∑ = 0
(RVA + RVB) – (q1 ( 9) + P1 + P2 +P3 +P4 sin 35o + P5 )
= 0
(15,566 + 13,294) – (2( 9) + 3,5 + 1,5 + 3 + P4 sin 35o + 2 )
= 0
28,86 – 28,86
= 0 … (ok)
Gaya Lintang ( D )
( Dari kiri )
DC = 0 t
DD1= DC = 0t
DD2 = DD1 – P1 = - 3,500 t
DA1 = DD2= - 3,500 t
DA2 = DA1+RVA = - 3,5+ 15,566 = 12,066 t
DE1 = DA2 – q ( 2 ) = 12,066 – 4 = 8,066 t
DE2 = DE1 – P2 = 8,066 – 1,5 = 6,566 t
DF1 = DE2 – q ( 2 ) = 6,56 – 4 = 2,566 t
DF2 = DF1 – P3 = 2,566 – 3 = - 0,434 t
DG1 = DF2 – q ( 3 ) = - 0,434 – 6 = - 6,434 t
DG2 = DG1 – P4sin 35O = - 6,434 – 1,5 sin 35O =- 7,294 t
DB1 = DG2 – q ( 2 ) = - 7,294 – 4 =- 11,294 t
DB2 = DB1 + RVB=- 11,294 +13,294 = 2,000t
DH1 = DB2 =2,000 t
DH2 = DH1 – P5 = 2 - 2 = 0 t
( Dari kanan )
DH1= 0 t
DH2= P5 = 2,000 t
DB1 = DH2= 2,000 t
DB2 = DB1 – RVB= 2 - 13,294 = - 11,294 t
DG1 = DB2+ q ( 2 ) = - 11,294 + 4 = - 7,294 t
DG2 = DG1+P4 sin 35O = 7,294 – 1,5 sin 35O = - 6,434 t
DF1 = DG2+q ( 3 ) = - 6,434+ 6 = - 0,434 t
DF2 = DF1+P3 = - 0,434+ 3 = 2,566 t
DE1 = DF2+q ( 2 ) = 2,566+4 = 6,566 t
DE2 = DE1+P2 = 6,566+1,5 = 8,066 t
DA1 = DE2+q ( 2 ) = 8,066+ 4 =12,066 t
DA2 = DA1 – RVA = 12,066 – 15,566 =- 3,500 t
DD1 = DA2= - 3,500 t
DD2 = DD1 + P1 = - 3,5 +3,5 = 0 t
DC = DD2 = 0 t
Momen ( M )
( Dari Kiri )
MC = 0 t m
MD = 0 t m
MA = – P1(1,25) = - 4,375 t m
ME = – P1(3,25)+RVA(2) – 0,5 (q) (2)2 = 15,758 t m
MF = – P1 (5,25)+ RVA (4) – 0,5 (q) (4)2 – P2 (2) = 24,889 t m
MG = – P1 (8,25)+ RVA (7) – 0,5 (q) (7)2 – P2 (5) – P3 (3) = 14,588 t m
MB = – P1 (10,25)+ RVA (9) – 0,5 (q) (9)2 – P2 (7) – P3 (5) = – P4 sin 35o( 2 ) = 4,000t m
MH = – P1 (12,25)+ RVA (11) – (q) (9) (6,5) – P2 (9) – P3 (7) = – P4 sin 35o( 4 ) = 0 t m
( Dari Kanan )
MH = 0 t m
MB= P5 (2) = 4,000 t m
MG = P5 (4) – RVB (2) + 0,5 q (2) 2= - 14,588 t m
MF = P5 (7) – RVB (5) + 0,5 q (5)2 + P4 sin 35o( 3 ) = - 24,889 t m
ME = – P1 (5,25)+ RVA (4) – 0,5 (q) (4)2 – P2 (2) = 24,889 t m
MA = – P1 (8,25)+ RVA (7) – 0,5 (q) (7)2 – P2 (5) – P3 (3) = 14,588 t m
MD = – P1 (10,25)+ RVA (9) – 0,5 (q) (9)2 – P2 (7) – P3 (5) = – P4 sin 35o( 2 ) = 4,000 t m
MC = – P1 (12,25)+ RVA (11) – (q) (9) (6,5) – P2 (9) – P3 (7) = – P4 sin 35o( 4 ) = 0 t m
Gaya Normal ( N )
N = – P4 cos 35o = - 1,229 t m
Menentukan Ix danIy
Koordinat
Bag.
A . X²
A . y²
A.x.y
x ( cm )
y ( cm )
I
1.875
35.625
1406.25
507656.25
II
16.875
5.625
142382.81
III
-13.125
-24.375
∑
IX
Ixo
Iyo
26718.75
3333.333
53333.333
15820.31
47460.94
104166.67
4166.667
120585.94
415898.44
223945.31
5833.333
285833.333
264375
939375
298125
113333.33
343333.333
= ∑ IXO + ∑ AY2 = 113333,33 + 939375 = 1052708.333 cm4
IY
Terhadap sumbu sendiri
= ∑ IYO + ∑ AX2 = 343333.333 + 264375 = 607708.333 cm
IXY = ∑ IXOYO + ∑ AXY 4
= 0 + 298125 = 298125cm
4
Gambar
1. Letakgarisnetralpenampangsecaragrafisdananalitis
