The document discusses beams supported on two supports. It contains 5 topics: 1) A beam carrying a concentrated central load. 2) A beam carrying eccentric central loads. 3) A beam carrying a uniformly distributed load. 4) A beam carrying a triangular load. 5) A beam carrying a combination of loads. The purpose is for students to understand internal forces (bending moment and shear force) in beams supported on two supports under different load conditions and to calculate and illustrate them. Detailed calculations are shown for each load condition.

1. STA
TIKA I
MODUL 3
BALOK DIATAS DUA PERLETAKAN
Materi Pembelajaran :
1. Balok Diatas Dua Perletakan Memikul Sebuah Muatan Terpusat.
2. Balok Diatas Dua Perletakan Memikul Muatan Terpusat Sembarang.
3. Balok Diatas Dua Perletakan Memikul Muatan Terbagi Rata.
4. Balok Diatas Dua Perletakan Memikul Muatan Segi Tiga.
5. Balok Diatas Dua Perletakan Memikul Muatan Campuran.
WORKSHOP/PELATIHAN
Tujuan Pembelajaran :
• Mahasiswa memahami dan mengetahui tentang gaya-gaya dalam dari struktur balok
terletak diatas dua perletakan dengan beban-beban terpusat, beban terbagi rata dan
segitiga, dan mampu melakukan perhitungan gaya-gaya dalam (M,D,N) dan mampu
menggambarkannya.
1

2. BALOK DIATAS
DUA PERLETAKAN
1. Balok Diatas Dua Perletakan Memikul Sebuah Muatan Terpusat.
A
R
AV
a = 2 m
P = 10 ton
C
L = 6 m
b = 4 m
B
R
BV
D
A-C
= + R
AV
(+)
P
(-)
D
B-C
= - R
BV
Bidang gaya lintang
(+)
MC
= P.a.b/L
Bidang momen
Penyelesaian :
a. Reaksi Perletakan.
Σ MB
= 0,
R
AV
. L - P . b = 0
R
AV
= P . b/L
= (10 t) x (4 m)/(6 m)
RAV
= + 6,667 ton ()
Σ MA
= 0,
- R
BV
. L + P . a = 0
R
BV
= P . a/L
= (10 t) x (2 m)/(6 m)
RBV
= + 3,333 ton ().
Kontrol :
Σ V = 0,
R
AV
+ R
BV
– P = 0
6,667 t + 3,333 t - 10 t = 0 …..(memenuhi)
b. Gaya lintang.
DA-C
= + RAV
= + 6,667 ton.
DC-A
= + DA-C
= + 6,667 ton.
2

3. D
C-B
= D
C-A
– P = 6,667 – 10 = - 3,333 ton.
3

4. DB-C
= DC-B
= - RBV
= - 3,333 ton.
c. M o m e n .
MA
= 0
M
C
= + R
AV
. a = + 6,667 t x 2 m = + 13,334 ton.m’, atau
M
C
= P.a.b/L
Lihat gambar bidang gaya lintang dan momen diatas.
2. Balok Diatas Dua Perletakan Memikul Muatan Terpusat Sembarang.
R
AH
P1 = 2 t
A
C
P2 = 3 t
45o
D
P3 = 4 t
60
o
B
E
RAV
RBVL = 6 m
a1 = 1 m
b1
a2 = 3 m b2
a3 = 4 m
b3
+ 3,883 t + 1,883 t
+ 0,121 t
- 0,238 t
- 3,702 t
Bid. D
Bid. N
+ 7,649 t.m
- 2 t
+ 3,883 t.m
+ 7,411 t.m
Bid. M
Penyelesaian :
a. Reaksi Perletakan.
Σ MB = 0,RAV
. L - P1.(L –a1) – P2 sin 45o
.(L – a2) – P3 sin 60o
.(L – a3) = 0
R
AV
= P1.(L – a1)/L + P2 sin 45o.(L – a2)/L + P3 sin 60o.(L – a3)/L
= 2 x (6 - 1)/6 + 3 x ½√2 x (6 – 3)/6 + 4 x 0,866 x (6 – 4)/6
= 1,667 + 1,061 + 1,155
RAV
= + 3,883 ton ()
Σ MA = 0,- RBV
. L + P1.(a1) + P2 sin 45o
.(a2) + P3 sin 60o
.(a3) = 0
R
BV
= P1.(a1)/L + P2 sin 45o.(a2)/L + P3 sin 60o.(a3)/L
= 2 x (1)/6 + 3 x ½√2 x (3)/6 + 4 x 0,866 x (4)/6
= 0,333 + 1,061 + 2,309

