1

STATIKA I

MODUL 3BALOK 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.

2

BALOK DIATAS DUA PERLETAKAN

1. Balok Diatas Dua Perletakan Memikul Sebuah Muatan Terpusat.

A

RAV

a = 2 m

P = 10 ton

C

L = 6 m

b = 4 m

B

RBV

DA-C = + RAV

(+)

P(-)

DB-C = - RBV

Bidang gaya lintang

(+)

MC = P.a.b/L

Bidang momen

Penyelesaian :a. Reaksi Perletakan.

MB = 0,RAV . L - P . b = 0RAV = P . b/L

= (10 t) x (4 m)/(6 m)RAV = + 6,667 ton ()

MA = 0,- RBV . L + P . a = 0RBV = P . a/L

= (10 t) x (2 m)/(6 m)RBV = + 3,333 ton ().

Kontrol : V = 0,RAV + RBV – P = 06,667 t + 3,333 t - 10 t = 0 …..(memenuhi)

b. Gaya lintang.DA-C = + RAV = + 6,667 ton. DC-A = + DA-C = + 6,667 ton.DC-B = DC-A – P = 6,667 – 10 = - 3,333 ton.

3

DB-C = DC-B = - RBV = - 3,333 ton.

c. M o m e n .MA = 0MC = + RAV . a = + 6,667 t x 2 m = + 13,334 ton.m’, atauMC = P.a.b/L

Lihat gambar bidang gaya lintang dan momen diatas.

2. Balok Diatas Dua Perletakan Memikul Muatan Terpusat Sembarang.

RAH

P1 = 2 t

A

C

P2 = 3 t

45o

D

P3 = 4 t

60o

B

E

RAV RBVL = 6 ma1 = 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

Bid. M

Penyelesaian :a. Reaksi Perletakan.

MB = 0,RAV . L - P1.(L –a1) – P2 sin 45o.(L – a2) – P3 sin 60o.(L – a3) = 0RAV = P1.(L – a1)/L + P2 sin 45o.(L – a2)/L + P3 sin 60o.(L – a3)/L

= 2 x (6 - 1)/6 + 3 x ½2 x (6 – 3)/6 + 4 x 0,866 x (6 – 4)/6= 1,667 + 1,061 + 1,155

RAV = + 3,883 ton ()

MA = 0,- RBV . L + P1.(a1) + P2 sin 45o.(a2) + P3 sin 60o.(a3) = 0RBV = P1.(a1)/L + P2 sin 45o.(a2)/L + P3 sin 60o.(a3)/L

= 2 x (1)/6 + 3 x ½2 x (3)/6 + 4 x 0,866 x (4)/6

+ 3,883 t.m+ 7,411 t.m

4

= 0,333 + 1,061 + 2,309

5

RBV = + 3,703 ton ()

H = 0,RA-H + P2 cos 45o – P3 cos 60o = 0RA-H = – P2 cos 45o + P3 cos 60o = – 3 x ½2 + 4 x ½ = – 2,121 + 2

= – 0,121 ton ()

Kontrol : V = 0,RAV + RBV – P1 – P2 sin 45o – P3 sin 60o = 03,883 t + 3,703 t – 2 t – 2,121 t – 3,464 t = 07,586 t – 7,585 t = 0,001 0 …..(memenuhi)

b. Gaya Lintang.DA-C = + RAV = + 3,883 ton.DC-D = + RAV – P1 = 3,883 – 2 = + 1,883 ton.DD-E = + RAV – P1 – P2 sin 45o = 3,883 – 2 – 3 x ½2

= 3,883 – 2 – 2,121 = – 0,238 ton.DE-B = + RAV – P1 – P2 sin 45o – P2 sin 60o

= 3,883 – 2 – 3 x ½2 – 4 x 0,866 = 3,883 – 2 – 2,121 – 3,464DE-B = – 3,702 ton.DE-B = – RBV = – 3,703 ton.

c. Gaya Normal .NA-D = + RAH = + 0,121 ton (tarik).ND-E = + RAH – P2 cos 45o = + 0,121 – 3 x ½2 = + 0,121 – 2,121

