TUGAS BESAR BAJA
Direncanakan Bangunan gedung berikut sambungan serta gambar kerjanya dengan data – data sebagai berikut 1. Beban (P) 2. Bentang kuda – kuda () !. "um#a$ kuda – kuda (n) %. "arak antar kuda – kuda 5. Panjang Bangunan . *1 +. ,udut kemiringan . Beban /ngin . "eni "eniss /ta ta 10. Di D inding ,aming 11. katan /ngin Dinding 12. 6utu Baja 1!. "e " enis ,ambungan
Tinggi /ta
= 15 Ton = 15.000 kg = 10 meter = % kuda & kuda = !'5 m = % m ! '5 = 1% 1% m =5m = !0= %0 kgm2 = 3a#4 3a#4a# a#um um = Terbuka = rangka tersusun = /! = Baut (/!25)
1 tan α . .b 2 = 1 tan !0 0. .10 2 = = 2'meter
1
TUGAS BESAR BAJA
Panjang sisi miring
= =
2 2 a + t
5 2 + 2' 2
= 5'meter
Perhitungan Beban Pada Gording 1. Beban Mati
Jarak antar gording
=
1,6 m.
Berat penutup atap (Galalum!
=
".1# kg$m%.
Berat gording (a&um&i! (a&um&i!
=
' kg$m.
Berat atap,
=
6.'6 kg$m
=
' kg$m *
=
11.'6 kg$m = 1% kg$m
= (1.6!)(".1#!
Berat gording, + 1
+ =1%kg$m
=> R +
=2
=> 2 +
1
=
2
x 12 x 10
=
60 kg
R +)
= (6#!.-o& #/
=
'1.06 kg
R +
= (6#!.&in #/
=
# kg
1
+ =1%kg$m
.( + . ,
=
.( + . ,2
1
=
8
x 12 x 10² =150 kg
2 +) = (1'#!.-o& #/
=
1%0,0# kgm
2 + = (1'#!.&in #/
=
3' kgm
2
TUGAS BESAR BAJA
Panjang sisi miring
= =
2 2 a + t
5 2 + 2' 2
= 5'meter
Perhitungan Beban Pada Gording 1. Beban Mati
Jarak antar gording
=
1,6 m.
Berat penutup atap (Galalum!
=
".1# kg$m%.
Berat gording (a&um&i! (a&um&i!
=
' kg$m.
Berat atap,
=
6.'6 kg$m
=
' kg$m *
=
11.'6 kg$m = 1% kg$m
= (1.6!)(".1#!
Berat gording, + 1
+ =1%kg$m
=> R +
=2
=> 2 +
1
=
2
x 12 x 10
=
60 kg
R +)
= (6#!.-o& #/
=
'1.06 kg
R +
= (6#!.&in #/
=
# kg
1
+ =1%kg$m
.( + . ,
=
.( + . ,2
1
=
8
x 12 x 10² =150 kg
2 +) = (1'#!.-o& #/
=
1%0,0# kgm
2 + = (1'#!.&in #/
=
3' kgm
2
TUGAS BESAR BAJA
!
