2#
2$E#24 #06
T mm
>m²
30#2
>m =
Q #
Baut =
62#7
2$E#24
0#7 e 0 ⁶
=
)
½
0#7
#
2#33E666
U
Q
200
)
>m
untuk R U maka
= =
2#34 2#34
e'k = 30#2 2 = 2$#7 >m² 25.7 e'k # V σ = =
(
#$
# $#$60
Pro'i"
= = = =
Di>o&a menggunakan pro'i" ,@ = 6#0 >m⁴ = $#$ >m² i@ = 2#7 >m s = 2$# mm →
LK i min
62#7 2#7
σ =
2#$
73E6#27$
77
Kg
D D2 D3 44#$ Kg 6#27 m 600 Kg#>mH
(
#$
272#$32
E0 # E0 # E
Kg#>mH
Baut = 3
62#7
mm
>m²
>m =
227#EE4
#
#$
I
Pro'i"
=
Kg
"
2 =
44#$ 3
=
$#$$E$$6
)
E0 # E0 # E
=
3 2 26#$ >m²
P e'k
#$
=
&isa dipakai#
B++/G *K+/ /ama &atang Gaya &atang
e'k = =
#
N N
Kg
40 # 40 # 0
R=
Rs 2#33E666
600
I 73E6#27$
# #
)
=
272#$32
J J
S 600
&isa dipakai#
Kg#>mH
>m
B++/G +.,K /ama &atang Gaya &atang
= = = =
Di>o&a menggunakan pro'i" ,@ = 02#0 >m⁴ = 7#E >m² i@ = 2#3E >m s = 2# mm →
LK i min
$0#E 2#3E
R= e'k = =
3$#4 3#3
P e'k
σ =
40 # 40 # 2 2 = 2#
= -
2
= (
>m
20 mm
>m²
3$#4
226#33
#$
#
3# 3#3
=
Pro'i"
Kg#>mH
40 # 40 # 2
#$
)
LK i min
72#3 2#3E
untuk R U maka
J J
S 600
2 = 2#
=
e 2#
= Q #
= =
e'k = 3$#4 2 = 3#3 >m² 31.3 e'k # V σ = =
#$
# 6#72
40 # 40 # 2
Kg
mm
>m²
3$#4
0#7 e 0 ⁶
)
½
0#7
#
=
2#743364
U
# #
Rs 2#743364
#
#$
600
I 7737#$4
>m
30#774
2#34 2#34 (
Baut =
72#3
>m
30#774 #06
=
Kg#>mH
92 93 9 66$#4 Kg 7#23 m 600 Kg#>mH
40 # 40 # 2
( R g = ' = 3# ( = #06
R
3#773
= = = =
Di>o&a menggunakan pro'i" ,@ = 02#0 >m⁴ = 7#E >m² i@ = 2#3E >m s = 2# mm →
Rg
=
&isa dipakai#
B++/G *K+/ /ama &atang Gaya &atang
Pro'i"
Baut =
$0#E
>m²
3#773
R s =
Kg m Kg#>mH
>m
I
R=
9$ 96 3# $#0E 600
N N
&isa dipakai#
Q
200
)
=
6#7236
) =
7737#$4
" 666
Kg
Kg
Pro'i" se"uru kuda-kuda te&a" p"at diameter &aut diameter "u&ang tegangan geser iMin V /g
= = =
/tp = = =
0#2$ 0#2$
'
# # 2426
d
#
# 3#
=
d H # V # #$
$ mm
H #
600
Kg t #
#$
00 # 00 # 0 0 mm mm ; 600 Kg#>mH
= = = =
#
Vt #0
# #
V 0#E
/g #
600
2426
N N
&* 260
260
!
=
gaya + /g
=
27$#7 2426
=
=
gaya B /g
=
26$0$#3 2426
=
#6 →
$
&
→
7
&
Batang B n
E#37E
!
=
gaya + - gaya +2 /g
27$#7
=
+
$33#
2426
=
2#$2
→
3
&
=
#0
→
2
&
Batang D n
=
gaya D /g
=
46E#2 2426
=
#723
→
2
&
=
gaya 9 /g
=
0 2426
=
0
→
2
&
Batang 9 n
!
=
gaya +2 - gaya +3 /g
$33#
=
+
2232
2426
Batang 92 n
=
gaya 92 /g
=
23#6 2426
=
0#462
→
2
&
!