5. RBV
= + 3,703 ton ()
Σ H = 0,
RA-H
+ P2 cos 45o
– P3 cos 60o
= 0
RA-H
= – P2 cos 45o
+ P3 cos 60o
= – 3 x ½√2 + 4 x ½ = – 2,121 + 2
= – 0,121 ton (¬ )
Kontrol :
Σ V = 0,R
AV
+ R
BV
– P1 – P2 sin 45o – P3 sin 60o = 0
3,883 t + 3,703 t – 2 t – 2,121 t – 3,464 t = 0
7,586 t – 7,585 t = 0,001 ≅ 0 …..(memenuhi)
b. Gaya Lintang.
DA-C
= + RAV
= + 3,883 ton.
D
C-D
= + R
AV
– P1 = 3,883 – 2 = + 1,883 ton.
DD-E
= + RAV
– P1 – P2 sin 45o
= 3,883 – 2 – 3 x ½√2
= 3,883 – 2 – 2,121 = – 0,238 ton.D
E-B
= + R
AV
– P1 – P2 sin 45o – P2 sin 60o
= 3,883 – 2 – 3 x ½√2 – 4 x 0,866 = 3,883 – 2 – 2,121 – 3,464
D
E-B
= – 3,702 ton.
D
E-B
= – R
BV
= – 3,703 ton.
c. Gaya Normal .
NA-D
= + RAH
= + 0,121 ton (tarik).
ND-E
= + RAH
– P2 cos 45o
= + 0,121 – 3 x ½√2 = + 0,121 – 2,121
= – 2 ton (tekan).
NE-B
= + RAH
– P2 cos 45o
– P3 cos 60o
= + 0,121 – 2 x ½√2 – 4 x ½
= 0,121 – 2,121 – 2
NE-B
= 0 ton.
c. M o m e n .
MC
= + RAV
. a1 = + 3,883 x 1 = + 3,883 ton.m’.
MD
= + RAV
. a2 – P1 . (a2 – a1) = + 3,883 x 3 – 2 x (3 – 1) = 7,649 t.m’.
ME
= + RAV
. a3 – P1 . (a3 – a1) – P2 sin 45o
. (a3 – a2)
= + 3,883 x 4 – 2 x (4 – 1) – 3 x ½√2 x (4 – 3)
= + 15.532 – 6 – 2,121
ME
= + 7,411 t.m’.

6. 3. Balok Diatas Dua Perletakan Memikul Muatan Terbagi Rata.
q = 3 t/m’
A B
R
AV
X
QR
x
L = 6 m
R
BV
Dx
+ ½ q L
Mx
+ 1/8 q L2
- ½ q L
Bid. D
Bid. M
Penyelesaian :
a. Reaksi Perletakan.
QR
= q . L = (3 t/m’) x (6 m) = 18 ton.
Σ M
B
= 0,
RAV
. L - QR
. ½ L = 0
R
AV
= ½ q . L2/L
R
AV
= ½ q . L …..(1)
= ½ x (3 t/m’)/(6 m)
RAV
= + 9 ton ()
R
BV
= R
AV
= ½ q . L = 9 ton. (simetris)
b. Gaya lintang.
DA-B
= + RAV
= + 7 ton.
D
B-A
= + R
AV
– q . L = - R
BV
= - 9 ton.
c. M o m e n .
Momen maksimum terjadi ditengah bentang,
Mmaks.
= + RAV
. ½L – q . ½L . ¼L
= ½ q L . ½L – 1/8 q L
2
= ¼ q L
2
– 1/8 q L
2
Mmaks. = + 1/8 q L
2
M
maks.
= + 1/8 x (3 t/m’) x (6 m’)2 = + 13,5 t.m’.
d. Tinjau tampang X.
Momen pada tampang X, dihitung dari kanan kekiri,
Mx = RAV
. x – q . x . ½ x
Mx = R
AV
. x – ½ q x2 …..(2)
Momen maksimum terjadi apabila gaya lintang sama dengan nol,
Dx = d(Mx)/dx = 0
(R
AV
. x – ½ q x2)/dx = 0
R
AV
– q . x = 0

