hitunganstructure_pbi.xls

PERHITUNGAN KONSTRUKSI BETON

DATA :- Mutu Beton K - 225

225 kg/cm2 6.5 kg/cm2 beban tetap, tanpa tulangan geser75 kg/cm2 10 kg/cm2 beban sementara, tanpa tulangan geser

n 21 16 kg/cm2 beban tetap, dengan tulangan geser20 kg/cm2 beban tetap, dengan tulangan geser & puntir

- Mutu Baja U24

2080 kg/cm21400 kg/cm2

1400 = 0.88921 75

- BJ Beton Bertulang = 2400 kg/m3- Berat Dinding = 250 kg/m2- Berat Spesi = 21 kg/m2 tebal 3 cm- Berat Lantai = 24 kg/m2 tebal 2 cm- Berat Guna = 250 kg/m2- Berat Tanah = 2000 kg/m2- Berat MEP = kg/m2

PENYELESAIAN :x y

Ukuran Plat Lantai : 2.5 5 m

Tebal Plat Lantai ht : 12 cm 0.12 cmh : 12 - 2 = 10 cm

Beban yang terjadi :- BS. Plat = 0.12 x 2400 x 1 = 288 kg/m1- B. Spesi = 3 x 21 x 1 = 63 kg/m1- B. Lantai = 2 x 24 x 1 = 48 kg/m1- B. Guna = 250 x 1 = 250 kg/m1- B. Tanah = 2000 x 0.6 = 1200 kg/m1- B. MEP = 0 x 1 = 0 kg/m1

q = 1849 kg/m1q/m' = 1849 kg/m1

diambil q = 2000 kg/m1

Perhitungan Penulangan :Ly = 5.5 = 1.00 Tabel M Lx = 21 -Mtx = 52Lx 5.5 M Ly = 21 -Mty = 52

M Lx = 0.001 x 2000 x 30.25 x 21.00 = 1270.50 kgm-Mtx = 0.001 x 2000 x 30.25 x 52.00 = 3146.00 kgmM Ly = 0.001 x 2000 x 30.25 x 21.00 = 1270.50 kgm-Mty = 0.001 x 2000 x 30.25 x 52.00 = 3146.00 kgm

Arah X - X

Ca = 10 = 1.456 d = 0.2

Tbk TbtTb' Tbs

Tbm,tTbm,t

T*auTa'

- Æ o

Tabel Æ = 0.67021 3146.0 100nw = 49.431 1400 nw = 0.49435

A = 0.49435 x 100 x 10 = 23.54 cm221

Amin = 0.25 x 100 x 12 = 3.00 cm2100

A' = 0.2 x 23.5403 = 4.71 cm2

Dipakai besi Æ 12 - 10 = 11.31 cm2 > 23.5403 cm2 no ok!12

Arah Y - Y

Ca = 10 = 2.291 d = 0Tabel Æ = 1.182

21 1270.50 100nw = 18.4591 1400 nw = 0.1846

A = 0.18459 x 100 x 10 = 8.79 cm221

Amin = 0.25 x 100 x 12 = 3.00 cm2100

A' = 0 x 8.7902 = 0.00 cm2

Dipakai besi Æ 12 - 10 = 11.31 cm2 > 8.7902 cm2 ok!





- Mutu Baja U32

2780 kg/cm21850 kg/cm2

1850 = 1.17521 75

- BJ Beton Bertulang = 2400 kg/m3- Berat Dinding = 250 kg/m2

PENYELESAIAN :b ht

Ukuran Balok I : 30 40 cm0.3 0.4 m

Bentang Balok : 5 mh : 40 - 5 = 35 cm

Beban yang terjadi :

- BS. Balok = 0.3 x 0.28 x 2400 = 201.6 kg/m1- B. Dinding = 0 x 250 = 0.0 kg/m1- B. Plat = 2000 x 0 x 1 = 0.0 kg/m1

MB = 0RA 1 = 6220

MA = 0RB 1 + 2 = 6220

Mmax (tump1) = 5650.000

Mmax (lap) = 2960.000

Mmax (tump2) = 0.000

M Tumpuan 1

Ca = 35 = 2.394 d = 0.2Tabel Æ = 1.248 >

21 5650.00 100nw = 20.340.3 1850 nw = 0.203396

A = 0.203396 x 30 x 35 = 10.17 cm221 5.06

Amin = 12 x 30 x 35 = 4.53 cm2

Tbk TbtTb' Tbs

Tbm,tTbm,t

T*auTb'