TUGAS BESAR BAJA
2. Beban ban Hidup
Berat peker4a = 1## kg 1 5 = 1## kg => R
=2
. 5 ,
1
=
2
.(100)
= '# kg
R )
= (' ('#!.-o& #/
= ".# kg
R
= ('#!.&in #/
= %' kg
1 5 = 1## kg => 2
=%
. 5 , . ,
1
=
4
x 50 x 5
2 ) =(1 =(1%'!. '!.-o& # #/ =
1#3.0 kgm
2 =(1 =(1%'!.&in #/ =
% kgm
= 1%' kgm kgm
%
TUGAS BESAR BAJA
3. Beban Angin /ngin tekan' * = 7(#.#%!)8 9 (#."!:) angin ) (4arak gording! = 7(#.#%!)(#! 9 (#."!:)("#!)(1.'! = 1% kg$m 1 1 .(. , x 12 x 6 =36 kgm * 2 R; = = 2 1 * ;
2
=
=
/ngin $isa' < = (< #."!.angin ) (4arak gording! = (< #."!.("#!.(1.'! = <%" kg$m 1 1 .. x ( < 2 R; = = 2 1 < ;
2
=
1
.(. ,2 =
1
.(. ,2
8
24
−
) x 6
8
x 12 x 6²
54 kgm
=
72 kgm
=−
x 12 x 6² =−108 kgm
Kombinasi Pembebanan A. Pembebanan Sementara 1. Arah Tegak Lurus Bidang Atap
RU)1
= (1,%!.(R +) ! * (#,'!.(R ) ! * (1,!.(R; ! 5
TUGAS BESAR BAJA
= (1,%!.(63.''! * (#,'!.(".#! * (1,!.(6! = 1"0.'1 kg RU)%
= (1,%!.(R +) ! * (#,'!.(R ) ! * (1,!.(R; ! = (1.%!.(63.''! * (#.'!.(".#! * (1.!.(<3%! =
0.11 kg
2. Arah Seaar Bidang Atap
RU
= (1,%!.(R + ! * (#,'!.(R ! = (1,%!.(0.##! * (#,'!.(%'! = '0. kg
B. Pembebanan Tetap 1. Arah Tegak Lurus Bidang Atap
RU)
= (1,%!.(R +) ! * (#,'!.(R ) ! = (1,%!.(63.''! * (#,'!.(".#! = 1#%.31 kg
2. Arah Seaar Bidang Atap
RU
= (1,%!.(R + ! * (#,'!.(R ! = (1.%!.(0.##! * (#.'!.(%'! = '0.# kg
"adi tia gording menerima beban sebesar' ara tegak luru& bidang atap
= %.(1#%.31! = %#'."% kg
ara &e4a4ar bidang atap
= %.('0.#! =
11>.6 kg
Kombinasi Momen A. Momen Akibat Beban Sementara 1. Arah Tegak Lurus Bidang Atap
2 U)1
= (1.%!.(2 +) ! * (#.'!.(2 ) ! * (1.!.(2 ; !
TUGAS BESAR BAJA
= (1.%!.(%#%.6'! * (#.'!.(1#3.0! * (1.!.('"! = 63.#> kgm 2 U)%
= (1.%!.(2 +) ! * (#.'!.(2 ) ! * (1.!.(2 ; ! = (1.%!.(%#%.6'! * (#.'!.(1#3.0! * (1.!.(<1#>! =
1'6."> kgm
2. Arah Seaar Bidang Atap
2 U
= (1.%!.(2 + ! * (#.'!.(2 ! = (1.%!.(113! * (#.'!.(%! = 1'6." kgm
B. Momen Akibat Beban Tetap 1. Arah Tegak Lurus Bidang Atap
2 U)
= (1.%!.(2 +) ! * (#.'!.(2 ) ! = (1.%!.(%#%.6'! * (#.'!.(1#3.0! = %06.>3 kgm
2. Arah Seaar Bidang Atap
2 U
= (1.%!.(2 + ! * (#.'!.(2 ! = (1.%!.(113! * (#.'!.(%! =
1'6." kgm
+ipa&ang &agrod di tenga bentang &eingga, 1
2 U
=
2 .(1'6."!
=
3>.% kgm
Keterangan
Beban Tetap
Beban Sementara Angin Tekan Angin Hisap
2omen 2 u)
%06.>3 kgm
63.#> kgm
1'6."> kgm +
TUGAS BESAR BAJA
2 u Reak&i
1'6." kgm
Ru)
1#%.31 kg
Ru
'0.# kg
1'6." kgm 1"0.'1 kg
0.11 kg
'0. kg
Tabe! hasi! perhitungan kombinasi momen dan reaksi.
