4. balok menganjur dan balok gerber

Download 4. Balok Menganjur Dan Balok Gerber

Post on 09-Jul-2016

61 views

Category:

Documents

36 download

Embed Size (px)

DESCRIPTION

materi kuliah mekanika statika

TRANSCRIPT

  • BALOK MENGANJURQ1 = 20 * 14 = 280 KN Q2 = 20 * 6 = 120 KN

    Perhitungan Reaksi tumpuan

    SH =0 RAH = 250 cos 60 = 125 KN

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

    SMA = 0 -{ -RBV*14+P4*20+P3*14+P2sina*6+P1*2+Q1*7+Q2*17}=0 14 RBV - 3000 - 4200 -1299.036-400 960 -2040 = 0 RBV = 921.3597 KN

    Kontrol SV = 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

    E300 KN150 KN200 KN250 KN60o2,0 4,0 8,0 6,0 A C D B +-BID MBID DBID NQ1Q2RBVRAHRAV20 KN/m

  • Perhitungan Gaya Dalam

    MOMEN MA = 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 segmental Bentang AC

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

    X012 MX0335.1463650.2926 Dx345.1463325.1463305.1463

  • Bentang CD ( 0 < x < 4 )

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

    Dx = dMx / dx = - 20 x + 105.1463 2.00xq = 20 KN/m Q = q*2Qx = q x RAVC200 KN

    X01234 (D) MX650.2926745.4389820.5852875.7315910.8778 Dx105.146385.146365.146345.146325.1463

  • Bentang DB ( 0 < x < 8 )

    Mx = RAV ( 6 + x ) q * 6 * ( 3 + x ) P1 (4 + x) P2 sin 60 * x - q x . x Mx = 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 Dx = dMx / dx = - 20 x - 191.3597

    Mx = 0 - 10 x^2191.3597 x + 910.8778 = 0 dgn rumus abc diperoleh x = 3,9462 m Mmax bila Dx = 0 - 20 x + 191.3597 = 0 x = 9.568 m ( tidak ada Mmax pada bentang DB )

    X012345678Mx910.878709.518488.158246.799-14.561-295.921-597.280-918.640-1260.000 Dx-191.360-211.360-231.360-251.360-271.360-291.360-311.360-331.360-351.360

  • Bentang EB ( 0 < x < 8 )

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

    GAYA dan BIDANG NORMAL

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

    X0123456MX0.000-160.000-340.000-540.000-760.000-1000.000-1260.000 Dx-150.000-170.000-190.000-210.000-230.000-250.000-270.000

  • BALOK GERBERBalok Gerber adalah balok menerus diatas beberapa buah perletakan dan diberi sendi diantara perletakan tersebut.

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

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

    Beberapa alternatif

    1

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

    S1P4 = Q3S2P3 = Q3+1/2 Q2P2 = Q1+1/2 Q2P1 = Q1RS2RS2RS1RS1

  • 2

  • Contoh Soal PEMBEBANANBeban miring P1V = 200 sin 60 = 173.2051 KN P1H = 200 cos 60 = 100 KNBeban terbagi rata Q1 = 50 * 6 = 300 KN Q2 = 50 * 8 = 400 KNBeban tidak langsungQa = 240 KN . Qb = 240 KN Qc = 240 KNDirubah jadi beban langsung Pb = 120 KN Pd= 440 KN Pe = 440 KNPc = 120 KNPERHITUNGAN REAKSI TUMPUANBagian A S RAH = 100 KN MS = 0 RAV. 6 Q1 . 3 = 0 RAV = 150 KN MA = 0 -{ Q1. 3 + P1v.*6 Rs. 6 = 0 - 900 - 1039.23 + 6 Rs = 0 Rs = 323.2051 KN V = 0 RAV + RS Q1 P1V = 0 OKJ

    Q2RAHRBVRCVSP1HP1VCBDESPb= 120 KNPd = 440 KNPe = 440 KNPc = 120 KN

  • Bentang SB

    MB = 0

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

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

    MC = 0 -Rs 20 - Q2.16120 * 12 + RBv 12 440 * 8 440*4 = 0

    RBv = 1632.008

    KONTROL

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

  • PERHITUNGAN REAKSI TUMPUANBagian S B - C SMC = 0 RS*20-Q2.16+RBV.12Pb*12-Pd*8Pe*4=0 RBV = 1632.008 KN MS = 0 -{-RCV*20+Pc*20+Pe*16+Pd*12+Pb*8-RBV*8 + Q2. 4 = 0 RCV = 211.1966 KN ( ) SV = 0 Tinjau Seluruh Konstruksi RAV+RBV+RCV-Q1-P1v-Q2-Pb-Pd-Pe-Pc=0 150+1632.008+211.1966 -300-173.2051- 400- 120-440-440-120 = 0 0 = 0 OK

  • Gaya gaya dalamBentang AS

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

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

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

    Bid N

    NAS = RAH = 100 KN

    X0246MX02002000 Dx15050-50-150

  • Bentang SB

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

    MB = - { -RCV.12 +Pc*12+Pe*8+Pd*4} = -4185.64 KN mMD = - { - RCV. 8 + Pc..8 + Pd.4 } = -2425,64 KN mME = - { - RCV. 4 Pc * 4 } = -364.7865 KN m

    BID D DBKr = -723.205 KN DBKn = -723.205-120+1632.008 = 788.8034 KN DDKn = 788.8034 440 = 348.8034 KN DEKn = 348.8034 440 = - 91.1966 KN DCKr = -91.1966 120 = -211.1966 DC Kn = - 211.1966 + RCV = 0

    X02468MX0-746.41-1692.82-2839.23-4185.64 Dx-323.205-423.205-523.205-623.205-723.205