- Æ o

2780 2.25

A' = 0.2 x 10.17 = 2.03 cm21.01

Dipakai besi 6 Æ 16 = 12.07 cm2 > 10.17 cm2 ATASDipakai besi 2 Æ 16 = 4.02 cm2 > 2.03 cm2 BAWAH

Æ 16 = 2.011429 cm2

M Lapangan

Ca = 35 = 3.307 d = 0.2Tabel Æ = 1.843 >

21 2960.00 100nw = 10.350.3 1850 nw = 0.103465

A = 0.103465 x 30 x 35 = 5.17 cm221 2.57

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 0.2 x 5.17 = 1.03 cm20.51

Dipakai besi 3 Æ 16 = 6.03 cm2 > 5.17 cm2 BAWAHDipakai besi 2 Æ 16 = 4.02 cm2 > 1.03 cm2 ATAS

Æ 16 = 2.011429 cm2

M Tumpuan 2

Ca = 35 = #DIV/0! d = 0Tabel Æ = #DIV/0! >

21 0.00 100nw = #DIV/0!0.3 1850 nw = #DIV/0!

A = #DIV/0! x 30 x 35 = #DIV/0! cm221 #DIV/0!

Amin = 12 x 30 x 35 = 4.53 cm22780 4.01

A' = 0 x #DIV/0! = #DIV/0! cm2#DIV/0!

Dipakai besi 2 Æ 12 = 2.262857 cm2 > #DIV/0! cm2 ATASDipakai besi 2 Æ 12 = 2.262857 cm2 > #DIV/0! cm2 BAWAH

Æ 12 = 1.131429 cm2

Kontrol Tegangan Geser :

Q Tumpuan 1 = 1 x 6220 = 6220.000 kg

t = 8 x 6220 = 6.770068 < 6.57 x 30 x 35 < 10

< 16< 20

Diambil harga tegangan Tumpuan 1 : 0.270068Seluruh tegangan geser dipikul oleh tulangan geser berupa sengkang atau dan tulangan miringDipakai :

8 = 0.5 cm28 = 0 cm2

- Misalkan Tulangan Sengkang menahan 50% 0.2701 = 1 x 2780

30 x 35

as = 0.003 diambil 20 cm8 - 200 mm

t = 1 x 2780 = 3.97142920 x 35

- Misalkan Tulangan Miring menahan 50% 0.2701 = 0 x 2780 ( sin 45 + cos 45 )

am x 350.2701 = 0 x ( 0.850904 + 0.525322 )

am x 35am = #DIV/0! diambil 15 cm

8 - 150 mm

= 0 x 2780 x 1.376226 = 0.000 cm15 x 35

= 3.971429 + 0.000 = 3.971429 kg/cm2 > 0.270068 kg/cm2

- Tul. Sengkang As Æ- Tul. Miring Am Æ

tm

Dipakai sengkang Æ

tm

Diberi tul. miring Æ

tm

Total t

beban tetap, tanpa tulangan geserbeban sementara, tanpa tulangan geserbeban tetap, dengan tulangan geserbeban tetap, dengan tulangan geser & puntir

1.175

1.175

1.175

kg/cm2kg/cm2kg/cm2kg/cm2

ok!





- Mutu Baja U32

2780 kg/cm21850 kg/cm2

1850 = 1.17521 75


PENYELESAIAN :b ht





MB = 0RA 1 = 11440

MA = 0RB 1 + 2 = 11440

Mmax (tump1) = 12230.00

Mmax (lap) = 9850.000

Mmax (tump2) =

M Tumpuan 1

Ca = 35 = 1.627 d = 1Tabel Æ = 1.283 >

21 12230.00 100nw = 42.890.3 1850 nw = 0.42887

A = 0.42887 x 30 x 35 = 21.44 cm221 10.66

Tbk TbtTb' Tbs

Tbm,tTbm,t

T*auTb'

- Æ o

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 1 x 21.44 = 21.44 cm210.66


Æ 16 = 2.011429 cm2

M Lapangan

Ca = 35 = 1.813 d = 0.8Tabel Æ = 1.257 >

21 9850.00 100nw = 34.760.3 1850 nw = 0.347628

A = 0.347628 x 30 x 35 = 17.38 cm221 8.64

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 0.8 x 17.38 = 13.91 cm26.91


Æ 16 = 2.011429 cm2

M Tumpuan 2

Ca = 35 = #DIV/0! d = 0.8Tabel Æ = #DIV/0! >

21 0.00 100nw = #DIV/0!0.3 1850 nw = #DIV/0!

