4. Balok Menganjur Dan Balok Gerber

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BALOK MENGANJUR BALOK MENGANJUR Q1 = 20 * 14 = 280 KN Q2 = 20 * 6 = Q1 = 20 * 14 = 280 KN Q2 = 20 * 6 = 120 KN 120 KN Perhitungan Reaksi tumpuan Perhitungan Reaksi tumpuan =0 RAH = 250 cos 60 = 125 KN RAH = 250 cos 60 = 125 KN MB = 0 RAV*14–P1*12 - P2 sin *8– P3*0– Q1*7+ P4*6 +Q2*3= 0 14 RAV - 2400 – 1732.048 – 0 – 1960 + 900 + 360 = 0 RAV = 345..1463 KN MA = 0 -{ - R BV *14+P4*20+P3*14+P2sin*6+P1*2+Q1*7+Q2*17} =0 14 RBV - 3000 - 4200 -1299.036-400 – 960 - 2040 = 0 RBV = 921.3597 KN E 300 KN 150 KN 200 KN 250 KN 60 o 2,0 4,0 8,0 6,0 A C D B + - BID M BID D BID N Q1 Q2 RBV RAH RAV 20 KN/m’

description

materi kuliah mekanika statika

Transcript of 4. Balok Menganjur Dan Balok Gerber

BALOK MENGANJURBALOK MENGANJUR

Q1 = 20 * 14 = 280 KN Q2 = 20 * 6 = 120 KNQ1 = 20 * 14 = 280 KN Q2 = 20 * 6 = 120 KN

Perhitungan Reaksi tumpuanPerhitungan Reaksi tumpuan

=0 RAH = 250 cos 60 = 125 KN RAH = 250 cos 60 = 125 KN

MB = 0 RAV*14–P1*12 - P2 sin *8– P3*0– Q1*7+ P4*6 +Q2*3= 0 14 RAV - 2400 – 1732.048 – 0 – 1960 + 900 + 360 = 0 RAV = 345..1463 KN

MA = 0 -{ -RBV*14+P4*20+P3*14+P2sin*6+P1*2+Q1*7+Q2*17}=0 14 RBV - 3000 - 4200 -1299.036-400 – 960 - 2040 = 0 RBV = 921.3597 KN

Kontrol V = 0 RAV + RBV – P1 – P2 sin 60 – P3 – P4 - Q1 – Q2= 0 345.1463+921.3597– 200 -216.5064 -300-150-280 –120 =0 1266.506– 1266.506 = 0 0 = 0 OK

E

300 KN

150 KN

200 KN 250 KN

60o

2,0 4,0 8,0 6,0

A C D

B

+

-

BID M

BID D

BID N

Q1 Q2

RBV

RAH

RAV

20 KN/m’

Perhitungan Gaya DalamPerhitungan Gaya Dalam

MOMEN MOMEN MA = 0MA = 0MC = RAV * 2 - 20 * 2 * 1 = 650.2926 KN m MD = RAV * 6 - 0.5 * 20 * 6 ^ 2 - 200 * 4 = 910.8777 KN m MB = RAV * 14 - 20 * 14 * 7 - 200 * 12 - 216.506 . 8 = -1260 KN m ME = RAV * 20 - 20 * 14 * 13 - 200 * 18 - 216.506 *14 - 150 *6 + RBV*6 -20*6 * 3 = 0 KN m

DA = RAV = 345.1463 KN DC KR = RAV - 20 * 2 = 305.1463 KN DC KN = DCKR - P1 = 305.1463 - 200 = 105.1463 KN DDKR = DC KN - 20 * 4 = 25.14629 KN DDKN = DDKR - P2 sin 60 = 25.1463 - 216.506 = -191.36 KN DBKR = DDKN - 20 * 8 = -351.36 KN DBKN = DBKR - P3 + RBV = -351.6 - 300 + 921.3597 = 270 KN DEKR = DB KN - 20 * 6 = 150 KN DEKN = DEKR - P4 = 150 - 150 = 0 KN

Atau Secara segmentalAtau Secara segmental Bentang AC Bentang AC

Mx = Rav. X - ½ q X^2 = 345.1463 x - 10 x*2Mx = Rav. X - ½ q X^2 = 345.1463 x - 10 x*2 Dx = dMX / Dx = 345.1463 - 20 xDx = dMX / Dx = 345.1463 - 20 x