7. x = R
AV
/q …..(3)

8. = ½ q L/q = ½ L = ½ x 6
x = 3 m (ditengah bentang).
Substitusikan (3) dan (1) kedalam (2), maka momen maksimum,
Mmaks.
= RAV
. (RAV
/q) – ½ q (RAV
/q)2
= (½ q L) . (½ q L/q) – ½ q . (½ q L/q)
2
= ¼ q L
2
– 1/8 q L
2
M
maks.
= 1/8 q L2 …..(4)
Untuk x = 1 m dan x = 3 m dari perletakan A, besar momen,
MX=1m
= 9 x 1 – ½ x 3 x (12
) = + 7,5 t.m’.
M
X=3m
= M
maks
= 9 x 3 – ½ x 3 x (32) = + 13,5 t.m’.
Gaya lintang,
Dx = d(Mx)/dx = RAV
– q . x …..(5)
Untuk x = 1 m dan x = 3 m dari perletakan A,
DX=1m
= 9 – 3 x (1) = + 6 t.m’.
D
X=3m
= 9 – 3 x (3) = + 0 t.m’.
10.000
Gaya Lintang (Ton)
Dx = 1/2 q L – q x
5.000
Dx
0.000
X
0
0
.5
1
.0
1
.5
2
.0
2
.5
3
.0
3
.5
4
.0
4
.5
5
.0
5
.5
6
.0
-5.000
-10.000
A B
16.0000
14.0000
12.0000
10.0000
8.0000
Mx
6.0000
4.0000
2.0000
0.0000

9. M
o
m
e
n
(
t
o
n
.
m
'
)
M
x
=
1/2 q L x – 1/2 q x2
X
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
A B

10. 4. Balok Diatas Dua Perletakan Memikul Muatan Segi Tiga.
2/3 x
1/3 x
qx (t/m’)
q = 3 t/m’
A
B
QX
X Q
R
x
RAV
RBV
2/3 L
L = 6 m
1/3 L
Penyelesaian :
a. Reaksi Perletakan.
QR
= q . ½ L = (3 t/m’) x ½ x (6 m) = 9 ton.
Σ M
B
= 0,
R
AV
. L – Q
R
. 1/3 L = 0
R
AV
= + 1/3 Q
R
= + 1/3 q . ½L
R
AV
= 1/6 q L
= 1/6 x (3 t/m’)/(6 m)
RAV
= + 3 ton ()
Σ M
B
= 0,
– R
BV
. L + Q
R
. 2/3 L = 0
R
BV
= + 2/3 Q
R
= + 2/3 q . ½L
R
BV
= 1/3 q L
= 1/3 x (3 t/m’)/(6 m)
RBV
= + 6 ton ()
Kontrol :
Σ V = 0,
R
AV
+ R
BV
– Q
R
= 0
3 ton + 6 ton – 9 ton = 0 ……(memenuhi)
b. Gaya lintang.
DA-B
= + RAV
= + 1/6 q L = + 3 ton.
D
B-A
= + R
AV
– Q
R
= 1/6 q L – ½ q L = – 1/3 q L = – R
BV
= – 6 ton.
c. Tinjau tampang X.
Tampang X terletak sejauh x dari perletakan A, momen pada tampang X, dihitung dari
kanan kekiri,
qx = q . x/L
QX
= qx
. ½ x = (q . x/L) . ½ x
= ½ q x2
/L
Mx = RAV
. x – QX
. 1/3 x
= (1/6 q L) . x – (½ q x
2
/L) . 1/3 x
Mx = 1/6 q L x – 1/6 q x3
/L …..(1)
2/3x
x
QX
1/3x
q
X
(t/m’)

