format research_kolom_diaginteraksi ( siap print) baru.pdf

8/17/2019 format research_kolom_diaginteraksi ( siap print) baru.pdf

1/24

Design by : Sondra Raharja, ST

Check Kapasitas Kolom 400 x 400 tulangan 16 D19

Diketahui Beban luar : Pu = 270 kNMu = 200 kN.m

Properties Penampang

Dimensi b = 400 mm

h = 400 mm

d' = dc = 50 mm

d = h - d' d = 350 mm

Beton Baja

Mutu beton, fc' = 27 MPa Mutu baja, fy = 400 MPa

fc' = 27 N/mm2 fy = 400 N/mm2

= 0.85

Modulus Elastisitas Beton Modulud Elastisitas Baja, Ey

Ec = 4700 ( fc' ) Ey = 200,000 MPa

Ec = 24,422 N/mm2 Regangan Leleh Baja : 0y = fy / Ey

0y = 0.00200

Regangan beton, 0c = 0.003

Diameter Tulangan yang dipakai

D = 19 mm

As = 283.5287 mm2

Jumlah kolom tulangan, N

N = 5 deret kolom tulangan

As1 = 5 D 19 ===> As1 = 1417.6437 mm2

As2 = 2 D 19 ===> As2 = 567.05747 mm2

As3 = 2 D 19 ===> As3 = 567.05747 mm2

As4 = 2 D 19 ===> As4 = 567.05747 mm2

As5 = 5 D 19 ===> As5 = 1417.6437 mm2

Posisi Tulangan :

Untuk menentukan posisi tulangan dapat dengan salah satu cara berikut ini :

- Cara Grafis

- Cara analitis = Y = A*y/A

- Dihitung dengan Integral pada interval tertentu.

- Dihitung dengan rumus :

di = dc + ( i - 1)( h - 2dc ) / (N - 1)

h = 400

b = 400

Hal 1 dari 24 Hal


2/24


Dari rumus umum diatas didapat sbb :

Tul Kolom 1 (d1) d1 = 50 mm





POSISI CENTROID PLASTIS (d")

Menghitung centroid plastis (d")

Statis Momen terhadap As3 :

Zc = d3 - h/2 Zc = 0 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

d" =

= - mm dari As3-

5,486,583.92

(0.85 x fc' x h x b x Zc) + (As1x fy x Z1) + (As2x fy x Z2) + (As3x fy x Z3) - (As4x fy x Z4) - (As5x fy x Z5)

(0.85 x fc' x h x b) + (As1+ As2 + As3 + As4 + As5) x fy

centroid plastis

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

0.85fc'

Hal 2 dari 24 Hal


3/24


KONDISI I

Penampang kolom mengalami tekan semua sehingga ;

y1 = ~ mm

0c = 0.003 0s5 0s4 0s3 0s2 0s1

h = 400

b = 400

a

DAERAH TEKAN

0.85 f c'

f s5 f s4 f s3 f s2 f s1

h = 400

b = 400

a

DAERAH TEKAN

GBR TEGANGANGBR REGANGAN

0.85f c'

h = 400

b = 400

aCc

DAERAH TEKAN

GBR GAYA

Hal 3 dari 24 Hal


4/24


3 GAYA AKSIAL

Maka gaya aksial (baja dan beton) yang terjadi akibat regangan tersebut adalah ;

Catt : Gaya aksial beton dihitung berdasarkan tegangannya.

Besar gaya yang ada adalah ;

Akibat baja :

Cs1.I = As1 x fs1.I ====> Cs1.I = 567,057.47 N

Cs2.I = As2 x fs2.I ====> Cs2.I = 226,822.99 NCs3.I = As3 x fs3.I ====> Cs3.I = 226,822.99 N

Cs4.I = As4 x fs4.I ====> Cs4.I = 226,822.99 N

Cs5.I = As5 x fs5.I ====> Cs5.I = 567,057.47 N

Akibat beton :

Cc.I = 0.85 x fc' x a.1 x b Cc.I = 3,672,000 N

Jadi …

P. I = Cc. I + Cs 1. I + C s2. I + C s3.I + Cs4.I + Cs5.I

P. I = N

= kN

4 MOMEN KAPASITAS

Dihitung dengan Statis Momen ke Titik Berat Centroid Plastis

Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm

Hitung Lengan Momen ( Z ) :

