Tentukan garis pengaruh RA
X adalah nilai NPM terakhir. Contoh :
.... 235 diambil angka 5
.....010 diambil angka 1
..... 200 diambil angka 2
Terakhir dijawab tanggal 11 Mei 2009
Sunday, May 3, 2009
Soal 1. Mekanika Rekayasa III
Tentukan garis pengaruh dari struktur dibawah ini.
Tentukan garis pengaruh RA
Tentukan garis pengaruh RA
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37 comments:
1. Perletakan
RB = 4.5 t ( ↑ )
RC = 3.5 t ( ↓ )
2. Segmen B – A → 0 < x1 < 7
Mx1 = 4.5x + 4.5
y’’= - = - = - ( 4.5x +4.5 )
Rotasi : y’ = ( - - + C1 )
Defleksi : y = ( - - + C1X + C2 )
3. Segmen C – B → 0 < x2 < 1
Mx2 = - 7X2
y’’= - = ( - )
Rotasi : y’ = ( + C3 )
Defleksi : y = ( - + C3X + C4 )
4. Syarat Batas
Titik B : X=0 ; y = 0 → C2 = 0
Titik C : X=0 ; y = 0 → C4 = 0
5. Defleksi di Titik A
7C1 – 2C3 = 185.32
C1 – C3 = 73.21
6. Subtitusi
C1 = 7.78
C3 = - 65.43
7. Faktor Pembagi
AA = -129.3/EI
GPRA
BA = y0 = -1/129.3
( - 4.5/12X3
- 4.5/4 X2+7.78X )
8. Defleksi Masing-masing titik
X1 = 3.5 m → y0 = 0.22
X2 = 7 m → y0 = 1
SEPTIAN CRIPSI PRATAMA
( 0753010010 )
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=-0.05
X=2, Y=-0.06
X=3, Y=-0.02
X=4, Y=0.8
X=5, Y=0.28
X=6, Y=0.58
X=7, Y=1
Segmen C-B;0 < X < 1
X=0, Y=0
X=0.5, Y=0.25
X=1, Y=0.5
Thomas Arya Pideksa
( 0753010053 )
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=0.02
X=2, Y=0.08
X=3, Y=0.18
X=4, Y=0.33
X=5, Y=0.51
X=6, Y=0.73
X=7, Y=1
Segmen B-C;0 < X < 3
X=0, Y=0
X=1, Y=0.19
X=2, Y=0.15
X=3, Y=0
Alfian Eka.H
( 0753010045 )
X1=2, Y1=0.39
X1=4, Y1=0.77
X1=5, Y1=0.96
X1=7, Y1=1.39
X2=2, Y2=0.41
X2=4, Y2=0.81
X2=5, Y2=1
Achmad Hannafi
( 0753010005 )
X1=2, Y1=0.39
X1=4, Y1=0.77
X1=5, Y1=0.96
X1=7, Y1=1.39
X2=2, Y2=0.41
X2=4, Y2=0.81
X2=5, Y2=1
Intan Timoer Sanrozi
(0753010055)
X1=2, Y1=0.39
X1=4, Y1=0.77
X1=5, Y1=0.96
X1=7, Y1=1.39
X2=2, Y2=0.41
X2=4, Y2=0.81
X2=5, Y2=1
segmen b-a;0>x1>7
y1=15/686*(-1/6*x1+14.7x1)
x1=1.75 ---- y1=0.543
x1=3.5 ---- y1=0.969
x1=5.25 ---- y1=1,160
x1=7 ---- y1=1
segmen c-a;0>x2>14
y1=15/686*(-1/30*x2+9.8x2)
x2=1.75 ---- y2=0.371
x2=3.5 ---- y2=0.719
x2=5.25 ---- y2=1.020
x2=7 ---- y2=1.250
x2=8.75 ---- y2=1.387
x2=10.5 ---- y2=1.406
x2=12.25 ---- y2=1.285
x2=14 ---- y2=1
dedik suhendrik p.
(0753010013)
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=0.02
X=2, Y=0.08
X=3, Y=0.18
X=4, Y=0.33
X=5, Y=0.51
X=6, Y=0.73
X=7, Y=1
Segmen B-C;0 < X < 3
X=0, Y=0
X=1, Y=0.19
X=2, Y=0.15
X=3, Y=0
segmen b-a;0>x1>7
y1=1/178.57*(-3/28*x1+30.76x1)
x1=1.75 ---- y1=0.298
x1=3.5 ---- y1=0.577
x1=5.25 ---- y1=1,818
x1=7 ---- y1=1
segmen c-a;0>x2>16
y1=1/178.57*(-8/105*x2+30.66x2)
x2=4 ---- y2=0.659
x2=8 ---- y2=1.55
x2=12---- y2=1.323
x2=16---- y2=1
GP RA; Btng AB/BA adlh
=>Y1^0 =>1/926.68*(1/12X1^3-128.30X^2)
Masukan nilai x;
x1=1.75,=> y1^0=0.24180
x2=3.5, => y1^0=0.4807
x3=5.25,=> y1^0=0.7138
x4=7, => y1^0=0.998, mendekati 1 =1.
GP RA; Btng CBA adlh
=>Y2^0 =>1/926.68*(0.229X2^3 + 3.75X^2)
Masukan nilai x;
x1=2,=> y2^0=0.0181
x2=4,=> y2^0=0.0807
x3=6,=> y2^0=0.199058
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=-0.05
X=2, Y=-0.06
X=3, Y=-0.02
X=4, Y=0.8
X=5, Y=0.28
X=6, Y=0.58
X=7, Y=1
Segmen C-B;0 < X < 1
X=0, Y=0
X=0.5, Y=0.25
X=1, Y=0.5
X=8 m.
