tugas anstruk 3 - cover & no 1
DESCRIPTION
ANALISA STRUKTURTRANSCRIPT
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TUGAS
ANALISA STRUKTUR
Nama Dosen Pengajar :Ir. F. X. Ferry Munaf MT.
Nama Ma!as"s#a : Ase$ Su%ra
NPM : &'( &'( )**+
PR,GRAM K-LAS R-GUL-R K
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FAKULTAS T-KNIK SIPIL
S-M-ST-R GASAL TAUN AKAD-MI
UNI1-RSITAS PAN2ASIL
AKARTA
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3
a%
USUS
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K '*(& / '*(0
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1. Suatu rangka batang statis tertentu dibebani muatan terpusat dengan ketentuan
sebagai berikut :
P = 250 kg
a = 3.00 m
h= 2.00 m
Ditanyakan : Hitung besarnya gaya batang dengan menggunakan metode fe!ibiitasP P
C 2 D 3 B
1 " # $ % 2.00
A & E 5 F
P P& m & m 3 m
Penyeesaian:
sin ' = 0.55%"
' (os ' = 0.#321
1) . Gaya P = 1 di titik C, D, E & F
! y )
C 2 D 3 B
1 " # $ % 2
A & E 5 F
! y )
& m & m 3 m
a. *eseimbangan +uar
, ! = 0 0 - /
0 500.00 - /
0 500.00 - /
b. *eseimbangan daam : - (ara keseimbangan titik /
S1 sin ' S1
, = 0
0
' S& S1 = -901.39 kg (Tekan)
A S1 (os '
4
H =, 6
4=
=
, 6=
4 =
7 8itik
9 S1 sin ' =
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, = 0
S& 9 S1 (os ' = 0
S = !"0.00 kg (Ta#ik)
![Page 7: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/7.jpg)
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P = 250
!
C S2 7 4atang S2 di titik
D1 S" 2 m 0
d S$ = -11$".00
A S& E
! ! 7 4atang S& di titik ;
& m
0
d = 5."2 m S = !"0.00
7 4atang S" di titik D
. 3<2 ! P. ! S& .h S" d =
S! = $$.11
P = 250
y
C S2 D
S# 2 m
S5
A & E y F
P = 250
! 1<2 !
& m 3 m
7 4atang S5 di titik D
0
S" = 11$".00 kg (Ta#ik)
7 4atang S# di titik >
0
S% = 0.00 kg
7 4atang S2 di titik
0
7 Potongan
, 6 = 0
. ! P . 1<2! 9 S2 h =
, 6 = 0
.1<2! S& h =
, 6 = 0
7 Potongan
, 6 = 0
. 3<2! P . ! P 1<2! S5 h =
, 6 = 0
. 2! P 3<2! P .! 9 S2. h9 S#. d =
, 6 = 0
. ! P. 1<2! 9 S2 . h=
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S$ = -11$".00 kg (Tekan)
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P = 250 P = 250
C D ) S3 B
S$ 2 m
S5
A E ) F
P = 250
& m & m 3 m
7 4atang S3 di titik >
0
S3 = -!"0.00 kg (Tekan)
7 4atang S$ di titik 4
0
S$ = 2&2.11 kg -8ekan/
S3 S% (os ' B'
2 m
S% S% sin '
, = 0
0
S = 901.39 kg (Tekan)
, = 0
S3 S% (os ' = 0
S3 = -!"0.00 kg (Ta#ik)