Zc = h/2 - a/2 Zc = 0 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.I = (Cc.I x Zc) + (Cs1.I x Z1) + (Cs2.I x Z2) + (Cs3.I x Z2) - (Cs4.I x Z2) - (Cs5.I x Z2)

"jika nilai negatif (-), maka tanda negatif

menunjukkan arah gaya berlawanan dg

yang lain"

5,486,583.92

5,486.58

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

centroid plastis

GBR MOMEN

0.85fc'

Hal 4 dari 24 Hal


5/24


Cc.I x Zc = N.mm

Cs1.I x Z1 = N.mm

Cs2.I x Z2 = N.mm

Cs3.I x Z3 = N.mm

Cs4.I x Z4 = N.mm

Cs5.I x Z5 = N.mm

Mn.I = N.mm

Mn.I = kN.m

KONDISI II

y2 = 400 mm

a2 = .y2

a2 = 340 mm

1 REGANGAN YANG TERJADI

Diketahui : 0c = 0.003

0y = 0.00200

Dari gambar diatas dapat dihitung masing -masing regangan baja

0.00

-17,011,724.22

-85,058,621.10

0.000

0.00

0.000

85,058,621.10

17,011,724.22

DAERAH TEKAN

0

c= 0.003

0s50s4 0s3 0s2

0s1

h = 400

b = 400

y

GBR REGANGAN

DAERAH TEKAN

GBR TEGANGAN

0.85


h = 400

b = 400

a

y

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

h = 400

a

0.85 f c'

h = 400

a

Cc

DAERAH TEKAN

GBR GAYA

y

0s0s

0s0s

0s0c = 0.003

d1

d2

d3

d4

d5

y

y

d y

y

d y

y

d y

y

d y

a selanjutny

y

d y

y

d y

yd y

s

s

s

s

s

c

s

c s

)(003.0

)(003.0

)(003.0

)(003.0

:

)(.003.0

).(

55

44

33

22

11

1

1

1

1

=

=

=

=

=

=

=

y

d y

y

d y

y

d y

y

d y

y

d y

c

s

c

s

c

s

c

s

c

s

55

44

33

22

11

;sbbdituliskandapatatau

=

=

=

=

=

Hal 5 dari 24 Hal


6/24


Regangan baja ;

s1.II = 0.003 x ( y2 - d1 )/y2

s2.II = 0.003 x ( y2 - d2 )/y2

s3.II = 0.003 x ( y2 - d3 )/y2

s4.II = 0.003 x ( y2 - d4 )/y2

s5.II = 0.003 x ( y2 - d5 )/y2

0.002625 0s1.II = 0.002625 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.002063 0s2.II = 0.0020625 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.0015 0s3.II = 0.0015 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya



2 TEGANGAN YANG TERJADI

Maka tegangan baja yang terjadi akibat regangan tersebut adalah ;

Catt : tegangan beton sesuai dengan mutu betonnya.

Diketahui : f c' = 27 N/mm2

f y = 400 N/mm2

Ey = 200,000 N/mm2Maka tegangan baja yang terjadi akibat regangan tersebut adalah ;

====> fs1.II = 400 N/mm2

====> fs2.II = 400 N/mm2

====> fs3.II = 300 N/mm2

====> fs4.II = 187.50 N/mm2

====> fs5.II = 75 N/mm2

3 GAYA AKSIAL




Akibat baja :

Cs1.II = As1 x fs1.II ====> Cs1.II = 567,057.47 N





Akibat beton :

Cc.II = 0.85 x fc' x a. 2 x b Cc.II = 3,121,200 N

Jadi …

P. II = Cc. II + Cs 1. II + Cs2. II + Cs3.II + Cs4.II + Cs5.II

P. II = N

= kN

4 MOMEN KAPASITAS


….. MOMEN KAPASITAS

fs1.II = fy

fs2.II = fy

fs3.II = 0s3.II x Ey





yang lain"

4,297,844.26

4,297.84

Hal 6 dari 24 Hal


7/24


4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 30 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.II = (Cc.II x Zc) + (Cs1.II x Z1) + (Cs2.II x Z2) + (Cs3.II x Z2) - (Cs4.II x Z2) - (Cs5.II x Z2)

