Bcj03

Perhitungan "Joint Shear" pada ETABS

vi )

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Arah momen ujung balok B18 dan B22 akibat gaya lateral

Story 4

Story 3

Story 2 C12 Story 1 B18

B22 C12 Base

(a) Arah momen ujung B18 dan B22 akibat gaya lateral

Story 4

Story 3

Story 2 C12 Story 1 B18

B22 C12 Base

(b) Arah momen ujung B18 dan B22 akibat berbaliknya arah gaya lateral Gambar 11 Arah momen-momen ujung balok akibat gaya lateral ( Portal YZ )

Ir. Nityananda Permadi Tjokrodimurti IP-Md

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Perhitungan "Joint Shear" pada ETABS

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Beam Column Joint yang akan dibahas Gambar 12 Hasil perhitungan tulangan pada balok dan kolom portal As-C vi ) Perhitungan kapasitas momen balok B18 Mutu beton dan besi beton : f'cr := 350 ⋅

Mutu beton K350, kuat tekan silinder uji :

f'c := 0.83 ⋅ f'cr

Kuat tekan beton :

kgf 2

cm

f'c = 290.50 ⋅

kgf 2

cm

f'c = 28.49 ⋅ MPa Mutu besi beton BjTD40, tegangan leleh :

fy := 3900 ⋅

kgf 2

cm

Faktor reduksi kekuatan dan koefisien blok tegangan tekan : ϕ := 0.8 β1 :=

0.85 if f'c ≤ 30 ⋅ MPa

⎛ f'c

0.85 − 0.008 ⋅ ⎜

⎝ 1 ⋅ MPa

β1 = 0.8500

⎞

− 30⎟ otherwise

⎠

untuk f 'c < 30 MPa

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Perhitungan "Joint Shear" pada ETABS

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Properti penampang : Stirrups diameter : ϕstirrups := 1.0 ⋅ cm Concrete cover : dcover := 2.5 ⋅ cm Rebar diameter ϕbar := 1.932050733101 ⋅ cm ( diameter tulangan dibuat sedemikian supaya menghasilkan luas tulangan sesuai dengan hasil perhitungan ETABS, lihat Gambar 14 ) 1 2 2 As := 4 ⋅ ⋅ π ⋅ ( 1.932050733101cm) = 11.727 ⋅ cm 4 1 2 2 A1 := ⋅ π ⋅ ( 1.932050733101cm) = 2.932 ⋅ cm 4 b := 30 ⋅ cm

h := 50 ⋅ cm

1 ⎛ ⎞ d := h − ⎜ dcover + ϕstirrups + ⋅ ϕbar⎟ 2 ⎝ ⎠ d = 45.534 ⋅ cm 1 d' := dcover + ϕstirrups + ⋅ ϕbar = 4.466 ⋅ cm 2 Luas tulangan : 2

As := 11.727cm

( Lihat Gambar 14 )

2

As' := 5.651cm

Asumsikan tulangan tekan leleh : f's := fy

kgf

f's = 3900 ⋅

2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ f's

a = 3.199 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = −0.000560 Tegangan tulangan tekan : Es := 2000000 ⋅

f's :=

kgf 2

cm

fy fy if ε's ≥ Es Es ⋅ ε's otherwise

f's = −1120 ⋅

kgf 2

cm

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Perhitungan "Joint Shear" pada ETABS

fs' :=

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0 ⋅ f's if f's < 0 f's otherwise kgf

fs' = 0 ⋅

2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ fs'

a = 6.174 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠

ε's :=

0 otherwise ε's = 0.001155 Tegangan tulangan tekan : f's :=

fy fy if ε's ≥ Es Es ⋅ ε's otherwise

f's = 2311 ⋅ fs' :=

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise

fs' = 2311 ⋅

kgf 2

cm

Regangan tulangan tarik :