' Batang Ta#ik - Tekan
1 S1 $01.3$
2 S2 1125.00
3 S3 "50.00
% S% $01.3$
7 Potongan ? ?
, 6 = 0
. 2! P . 3<2! P . ! P . 1<2! 9 S3 h =
, 6 = 0
. 5<2! P . 2! P . ! P . 1<2! S5 h S$ . d=
4
4 S% sin ' =
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5 S5 1125.00
& S& "50.00
" S" 2&2.11
# S# 0.00
$ S$ 2&2.11
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m
m
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kg (Tekan)
kg (Ta#ik)
0
kg (Ta#ik)
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$) . Pe*+e#ian gaya P = 1 atan di titik C
P = 1 atan
! y )
C 2 D 3 B
1 " # $ % 1.#
A & E 5 F
! y )
3.# m 3.# m 1.$ m
a. *eseimbangan +uar
, ! = 0 0 - /0 0.#0 - /
0 0.20 - /
b. *eseimbangan daam : - (ara keseimbangan titik /
S1 sin ' S1
, = 0
0
' S& S1 = -1.1 kg (Tekan)
A S1 (os '
, = 0
S& 9 S1 (os ' = 0
S = 0.% kg (Ta#ik)
4
H =, 6
4=
=
, 6=
4 =
7 8itik
9 S1 sin ' =
![Page 18: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/18.jpg)
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P = 1 satuan
!
C S2 7 4atang S2 di titik
D1 S" 1.# m 0
d S$ = -0.3
A S& E
! ! 7 4atang S& di titik ;
3.#m
0
d = 3.5" m S = 0.%
7 4atang S" di titik D
. 3<2 ! P. ! S& .h S" d =
S! = -0.$1
P = 1 satuan
y
C S2 D
S# 1.# m
A & E S5 F
y
! 1<2 !
3.# m 1.$ m
7 4atang S5 di titik D
0
S" = 0.$$$$$$$$$ kg (Ta#ik)
7 4atang S# di titik >
0
S% = 0.$1 kg (Ta#ik)
7 4atang S2 di titik
0
7 Potongan
, 6 = 0
. ! P . 1<2! 9 S2 h =
, 6 = 0
.1<2! S& h =
, 6 = 0
7 Potongan
, 6 = 0
. 3<2! P . ! S5 h =
, 6 = 0
. 2! P 3<2! 9 S2. h9 S#. d =
, 6 = 0
. ! P. 1<2! 9 S2 . h=
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S$ = -0.3 kg (Tekan)
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P = 1 satuan
)
C D S3 B
S$ 1.# m
S5
A E F
)
3.# m 3.# m 1.$ m
7 4atang S3 di titik >
0
S3 = -0.$1 kg (Tekan)
7 4atang S$ di titik 4
0
S9 = -0.$1 kg (Tekan)
S3 S% (os ' B'
1.#m
S% S% sin '
, = 0
0
S = 0.$9 kg (Tekan)
, = 0
S3 S% (os ' = 0
S3 = -0.$1 kg (Ta#ik)
@o 4atang 9 8arik 8ekan
1 S1 1.1&
2 S2 0.&3
3 S3 0.21
% S% 0.2$
7 Potongan ? ?
, 6 = 0
. 2! P . 3<2! 9 S3 h =
, 6 = 0
. 5<2! P . 2! S5 h S$ . d=
4
4 S% sin ' =
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5 S5 0.%222222222
& S& 0.#%
" S" 0.21
# S# 0.21
$ S$ 0.21
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m
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kg (Tekan)
kg (Ta#ik)
0
kg (Tekan)
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3) . Pe*+e#ian gaya P = 1 atan di titik D
P= 1 atan
! y )
C 2 D 3 B
1 " # $ % 1.#
A & E 5 F
! y )
3.# m 3.# m 1.$ m
a. *eseimbangan +uar
, ! = 0 0 - /
0 0.%0 - /
0 0.&0 - /
b. *eseimbangan daam : - (ara keseimbangan titik /
S1 sin ' S1
, = 0
0
' S& S1 = -0."% kg (Tekan)
A S1 (os '
, = 0
S& 9 S1 (os ' = 0
S = 0.$ kg (Ta#ik)
4
H =
, 64= =, 6
=
4 =
7 8itik
9 S1 sin ' =
![Page 28: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/28.jpg)
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!