Cc.II x Zc = N.mm

Cs1.II x Z1 = N.mm

Cs2.II x Z2 = N.mm

Cs3.II x Z3 = N.mm

Cs4.II x Z4 = N.mm

Cs5.II x Z5 = N.mm

Mn.II = N.mm

Mn.II = kN.m

93,636,000.00

85,058,621.10

17,011,724.22

0.00

-7,974,245.73

-15,948,491.46

171,783,608.132

171.784

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 7 dari 24 Hal


8/24


KONDISI III

y3 = 300 mm

a3 = .y3

a3 = 255 mm



0y = 0.00200

Dari gambar disamping dapat dihitung masing -masing regangan baja sbb:

b = 400

DAERAH TEKAN

GBR TEGANGAN

0.85f c'


h = 400

a

y

DAERAH TARIK

b = 400

DAERAH TEKAN

GBR GAYA

0.85f c'

Cs5 Cs4 Cs3 Cs2 Cs1

h = 400

a

y

DAERAH TARIK

Cc

0c = 0.003

DAERAH

TEKAN

0s5 0s4 0s3

h = 400

b = 400

y

GBR REGANGAN

0s20s1

DAERAH

TARIK

0s0s

0s

0s

0s0c = 0.003

d1

d2

d3

d4

d5

y

0s0c = 0.003

d5

y

0s

y

yd

y

d y

y

d y

y

d y

y

d y

c

s

c

s

c

s

c

s

c

s

=

=

=

=

=

55

44

33

22

11


y

d y

yd y

y

d y

yd y

y

d y

yd y

y

d y

yd y

sc s

s

c s

sc s

sc s

)(003.0

)(003.0

)(003.0

)(003.0

44

4

4

3

33

3

22

2

2

11

1

1

= =

= =

= =

= =

y yd

yd y

yd yd d

yd d

d

yd

c s

cc s

cc s s s

c s s

c s

s

)(

)).((

55

55

555555

5555

5

5

5

5

=

××=×

××+××=×

+=×

=

+

Hal 8 dari 24 Hal


9/24


Regangan baja ;

s1.III = 0.003 x ( y3 - d1 )/y3

s2.III = 0.003 x ( y3 - d2 )/y3

s3.III = 0.003 x ( y3 - d3 )/y3

s4.III = 0.003 x ( y3 - d4 )/y3

s5.III = 0.003 x ( y3 - d5 )/y3

0.0025 0s1.III = 0.002500 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.00175 0s2.III = 0.0017500 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya



0.0005 0s5.III = -0.0005 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya





f y = 400 N/mm2

Ey = 200,000 N/mm2Maka tegangan baja yang terjadi akibat regangan tersebut adalah ;

====> fs1.III = 400 N/mm2

====> fs2.III = 350 N/mm2

====> fs3.III = 200 N/mm2

====> fs4.III = 50 N/mm2

====> fs5.III = -100 N/mm2

3 GAYA AKSIAL




Akibat baja :

Cs1.III = As1 x fs1.III ====> Cs1.III = 567,057.47 N




Cs5.III = As5 x fs5.III ====> Cs5.III = -141764.368 N

Akibat beton :

Cc.III = 0.85 x fc' x a.3 x b Cc.III = 2,340,900 N

Jadi …

P. III = Cc. III + Cs 1. III + Cs2. III + Cs3.III + Cs4.III + Cs5.III

P. III = N

= kN

4 MOMEN KAPASITAS


…… MOMEN KAPASITAS

fs1.III = fy

fs2.III = 0s2.III x Ey






yang lain"

3,106,427.59

3,106.43

Hal 9 dari 24 Hal


10/24


4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 72.5 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.III = (Cc.III x Zc) + (Cs1.III x Z1) + (Cs2.III x Z2) + (Cs3.III x Z2) - (Cs4.III x Z2) + (Cs5.III x Z2)

Cc.III x Zc = N.mm

Cs1.III x Z1 = N.mm

Cs2.III x Z2 = N.mm

Cs3.III x Z3 = N.mm

Cs4.III x Z4 = N.mmCs5.III x Z5 = N.mm

Mn.III = N.mm

Mn.III = kN.m

169,715,250.00

85,058,621.10

14,885,258.69

0.00

-2,126,465.5321,264,655.27

288,797,319.534

288.797

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 10 dari 24 Hal


11/24


KONDISI IV

y4 = 200 mm

a4 = .y4

a4 = 170 mm

1 REGANGAN YANG TERJADIDiketahui : 0c = 0.003

0y = 0.00200

b = 400

DAERAH TEKAN

GBR TEGANGAN

0.85f c'

f s5 f s4 f s2 f s1

h = 400

a

y

DAERAH TARIK

0c = 0.003

DAERAH

TEKAN

0s50s4

h = 400

b = 400

y

GBR REGANGAN

0s20s1

DAERAH

TARIK

GBR GAYA

0.85f c'

