Tugas Beton Kolom

7/25/2019 Tugas Beton Kolom

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Hitung kolom persegi (500 x 500) mm dengan 8D32. =0,0257, dan fc = 27,5 Mpa, Fy = 420 Mpa.Gambar diagram interaksi.

Ast = 8.(804,2)=6434 mm2

Ag = (500 mm)2=250000 mm

2

=0,0257

tinggi efektif, d = 500 mm60 mm = 440 mm

d1= 60 mm

1. Point 1

0 c g st y stP = 0, 85.f ' A - A + f .A

0P = 0,85.27,5 250000- 6434 +420.6434

0P =8395,6KN

n 0P = P

nP =0,8.8395,6KN

nP =6716,48KN

2. Point 2 (c=500mm)

a =.C = 0,85.500mm= 425mm

Menghitung regangan tulangan baja

s1 cu

C - 60 500 - 60 = = 0,003 = 0,00264

C 500

s2 cu

C - 250 500 -250 = = 0,003 = 0,0015

C 500

s3 cu

C - 440 500 - 440 = = 0,003 = 0,00036

C 500

Menghitung tegangan tulangan baja


2/12

s1 s s1f =E . = 200000.0,00264 =528MPa= 420MPa Compression

s2 s s2f =E . = 200000.0,0015 = 300MPa Compression


Menghitung gaya pada penampang kolom

c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.500

C = 4967,19KN

3.

s1 s1 s1 c

s1

s1

C = A f - 0,85f '

C = 3 804,2 420 -0,85.27,5

C = 956.95KN

s2 s2 s2 c

s2

s2

C = A f - 0,85f '

C = 2 804,2 300- 0,85.27,5

C = 444,92KN

s2 s2 s2 c

s2

s2

C = A f - 0,85f '

C = 3 804,2 72- 0,85.27,5

C =117,3KN

n c s1 s2 s3

n

n

P = C +C + C C

P = 4967,19KN+956.95KN+ 444,92KN+117,32KN

P =6486,41KN

Menghitung momen terhadap titik pusat penampang kolom

n c s1 1 s3 1h a h h

M = C - + C - d C - d2 2 2 2

n

0,85 500500 500 500M = 4967,19 - +956.95 - 60 -117,3 - 60

2 2 2 2

nM = 345,8KNmMenghitung eksentrisitas (eksesntrisitas dari titik pusat)

M 345,8e = = = 53,31mm

P 6486,41



3/12

a =.C=0,85.400mm=374mm


s1 cu

C - 60 440 - 60 = = 0,003 = 0,002591

C 440

s2 cuC - 250 440 - 250 = = 0,003 = 0,001295

C 440


s1 s s1f =E . = 200000.0,002591=518,2MPa= 420MPa Compression

s2 s s2f =E . = 200000.0,001295= 259MPa Compression


c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.440C = 4371,13KN

4.

s1 s1 s1 c

s1

s1

C = A f - 0,85f '

C = 3 804,2 420 -0,85.27,5

C = 956.95KN

s2 s2 s2 c

s2

s2

C = A f - 0,85f '

C = 2 804,2 259- 0,85.27,5

C = 379,15KN

s s

s

s

T =A .fy

T =3 804,2 .0

T =0KN


4/12

n c s1 s1 s

n

n

P = C +C +C - T

P = 4371,13KN+956.95KN+379,15KN

P =5707,22KN


n c s1 1

h a hM = C - +C - d

2 2 2

440

n

0,85500 500M = 4371,13 - +956.95 - 60

2 2 2

nM = 457,2KNmMenghitung eksentrisitas (eksesntrisitas dari titik pusat)