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d − ε s :=

⎛⎜ As ⋅ fy ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy

if f's < 0

0.85⋅ f' c⋅ b

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d −

⎛⎜ As ⋅ fy − As' ⋅ fs' ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy− As'⋅ fs'

otherwise

0.85⋅ f' c⋅ b

ε s = 0.023322 ρ :=

As b⋅d

>

ρ = 0.008585

Ir. Nityananda Permadi Tjokrodimurti IP-Md

fy Es

= 0.001950 ρ' :=

As' b⋅d

ρ' = 0.004137 26/45

Perhitungan "Joint Shear" pada ETABS

ρmin :=

1.4 ⋅ MPa fy

www.itsindo.org

ρmin = 0.003661

Rasio penulangan maksimum yang diijinkan :

⎛ 0.85 ⋅ f'c ⋅ β1

ρb075 := 0.75 ⋅ ⎜ ⎜

fy

⎝

ρb075 = 0.026913 >

⋅

⎞⎟ ρ'⋅ fs' + 0.003 ⋅ Es + fy ⎟ fy ⎠ 0.003 ⋅ Es

(OK)

ρ = 0.008585

Momen kapasitas yang dapat dipikul oleh penampang dengan faktor overstrength 1.25 : Mkap :=

⎡ ( ⎣

) ⎛⎝

1.25 ⋅ ⎢ As ⋅ f y − As' ⋅ fs' ⋅ ⎜ d −

⎡ ⎣

⎛ ⎝

1.25 ⋅ ⎢0.85 ⋅ f'c ⋅ a ⋅ b ⋅ ⎜ d −

a⎞

⎤ ⎟ + As' ⋅ fs' ⋅ ( d − d')⎥ if εs > 2⎠ ⎦

fy Es

a ⎞⎤

⎟⎥ otherwise

2 ⎠⎦

Mkap = 2404146.60 ⋅ kgf ⋅ cm

Tabel 3

Output post processor ETABS untuk kapasitas balok

Hasil perhitungan ETABS lihat Gambar 15 pada bagian "Beam Capacities and Angles" baris Beam 3 kolom Capacity -veM = 2383417.859 kgf-cm bandingkan dengan Mkap = 2404146.60 kgf-cm.Perbedaaan dapat terjadi karena adanya perbedaan asumsi formasi tulangan As yang menyebabkan perbedaan jarak titik berat tulangan As dan juga akibat perbedaan pembulatan. vii ) Perhitungan kapasitas momen balok B18 akibat berbaliknya arah gaya lateral Properti penampang : Stirrups diameter : ϕstirrups := 1.0 ⋅ cm Concrete cover : dcover := 2.5 ⋅ cm Rebar diameter ϕbar := 1.932050733101 ⋅ cm ( diameter tulangan dibuat sedemikian supaya menghasilkan luas tulangan sesuai dengan hasil perhitungan ETABS, lihat Gambar 14 ) b := 30 ⋅ cm

h := 50 ⋅ cm

1 ⎛ ⎞ d := h − ⎜ dcover + ϕstirrups + ⋅ ϕbar⎟ 2 ⎝ ⎠ d = 45.534 ⋅ cm 1 d' := dcover + ϕstirrups + ⋅ ϕbar = 4.466 ⋅ cm 2 Luas tulangan : 2

As := 5.651cm

2

As' := 11.727cm

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Perhitungan "Joint Shear" pada ETABS

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Asumsikan tulangan tekan leleh : f's := fy

kgf

f's = 3900 ⋅

2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ f's

a = −3.199 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = 0.000000 Tegangan tulangan tekan : Es := 2000000 ⋅

kgf 2

cm

fy fy if ε's ≥ Es

f's :=

Es ⋅ ε's otherwise f's = 0 ⋅

fs' :=

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise

fs' = 0 ⋅

kgf 2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ fs' 0.85 ⋅ f'c ⋅ b

a = 2.975 ⋅ cm

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = −0.000828 Tegangan tulangan tekan : f's :=

fy fy if ε's ≥ Es Es ⋅ ε's otherwise

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Perhitungan "Joint Shear" pada ETABS

f's = −1656 ⋅ fs' :=

www.itsindo.org

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise kgf

fs' = 0 ⋅

2

cm

Regangan tulangan tarik :

