Momen rencana balok tengah - TUGAS STRUKTUR BETON BERTULANG

Dipilih momen yang terbesar diantara kombinasi pembebanan berikut:

Tabel 6.1Output momen SAP 2000 v.14 untuk perencanaan balok tengah

Jenis Balok

Gaya Geser

Momen Balok (kNm) Momen

Puntir

Tumpuan Lapangan

(kN) M (+) M (-) M (+) M (-) (kN)

Balok Atap 90,68 103,1429 -104,01 54,1019 -90,4173 -

300 X 400 COMB7 COMB3 COMB2 COMB6 COMB9

Balok Lantai 3 115,411 234,048 -250,123 163,616 -156,991 -

300 X 500 COMB8 COMB7 COMB6 COMB8 COMB7

Balok Lantai 2 114,901 232,835 -257,474 109,981 -128,504 -

300 X 500 MODAL MODAL MODAL MODAL MODAL

Balok Lantai 1 116,013 218,590 -237,357 151,861 -148,731 -

300 X 500 MODAL MODAL MODAL COMB 8 MODAL

Balok Lantai Dasar 42,217 95,794 -94,438 54,433 -52,637 -

300 X 500 COMB2 MODAL COMB 7 COMB1 COMB1

Balok Anak 7,351 17,3826 -19,6881 8,9442 -8,7435 -

200 X 300 COMB3 MODAL MODAL MODAL MODAL

Balok Tepi Atap 72,406 121,564 -121,995 63,0405 -63,4088 11,677

300 X 500 MODAL MODAL COMB2 COMB2 COMB1

6 COMB2

Balok Tepi Lantai 3 78,944 131,721 -131,536 67,655 -69,713 21,76

300 X 500 COMB2 COMB16 COMB2 COMB2 COMB4 COMB2

Balok Tepi Lantai 2 115,961 225,445 -229,582 121,167 -138,484 88,765

300 X 500 COMB2 COMB16 COMB4 COMB1 COMB4 COMB2

Balok Tepi Lantai 1 111,551 221,484 -203,663 159,182 -107,358 82,0502

300 X 500 COMB2 COMB16 COMB4 COMB1 COMB1

6 COMB2

Balok Tepi Lantai Dasar 63,075 152,415 -142,143 153,288 -83,089 39,764

300 X 500 COMB3 COMB12 COMB2 COMB2 COMB2 COMB7

Balok Borders 10,65 37,372 -37,176 18,735 -18,539 39,467

200 X 300 COMB16 COMB8 COMB8 COMB8 COMB8 COMB3

Momen rencana balok tengah:

Momen tumpuan Mu+ = 234,048 kNm Mu- = 257,474 kNm Momen lapangan

Mu+ = 163,616 kNm Mu- = 156,991 kNm Gaya geser

Vu = 116,013 kN

VI.1.3 Perencanaan tulangan lentur balok tengah lantai 1,2 dan 3 a. Perancangan tulangan lentur pada tumpuan, momen negatif (Mu-)

Gambar 6.2 Rencana tulangan lentur pada tumpuan, momen negatif (Mu-) Mu- = 257,474 kNm (Tabel 6.1)

n- = 257,474 / 0,8 = 321,843 kNm diambil:

d’ : 61,5 mm

d : 500 – 61,5 = 438,5 mm Rasio penulangan

ρmin = 1,4/fy = 1,4/370 = 0,00378 ρmax = 0,75 × ρb

ρ b=0,85× f c^'× β1

fy × 600

600+fy

¿0,85×31×0,83

370 × 600

600+370=0,0365 ρ max = 0,75 × 0,0365 = 0,0274

ρ_perlu=1

(

¹⁻

√

¹⁻²^{m R}^fy ⁿ

⁾

dengan :

m= fy

0,85× f c^'= 370

0,85×31=14,042 M^−¿_n

b× d²= 321,834

300×438,5²=5,579 R_n=¿

ρ_perlu= 1

14,042

(

¹⁻

√

¹⁻²^×^14,042³⁷⁰^×5,5794

⁾

^=0,017

ρperlu < ρmax, sehingga tidak diperlukan tulangan rangkap dipakai ρ = 0,017

As perlu = ρperlu × b × d

= 0,017 × 300 × 438,5

= 2255,100 mm²

dicoba tulangan 10D19 (As = 283,528 mm²) As ada = 10 × 283,528

= 2835,287 mm²

As perlu = 2255,100 mm² < As ada = 2835,287 mm² → OK Maka digunakan tulangan 10D19