A = #DIV/0! x 30 x 35 = #DIV/0! cm221 #DIV/0!

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 0.8 x #DIV/0! = #DIV/0! cm2#DIV/0!


Æ 16 = 2.011429 cm2


Q Tumpuan 1 = 1 x 11440 = 11440.00 kg

t = 8 x 11440 = 12.4517 < 6.57 x 30 x 35 < 10

< 16< 20

Diambil harga tegangan Tumpuan 1 : -5.951701Seluruh tegangan geser dipikul oleh tulangan geser berupa sengkang atau dan tulangan miringDipakai :

8 = 0.5 cm28 = 0 cm2


30 x 35

as = 0.075 diambil 12.5 cm8 - 125 mm

t = 1 x 2780 = 6.35428612.5 x 35


am x 355.9517 = 0 x ( 0.850904 + 0.525322 )


8 - 150 mm

= 0 x 2780 x 1.376226 = 0.000 cm15 x 35

= 6.354286 + 0.000 = 6.354286 kg/cm2 > 5.951701 kg/cm2


tm

Dipakai sengkang Æ

tm


tm

Total t


1.175

1.175

1.175


ok!





- Mutu Baja U32

2780 kg/cm21850 kg/cm2

1850 = 1.17521 75


PENYELESAIAN :b ht





MB = 0RA 1 = 28410

MA = 0RB 1 + 2 = 28410

Mmax (tump1) = 15970.00

Mmax (lap) = 11930.00

Mmax (tump2) =

M Tumpuan 1

Ca = 35 = 1.424 d = 1Tabel Æ = 1.185 >

21 15970.00 100nw = 53.650.3 1850 nw = 0.536493

A = 0.536493 x 30 x 35 = 26.82 cm221 13.34

Tbk TbtTb' Tbs

Tbm,tTbm,t

T*auTb'

- Æ o

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 1 x 26.82 = 26.82 cm213.34


Æ 16 = 2.011429 cm2

M Lapangan

Ca = 35 = 1.647 d = 1Tabel Æ = 1.293 >

21 11930.00 100nw = 41.810.3 1850 nw = 0.418102

A = 0.418102 x 30 x 35 = 20.91 cm221 10.39

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 1 x 20.91 = 20.91 cm210.39


Æ 16 = 2.011429 cm2

M Tumpuan 2

Ca = 35 = #DIV/0! d = 1Tabel Æ = #DIV/0! >

21 0.00 100nw = #DIV/0!0.3 1850 nw = #DIV/0!

A = #DIV/0! x 30 x 35 = #DIV/0! cm221 #DIV/0!

Amin = 12 x 30 x 35 = 4.53 cm22780 2.25

A' = 1 x #DIV/0! = #DIV/0! cm2#DIV/0!


Æ 16 = 2.011429 cm2


Q Tumpuan 1 = 1 x 28410 = 28410.00 kg

t = 8 x 28410 = 30.92245 < 6.57 x 30 x 35 < 10

< 16< 20

Diambil harga tegangan Tumpuan 1 : -24.42245Seluruh tegangan geser dipikul oleh tulangan geser berupa sengkang atau dan tulangan miringDipakai :

8 = 0.5 cm28 = 0 cm2


30 x 35

as = 0.307 diambil 12.5 cm8 - 125 mm

t = 1 x 2780 = 6.35428612.5 x 35


am x 3524.4224 = 0 x ( 0.850904 + 0.525322 )


8 - 150 mm

= 0 x 2780 x 1.376226 = 0.000 cm15 x 35

= 6.354286 + 0.000 = 6.354286 kg/cm2 > 24.42245 kg/cm2


tm

Dipakai sengkang Æ

tm


tm

Total t


1.175

1.175

1.175


no ok!





