1. Dikethaui :a = 176 cmb = 54 cmc = 26 cmd = 26 cme = 124 cmf = 26 cmg = 176 cmh = 33 cmi = 8 cmj = 54 cmL = 24 mP = 11625 kg = 11.625 tonQ = 1400 kg/m = 1.4 ton/m

Bj Beton pracetak = 2.5 ton/mBj Beton Bertulang = 2.5 ton/m

Mutu Beton Pracetak = fc' 45Mutu Beton Slab = fc' 21

Po awal = 450 tonPo akhir = 450 ton x 0.8 = 360 tonTegangan Awal = 0.38 fc'

Tegangan Akhir = 0.33 fc'

2. Gambar RencanaP

Q

L

a

b

cYa' d

Ya88

I

II

he g

Yb'Yb88

i f

J

3. Gambar Rencana

F1 = 54 x 26 = 1404 = 163 = 228852F2 = 33 x 124 = 4092 = 88 = 360096F3 = 54 x 26 = 1404 = 13 = 18252

F = 6900 Fy = 607200

4. Garis Netral Dan Momen Inersiaa. Garis Netral- Yb = F.y/F = 607200 = 88.00 = 88 cm

6900- Ya = 176 - 88 = 88 cm

b. Momen Inersia= 0.0833 x 54 x 17576 + 1404 x 5625

79092.0 + 7897500.0 = 7976592.00= 0.0833 x 33 x 1906624 + 4092 x 0

5243216 + 0 = 5243216.00= 0.0833 x 54 x 17576 + 1404 x 5625

79092.0 + 7897500 = 7976592.00

Total "I" = + + = 21196400.00

5. Kern dan Komposite

- = I/F = 21196400 = 3071.9426900

- Ka = = 3071.942 = 34.908 cm88

- Kb = = 3071.942 = 34.908 cm88

a. Menghitung section properties dari balok composite :- Balok pracetak = fc = 45 dengan Eb = 2 x 100000

= 200000- Beton Bertulang = fc = 22.5 dengan Eb = 1.5 x 100000

= 150000

n = 150000 = 0.75200000

I1

I3

I5

I1 I2 I3

i2 cm2

i2/yb

i2/ya

II

III

- Jadi lebar pengganti = 0.75 x 176 = 132 cm

Fb = 132 x 26= 3432

BE = 132 (Lebar Efektif )

y= C

+ G= 26

+ 176=

189 cm (Terhadap Alas )2 2

cm2

- Jadi jumlah F = 6900 + 3432= 10332

yb' = F.y/F = ( 607200 + ( 3432 x 189 ))10332

yb' = 121.54936 cm = 122 cm ( Dari alas )ya' = 176 + 26 = 202 - 122 = 80 cm

"I" composit = + I b + Fb x= 21196400.00 + 6900 x ( 122 - 88= 29172800

= 0.0833 x 132 x 17576 + 3432 x 4489= 15599584

"I" composit = 29172800 + 15599584= 44772384

- Jari-jari lembam = = I/F = 44772384 = 4333.370510332

b. KernKa' = = 4333.3705 = 35.519 cm

yb' 122

Kb' = = 4333.3705 = 54.167 cmya' 80

6. Menentukan daerah aman kabel- Untuk menentukan daerah aman kabel, maka harus ditinjau beberapa keadaan sbb :a. Keadaan awal ( Balok Pracetak )b. Keadaan setelah kehilangan prategangc. Keadaan setelah plat beton bertulang di cor ( balok pracetak )d. Keadaan setelah beban luar bekerja ( beton komposite )

a = M/P ---> dimana a = Jarak titik berat kabel ke KbM = Momen yang bekerja pada setiap penampangP = Gaya prestress

7. Tahapan MenghitungUntuk menghitung besarnya momen-momen ditampang 1,2 dan 3 dipergunakan garispengaruh men titik 1,2, dan 3