Cs5 Cs4 Cs2

b=400

h = 400

Cc

a

Cs1

DAERAH TEKANDAERAH TARIK

0s

0s

0s

0s

0c = 0.003

d1

d2

d3

d4

d5

y

0s

0s

0c = 0.003

d4

yd5

0s

0s

dari gambar disamping dapat dibuatkan rumus sbb :

0s

0s

0c = 0.003

d4

yd5

0s

0s

y

yd

y

yd

yd y

yd yd d

yd d

d

yd

c s

c s

cc s

cc s s s

c s s

c s

s

)(

;sbbdituliskandapatuntuk

)(

)).((

44

s4

55

55

555555

5555

5

5

5

5

=

=

××=×

××+××=×

+=×

=

+

y

d y

yd y

y

d y

yd y

sc s

sc s

)(003.0

)(003.0

22

2

2

11

1

1

= =

= =

Hal 11 dari 24 Hal


12/24


Regangan baja ;

s1.IV = 0.003 x ( y4 - d1 )/y4

s2.IV = 0.003 x ( y4 - d2 )/y4

s3.IV = 0.003 x ( y4 - d3 )/y4

s4.IV = 0.003 x ( y4 - d4 )/y4

s5.IV = 0.003 x ( y4 - d5 )/y4

0.00225 0s1.IV = 0.002250 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.001125 0s2.IV = 0.0011250 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya0 0s3.IV = - < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.001125 0s4.IV = -0.001125 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.00225 0s5.IV = -0.00225 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy





f y = 400 N/mm2

Ey = 200,000 N/mm2


====> fs1.IV = 400 N/mm2

====> fs2.IV = 225 N/mm2

====> fs3.IV = 0 N/mm2

====> fs4.IV = -225 N/mm2

====> fs5.IV = -400 N/mm2

3 GAYA AKSIAL




Akibat baja :

Cs1.IV = As1 x fs1.IV ====> Cs1.IV = 567,057.47 N

Cs2.IV = As2 x fs2.IV ====> Cs2.IV = 127,587.93 N

Cs3.IV = As3 x fs3.IV ====> Cs3.IV = 0.00 N

Cs4.IV = As4 x fs4.IV ====> Cs4.IV = -127,587.93 N

Cs5.IV = As5 x fs5.IV ====> Cs5.IV = -567,057.47 N

Akibat beton :

Cc.IV = 0.85 x fc' x a.4 x b Cc.IV = 1,560,600 N

Jadi …

P. IV = Cc. IV + Cs1. IV + Cs2. IV + Cs3.IV + Cs4.IV + Cs5.IVP. IV = N

= kN

fs1.IV = fy

fs2.IV = 0s2.IV x Ey

fs5.IV = -fy





yang lain"

1,560,600.00

1,560.60

0s

0s

0s

0s

0c = 0.003

d1

d2

d3

d4

d5

y

y

yd

y yd

y

d y

y

d y

c

s

c

s

s

c

s

c

s

=

=

=

=

=

55

44

3

22

11

netral)grs padatepatreg.(karena0


Hal 12 dari 24 Hal


13/24


4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 115 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.IV = (Cc.IV x Zc) + (Cs1.IV x Z1) + (Cs2.IV x Z2) + (Cs3.IV x Z2) + (Cs4.IV x Z2) + (Cs5.IV x Z2)

