Diagram Interaksi Kolom Tak Berdimensi.xls

DIAGRAM INTERAKSI KOLOM TAK BERDIMENSI PENAMPANG KOLOM LINGKARAN

Source : www.ilmutekniksipil.com CARA MENENTUKAN LUAS TULANGAN KOLOM

Diketahui :

Pu = "a!a ak#ial tekan ren'ana aki&at &e&an ter(aktor$ Mu = momen ren'ana aki&at &e&an ter(aktor$ ter(aktor$ D = diameter kolom, ( ') = kuat tekan &eton, ( ! = te"an"an leleh &a*a$

Tetapkan nilai d  D !an" #e#uai$ %itun" : A" = p  +  D

f  Mn  . ( ')  A"  D / = Mu  . ( ')  A"  D/

→

#e&a"ai a&#i#

f  Pn  . ( ')  A" / = Pu  . ( ')  A" /

→

#e&a"ai ordinat

Plot nilai ter#e&ut ke dalam dia"ram interak#i !an" #e#uai, Lua# tulan"an !an" diperlukan,

→

A# = r  A"

→

diperoleh : r

4 0$13 eD=0$04 0$1

eD=0$03

eD=0$4 0

r = 38

eD=0$43

0$73 0$7 0$63

eD=0$-0

r = +8

0$6 0$53

r = 28

0$5 0$33

eD=0$20

r = -8

0$3 0$+3 0$+

r = 48

0$23

eD=0$30

0$2 0$-3 0$0$43 0$4

eD=4$00

0$03 0

eD=-$00

f =0$53 f = 0$70

e DIAGRAM INTERAKSI KOLOM

( ! = 210 MPa

dD = 0$10

4 eD=0$04 0$13 0$1r = 38 0$73

eD=0$03

eD=0$40 eD=0$43 eD=0$-0

0$7r = +8 0$63 0$6r = 28 0$53 0$5

r = -8

eD=0$20

0$33

0$3 0$+3r = 48 0$+ 0$23

eD=0$30

0$2 0$-3 0$0$43

eD=4$00

0$4 0$03 0

eD=-$00

f =0$53 f = 0$70


( ! = 210 MPa

dD = 0$73

4 eD=0$03 0$13 eD=0$04 = 38 r 0$1

eD=0$40 eD=0$43

0$73 eD=0$-0

0$7

r = +8

0$63

0$6

r = 28

0$53

0$5 eD=0$20

0$33 r = -8 0$3 0$+3

r = 48 0$+ 0$23

eD=0$30

0$2 0$-3 0$0$43 0$4

eD=4$00

0$03 0 eD=-$00

f =0$53 f = 0$70


( ! = 210 MPa

dD = 0$70

4 eD=0$03

eD=0$04 0$13

r = 38 0$1 0$73 0$7r =

eD=0$40 eD=0$43

eD=0$-0

+8

0$63 0$6r =

28

0$53 0$5

r = -8

eD=0$20

0$33 0$3

r = 48

0$+3

0$+

eD=0$30

0$23 0$2 0$-3 0$-

eD=4$00

0$43 0$4 0$03 0

eD=-$00

f =0$53 f = 0$70


( ! = 210 MPa

dD = 0$63

4 eD=0$04 0$13 0$1

eD=0$03

eD=0$40 eD=0$43

r = 38

0$73 0$7

eD=0$-0

r = +8

0$63 0$6 0$53

r = 28

0$5 0$33

eD=0$20

r = -8

0$3 0$+3 0$+

r = 48

0$23

eD=0$30

0$2 0$-3 0$0$43 eD=4$00

0$4 0$03 0

eD=-$00

f =0$53 f = 0$70


( ! = 210 MPa

dD = 0$60

DIAGRAM INTERAKSI KOLOM TAK BERDIMENSI PENAMPANG KOLOM 9;R
CARA MENENTUKAN LUAS TULANGAN KOLOM

Diketahui :

Pu = "a!a ak#ial tekan ren'ana aki&at &e&an ter(aktor$ Mu = momen ren'ana aki&at &e&an ter(aktor$ D = diameter kolom, ( ' = kuat tekan &eton, ( ! = te"an"an leleh &a*a$