a. Garis Pengaruh M1- Ordinat Maksimum = 10.0

meter3

= (( 24 - 4.00 ) x 4.00 = 20 x 4.00 = 80 = 10.0 = 3.33324 24 24 3

= Luas bidang pengaruh = 0.5 x 24 x 3.333 = 40.0

cm2

"I" Pracetak + F Pracetak ( yb'-yb)2 Z2

)2

I2 cm2

I2

I2

m2

b. Garis Pengaruh M2- Ordinat Maksimum = 16

meter3

= (( 24 - 8 ) x 8 = 16 x 8 = 128 = 16 = 5.33324 24 24 3

= Luas bidang pengaruh = 0.5 x 24 x 5.333 = 64

c. Garis Pengaruh M3- Ordinat Maksimum = 12

meter2

= (( 24 - 12.0 ) x 12 = 12 x 12 = 144 = 12 = 6.00024 24 24 2

= Luas bidang pengaruh = 0.5 x 24 x 6.000 = 72

8. Keadaan "a"Keadaan awal ; balok pracetak :

P = Po = 45 tonBeban = berat sendiri balok pracetakqbs = F pracetak x Bj beton pracetak = 0.6900 x 2.5

= 1.725 ton/m

a. Momen-momen

Ma1 = Luas gp M1 x qbs = 40 x 1.725 = 69.000 ton/mMa2 = Luas gp M2 x qbs = 64 x 1.725 = 110.400 ton/mMa3 = Luas gp M3 x qbs = 72 x 1.725 = 124.200 ton/m

b. Jarak = Momen / betonaa1 = Ma1/Po = 69.000 = 1.533 m = 153 cm

45

aa2 = Ma2/Po = 110.400 = 2.453 m = 25 cm45

aa3 = Ma3/Po = 124.200 = 2.760 m = 28 cm45

9. Keadaan "b"Setelah kehilangan tegangan

P = 0,8 x Po = 360 tonBalok pracetakbeban berat sendiri balok pracetak q = 1.725 ton/m

a. Jarak = Momen / betonab1 = Mb1/Po = 69.000 = 0.192 m = 19 cm

360

m2

m2

ab2 = Mb2/Po = 110.400 = 0.307 m = 31 cm360

ab3 = Mb3/Po = 124.200 = 0.345 m = 35 cm360

catatan =pada keadaan b , pembebanan sama dengan keadaan a , jadi besarnyaMb1 = Ma1 = 69.000 ton/mMb2 = Ma2 = 110.400 ton/mMb3 = Ma3 = 124.200 ton/m

10. Keadaan "c"- Setelah plat beton bertulang selesai di cor- P = 360 ton- Balok pracetak- Beban : berat sendiri balok pracetak + berat sendiri beton bertulang- q = 1.725 + 1.76 x 0.26 x 2.5

= 2.87 ton/m

a. keadaan "c"

Mc1 = Luas gp M1 x qbs = 40 x 2.869 = 114.760 ton/mMc2 = Luas gp M2 x qbs = 64 x 2.869 = 183.616 ton/mMc3 = Luas gp M3 x qbs = 72 x 2.869 = 206.568 ton/m

b. Jarak = Momen / betonac1 = Mc1/Po = 114.760 = 0.319 m = 32 cm

360

ac2 = Mc2/Po = 183.616 = 0.510 m = 51 cm360

ac3 = Mc3/Po = 206.568 = 0.574 m = 57 cm360

11. Keadaan "d"- Setelah beban luar bekerja- P = 360 ton- Balok composite- Beban : berat sendiri balok composite + beban luar- q = 2.87 + 1.400

= 4.269 ton/m- p = 11.625 ton

a. keadaan "d"

Md1 = Luas gp M1 x qbs = 40 x 4.269 + 3.333 x 12= 209.510 ton/m

Md2 = Luas gp M2 x qbs = 64 x 4.269 + 5.333 x 12= 335.216 ton/m

Md3 = Luas gp M3 x qbs = 72 x 4.269 + 6.000 x 12= 377.118 ton/m

b. Jarak = Momen / betonad1 = Md1/Po = 209.510 = 0.582 m = 58 cm

360

ad2 = Md2/Po = 335.216 = 0.931 m = 93 cm360

ad3 = Md3/Po = 377.118 = 1.05 m = 105 cm360

adc.g.c'

ab 1=19

ab

c.g.c'