= – 2 ton (tekan).NE-B = + RAH – P2 cos 45o – P3 cos 60o = + 0,121 – 2 x ½2 – 4 x ½

= 0,121 – 2,121 – 2NE-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 ½2 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

RAV

X QR

x

L = 6 m

RBV

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. MB = 0,RAV . L - QR . ½ L = 0RAV = ½ q . L2/LRAV = ½ q . L …..(1)

= ½ x (3 t/m’)/(6 m)RAV = + 9 ton ()RBV = RAV = ½ q . L = 9 ton. (simetris)

b. Gaya lintang.DA-B = + RAV = + 7 ton.DB-A = + RAV – q . L = - RBV = - 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 L2 = ¼ q L2 – 1/8 q L2

Mmaks. = + 1/8 q L2

Mmaks. = + 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 xMx = RAV . x – ½ q x2 …..(2)

Momen maksimum terjadi apabila gaya lintang sama dengan nol, Dx = d(Mx)/dx = 0(RAV . x – ½ q x2)/dx = 0

7

RAV – q . x = 0x = RAV/q …..(3)

8

= ½ q L/q = ½ L = ½ x 6x = 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 L2 – 1/8 q L2

Mmaks. = 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’.MX=3m = Mmaks = 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’.DX=3m = 9 – 3 x (3) = + 0 t.m’.

10.000

Gaya Lintang (Ton)

Dx = 1/2 q L – q x

5.000

0.000 X.0

-5.000

-10.000

A B

16.0000

14.0000

12.0000

10.0000

8.0000

6.0000

4.0000

2.0000

0.0000

M o m e n (ton.m')

Mx = 1/2 q L x – 1/2 q x2

X0 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

0 0.5 1.0 1.5 2 .0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6

Mx

Dx

9

4. Balok Diatas Dua Perletakan Memikul Muatan Segi Tiga.

2/3 x 1/3 x

qx (t/m’)

q = 3 t/m’

A BQX

X QR

RAV RBV

2/3 LL = 6 m

1/3 L

Penyelesaian :a. Reaksi Perletakan.

QR = q . ½ L = (3 t/m’) x ½ x (6 m) = 9 ton. MB = 0,RAV . L – QR . 1/3 L = 0RAV = + 1/3 QR = + 1/3 q . ½LRAV = 1/6 q L

= 1/6 x (3 t/m’)/(6 m)RAV = + 3 ton () MB = 0,– RBV . L + QR . 2/3 L = 0RBV = + 2/3 QR = + 2/3 q . ½LRBV = 1/3 q L

= 1/3 x (3 t/m’)/(6 m)RBV = + 6 ton ()

Kontrol : V = 0,RAV + RBV – QR = 03 ton + 6 ton – 9 ton = 0 ……(memenuhi)

b. Gaya lintang.DA-B = + RAV = + 1/6 q L = + 3 ton.DB-A = + RAV – QR = 1/6 q L – ½ q L = – 1/3 q L = – RBV = – 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/LQX = qx . ½ x = (q . x/L) . ½ x

= ½ q x2/LMx = RAV . x – QX . 1/3 x

= (1/6 q L) . x – (½ q x2/L) . 1/3 xMx = 1/6 q L x – 1/6 q x3/L …..(1)

2/3xx

QX

1/3x

qX (t/m’)

x

10

Momen maksimum terjadi apabila gaya lintang sama dengan nol, Dx = d(Mx)/dx = 0

= d(1/6 q L x – 1/6 q x3/L )/dxDx = 1/6 q L – ½ q x2/L …..(2)1/6 q L – ½ q x2/L = 0 x2 = 1/6 q L . 2 L /qx = (1/3 L2)x = 1/3 L3 …..(3)

= 1/3 . (6 m) .3x = 3,464 m (dari perletakan A).

Substitusikan pers.(3) kedalam (1), maka diperoleh momen maksimum, Mmaks = 1/6 q L . (1/3 L3) – 1/6 q (1/3 L3)3/L

= 1/6 q L2 {1/33 – 1/93}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 lintangX Dx Mxm ton ton.m'.