Checking (i) d^'_act =

(

⁴⁰⁺¹²⁺¹⁹2

)

⁺⁴

(

⁴⁰⁺¹²⁺¹⁹⁺³¹⁺¹⁹2

)

⁺10^¿

¿ 40+10+19+31+19+31+19

2 2¿¿

d^'act = 500 – 61,266 = 438,733 mm > d awal = 438,5 mm (ii) ρ_act=¿ As ada

b x d^'act

= 2835,287

300x438,733 = 0,021 < ρmax = 0,027 (iii) Jarak tulangan:

¿300−(2×40)−(2×12)−(4×19)

(4−1) =40mm

spasi bersih antar tulangan = 40 mm > Smin = 25 mm

b. Perancangan tulangan lentur pada tumpuan, momen positif (Mu+)

Gambar 6.3 Rencana tulangan lentur pada tumpuan momen positif (Mu+) Mu+ = 234,048 kNm (Tabel 6.1)

Mn+ = 234,048 / 0,8 = 292,560 kNm diambil:

d’ : 61,5 mm d : 438,5 mm Rasio penulangan ρmin = 0,00378 ρmax = 0,0274

m=14,042 M_n^+¿

b× d²= 292,560

300×438,5²=5,072 R_n=¿

ρ_perlu= 1

14,042

(

¹⁻

√

¹⁻²^×^14,042³⁷⁰^×5,072

⁾

⁼^0,0154

ρperlu < ρmax, sehingga tidak diperlukan tulangan rangkap Dipakai ρ = 0,0154

As perlu = ρperlu × b × d

= 0,0154 x 300 × 438,5

= 2021,238 mm²

Dicoba tulangan 8D19 (As=283.5287 mm²) As ada = 8 × 283.5287

= 2268,230 mm²

As perlu = 2021,238 mm² < As ada = 2268,230 mm² → OK Maka digunakan tulangan 8D19

Checking (i) d^'_act =

(

⁴⁰⁺¹²⁺¹⁹2

)

⁺⁴

(

⁴⁰⁺^12+19+31+¹⁹2

)

8 =86,5mm

d_act = 500 – 86,5 = 413,5 mm (ii) ρ_act=¿ As ada

b . d_act = 2268,230

300×413,5 = 0,0183 < ρmax = 0,0274 (iii) Jarak tulangan:

¿300−(2×40)−(2×12)−(4×19)

(4−1) =40mm

spasi bersih antar tulangan = 40 mm > Smin = 25 mm c. Checking kapasitas momen tumpuan negatif (Mu-)

Gambar 6.4 Analisis tulangan rangkap pada tumpuan momen negatif (Mu-) As 10D19 = 2835,287 mm²

As’ 8D19 = 2268,230 mm²

d’ = 61,5 mm

d = 438,5 mm

Letak garis netral

(0,85 × f’c× β1 × b)c² + (600 × As’ – As × fy)c – (600 × d’ × As’) = 0

(0,85 × 31 × 0,83 × 300)c² + (600 × 2268,230 – 2835,287 × 370) c – (600 × 61,5 × 2268,230 ) = 0

6549,857c² + 311881,611 – 83697683,158 = 0 Digunakan nilai c = 91,71 mm

a = 75,99 mm Cek terhadap regangan :

ɛ

f_y

E = 370

200000 = 0,0019 Regangan yang terjadi pada tarik:

ɛ

^s1⁼ ^d−c_c ^{x 0,003 =} ^438,5−_91.71^91,71 x 0,003 = 0,0113

ɛ

^s1^>

ɛ

y tulangan tarik leleh.

Maka f s1 = fy = 370 MPa

ɛ

^s2⁼ ^d−c_c⁻⁵⁰ ^{x 0,003 =} 438,5−91,71−50

91,71 x 0,003 = 0,0097

ɛ

^s2^>

ɛ

y tulangan tarik leleh.

Maka f s2 = fy = 370 MPa

ɛ

^s3⁼ ^d−c−100_c x 0,003 = 438,5−91,71−100

91,71 x 0,003 = 0,0081

ɛ

^s3^>

ɛ

y tulangan tarik leleh.

Maka f s3 = fy = 370 MPa

Regangan yang terjadi pada tulangan tekan:

ɛ

^s4⁼ ^d−c−327_c ^{x 0,003 =} ^438,5−_91,71^91,71−327 x 0,003 = 0,00065

ɛ

s4 >

ɛ

y tulangan tekan belum leleh.