X 0 1 2 MX 0 335.1463 650.2926 Dx 345.1463 325.1463 305.1463

Bentang CD ( 0 < x < 4 ) Bentang CD ( 0 < x < 4 )

Mx = RAV ( 2 + x ) – q * 2 * ( 1 + x ) – P1 x – q.x. ½ xMx = RAV ( 2 + x ) – q * 2 * ( 1 + x ) – P1 x – q.x. ½ x Mx = 345.1463 (2+ x)- 20*2*(1+x ) - 200 x - 1/2 * 20 x^2Mx = 345.1463 (2+ x)- 20*2*(1+x ) - 200 x - 1/2 * 20 x^2 = - 10 x^2 + 105.1463 x + 650.2926 = - 10 x^2 + 105.1463 x + 650.2926

Dx = dMx / dx = - 20 x + 105.1463 Dx = dMx / dx = - 20 x + 105.1463

X 0 1 2 3 4 (D) MX 650.2926 745.4389 820.5852 875.7315 910.8778 Dx 105.1463 85.1463 65.1463 45.1463 25.1463

2.00 xq = 20 KN/m

Q = q*2 Qx = q x RAV

C

200 KN

Bentang DB ( 0 < x < 8 ) Bentang DB ( 0 < x < 8 )

Mx = RAV ( 6 + x ) – q * 6 * ( 3 + x ) – P1 (4 + x) – P2 sin 60 * x - q x . ½ xMx = RAV ( 6 + x ) – q * 6 * ( 3 + x ) – P1 (4 + x) – P2 sin 60 * x - q x . ½ x Mx = 345.1463 (6+ x)- 120 *(3+x ) - 200 (4 +x ) – 216.5064 x - 1/2 * 20 x^2Mx = 345.1463 (6+ x)- 120 *(3+x ) - 200 (4 +x ) – 216.5064 x - 1/2 * 20 x^2 = - 10 x^2 - 191.3597 x + 910.8778 = - 10 x^2 - 191.3597 x + 910.8778 Dx = dMx / dx = - 20 x - 191.3597 Dx = dMx / dx = - 20 x - 191.3597

Mx = 0 - 10 x^2–191.3597 x + 910.8778 = 0 dgn rumus abc diperoleh x = 3,9462 mMx = 0 - 10 x^2–191.3597 x + 910.8778 = 0 dgn rumus abc diperoleh x = 3,9462 m Mmax bila Dx = 0 - 20 x + 191.3597 = 0 Mmax bila Dx = 0 - 20 x + 191.3597 = 0 x = 9.568 m ( tidak ada Mmax pada bentang DB )x = 9.568 m ( tidak ada Mmax pada bentang DB )

X 0 1 2 3 4 5 6 7 8Mx 910.878 709.518 488.158 246.799 -14.561 -295.921 -597.280 -918.640 -1260.000

Dx -191.360 -211.360 -231.360 -251.360 -271.360 -291.360 -311.360 -331.360 -351.360

2.00 4,00 q = 20 KN/m

Q = q*6Qx = q x

RAV

x250216.5064 KN

3,00DC

Bentang EB ( 0 < x < 8 )

Mx = - { q x * ½ x + P4 x } = - ½ q x^2 - 150 x = - 10 x^2 – 150 x Dx = dMx/dx = - 20 x

X 0 1 2 3 4 5 6MX 0.000 -160.000 -340.000 -540.000 -760.000 -1000.000 -1260.000 Dx -150.000 -170.000 -190.000 -210.000 -230.000 -250.000 -270.000

Q = q*x

x

EB

P4

GAYA dan BIDANG NORMAL

NA = RAH = 125 KN NAD = RAH = 125 KN NAD = RAH - P2 cos 60 = 0

2.00 4,00

RAV

250

DC

BALOK GERBERBALOK GERBER

Balok Gerber adalah balok menerus diatas beberapa buah perletakan dan diberi sendi diantara Balok Gerber adalah balok menerus diatas beberapa buah perletakan dan diberi sendi diantara perletakan tersebut.perletakan tersebut.