Tegangan Pada Pro"i! #
Dicoba dengan ro7i# 8150520 dan teba# = !.0 mm Dari tabe# didaat ni#ai 9 A = 0.#1 -m?, @) = "1.> -m, @ = 11.6 -m, r) = '.0# -m, r = %.> -m, l) = 1" -m" , l = '1 -m" , = %.11 -m, Co = '.11 -m, J = %3#% -m, ; = %">% -m
2 U) D
=
θ )
2 U1
θ 1
( 296.87 x 100 ) ( 156.4 x 100) + (0.9 x 41.8 ) (0.9 x 11.6 )
= =
+
%%>3.%1 kg$-m% F %"## kg$-m% (ke!
Lendutan Pada Pro"i! # , ∆ma)
=
2%0 500
=
2%0
= %,#> -m =#,#%> m
+
5 +
= 1 kg$m => +)
=(1!.-o& #/
= 11.%6 kg$m
+
= (1!.&in #/
= 6.' kg$m
= 1## kg => 5 +) 5 +
=(1##!.-o& #/ = >6.6# kg = (1##!.&in #/ = '# kg
endutan ter$ada sumbu ' )
= (1.%!. +) = (1.%!.(11.%6!
=
1.'1 kg$m
TUGAS BESAR BAJA
5 )
= (1.%!.5 +) = (1.%!.(>6.6#!
( ) . ,%
1
∆ )
= =
. !% E . H )
(
*
= 1#.0% kg
! 1 5 ) . , . % E . H )
)(
( 13.51 x 5 4 ) x + 384 ( 2.1 x 10 10) x ( 0.00000314 ) 1
( 103.92 x 5 4) x 48 ( 2.1 x 10 10) x ( 0.00000314 ) 1
)
= #.### * #.#%#' = #.#%#> m endutan ter$ada sumbu y'
= (1.%!. + = (1.%!.(6.'!
= 3.> kg$m
5
= (1.%!.5 + = (1.%!.('#!
= 6# kg
( 1 . ,%
1
∆
= =
. !% E . H 1
(
*
! 1 5 1 . , . % E . H 1
)(
( 7.8 x 5 4) + x 384 ( 2.1 x 10 10) x ( 0.0000051 ) 1
( 60 x 54 ) x 48 ( 2.1 x 1010 ) x ( 0.0000051 ) 1
)
= #.###110 * #.##3%0 = #.##3"1 m
∆
= =
( ∆ ) ) 2
+ ( ∆ 1 )
(0.020) 2
= #.#%% m
+
F
2
( 0.00+%1) 2 ∆ma)
= #.#%> m (ke rek!
TUGAS BESAR BAJA
:
;
"adi gording dengan ro7i# ro7i# 1'#)6')%# dan tebal = .# mm daat digunakan karena te#a$ memenu$i ersyaratan. A = 0.#1 -m? =3.#3 kg$m @) = "1.> -m @ = 11.6 -m r) = '.0# -m r = %.> -m l) = 1" -m" l = '1 -m" = %.11 -m Co = '.11 -m J = %3#% -m ; = %">% -m
10
TUGAS BESAR BAJA
Peren$anaan Tra$k Stang
Dari tabe# didaat ni#ai r = '.0# cm dan ry = %.> cm
λ = =
5
=
,k r1
< 2%0
500 2.!
