Perhitungan Manual Balok Anak

Lx

Ly

A.

P1

× =½×

Mmax

= P1

P

b/2

=½

P1

P1

qa

×

P=¼× P

( b/2 – 1/3

×

=¼(½

P1

×

b/2

×

=(¼ P1

b

P

×

× b

×

Mmax

qeq

b ) ( b/3 )

×

qa )

×

= Meq-max

1/24 × qa × b2 = ⅛ × qeq × b2 qeq

Lx=b

= ⅓ × qa × b

= ⅓ × qa × L x

qeq

Ly-b

b/2

x

P1+P2

P1

= (P1 + P2)

×

Dima Dimana na : P1

P3

P2

Ly/2 – P1 (Ly/2 Ly - ⅔ =½

⋅

b/2

P

⋅

b/2

P4

⋅

P1+P2

b/2) – P2 (Ly –

b

b/3 )

Meq max = ⅛ × qeq × b Lx=b

×

b

×

b/3

P2

P ⋅ ( Ly 2

=

Meq-max = ⅛ Mmax

⋅

− b)

qeq

⋅

Ly2

⋅

Lx2/Ly2)

= Meq-max

qeq = ½ × qa × Lx (1 - ⅓

4.2. 35/55

Lx = 350 – (35/2 + 25/2) = 320 cm

Ly = 500 – (35/2 + 35/2) = 465 cm        0        4        /        5        2

       5        5        /        5        3

       5        5        /        5        3

3.50

       0        0   .        5

3.50

Dari data perhitungan pembebanan plat didapat : qd = 402 kg/m2 ql = 100 kg/m2

Beban mati (qd)



Berat sendiri balok= 0.25 x 0.28 x 2400 Beban mati plat = 2

×

½

×

402

×

= 168

3.20 × (1 - ⅓

⋅

(3.20/4.65)2) = 1083.33 kg/m2 qd

 Beban

×

100

×

3.20

×

(1 - ⅓

⋅

(3.20/4.65) 2)

= 269.48 kg/m2  Beban

= 1251.33 kg/m2

hidup (ql)

Beban hidup plat = 2 × ½

kg/m2

berfaktor (qu)

qu = 1.2 x qd + 1.6 x ql = 1.2 x 1251.33 + 1.6 x 269.48 = 1520.81 kg/m2

Koefisien momen dan gaya lintang (sesuai SNI 03-2847-2002 Pasal



10.3.3.5) -1 / 16

A

- 1 / -1 11 / 1 1

- 1 / -1 10 / 1 1 +1 / 14

+1 / 14

B

C

- 1 / 1- 11 / 1 1 +1 / 14

- 1 / 1- 11 / 1 0 +1 / 14

D

E

-1 / 11 +1 / 14

 Momen o

Bentang Ujung

M A 

qu ⋅ l 2

=

M A-B

=

16 qu ⋅ l 2

=

=

14 qu ⋅ l 2

=

1520 .81 x 4.65 2 16 1520 .81 x 4.65 2 14

1520 .81 x 4.65 2

MB

=

•

Bentang Dalam

MB

10

qu ⋅ l 2

=

MB-C

MC

=

 Contoh

=

11 qu ⋅ l 2

=

=

16 qu ⋅ l 2

=

11

10

11 1520 .81 x 4.65 2 16

1520 .81 x 4.65 2 11

2055.232

kgm

=

2348.837

kgm

=

3288.371

kgm

=

2989.429

kgm

=

2055.232

kgm

=

2989.429 kgm

Perhitungan Balok Anak Atap :

Data-data : - b

 ρ b =

1520 .81 x 4.65 2

=

= 250 mm

d = 400 – (40 + 8 + 1/2 * 12) = 346 mm

-

h

= 400 mm

Tulangan Utama = D 12 mm

-

fc’

= 35 MPa

Tulangan Sengkang = D 8 mm

-

fy

= 400 MPa

0.85β 1 fc '   600

       = + 600  fy    

 fy

0.85 x 0.81 x 35   400

 ρ  max = 0.75 ρ b = 0.75 x 0.036 = 0.027  ρ min

=

1.4  fy

1.4 = 

400

= 0.0035

600  = 0.036     600 + 400  

F

m

=

 fy 0.85 fc '