- Mutu Baja U32

2780 kg/cm21850 kg/cm2

1850 = 1.17521 75

- BJ Beton Bertulang = 2400 kg/m3

PENYELESAIAN :b ht

Ukuran Kolom : 30 30 cm0.3 0.3 m

Tinggi Kolom 3 mh : 30 - 5 = 25 cm 0.25 m

Beban yang terjadi :- BS. Kolom = 0.3 x 0.3 x 3 x 2400 = 648.00- R. Kolom atas 1 = 0.00 = 0.00- R. Balok 1 = 9216.00 x 1 = 9216.00

P = 9864.00

eo1 = 3456.00 = 0.3503659864

eo2 = 1 0.3 = 0.01 m minimum 0.02 m30

eo = 0.350365 + 0.02 = 0.37 m

eo = 0.37 = 1.23455 Tabel C2 = 7 C1 = 1ht 0.3e1 = 7 1 x 3 ^2 x 0.3 = 0.0210

100 x 0.3e2 = 0.15 x 0.3 = 0.045e = 0.37 + 0.0210 + 0.0450 = 0.436

ea = 0.436 + 0.5 x 0.3 - 0.05 = 0.5364Nea = 9864.00 x 0.5364 = 5290.703 kgm

Nea = kgm

Ca = 25 = 1.767 d = 0.2Tabel Æ = 0.861 >

21 5290.703 100nw = 30.450.3 1850 nw = 0.3045

Tbk TbtTb' Tbs

Tbm,tTbm,t

T*auTb'

- Æ o

d = 1 - 7 0.25 = 0.592 N bekerja8 0.5364 didalam kern

A = 0.3045 x 30 x 25 = 10.87 cm221 5.41

Amin = 1 x 30 x 30 = 9.00 cm2100 4.47

1/2Amin = 0.5 x 10.8742 = 5.4370982.70

A' = 0.2 x 10.8742 = 2.17 cm21.08

Dipakai besi 5 Æ 16 = 10.05714 cm2 > 10.87 cm2Dipakai besi 3 Æ 16 = 6.034286 cm2 > 10.87 cm2

Æ 16 = 2.011429 cm2


kgkgkg

kg

1.175

PERHITUNGAN ATAP

Bentang = 10 mJarak kuda2 = 11 mGenteng = 40 kg/m2Angin = 35 kg/m2B. Guna = 100 kgMutu Baja = bj 37 t = 1600 kg/cm2

Sudut a = 30 cos a = 0.8660 sin a = 0.5000Bentang miring = 5.77 m Tan a = 0.5774Jumlah Gording = 4 Vertikal = 2.887

Penyelesaian :1. Berat Gording

- Jarak Gording 1.443 m horisontal 1.25- Dipakai Gording C 150 65 20- Beban Gording 7.51 kg/m1 = 8.261 kg/m1

- G cos a = 8.261 0.8660 = 7.154 kg/m1= 0.072 kg/cm1

- G sin a = 8.261 0.5000 = 4.130 kg/m1= 0.041 kg/cm1

2. Beban Penutup Atap- Beban thd gording = 40 1.443 = 57.735 kg/m1

- G cos a = 57.735 0.8660 = 50.000 kg/m10.500 kg/cm1

- G sin a = 57.735 0.5000 = 28.868 kg/m10.289 kg/cm1

3. Beban Angin- Angin tekan ( W ) = 0.02 30 - 0.4 35

= 7 kg/m2- Beban thd gording = 7 1.443 = 10.104 kg/m1

o

- Angin isap ( W ) = - 0.4 35= -14 kg/m2

- Beban thd gording = -14 1.443 = -20.207 kg/m1

4. Beban guna- P cos a = 100 0.8660 = 86.6025404 kg- P sin a = 100 0.5000 = 50 kg

5. Momen yang terjadiMx1 = 0.125 7.154 121 = 108.207817 kgmMx2 = 0.125 50.000 121 = 756.25 kgmMx3 = 0.125 10.104 121 = 152.817399 kgmMx4 = 0.25 86.603 11 = 238.156986 kgm

My1 = 0.125 4.130 121 = 62.4738125 kgmMy2 = 0.125 28.868 121 = 436.621141 kgmMy3 = 0 kgmMy4 = 0.25 50 11 = 137.5 kgm