Tetapkan nilai h)  h !an" #e#uai$ %itun" : A" = &  h

f  Mn  . ( ')  A"  h / = Mu  . ( ')  A"  h /

→

#e&a"ai a&#i#

f  Pn  . ( ')  A" / = Pu  . ( ')  A" /

→

#e&a"ai ordinat


4

eh = 0,0- eh = 0,03 r = 0,03

0$13

→

diperoleh : r

A# = r  A"

eh = 0,4 eh=0$43 eh = 0,-

0$1

) g A . ' c f ( / n P . f

→

0$73

r = 0,0+

0$7 0$63

eh = 0,-3

r = 0,02

0$6 0$53

r = 0,0-

eh = 0,22

0$5 0$33 r = 0,04

0$3

eh = 0,3

0$+3 0$+ 0$23 0$2 0$-3

eh = 4

0$0$43 0$4 0$03 0

eh = 40

f.Mn/(fc'.Ag.h) 0

0$04 0$0- 0$02 0$0+ 0$03 0$05 0$06 0$07 0$01 0$4 0$44 0$4- 0$42 0$4+ 0$43 0$45 0$46 0$47 0$41 0$- 0$-4 0$-- 0$-2 0$-+ 0$-3 0$-5 0$-6 0$-7 0$-1 0$2

DIAGRAM INTERAKSI KOLOM

( ! = 210 MPa

4 0$13

) g A . ' c f ( / n P . f

r = 0,03

eh = 0,03

eh = 0,4

0$1 0$73

r = 0,0+

eh = 0,-

0$7 0$63

r = 0,02

eh = 0,-3

0$6 0$53

r = 0,0-

0$5 0$33

r = 0,04

eh = 0,22

0$3 eh = 0,3

0$+3 0$+ 0$23 0$2

eh = 0,0-

h)h = 0$73

eh = 4

0$-3 0$0$43 0$4 0$03

eh = 40

f.Mn/(fc'.Ag.h)

0 0

0$04 0$0- 0$02 0$0+ 0$03 0$05 0$06 0$07 0$01 0$4 0$44 0$4- 0$42 0$4+ 0$43 0$45 0$46 0$47 0$41 0$- 0$-4 0$-- 0$-2 0$-+ 0$-3 0$-5 0$-6 0$-7 0$-1 0$2

( ! = 210 MPa

DIAGRAM INTERAKSI KOLOM 4

eh = 0,0-

0$13

eh = 0,4

r = 0,03

0$1

) g A . ' c f ( / n P . f

eh = 0,03

h)h = 0$63

0$73

r = 0,0+

0$7

eh = 0,-

0$63 r = 0,02

0$6

eh = 0,-3

0$53 r = 0,0-

0$5 0$33

eh = 0,22 r = 0,04

0$3 0$+3

eh = 0,3

0$+ 0$23 0$2 0$-3

eh = 4

0$0$43 0$4 0$03 0

f.Mn/(fc'.Ag.h) 0

eh = 40

0$04 0$0- 0$02 0$0+ 0$03 0$05 0$06 0$07 0$01 0$4 0$44 0$4- 0$42 0$4+ 0$43 0$45 0$46 0$47 0$41 0$- 0$-4 0$-- 0$-2 0$-+ 0$-3 0$-5 0$-6 0$-7 0$-1 0$2


( ! = 210 MPa

4 0$13

eh = 0,0- eh = 0,03 r = 0,03

eh = 0,4

0$1 ) g A . ' c f ( / n P . f

0$73

r = 0,0+

0$7 0$63

r = 0,02

eh = 0,-3

0$6 0$53

eh = 0,-

r = 0,0-

0$5 eh = 0,22

0$33 0$3

r = 0,04

0$+3 0$+ 0$23

eh = 0,3

h)h = 0$60

0$2 eh = 4

0$-3 0$0$43 0$4 0$03 0

eh = 40

f.Mn/(fc'.Ag.h) 0

0$04 0$0- 0$02 0$0+ 0$03 0$05 0$06 0$07 0$01 0$4 0$44 0$4- 0$42 0$4+ 0$43 0$45 0$46 0$47 0$41 0$- 0$-4 0$-- 0$-2 0$-+ 0$-3 0$-5 0$-6 0$-7 0$-1 0$2


( ! = 210 MPa

h)h = 0$50

DIAGRAM INTERAKSI KOLOM TAK BERDIMENSI PENAMPANG EMPAT PER
CARA MENENTUKAN LUAS TULANGAN KOLOM

Diketahui :