C S2 7 4atang S2 di titik
D1 S" 1.# m 0
d S$ = -0.%
A S& E
! ! 7 4atang S& di titik ;
3.#m
0
d = 3.5" m S = 0.$
7 4atang S" di titik D
. 3<2 ! S& .h S" d =
S! = 0.3
y
C S2 D
S# 1.# m
A & E S5 F
y
! 1<2 !
3.# m 1.$ m
7 4atang S5 di titik D
0
S" = 1.$! kg (Ta#ik)
7 4atang S# di titik >
0
S% = -0.3 kg (Tekan)
7 4atang S2 di titik
7 Potongan
, 6 = 0
. ! 9 S2 h =
, 6 = 0
.1<2! S& h =
, 6 = 0
7 Potongan
, 6 = 0
. 3<2! S5 h =
, 6 = 0
. 2! 9 S2. h9 S#. d =
, 6 = 0
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0
S$ = -0.% kg (Tekan)
. ! 9 S2 . h=
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P = 1 satuan
)
C D S3 B
S$ 1.# m
S5
A E F
)
3.# m 3.# m 1.$ m
7 4atang S3 di titik >
0
S3 = -0.3 kg (Tekan)
7 4atang S$ di titik 4
0
S9 = -0. kg (Tekan)
S3 S% (os ' B'
1.#m
S% S% sin '
, = 0
0
S = 0.%! kg (Ta#ik)
, = 0
S3 S% (os ' = 0
S3 = -0.3 kg (Tekan)
@o 4atang 9 8arik 8ekan
1 S1 0.5#
2 S2 0.#%
3 S3 0.&3
% S% 0.#"
7 Potongan ? ?
, 6 = 0
. 2! P . 1<2! 9 S3 h =
, 6 = 0
. 5<2! P . ! S5 h S$ . d=
4
4 S% sin ' =
![Page 31: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/31.jpg)
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5 S5 1.2&&&&&&&&"
& S& 0.%2
" S" 0.%3
# S# 0.%3
$ S$ 0.&%
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m
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kg (Tekan)
kg (Ta#ik)
0
kg (Ta#ik)
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) . Pe*+e#ian gaya P = 1 atan di titik F
! y )
C 2 D 3 B
1 " # $ % 1.#
A & E 5 F
! y ) P= 1
3.# m 3.# m 1.$ m
a. *eseimbangan +uar
, ! = 0 0 - /
0 0.20 - /
0 0.#0 - /
b. *eseimbangan daam : - (ara keseimbangan titik /
S1 sin ' S1
, = 0
0
' S& S1 = -0.$9 kg (Tekan)
A S1 (os '
, = 0
S& 9 S1 (os ' = 0
S = 0.$1 kg (Ta#ik)
4
H =
, 64= =, 6
=
4 =
7 8itik
9 S1 sin ' =
![Page 38: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/38.jpg)
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!
C S2 7 4atang S2 di titik
D1 S" 1.# m 0
d S$ = -0.$
A S& E
! ! 7 4atang S& di titik ;
3.#m
0
d = 3.5" m S = 0.$1
7 4atang S" di titik D
. 3<2 ! S& .h S" d =
S! = 0.$1
y
C S2 D
S# 1.# m
A & E S5 F
y
! 1<2 !
3.# m 1.$ m
7 4atang S5 di titik D
0
S" = 0.333333333 kg (Ta#ik)
7 4atang S# di titik >
0
S% = -0.$1 kg (Tekan)
7 4atang S2 di titik
7 Potongan
, 6 = 0
. ! 9 S2 h =
, 6 = 0
.1<2! S& h =
, 6 = 0
7 Potongan
, 6 = 0
. 3<2! S5 h =
, 6 = 0
. 2! 9 S2. h9 S#. d =
, 6 = 0
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0
S$ = -0.$ kg (Tekan)
. ! 9 S2 . h=
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)
C D S3 B
S$ 1.# m
S5
A E F
)
3.# m 3.# m 1.$ m
7 4atang S3 di titik >
0
S3 = -0.% kg (Tekan)
7 4atang S$ di titik 4
0
S9 = 0.$1 kg (Ta#ik)
S3 S% (os ' B'
1.#m
S% S% sin '
, = 0
0
S = 1.1 kg (Ta#ik)
, = 0
S3 S% (os ' = 0
S3 = -0.% kg (Tekan)