6. Kombinasi PembebananMx = 108.207817 756.25 152.817399 = 1017.27522 kgmMx' = 108.207817 756.25 238.156986 = 1102.6148 kgm

My = 62.4738125 436.621 0 = 499.094954 kgmMy' = 62.4738125 436.621 137.5 = 636.594954 kgm

7. Kontrol TeganganDipakai Profil C 150 65 20- Data profil :

a = 150 mm ix = 5.89 cmb = 65 mm iy = 2.37 cmc = 20 mm Wx = 44.3 cm3t = 3.2 mm Wy = 12.2 cm3

Ix = 332 cm4

Iy = 53.8 cm4

t = 1102.6148 + 636.594954 x 444.3

t = 1160.09516 < 1600 kg/cm2 ok!

8. Kontrol Lendutan

Fx = 5 0.072 14641 + 86.6025404 x 1331384 2.10E+06 332 48 2.10E+06 332

= 1.96E-08 + 3.44E-06= 3.46E-06

Fy = 5 0.041 14641 + 50 x 1331384 2.10E+06 53.8 48 2.10E+06 53.8

= 6.97E-08 + 1.23E-05= 1.23E-05

f = 1 1100 = 5.5 cm200

Fijin = 1.20E-11 + 1.52E-10

Fijin = 1.28E-05 < 5.5 cm ok!

9. Pembebanana.Beban atap dan gording- B. Atap = 40 1.443 11 = 635.085 kg- B. Gording = 7.51 11 = 82.610 kg

P2 = 717.695 kg

b.Beban Langit-langit + penggantung- Berat gypsum = 11 kg/m2- B. penggantung = 7 kg/m2

P2 = 18 1.250 11 = 247.500 kg

Total B. mati (P) = 717.695 247.500 100 = 1065.195 kg

LENDUTAN BALOK BETON

A. Balok Lantaib ht

Balok 15 25Selimut 3 cmBentang 6.3

Lendutan ijin f < 1 lo250

f < 0.004 630f < 2.52 cm

lo = 6.3 m630 cm

Lendutan yang terjadi f = fe g' k1g

f = 2.3888805 325 2315

f = 2.3888805 425315

f = 2.3888805 1.349206349206

f = 3.2230928 cm <

g = beban mati terbagi rata per m2 pada pelat dan per m1 pada balok

p = beban hidup terbagi rata per m2 pada pelat dan per m1 pada balok

g' = beban jangka panjang terbagi rata per m2 pada pelat dan per m1 pada balokyaitu beban mati + bagian dari beban hidup yang bersifat tetap

p' = beban jangka pendek terbagi rata per m2 pada pelat dan per m1 pada balokyaitu bagian dari beban hidup yang bersifat sementara, yaitu bekerja kontinyuselama kurang dari 24 jam dan ditentukan lebih lanjut sbb:

k1 = koefisien yang bergantung pada bentuk penampang dan harus diambil sbb :- balok persegi, balok I dan balok kotak

k2 = 1 =1 + A'/A

= 0.5

A' = 4.02

A = 4.02

fe = Ta lo ^2 10 ^-6 (cm)a (1 - Z) h

= 1850 396900 10 ^-622.4 0.6237172 22

= 734265000 10 ^-6 = 2388880.5307.367822222222

= 2.3888805

Ta = 1850 kg/cm2 u32

lo = 630 cm

h = 22 cm

a = 22.4 profil 22.4polos 20.8

Z = 2.5 A . T*au = 2.5b . h . T'bk

= 2.5 11175.6 = 0.376282874250

A = 4.02T*au = 2780 u32T'bk = 225

A. Balok Jepit - Jepit

DiketahuiBalok 0.3 0.2 M

30 20 CmBJ beton 2400 Kg/m3

0.00240 Kg/cm3q 1000.00 Kg/m

10.00 Kg/cmP 2000.00 kg

BebanB. Balok = 30 x 20 xq =

Q = 11.44 kg/cmP = 2000.00l = 300 cmE = 96000 kg/cm2 tergantung pada modulus sekan beton :