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d − ε s :=

⎛⎜ As ⋅ fy ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy

if f's < 0

0.85⋅ f' c⋅ b

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d −

⎛⎜ As ⋅ fy − As' ⋅ fs' ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

otherwise

As⋅ fy− As'⋅ fs' 0.85⋅ f' c⋅ b

ε s = 0.036028 ρ :=

As b⋅d

ρmin :=

fy

>

Es

= 0.001950

ρ = 0.004137 1.4 ⋅ MPa fy

ρ' :=

As' b⋅d

ρ' = 0.008585

ρmin = 0.003661

Rasio penulangan maksimum yang diijinkan :

⎛ 0.85 ⋅ f'c ⋅ β1

ρb075 := 0.75 ⋅ ⎜ ⎜

fy

⎝

ρb075 = 0.024462 >

⋅

⎞⎟ ρ'⋅ fs' + 0.003 ⋅ Es + fy ⎟ fy ⎠ 0.003 ⋅ Es

(OK)

ρ = 0.004137

Momen kapasitas yang dapat dipikul oleh penampang dengan faktor overstrength 1.25 : Mkap :=

⎡ ( ⎣

)

⎛ ⎝

1.25 ⋅ ⎢ As ⋅ f y − As' ⋅ fs' ⋅ ⎜ d −

⎡ ⎣

⎛ ⎝

1.25 ⋅ ⎢0.85 ⋅ f'c ⋅ a ⋅ b ⋅ ⎜ d −

a⎞

⎤ ⎟ + As' ⋅ fs' ⋅ ( d − d')⎥ if εs > 2⎠ ⎦

fy Es

a ⎞⎤

⎟⎥ otherwise

2 ⎠⎦

Mkap = 1213418.24 ⋅ kgf ⋅ cm Hasil perhitungan ETABS lihat Gambar 15 pada bagian "Beam Capacities and Angles" baris Beam 3 kolom Capacity +veM = 1203521.84 kgf-cmbandingkan dengan Mkap = 1213418.24 kgf-cm. Perbedaaan dapat terjadi karena adanya perbedaan asumsi formasi tulangan As yang menyebabkan perbedaan jarak titik berat tulangan As dan juga akibat perbedaan pembulatan.

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Perhitungan "Joint Shear" pada ETABS

viii )

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Perhitungan kapasitas momen balok B22

Properti penampang : Stirrups diameter : ϕstirrups := 1.0 ⋅ cm Concrete cover : dcover := 2.5 ⋅ cm Rebar diameter ϕbar := 2.004051380897 ⋅ cm ( diameter tulangan dibuat sedemikian supaya menghasilkan luas tulangan sesuai dengan hasil perhitungan ETABS, lihat Gambar 14 ) b := 30 ⋅ cm

h := 50 ⋅ cm

1 ⎛ ⎞ d := h − ⎜ dcover + ϕstirrups + ⋅ ϕbar⎟ 2 ⎝ ⎠ d = 45.498 ⋅ cm 1 d' := dcover + ϕstirrups + ⋅ ϕbar = 4.502 ⋅ cm 2 Luas tulangan : 2

As := 5.731cm

( Lihat Gambar 14 )

2

As' := 9.463cm

Asumsikan tulangan tekan leleh : f's := fy

kgf

f's = 3900 ⋅

2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ f's

a = −1.965 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = 0.000000 Tegangan tulangan tekan : Es := 2000000 ⋅

f's :=

kgf 2

cm

fy fy if ε's ≥ Es Es ⋅ ε's otherwise

f's = 0 ⋅ fs' :=

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise

fs' = 0 ⋅

kgf 2

cm

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Perhitungan "Joint Shear" pada ETABS