11. Momen maksimum terjadi apabila gaya lintang sama dengan nol,
Dx = d(Mx)/dx = 0
= d(1/6 q L x – 1/6 q x
3
/L )/dx
Dx = 1/6 q L – ½ q x
2
/L …..(2)
1/6 q L – ½ q x2
/L = 0 x2
= 1/6 q L . 2 L /q
x = √ (1/3 L
2
)
x = 1/3 L√3 …..(3)
= 1/3 . (6 m) .√3
x = 3,464 m (dari perletakan A).
Substitusikan pers.(3) kedalam (1), maka diperoleh momen maksimum,
Mmaks
= 1/6 q L . (1/3 L√3) – 1/6 q (1/3 L√3)3
/L
= 1/6 q L
2
{1/3√3 – 1/9√3}
Mmaks
= 1/27 q L2
√3 …..(4)
Mmaks
= 1/27 x 3 x 62
x √3 = 6,9282 t.m’.
Tabel nilai momen dan gaya lintang
X D
x
M
xm t
o
ton.m'
.0 3.00
0
0.000
00.5 2.93
8
1.489
61.0 2.75
0
2.916
71.5 2.43
8
4.218
82.0 2.00
0
5.333
32.5 1.43
8
6.197
93.0 0.75
0
6.750
03.5 -
0.06
6.927
14.0 -
1.00
6.666
74.5 -
2.06
5.906
35.0 -
3.25
4.583
35.5 -
4.56
2.635
46.0 -
6.00
0.000
0
4.000
Gaya Lintang (Ton)
Dx = 1/6 q L – ½ q x2
/L
2.000
0.000
X
0
0
.5
1
.0
1
.5
2
.0
2.
5
3.
0
3.
5
4.
0
4.
5
5.
0
5.
5
6.
0
Dx
-2.000

12. -4.000
-6.000
-8.000
A
B

13. 8.0000
7.0000
6.0000
5.0000
4.0000
Mx
3.0000
2.0000
1.0000
0.0000
M o m e n (ton.m')
Mx = 1/6 q L x – 1/6 q x3
/L
X
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
A B
5. Balok Diatas Dua Perletakan Memikul Muatan Campuran.
P1 = 4 t P2 = 2 t
QR
q = 3 t/m’
A
R
AV
C
D E
1 m 1 m 1 m
3 m
B
R
BV
+ 4,250 t
+ 0,250 t
L = 6 m
+ 2,250 t
Dx = 0
x
Bid. D
+ 4,500 t.m
+ 4,250 t.m
- 6,750 t
+ 6,750 t.m
Penyelesaian :
a. Reaksi Perletakan.
QR
= (3 t/m’) x (3 m) = 9 ton.
Σ MB
= 0,
Mmaks = 7,59375 t.m’
Bid. M

14. RAV
. (6 m) – P1 . (5 m) + P2 . (4 m) – q . (3 m) . ½ .(3 m) = 0
R
AV
x 6 – 4 x 5 + 2 x 4 – ½ x 3 x 32 = 0
R
AV
= (20 – 8 + 13,5)/6
R
AV
= + 4,250 ton ()
Σ MA
= 0,
– R
BV
. (6 m) + P1 . (1 m) – P2 . (2 m) + q . (3 m) . (4,5 m) = 0
– R
BV
x 6 + 4 x 1 – 2 x 2 + 3 x 3 x 4,5 = 0
R
BV
= (4 – 4 + 40,5)/6
R
BV
= + 6,750 ton ()