Cc.IV x Zc = N.mm

Cs1.IV x Z1 = N.mm

Cs2.IV x Z2 N.mm

Cs3.IV x Z3 N.mm

Cs4.IV x Z4 N.mm

Cs5.IV x Z5 N.mm

Mn.IV = N.mm

Mn.IV = kN.m

9,569,094.87

85,058,621.10

179,469,000.00

85,058,621.10

9,569,094.87

0.00

368,724,431.938

368.724

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 13 dari 24 Hal


14/24


KONDISI V

y5 = 150 mm

a5 = .y5

a5 = 127.5 mm

162.51 REGANGAN YANG TERJADI


0y = 0.00200

0c = 0.003

DAERAH

TEKAN

0s50s4

h = 400

b = 400

y

GBR REGANGAN

0s2

0s1

DAERAH

TARIK

0s3

0s2

GBR TEGANGAN

b = 400

DAERAH

TEKAN

0.85 f c'

f s5 f s4 f s2 f s1

h = 400

a

y

DAERAH TARIK

f s3

0s

0s

0s

0s

0c = 0.003

d1

d2

d3

d4

d5

y

0s



b = 400

DAERAH

TEKAN

0.85f c'

Cs 5 Cs4 Cs2 Cs1

h = 400

a

y

DAERAH TARIK

Cs3

GBR GAYA

Cc

y

d y

yd y

y

d y

yd y

sc s

sc s

)(003.0

)(003.0

22

2

2

11

1

1

= =

= =

y yd

y

yd

y

yd

yd y

yd yd d

yd d

d

yd

c

s

c

s

c

s

cc s

cc s s s

c s s

c s

s

)(

)(

;sbbdituliskandapatdanuntuk

)(

)).((

3

3

4

4

s3s4

5

5

55

555555

5555

5

5

5

5

=

=

=

××=×

××+××=×

+=×

=

+

Hal 14 dari 24 Hal


15/24


Regangan baja ;

s1.V = 0.003 x ( y5 - d1 )/y5

s2.V = 0.003 x ( y5 - d2 )/y5

s3.V = 0.003 x ( y5 - d3 )/y5s4.V = 0.003 x ( y5 - d4 )/y5

s5.V = 0.003 x ( y5 - d5 )/y5

0.002 0s1.V = 0.00200 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.0005 0s2.V = 0.00050 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.001 0s3.V = -0.00100 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.0025 0s4.V = -0.00250 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.004 0s5.V = -0.00400 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy


Maka tegangan baja yang terjadi akibat regangan tersebut adalah ;Catt : tegangan beton sesuai dengan mutu betonnya.


f y = 400 N/mm2

Ey = 200,000 N/mm2


====> fs1.V = 400 N/mm2

====> fs2.V = 100 N/mm2

====> fs3.V = -200 N/mm2

====> fs4.V = -400 N/mm2

====> fs5.V = -400 N/mm2

3 GAYA AKSIAL




Akibat baja :

Cs1.V = As1 x fs1.V ====> Cs1.V = 567,057.47 N

Cs2.V = As2 x fs2.V ====> Cs2.V = 56,705.75 N

Cs3.V = As3 x fs3.V ====> Cs3.V = -113,411.49 N

Cs4.V = As4 x fs4.V ====> Cs4.V = -226,822.99 N

Cs5.V = As5 x fs5.V ====> Cs5.V = -567,057.47 N

Akibat beton :

Cc.V = 0.85 x fc' x a.5 x b Cc.V = 1,170,450 N

fs1.V = fy

fs2.V = 0s2.V x Ey

fs3.V = 0s3.V x Ey

fs4.V = -fy

fs5.V = -fy



yang lain"

0s

0s

0c = 0.003

d3d4

d5

y

0s 0s

0s

0s

y

yd

y

yd

y

yd

y

d y

y

d y

c

s

c

s

c

s

c

s

c

s

=

=

=

=

=

55

44

33

22

11


Hal 15 dari 24 Hal


16/24


Jadi …

P. V = Cc. V + Cs1. V + C s2. V + C s3.V+Cs4.V+Cs5.V

P. V = N

= kN

4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 136.25 mm

Z1 = d3 - d1 Z1 = 150 mmZ2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.V = (Cc.V x Zc) + (Cs1.V x Z1) + (Cs2.V x Z2) + (Cs3.V x Z2) + (Cs4.V x Z2) + (Cs5.V x Z2)

