Beton 2 Euy...Ngulang Boww..

Diketahui :

Diketahui : Fungsi bangunan: Perpustakaan (AC)

Letak bangunan: Biak

fc = 22.5 Mpa

fy = 300 Mpa ( BJTP

fy = 400 Mpa ( BJTD

DENAH BANGUNAN (Balok anak arah y)

No. mahasiswa : 0 3 5 1 1 0 2 3

x x x x x x A B

L1 = 7.A =7.2 m

L2 = 5.B = 5.3 m

POTONGAN PORTAL 2

PEMBEBANAN PLAT

a) Plat Atap

Beban Mati (Qd)

Lapisan kedap air0.03 x 2400 kg/m3= 72 kg/m2Plat atap

0.1 x 2400 kg/m3= 240 kg/m2Pengantung

= 7 kg/m2Plafon

= 11 kg/m2Ducting AC

= 20 kg/m2

Qd= 350 kg/m2

= 0.35 t/m2 Beban Hidup (Ql)Ql plat atap

= 100kg/m2= 0.1 t/m2

b) Plat Lantai

Beban Mati (Qd)

Keramik

0.01 x 1500 kg/m3= 15 kg/m2Spesi

0.03 x 2000 kg/m3= 60 kg/m2Pasir

0.05 x 1800 kg/m3= 90 kg/m2Plat

0.12 x 2400 kg/m3= 288 kg/m2Penggantung

= 7 kg/m2Plafon

= 11 kg/m2Ducting AC

= 20 kg/m2

Qd= 491 kg/m2

= 0.491 t/m2 Beban Hidup (Ql)

Ql plat lantai

400 kg/m2= 0.4 t/m2PEMBEBANAN PORTAL

BEBAN MERATAa) Balok Atap

Untuk Lx = 7.2 m

P1 = (0.5 x 3.6 x 1.8) x 0.35= 1.134 t

P2 = (3.6 x 1.8) x 0.35 = 2.268 t

RA= 3.402 t

MT = (RA x 3.6) + (P1 x 2.4) + (P2 x 0.9)

= (3.402 x 3.6) + (1.134 x 2.4) + (2.268 x 0.9)

= 7.484 tm

MT= . Qeq . L27.484= x Qeq x 7.22Qeq= 1.155 t/m

Berat sendiri balok (40/80)

Qb = (0.4 x 0.7 x 1) x 2.4 t/m3= 0.672 t/m

Total beban terbagi rata :

Qtotal= Qeq + Qb

=1.155 + 0.672

= 1.827 t/m

Ql= x 1.155 = 0.33 t/m

Untuk Lx = 5.3 m

P1 = (0.5 x 3.6 x 1.8) x 0.35= 1.134 t

P2 = (0.85 x 3.6) x 0.35 = 1.071 t

RA= 2.205 t

MT = (RA x 2.65) + (P1 x 1.45) + (P2 x 0.425)

= (2.205 x 2.65) + (1.134 x 1.45) + (1.071 x 0.425)

= 3.744 tm



Qb = (0.4 x 0.7 x 1) x 2.4 t/m3= 0.672 t/m


Qtotal = Qeq + Qb

=1.066 + 0.672

= 1.738 t/m

Ql= x 1.066 = 0.3 t/m

b) Balok Lantai (Lantai 1 4)

Untuk Lx = 7.2 m

P1 = (0.5 x 3.6 x 1.8) x 0.491= 1.590 t

P2 = (3.6 x 1.8) x 0.491 = 3.181 t

RA= 4.771 t

MT = (RA x 3.6) + (P1 x 2.4) + (P2 x 0.9)

= (4.771 x 3.6) + (1.590 x 2.4) + (3.181 x 0.9)

= 10.497 tm



Qb = (0.4 x 0.68 x 1) x 2.4 t/m3= 0.65 t/m

Berat sendiri tembok per m balok

Qt= (0.15 x (4-0.8) x 1.8 t/m3 = 0.864 t/m


Qtotal = Qeq + Qb + Qt

=1.62 + 0.65 + 0.864

= 3.134 t/m

Ql= x 1.62 = 1.32 t/m

Untuk Lx = 5.3 m

P1 = (0.5 x 3.6 x 1.8) x 0.491= 1.590 t

P2 = (1.8 x 3.6) x 0.491 = 1.502 t

RA= 3.092 t

MT = (RA x 2.65) + (P1 x 1.45) + (P2 x 0.425)

= (3.092 x 2.65) + (1.590 x 1.45) + (1.502 x 0.425)

= 5.25 tm



Qb = (0.4 x 0.68 x 1) x 2.4 t/m3= 0.65 t/m

Berat sendiri tembok per m balok

Qt= (0.15 x (4-0.8) x 1.8 t/m3 = 0.864 t/m


Qtotal = Qeq + Qb + Qt

=1.495+ 0.65 + 0.864

= 3.009 t/m

Ql= x 1.495 = 1.218 t/m

BEBAN TITIK

a) Atap

- Plat

= 2 x (((3.6 + 1.8) x 0.5 x 1.8) + (1.8 x 1.8) x 0.35 t/m2 = 5.67 t

- Balok anak= 2 x ((0.25 x 0.5) x 3.6 x 2.4 t/m3

= 2.16 t

PD1 = 7.83 t

PL1= 2 x (((3.6 + 1.8) x 0.5 x 1.8) + (1.8 x 1.8)) x 0.1 t/m2 = 1.62 t- Plat = 5.67 + 2 x (((3.6 + 1.8) x 0.5 x 1.8) + ( 0.85 x 1.8)) x 0.35 =10.143 t- B. anak = 2.16 + 2 x ((0.25 x 0.5) x 2.65 x 2.4 t/m2