15. Kontrol :
Σ V = 0,
R
AV
+ R
BV
– P1 + P2 – Q
R
= 0
4,250 t + 6,750 t – 4 t + 2 t – 9 t = 0
13 t – 13 t = 0 …..(memenuhi)
b. Gaya Lintang.
DA-C
= + RAV
= + 4,250 ton.
D
C-D
= D
A-C
– P1 = + R
AV
– P1 = 4,250 – 4 = + 0,250 ton.
D
D-E
= D
C-D
+ P2 = + R
AV
– P1 + P2
= 0,250 + 2 = + 2,250 ton.
DE-B
= DD-E
– QR
= + RAV
– P1 + P2 – QR
= 2,250 – 9 = – 6,750 t.
DE-B
= – RBV
(memenuhi).
c. M o m e n .
MC
= + RAV
. (1 m) = + 4,250 x 1 = + 4,250 ton.m’.
M
D
= + R
AV
. (2 m) – P1 . (1 m) = + 4,250 x 2 – 4 x 1 = 4,500 t.m’.
M
E
= + R
AV
. (3 m) – P1 . (2 m) + P2 . (1 m)
= + 4,250 x 3 – 4 x 2 + 2 x 1 = 6,750 t.m’.
MB
= + RAV
. (6 m) – P1 . (5 m) + P2 . (4 m) – q . (3 m) . ½ .(3 m)
= + 4,250 x 6 – 4 x 5 + 2 x 4 – 3 x 3 x ½ x 3
= + 25,500 – 20 + 8 – 13,5
MB
= 0 t.m’ (memenuhi).
d. Tinjau titik dimana gaya lintang sama dengan nol.
Gaya lintang dihitung dari kanan kekiri,
Dx = – RBV
+ q . x = 0
x = R
BV
/q = (6,75 t)/(3 t/m’)
= 2,25 m (dari perletakan B).
Momen pada titik x = 2,25 m dari B,
Mx = RBV
. x – ½ q . x2
= 6,75 x 2,25 – ½ x 3 x (2,25)2
Mx = + 7,59375 t.m’ (maksimum).

16. WORKSHOP/PELATIHAN
P1 P2 P3 P4 P5
A C D E F G
B
a1 b1
a2
b2
a3
b3
a4 b4
a5 b5
L = 4 + X/2
Diketahui : Struktur seperti tergambar
Memikul gaya-gaya,
P1 = 1 ton ; P2 = 1,5 ton ; P3 = 4 ton ; P4 = 3 ton ; P5 = 2 ton.
a1 = 0,2 L ; a2 = 0,3 L ; a3 = 0,6 L ; a4 = 0,7 L ; a5 = 0,8 L
L = 4 + X/2 (meter).
X = Satu angka terakhir No.Stb.
Misal, No.Stb. 08101012, maka X = 2 meter.
Diminta : Gambarkan bidang-bidang momen (M) dan gaya lintang (D).
Penyelesaian :
a). Data.
Misal X = -1, maka L = 4 – 1/2 = 3,5 meter
(Bilangan negatip jangan ditiru).
a1 = 0,2 x (3,5 m) = 0,70 m. b1 = 3,5 m – 0,70 m = 2,80 m.
a2 = 0,3 x (3,5 m) = 1,05 m. b2 = 3,5 m – 1,05 m = 2,45 m.
a3 = 0,6 x (3,5 m) = 2,10 m. b3 = 3,5 m – 2,10 m = 1,40 m.
a4 = 0,7 x (3,5 m) = 2,45 m. b4 = 3,5 m – 2,45 m = 1,05 m.
a5 = 0,8 x (3,5 m) = 2,80 m. b5 = 3,5 m – 2,60 m = 0,70 m.
b). Reaksi perletakan.
Σ MB
=
0
R
AV
= P1.b1/
L
+ P2.b2/L + P3.b3/
L
+ P4.b4/
L
+ P5.b5
/L= (1 t).(2,8/3,5) + (1,5 t).(2,45/3,5) + (4 t).(1,4/3,5) + (3 t).(1,05/3,5) + (2 t).(0,7/3,5)
= 0,800 t + 1,050 t + 1,600 t + 0,900 t + 0,400 t
R
AV
= 4,750 ton.
Σ MA
= 0
R
BV
= P1.a1/L + P2.a2/L + P3.a3/L + P4.a4/L + P5.a5/L
= (1 t).(0,7/3,5) + (1,5 t).(1,05/3,5) + (4 t).(2,1/3,5) + (3 t).(2,45/3,5) + (2 t).(2,8/3,5)
= 0,200 t + 0,450 t + 2,400 t + 2,100 t + 1,600 t
RBV
= 6,750 ton.
Kontrol :
Σ V = 0,
R
AV
+ R
BV
– P1 – P2 – P3 – P4 – P5 = 0
4,750 t + 6,750 t – 1 t – 1,5 t – 4 t – 3 t – 2 t = 0
16