M 457,2e = = = 80,11mm

P 5707,22


a =.C=0,85.320mm=272mm


s1 cu

C - 60 320 - 60 = = 0,003 = 0,0024375

C 320

s2 cu

C -250 320 -250 = = 0,003 = 0,00065625

C 320

s cu

440 -C 440 - 320 = = 0,003 = 0,001125

C 320


s1 s s1f =E . = 200000.0,0024375= 487,5MPa= 420MPa Compression

s2 s s2f =E . = 200000.0,00065625=131,25MPa Compression

s s sf =E . = 200000.0,001125= 225MPa Tension



5/12

c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.320

C = 3179KN

5.

s1 s1 s1 c

s1

s1

C = A f - 0,85f 'C = 3 804,2 420 -0,85.27,5

C = 956.95KN

s2 s2 s2 c

s2

s2

C = A f - 0,85f '

C = 2 804,2 131,25- 0,85.27,5

C =173,52KN

s s

s

s

T = A .fy

T =3 804,2 .225

T =542,87KN

n c s1 s2 s

n

n

P = C +C + C - T

P = 3179KN+956.95KN+173,52KN- 542,87KN

P =3766,60KN


n c s1 1

h a h hM = C - +C - d + Ts d-

2 2 2 2

n

0,85 320500 500 500M = 3179 - +956.95 - 60 +542,87 440 -

2 2 2 2

nM = 647,4KNm

Menghitung eksentrisitas (eksesntrisitas dari titik pusat)M 647,4

e = = =171,87mmP 3766,60

5. Point 5 (Balanced Point)Menghitung nilai cb, dimana nilai d = 500 mm - 60 mm = 440 mm

bc 440=0,003 0,003+0,0021

bC = 258,8mm


6/12

a =.C=0,85.258,8mm=219,98mm

sy

E 420 = = = 0,0021

fy 200000


bs1 cu

b

C -6 0 258,8-60 = = 0,003 = 0,002305

C 258,8

bs2 cu

b

C -250 258,8-250 = = 0,003 = 0,000102

C 258,8


s1 s s1f =E . = 200000.0,002305 = 461MPa= 420MPa Compression

s2 s s2f =E . = 200000.0,000102= 20,5MPa Compression


c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.258,8

C = 2571,25KN

s1 s1 s1 c

s1

s1

C = A f - 0,85f '

C = 3 804,2 420 -0,85.27,5

C = 956.95KN

s2 s2 s2 c

s2

s2

C = A f - 0,85f '

C = 2 804,2 20,5- 0,85.27,5

C =-4,7KN


7/12

s s

s

s

T =A .fy

T =3 804,2 .420

T =1013,35KN

n c s1 s1 s

n

n

P = C +C +C - TP = 2571,25KN+956.95KN-1013,35KN

P =2514,85KN


n c s1 1

h a h hM = C - +C - d + Ts d-

2 2 2 2

n

0,85 258,8500 500 500M = 2571,25 - +956.95 - 60 +1013,35 440 -

2 2 2 2

nM = 734,33KNm

Menghitung eksentrisitas (eksesntrisitas dari titik pusat)

M 734, 33e = = = 292mm

P 2514,85


a =.C=0,85.200mm=170mm


s1 cu

C - 60 200 - 60 = = 0,003 = 0,0021

C 200

s2 cu

C - 250 200 -250 = = 0,003 = -0,00075

C 200

s cu

d- C 440 - 200 = = 0,003 = 0,0036

C 200


s1 s s1f =E . = 200000.0,0021= 420MPa Compression

s2 s s2f =E . = 200000.0,00075=150MPa Tension


8/12

s s sf = E . = 200000.0,0036 = 720MPa= 420MPa Tension


c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.200C =1986,87KN

7.

s1 s1 s1 c

s1

s1

C = A f - 0,85f '