@o 4atang 9 8arik 8ekan
1 S1 0.2$
2 S2 0.%2
3 S3 0.#%
% S% 1.1&
7 Potongan ? ?
, 6 = 0
. 2! 9 S3 h =
, 6 = 0
. 5<2! S5 h S$ . d=
4
4 S% sin ' =
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5 S5 0.&333333333
& S& 0.21
" S" 0.21
# S# 0.21
$ S$ 0.21
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m
atan
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kg (Tekan)
kg (Ta#ik)
0
kg (Ta#ik)
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") . Pe*+e#ian gaya P = 1 atan di titik E
! y )
C 2 D 3 B
1 " # $ % 1.#
A & E 5 F
! y )
P= 1 atan
3.# m 3.# m 1.$ m
a. *eseimbangan +uar
, ! = 0 0 - /
0 0.&0 - /
0 0.%0 - /
b. *eseimbangan daam : - (ara keseimbangan titik /
S1 sin ' S1
, = 0
0
' S& S1 = -0.%! kg (Tekan)
A S1 (os '
, = 0
S& 9 S1 (os ' = 0
S = 0.3 kg (Ta#ik)
4
H =
, 64= =, 6
=
4 =
7 8itik
9 S1 sin ' =
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!
C S2 7 4atang S2 di titik
D1 S" 1.# m 0
d S$ = -1.$!
A S& E
! ! 7 4atang S& di titik ;
3.#m
0
d = 3.5" m S = 0.3
7 4atang S" di titik D
. 3<2 ! S& .h S" d =
S! = 0.
y
C S2 D
S# 1.# m
A & E S5 F
y
P = 1 satuan
! 1<2 !
3.# m 1.$ m
7 4atang S5 di titik D
0
S" = 0.% kg (Ta#ik)
7 4atang S# di titik >
0
S% = 0.3 kg (Ta#ik)
7 4atang S2 di titik
7 Potongan
, 6 = 0
. ! 9 S2 h =
, 6 = 0
.1<2! S& h =
, 6 = 0
7 Potongan
, 6 = 0
. 3<2! P .1<2 ! S5 h =
, 6 = 0
. 2! P.! 9 S2. h9 S#. d =
, 6 = 0
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0
S$ = -1.$! kg (Tekan)
. ! 9 S2 . h=
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)
C D S3 B
S$ 1.# m
S5
A E F
)
P = 1 satuan
3.# m 3.# m 1.$ m
7 4atang S3 di titik >
0
S3 = -0.$ kg (Tekan)
7 4atang S$ di titik 4
0
S9 = -0.3 kg (Tekan)
S3 S% (os ' B'
1.#m
S% S% sin '
, = 0
0
S = 0."% kg (Ta#ik)
, = 0
S3 S% (os ' = 0
S3 = -0.$ kg (Tekan)
