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materi kuliah mekanika statika
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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
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
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