= 3.75 t

PD2 =13.893 tPL2= 1.62 + 2 x (((3.6 + 1.8) x 0.5 x 1.8) + (0.85 x 1.8) x 0.1 = 2.898 t

b) Lantai 1 4

- Plat

= 2 x (((3.6 + 1.8) x 0.5 x 1.8) + (1.8 x 1.8) x 0.491 = 7.954 t

- Balok anak= 2 x ((0.25 x 0.5) x 3.6 x 2.4 t/m3

= 2.16 t

PD1 = 10.114 t

PL1= 2 x (((3.6 + 1.8) x 0.5 x 1.8) + (1.8 x 1.8)) x 0.4 t/m2 = 6.48 t

- Plat = 7.954 + 2x(((3.6 + 1.8) x 0.5 x 1.8) + (0.85 x 1.8) x 0.491= 14.229 t

- B. anak = 2.16 + 2 x ((0.25 x 0.5) x 2.65 x 2.4 t/m2

= 3.75 t

PD2 = 17.979 tPL2= 6.48 + 2(((3.6 + 1.8) x 0.5 x 1.8) + (0.85 x 1.8)) x 0.4 = 11.592 t BEBAN GEMPA

a) Berat Str. Atap (Watap) Beban mati (Wd)

- plat

= (7.2 + 5.3 + 7.2) x 7.2 x 0.35 t/m3

= 49.644 t

- balok induk= 0.4 x 0.8 x (7.2 + 5.3 + 7.2 + (4 x 7.2)) x 2.4 =37.248 t

- balok anak= 0.25 x 0.5 x (7.2 + 5.3 + 7.2) x 2.4 t/m3 = 5.91 t

- kolom= 0.5 x 0.5 x 2 x 2.4 x 4

= 4.8 t

Wd = 97.602 t Beban hidup (Wl)= (7.2 + 5.3 + 7.2) x 7.2 x 0.1 x 0.5 = 7.092 t

Watap = Wd + Wl = 97.602 + 7.092 = 104.694 t

b) Berat Str. Lantai 2 4 (W2-4) Beban mati (Wd)

- plat

= (7.2 + 5.3 + 7.2) x 7.2 x 0.491 t/m2

= 69.643 t

- balok induk= 0.4 x 0.8 x (7.2 + 5.3 + 7.2 + (4 x 7.2)) x 2.4 =37.248 t- balok anak = 0.25 x 0.5 x (7.2 + 5.3 + 7.2 ) x 2.4 t/m3 = 5.91 t

- kolom= 0.5 x 0.5 x 4 x 2.4 x 4

= 4.8 t

- tembok= 0.15 x (4 0.8) x 1.8 x 19.7

= 17.021 t

Wd= 134.622 t Beban hidup (Wl)= (7.2 + 5.3 + 7.2) x 7.2 x 0.4 x 0.5= 28.368 t

W2-4= Wd + Wl = 134.622 + 28.368 = 162.99 tc) Berat Str. Lantai 1 (W1) Beban mati (Wd)

- plat

= (7.2 + 5.3 + 7.2) x 7.2 x 0.491 t/m2

= 69.643 t

- balok induk= 0.4 x 0.8 x (7.2 + 5.3 + 7.2)) x 2.4 t/m3= 37.248 t

- balok anak = 0.25 x 0.5 x (7.2 + 5.3 + 7.2) x 2.4 t/m3= 5.91 t

- kolom= 0.5 x 0.5 x 7 x 2.4 x 4

= 16.8 t

- tembok= 0.15 (4 0.8) x 1.8 x 19.7

= 17.021 t

Wd= 146.622 t

Beban hidup (Wl)= (7.2 + 5.3 + 7.2) x 7.2 x 0.4 x 0.5= 28.368 t

W1= Wd + Wl = 146.622 + 28.368 = 174.99 t

Wtotal= Watap + W4 + W3 + W2 + W1

= 104.694 + 162.99 + 162.99 + 162.99 + 174.99

= 768.654 t

T = 0.06 H

= 0.06 x (21)

= 0.567

letak bangunan: Biak

Wilayah gempa: 4

Jenis tanah : Tanah sedang

Tc= 0.6 detik

Am= 0.70

Ar= 0.42

C= = = 0.7

V= x Wtotal ; I = 1.0

R = 8.5

V= x 768.654

= 63.3 t

F= x VLantaih (m)Wt (t)Wt x h (tm)V (t)F (t)

Atap21104.6942198.57463.314.76

417162.992770.83063.318.60

313162.992118.87063.314.22

29162.991466.91063.39.85

15174.99874.95063.35.87

768.6549430.134

REDISTRIBUSI MOMENT Lantai 1

Bentang kiri = bentang kanan

M = 610.23 + 679.92 + 318.22 + 160.35 = 1768.72 KNm

Mu- = 580 KNm

Mu+ = = 304.36 KNm > 50% x Mu- Ok!