24 m

ab 2 = 31

ac2 = 51

Kb'

ac

Ka'

aa

c.g.cc.g.c

Kb

ac1 = 32

Ka

aa2 = 25ab3 = 35

ad1 = 58 ad1 = 58ac2 = 51

aa3 = 28

ad2 = 98

ac3= 57

aa2 = 25aa1 = 15ab 1=19

ab 2 = 31

aa 1 = 15

ad2 = 98

ac1 = 32

ad3 = 105

17654

26Ya' 26

80 Ya88

33124 176

Yb'122 Yb e' = 70

88

18 26

J

* KONDISI AWAL

'-Plat Lantai Belum di Cor

Tegangan Awal = 0.38 x fc'= 0.38 x 450 kg/cm2= 171 kg/cm2

24

Berat Sendiri Beton Pracetak :q = F x BJ Beton Pracetak

= 0.6900 x 2.5= 1.72 t/m= 1725 kg/m

Momen maximum terjadi di tengah bentangMmax = 1/8

= 1/8 x 1725 x 24 2

= 124200 kg/m= 12420000 kg/cm

Kondisi awal ideal bila diagram tegangan sedagai berikut := 0 → (di anggap tidak menerima tarik )

ql²

σa

I

II

III

σb = σ ijin awal = -171 kg/cm2

0 = - P/F + (P.e'.Ya)/I-(M.Ya/l)

0 =-

P+

P x 70 x 88-

12420000 x 886900 21196400 21196400

- P+

6160 P-

10929600006900 21196400 21196400

=- 3071.942029 P

+6160 P

-1092960000

21196400 21196400 21196400

0 =3088.057971 P

+1092960000

21196400 21196400

P =1092960000

/3088.05797101449

21196400 21196400

P =51.5634730425921

=353931

=353.93118 Ton

~354

0.000145687851287

σb -171 = - P/F + (P.e'.Yb)/I-(M.Yb/l)

-171 =-

P-

P x 70 x 88+

12420000 x 886900 21196400 21196400

-171=

- P-

6160 P+

10929600006900 21196400 21196400

-171=

-3071.942029 P-

6160 P+

109296000021196400 21196400 21196400

-171 =9231.942029 P

+1092960000

21196400 21196400

P=

1092960000+

171/

9231.942028985521196400 1 21196400

-1092960000+

3624584400/

9231.942028985521196400 21196400 21196400

P =119

= 274224 = 274.224469 Ton ~0.000435542923751

Jadi yang menentukan P Max = 274 ton

A Kondisi Awal : Gaya Prestress Awal P0 = 450 ton Balok Pracetak , Lantai belum dicor

= - P/F + (P.e'.Ya)/I-(M.Ya/l)

σa =-

450000+

450000 x 70 x 88-

124200006900 21196400 21196400

- 450000+

2772000000-

10929600006900 21196400 21196400

=- 1382373913.04348

+2772000000

-1092960000

σa

σa

=21196400

+21196400

-21196400

P =296666087

=13.996 kg/cm2 < Teg. Awal -171 kg/cm2

21196400

σb = - P/F - (P.e'.Yb)/I+(M.Yb/l)

σb =-

450000-

450000 x 70 x 88+

124200006900 21196400 21196400

- 450000-

2772000000+

10929600006900 21196400 21196400

=- 1382373913.04348

-2772000000

+1092960000

21196400 21196400 21196400

P =-3061413913

=-144 kg/cm2 < Teg. Awal -171 kg/cm2

21196400

B Setelah Kehilangan tegangan 20 % = 360 ton Balok Pracetak , Lantai belum dicorTegangan Akhir = 0.33 x 450 = 149 kg/cm2

= - P/F + (P.e'.Ya)/I-(M.Ya/l)

σa =-

36000+

36000 x 70 x 88-

124200006900 21196400 21196400

- 36000+

221760000-

10929600006900 21196400 21196400

=- 110589913.043478

+221760000

-1092960000

21196400 21196400 21196400

P =-981789913

=-46.319 kg/cm2 < Teg. Awal -149 kg/cm2

21196400

σb = - P/F - (P.e'.Yb)/I+(M.Yb/l)

σb =-

36000-

36000 x 70 x 88+

124200006900 21196400 21196400

- 36000-

221760000+

10929600006900 21196400 21196400

=- 110589913.043478

-221760000

+1092960000

21196400 21196400 21196400

P =760610087

=36 kg/cm2 < Teg. Awal -149 kg/cm2

21196400

σa

C. Setelah Plat Beton Bertulang selesai di cor ( Beton Muda )Balok Pracetak dengan beban sebagai berikut :

qbs = qbs balok paracetak + qbs Beton Bertulang= 1725 + 0.26 x 1.76 x 2500= 2869 kg/m