Cc.V x Zc = N.mm

Cs1.V x Z1 = N.mm

Cs2.V x Z2 = N.mm

Cs3.V x Z3 = N.mm

Cs4.V x Z4 = N.mm

Cs5.V x Z5 = N.mm

Mn.V = N.mm

Mn.V = kN.m

886,921.26

886.92

159,473,812.50

85,058,621.10

4,252,931.05

0.00

17,011,724.22

85,058,621.10

350,855,709.966

350.856

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 16 dari 24 Hal


17/24


KONDISI VI

y6 = 130 mm

a6 = .y6

a6 = 110.5 mm



0y = 0.00200

0c = 0.003

DAERAH

TEKAN

0s50s4

h = 400

b = 400

y

GBR REGANGAN

0s1

DAERAH

TARIK

0s3

0s2

GBR TEGANGAN

b = 400

DAERAH

TEKAN

0.85 f c'

f s5 f s4 f s2 f s1

h = 400

a

y

DAERAH TARIK

f s3

b = 400

DAERAH

TEKAN

0.85 f c'

Cs 5 Cs4 Cs2 Cs1

h = 400

a

y

DAERAH TARIK

Cs3

GBR GAYA

Cc

0s

0s

0s

0s

0c = 0.003

d1

d2

d3

d4

d5

y

0s



y

d y

yd y

y

d y

yd y

sc s

sc s

)(003.0

)(003.0

22

2

2

11

1

1

= =

= =

y yd

y

yd

y

yd

yd y

yd yd d

yd d

d yd

c s

c s

c s

cc s

cc s s s

c s s

c s

s

)(

)(


)(

)).((

33

44

s3s4

55

55

555555

5555

5

5

5

5

=

=

=

××=×

××+××=×

+=×

=

+

Hal 17 dari 24 Hal


18/24


Regangan baja ;

s1.VI = 0.003 x ( y6 - d1 )/y6

s2.VI = 0.003 x ( y6 - d2 )/y6

s3.VI = 0.003 x ( y6 - d3 )/y6

s4.VI = 0.003 x ( y6 - d4 )/y6

s5.VI = 0.003 x ( y6 - d5 )/y6

0.001846 0s1.VI = 0.00185 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.000115 0s2.VI = 0.000115 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.001615 0s3.VI = -0.001615 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.003346 0s4.VI = -0.003346 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy

0.005077 0s5.VI = -0.0050769 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy



Catt : tegangan beton sesuai dengan mutu betonnya.Diketahui : f c' = 27 N/mm2

f y = 400 N/mm2

Ey = 200,000 N/mm2


====> fs1.VI = 369.231 N/mm2

====> fs2.VI = 23.0769 N/mm2

====> fs3.VI = -323.0769 N/mm2

====> fs4.VI = -400 N/mm2

====> fs5.VI = -400 N/mm2

3 GAYA AKSIAL




Akibat baja :

Cs1.VI = As1 x fs1.VI ====> Cs1.VI = 523,437.67 N

Cs2.VI = As2 x fs2.VI ====> Cs2.VI = 13,085.94 N

Cs3.VI = As3 x fs3.VI ====> Cs3.VI = -183,203.18 N



Akibat beton :

Cc.VI = 0.85 x fc' x a.6 x b Cc.VI = 1,014,390 N

fs1.VI = 0s1.VI x Ey



fs4.VI = -fy

fs5.VI = -fy



yang lain"

0s

0s

0c = 0.003

d3d4

d5

y

0s 0s

0s

0s

y

yd

y

yd y

yd

y

d y

y

d y

c

s

c

s

c

s

c

s

c

s

=

=

=

=

=

55

44

33

22

11


Hal 18 dari 24 Hal


19/24


Jadi …

P. VI = Cc. VI + Cs1. VI + Cs2. VI + Cs3.VI + Cs4.VI + Cs5.VI

P. VI = N

= kN

4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 144.75 mm

Z1 = d3 - d1 Z1 = 150 mmZ2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.VI = (Cc.VI x Zc) + (Cs1.VI x Z1) + (Cs2.VIx Z2) + (Cs3.VI x Z2) + (Cs4.VI x Z2) + (Cs5.VI x Z2)

Cc.VI x Zc = N.mm

Cs1.VI x Z1 = N.mm

Cs2.VI x Z2 N.mm

Cs3.VI x Z3 N.mm

Cs4.VI x Z4 N.mm

Cs5.VI x Z5 N.mm

Mn.VI = N.mm

Mn.VI = kN.m

573,829.96

573.83

146,832,952.50

78,515,650.24

981,445.63

0.00

17,011,724.22

85,058,621.10

328,400,393.686

328.400

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 19 dari 24 Hal


20/24


KONDISI VII

y7 = 100 mm

a7 = .y7

a7 = 85 mm



0y = 0.00200

0c = 0.003

DAERAH

TEKAN

0s5

0s4

h = 400

b = 400

y

GBR REGANGAN

0s1

DAERAH

TARIK

0s3

0s2

GBR TEGANGAN

b = 400

DAERAH

TEKAN

0.85 f c'

f s5 f s4 f s2 f s1

h = 400

a

y

DAERAH TARIK

f s3

b = 400

DAERAH

TEKAN

0.85f c'