17. 11,5 t – 11,5 t = 0 (memenuhi).
c). Gaya Lintang.
17

18. Da-c = + RAV
= + 4,750 t.
Dc-d = + R
AV
– P1 = 4,750 t – 1 t = + 3,750 t.
Dd-e = + R
AV
– P1 – P2 = 4,750 t – 1 t – 1,5 t = + 2,250 t.
De-f = + R
AV
– P1 – P2 – P3 = 4,750 t – 1 t – 1,5 t – 4,0 t = – 1,750 t.
Df-g = + R
AV
– P1 – P2 – P3 – P4 = 4,750 t – 1 t – 1,5 t – 4,0 t – 3,0 t = – 4,750 t.
Dg-b = + R
AV
– P1 – P2 – P3 – P4 – P4 – P5
= 4,750 t – 1 t – 1,5 t – 4,0 t – 3,0 t – 2,0 t = – 6,750 t.
Dg-b = – RBV
d). Momen.
Perhitungan momen lentur dari kiri ke kanan, diperoleh,
Ma = 0 t.m’.
Mc = + R
AV
. a1 = + 4,750 t x 0,7 m = + 3,325 t.m’.
Md = + R
AV
. a2 – P1 . (a2 – a1) = + 4,750 t x 1,05 m – 1 t .(1,05 m – 0,7 m)
= + 4,638 t.m’.
Me = + R
AV
. a3 – P1 . (a3 – a1) – P2 . (a3 – a2)
= + 4,750 t x 2,1 m – 1 t .(2,1 m – 0,7 m) – 1,5 t .(2,1 m – 1,05 m)
= + 7,000 t.m’.
Mf = + R
AV
. a4 – P1 . (a4 – a1) – P2 . (a4 – a2) – P3 . (a4 – a3)
= + 4,750 t x 2,45 m – 1 t .(2,45 m – 0,7 m) – 1,5 t .(2,45 m – 1,05 m)
– 4,0 t .(2,45 m – 2,1 m)
= + 6,388 t.m’.
Mg = + RAV
. a5 – P1 . (a5 – a1) – P2 . (a5 – a2) – P3 . (a5 – a3) – P4 . (a5 – a4)
= + 4,750 t x 2,8 m – 1 t .(2,8 m – 0,7 m) – 1,5 t .(2,8 m – 1,05 m)
– 4,0 t .(2,8 m – 2,1 m) –3,0 t .(2,8 m – 2,45 m)
= + 4,725 t.m’.
Mb = 0 t.m’.
Perhitungan momen lentur dari kanan ke kiri, persamaannya,
Mb = 0 t.m’.
Mg’ = + R
BV
. b5
Mf’ = + R
BV
. b4 – P5 . (b4 – b5)
Me’ = + R
BV
. b3 – P5 . (b3 – b5) – P4 . (b3 – b4)
Md’ = + R
BV
. b2 – P5 . (b2 – b5) – P4 . (b2 – b4) – P3 . (b2 – b3)
Mc’ = + R
BV
. b1 – P5 . (b1 – b5) – P4 . (b1 – b4) – P3 . (b1 – b3) – P2 . (b1 – b2)
Ma = 0 t.m’.
Keseimbangan mengharuskan,
Ma = Mb = 0
Mc = Mc’ ; Md = Md’ ; Me = Me’ ; Mf = Mf’ ; Mg = Mg’
18

19. P1 P2 P3 P4 P5
A C D E F G
B
a1 b1
a2
b2
a3
b3
a4 b4
a5 b5
L = 4 + X/2
Bidang gaya
lintang (D)
+ 4,750 t
+ 3,750 t +2,250 t
- 1,750 t
Bidang
Momen (M)
+ 7,000 t.m
- 4,750 t
+ 6,388 t.m
- 6,750 t
+ 4,638 t.m
+ 4,725 t.m
+ 3,325 t.m
0 t.m 0 t.m
Kunci Jawaban
No
.
St
P
1
P
2
P
3
P
4
P
5
-
1
1.00
0
1.50
0
4.00
0
3.00
0
2.00
00 1.00
0
1.50
0
4.00
0
3.00
0
2.00
01 1.00
0
1.50
0
4.00
0
3.00
0
2.00
02 1.00
0
1.50
0
4.00
0
3.00
0
2.00
03 1.00
0
1.50
0
4.00
0
3.00
0
2.00
04 1.00
0
1.50
0
4.00
0
3.00
0
2.00
05 1.00
0
1.50
0
4.00
0
3.00
0
2.00
06 1.00
0
1.50
0
4.00
0
3.00
0
2.00
07 1.00
0
1.50
0
4.00
0
3.00
0
2.00
08 1.00
0
1.50
0
4.00
0
3.00
0
2.00
09 1.00
0
1.50
0
4.00
0
3.00
0
2.00
0
19