= 210.1 < 2%0 ( dapat digunakan)
5.!0 kg <
5u = 2 )5 = 2 )5.!0 = 11. kg
11
TUGAS BESAR BAJA
φ 5n = φ )A)D1 > 5u 5u < φ )A)D1 A > A = d 2 =
5u
φ )D1 1
=
11. 0. ) 2%00
= 0.055 -m 2
π d 2
% % )A
π
=
% ) 0.055 !.1%
= 0.0+ -m 2
d = 0.0+ = 0.25 -m ≈ 0.! -m digunakan tra-k − & tan g φ ! mm
Perhitungan %katan Angin &bra$ing'
1. Pada Atap Tinggi atap (= ) = (tan α ) )0'5b
= (tan !0°) )(0'5 )12) = !.5 m 1
ua& bidang berangin, A =
2
.(12 ).( !.5)
= %1 m% 5 = (*#,0!.(%1!.("#! = 3'6 kg 1
Tiap &impul menerima beban &ebe&ar, 5 =
%
.(+5)
= 1>0 kg 12
TUGAS BESAR BAJA
∑ I = 0
P = 1 kg
5 − S o&α = 0 S =
5u
5 o&α
=
1
o& !5
= 2!0.+ kg
2!0.+ Jg
=
P = 1 kg
tan α =
!.5
= 0 .+ 5 α 1 = Ar- tan 0.+ = !5°
Ra.5 = 1 ) !.5 Ra = 10.kg φ 5n = φ )A) D1 > 5u
5u < φ )A) D1
A
>
A 2
=
d
=
d
=
5u
φ )D1 1
=
2!0.+ 0' ) 2%00
= 0'11 -m
2
π d 2
% % )A
π
=
0'1%
% ) 0'11
=
!'1%
= 0'1%
-m
2
0'!+ -m ≈ 0'% -m
digunakan bra-ing φ % mm
2. Pada (inding
Diakai rangka engaku. uas bidang berangin' / =
P
(20).()
=
120 m2
=
(0').(120).(%0)
=
%!20 kg 1
Tia simu# menerima beban sebesar' P = =
2
.( %!20)
210 kg 1!
TUGAS BESAR BAJA
1 m
1 m
1 m
1 m
1 m
1 m
1 m
100 kg
0.5 m
100 kg
= /
= B
Pro7i# batang $orisonta# dicoba memakai ro7i# ia' D y
(
(
y
t
d t
=
2'! mm
>
=
1'0 kgm
/
=
2'21 cm2
=
2' cm%
?
=
1'+0 cm!
i
=
1'12 cm
=
!%'0 mm
Pro7i# batang diagona# dicoba memakai tu#angan baja biasa' d
=
0'5 cm
/
=
(0'25).@.(0'5)2
=
0'20 cm2
Per$itungan ke#angsingan komonen tersusun' (,A 0! & 1+2 & 2002) A =
#uas enamang komonen struktur tersusun' mm
Ad =
#uas enamang satu unsur diagona#' mm
d =
anjang unsur diagona#' mm
l =
anjang komonen struktur ada kedua ujungnya yang dibatasi o#e$
2
2
unsur eng$ubung' mm a
=
jarak antara dua usat titik berat e#emen komonen struktur' mm
=
konstanta
1%
TUGAS BESAR BAJA
25cm
50 cm
λ l
π . =
y
π . =
A. ,d ! . Ad . ,1 .a 2
(2'21).(+1) ! 2.(0'20).(100).(50) 2 ≤ 50
=
=
2./.y2 2.y1
=
2.(2'21).(25)2 2.(2')
=
2'5! cm%
15
TUGAS BESAR BAJA
H 1 iy
2. A
=
2'5! 2.(2'21)
= =
25'0! cm ,k
λ 1
i1
=
+00
λ i1
=
λ 1
=
25'0!
=
2+'+
2
+
m 2
.λ l
( 2+'+) 2
2
+
= =
2'!
2 2
C
.() 2
200
Perhitungan )angka Horisonta!
Dari er$itungan statika dengan menggunakan rogram staad ro' diero#e$ gaya tarik maksimum terjadi ada batang 1 yaitu sebesar 1000 kg. 5
σ =
φ . A 1000
=
(0').(2'21)
=
%%' kgcm2
C
2%00 kgcm2
Dari er$itungan statika dengan menggunakan rogram staad ro' diero#e$ gaya tekan maksimum terjadi ada batang 1 yaitu sebesar 100 kg.
lk λ -
=
π .i min
.
D1 E
1
TUGAS BESAR BAJA
50 = =
π .(1'12)
.
!00 2'1.10 (0'25 ≤
0'5
λ - ≤ 1'25)
1'%!
ω =
1' − 0'+.λ -
2
1'%!
=
1' − (0'+).(0'5) 2
=
1'05
σ =
ω .