Cs 5 Cs4 Cs2 Cs1

h = 400

a

y

DAERAH TARIK

Cs3

GBR GAYA

Cc

0s

0s

0s

0s

0c = 0.003

d1

d2

d3

d4

d5

y

0s

untuk reg. 2, 3, 4 dan 5 dapat dibuatkan rumus sbb :


y

d y

yd y sc s

)(003.0 11

1

1 = =

y

yd y

yd

y

yd

y

yd

yd y

yd yd d

yd d

d

yd

c s

c s

c s

c s

cc s

cc s s s

c s s

c s

s

)(

)(

)(


)(

)).((

22

33

44

s3s4

55

55

555555

5555

5

5

5

5

=

=

=

=

××=×

××+××=×

+=×

=

+

Hal 20 dari 24 Hal


21/24


Regangan baja ;

s1.VII = 0.003 x ( y7 - d1 )/y7

s2.VII = 0.003 x ( y7 - d2 )/y7

s3.VII = 0.003 x ( y7 - d3 )/y7

s4.VII = 0.003 x ( y7 - d4 )/y7

s5.VII = 0.003 x ( y7 - d5 )/y7

0.0015 0s1.VII = 0.00150 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.00075 0s2.VII = -0.000750 < 0y ===> Baja belum leleh, maka hitung dulu tegangannya sesuai dengan regangannya

0.003 0s3.VII = -0.003000 > 0y ===> Baja leleh, maka untuk tegangan baja pakai fy







f y = 400 N/mm2

Ey = 200,000 N/mm2


====> fs1.VII = 300.00 N/mm2

====> fs2.VII = -150.00 N/mm2

====> fs3.VII = -400 N/mm2

====> fs4.VII = -400 N/mm2

====> fs5.VII = -400 N/mm2

3 GAYA AKSIAL



…… Besar gaya yang ada adalah ;

fs1.VII = 0s1.VII x E

fs2.VII = 0s2.VII x E

fs3.VII = -fy

fs4.VII = -fy

fs5.VII = -fy

0s

0s

0s

0s

0c = 0.003

d2

d4

y

0s

d3

d5

0s

0s

0s

0s

y

yd

y

yd

y

yd

y

yd

y

d y

c

s

c

s

c

s

c

s

c

s

=

=

=

=

=

55

44

33

22

11


Hal 21 dari 24 Hal


22/24



Akibat baja :

Cs1.VII = As1 x fs1.VII ====> Cs1.VII = 425,293.11 N

Cs2.VII = As2 x fs2.VII ====> Cs2.VII = -85,058.62 N




Akibat beton :

Cc.VII = 0.85 x fc' x a.7 x b Cc.VI = 780,300 N

Jadi …

P. VII = Cc. VII + Cs 1. VII + Cs2. VII + Cs3.VII + Cs4.VII + Cs5.VII

P. VII = N

= kN

4 MOMEN KAPASITAS


Diketahui :

d1 = 50 mm

d2 = 125 mm

d3 = 200 mm

d4 = 275 mm

d5 = 350 mm


Zc = h/2 - a/2 Zc = 157.5 mm

Z1 = d3 - d1 Z1 = 150 mm

Z2 = d3 - d2 Z2 = 75 mm

Z3 = d3 - d3 Z3 = 0 mm

Z4 = d4 - d3 Z4 = 75 mm

Z5 = d5 - d3 Z5 = 150 mm

Mn.VII = (Cc.VII x Zc) + (Cs1.VII x Z1) - (Cs2.VII x Z2) + (Cs3.VII x Z2) + (Cs4.VII x Z2) + (Cs5.VII x Z2)



yang lain"

99,831.03

99.83

h = 400

b = 400

Cs5 Cs4 Cs3 Cs2 Cs1

Cc

d1

d2

d3

d4

d5

Z1

Z2Z4

Z5

GBR MOMEN

a

centroid plastis

0.85fc'