=

400 0.85 x 35

= 13.45

•  Tumpuan

Mn

=

Rn

=

ρperlu

=

A 

2055.232 × 9,81 × 1000 Mu = = 25202282.4 Nmm φ 0.8 Mn bd

2

=

25202282.4 250 × 346

= 0.842 N/mm2

2

      1   1 − 1 − 2m × Rn   = 1 1 − 1 − 2 × 13.45 × 0.842     13.45    m   fy 400      

= 0.0021

ρpakai  Aspakai

× = ρ⋅ b ⋅ = 1.33

0.0021

= 0.0028 <

d = 0.0035 × 250

×

ρmin

346

= 302.75 mm2 Maka dipasang tulangan 4-D12

       0        6        0        4        4        3 2-D12

250

Lapangan A – B



Mn =

Rn

ρperlu

2348.837 × 9.81 × 1000 Mu = = 28802613.71 N-mm φ 0 .8

=

Mn bd

2

=

28802613.7 1 250 × 346

2

= 0.0.962 N/mm2

1   2 × 13.45 × 0.962   1   2m × Rn         1 1 − − 1 1 − − =   = 13.45    m   fy 400      

= 0.0024

ρpakai  Aspakai

= 1.33

×

0.0024

ρ⋅

⋅

d = 0.0035× 250

=

b

= 0.0032 <

×

ρmin

346

= 302.75 mm2 Maka dipasang tulangan

2-D12

       0        6        0        4        4        3

4-D12

250

Tumpuan B



Mn =

3288.371 × 9.81 × 1000 Mu = = 40323649.39 N-mm φ 0.8

Rn

ρperlu

=

=

Mn

=

bd2

40323649.3 9 250 × 346

2

= 1.347 N/mm2

    1   1 − 1 − 2m × Rn   = 1 1 − 1 − 2 × 13.45 × 1.347   13.45  m   fy 400     = 0.0034

ρpakai  Aspakai

= 1.33

×

0.0034

ρ⋅

⋅

d = 0.0045 × 250

=

b

= 0.0045 >

×

ρmin

346

= 389.25 mm2 Maka dipasang tulangan 4-D12

       0        6        0        4        4        3 2-D12

250

       

o

 Tumpuan

Mn

=

Rn

=

ρperlu

=

B

2989.429 × 9,81 × 1000 Mu = = 36657873.11 Nmm φ 0.8 Mn bd

=

2

36657873.11 250 × 346

2

= 1.225 N/mm2

      1   1 − 1 − 2m × Rn   = 1 1 − 1 − 2 × 13.45 × 1.225     13.45    m   fy 400      

= 0.0031

ρpakai  Aspakai

= 1.33

×

0.0031

ρ⋅

⋅

d = 0.004 × 250

=

b

= 0.004 >

×

ρmin

346

= 346 mm2 Maka dipasang tulangan 4-D12

       0        6        0        4        4        3 2-D12

250

Lapangan B - C



Mn

=

Rn

=

ρperlu

=

2055.232 × 9.81 × 1000 Mu = = 25202282.4 N-mm φ 0.8 Mn bd

=

2

  1 − m   1

25202282.4 250 × 346

1−

2

2m × Rn fy

= 0.842 N/mm2

       

=

  1 − 13.45   1

= 0.0021

ρpakai  Aspakai

= 1.33

×

0.0021

ρ⋅

⋅

d = 0.0035 × 250

=

b

= 302.75 mm2

= 0.0028 <

×

ρmin

346

1−

2 × 13.45 × 0.842   400

     

Maka dipasang tulangan

2-D12

       0        6        0        4        4        3

4-D12

250

 Tumpuan

Mn

=

Rn

=

ρperlu

=

C

2989.429 × 9,81 × 1000 Mu = = 36657873.11 Nmm φ 0.8 Mn bd

=

2

36657873.1 1 250 × 346

2

= 1.225 N/mm2

      1   1 − 1 − 2m × Rn   = 1 1 − 1 − 2 × 13.45 × 1.225     13.45    m   fy 400      

= 0.0031

ρpakai  Aspakai

= 1.33

×

0.0031

ρ⋅

⋅

d = 0.004 × 250

=

b

= 0.004 >

×

ρmin

346

= 346 mm2 Maka dipasang tulangan 4-D12

       0        6        0        4        4        3 2-D12

250