Maka f _s4=

ɛ

s4 × E = 0,00065 × 200000 = 129,442 MPa

ɛ

^’^s5⁼ ^c−d_c ^' x 0,003 = 91,71−61,5

91,71 x 0,003 = 0,00098

ɛ

^’^s5^>

ɛ

y tulangan tekan belum leleh.

Maka ^f s5 =

ɛ

’s5 × E = 0,00098 × 200000 = 197,662 MPa Ts1 = 4D19 × fy = 1134,1149 × 370 = 419622,5307 Nmm Ts2 = 4D19 × fy = 1134,1149 × 370 = 419622,5307 Nmm Ts3 = 2D19 × fy = 567,0574 × 370 = 209811,2654 Nmm

Ts4 = 4D19 × f s4 = 1134,1149 × 129,442 = 146802,0080 Nmm Cs5 = 4D19 × f s5= 1134,1149 × 197,662 = 224171,5242 Nmm

Cc = 0,85 × f’c × a × b = 0,85 × 31 × 75,99 × 300 = 600713,2785 Nmm Total = 2020743,138 Nmm

Mn1 = Ts1 × Z1

= 419622,5307 × (0)

= 0 Nmm Mn2 = Ts2 × Z2

= 419622,5307 × (50)

= 20981126,54 Nmm Mn3 = Ts3 × Z3

= 209811,2654 × (100)

= 20981126,54 Nmm Mn4 = Ts4 × Z4

= 146802,0080 × (438,5 - 61,5 - 50)

= 48004256,63 Nmm

Mn5 = Cs5 × Z5

= 224171,5242 × (438,5 - 61,5)

= 84512664,61 Nmm Mn6 = Cc × Z6

= 600713,2785 × (438,5 - ½ × 61,5)

= 240588203,19 Nmm

Mn = Mn1 +Mn2 + Mn3 + Mn4 + Mn5 + Mn6

= 0 + 20981126,54 + 20981126,54 + 48004256,63 + 84512664,61 + 240588203,19

= 415067377,499 Nmm = 415,067 kNm ØMnb = 0,8 × Mn

= 0,8 × 415,067

= 332,054 kNm

Hasil analisis pada tumpuan momen negatif dapat dilihat pada Tabel 6.2 berikut:

Tabel 6.2 Hasil analisis pada tumpuan momen negatif (Mu-)

Gaya (N) Jarak (mm) Momen (Nmm) Ts1 = 4D19. fy 419622,531 Z1= 0 0 Mn1 0

Ts2 = 4D19. fy 419622,531 Z2= 50 50 Mn2 20981126,54 Ts3 = 2D19. fy 209811,265 Z3= 100 100 Mn3 20981126,54 Ts4 = 4D19. f s4 146802,008 Z4= d-d'-50 327 Mn4 48004256,63 Cs5 = 4D19. f s5 -224171,524 Z5= d-d' 377 Mn5 -84512664,61 Cc = 0,85.f’c.a.b -600713,279 Z6= d-½.a 401 Mn6 -240588203,2

Pnb 2020743,138 - - Mnb 415067377,499

ØPnb 1616594,510 - - ØMnb 332053902

Kontrol hasil analisis pada tumpuan momen negatif:

ØMnb = 332,054 kNm > Mu- = 257,474 kNm→ OK

d. Checking kapasitas momen tumpuan positif (Mu+)

Gambar 6.5 Analisis tulangan rangkap pada tumpuan momen positif (Mu+) As’ 10D19 = 2835,287 mm²

As 8D19 = 2268,230 mm²

d’ = 61,5 mm

d = 438,5 mm

Letak garis netral

(0,85 × fc’× β1 × b)c² + (600 × As’ – As × fy)c – (600 × d × As’) = 0

(0,85 × 31 × 0,83 × 300)c² + (600 × 2835,287 – 2268,230 × 370) c – (600 × 438,5 × 2835,287) = 0

6549,857c² + 311881,611 – 104622103,948 = 0 didapat c = 104,80 mm

a = β1 × c = 0,85 × 104,80 ‘ = 86,83 mm

Cek terhadap regangan :

ɛ

f_y

E = 370

200000 = 0,0019 Regangan yang terjadi pada tarik:

ɛ

’s1 = c−d^'

c x 0,003 =

104,80−61,5

104,80 x 0,003 = 0,0012

ɛ

^’^s1^>

ɛ

y tulangan tarik belum leleh.