0 3.000 0.00000.5 2.938 1.48961.0 2.750 2.91671.5 2.438 4.21882.0 2.000 5.33332.5 1.438 6.19793.0 0.750 6.75003.5 -0.063 6.92714.0 -1.000 6.66674.5 -2.063 5.90635.0 -3.250 4.58335.5 -4.563 2.63546.0 -6.000 0.0000

4.000

Gaya Lintang (Ton)

Dx = 1/6 q L – ½ q x2/L

2.000

0.000 X0

-2.000

-4.000

-6.000

-8.000

A B

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.

Dx

11

8.0000

7.0000

6.0000

5.0000

4.0000

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

X0 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

RAV

C D E

1 m 1 m 1 m 3 m

B

RBV

+ 4,250 t

+ 0,250 t

L = 6 m

+ 2,250 t

Dx = 0

x

Bid. D

- 6,750 t

+ 6,750 t.m

Penyelesaian :a. Reaksi Perletakan.

QR = (3 t/m’) x (3 m) = 9 ton. MB = 0,

Mmaks = 7,59375 t.m’

Bid. M

RAV . (6 m) – P1 . (5 m) + P2 . (4 m) – q . (3 m) . ½ .(3 m) = 0RAV x 6 – 4 x 5 + 2 x 4 – ½ x 3 x 32 = 0RAV = (20 – 8 + 13,5)/6RAV = + 4,250 ton ()

MA = 0,– RBV . (6 m) + P1 . (1 m) – P2 . (2 m) + q . (3 m) . (4,5 m) = 0– RBV x 6 + 4 x 1 – 2 x 2 + 3 x 3 x 4,5 = 0

+ 4,500 t.m+ 4,250 t.m

Mx

12

RBV = (4 – 4 + 40,5)/6RBV = + 6,750 ton ()

13

Kontrol : V = 0,RAV + RBV – P1 + P2 – QR = 04,250 t + 6,750 t – 4 t + 2 t – 9 t = 013 t – 13 t = 0 …..(memenuhi)

b. Gaya Lintang.DA-C = + RAV = + 4,250 ton.DC-D = DA-C – P1 = + RAV – P1 = 4,250 – 4 = + 0,250 ton.DD-E = DC-D + P2 = + RAV – 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’.MD = + RAV . (2 m) – P1 . (1 m) = + 4,250 x 2 – 4 x 1 = 4,500 t.m’.ME = + RAV . (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 = 0x = RBV/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).

1

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

Di k e t ahui : Struktur seperti tergambarMemikul 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 LL = 4 + X/2 (meter).X = Satu angka terakhir No.Stb.Misal, No.Stb. 08101012, maka X = 2 meter.

Di m i n t a : Gambarkan bidang-bidang momen (M) dan gaya lintang (D). P en y e l esa i an :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. MB = 0RAV = 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

RAV = 4,750 ton.

MA = 0RBV = 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 : V = 0,RAV + RBV – P1 – P2 – P3 – P4 – P5 = 04,750 t + 6,750 t – 1 t – 1,5 t – 4 t – 3 t – 2 t = 011,5 t – 11,5 t = 0 (memenuhi).

2

c). Gaya Lintang.

3

Da-c = + RAV = + 4,750 t.Dc-d = + RAV – P1 = 4,750 t – 1 t = + 3,750 t.Dd-e = + RAV – P1 – P2 = 4,750 t – 1 t – 1,5 t = + 2,250 t.De-f = + RAV – P1 – P2 – P3 = 4,750 t – 1 t – 1,5 t – 4,0 t = – 1,750 t.Df-g = + RAV – P1 – P2 – P3 – P4 = 4,750 t – 1 t – 1,5 t – 4,0 t – 3,0 t = – 4,750 t.Dg-b = + RAV – 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 = + RAV . a1 = + 4,750 t x 0,7 m = + 3,325 t.m’.Md = + RAV . a2 – P1 . (a2 – a1) = + 4,750 t x 1,05 m – 1 t .(1,05 m – 0,7 m)

= + 4,638 t.m’.Me = + RAV . 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 = + RAV . 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’ = + RBV . b5Mf’ = + RBV . b4 – P5 . (b4 – b5)Me’ = + RBV . b3 – P5 . (b3 – b5) – P4 . (b3 – b4)Md’ = + RBV . b2 – P5 . (b2 – b5) – P4 . (b2 – b4) – P3 . (b2 – b3)Mc’ = + RBV . b1 – P5 . (b1 – b5) – P4 . (b1 – b4) – P3 . (b1 – b3) – P2 . (b1 – b2)Ma = 0 t.m’.