Banyaknya sendi yang diberikan agar struktur menjadi statis tertentu, adalah :Banyaknya sendi yang diberikan agar struktur menjadi statis tertentu, adalah :

S = n - 2 S = banyaknya sendi tambahanS = n - 2 S = banyaknya sendi tambahan n = banyaknya perletakann = banyaknya perletakan Untuk contoh diatas S = 2Untuk contoh diatas S = 2

Beberapa alternatif Beberapa alternatif

S1 S2Q1 Q2 Q3

1

S1 S2Q1 Q2 Q3

S1 S2Q1 Q2 Q3

Penyelesainnya adalah dengan memotong struktur pada beberapa bagian dan menambahkan satu Penyelesainnya adalah dengan memotong struktur pada beberapa bagian dan menambahkan satu persamaan MS = 0, kemudian menyelesaikan satu demi satu segmen potonganpersamaan MS = 0, kemudian menyelesaikan satu demi satu segmen potongan

S1 P4 = ½ Q3

S2

P3 = ½ Q3+1/2 Q2P2 = ½ Q1+1/2

Q2

P1 = ½ Q1RS2

RS2

RS1

RS1

S1 S2Q1 Q2 Q3

S1

S2

P4 = ½ Q3

P3 = ½ Q3+1/2 Q2

P2 = ½ Q1+1/2 Q2

RS2

RS1 RS2

S1

2

Contoh SoalContoh Soal

PEMBEBANANPEMBEBANAN Beban miringBeban miring P1V = 200 sin 60 = 173.2051 KNP1V = 200 sin 60 = 173.2051 KN P1H = 200 cos 60 = 100 KNP1H = 200 cos 60 = 100 KN Beban terbagi rata Beban terbagi rata Q1 = 50 * 6 = 300 KNQ1 = 50 * 6 = 300 KN Q2 = 50 * 8 = 400 KNQ2 = 50 * 8 = 400 KN Beban tidak langsungBeban tidak langsung- Qa = 240 KN . Qb = 240 KN Qc = 240 KNQa = 240 KN . Qb = 240 KN Qc = 240 KN- Dirubah jadi beban langsung Dirubah jadi beban langsung - Pb = 120 KN Pd= 440 KN Pe = 440 KNPb = 120 KN Pd= 440 KN Pe = 440 KN- Pc = 120 KNPc = 120 KNPERHITUNGAN REAKSI TUMPUANPERHITUNGAN REAKSI TUMPUANBagian A – S Bagian A – S RAH = 100 KNRAH = 100 KN MS = 0 RAV. 6 – Q1 . 3 = 0MS = 0 RAV. 6 – Q1 . 3 = 0 RAV = 150 KNRAV = 150 KN MA = 0 -{ Q1. 3 + P1v.*6 – Rs. 6 = 0MA = 0 -{ Q1. 3 + P1v.*6 – Rs. 6 = 0 - 900 - 1039.23 + 6 Rs = 0- 900 - 1039.23 + 6 Rs = 0 Rs = 323.2051 KN Rs = 323.2051 KN V = 0 RAV + RS – Q1 – P1V = 0 OKJ V = 0 RAV + RS – Q1 – P1V = 0 OKJ

Q2

RAH

RBV RCV

Qa Qb Qc

Q1

Q2

q2 = 50 KN/m’

P1 = 200 KN

60o

P1 =400KN q2 = 60 KN/m’

6.00 8.00 4.00 4.00 4.00

SA

B D EC

Q1

S P1H

P1V

RAV

A

CB

D ES

Pb= 120 KN

Pd = 440 KN

Pe = 440 KN

Pc = 120 KN

Q2

RBV RCV

CB

D E

Pd = 440 KN

S

323,2051 KN

Bentang SBBentang SB

MB = 0MB = 0

- Rs. 8 – Q2.4+ Pd * 4 + Pe. 8 + Pc*12 – Rvc. 12=0. - Rs. 8 – Q2.4+ Pd * 4 + Pe. 8 + Pc*12 – Rvc. 12=0.

-323.2051*8-1600+1760+3520+1440- 12 Rcv = 0-323.2051*8-1600+1760+3520+1440- 12 Rcv = 0 Rcv = 211.1966 KNRcv = 211.1966 KN

MC = 0MC = 0 -Rs 20 - Q2.16–120 * 12 + RBv 12– 440 * 8 – 440*4 = -Rs 20 - Q2.16–120 * 12 + RBv 12– 440 * 8 – 440*4 =