5
φ . A
(1'05). = =
100 (0').(2'21)
5%' kgcm2
C
2%00 kgcm2
Perhitungan )angka (iagona!
Dari er$itungan statika dengan menggunakan rogram staad ro' diero#e$ gaya tarik maksimum terjadi ada batang 1' 20' 22' 2%' 2' 2' !0 yaitu sebesar 10'0 kg. 5
7
=
φ . Ag 10'0
=
(0').(0'25).π .(0'5) 2
=
1+'! kgcm2
C
2%00 kgcm2
10'0) kg
10'0) kg
D 5 mm
1+
TUGAS BESAR BAJA
1
TUGAS BESAR BAJA
Perhitungan Beban Pada Kuda*kuda 1. Beban Tetap
P
=
E(1'2).D (0'5). F.2
=
E(1'2).(52'5) (0'5).(50)F.2
=
1+ kg
2. Beban Angin
/ngin tekan'
P
/ngin $isa'
P
=
(1'!).G .2
=
(1'!).(&21).2
=
&5%' kg
=
(1'!).G .2
=
(1'!).(&%).2
=
&21'% kg
Peren$anaan Kuda*kuda
Dari $asi# er$itungan staad ro ada kuda&kuda didaatkan ni#ai&ni#ai sebagai berikut9 6ma
=
52!0 kg.m
Ama
=
2100 kg
Hma
=
1+00 kg
Digunakan ro7i# IJ !00.!00.10.15 t1
=
10 mm
t2 r t1 * r
t2 B
t2 r
=
1 mm
/
=
11' cm2
>
=
% kgm
=
20%00 cm%
=
15 mm
1
TUGAS BESAR BAJA
y
=
+50 cm%
i
=
1!'1 cm
iy
=
+'15 cm
?
=
1!0 cm!
?y
=
%50 cm!
Tekuk #oka# ada saya B D
λ = λ p
!00
2.t D
=
(0'!).
E D1
=
2.15
(0'!). =
=
10
=
10'+
=
!0
=
10'5%
2.10 5 2%0
(
λ ≤ λ p
)
(
λ ≤ λ p
)
Tekuk #oka# ada badan !00
I
λ = λ p
t
=
(!'+). =
E D1
10
(!'+). =
2.10 5 2%0
"adi tidak memer#ukan engaku 7#ens atau Geb.
8ek tegangan
52!0.100
2
σ =
φ .@
2
σ =
φ . @
±
=
(0').(1!0)
=
%2+'2 kgcm2
52!0.100
K
φ . A
±
=
(0').(1!0)
=
%%'++ kgcm2
≤ 2%00 kgcm2
2100 (0').(11') ≤ 2%00 kgcm2
20
TUGAS BESAR BAJA
!00 mm 15 mm e1
e2
!00 mm
10 mm
8ek tegangan geser ,
=
A1 .e1
+ A2 .e2
=
(!0).(1'5).(1%'25) (1).(1!'5).('+5)
=
+!2'! cm!
(1+00).(+!2'!)
L .S
τ =
b.H
(1).(20%00)
= =
1'0! kgcm2
≤ 0'5. D 1 = 1!%% kgcm2
8ek #endutan , ∆
=
!0
10!50
= ∆ ma
=
!0
=
1'%0 mm
2'+5 mm C
2'+5 mm
Perhitungan Beban Pada #rane
(ead Load
Li"e Load
= meter
= meter
P
= 10 ton
P
= 100 kg
= 10.000 kg
21
TUGAS BESAR BAJA
K6
= ( ) (P 12)
K6
= ( ) (10.000 %)
= ( ) (P 12) = ( ) (100 %)
D
= 5000 kg
6D
= LP.
6
= 50 kg = L P.
= L . 10.000 .
= L . 100 .
= 20.000 kg.m
= 200 kg.m
,ca#e Jactor M
= 1.2 D 1.
6M
= 1.2 D 1.
= 1.2 5000 1. 50
= 1.2 20.000 1. 200
= 00 kg
= 2%.!20 kg
Asumsi Berat Batang 2+3 kg,m 2 Arah -
Dead oad 6D
= LP.