Bentang tengah

M = 537.17 + 537.17 + 203.13 + 203.13 = 1480.6 KNm

Mu- = 490 KNm

Mu+ = = 250.3 KNm > 50% x Mu- Ok!

Lantai 2 4


M = 520.01 + 610.57 + 171.04 + 109.59 = 1411.21 KNm

Mu- = 470 KNm

Mu+ = = 235.605 KNm > 50% x Mu- Ok!

Bentang tengah

M = 477.62 + 477.62 + 172.72 + 172.72 = 1300.68 KNm

Mu- = 430 KNm

Mu+ = = 220.34 KNm > 50% x Mu- Ok!

Lantai atap


M = 137.15 + 186.47 = 323.62 KNm

Mu- = 110 KNm

Mu+ = = 3.69 KNm > 50% x Mu-

Dipakai Mu+ = 55 KNm Bentang tengah

M = 2 x 129.84 = 259.68 KNm

Mu- = 100 KNm

Mu+ = = 1.53 KNm > 50% x Mu-

Dipakai Mu+ = 50 KNmPENULANGAN LENTUR BALOKLantai 1 (bentang kiri = bentang kanan)

Diketahui :

Mu-= 580 KNm = 59.16 tm

Mu+= 304.36 KNm = 31.045 tm

D= 25 mm

Ad= x x (2.5)2fc= 22.5 Mpa= 229.5 Kg/cm2fy= 400 Mpa= 4080 Kg/cm2Es= 2100000 Kg/cm2c= 0.003= 0.85

d= 4 + 1 + 2.5 + (0.5 x 2.5) = 8.75 cm

d= 4 + 1 + (0.5 x 2.5) = 6.25 cm

y= = b = 0.001943

Perhitungan Dimensi Balok

m= = = 20.915

b= x = x = 0.025

m= 0.75 b = 0.75 x 0.025 = 0.01875

Rb= b x fy x (1 - (0.5 x b x m))

= 0.025 x 4080 x (1 (0.5 x 0.025 x 20.915))

= 74.67 Kg/cm2

Rm= 0.75 x Rb = 0.75 x 74.67 = 56 Kg/cm

Mn = Rm x b x h2

; h = 2b

= 56 x b x b2

7020000 = 224 x b3

b = = 32.078 cm

dipakai :

b = 30 cm

h = 61.25 cm

ht = 61.25 + 8.75 = 70 cm

Komponen Tulangan SebelahR1 = 0.3 Rb = 0.3 x 74.67 = 22.4037 Kg/cm2M1 = R1 x b x h2 = 22.4037 x 30 x (61.25)2 = 2521464 Kg cmM1 = 0.85 fc .a .b .(h a/2)

2521464 = 0.85 x 229.5 x a x 30 x (61.25 a/2)

2521464 = 258450.3125 a 2926.125 a2a2 -122.5a +861.7 = 0

a = = 7.49 cm

c = a/ = 7.49/0.85 = 8.81 cm

s = x c = x 0.003 = 0.000873 < 0.003

Baja desak belum lelehCc = 0.85 x 229.5 x 7.49 x 30 = 43848.74 KgTs1 = Cc = As1 x fy

As1 = = = 10.747 cm2n1 = = = 2.19 dipakai 3 buah

As1 = 3 x 4.906 = 14.718 cm2Ts1 = As1 x fy = 14.718 x 4080 = 60049.44 Kg

Ts1 = Cc = 0.85 fc .a .b a = = = 10.26 cm

c = a/ = 10.26/0.85 = 12.07 cm

s = x c = x 0.003 = 0.001447 < 0.003

Sekali lagi baja desak belum leleh

M1 = Cc.(h-(a/2)) = 0.85 x 229.5 x 10.26 x 30 x (61.25(10.26/2)) = 3370103 Kg cm

Komponen Tulangan Rangkap

M2 = Mn M1 = (73.95 x 105) (33.70 x 105) = 40.25 x 105 Kg cm

Ts2 = Cs = = = 76664.7 KgTs2 = As2 x fy

As2 = = = 18.79 cm2n2 = = = 3.83 dipakai 4 buah

Sehingga :

Kontrol jarak antar tulangan :

S = 0.85 fc .a .b + As = = 3.33 cm > 2.5 cm Ok!

Analisis Balok Tulangan Rangkap dengan Baja Desak Belum Leleh

Keseimbangan gaya-gaya horizontalTs1 + Ts2 = Cc + Cs

(Ast) fy = 0.85 fc .a .b + As .fs

(Ast) fy = 0.85 fc .a .b + As .s .Es

(Ast) fy = 0.85 fc .a .b + As x x c x Es(7x4.906)x4080 = (0.85x229.5xax 30) + (4x4.906) x x0.003x2100000

140115.36 = 5852.25 a +

140115.36 a = 5852.25 a2 +123631.2 a 656790.75

a2 2.82 a 112.234 = 0

a = = 12.09 cmc = a/ = 12.09/0.85 = 14.23 cm

s = x c = x 0.003 = 0.00168 < 0.001943

maka betul Baja desak belum leleh

fs = s .Es = 0.00168 x 2100000 = 3533 Kg/cm2Moment nominal yang dapat dikerahkan :

M1 = 0.85 x 229.5 x 12.09 x 30 x ( 61.25 (12.09/2)) = 3907604 Kg cm

M2 = (4 x 4.906) x 3533 x (61.25 6.25)

= 3640104 Kg cm

Moment nominal, Mn = 7547708 Kg cm

= 75.477 tmMt = x Mn = 0.8 x 75.477 = 60.38 tm > 59.16 tm

Desain tulangan moment negatif sukses.

