7/26/2019 Contoh 5 (Gerber)
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MEKANIKA TEKNIK STATISTERTENTU
(BALOK GERBER)
Oleh :
Ir. Rusman Asri, MM
FAKULTAS TEKNIK URUSAN SI!ILUNI"ERSITAS IN#O GLOBAL MAN#IRI
!ALEMBANG
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Lihat Gambar :
A BS
1
Rs1
Rs2
P
S1
S2
S2
C D
S1
S2
a aP
A B C D
Rs1
Rs2
RA
RB
-
RDRC
-
-
+
-
+
P
+
+
-
7/26/2019 Contoh 5 (Gerber)
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1. Hitung besarnya reaksi peretakan! m"men! gaya intang #an gaya
n"rma #ari k"nstruksi #iba$ah ini! #an gambarkan hasi hitungan
tersebut.
AB
$S
% m & m & m
' m
&*+m &*+m
! &
*n
%enyeesaian :
a&. Reaksi %eretakan :
Rs ' RC ' ()2 ' *)2 ' 2 t"n ↑
&b&. ,"men :
, ' Rs. / 02 ' 2. / 2&2
, ' 2. 2
' ,s '
' ! , ' !3 t"n.m
' 1!
, ' 1! t"n.m ' 1! , ' !3 t"n.m
' 2! ,4 ' t"n.m
b&. Gaya Lintang :
D ' #,&)# ' 2 2
' Ds ' 2! t"n
' ! D ' 1! t"n ' 1! D ' -
1! t"n ' 1! D ' ! ' 2! D4 ' -
s $
& *+m
Rs &*n
R$
&*n
-,/
-,/',--
&,--
&,--',--
',--
1&. Ba"k %enggantung S - C: 5 5 2 m
( ' 0. L ' 2. 2 ' * t"n
0 % Tn
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2&. Ba"k 6tama A S : 5 5 3 m
%a#a perhitungan ba"k utama!
,aka Rs ' 2 t"n men7a#i beban #i S
a&. Reaksi %eretakan :
( ' 0. * ' 2. * ' 8 t"n
,B '
RA. 9 (. * %. 2 + Rs. 1 '
RA ' *( + 2% Rs& ) 9
RA ' *. 8 + 2. 2 2&)9
RA ' !93 t"n ↑&
'
RA + RB ( % Rs '
RB ' ( + % + Rs RA
RB ' 8 + 2 + 2 !93
RB ' 9!;; t"n ↑&
Rs &*n
(
RA /,1*n
RB 1,22*n
AB S
% m & m ' m
&*+m
! &*n
#
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b&. ,"men :
,e#an A D : 5 5 * m
, ' RA. / 0.2
, ' !93. / 2& 2
, ' !93. 2
' ,A '
' 1! , ' *!93 t"n.m
' 2! , ' 3!;* t"n.m
' ;! , ' 8!1 t"n.m
' *! ,D ' 9!93 t"n.m
,e#an D B : 5 5 2 m, ' RA. +*& (+2& %.
, ' !93 +*& 8+2& 2
, ' *!;;. + 9!93
' ,D ' 9!93 t"n.m
' 1! , ' 2!;* t"n.m
' 2! ,B ' - 2! t"n.m
,e#an S B : 5 5 1 m
, ' Rs. ' - 2.
' ,s '
' ! , ' - 1! t"n.m
' 1! ,B ' - 2! t"n.m
AB
$S
% m & m & m' m
&*+m
&*+m
! &*n
+
-
M
4,67
7,34 8,01 6,67
2,23
2,00
RA RB R$
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4&. Gaya Lintang :
D ' #,&)#
,e#an A D : 5 5 * m
, ' !93. 2
D ' #!93. 2
&)#D ' !93 2
' DA ' !93
t"n
' 1! D ' 2!93
t"n
' 2! D ' 1!;*
t"n ' ;! D ' - !;;
t"n
' *! DD ' - 2!;;
t"n
,e#an D B : 5 5 2 m, ' *!;;. + 9!93
D ' *!;;
' DD ' - *!;;
t"n
' 1! D ' - *!;;
t"n
' 2! DB ' - *!;;t"n
,e#an S B : 5 5 1 m
D ' #-2.&)#
D ' - 2
' Ds ' - 2! t"n
' ! D ' - 2! t"n
' 1!
DB ' - 2! t"n
AB
$S
% m & m & m' m
&*+m
&*+m
! &*n
4,33 4,33
2,002,00
-
++
#-
5,672,67
1,34
0,332,33
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+
+
-
4,67
7,34 8,01 6,67
2,23
2,00
1,00 1,002,00
M
--
++
5,672,67
1,34
0,332,33
4,33 4,33
2,00 2,00
2,00
#
AB
$S
% m & m & m' m
&*+m
&*+m
! &*n
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2. Hitung Reaksi peretakan! m"men! gaya intang #an gaya n"rma #ari
k"nstruksi berikut ini! #an gambarkan bi#ang masing-masing.