Momen maximum terjadi di tengah bentangMmax = 1/8 x 2869 x 576

= 206568 kg/m= 20656800 kg/cm

= - P/F + (P.e'.Ya)/I-(M.Ya/l) h

σa =-

36000+

36000 x 70 x 88-

206568006900 21196400 21196400

- 36000+

221760000-

18177984006900 21196400 21196400

=- 110589913.043478

+221760000

-1817798400

21196400 21196400 21196400

P =-1706628313

=-80.515 kg/cm2 < Teg. Awal -149 kg/cm2

21196400

σb = - P/F - (P.e'.Yb)/I+(M.Yb/l)

σb =-

36000-

36000 x 70 x 88+

206568006900 21196400 21196400

- 36000-

221760000+

18177984006900 21196400 21196400

=- 110589913.043478

-221760000

+1817798400

21196400 21196400 21196400

P =1485448487

=70.080 kg/cm2 < Teg. Awal -149 kg/cm2

21196400

D. Setelah dimuati beban hidup : Balok Komposit

TEGANGAN Izin Akhir = 0.33 x 210 = -69.3 kg/cm2BebanTerbagi rata penuh q = 1.4 t/mBeban terpusat P = 11.625 t , memberikan momen maksimum bila bekerja di tengah bentang

Mmax = 1/8 + 1/4 PL

= 1/8 x 1400 x 24 2 + 1/4 x 11625 x 24= 100800 + 69750 = 170550 kg/m

σa

qL2

= 17055000 kg/cm

24

= -M . Ya' / I= -17055000 x 80

44772384= -30.474 kg/cm2 < -69.3 kg/cm2

σb = M . Yb' / I= 17055000 x 122

44772384= 46.473 kg/cm2

σa 80 : 54 = -30.474 : xx = 54 x -30.474

= ( 80= -20.570046 kg/cm2

Superposisiσa Total = -80.515 - -20.57004604

= -101.085 kg/cm2

σb Total = 70 + 46.473= 116.55 kg/cm2

σaa

-144

cgc

-46.319

cgc'

=

46.473

-30.474

Muatan Hidup

70.080

-101.085

(D)

-80.515

(B)

13.996

(A)

116.553

Skala Balok = 1:20 cmSkala Diagram = 1 : 50 kg

-20.570

(C)

-30.474

36

-144

cgc

-46.319

cgc'

=

46.473

-30.474

Muatan Hidup

70.080

-101.085

(D)

-80.515

(B)

13.996

(A)

116.553

Skala Balok = 1:20 cmSkala Diagram = 1 : 50 kg

-20.570

(C)

-30.474

36

0.18

= -P /

3071.94203= -P 2240 + P 3055

3071.94203 10987254 10980254

+ P5295

+896870000

10987254 10980254

Ton= 2774493434

/10987254

10980254 5295

= 524361.820454 2075.198992= 524.361820454

274 Ton

x 8821196400

kg/cm2

x 8821196400

kg/cm2

x 8821196400

kg/cm2

x 8821196400

kg/cm2

x 8821196400

kg/cm2

1 " = 2.54 cm1 "2 = 6.4516 cm2

2.54

x 8821196400

kg/cm2

, memberikan momen maksimum bila bekerja di tengah bentang

-144

cgc

-46.319

cgc'

=

46.473

-30.474

Muatan Hidup

70.080

-101.085

(D)

-80.515

(B)

13.996

(A)

116.553

Skala Balok = 1:20 cmSkala Diagram = 1 : 50 kg

-20.570

(C)

-30.474

36

-144

cgc

-46.319

cgc'

=

46.473

-30.474

Muatan Hidup

70.080

-101.085

(D)

-80.515

(B)

13.996

(A)

116.553

Skala Balok = 1:20 cmSkala Diagram = 1 : 50 kg

-20.570

(C)