Kontrol Kuat Lentur Moment Positif

Keseimbangan gaya-gaya horizontal :Ts = Cc + Cs

As .fy = 0.85 fc .a .b + As x x c x Es

(4x4.906)x4080 = (0.85x229.5xax 30) + (7x4.906) x x0.003x2100000

80065.92 = 5852.25 a +

80065.92 a = 5852.25 a2 +216354 a 16091.37

a2 23.28 a 288.65 = 0

a = = 8.95 cm ; c = a/ = 8.95/0.85 = 10.53 cm

s = x c = x 0.003 = 0.0005 < 0.001943 betul Baja desak belum leleh

fs = s .Es = 0.0005 x 2100000 = 1066.27 Kg/cm2Moment nominal yang dapat dikerahkan :

M1 = 0.85 x 229.5 x 8.95 x 30 x ( 63.75 (8.95/2))= 3105567.309 Kg cm

M2 = (7 x 4.906) x 1066.27 x (63.75 8.75)

= 2014092.824 Kg cm

Moment nominal, Mn= 5119660.133 Kg cm

= 51.196 tm

Mt = x Mn = 0.8 x 51.196 = 40.96 tm > 31.04 tm

Desain tulangan moment positif sukses.

Desain balok tulangan rangkap SUKSES MOMENT KAPASITAS BALOK

a. Moment kapasitas moment negative

Ts = Ts + Cc

(As .fy) . o = As .fy + 0.85 .fc .a .b

a = = = 19.838 cm

c = a/ = 19.838/0.85 = 23.34 cm

s = x c = x 0.003 = 0.00219

Baja desak sudah leleh, tetapi belum mencapai over strength

M1 = 0.85 x 229.5 x 19.838 x 30 x ( 61.25 (19.838/2))= 5959560.819 Kg cm

M2 = (4 x 4.906) x 4080 x (61.25 6.25)

= 4403850.000 Kg cm

Mkap- = M1 + M2 = 5959560.819 + 4403850.000 = 10363410.82 Kg cm =103.63 tmb. Moment kapasitas moment positif

Ts = Ts + Cc

(As .fy) . o = As . .c .Es + 0.85 .fc .a .b

(4x4.906)x4080x1.35=((7x4.906) x)x0.003x2100000+(0.85x229.5xax30)

108088.992 = + 5852.25 a108088.992 a = 216354.6 a 1609137.338 + 5852.25 a2a2 + 17.82 a 274.97 = 0

a = = 9.915 cm

c = a/ = 9.915/0.85 = 11.66 cm

s = x c = x 0.003 = 0.000745

M1 = 0.85 x 229.5 x 9.915 x 30 x ( 63.75 (9.915/2))= 3774234.337 Kg cm

M2 = (4 x 4.906) x 4080 x (63.75 8.75)

= 3311825.494 Kg cm

Mkap+ = M1 + M2 = 3774234.337 + 3311825.494 = 7086059.831 Kg cm =70.86 tm

PENULANGAN GESER BALOKBalok Lantai 1 (bentang kiri = bentang kanan)

Menghitung Gaya Geser pada Balok

a. Akibat beban gempa

Vu gempa=

=

= 16.96 ton

= = 28.27384 ton = 28273.84 kg

b. Akibat berat sendiri

Vu sendiri= 1.05 * ( Vd + Vl )

= 1.05 * ( 14.491 + 4.827 )

= 20.28 ton

= = 33.8065 ton = 33806.5 kg

Vcn = x x b x h

= x x 300 x 612.5

= 145267.13 N

= 14817.25 kg

Dipakai sengkang P10 Ad = d2 = . . (1)2 = 0.785 cm2

X1 = = 5164.88 kg

X2 = = 10329.76 kg

X3 = 33806.5 10329.76 = 23476.74 kg

Daerah Sendi Plastis

Vsn= Vu sendiri + Vu gempa

= 33806.5 + 28273.84

= 62080.34 kg

Dipakai sengkang 2 kaki

As= 2 x 0.785 = 1.57 cm2

S= = = 6.89 cm

Dipakai :

S = 65 mm

P10 - 65 mm

Daerah Luar Sendi Plastis

Vsn= 23476.74 + 28273.84 14817.25

= 36933.33 kg


As= 2 x 0.785 = 1.57 cm2

S= = = 10.62 cm

Dipakai :

S = 100 mm

P10 - 100 mm

MOMENT PERLU KOLOM ( Mu,k )

Kolom Fa dan Fb

Diketahui :Ukuran balok 30/70

ha= 4 m

Ukuran kolom 50/50

hb = 4 m

Mkap,i = 89.25 tm

Lbi= 7.2 m

Mkap,a = 53.73 tm

Lba= 5.3 ma= = = 0.5b= a = 0.5

= 1.3

= 0.7

Tinggi bersih tingkat, hk = 4 (2*0.5*0.7) = 3.3m

Bentang bersih balok kiri,Lbi

= 7.2 (0.5*0.5) (0.5*0.5) = 6.7 m

Bentang bersih balok kanan, Lba= 5.3 (0.5*0.5) (0.5*0.5) = 4.8 m

Mu,ka= * * a * * ((* Mkap,i) + ( * Mkap,a))