C = 3 804,2 420 -0,85.27,5

C = 956.95KN

.s2 s2 s2

s2

s2

C = A f

C =2 804,2 150C = 241,27KN

s s

s

s

T = A .fy

T =3 804,2 .420

T =1013,35KN

n c s1 s2 s

n

n

P = C + C - T - TP =1490,16KN+ 956.95KN- 241,27KN-1013,35KN

P =1689,2KN


n c s1 1

h a h hM = C - +C - d + Ts d-

2 2 2 2

n

0,85 200500 500 500M =1986,87 - +956.95 - 60 +1013,35 440 -

2 2 2 2

nM = 702,2KNm


M 702,2e = = = 415,7mm

P 5707,22


a =.C= 0,85.150mm=127,5mm


9/12


s1 cu

C - 60 150 - 60 = = 0,003 = 0,0018

C 150

s2 cu

C -250 150 - 250 = = 0,003 = -0,002

C 150

s3 cu

C - 400 150 -400 = = 0,003 = -0,0058

C 150



s2 s s2f =E . = 200000.- 0,002 = -400MPa Tension

s3 s s3f =E . = 200000.- 0,0058= -420MPa Tension

Menghitung gaya pada penampang kolomc 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.150

C =1490,16KN

s1 s1 s1 c

s1

s1

C = A f - 0,85f '

C = 3 804,2 360- 0,85.27,5

C = 812,2KN

s2 s2 s2

s2

s2

C = A f

C = 2 804,2 400

C =643,4KN

s s

s

s

T =A .fy

T =3 804,2 .420

T =1013,35KN


10/12

n c s1 s1 s

n

n

P = C + C - C - T

P =1490,16KN+812,2 - 643,4KN-1013,35KN

P =645,6KN


n c s1 1

h a h hM = C - +C - d + Ts d-

2 2 2 2

2

n

0,85 150500 500 500M =1490,16 - +812,2 - 60 +1013,35 440 -

2 2 2

nM = 624,4KNm


M 624, 4

e = = = 967,16mmP 645, 6

8. Point 8 (C=109,68 mm)

a =.C=0,85.109,68mm=93,23mm


s1 cu

C -60 109,68- 60 = = 0,003 = 0,001359

C 109,68

s2 cu

C - 250 109,68 - 250 = = 0,003 = -0,003838

C 109,68

s3 cu

d- C 440 -109,68 = = 0,003 = 0,009035

C 109,68


s1 s s1f =E . = 200000.0,001359 = 271,77MPa Compression

s2 s s2f =E . = 200000.- 0,003838 = -767,62 = 420MPa Tension

s3 s s3f =E . = 200000.0,009035=1807,002MPa= 420MPa Tension



11/12

c 1

c

c

C = 0,85.fc'.b. .c

C = 0,85.27,5.500.0,85.109,68

C =1089,60KN

s1 s1 s1 c

s1

s1

C = A f - 0,85f 'C = 3 804,2 271,77- 0,85.27,5

C = 599,32KN

s2 s2 s2

s2

s2

T = A f

T =2 804,2 420

T = 675,57KN

s s

s

s

T =A .fy

T =3 804,2 .420

T =1013,35KN

n c s1 s2 s

n

n

P = C +C - T - T

P =1089,60KN+599,32KN- 675,57KN-1013,35KN

P =0KN


n c s1 1

h a h hM = C - +C - d + Ts d-

2 2 2 2

2

n

0,85 109,68500 500 500M =1089,60 - +599,32 - 60 +1013,35 440 -

2 2 2

nM = 528,02KNm

Menghitung eksentrisitas (eksesntrisitas dari titik pusat)M 528,02

e = = = InfinityP 0

9. Point 9 (Tarik Maksimum)

n s

n

n

P =A .fy

P =8 804,2 .420

P =2702,3KN

Rekap Hasil Perhitungan Kemampuan Kolom


12/12

Point C (mm) Pn (KN) Mn (KNm) E (mm)

1 - 6716,5 0 0

2 500 6486,4 345,8 53,3

3 440 5707,2 457,2 80,1

4 320 3766,6 647,4 171,9

5 258,8 2514,9 734,3 2926 200 1689,2 702,2 415,7

7 150 645,6 624,4 967,2

8 109,68 0 528 infinity

9 0 -2702,3 0

Tugas Beton Kolom

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