=
gaya +3 - gaya + /g
2232
=
+
-47E#
2426
=
#0
→
3
&
=
0#000
→
2
&
=
#E2
→
2
&
Batang D3 n
=
gaya D3 /g
=
44#$ 2426
=
2#EE0
→
3
&
=
gaya 93 /g
=
47#E 2426
=
#72$
→
2
&
gaya D2 /g
=
77# 2426
=
2#$34
→
3
&
Batang 93 n
Batang D2 n
=
!
=
gaya + /g
=
-47E# 2426
=
-0#3
→
2
&
=
gaya +$ /g
=
4#$ 2426
=
0#$E
→
2
&
=
gaya 9 /g
=
66$#4 2426
=
2#3
→
3
&
Batang +$ n Batang 9 n
!
=
gaya B - gaya B2 /g
26$0$#3
=
+
26$0$#3
2426
Batang 9 n
=
gaya 9 /g
=
0 2426
=
0#000
→
2
&
!
=
gaya B2 - gaya B3 /g
26$0$#3
=
+
22244#
2426
Batang D n
=
gaya D /g
=
46E#2 2426
=
#723
→
2
&
=
gaya 92 /g
=
23#6 2426
=
0#462
→
2
&
gaya D2 /g
=
77# 2426
=
2#$34
→
3
&
Batang 92 n
Batang D2 n
=
!
gaya B3 - gaya B 0
=
22244#
=
+
406#4
2426
Batang 93 n
=
gaya 93 /g
47#E 2426
=
=
#72$
→
2
&
!
=
gaya B /g
=
406#4 2426
=
6#3E
→
7
&
=
gaya B$ /g
=
3$E# 2426
=
#763
→
$
&
=
gaya 9 /g
=
66$#4 2426
=
2#3
→
3
&
=
gaya D3 /g
=
44#$ 2426
=
2#EE0
→
3
&
=
gaya D /g
=
76#6 2426
=
0#26
→
3
&
Batang B$ n Batang 9 n Batang D3 n Batang D n
<*/*/K+/ KP*L jum"a pe"at koppe" te&a" pe"at ( ; ) gaya ma@ "e&ar pe"at
= = = =
4 4 $43E# 4
gaya &atang yang dipiku" pe"at koppe" ( D ) : D = 0#0$ # P = 0#0$ # $43E#
& mm Kg >m
=
47#$47
Kg
=
#E6
→
2
&
/g
= = =
/tp = = =
0#2$ 0#2$ $02 d
'
# #
#
d H # # 2
3#
V H #
600
V 2
#
600
=
0#0E
Kg t #
#4
#
#
Vt
# #
0#4
604
/g N /tp maka /tp "a yang menentukan
n
=
D /tp
47#$47 604
=
→
2
&
P*.1,/G+/ KL< N
N
Data - data Konstruksi : Berat gording Kemiringan atap ( α ) Bentang Kuda - kuda ( B ) !arak &agin Kuda - kuda !arak antar ko"om Penutup atap Penggantung P"a'ond inggi ko"om ( ) !arak gording
= = = = = = = = =
3# 30 ˚ 2$ m 3#2$ m #$ m +s&es Ge"om&ang 4 Kg#>mH $m + 3#60E = L #$
0#3 Kg#>mH
=
0#402
Pem&e&anan &e&an gording = #
3#
#
#$
=
2#200 Kg#>mH
&e&an atap dan p"a'ond = 24 # 3#60E
#
#$
=
$E#$4 Kg#>mH
&e&an kuda - kuda = 27 # 3#60E
#
#$
=
34#34 Kg#>mH
=
3#$3 Kg#>mH
= =
00#000 Kg#>mH 370#72 Kg#>mH
sam&ungan dan per"engkapan = 30 34#34 &e&an idup P
;
m
Peritungan reaksi ko"om :
P P
P
P
P
P
P
P
.+
.+ 2$
.