= * 1.3 * 0.5 * 0.7 * ((* 89.25) + ( * 53.73))

= 58.27 tm

Mu,kb= * * a * * ((* Mkap,i) + ( * Mkap,a))

= * 1.3 * 0.5 * 0.7 * ((* 89.25) + ( * 53.73))

= 58.27 tm

Mu,k = Mu,ka + Mu,kb = 58.27 + 58.27 = 116.54 tm

Mu-,bi = 47.94 tm

Mu+,ba = 22.47 tm

Mu,b = Mu-,bi + Mu+,ba = 47.94 +22.47

Mu,k = 1.66 Mu,b

Jadi yang dipakai adalah, Mu,k = 58.27 tm

Kolom aAMu,aA = 1.05 (MD + ML + ME)

= 1.05 (21.53 + 7.67 + 399.13)

= 449.75 KNm

= 45.87 tm

Kolom aB

Mu,aB = 1.05 (MD + ML + ME)

= 1.05 (6.98 + 2.21 + 446.67)

= 478.65 KNm

= 48.82 tm

Kolom aC

Mu,aC = 1.05 (MD + ML + ME)

= 1.05 (6.98 + 2.21 + 446.11)

= 478.07 KNm

= 48.76 tm

Kolom aD

Mu,aD = 1.05 (MD + ML + ME)

= 1.05 (21.53 + 7.67 + 397.41)

= 447.94 KNm

= 45.69 tm

GAYA AKSIAL KOLOM ( Nu,k )Nu,ki= Rv x 0.7 x x 1.05 Ng,kRv = 1.1 (0.025 x 5) = 0.975

Kolom I = IV Kolom aA = aD

(Nu,k)aA = 0.975 x 0.7 x 89.54 + 1.05 (126.879 + 48.713) = 245.48 ton Kolom Aa = Da

(Nu,k)Aa = 0.975 x 0.7 x 65.31 + 1.05 (100.151 + 37.556) = 189.17 ton Kolom aE = aH

(Nu,k)aE = 0.975 x 0.7 x 45.46 + 1.05 (73.516 + 26.179) = 135.71 ton

Kolom Eb = Hb

(Nu,k)Eb = 0.975 x 0.7 x 25.61 + 1.05 (46.675 + 14.695) = 81.92 ton

Kolom bI = bL

(Nu,k)bI = 0.975 x 0.7 x 5.76 + 1.05 (19.673 + 3.158) = 27.90 ton

Kolom II = III

Kolom aB = aC

(Nu,k)aB = 0.975 x 0.7 x 24.31 + 1.05 (210.234 + 88.578) = 330.34 ton

Kolom Ba = Ca

(Nu,k)Ba = 0.975 x 0.7 x 21.56 + 1.05 (164.749 + 68.932) = 260.08 ton

Kolom aF = aG

(Nu,k)aF = 0.975 x 0.7 x 14.43 + 1.05 (120.371 + 49.504) = 188.22 ton

Kolom Fb = Gb

(Nu,k)Fb = 0.975 x 0.7 x 7.3 + 1.05 (76.201 + 30.185) = 116.69 ton

Kolom bJ = bK

(Nu,k)bJ = 0.975 x 0.7 x 0.17 + 1.05 (32.190 + 4.531) = 38.67 ton

DIAGRAM ITERASIDiketahui : ukuran kolom = 500/500 mm

fy = 400 Mpa = 4080 kg/cm2

fc = 22.5 Mpa = 229.5 kg/cm2

c = 0.003

Es = 2100000 kg/cm2

s = = = 0.00194

d = d = 4 + 1 + 1.25 = 6.25 cm

tulangan = 25 mm = 2.5 cm

Ad = 0.25 (2.5)2 = 4.908 cm2

Jumlah tulangan = 6 buah (1%)

As = As = 6 x Ad = 6 x 4.908 = 29.45 cm2 Kolom Pendek dengan Beban Sentris ( Mno = 0 )

Pno = (0.85 . fc . b . ht) + (As + As) . (fy 0.85 . fc)

= (0.85 x 229.5 x 50 x 50) + (29.45 + 29.45) x (4080 0.85 . 229.5)

= 716509.58 kg

= 716.5 t

Mno= 0

Jadi, Mno = 0 ; Pno = 716.5 t

Kondisi Balance pada Kolom Pendek

Cb = h = (43.750) = 26.57 cmab = 0.85 x Cb = 0.85 x 26.57 = 22.58 cms = x c = x 0.003 = 0.0023 > 0.00194

Baja desak leleh

Cc = 0.85 fc . ab . b

= 0.85 x 229.5 x 22.58 x 50

= 220239.67 kg

Cs= As (fy 0.85 fc)

= 29.45 ( 4080 0.85 x 229.5)

= 114411.04 kg

Ts= As x fy

= 29.45 x 4080

= 120156 kg

Pb = Cc + Cs Ts

= 220239.67 + 114411.04 120156

= 214494.04 kg

= 214.49 t

Pb x eb = Cc (0.5ht a/2) + Cs (0.5ht d) Ts (0.5ht d)