Keseimbangan mengharuskan, Ma = Mb = 0Mc = Mc’ ; Md = Md’ ; Me = Me’ ; Mf = Mf’ ; Mg = Mg’

4

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

BidangMomen (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. Stb.

P1 ton

P2 ton

P3 ton

P4 ton

P5 ton

-1 1.000 1.500 4.000 3.000 2.0000 1.000 1.500 4.000 3.000 2.0001 1.000 1.500 4.000 3.000 2.0002 1.000 1.500 4.000 3.000 2.0003 1.000 1.500 4.000 3.000 2.0004 1.000 1.500 4.000 3.000 2.0005 1.000 1.500 4.000 3.000 2.0006 1.000 1.500 4.000 3.000 2.0007 1.000 1.500 4.000 3.000 2.0008 1.000 1.500 4.000 3.000 2.0009 1.000 1.500 4.000 3.000 2.000

5

No. Stb.

Lm

a1 m

a2 m

a3 m

a4 m

a5 m

b1 m

b2 m

b3 m

b4 m

b5 m

-1 3.500 0.700 1.050 2.100 2.450 2.800 2.800 2.450 1.400 1.050 0.7000 4.000 0.800 1.200 2.400 2.800 3.200 3.200 2.800 1.600 1.200 0.8001 4.500 0.900 1.350 2.700 3.150 3.600 3.600 3.150 1.800 1.350 0.9002 5.000 1.000 1.500 3.000 3.500 4.000 4.000 3.500 2.000 1.500 1.0003 5.500 1.100 1.650 3.300 3.850 4.400 4.400 3.850 2.200 1.650 1.1004 6.000 1.200 1.800 3.600 4.200 4.800 4.800 4.200 2.400 1.800 1.2005 6.500 1.300 1.950 3.900 4.550 5.200 5.200 4.550 2.600 1.950 1.3006 7.000 1.400 2.100 4.200 4.900 5.600 5.600 4.900 2.800 2.100 1.4007 7.500 1.500 2.250 4.500 5.250 6.000 6.000 5.250 3.000 2.250 1.5008 8.000 1.600 2.400 4.800 5.600 6.400 6.400 5.600 3.200 2.400 1.6009 8.500 1.700 2.550 5.100 5.950 6.800 6.800 5.950 3.400 2.550 1.700

No. Stb.

RAVton

RBVton

Da-c ton

Dc-d ton

Dd-e ton

De-f ton

Df-g ton

Dg-b ton

-1 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7500 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7501 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7502 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7503 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7504 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7505 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7506 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7507 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7508 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.7509 4.750 6.750 4.750 3.750 2.250 -1.750 -4.750 -6.750

No. Stb.

Mc ton.m'

Md ton.m'

Me ton.m'

Mf ton.m'

Mg ton.m'

Mg' ton.m'

Mf' ton.m'

Me' ton.m'

-1 3.325 4.638 7.000 6.388 4.725 4.725 6.388 7.0000 3.800 5.300 8.000 7.300 5.400 5.400 7.300 8.0001 4.275 5.963 9.000 8.213 6.075 6.075 8.213 9.0002 4.750 6.625 10.000 9.125 6.750 6.750 9.125 10.0003 5.225 7.288 11.000 10.038 7.425 7.425 10.038 11.0004 5.700 7.950 12.000 10.950 8.100 8.100 10.950 12.0005 6.175 8.613 13.000 11.863 8.775 8.775 11.863 13.0006 6.650 9.275 14.000 12.775 9.450 9.450 12.775 14.0007 7.125 9.938 15.000 13.688 10.125 10.125 13.688 15.0008 7.600 10.600 16.000 14.600 10.800 10.800 14.600 16.0009 8.075 11.263 17.000 15.513 11.475 11.475 15.513 17.000