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Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ fs'

a = 3.017 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠

ε's :=

0 otherwise ε's = −0.000805 Tegangan tulangan tekan : fy fy if ε's ≥ Es

f's :=

Es ⋅ ε's otherwise

f's = −1610 ⋅ fs' :=

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise kgf

fs' = 0 ⋅

2

cm

Regangan tulangan tarik :

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d − ε s :=

⎛⎜ As ⋅ fy ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy

if f's < 0

0.85⋅ f' c⋅ b

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d −

⎛⎜ As ⋅ fy − As' ⋅ fs' ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy− As'⋅ fs'

otherwise

0.85⋅ f' c⋅ b

ε s = 0.035452 ρ :=

As b⋅d

ρmin :=

>

ρ = 0.004199 1.4 ⋅ MPa fy

fy Es

= 0.001950 ρ' :=

As' b⋅d

ρ' = 0.006933

ρmin = 0.003661

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Rasio penulangan maksimum yang diijinkan :

⎛ 0.85 ⋅ f'c ⋅ β1

ρb075 := 0.75 ⋅ ⎜ ⎜

fy

⎝

ρb075 = 0.024462 >

⋅

⎞⎟ ρ'⋅ fs' + 0.003 ⋅ Es + fy ⎟ fy ⎠ 0.003 ⋅ Es

ρ = 0.004199

(OK)

Momen kapasitas yang dapat dipikul oleh penampang dengan faktor overstrength 1.25 : Mkap :=

⎡ ( ⎣

) ⎛⎝

1.25 ⋅ ⎢ As ⋅ f y − As' ⋅ fs' ⋅ ⎜ d −

⎡ ⎣

⎛ ⎝

1.25 ⋅ ⎢0.85 ⋅ f'c ⋅ a ⋅ b ⋅ ⎜ d −

fy ⎤ ⎟ + As' ⋅ fs' ⋅ ( d − d')⎥ if εs > Es 2⎠ ⎦ a⎞

a ⎞⎤

⎟⎥ otherwise

2 ⎠⎦

Mkap = 1229002.19 ⋅ kgf ⋅ cm Hasil perhitungan ETABS lihat Gambar 15 pada bagian "Beam Capacities and Angles" baris Beam 4 kolom Capacity +veM = 1144047.22 kgf-cm. Perbedaaan dapat terjadi karena adanya perbedaan asumsi formasi tulangan As yang menyebabkan perbedaan jarak titik berat tulangan As dan juga akibat perbedaan pembulatan. ix )

Perhitungan kapasitas momen balok B22 akibat berbaliknya arah gaya lateral Properti penampang : Stirrups diameter : ϕstirrups := 1.0 ⋅ cm Concrete cover : dcover := 2.5 ⋅ cm Rebar diameter ϕbar := 2.004051380897 ⋅ cm ( diameter tulangan dibuat sedemikian supaya menghasilkan luas tulangan sesuai dengan hasil perhitungan ETABS, lihat Gambar 14 ) 1 2 2 As := 3 ⋅ ⋅ π ⋅ ( 2.004051380897cm) = 9.463 ⋅ cm 4 1 2 2 A1 := ⋅ π ⋅ ( 2.004051380897cm) = 3.154 ⋅ cm 4 b := 30 ⋅ cm

h := 50 ⋅ cm

1 ⎛ ⎞ d := h − ⎜ dcover + ϕstirrups + ⋅ ϕbar⎟ 2 ⎝ ⎠ d = 45.498 ⋅ cm 1 d' := dcover + ϕstirrups + ⋅ ϕbar = 4.502 ⋅ cm 2 Luas tulangan : 2

As := 9.463cm

( Lihat Gambar 14 )

2

As' := 5.731cm

Asumsikan tulangan tekan leleh : f's := fy

f's = 3900 ⋅

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kgf 2

cm

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Perhitungan "Joint Shear" pada ETABS