20. No
.
St
L
m
a
1
a
2
a
3
a
4
a
5
b
1
b
2
b
3
b
4
b
5
-
1
3.50
0
0.70
0
1.05
0
2.10
0
2.45
0
2.80
0
2.80
0
2.45
0
1.40
0
1.05
0
0.70
00 4.00
0
0.80
0
1.20
0
2.40
0
2.80
0
3.20
0
3.20
0
2.80
0
1.60
0
1.20
0
0.80
01 4.50
0
0.90
0
1.35
0
2.70
0
3.15
0
3.60
0
3.60
0
3.15
0
1.80
0
1.35
0
0.90
02 5.00
0
1.00
0
1.50
0
3.00
0
3.50
0
4.00
0
4.00
0
3.50
0
2.00
0
1.50
0
1.00
03 5.50
0
1.10
0
1.65
0
3.30
0
3.85
0
4.40
0
4.40
0
3.85
0
2.20
0
1.65
0
1.10
04 6.00
0
1.20
0
1.80
0
3.60
0
4.20
0
4.80
0
4.80
0
4.20
0
2.40
0
1.80
0
1.20
05 6.50
0
1.30
0
1.95
0
3.90
0
4.55
0
5.20
0
5.20
0
4.55
0
2.60
0
1.95
0
1.30
06 7.00
0
1.40
0
2.10
0
4.20
0
4.90
0
5.60
0
5.60
0
4.90
0
2.80
0
2.10
0
1.40
07 7.50
0
1.50
0
2.25
0
4.50
0
5.25
0
6.00
0
6.00
0
5.25
0
3.00
0
2.25
0
1.50
08 8.00
0
1.60
0
2.40
0
4.80
0
5.60
0
6.40
0
6.40
0
5.60
0
3.20
0
2.40
0
1.60
09 8.50
0
1.70
0
2.55
0
5.10
0
5.95
0
6.80
0
6.80
0
5.95
0
3.40
0
2.55
0
1.70
0
No
.
St
RA
V
ton
RB
V
ton
D
a
-c
Dc
-
d
Dd
-e
to
D
e
-f
Df
-
g
Dg
-b
to-
1
4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.750 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.751 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.752 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.753 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.754 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.755 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.756 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.757 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.758 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.759 4.75
0
6.75
0
4.75
0
3.75
0
2.25
0
-
1.75
-
4.75
-
6.75
No
.
St
M
c
ton.
M
d
ton.
M
e
ton.
M
f
ton.
M
g
ton.
M
g'
ton.
Mf
'
ton.
M
e'
ton.-
1
3.32
5
4.63
8
7.00
0
6.38
8
4.72
5
4.72
5
6.38
8
7.00
00 3.80
0
5.30
0
8.00
0
7.30
0
5.40
0
5.40
0
7.30
0
8.00
01 4.27
5
5.96
3
9.00
0
8.21
3
6.07
5
6.07
5
8.21
3
9.00
02 4.75
0
6.62
5
10.00
0
9.12
5
6.75
0
6.75
0
9.12
5
10.00
03 5.22
5
7.28
8
11.00
0
10.03
8
7.42
5
7.42
5
10.03
8
11.00
04 5.70
0
7.95
0
12.00
0
10.95
0
8.10
0
8.10
0
10.95
0
12.00
05 6.17
5
8.61
3
13.00
0
11.86
3
8.77
5
8.77
5
11.86
3
13.00
06 6.65
0
9.27
5
14.00
0
12.77
5
9.45
0
9.45
0
12.77
5
14.00
07 7.12
5
9.93
8
15.00
0
13.68
8
10.12
5
10.12
5
13.68
8
15.00
08 7.60
0
10.60
0
16.00
0
14.60
0
10.80
0
10.80
0
14.60
0
16.00
09 8.07
5
11.26
3
17.00
0
15.51
3
11.47
5
11.47
5
15.51
3
17.00
0