6
= LP.
6/
=LP.
= L . 2!.
= L . 10.000 .
= L . 100.
= 5 kg.m
= 20.000 kg.m
= 200 kg.m
,ca#e Jactor 6M;
= 1.2 6D 1. 6 0.5 6/ = 1.2 5 1. 20.000 0.5 200 = !2.++'2 kg.m
Arah
Dead oad 6D
=LP.
6
= LP.
6/
=LP.
= L . 2!. 5
= L . 10.000 . 5
= L . 100. 5
= !5!'+5 kg.m
= 12.500 kg.m
= 125 kg.m
,ca#e Jactor 6M;
= 1.2 6D 1. 6 0.5 6/ = 1.2 !5!'+5 1. 12.500 0.5 125 22
TUGAS BESAR BAJA
= 20.%+ kg.m Tegangan Pada Pro"i! /0
σ =
M < σ ( A 36 ) ∅ W
σ =
Mux Muy σ + < n ∅ Wx ∅ Wy
σ
=
32.767,2 0,85 x 0,0055
1.%+.22'51 kgm2 C
+
20.487 0,85 x 0,00193
<
3600 1,5
2%.000.000 kgm2 (ter#a#u aman rek)
6aka digunakan ro7i# IJ (%00%00) dengan asumsi berat 2! kgm2
Peren$anaan Ko!om
Dari $asi# er$itungan staad ro ada ko#om didaatkan ni#ai&ni#ai sebagai berikut9 6ma
=
52!0 kg.m
Ama
=
200 kg
Hma
=
100 kg
Digunakan ro7i# IJ !00.!00.10.15 t1
=
10 mm
2!
TUGAS BESAR BAJA
t2 r t1 * r
t2 B
t2 r
=
1 mm
/
=
11' cm2
>
=
% kgm
=
20%00 cm%
y
=
+50 cm%
i
=
1!'1 cm
iy
=
+'15 cm
?
=
1!0 cm!
?y
=
%50 cm!
=
15 mm
Ne#angsingan ter$ada sumbu 00.
#k =
1 2
lk
λ )
=
i)
2
=
%2%'2 mm
=
!2'!
C
200
=
5'!%
C
200
%2%'2 =
1!'1
Ne#angsingan ter$ada sumbu y lk
λ 1
=
i1
%2%'2 =
+'15
( λ = 5) dari tabe# didaatkan ni#ai ω ada#a$ 1'!2.
2%
TUGAS BESAR BAJA
8ek tegangan
ω .
σ =
K
(1'!2).
φ . A
= =
ω .
σ =
K
φ . A
+
2 = =
(0').(11') ≤ 2%00 kgcm2
!5'+2 kgcm2
(1'!2).
φ .@
(200)
(200) (0').(11')
+
(52!0).(100) (0').(1!0)
≤ 2%00 kgcm2
%! kgcm2
!00 mm 15 mm e2
!00 mm
e1
10 mm
8ek tegangan geser ,
=
A1 .e1
+ A2 .e2
=
(!0).(1'5).(1%'25) (1).(1!'5).('+5)
=
+!2'! cm!
(100).(+!2'!)
L .S
τ =
b.H
(1).(20%00)
= =
5+'%% kgcm2
≤ 0'5. D 1 = 1!%% kgcm2
Tekuk #oka# ada saya B D
λ = λ p
!00
2.t D
(0'5).
=
E
=
D1
2.15
(0'5). =
=
10
=
1'1+
=
!0
2.10 5 2%0
(
λ ≤ λ p
)
Tekuk #oka# ada badan !00
I
λ =
t
=
10
25
TUGAS BESAR BAJA
λ p
(1'%). =
E D1
(1'%). =
2.10 5 2%0
=
%!