Maka f _s1=

ɛ

^’s1 × E = 0,0012 × 200000 = 247,900 MPa

ɛ

^s2⁼ ^d−c−327_c ^{x 0,003 =} 438,5−104,80−327

104,80 x 0,003 = 0,000192

ɛ

s2 >

ɛ

y tulangan tarik belum leleh.

Maka ^f s2 =

ɛ

s2 × E = 0,000192 × 200000 = 38,360 MPa

ɛ

s3 =

d−c−277

c x 0,003 = 438,5−104,80−277

104,80 x 0,003 = 0,0016

ɛ

s3 >

ɛ

y tulangan tarik belum leleh.

Maka f _s3=

ɛ

^s3 × E = 0,0016 × 200000 = 324,621 MPa Regangan yang terjadi pada tulangan tekan:

ɛ

^s4⁼ ^d−c_c⁻⁵⁰ ^{x 0,003 =} 438,5−104,80−50

104,80 x 0,003 = 0,0081

ɛ

^s4^>

ɛ

y tulangan tekan leleh.

Maka f s4 = fy = 370 MPa

ɛ

^s5⁼ ^d−c_c x 0,003 = 438,5−104,80

104,80 x 0,003 = 0,0096

ɛ

^s5^>

ɛ

y tulangan tekan leleh.

Maka f s5 = fy = 370 MPa

Cc = 0,85 × f’c × a × b = 0,85 × 31 × 86,83 × 300 = 686423,350 Nmm Cs1 = 4D19 × f s1 = 1134,1149 × 247,900 = 281146,985 Nmm Ts2 = 4D19 × f s2 = 1134,1149 × 38,360 = 43505,035 Nmm Ts3 = 2D19 × f s3 = 567,0574 × 324,621 = 184078,527 Nmm

Ts4 = 4D19 × fy = 1134,1149 × 370 = 419622,531 Nmm Ts5 = 4D19 × fy = 1134,1149 × 370 = 419622,531 Nmm Total = 2034398,958 Nmm

Mn1 = Cc × Z1

= 686423,350 × (438,5 - ½ × 86,83)

= 271194171,083 Nmm Mn2 = Cs1 × Z2

= 281146,985 × (438,5 - 61,5)

= 105992413,207 Nmm Mn3 = Ts2 × Z3

= 43505,035 × (438,5 - 61,5 - 50)

= 14226146,441 Nmm Mn4 = Ts3 × Z4

= 184078,527 × (438,5 - 61,5 - 100)

= 50989752,063 Nmm Mn5 = Ts4 × Z5

= 419622,531 × (50)

= 20981126,537 Nmm Mn6 = Ts5 × Z6

= 419622,531 × (0)

= 0 Nmm

Mn = Mn1 +Mn2 + Mn3 + Mn4 + Mn5 + Mn6

= 271194171,083 + 105992413,207 + 14226146,441 + 50989752,063 20981126,537 + 0

= 463383609,331 Nmm = 463,384 kNm ØMnb = 0,8 × Mn

= 0,8 × 463,384 = 370,707 kNm

Hasil analisis pada tumpuan momen positif dapat dilihat pada Tabel 6.3 berikut:

Tabel 6.3 Hasil analisis pada tumpuan momen positif (Mu+)

Gaya (N) Jarak (mm) Momen (Nmm)

Cc = 0,85.f’c.a.b -686423,350 Z1 = d - ½.a 395,083 Mn1 -271194171,083 Cs1 = 4D19 .

f s1

-281146,985

Z2 = d - d' 377 Mn2 -105992413,207 Ts2 = 4D19 .

f s2

43505,035

Z3 = d-d'-50 327 Mn3 14226146,441 Ts3 = 2D19 .

f s3

184078,527

Z4 = d-d'-100 277 Mn4 50989752,063 Ts4 = 4D19 . fy 419622,531 Z5 = 50 50 Mn5 20981126,537 Ts5 = 4D19 . fy 419622,531 Z6 = 0 0 Mn6 0

Pnb 2034398,958 - - Mnb 463383609,331

ØPnb 1627519,166 - - ØMnb 370706887,5

Kontrol hasil analisis pada tumpuan momen positif:

ØMnb = 370,707 kNm > Mu+ = 234,048 kNm → OK

e. Perancangan tulangan lentur pada lapangan, momen positif (Mu+)

Gambar 6.6 Rencana tulangan lentur pada lapangan momen positif (Mu+) Mu+ = 163,616 kNm (Tabel 6.1)

Mn+ = 163,616 / 0,8 = 204,520 kNm diambil:

d’ : 61,5 mm

d : 500 – 61,5 = 438,5 mm Rasio penulangan

ρmin = 0,00378

ρ b=0,0365 ρmax 0,0274 dengan:

m=14,0417 M_n^+¿

b× d²= 204,520

300×438,5²=3,5455 R_n=¿

ρ_perlu= 1

14,0417

(

¹⁻

√

¹⁻²^×^14,0417³⁷⁰^×^3,5455

⁾

^=0,0103

ρperlu < ρmax, sehingga tidak diperlukan tulangan rangkap Dipakai ρ = 0,0103

As perlu = ρperlu × b × d

= 0,0103 x 300 × 438,5

= 1359,154 mm²

Dicoba tulangan 8D19 (As = 283,5287 mm²) As ada = 8 × 283,5287

= 2268,230 mm²

As perlu = 1359,154 mm² < As ada = 2268,230 mm² → OK Maka digunakan tulangan 8D19

Checking (i) d^'act =

(

⁴⁰⁺¹²⁺¹⁹2

)

⁺⁴

(

⁴⁰⁺^12+19+31+¹⁹2

)

8 =86,5mm

d_act = 500 – 86,5 = 413,5 mm (ii) ρ_act=¿ As ada

b . d_act = 2268,230

300.413,5 = 0,01828 < ρmax = 0,0274 (iii) Jarak tulangan:

¿300−(2×40)−(2×12)−(4×19)

(4−1) =40mm

spasi bersih antar tulangan = 40 mm > Smin = 25 mm

f. Perancangan tulangan lentur pada lapangan, momen negatif (Mu-)

Gambar 6.7 Rencana tulangan lentur pada lapangan momen negatif (Mu-) Mu- = 156,9909 kNm (Tabel 6.1)

Mn- = 156,9909 / 0,8 = 196,238 kNm diambil:

d’ : 61,5 mm d : 438,5 mm Rasio penulangan ρmin = 0,00378

ρ b=¿ 0,0365ρ max = 0,0274 m=14,042

M^−¿_n

b× d²= 196,238

300×438,5²=3,402 R_n=¿

ρ_{per lu}= 1

14.042

(

¹⁻

√

¹⁻^2×^14.042³⁷⁰^×3,402

⁾

^=0,0099

ρperlu < ρmax, sehingga tidak diperlukan tulangan rangkap dipakai ρ = 0,0099

As perlu = ρperlu × b × d

= 0,0099× 300 × 438,5

= 1299,671 mm²

dicoba tulangan 6D19 (As = 283.5287 mm²) As ada = 6 × 283,5287

= 1701,172 mm²

As perlu = 1299,671 mm² < As ada = 1701,172 mm²→ OK Maka digunakan tulangan 6D19

Checking

(i) d^'_act=¿ 4

(

⁴⁰⁺¹²⁺¹⁹2

)

⁺²

(

^40+12+19+³¹⁺¹⁹2

)

¿74,167mm

d_act = 500 – 74,167

= 425,833 mm (ii) ^ρact=¿ As ada

b . d^'_act

¿ 1701,172 300 .444,375

= 0,01276 < ρmax = 0,0274 (iii) Jarak tulangan:

¿300−(2×40)−(2×12)−(4×19) (4−1)

¿40mm

spasi bersih antar tulangan = 40 mm > Smin = 25 mm

g. Checking kapasitas momen lapangan momen positif

Gambar 6.8 Analisis tulangan rangkap pada lapangan momen positif (Mu+) As’ 6D19 = 1701,172 mm²

As 8D19 = 2268,230 mm²

d’ = 61,5 mm

d = 438,5 mm

Letak garis netral

(0,85 × fc’× β1 × b)c² + (600 × As’ – As × fy)c – (600 × d’ × As’) = 0 (0,85 × 31 × 0,83 × 300)c² + (600 × 1701,172 – 2268,230 × 370) c – (600 × 61,5 × 1701,172) = 0

6549,857c² + 181558,392 c – 62773262,369 = 0 didapat c = 85,02 mm

a = β1 × c = 0,83 × 85,02 = 70,45 mm Cek terhadap regangan:

ɛ

^y⁼ ^fy_E ⁼ ₂₀₀₀₀₀³⁷⁰ ^{= 0,0019}

Regangan yang terjadi pada tulangan tarik:

ɛ

^’^s1⁼ ^c−d_c ^' x 0,003 = 85,02−61,5

85,02 x 0,003 = 0,00083

ɛ

^’s1 >

ɛ

ymaka tulangan tarik belum leleh

Maka f _s1=

ɛ

’s1 × E = 0,00083 × 200000 = 165,987 MPa

ɛ

s2 = d−c−327

c x 0,003= 438,5−85,02−327

85,02 x 0,003 = 0,00093

ɛ

s2 >

ɛ

ymaka tulangan tarik belum leleh

Maka f _s2=

ɛ

^s2 × E = 0,00093 × 200000 = 186,869 MPa Regangan yang terjadi pada tulangan tekan:

ɛ

s3 = d−c−50

c x 0,003 =

438,5−85,02−50

85,02 x 0,003 = 0,0107

ɛ

s3 >

ɛ

ymaka tulangan tekan leleh Maka f s3 = fy = 370 MPa

ɛ

s4 = d−c

c x 0,003 =

438,5−85,02

85,02 x 0,003 = 0,0125

ɛ

^s4^>

ɛ

^y^maka tulangan tekan leleh Maka f s4 = fy = 370 MPa

Cc = 0,85 × f’c × a × b = 0,85 × 31 × 70,45 × 300 = 556872,293 Nmm Cs1 = 4D19 × f s1 = 1134,1149 × 165,987 = 188248,513 Nmm Ts2 = 2D19 × f s2 = 567,0574 × 186,869 = 10596,360 Nmm Ts3 = 4D19 × fy = 1134,1149 × 370 = 419622,531 Nmm Ts4 = 4D19 × fy = 1134,1149 × 370 = 419622,531 Nmm Total = 1133458,934 Nmm

Mn1 = Cc × Z1

= 556872,293 × (438,5 - ½ × 70,45)

= 224573905,112 Nmm Mn2 = Cs1 × Z2

= 188248,513 × (438,5 - 61,5)

= 70969689,234 Nmm

Mn3 = Ts2 × Z3

= 10596,360 × (438,5 - 61,5 - 50)

= 34650672,742 Nmm Mn4 = Ts3 × Z4

= 419622,531 × (50)

= 20981126,537 Nmm Mn5 = Ts4 × Z5

= 419622,531 × (0)

= 0 Nmm

Mn = Mn1 + Mn2+ Mn3+ Mn4+ Mn5

= 224573905,11 + 70969689,23 + 34650672,74 + 20981126,52 + 0

= 351175393,625 Nmm = 351,175 kNm ØMnb = 0,8 × Mn

= 0,8 × 351,175 = 280,940 kNm

Hasil analisis pada lapangan momen positif dapat dilihat pada Tabel 6.4 berikut:

Tabel 6.4 Hasil analisis pada lapangan momen positif (Mu+)

Gaya (N) Jarak (mm) Momen (Nmm)

Cc = 0,85 . fc’. a.b -556872,293 Z1 = .(d- ½ a) 403,277 Mn1 -224573905,112 Cs1 = 4D19 . f s1 -188248,513 Z2 = (d - d’) 377 Mn2 -70969689,234 Ts2 = 2D19 . f s2 105965,360 Z3 = (d - d’- 50) 327 Mn3 34650672,742 Ts3 = 4D19 . fy 105965,360 Z4 = 50 50 Mn4 20981126,537 Ts4 = 4D19 . fy 419622,531 Z5 = 0 0 Mn5 0

Pnb 1133458,934 - - Mnb 351175393.,625

ØPnb 906767,147 - - ØMnb 280940314,9

Kontrol hasil analisis pada lapangan momen positif:

ØMnb = 280,940 kNm > Mu+ = 163,616 kNm → OK

h. Checking kapasitas momen lapangan negatif

Gambar 6.9 Analisis tulangan rangkap pada lapangan momen negatif (Mu-) As 6D19 = 1701,172 mm²

As’ 8D19 = 2268,230 mm²

d’ = 61,5 mm

d = 438,5 mm

Letak garis netral

(0,85 × fc’× β1 × b)c² + (600 × As’ – As × fy)c – (600 × d’ × As’) = 0

(0,85 × 31 × 0,83 × 300)c² + (600 × 2268,230 – 1701,172 × 370) c – (600 × 61,5 × 2268,230) = 0