I = 45000 kg/cm4 tergantung pada :

f = q l^4 + P l^3384EI 192 EI

= 11.44 8.1E+09 +384 96000 45000

f = 9.27E+10 + 5.40E+10 =1658880000000 829440000000

k2 + p'+ p

0.5 + 100+ 0

2.52 NO OK!

= 315

= 0

beban jangka panjang terbagi rata per m2 pada pelat dan per m1 pada balok = 325

beban jangka pendek terbagi rata per m2 pada pelat dan per m1 pada balok = 100yaitu bagian dari beban hidup yang bersifat sementara, yaitu bekerja kontinyu

koefisien yang bergantung pada bentuk penampang dan harus diambil sbb := 2

11 + 1

cm

10 ^-6 cm

cm

4.02 278015 22 225

0.00240 = 1.440 kg/cm= 10.000 kg/cm

Q = 11.440 kg/cm

tergantung pada modulus sekan beton : Mutu beton = 225beban tetap = 96000 kg/cm2beban sementara = 144000 kg/cm2

b = 20h = 30

2000.00 2.70E+07192 96000 45000

0.1210 cm

LENDUTAN BALOK BAJA

Fx = 5 qx Lx^4 + Px384 E Ix 48

1030= 5 3.1695 1.125509E+12 + 1625

384 2100000 4050 48

= 0.01302083333 3.5673E+12 + 1.776E+128505000000 4.082E+11

= 0.01302083333 419.435646478 + 4.3496016

= 5.46140164685 + 4.34960164364

= 9.81100329049 cm

Fy = 5 qy 0,5 L^4 + Py384 E Iy 48

= 5 0 5627.54405 + 0384 2100000 117 48

= 0.01302083333 0 + 0

= 0 + 0

= 0

Fx + Fy = 9.81100329049 + 0

= 9.81100329049

= 3.13225211158 cm >

F ijin = 0.00333333333 10.3= 0.03433333333 = 3.43333333333 cm

Lx^3E Ix

1092727000 beton 126 1.262100000 4050

baja 29.6 0.2963.1695

38.5 0.3851228.5 1.2285

0,5 L^3E ly

546.36352100000 117

3.433333333 ok!

PORTAL

Dimensi b h Bentang / TinggiBalok 20 30 cm 3 mKolom 30 30 cm 3 m

I balok = 1 x 20 x 30 ^3 = 45000 cm312

I kolom = 1 x 30 x 30 ^3 = 67500 cm312

Bebanq = 6000 kg/mBS. Balok = 0.20 x 0.30 x 2400 = 144 kg/m

Koef. Dist.

Titik 2 = K 21 : K 23 = 4 67500 : 4 450003 3

= 90000 : 60000

K 21 = 90000 = 0.60150000

K 23 = 60000 = 0.40150000

Titik 3

K 32 = 0.40

K 34 = 0.60

Momen Primer

M 23 = 1 x 6144 x 9 = 4608 kgm12

M 32 = -1 x 6144 x 9 = -4608 kgm12

Perataan Momen

Titik 2 3Batang 12 21 23 32 34 43Koef. Dist. 0.6 0.4 0.4 0.6M.P. 4608 -4608

-1382.400 -2764.800 -1843.200 -921.6001105.920 2211.840 3317.76 1658.880

-331.776 -663.552 -442.368 -221.18444.237 88.474 132.710 66.355

-13.271 -26.542 -17.695 -8.8471.769 3.539 5.308 2.654

0.212 0.425 -0.708 -0.3540.071 0.142 -0.085 -0.042

-0.021 -0.042 -0.028 -0.0140.003 0.006 0.008 0.004

Momen -1727.256 -3454.5 3456.00 -3456.000 3455.70 1727.851

ReaksiM 3 = 0

R 23 x 3 - 6144 x 3.000 x 1.50 - 3456.00 +

R 23 x 3 - 27648 - 3456.001 + 3456.0

R 23 = 27648 = 9216.00 kg3

R 32 = 9216.00 kg

Momen

Mx = 9216 x X - 0.5 x 6144 x X^2 -

dMx = 0dx

dMx = 9216 - 6144 X = 0dx

X = 9216 = 1.50 m6144

Mmax = 9216 x 1.50 - 0.5 x 6144 x 2.25 -

= 13824.00 - 6912.00 - 3456.00

Mmax = 3456.00 kgm = 3.45600 tm

LENDUTAN

f maks = q x L^4384 x EI

3456.00

M 23

3456.00

hitunganstructure_pbi.xls

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