(
λ ≤ λ p
)
2
TUGAS BESAR BAJA
Peren$anaan Pe!at Kaki Ko!om Angkur 5 cm
!0 cm
5 cm
5 cm
!0 cm
5 cm
6ma = Pma
=
200 kg
Hma
=
100 kg
52!0 kg.m
Direncanakan e#at kaki ko#om ukuran %0 cm ; %0 cm 5
σ
=
b.=
±
2 . b.= 2
50.%
%+.22
20.+% cm
1.2) cm
= 200 %0.%0
σ ma
±
(52!0).100. %0.%0 2 ±
=
1'1
%'0!
=
50'% kgcm2
σ min = &%+'22 kgcm2 σ min .(%0 & ) 1
=
σ ma
(%+'22).(%0 & 1)
=
(50'%).1
1' – (%+'22). 1
=
(50'%).1
1 =
1'2 cm
.1
6omen ada e#at kaki ko#om.
2+
TUGAS BESAR BAJA
1
6
.(%+'22).%0.5 2
=
2
=
2!10 kgcm
2
TUGAS BESAR BAJA
Ditentukan teba# e#at kaki ko#om ada#a$ 15 mm. Nontro# tegangan yang terjadi. 2u
σ
=
φ .@
=
2!10 1 φ . .b.t !
=
2!10 1 (0'). .%0.(1'5) !
=
11 kgcm2
≤ 2%00 kgcm2
Peren$anaan angkur 1
Tegangan J
2
=
1
.σ min .b.. )1
.(%+'22).%0.(1'2)
=
2
=
11'1% kg 11'1%
J
2
= =
0%'5+ kg
Ditentukan diameter angkur sebesar 10 mm dengan anjang 200 mm. µ =
(!5 – 0 O).7y
=
!5O.(2%00)
=
%0 kgcm2
Pn =
teg. #ekatan ke#i#ing
= µ .π .d ., .% =
(%0). π .1.(20).% 2
TUGAS BESAR BAJA
=
211 ton
5 cm
!0 cm
15 cm
5 cm
5 cm
!0 cm
5 cm
Nontro# ter$ada momen. 2u =
φ .2. 5n.e
=
(0'+5).2.(211).(0'15)
=
%+'% ton.m
5'2! ton.m
Nontro# ter$ada geser.
L
=
φ .r 1 . Du. A.% 1
=
(0'+5).(0'%).(!+00).( %
=
!'% ton
2
.π .1
).%
1' ton
!0
TUGAS BESAR BAJA
Peren$anaan Sambungan Ko!om dan Kuda*kuda
6ma
=
52!0 kg.m
Pma
=
200 kg
Hma
=
1+00 kg
Digunakan sambungan baut /!25' 7y =
000 kgcm2
7u =
250 kgcm2
Teba# e#at sambung 2 cm' jum#a$ baut direncanakan bua$ ada sisi ko#om dan ba#ok. Baut tei
9 1'5d C
Baut tenga$ 9 !d
C
,
C
!d
M C
+d
Teba# e#at2 cm ) cm
T
1% cm Q
T
1% cm ) cm
+cm 1) cm +cm
BautD2cm mutu/!25
/#ob.y
=
n.d.t !1
TUGAS BESAR BAJA
=
!.2.2
=
12 cm2
/gross.y =
#.t
=
%0.2
=
0 cm2
8ek /#ob.
≤ 15O.(/gross.)
12
≤ 15O.(0)
12 cm2 ≤ 12 cm2
!2
TUGAS BESAR BAJA
Nuat tarik baut T
6n
=
φ .(0'+5). Du. Ab.n
=
2 (0'+5).(0'+5).(250).(0'25).π .2 .2
=
215 kg
=
T.Q
=
(215).(0'2)
=
1% kg.m
52!0 kg.m
Nuat geser baut ada sisi ko#om
φ . Rn =
Hn
φ .r 1 . Du.m. Ab
=
(0'+5).(0'5).(250).1.(0'25).π .2 2
=
+1 kg
=
n. φ .Rn
=
.(+1)
=
5!1% kg
1+00 kg
Nuat geser baut ada sisi ba#ok
φ . Rn =
Hn
φ .r 1 . Du.m. Ab
=
(0'+5).(0'5).(250).1.(0'25).π .2 2
=
+1 kg
=
n. φ .Rn
=
.(+1)
=
5!1% kg
200 kg
Naasitas desak e#at
φ . Rn =
( 2'%).φ .d .t . Du
=
(2'%).(0'+5).2.2.!+00
=
2%0 kg
!!