Pu = "a!a ak#ial tekan ren'ana aki&at &e&an ter(aktor$ Mu = momen ren'ana aki&at &e&an ter(aktor$ D = diameter kolom, ( ' = kuat tekan &eton, ( ! = te"an"an leleh &a*a$

Tetapkan nilai h)  h !an" #e#uai$ %itun" : A" = &  h f  Mn  . ( ')  A"  h / = Mu  . ( ')  A"  h /

→

#e&a"ai a&#i#

f  Pn  . ( ')  A" / = Pu  . ( ')  A" /

→

#e&a"ai ordinat


→

diperoleh : r

A# = r  A"

→

4 0$13

eh=0$04

r = 38

eh=0$4 eh=0$03 0

eh=0$43

0$1 0$73

eh=0$-0

r = +8

0$7 0$63

eh=0$20

r = 28

0$6 0$53

r = -8

0$5 0$33

eh=0$30

r = 48

0$3 0$+3 0$+ 0$23

eh=4$00

0$2 0$-3 0$eh=-$00

0$43 0$4 0$03

f =0$53

0 0

0$0-

0$0+

0$05

0$07

0$4

0$4-

0$4+

0$45

0$47

0$-

0$--

0$-+

0$-5

0$-7

0$2

0$2-

0$2+

0$25

0$27

0$+

f = 0$70

e ( ! = 210 MPa


4 eh=0$04 0$13 r =

eh=0$03

38

h)h = 0$10

eh=0$40 eh=0$43

0$1 0$73

r = +8

eh=0$-0

0$7 0$63

r = 28

eh=0$20

0$6 0$53

r = -8

0$5 0$33

r = 48

eh=0$30

0$3 0$+3 0$+

eh=4$00

0$23 0$2 0$-3 eh=-$00

0$0$43 0$4

f =0$53

0$03

f = 0$70

0 0

0$0-

0$0+

0$05

0$07

0$4

0$4-

0$4+

0$45

0$47

0$-

0$--

0$-+

0$-5

0$-7

0$2

0$2-

0$2+

0$25

0$27

0$+

e ( ! = 210 MPa


h)h = 0$73

4 eh=0$04 0$13

eh=0$03

eh=0$40 eh=0$43

r = 38

0$1 0$73

eh=0$-0

r = +8

0$7 0$63

eh=0$20

r = 28

0$6 0$53

r = -8

0$5 0$33

r = 48

0$3 0$+3 0$+ eh=4$00

0$23 0$2 0$-3 0$0$43

eh=-$00

eh=0$30

0$4 0$03

f =0$53

0 0

0$0-

0$0+

0$05

0$07

0$4

0$4-

0$4+

0$45

0$47

0$-

0$--

0$-+

0$-5

0$-7

0$2

0$2-

0$2+

0$25

0$27


( ! = 210 MPa

4 0$13

eh=0$04

eh=0$03

r = 38

eh=0$40 eh=0$43

0$1 0$73

r = +8

eh=0$-0

0$7 0$63

r = 28

eh=0$20

0$6 0$53

r = -8

0$5 0$33 0$3 0$+3

r = 48 eh=0$30

0$+

f = 0$70

h)h = 0$70

$ 0$+

eh=4$00

0$23 0$2 0$-3 eh=-$00

0$0$43 0$4

f =0$53

0$03

f = 0$70

0 0

0$0-

0$0+

0$05

0$07

0$4

0$4-

0$4+

0$45

0$47

0$-

0$--

0$-+

0$-5

0$-7

0$2

0$2-

0$2+

0$25

0$27

0$+

e ( ! = 210 MPa

DIAGRAM INTERAKSI KOLOM 4 eh=0$04 0$13

eh=0$03

h)h = 0$63

eh=0$40 eh=0$43

r = 38

0$1 0$73

eh=0$-0

r = +8

0$7 0$63

eh=0$20

r = 28

0$6 0$53

r = -8

0$5 0$33

r = 48

0$3

eh=0$30

0$+3 0$+ 0$23

eh=4$00

0$2 0$-3 0$eh=-$00

0$43 0$4 0$03

f =0$53

0 0

0$0-

0$0+

0$05

0$07

0$4

0$4-

0$4+

0$45

0$47

0$-

0$--

0$-+

0$-5

0$-7

0$2

0$2-

0$2+

0$25

0$27

0$+

f = 0$70


( ! = 210 MPa

h)h = 0$60

Diagram Interaksi Kolom Tak Berdimensi.xls

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