,-A 0 - .B # 2$
;
P
#
# E# ;
P
#
6
;
- .B
#
2$
;
P
(
2$
;
- .B
#
2$
;
P
(
2#$0
)
P
#
2#$0
=
2#$0 P 2$
P
.B =
2$
.+ = .B = = =
#$ #$
# # 664#23
Pendimensian ko"om tinggi ko"om ( ) V ko"om Bj 37 /
= = =
Di>o&a menggunakan pro'i" t = E t2 = + = 0#$ i@ = #6 iy = 6 ?@ = 240 ?y = 2E2
mm mm >m² >mH >mH >m³ >m³
ara tegak "urus sum&u @ k = L ky R@ = i@ R g = ' = 3# = #06
Rs=
R@ Rg
2#E
;
P
#
E
;
P
P
#
3#
;
=
0
;
E
;
2#E
=
#$
#
$#6
$#6
;
P
P 370#72
$m → 600 Kg#>mH 664#23
$00
mm
? 3$0 # 2$0
= ( (
=
k
L i@
e 2#
Q #
3#27 #06
=
# $00 #6
0#7 e ) 0 ⁶ Q
=
=
3#266
½
0#7 #
0#304
200
)
3
;
P
#
3
;
;
E# ;
6
; 3# )=0
untuk
0#43
?@ =
J
Rs
J
# -
#$E3
maka =
Rs
# -
#$E3
=
0#304
#0E4
T n@
=
n@ n@ - B@
= = =
<@ = = =
'²
*
+
/
#
R @²
=
2E0#$02 2E0#$02 -
0#6 0#6 0#6
; ;
0# 0#
ara tegak "urus sum&u @ k = L ky Ry = iy
Ry Rg
Rs= untuk
0#43
?y =
# 664#23
L
e 2#
Rs
J
# -
#$E3
#
2*;0E 7*;06
= 2E0#$02
Q #
Rs
$00
#
$00
=
6
0#7 e ) ⁶ 0 Q
43#333 #06
= J
=
L
=
iy
( (
0#$ ² 3#27
# →
#003
0 ⁶
#
<@Q<@2 0
k
=
R g = ' = 3# = #06
=
/
2#
664#23
# #
0#2$ # 0#2$ # 34$$#04E20323
² #
3#
=
=
43#3333
½
0#7 #
200
)
0#7$
maka =
# -
#$E3
=
0#7$
#67
T ny
=
ny ny -
'²
*
+
/
#
R @² E#062 E#062 -
=
ek persamaan interaksi / a) ?@ + #0E4 27#7
;
ny ny - ; J
#
0 ⁶
0#$ ² 43#333
# →
@
<@ ?@
J
V
0#6
@
34$$#04E J 240
#02
664#23 0#$ Kg#>mH
2#
664#23 =
B@
² #
3#
=
600
Kg#>mH
=
2*;0E *;07
600
=
Kg#>mH
E#062
&)
?@
/ +
J
#67 0#723 >)
/ +
V 664#23 0#$
J
600
Kg#>mH
J
600
<@ ?@
;
J
664#23 0#$
Pro'i"
34$$#04E2032 240
Kg#>mH
? 3$0 # 2$0
PW p = = =
0#3$ 0#3$ #0$
+ per"u = = =
P Q P
J
# >m²
= =
600
0 ²
664#23 3 ksi # #
J
600 Kg#>mH
&isa dipakai#
Pe"at dasar ko"om pada pondasi kip = #$36 m² = 6#$2 Be&an aksia" ( P )
Kg#>mH
V
;
36#E$
Kg#>mH
Kg
=
3$E#43E
PW 3
kgi
3$E#43E 2E$#04$
Pro'i" yang digunakan ada"a >m² + = 0#$ d = 3 >m & = 2$ >m
Q
#0$
? 3$0 # 2$0
menentukan "e&ar ( B ) dan panjang ( / ) pe"at dimana arus memenui syarat : B @ / U
$$E0#46
>mH
Kg#>mH
untuk pemi"ian B dan / sedemikian ingga per&edaan ( se"isi ) diantara keduanya mendekati : 0#E2 d - 0#40 & 0#E2 # 3
-
0#4
#
2$
=
#24
>m
di>o&a &er&agai gaya B dan / se&agai &erikut : B @ / U
$$E0#46
7$ 7$ 7 70
>mH
76 74 40 43
$700 $6E $640 $70
gunakan B = 70 dan / = tekanan aktua" ada"a " p = B # /
3 E 3
43 664#23 70 #
=
=
43
#062
Kg#>mH
itung < dan / <
=
/ - 0#E$ d 2
=
/
=
/ - 0#40 d 2
=
itung te&a" pe"at ( tp ) tp = ( = 2$ ( = #0$2
P →
43
-
0#E$ 2
#
3
70
-
0#40 2
#
2$
Q 0#2$ y #062 Q >m
) 0#2$
pe"at "andasan yang dipakai ada"a
=
2$#3$
>m
=
2$
>m
½
#
200
)X
70
#
#
2$0 )
# =
E
43
= = =
(
? ;
# E
3$0 Q
23
;
=
2 mm
2
=
4
2
=
72
0#2$ # 0#2$ # 723#$6
'
# 3#
Kg
d H # # 2#
V H #
# 2
600
# 23
mm
