Soal no: 1= 06 RA= . q . L2 6 RA= . 4 . 62 RA= = 12 ton = 0RB .
6= . q . L2 RB= = 12 tonKontrol : = 72 72 = 0MomenMA = 0
M max= RA . X - . q . X2 = 12 . 3 - . 4 . 32= RA qX = 18 tm = 12
4 X X = = 3 mMB= RA . 6 4 . . q . L2= 0Gaya Lintang
DA1= 0DA2= RA= 12 tonDB1= DA2 q . 6= 12 24= - 12 tonDB2= DB1 +
RB= -12 + 12= 0
Gambar diagram M .D .N
PERHITUNGAN LUAS dan TITIK BERAT BIDANG MOMEN
Cara PendekatanL1 = 2/3 x 3 x 18 = 36 Tm2Titik berat L1 terhadap
A= 5/8 x 3 = 1,875 m
Cara INTEGRALL1 = = L1= 36 Tm2TITIK BERAT
= TBLI = = 1,875 m
Cara pendekatanL2 = 2/3 x 3 x 18= 36 Tm2Titik berat L2 terhadap
A= (3/8 x 3 )+ 3 =4,125m
Cara INTEGRAL
L2 = = L2 = 36 Tm2
TITIK BERAT L2 dari A
= = 216 67,5TB L2 = = 4,125 m
Soal no : 2
= 0= 2 . 3 RB . 6 + . 4 . 62RB = = 13 tonRB= 13 ton= - . 4 . 62
2 . 3 + RA . 6 RA = 13 ton
= 0 = 26 26 = 0 .. ( ok )Moment MA= 0MC= RA . 3 - . 4 . 32 = 21
tmMB= RA . 6 - . 42 . 62 2 . 3= 0 tmGaya Lintang
DA1= 0DA2= RADC1= DA2 q . 3= 1 tonDC2 = DC1 P= 1 2= -1 tonDB1=
DC2 4 . 3= - 13 tonDB2= DB1 + RB= - 13 + 13= 0 ton
Gambar diagram M. D .N
PERHITUNGAN LUAS BIDANG MOMEN
PERHITUNGAN LUAS BIDANG MOMEN
Cara biasa.L 1 = 2/3 x 3 x 21= 42 Tm2TITIK BERAT L 1 thd. ATB L1
= 5/8 x 3 = 1,875 mCara integralMx = RA .X P . ( X-3 ) 4/2 .q2L 1
=== 40,5 Tm2TITIK BERAT L 1 TERHADAP A.TB L 1 = .TB L 1= .TB L 1=
40,5.TB L 1= 153/2TB L1= 17/9 m
PERHITUNGAN LUAS BIDANG MOMEN
Metode PendekatanL 2= 2/3 x 3 x 21= 42 Tm2TITIK BERAT L 2 thd.
ATB L1 = (3/8 x 3)+ 3 = 4,125 m
L 2= == 81 40,5L2= 40,5 Tm2TITIK BERAT L2 TERHADAP A.TB L 1 =
.TB L 1= .TB L 1= .TB L 1= 252 171/2.TB L 1= 333/2TB L 1= 37/9
m
Soal no 3 :
= 0 =2 . 2 + 2 . 4 + q . L2 RB. 6RB= = 14 ton = 0 =-2 . 2 - 2 .
4 - q . L2 + RB . 6RB= = 14 ton
= 0= 14 + 14 - 28 0 = 0 ( ok )MomentMA= 0MC= RA . 2 - 4 . 22= 20
tmMD= RA . 4 - . 4 . 42 2 . 4= 20 tmMB= RA . . 4 . 62 2 . 4 2 . 2=
0 tm
Momen max
Mx= RA . X P1 . ( X 2 ) - . q . X2 14 . 3 2 ( 3 2 ) - . 4 . 32 =
RA P q X = 22 tm= 3 mGaya Lintang
DA1= 0DA2= RADC1= DA2 4 . 2= 6 tonDC2= DC1 2 = - 4 tonDD1= DC2 4
. 2= - 4 tonDD2= DD1 2= - 6 tonDB1= DD2 4 . 2= - 14 tonDB2= DB1 +
DB2= - 14 14= 0 ton
Gambar diagram M ,D ,N
Perhitungan LUAS BIDANG MOMEN
Perhitungan LUAS 1
L1 = 2/3 x 2 x 20=80/3 Tm2Titik berat L1 terhadap AL1= 5/8 x 2=
10/8 = 1,25 m
Metode IntegralMx = RA . X . q . X2 P1 ( X -2 ) P2 ( X 4 )L1 = =
L1= = Tm2L1Titik berat L1 tehadap A
= = / = 22/17 = 1,294 m
Perhitungan LUAS 2
L2= 1 X 20= 20 Tm2Titik berat L2 terhadap AL2= ( x 1 ) + 2= 2,5
mPerhitungan LUAS 3
L3= 2/3 x 1x (Mmax MC)= 2/3 x 1 x ( 22 20 )= 4/3 Tm2Titik berat
L3 terhadap AL3= 2 + (5/8 x 1 )= 2,625 m
Perhitungan LUAS 4
L4= 2/3 x (Mmax MD )= 2/3 x ( 22 -20 )= 4/3 Tm2Titik berat L4
terhadap AL4= 3 + ( 3/8 x 1 )= 3,375 mPerhitungan LUAS 5
L5= 1 X 20= 20 Tm2Titik berat L5 terhadap AL5= 3 + 0,5= 3,5
m
Perhitungan LUAS 6
Metode pendekatanL6 = 2/3 x 2 x 20=80/3 Tm2
Titik berat L6 terhadap AL6= 4 + (3/8 x 2 )= 4,75 mMetode
IntegralL6= = = RA/2 X2 q/6 X3 P1/2.(X-2)2 P2/2 ( X-4 )= 88
196/3L6= Tm2Titik berat L6 terhadap A 68/3 TB L6= = = = 288 - = / =
= 4.706 m