6549,857c² + 731504,141 – 83697683,158 = 0 didapat c = 70,24 mm

a = β1 × c = 0,83 × 70,24 = 58,20 mm

Cek terhadap regangan:

ɛ

^y⁼ ^fy_E ⁼ ₂₀₀₀₀₀³⁷⁰ ^{= 0,0019}

Regangan yang terjadi pada tulangan tarik:

ɛ

^s1⁼ ^d−c_c x 0,003 = 438,5−70,24

70,24 x 0,003 = 0,0157

ɛ

^s1^>

ɛ

^y^maka tulangan tarik leleh.

Maka f s1 = fy = 370 MPa

ɛ

^s2⁼ ^d−c−50_c ^{x 0,003=} 438,5−70,24−50

70,24 x 0,003 = 0,0136

ɛ

^s2^>

ɛ

^y^maka tulangan tarik leleh.

Maka f s2 = fy = 370 MPa

Regangan yang terjadi pada tulangan tekan:

ɛ

s3 =

d−c−327

c x 0,003 =

438,5−70,24−327

70,24 x 0,003 = 0,00176

ɛ

s3 >

ɛ

ymaka tulangan tekan belum leleh.

Maka f _s3=

ɛ

^s3 × E = 0,00176 × 200000 = 352,432 MPa

ɛ

^s4⁼ ^c−d_c ^' ^{x 0,003 =} ^70,24−61,5_70,24 x 0,003 = 0,000373

ɛ

^s4^>

ɛ

^y^maka tulangan tekan belum leleh.

Maka f _s4=

ɛ

s4 × E = 0,000373 × 200000 = 74,668 MPa Ts1 = 4D19 × fy = 1134,1149 × 370 = 419622,5307 Nmm Ts2 = 2D19 × fy = 567,0574 × 370 = 209811,2654 Nmm

Ts3 = 4D19 × f s3 = 1134,1149 × 352,432 = 399698,0278 Nmm Cs4 = 4D19 × f s4 = 1134,1149 × 74,668 = 84681,7913 Nmm

Cc = 0,85 × f’c × a × b = 0,85 × 31 × 86,83 × 300 = 460070,2135 Nmm Total = 1573883,829 Nmm

Mn1 = Ts1 × Z1

= 419622,5307 × (0)

= 0 Nmm

Mn2 = Ts2 × Z2

= 209811,2654 × (50)

= 10490563,268 Nmm Mn3 = Ts3 × Z3

= 399698,0278 × (438,5 - 61,5 - 50)

= 130701255,082 Nmm Mn4 = Cs4 × Z4

= 84681,7913 × (438,5 - 61,5)

= 31925035,322 Nmm Mn5 = Cc × Z5

= 460070,2135 × (438,5 - ½ × 58,20 )

= 188352768,295 Nmm Mn = Mn1 + Mn2+ Mn3+ Mn4+ Mn5

= 0 + 10490563,3 + 130701255,1 + 31925035,3 + 188352768,29

= 361469621,967 Nmm = 361,470 kNm ØMnb = 0,8 × Mn

= 0,8 × 361,470 = 289,176 kNm

Hasil analisis pada lapangan momen negatif dapat dilihat pada Tabel 6.5 berikut:

Tabel 6.5 Hasil analisis pada lapangan momen negatif (Mu-)

Gaya (N) Jarak (mm) Momen (Nmm)

Ts1 = (4D19).fy 419622,531 Z1 = 0 0 Mn1 0

Ts2 = (2D19).fy 209811,265 Z2 = 50 50 Mn2 10490563,268 Ts3 = (4D19). f

399698,028 Z3 = d - d' - 50 327 Mn3 130701255,082 Cs4 = (4D19). f

-84681,791 Z4 = d – d’ 377 Mn4 -31925035,322 Cc = 0,85.f’c.a.b -460070,214 Z5 = d - ½.a 409,400 Mn5 -188352768,295

Pnb 1573883,829 - - Mnb 361469621,967

ØPnb 1259107,063 - - ØMnb 289175697,6

Kontrol hasil analisis pada lapangan momen negatif:

ØMnb = 289,176 kNm > Mu- = 156,991 kNm→ OK

Dalam dokumen TUGAS STRUKTUR BETON BERTULANG (Halaman 114-135)