= 220239.67 (25 22.58/2) + 114411.04 (25 6.25) 120156 (25 - 6.25)

= 7417617.87 kgcm

= 74.18 tm

Mb = Pb x eb = 74.18 tmeb = = = 0.3458 m = 34.58 cm

Jadi, Mb = 74.18 tm ; Pb = 214.49 t

Kondisi Patah Desak (Compression Failure) C > Cb

Diasumsikan C = 35 cm , a = 0.85 x 35 = 29.75 cm

Cc = 0.85 fc . a . b

= 0.85 x 229.5 x 29.75 x 50

= 290174.06 kg

Cs= As (fy 0.85 fc)

= 29.45 ( 4080 0.85 x 229.5)

= 114411.04 kg

Ts= As x fs

= As x s x c

= As x x s x c

= 29.45 x x 0.003 x 2100000

= 46383.75 kg

Pn = Cc + Cs Ts

= 290174.06 + 114411.04 46383.75

= 358201.35 kg

= 358.201 t

Mn = Cc (0.5ht a/2) + Cs (0.5ht d) Ts (0.5ht d)

= 290174.06 (25 22.58/2) + 114411.04 (25 6.25) 46383.75 (25 - 6.25)

= 5952914.67 kgcm

= 59.53 tm

e = = = 0.1662 m = 16.62 cm

Jadi, Mn = 59.53 tm ; Pn = 358.201 t

Kondisi Lentur Murni ( Pn = 0 )

Cc = 0.85 fc . a . b = 0.85 x 229.5 x a x 50 = 9753.75 a

Ts = As x fy = 29.45 x 4080 = 120156 kg

Cs = As x fs

= As x s x c

= As x x s x c

= 29.45 x x 0.003 x 2100000

=

Keseimbangan gaya-gaya horizontal :

Cc + Cs Ts = 0

9753.75 a + - 120156 = 0

9753.75 a2 + 185535 a -985654.68 -120156a = 0

a2 + 6.7 a 101.05 = 0

a = = 7.25 cmc = a/ = 7.25/0.85 = 8.53 cm

s = x c = x 0.003 = 0.0008 > 0.00194

fs = s x Es = 0.0008 x 2100000 = 1683.94 kg/cm2 < fy = 4080 kg/cm2 Cc = 9753.75 a = 9753.75 x 7.25 = 70714.68 kg

Cs = = = 49582.63 kg

Ts = 120156 kg

Mn = Cc (h a/2) + Cs (h-d)

= 70714.68 (43.75 7.25/2) + 49582.63 (43.75 6.25)

= 4690760.29 kg.cm = 46.9 tm

e = = = ~

Jadi, Mn = 46.9 tm ; Pn = 0

Kondisi Patah Tarik (Tension Failure) C < Cb

Diasumsikan C = 20 cm , a = 0.85 x 20 = 17 cm

Cc = 0.85 fc . a . b

= 0.85 x 229.5 x 17 x 50

= 140941.68 kg

Cs= As (fy 0.85 fc)

= 29.45 ( 4080 0.85 x 229.5)

= 114411.04 kg

Ts= As x fy

= 29.45 x 4080

= 120156 kg

Pn = Cc + Cs Ts

= 140941.68 + 114411.04 120156

= 135196.72 kg

= 135.19 t

Mn = Cc (0.5ht a/2) + Cs (0.5ht d) Ts (0.5ht d)

= 140941 (25 17/2) + 114411.04 (25 6.25) 120156 (25 - 6.25)

= 6723669.72 kgcm

= 67.24 tm

e = = = 0.4974 m = 49.74 cm

Jadi, Mn = 67.24 tm ; Pn = 135.19 t

Kondisi Tarik Murni

Pt = (As + As) x fy

= (29.45 + 29.45) x 4080

= 240312 kg

= 240.31 t

Mt = 0

Jadi, Mt = 0 ; Pt = 240.31 t

PENULANGAN KOLOM

Kolom 1

Pn = = = 409.13 tm

Mn = = = 37.16 t

Dari diagram iterasi diperoleh % jumlah tulangan = 0.9 %

As = 0.009 x 50 x 43.75 = 19.68 cm2

Dipakai tulangan 5D25

As = 5 x (0.25 x x 2.52) = 24.54 cm2 > 19.68 cm2PENULANGAN GESER KOLOMKolom 1 = Kolom 16

Diketahui :

fc = 22.5 Mpa = 229.5 kg/cm2

fcy = 400 Mpa = 4080 kg/cm2

Ma = 48.42 tm

Mb = 45.87 tm

Balok atas 30/70

Kolom 50/50

Nu,k = 245.48 ton

Vd = 1.319 ton

Vl = 0.470 ton

Ve = 14.364 ton

hn = 5 (2 x 0.5 x 0.70) = 4.3 m

Vu,k = = 21.93 ton

Vu,k maks = 1.05 x (Vd + Vl + Ve)

= 1.05 x (1.319 + 0.470 + . 14.364)