%enyeesaian :
1. Ba"k %enggantung S C : 5 5 8m
a&. Reaksi %eretakan :
A
B $# S% m % m % m& m
! &√
&
&*n+m
RS % *n R$ % *nb&. Bi#ang ,"men :
, ' RS. /02 ' *. / 2& 2
, ' *. - 2
' ,S '
' 1 , ' ;! t"n.m
' 2 , ' *! t"n.m
' ; , ' ;! t"n.m ' * , '
D ' #*. - 2&)# ' * 2
' ,S ' *! t"n
' 1 , ' 2! t"n
' 2
, '
' ; , ' - 2! t"n
< ' * ,C ' - *! t"n
4&. Gaya Lintang :
2,--%,--2,--
%,--
%,--
&,--
&,--
(
$S
% m
&*n+m( ' 0. * ' 2*& ' 8 t"n
RS ' RC ' ()2 ' 8)2 ' * t"n ↑& ' 8 t"n
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AB
# S% m % m & m
! &√
&
B
A
# S% m % m & m
!. sin%/- !. 3s
%/-
RS
0
2. Ba"k 6tama A S : 5 5 1 m
a&. Reaksi %eretakan : ,B '
RA=. 8 %sin *. * + (.1 + RS.2 '
RA= ' %sin *. * (.1 RS.2&)8
RA= ' >2*& *1& *2&?)8
RA= ' ! t"n ↓&
% ' 2√2 t"n
%.sin * ' %.4"s *
%.sin * ' 2√2& /√2& ' 2 t"n
( ' 0.2 ' 2&2& ' * t"n
%a#a perhitungan ba"k utama!
,aka Rs ' * t"n men7a#i beban #i S RSRS
RA= ' - ! t"n
'
RA=. + RB %sin * (.1 RS '
RB ' %sin * + ( + RS Ra=
RB ' 2 + * + * !&
RB ' 1! t"n ↑&
RA= ' - ! t"n
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BA
# S
% m % m & m
!. sin%/- !. 3s
%/-
RS
0
RA= ' - ! t"n
RA= ' - ! t"n
b&. ,"men :
,e#an A D : 5 5 * m
, ' RA=. , ' !.
' ,A '
' 1 , ' ! t"n.m
' 2 , ' 1! t"n.m
' ; , ' 1! t"n.m
' * , ' 2! t"n.m
,e#an D B : 5 5 * m
, ' RA=. +*& %sin *.
, ' !+*& 2.
, ' 2!. 2 ' ,D ' 2! t"n.m
' 1 , ' *! t"n.m
' 2 , ' 3! t"n.m
' ; , ' @! t"n.m
' * ,B ' 12! t"n.m
,e#an S B : 5 5 2 m
, ' RS. / 0.2
, ' *. / 2&.2
, ' *. 2
' ,S '
' 1 , ' ! t"n.m
' 2 , ' 12! t"n.m
-0,50 -1,00 -1,50 -2,00
-4,50
-7,00
-9,50
-12,00
-5,00
-
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4&. Gaya Lintang :
,e#an A D : 5 5 * m
D ' # !.&)# ' !
' ,A ' ' 1 , ' ! t"n.m
' 2 , ' ! t"n.m
' ; , ' ! t"n.m
' * , ' ! t"n.m
,e#an S B : 5 5 2 m
D ' # *. 2&)# ' * 2
' DS ' *! t"n
' 1 D ' 9! t"n.m
' 2 DB ' 8! t"n.m
,e#an D B : 5 5 * mD ' #2!.2&)# ' 2!
' ,D ' 2! t"n.m
' 1 , ' 2! t"n.m
' 2 , ' 2! t"n.m
' ; , ' 2! t"n.m
' * ,B ' 2! t"n.m
BA
# S
% m % m & m
!. sin%/- !. 3s
%/-
RS
0
RA= ' - ! t"n
RA= ' - ! t"n
-,/&,/
4,-
%,-
#&. Bi#ang "rma :
,e#an A D : 5 5 * m
RAH ' %.4"s *
RAH ' 2 t"n ←& arik
-
+
&,-
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% m % m % m& m
AB
$# S
! &√
&
&*n+m
;! *! ;!
-
-! -1!-1! -2!
-*!
-3!
-@!
-12!
-!
M
*!
*!
2!
2!-
8!
*!
!2!
#
+
&,-N
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Dari S"a #iatas! 7ika ba"k #iren4anakan sebagaimana gambar #iba$ah
ini! maka ren4anakanah ba"k tersebut 7ika #an
peremahan akibat penger7aan 2 :
h ' *.b
b%ELSAFA :
,en#imensi ukuran ba"k! #iambi pa#a saat m"menmaksimum
,"men maksimum ,ma& a#aah m"men yang
teretak pa#a tumpuan B sebesar ,ma ' 12 t"n.m
,ma ' 12 t"n.m ' 1.2. kg.4m
2kg/cm1.400σ =
σ
' 1.* kg)4m2
akibat penger7aan ter7a#i peremahan sebesar 2!maka : σ ' !8 1.* kg)4m2 ' 1.12 kg)4m2
; 4m
3! 4m
$ ' ,)σ ' 1.2.)1.12 ' 1.31!*; 4m; ......1&
$ ' 1)9 bh2 #imana : h ' *.b
$ ' 1)9.b *b&2
$ ' 19)9.b; ....................................................................2&
%ers. 1& #an 2&
19)9.b; ' 1.31!*; cm7,5cm7,416
43)(6)(1.071, b 3 ≈==
b ' 3! 4m h ' *b ' *3!& ' ; 4m
,aka ukuran ba"k a#aah h ' ; 4m #an b ' 3! 4m
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