Hal 22 dari 24 Hal


23/24


Cc.VII x Zc N.mm

Cs1.VII x Z1 N.mm

Cs2.VII x Z2 N.mm

Cs3.VII x Z3 N.mm

Cs4.VII x Z4 N.mm

Cs5.VII x Z5 N.mm

Mn.VII = N.mm

Mn.VII = kN.m

Dari perhitungan-perhitungan diatas, maka diperoleh resume (dalam bentuk tabulasi) sbb :

Tabel.1 Tabulasi perhitungan P i dan M i

Kondisi (1) (2) (3) (4) (5) (6) (7)

y (cm) = 40 30 20 15 13 10 A

a (cm) 40 34 25.5 17 12.75 11.05 8.5 Tulangan

s1 0.003 0.002625 0.002500 0.002250 0.00200 0.00185 0.00150 1417.643685 ==> dalam mm

s2 0.003 0.002063 0.001750 0.001125 0.00050 0.00012 -0.00075 567.057474 ==> dalam mm

s3 0.003 0.001500 0.001000 - -0.00100 -0.00162 -0.00300 567.057474 ==> dalam mm

s4 0.003 0.000938 0.000250 -0.001125 -0.00250 -0.00335 -0.00525 567.057474 ==> dalam mm s5 0.003 0.000375 -0.000500 -0.002250 -0.00400 -0.00508 -0.00750 1417.643685 ==> dalam mm

f s1 400 400 400 400 400 369.231 300.00 N/mm2

f s2 400 400 350 225 100 23.077 -150.00 N/mm2

f s3 400 300 200 - -200 -323.077 -400.00 N/mm2

f s4 400 188 50 -225 -400 -400.000 -400.00 N/mm2

f s5 400 75 -100 -400 -400 -400.000 -400.00 N/mm2

Cs1 567.06 567.06 567.06 567.05747 567.06 523.44 425.29 kN

Cs2 226.82 226.82 198.47 127.58793 56.71 13.09 -85.06 kN

Cs3 226.82 170.12 113.41 0 -113.41 -183.20 -226.82 kN

Cs4 226.82 106.32 28.35 -127.5879 -226.82 -226.82 -226.82 kNCs5 567.06 106.32 -141.76 -567.0575 -567.06 -567.06 -567.06 kN

Csi 1,814.58 1,176.64 765.53 0 -283.53 -440.56 -680.47 kN

0.85.fc a.b 3,672 3,121.20 2,340.90 1,560.60 1,170.45 1,014.39 780.30 kN

Pn 5,486.58 4,297.84 3,106.43 1560.6 886.92 573.83 99.83 kN

Mn 0.000 171.784 288.797 368.724 350.856 328.400 282.38 kN.m

122,897,250.00

63,793,965.82

-6,379,396.58

0.00

17,011,724.22

85,058,621.10

282,382,164.555

282.382

Hal 23 dari 24 Hal


24/24


B. GRAFIK DIAGRAM INTERAKSI KOLOM

Resume Variasi Beban Ultimate :

PI = 5,486.58 kN MtotI = 0.000 kN.m

PII = 4,297.84 kN MtotII = 171.784 kN.m

PIII = 3,106.43 kN MtotIII = 288.797 kN.m

PIV = 1,560.60 kN MtotIV = 368.724 kN.m

PV =886.92

kN MtotV =350.856

kN.mPVI = 573.83 kN MtotVI = 328.400 kN.m

PVII = 99.83 kN MtotVII = 282.382 kN.m

I 5,486.58 0.000

II 4,297.84 171.784

III 3,106.43 288.797

IV 1,560.60 368.724

V 886.92 350.856

VI 573.83 328.400VII 99.83 282.382

ONDISI

Beban

Aksial,Pn

(kN)

Mn (kN.m)

Diagram Interaksi Kolom

0.000, 5,486.58

171.784, 4,297.84

288.797, 3,106.43

368.724, 1,560.60

350.856, 886.92328.400, 573.83

282.382, 99.83200, 270

0

1,000

2,000

3,000

4,000

5,000

6,000

0 100 200 300 400

Mn

P n

Kapasitas Kolom

Beban Luar

format research_kolom_diaginteraksi ( siap print) baru.pdf

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