I. Balok Melintang B2 25/35 (E-134) pada Lantai 2

qu = qu (beban kombinasi plat lantai 2 terbesar)

= 1172.8 kg/m2

q segitiga a = (1/3 x qu x Lx) x 2

= (1/3 * 1172,8 * 3)*2

= 2345.6 kg/m

q dinding = 0.15*1.0*1700*4

= 1020 kg/m

q balok = b*h*2400

= 0.25*0.35*2400

= 210 kg/m

q1 total = q segitiga + q dinding + q balok

= 3575.6 kg/m

q2 total = q segitiga + q balok

= 2555.6 kg/m

P1 = Akibat balok B1 + akibat plat arah memanjang

= b*h*L*2400 + q trapesium * n * L

= 1890 + 2775.75

= 4665.75 kg

P2 = Akibat balok B1 + akibat plat arah memanjang

= b*h*L*2400 + q trapesium * n * L

= 1890 + 5551.5

= 7441.5 kg

P3 = Akibat balok B1 + akibat plat arah memanjang

= b*h*L*2400 + q trapesium * n * L

= 1890 + 2775.75

= 4665.75 kg

input beban diSAP

Output hasil dari SAP, Vu maks = 5415.11 kg

Output hasil dari SAP, Momen maksimum.

Mu tumpuan = 2707.55 kgm

= 27075500 Nmm

Mu lapangan = 1353.78 kgm

= 13537800 Nmm

Perencanaan Penulangan Balok (B2 25/35)

B balok = 250 mm

H balok = 350 mm fy = 360 Mpa

Selimut = 40 mm f'c = 25 Mpa

Asumsi tul. Utama = 16 mm

Asumsi tul. Sengkang = 12 mm

Tinggi efektif (d) = H - selimut - tul. Sengkang - 1/2 * tul. Utama

= 290 mm

A. Tulangan Tumpuan :

Rn = Mu tumpuan

* b * d2

= 1.610

m = fy

0.85 * f'c

= 16.94

perlu =

= 0.00465

max = 0.75 * balance=

= 0.0235

min = 1.4 / fy= 0.00389

dipakai = 0.00465 min < perlu < max kondisi balance

As = * b * d

= 407.31 mm2

Digunakan tulangan 3D16 (As pasang = 602.88 mm2

)

Cek Perlu tulangan tekan (As') atau tidak

a = As * fy =

0.85*fc'*b

= 40.85 mm

Mn = As * fy * (d-a/2)

= 602.88*360*(290-40.85/2)

= 58507263 Nmm

Mn perlu = = 27075500

0.8

= 33844375 Nmm

Karena, Mn Penampang > Mn Perlu, 58507263 Nmm > 33844375 Nmm maka tidak perlu

tulangan tekan. Tetapi tetap perlu diadakan tulangan tekan dengan luasan 20% dari As

602.88 * 360

0.85 * 25 * 250

Mu(SAP2000)

As' = 20% * As

= 81.46 mm2

Digunakan tulangan 2D13 (As pasang = 132.665 mm2

)

B. Tulangan Lapangan :

Rn = Mu lapangan

* b * d2

= 0.805

m = fy

0.85 * f'c

= 16.94

perlu =

= 0.00228

max = 0.75 * balance=

= 0.0235

min = 1.4 / fy= 0.00389

dipakai = 0.00389 min < perlu < max kondisi balance

As = * b * d

= 340.28 mm2

Digunakan tulangan 2D16 (As pasang = 401.92 mm2

)

Cek Perlu tulangan tekan (As') atau tidak

a = As * fy =

0.85*fc'*b

= 8.99 mm

Mn = As * fy * (d-a/2)

= 602.88*360*(290-40.85/2)

= 13635547 Nmm

Mn perlu = = 27075500

0.8

= 33844375 Nmm

Karena, Mn Penampang > Mn Perlu, 58507263 Nmm > 33844375 Nmm maka tidak perlu

tulangan tekan. Tetapi tetap perlu diadakan tulangan tekan dengan luasan 20% dari As

As' = 20% * As

= 68.06 mm2

Digunakan tulangan 2D13 (As pasang = 132.665 mm2

)

C. Tulangan Geser

Mu(SAP2000)

602.88 * 360

0.85 * 25 * 250

Dari hasil perhitungan SAP 2000,

Vu maks = 5415.11 kg

= 54.15 kN

Vc = 1/6 * * fc' * b * d

= 1/6 * 0.85 * 25 *250 * 290

= 51354.1667 N

= 51.35 kN 2875

Vc = 0.6 * 60.42

= 30.8125 kN

Cek kondisi penampang geser pada balok :

Vu = 51.35 kN > Vc = 36.25 kN Maka perlu tulangan geser.

Spasi tulangan geser

S max = d/2

= 290/2 = 145 mm

diambil jarak tulangan geser = 140 mm

Av min = 0.062 * fc' * b * S / fy

= 30.14 mm2

Dipakai 8 - 140 ( 50.24 mm2)

Vs = Av * fy * d

s

= 37464.6857 N