-30.474

36

4906 + 47 x 65 ) / 10980254 ) - ( 13798000 x 65 ) / 10980254

3055 - 89687000010980254 10980254

12240 64212

896870000 1877623434 0.0058510980254 10980254 2238.12759885854

-816.872401141459-1E+06

109872545295

2075.198992 252.680259856

524362

10980254

Tegangan Izin beton

1. Pada Keadaan Awal (Ty Lin Load Blancing hal 17 )-0.6 fci = -0.6 x 0.8 x 450

= -216 kg/cm2

σ= 0.25 = 0.25 √ 0.8 x 450= 4.743 kg/cm2

2. Pada Keadaan Akhir0.45 Fc = -0.45 x 450

= -202.500 kg/cm20.5 = 0.5 √ 450

= 10.6066017177982 kg/cm2

Tegangan Ijin Baja

1. Pada Keadaan Awal 1 ksi = 70.307136 kg/cm2

σaa = 0.67 x 2E+04 = 12718.5609 kg/cm2 270 k grade = 270 ksi

2. Pada Keadaan Akhir 270 ksi = 18983 kg/cm2

0.58 x 2E+04 = 11010.0975 kg/cm2

Analisa Pembebanan

Panjang Bentang = 24 m

Berat Sendiri Akibat Penampang Precast

Bs, Precast = 0.6900 x 2500= 1725 kg/m

Mg = 1/8 q l 2= 1/8 x 1725 x 24 2= 124200 kg/m= 12420000 kg/cm

akibat berat max penampang precastBs, Precast = 1725 kg/mbs, Slab = 1.76 x 0.26 x 2500 = 1144 kg/mBekisting = 50 kg/m

q max precast = 2919.00 kg/m

MmaxPrecast = 1/8 Qmax L 2 = 1/8 x 2919 x 24

σbb =

√ f'ci

σba =

σbb = √ f'c

σak =

= 210168 kg/m= 21016800 kg/cm

Akibat Beban Hidup pada Balok KompositeMc = 1/8 1/4 PL = 1/8 x 1400 x 24

= 100800 + 69750 = 170550 kg/m= 17055000 kg/cm

Mt = 210168 + 170550 = 380718 kg/m= 38071800 kg/cm

Analisa Prategang

Gaya Pretegang Efektif

F =Mt

=38071800

0.65 ht 0.65 x 176F = 332795.454545455 kg

Ditafsir LOP 18 %

FO = F / ( 1 - 0.18 ) = 332795.4545454551 - 0.18

FO = 405848.11529933 kg

Luas baja yang di perlukan

= 405848.11529933512718.5609024

A 31.91 cm2

= 332795.45454545511010.0974976

A 30.23 cm2

Yang Menentukan As = 31.91 cm2

di rencanakan untuk menggunakan 2 ( dua ) tender baja prategang dengan As per Unit untuk stand 270 ksi dia 1/2", di gunakan Type ES 19/17 dengan A ada =jika 1 " 2.54 cm ---> maka in2 = 6.4516 x 2.601

= 16.781 cm2tegangan baja yang terjadi pada keadaan awal :

Q L2 +

A awal = FO / σaa

A akhir= F / σak

Aσs =405848.115299335

= 12092.76887438762 x 16.781

σs = 12092.769 kg/cm< σaa = 12718.5609024 ----> ok

Dia Selongsongan perunit Type ES - 19/17 = 3 " x 2.54 = 7.62

Koontrol penempatan kabel

e = kb + e1 + e2

kb = 34.908432 cm

= 4.74341649025257 x 21196400.0088 x 405848.115299335

e1 = 2.81518642699827 cm

e2 = Mg / FO = 12420000 405848.115299335

e2 = 30.60 cmMg = berat sendiri penampang precaste = kb + e1 + e2e = 68.33 cm

yb = 88 Dia selongsongan = a = 8 + Dia /2

= 8 + 7.62= 11.81

emax = Yb - a8 a = 88 -

= 76e < emax jadi kita gunakan e =

Koefisien Momen Ekivalen

mt = ( I / Ya ) / ( I' / Ya')mt = 21196400.00 / 88

44772384 / 80= 0.430

e1 = ( σba . I ) / ( ya . FO )

mt = ( I / Yb ) / ( I' / Yb')mb = 21196400.00 / 88

44772384 / 122= 0.656

GAYA PRA TEGANG

F = 21016800 + 0.656 x 17055000 - 10.6066 x76 + 34.908

F = 29655890.2733354111.098

F = 266933.472 Kg

KONTROL LOP

1. DENGAN METHODE AASTHOSHRIKAGE ( SK )SH = 0.07 ( 17000 - 150 RH )RH = Kelembaman relativeDi Asumsikan RH = 70%SH = 0.07 ( 17000 - 150 70 )SH = 455 kg/cm2

ELASTIC SHORTENING ( ES )---> Perpendekan Elastisitas

FO = 405848.11529933 Kg

= 405848.115299335 + 405848.115299335 x 76 2 - 12420000 6900 21196400.00 21196400.00