00

RBv = 1632.008RBv = 1632.008

KONTROLKONTROL

V = 0V = 0 RCV + RBV – RS – Q2 – Pb – Pd – Pe – Pc = 0RCV + RBV – RS – Q2 – Pb – Pd – Pe – Pc = 0 211.1966 + 1632.008 – 323.2051 – 400 – 120 – 440 – 211.1966 + 1632.008 – 323.2051 – 400 – 120 – 440 –

440 – 120 = 0440 – 120 = 0

PERHITUNGAN REAKSI TUMPUANPERHITUNGAN REAKSI TUMPUAN

Bagian S– B - C Bagian S– B - C MC = 0 MC = 0 RS*20-Q2.16+RBV.12–Pb*12-Pd*8–RS*20-Q2.16+RBV.12–Pb*12-Pd*8–

Pe*4=0Pe*4=0 RBV = 1632.008 KNRBV = 1632.008 KN MS = 0MS = 0 -{-RCV*20+Pc*20+Pe*16+Pd*12+Pb*8--{-RCV*20+Pc*20+Pe*16+Pd*12+Pb*8-

RBV*8 + Q2. 4 = 0RBV*8 + Q2. 4 = 0 RCV = 211.1966 KN ( )RCV = 211.1966 KN ( ) V = 0 Tinjau Seluruh KonstruksiV = 0 Tinjau Seluruh Konstruksi RAV+RBV+RCV-Q1-P1v-Q2-Pb-Pd-Pe-RAV+RBV+RCV-Q1-P1v-Q2-Pb-Pd-Pe-

Pc=0Pc=0 150+1632.008+211.1966 -300-150+1632.008+211.1966 -300-

173.2051- 400- 120-440-440-120 = 0173.2051- 400- 120-440-440-120 = 0 0 = 0 OK0 = 0 OK

Qa Qb QcQ1 Q2

q2 = 50 KN/m’

P1 = 200 KN

60o

P1 =400KN q2 = 60 KN/m’

6.00 8.00 4.00 4.00 4.00

S

Gaya – gaya dalamGaya – gaya dalam

Bentang ASBentang AS

Mx = RAV. X – ½ q x*2 = 150 x – 25 x^2Mx = RAV. X – ½ q x*2 = 150 x – 25 x^2 Dx = - 50 x + 150Dx = - 50 x + 150

Mmax dMx/dx = 0Mmax dMx/dx = 0 -50 x + 150 = 0-50 x + 150 = 0 x = 3x = 3

Mmax = 150 * 3 – 25 * 3*2 = 225 KN mMmax = 150 * 3 – 25 * 3*2 = 225 KN m

Bid N Bid N

NAS = RAH = 100 KN NAS = RAH = 100 KN

RAV

A

X 0 2 4 6MX 0 200 200 0 Dx 150 50 -50 -150

Q2

RBV RCV

CB

D E

Pd = 440 KN

Bentang SBBentang SB

Mx = - Rs x - ½ qx^2Mx = - Rs x - ½ qx^2 = - 323.2051 x – 25 x^2= - 323.2051 x – 25 x^2 Dx = - 50 x - 323.2051 Dx = - 50 x - 323.2051

MB = - { -RCV.12 +Pc*12+Pe*8+Pd*4}MB = - { -RCV.12 +Pc*12+Pe*8+Pd*4} = -4185.64 KN m= -4185.64 KN m MD = - { - RCV. 8 + Pc..8 + Pd.4 }MD = - { - RCV. 8 + Pc..8 + Pd.4 } = -2425,64 KN m= -2425,64 KN m

ME = - { - RCV. 4 – Pc * 4 } ME = - { - RCV. 4 – Pc * 4 } = -364.7865 KN m= -364.7865 KN m

BID DBID D DBKr = -723.205 KNDBKr = -723.205 KN DBKn = -723.205-120+1632.008 = 788.8034 KNDBKn = -723.205-120+1632.008 = 788.8034 KN DDKn = 788.8034 – 440 = 348.8034 KNDDKn = 788.8034 – 440 = 348.8034 KN DEKn = 348.8034 – 440 = - 91.1966 KNDEKn = 348.8034 – 440 = - 91.1966 KN DCKr = -91.1966 – 120 = -211.1966DCKr = -91.1966 – 120 = -211.1966 DC Kn = - 211.1966 + RCV = 0DC Kn = - 211.1966 + RCV = 0

X 0 2 4 6 8MX 0 -746.41 -1692.82 -2839.23 -4185.64 Dx -323.205 -423.205 -523.205 -623.205 -723.205