= 62.20 ton > Vu,k dipakai Vu,k = 21.93 ton

= = 36.55 ton = 36550 kg

Vc =

= 307589 N

= 31374.1 kg

= 31.374 ton

Dipakai sengkang 10mm As = 0.25..12 = 0.785 cm2 Daerah Lo = 1.5 h = 1.5 x 50 = 75 cm = 750 mmVsn = - Vc = 36550 31374.1 = 5172.46 kg


S = = = 81.27 cm

Control jarak sengkang :

S 8dl = 8 x 2.5 = 20 cm

S 0.25bc = 0.25 x 50 = 12.5 cm

S 10 cm

Dipakai tulangan 3P10 200 mm

Daerah diantara 2 Lo

Jarak sengkang :

S d/2 = 43.75/2 = 21.875 cm

S 60 cm


Daerah sendi plastis

Me = 40.69 tm

Ve = 14.36 ton

Mc,kap = x Mn = 1.4 x 57.34 = 80.27 tm

Vu,kb = 1.3 x 0.7 x x 14.36 = 25.78 ton

Vn = = = 42.964 ton = 42964 kgDipakai sengkang 4 kaki

S = = 13.04 cm dipakai S = 13 cm


Kolom 6 = Kolom 11

Diketahui :

fc = 22.5 Mpa = 229.5 kg/cm2

fcy = 400 Mpa = 4080 kg/cm2

Ma = 75.95 tm

Mb = 48.82 tm

Balok atas 30/70

Kolom 50/50

Nu,k = 330.34 ton

Vd = 0.437 ton

Vl = 0.138 ton

Ve = 17.327 ton

hn = 5 (2 x 0.5 x 0.70) = 4.3 m

Vu,k = = 29.02 ton

Vu,k maks = 1.05 x (Vd + Vl + Ve)

= 1.05 x (0.437 + 0.138 + . 17.327)

= 73.38 ton > Vu,k dipakai Vu,k = 29.02 ton

= = 48.366 ton = 18366 kg

Vc =

= 332895.54 N

= 33955.34 kg

= 33.95 ton

Dipakai sengkang 10mm As = 0.25..12 = 0.785 cm2 Daerah Lo = 1.5 h = 1.5 x 50 = 75 cm = 750 mm

Vsn = - Vc = 48366 33955.34 = 14410.66 kg


S = = = 29.17 cm

Control jarak sengkang :

S 8dl = 8 x 2.5 = 20 cm

S 0.25bc = 0.25 x 50 = 12.5 cm

S 10 cm


Daerah diantara 2 Lo

Jarak sengkang :

S d/2 = 43.75/2 = 21.875 cm

S 60 cm


Daerah sendi plastis

Me = 45.55 tm

Ve = 17.33 ton

Mc,kap = x Mn = 1.4 x 61.025 = 85.44 tm

Vu,kb = 1.3 x 0.7 x x 17.33 = 29.58 ton

Vn = = = 49.301 ton = 49301 kg


S = = 11.368 cm dipakai S = 11 cm


PENULANGAN BEAM COLUMN JOINTLantai 1 Tengah

fc = 22.5 Mpa = 229.5 kg/cm2fys = 400 Mpa = 4080 kg/cm2 Menghitung Vcol

Vcol=

=

= 26.55 ton

Menghitung Vjh dan Vjv

Vjh = Ts + Cc Vcol

Tsi = Cci =

=

= 141.3 ton

Tsi = Cci =

=

= 52.23 ton\

Vjh = 141.3 + 52.23 26.55 = 187.07 ton

Vjv = x Vjh = x 187.07 = 261.89 ton

Kontrol :

jh = = = 106.89 kg/cm2jh maks = 1.5 = 1.5 = 7.12 Mpa = 72.57 kg/cm2jh < jh maks ukuran joint/kolom tidak perlu diperbesar Gaya geser oleh beton Vc

Nu,k = 330.34 ton

bc = 50 cm

hc = 50 cm

= = 132.14 kg/cm2 > 0.1fc = 22.95 kg/cm2 (1 N = 0.102 kg)

Vch = x bba x htc

= x 500 x 500

= 189879 kg

= 189.879 ton

Gaya geser yang ditahan oleh sengkang,VsVs = Vjv Vch

= 261.89 189.879

= 72.012 ton

Jarak sengkang

Ash = = = 17.65 cm2Dipakai sengkang 10mm Ad = 0.25 12 = 0.785 cm2Jika dipakai sengkang 4 kaki

As = 4 x Ad = 4 x 0.785 = 3.14 cm2n = = = 5.62 6 buah

s = = 91.43 mm

dipakai tulangan sengkang kolom 6P10 90 mm

PERENCANAAN PONDASI

1. Pondasi Kolom K1 = Kolom K16Siklop adalah beton Bo dengan tegangan desak (fc) = 10 Mpa = 10000 KN/m2 tegangan ijin siklop ( q siklop )

siklop = 22 KN/m2 , h siklop = 1.5 m ; kedalaman pondasi = 2 m; hpondasi = 0.75 m

Tegangan tanah ijin 2.5 kg/cm2 = 245 KN/m2PuK = 2406.67 Kn P

MuK = 449.70 Kn M

a. Menghitung Luasan Telapak Siklop

q netto= q tanah ijin ( h siklop x siklop )-(D- h siklop- hpondasi )x(hpondasi x beton )

= 245-(1.5x22)-(3.5-1.5-0.75)x1.5-(0.75x24)