= 58.81856743 + 111.147 - 44.64= 125.322 kg/cm2

Mp + mb . Mc - σbb . Ka . Ab F = ---------------------------- e + ka

Fcir = ( FO / Ab ) + ((FO.e^2)/Ib ) - (( Mg.e)/Ib )

Es = Modulus Elastis Baja = 2.1 * 10^6Eci = Modulus elestis betan pada saat Transfer

= 303578.65537616 kg/cm

ES = 2.1 x 1000000x 125.322 = 866.91308622117 kg/cm

303578.655376164

CREEP OF COCRETE ( CRc )---> Rangkak

CREEP OF COCRETE ( CRc )---> Rangkak

Ms ( Momen Akibat Slap )= Mp - Mg= 21016800 - 12420000 = 8596800 kg/cm

Mc' ( momen akibat berat limpasan air hujan )= 1/8 x 100 x 24 2= 7200 kg/m= 720000 kg/cm

Fcds = 8596800 x 76+

720000 ( 76 + (21196400.00 44772384.00

= 30.90101112 + 1.77200302758057= 32.67301414 Kg/cm2

CRC = 12 fci - 7 fcds= 12 x 125.322 - 7 x 32.67301414= 1275.154 kg/cm2

RELAXATION OF PRESTRESSING STEEL ( CRs )

untuk ( 250 - 270 ) KSI strand

= 16000 √ ( 0.8 fc ) = 16000 √ 0.8 450 =

ES = ( Es / Eci ) . Fcir

Fcds = ( Ms * e / Ib ) + ( Mc' ( e + ( Yb' - Yb )) / Ib'

CRs ( t2 , te ) = 1407.43 - 0.4 ES - 0.2 ( SH + CRC )

CRs = 1407.43 - 0.4 * 866.913086221167 -0.2 ( 455 + 1275.154

CRs = 714.63406 Kg/cm

LOSS OF PRESTESSED ( LOP )

LOP = SH + CRC + CRS (t2,te)

LOP = 455 + 1275.154 + 714.6340607LOP = 2444.788 kg/cm2

% LOP = 2444.7876 * 100%% LOP = 2444.7876 / 12092.768874388 x 100%% LOP = 20.2 %

LOP tafsir < LOP hitung, Maka lop di coba lagi dan untuk mendapatkan yang akurat, maka tegangan -efektif yang di pakai adalah tegangan efektaif yang telah di koreksi dengan koofesien Momen ekivalen

Di tafsir LOP = 19 %

F = 266933.472 kgFO = 266933.47241792 / ( 1 - 0.19 )FO = 329547.49681225 kg

luas baja yang di perlukan :

329547.49681225 / 12718.5609= 25.911 cm2

Di rencankan Menggunakan tendon tunggal dengan

As 25.911 cm2 ~ 4.0161748 in2 lihat Tabel LIN hal 305Digunakan Type ES - 31 /27 dengan As ada = 4.131 In2 = 26.65156 cm2

e1 = (As * Ib )/(Ya * FO )e1 = 4.131 x 21196400.00

88 x 329547.50

/ σs

As = FO / σaa =

e1 = 3.0193719 cm

e2 = Mg / FO= 12420000

329547.50= 37.69 cme = Kb + e1 + e2 = 34.91 + 3.019371923 + 37.69= 75.616 cm

Dia ES - 31 / 27 3 3/4" = 9.525 ---> 2 BetonDia /2 = 4.7625 Jika Selimut Beton 8 cm maka…… 8

a = 8 + 4.76 = 12.763 cme max = Yb - a = 88 - 12.76

= 75.2375 cm

e > e max diambil e = 75.2375 < 75.616 -------> OK

2

2 + 1/4 x 11625 x 2469750

di rencanakan untuk menggunakan 2 ( dua ) tender baja prategang dengan As per Unit = 15.95495428 cm2 = 2.473022859 in22.601 in2

cm (TY Lin Hal 309 Jilid 1)

7.62 cm

/ 2

11.81

e < emax jadi kita gunakan e = 76 cm

34.91 x 690034.908

12420000 x 7621196400.00

kg/cm

2

122 - 88 )44772384.00

)

LOP tafsir < LOP hitung, Maka lop di coba lagi dan untuk mendapatkan yang akurat, maka tegangan -efektif yang di pakai adalah tegangan efektaif yang telah di koreksi dengan koofesien Momen ekivalen

yb 88

a