' Batang Ta#ik - Tekan
1 S1 0.#"
2 S2 1.2"
3 S3 0.%2
% S% 0.5#
7 Potongan ? ?
, 6 = 0
. 2! P . ! 9 S3 h =
, 6 = 0
. 5<2! P . 3<2! S5 h S$ . d=
4
4 S% sin ' =
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5 S5 0.#%%%%%%%%%
& S& 0.&3
" S" 0.&%
# S# 0.%3
$ S$ 0.%3
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m
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kg (Tekan)
kg (Ta#ik)
0
kg (Ta#ik)
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' BatangBe+an a# (P = 1)
P = 1 Sat
C D
Ta#ik - Tekan Ta#ik - Tekan Ta#ik - Tekan
1 S1 0.00 $01.3$ 0.00 1.1& 0.00 0.5#
2 S2 0.00 1125.00 0.00 0.&3 0.00 0.#%
3 S3 0.00 "50.00 0.00 0.21 0.00 0.&3
% S% $01.3$ 0.00 0.2$ 0.00 0.#" 0.00
5 S5 1125.00 0.00 0.%2 0.00 1.2" 0.00
& S& "50.00 0.00 0.#% 0.00 0.%2 0.00
" S" 2&2.11 0.00 0.00 0.21 0.%3 0.00
# S# 0.00 0.00 0.21 0.00 0.00 0.%3
$ S$ 0.00 2&2.11 0.00 0.21 0.00 0.&%
Aaya batang disusun daam bentuk matri! statis B PC
/P/ = 1.1& 0.5# 0.2$ 0.#" S1
0.&3 0.#% 0.%2 1.2" S2
0.21 0.&3 0.#% 0.%2 S3
0.2$ 0.#" 1.1& 0.5# S%
0.%2 1.2" 0.&3 0.#% S5
0.#% 0.%2 0.21 0.&3 S&
0.21 0.%3 0.21 0.&% S"
0.21 0.%3 0.21 0.%3 S#
0.21 0.&% 0.21 0.%3 $ ! % S$
; D >
= 1.1& 0.&3 0.21 0.2$ 0.%2 0.#%
0.5# 0.#% 0.&3 0.#" 1.2" 0.%2
0.2$ 0.%2 0.#% 1.1& 0.&3 0.21
0.#" 1.2" 0.%2 0.5# 0.#% 0.&3
// = 0.02 0 0 0 0 0
0 0.03 0 0 0 00 0 0.03 0 0 0
0 0 0 0.02 0 0
0 0 0 0 0.03 0
0 0 0 0 0 0.03
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
= 0.021 0.01$ 0.00& 0.005 0.013 0.0250.010 0.025 0.01$ 0.01& 0.03# 0.013
0.005 0.013 0.025 0.021 0.01$ 0.00&
/PT/
/P/T
//
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0.01& 0.03# 0.013 0.010 0.025 0.01$
/F/ = // /P/
0.0&# 0.0&3 0.03& 0.0"&
0.0&3 0.121 0.0"& 0.105
0.03& 0.0"& 0.0&# 0.0&30.0"& 0.105 0.0&3 0.121 % ! %
/P/T
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2 = 1.2
1.2
1.2
1.2 % 1
D = /F/ 2
0.2$ - /
0.%% - /
0.2$ - /
0.%% - /
Aaya 4atang BHC
/4/ = /P/ 2
3.%$
3.#0 =
2.53
3.%$
= 3.#0 8on
2.53
1.2#
0.00
1.2#
H1
D1
H2
D2
H3
D3
H%
D%
H5
H&
H"
H#
H$
![Page 60: Tugas Anstruk 3 - Cover & No 1](https://reader030.vdocuments.mx/reader030/viewer/2022020716/563db82d550346aa9a9140a4/html5/thumbnails/60.jpg)
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an di Titik
Tta5F E
Ta#ik - Tekan Ta#ik - Tekan
0.00 0.2$ 0.00 0.#" 2.$1
0.00 0.%2 0.00 1.2" 3.1"
0.00 0.#% 0.00 0.%2 2.11
1.1& 0.00 0.5# 0.00 2.$1
0.&3 0.00 0.#% 0.00 3.1"
0.21 0.00 0.&3 0.00 2.11
0.21 0.00 0.&% 0.00 1.0"
0.00 0.21 0.%3 0.00 0.00
0.21 0.00 0.00 0.%3 1.0"
0.21 0.21 0.21
0.%3 0.%3 0.&%
0.21 0.21 0.21
0.&% 0.%3 0.%3 % ! $
0 0 0
0 0 00 0 0
0 0 0
0 0 0
0 0 0
0.02 0 0
0 0.02 0
0 0 0.02 $ ! $
0.00% 0.00% 0.00%0.00# 0.00# 0.012
0.00% 0.00% 0.00%
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0.012 0.00# 0.00# % ! $
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0.2$ - /
0.%% - / (m
0.2$ - /
0.%% - /