= 192.125 Kn/m2

e

q netto

192.125

b. Perencanaan Luasan Telapak Footplat

q netto 1= q siklop -(D- h siklop- hpondasi )x tanah-(hpondasi x beton )

= 1000-(3.5-1.5-0.75)x1.5-(0.75-24)

= 963.25 Kn/m2q netto 1

963.25

c. Kontrol Geser Pada Pondasi

e

q maks q min d. kontrol geser 1 arah

d = hpondasi - Z = 0.75 0.08 = 0.67 m

Nilai q1 dicari dengan perbandingan segi 3

q 1= =

Vu1=

Vu1=

Vcl=

0.6 Vcl = 549.09 Kn > Vul = AMAN!!!!

e. Kontrol Geser 2 Arahqc = tegangan pada tengah tengah pondasi

qc = 0.5x(q maks+q min) = 0.5x(836.32 + 367.02) = 601.67 KN/m2

q2 = tegangan pada jarak d/2 dari muka kolom

dengan perbandingan segi 3 didapat :

q 2= =

Vu2 =

= 979.19 Kn

bo= 2x(h kolom + d)+2x(b kolom + d)

= 2x(0.5+0.67)+2x(0.5+0.67)

= 4.68 m

Vc2=

0.6 Vc2 = 1187.862 Kn > Vul = AMAN!!!!

f. Penulangan Lentur Pondasi

q3 = tegangan tepat dimuka

cara mencari q3 dengan persamaan segi 3

q3 =

=

= 202.22 KN/m2

Dari persamaan pangkat 2 diatas didapat a = 20.025 m

AS =

AS min = 0.0018xbx hpondasi = 0.0018x1000x750 = 1350 mm2

AS pakai = 1350 mm2 ( yang terbesar )Pakai tulangan D = 16 mm Ai = 200.96 mm2Jarak tulangan ( S ) =

pakai tulangan D16 145g. Penulangan susut

h = hpondasi - 100 mm = 750- 100 mm = 650 mm

As.susut = 0.002xbxh

= 0.002x1500x500 = 1500 mm2Pakai tulangan D = 16 mm Ai = 200.96 mm2 S.susut =

pakai tulangan D16 145 ( disamakan dengan tulangan lentur )

2. Pondasi Kolom K2 = Kolom K11

Siklop adalah beton Bo dengan tegangan desak (fc) = 10 Mpa = 10000 KN/m2 tegangan ijin siklop ( q siklop )

siklop = 22 KN/m2 , h siklop = 1.5 m ; kedalaman pondasi = 2 m; hpondasi = 0.75 m

Tegangan tanah ijin 2.5 kg/cm2 = 245 KN/m2PuK = 3238.63 Kn P

MuK = 478.63 Kn M

a. Menghitung Luasan Telapak Siklop

q netto= q tanah ijin ( h siklop x siklop )-(D- h siklop- hpondasi )x(hpondasi x beton )

= 245-(1.5x22)-(3.5-1.5-0.75)x1.5-(0.75x24)

= 192.125 Kn/m2

e

q netto

192.125

b. Perencanaan Luasan Telapak Footplat

q netto 1= q siklop -(D- h siklop- hpondasi )x tanah-(hpondasi x beton )

= 1000-(3.5-1.5-0.75)x15-(0.75-24)

= 963.25 Kn/m2q netto 1

963.25

c. Kontrol Geser Pada Pondasi

e

q maks q min d. kontrol geser 1 arah

d = hpondasi - Z = 0.75 0.08 = 0.67 m

Nilai q1 dicari dengan perbandingan segi 3

q 1= =

Vu1=

Vu1=

Vcl=

0.6 Vcl = 549.09 Kn > Vul = AMAN!!!!

e. Kontrol Geser 2 Arahqc = tegangan pada tengah tengah pondasi

qc = 0.5x(q maks+q min) = 0.5x(1059.84 + 559.47) = 809.66 KN/m2

q2 = tegangan pada jarak d/2 dari muka kolom

dengan perbandingan segi 3 didapat :

q 2= =

Vu2 =

= 1265.25 Kn

bo= 2x(h kolom + d)+2x(b kolom + d)

= 2x(0.5+0.67)+2x(0.5+0.67)

= 4.68 m

Vc2=

0.6 Vc2 = 1187.86 Kn > Vul = AMAN!!!!

f. Penulangan Lentur Pondasi

q3 = tegangan tepat dimuka

cara mencari q3 dengan persamaan segi 3

q3 =

=

= 262.89 KN/m2

Dari persamaan pangkat 2 diatas didapat a = 26.16 m

AS =

AS min = 0.0018xbx hpondasi = 0.0018x1000x750 = 1350 mm2

AS pakai = 1350 mm2 ( yang terbesar )Pakai tulangan D = 16 mm Ai = 200.96 mm2Jarak tulangan ( S ) =

pakai tulangan D16 145g. Penulangan susut

h = hpondasi - 100 mm = 750- 100 mm = 650 mm

As.susut = 0.002xbxh

= 0.002x1500x500 = 1500 mm2Pakai tulangan D = 16 mm Ai = 200.96 mm2 S.susut =

pakai tulangan D16 145 ( disamakan dengan tulangan lentur )

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Beton 2 Euy...Ngulang Boww..

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