TUGAS BESAR BAJA
Perencanaan ,ambungan as untuk Pro7i# Bentukan ada ,ambungan Ba#ok&No#om. /ra$ 4ertika#
P =1+00kg
%0cm
!0cm
Teba# #as sudut 1
a
2
=
1
2 )t
2 ) 2
=
2
=
1'%1 cm2
Tegangan #as
σ la&
5 2 a),
=
1+00
=
2 (1'%1).(%0)
=
15'0+ kgcm2
C
2%00 kgcm2
P=200kg
%0 cm
!0 cm
/ra$ diagona# Teba# #as sudut 1
a
=
2
2 )t
!%
TUGAS BESAR BAJA
1
2 ) 2
=
2
=
1'%1 cm2
Tegangan #as
σ la&
=
5 2 a), 200
=
2 (1'%1).(%0)
=
25'+1 kgcm2
C
2%00 kgcm2
!5
TUGAS BESAR BAJA
Peren$anaan Sambungan Kuda*kuda
6ma
=
2500 kg.m
Pma
=
110 kg
Hma
=
500 kg
Digunakan sambungan baut /!25' 7y =
000 kgcm2
7u =
250 kgcm2
Teba# e#at sambung 2 cm' diameter baut 2 cm jum#a$ baut ada sisi 4ertika# bua$ dan diagona# % bua$. Baut tei
9 1'5d C
Baut tenga$ 9 !d
C
,
C
!d
M C
+d
Teba# e#at 2 cm cm T1 1% cm cm cm
T2 Q2
Q1
1% cm cm
IJ !00.!00.10.15 Baut D2 cm mutu /!25 + cm 1) cm + cm
/#ob.y
=
n.d.t
=
%.2.(2)
=
1 cm2
/gross.y =
#.t
=
0.2
=
120 cm2
8ek !
TUGAS BESAR BAJA
/#ob.
≤ 15O.(/gross.)
1
≤ 15O.(120)
1 cm2 ≤ 1 cm2
!+
TUGAS BESAR BAJA
Nuat tarik baut T
6n
=
φ .(0'+5). Du. Ab.n
=
2 (0'+5).(0'+5).(250).(0'25).π .( 2) .2
=
215 kg
=
T.Q1 T.Q2
=
(215).(0'%%) 22 .(215).(0'1)
=
1%52 kg.m
2500 kg.m
Nuat geser baut ara$ 4ertika#
φ . Rn =
Hn
φ .r 1 . Du.m. Ab
=
(0'+5).(0'5).(250).1.( 0'25).π .( 2) 2
=
+1 kg
=
n. φ .Rn
=
.(+1)
=
+++52 kg
500 kg
Nuat geser baut ara$ diagona#
φ . Rn =
Hn
φ .r 1 . Du.m. Ab
=
(0'+5).(0'5).(250).1.( 0'25).π .( 2)
=
+1 kg
=
n. φ .Rn
=
%.(+1)
=
!+ kg
2
110 kg
Naasitas desak e#at
φ . Rn =
(2'%).φ .d .t . Du
=
(2'%).(0'+5).2.2.!+00
=
2%0 kg !
TUGAS BESAR BAJA
Perencanaan ,ambungan as untuk Pro7i# Bentukan ada ,ambungan Nuda&kuda.
P= 5 0 0 k g
)0 cm
!0 cm
/ra$ 4ertika# Teba# #as sudut 1
a
=
2 1
2 )t
2 ) 2
=
2
=
1'%1 cm2
Tegangan #as
σ la&
=
5 2 a), 500
=
2 (1'%1).(0)
=
2' kgcm2
C
2%00 kgcm2
!