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Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ f's

a = 1.965 ⋅ cm

0.85 ⋅ f'c ⋅ b

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = −0.002843 Tegangan tulangan tekan : Es := 2000000 ⋅

kgf 2

cm

fy fy if ε's ≥ Es

f's :=

Es ⋅ ε's otherwise f's = −5686 ⋅

fs' :=

kgf 2

cm

0 ⋅ f's if f's < 0 f's otherwise

fs' = 0 ⋅

kgf 2

cm

Tinggi blok tegangan tekan : a :=

As ⋅ fy − As' ⋅ fs' 0.85 ⋅ f'c ⋅ b

a = 4.982 ⋅ cm

Regangan tulangan tekan : ε's :=

a − β1 ⋅ d' ⎞ ⎛ ⎜ 0.003 ⋅ ⎟ if a > 0 a ⎝ ⎠ 0 otherwise

ε's = 0.000696 Tegangan tulangan tekan : f's :=

fy fy if ε's ≥ Es Es ⋅ ε's otherwise

f's = 1391 ⋅

kgf 2

cm

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Perhitungan "Joint Shear" pada ETABS

fs' :=

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0 ⋅ f's if f's < 0 f's otherwise

fs' = 1391 ⋅

kgf 2

cm

Regangan tulangan tarik :

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d − ε s :=

⎛⎜ As ⋅ fy ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

As⋅ fy

if f's < 0

0.85⋅ f' c⋅ b

⎡ ⎢⎣

0.003 ⋅ ⎢β1 ⋅ d −

⎛⎜ As ⋅ fy − As' ⋅ fs' ⎞⎟⎤⎥ ⎜⎝ 0.85 ⋅ f'c ⋅ b ⎟⎠⎥⎦

otherwise

As⋅ fy− As'⋅ fs' 0.85⋅ f' c⋅ b

ε s = 0.026706 ρ :=

As b⋅d

ρmin :=

fy

>

Es

= 0.001950

ρ = 0.006933 1.4 ⋅ MPa fy

ρ' :=

As' b⋅d

ρ' = 0.004199

ρmin = 0.003661

Rasio penulangan maksimum yang diijinkan :

⎛ 0.85 ⋅ f'c ⋅ β1

ρb075 := 0.75 ⋅ ⎜ ⎜

fy

⎝

ρb075 = 0.025960 >

⋅

⎞⎟ ρ'⋅ fs' + 0.003 ⋅ Es + fy ⎟ fy ⎠ 0.003 ⋅ Es

ρ = 0.006933

(OK)

Momen kapasitas yang dapat dipikul oleh penampang dengan faktor overstrength 1.25 : Mkap :=

⎡ ( ⎣

)

⎛ ⎝

1.25 ⋅ ⎢ As ⋅ f y − As' ⋅ fs' ⋅ ⎜ d −

⎡ ⎣

⎛ ⎝

1.25 ⋅ ⎢0.85 ⋅ f'c ⋅ a ⋅ b ⋅ ⎜ d −

a⎞

⎤ ⎟ + As' ⋅ fs' ⋅ ( d − d')⎥ if εs > 2⎠ ⎦

fy Es

a ⎞⎤

⎟⎥ otherwise

2 ⎠⎦

Mkap = 1963957.53 ⋅ kgf ⋅ cm

Tabel 4

Ir. Nityananda Permadi Tjokrodimurti IP-Md

Output post processor ETABS untuk kapasitas balok

34/45

Perhitungan "Joint Shear" pada ETABS

www.itsindo.org

Hasil perhitungan ETABS lihat Gambar 15 pada bagian "Beam Capacities and Angles" baris Beam 4 kolom Capacity -veM = 1957606.044 kgf-cm. Perbedaaan dapat terjadi karena adanya perbedaan asumsi formasi tulangan As yang menyebabkan perbedaan jarak titik berat tulangan As dan juga akibat perbedaan pembulatan.

Ir. Nityananda Permadi Tjokrodimurti IP-Md

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