Ra = 1.875 t (↑)
Rb = 0.875 t (↓)
Step 1.=> Batang CBA; 8 < X1 < 15 (mtr)lht kanan!
Mx1= 0.875X-1.875(X-8)
EI Y" = 1.375X -7.5
EI Y' = 0.6875X^2 - 7.5X +C3
EI Y = 0.229X^3-3.75X^2+C3X+C4
=> Batang AB/BA; 0 < X2 < 7 (mtr)lht kiri!
Mx2= -X
EI Y" = + X/2
EI Y' = 1/4X^2 +C1
EI Y = 1/12X^3 + C1X+C2
Step 2.=> Cek thdp rotasi&defleksi akibat GP 1 ton (virtuil) dititik A, pengaruh deformasi!
*)Deformasi ttk A,=> Y1 (7)= Y2 (15) ; didpt!
=> 7 C1 - 15 C2 = -99.45...(1)
*)Rotasi ttk A, => Y1 (7)= Y2 (15) ; didpt!
=> C3 - C1 = 29.9371...(2)
(1)->(2); didpt!
=>C3= -13.079, Subt ke pers (5)=>didpt =>C1=-13.82
Step 3.=> Fktr Pembagi thdp msng2 btng,Y1 (7)atauY2 (15)!
=> Delta ; btng AB=> -68.150/EI
=> Delta ; btng CBA=> -337.935/EI
Step 4.=> GP RA; Btng AB/BA adlh
=>Y1^0 =>1/68.150*(1/12X1^3-13.82X)
Masukan nilai x;
x1=1.75,=> y1^0=0.348
x2=3.5, => y1^0=0.0.657
x3=5.25,=> y1^0=0.887
x4=7, => y1^0=1.00.
Step 5.=> GP RA; Btng CBA adlh
=>Y2^0 =>1/337.935*(0.229X2^3 – 3.75X^2-13.079X)
Masukan nilai x;
x1=2,=> y2^0=0.116
x2=4,=> y2^0=0.288
x3=6,=> y2^0=0.485
Nama : Musa U. Katipana
Npm : 0853310088
Rama Ad Hadirawan
(0553010076)
x1=1/162.19*(-1/14*x1^3+26.67*x1)
X1=7, y1=1
x1=1/162.19*(-1/14*(7)^3+26.67*(7))
X1=3.5 y1=0.56
x2=1/162.19*(-1/13*x2^3+5.78*x2)
x2 =14 y2=1
x2=1/162.19*(-1/13*(14)2^3+5.78*(14))
x2=7 y2=0.12
Sanggra Umar Dani
(0753010038)
GP RA; Btng AB/BA adlh
=>Y1^0 =>1/926.68*(1/12X1^3-128.30X^2)
Masukan nilai x;
x1=1.75,=> y1^0=0.24180
x2=3.5, => y1^0=0.4807
x3=5.25,=> y1^0=0.7138
x4=7, => y1^0=0.998, mendekati 1 =1.
GP RA; Btng CBA adlh
=>Y2^0 =>1/926.68*(0.229X2^3 + 3.75X^2)
Masukan nilai x;
x1=2,=> y2^0=0.0181
x2=4,=> y2^0=0.0807
x3=6,=> y2^0=0.199058
Tito Kusuma
( 0753010032 )
RB = 4.5 t ( ↑ )
RC = 3.5 t ( ↓ )
2. Segmen B – A → 0 < x1 < 7
Mx1 = 4.5x + 4.5
y’’= - = - = - ( 4.5x +4.5 )
Rotasi : y’ = ( - - + C1 )
Defleksi : y = ( - - + C1X + C2 )
3. Segmen C – B → 0 < x2 < 1
Mx2 = - 7X2
y’’= - = ( - )
Rotasi : y’ = ( + C3 )
Defleksi : y = ( - + C3X + C4 )
4. Syarat Batas
Titik B : X=0 ; y = 0 → C2 = 0
Titik C : X=0 ; y = 0 → C4 = 0
5. Defleksi di Titik A
7C1 – 2C3 = 185.32
C1 – C3 = 73.21
6. Subtitusi
C1 = 7.78
C3 = - 65.43
7. Faktor Pembagi
AA = -129.3/EI
GPRA
BA = y0 = -1/129.3
( - 4.5/12X3
- 4.5/4 X2+7.78X )
8. Defleksi Masing-masing titik
X1 = 3.5 m → y0 = 0.22
X2 = 7 m → y0 = 1
setyo ajeng mahardika/0753010051
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=-0.05
X=2, Y=-0.06
X=3, Y=-0.02
X=4, Y=0.8
X=5, Y=0.28
X=6, Y=0.58
X=7, Y=1
Segmen C-B;0 < X < 1
X=0, Y=0
X=0.5, Y=0.25
X=1, Y=0.5
mei suci wulan sari (0753010049)
y1=1/178.57*(-3/28*x1+30.76x1)
x1=1.75 ---- y1=0.298
x1=3.5 ---- y1=0.577
x1=5.25 ---- y1=1,818
x1=7 ---- y1=1
segmen c-a;0>x2>16
y1=1/178.57*(-8/105*x2+30.66x2)
x2=4 ---- y2=0.659
x2=8 ---- y2=1.55
x2=12---- y2=1.323
x2=16---- y2=1
REZA ACHDA MARTA
(0753010018)
GP RA; Btng AB/BA adlh
=>Y1^0 =>1/926.68*(1/12X1^3-128.30X^2)
Masukan nilai x;
x1=1.75,=> y1^0=0.24180
x2=3.5, => y1^0=0.4807
x3=5.25,=> y1^0=0.7138
x4=7, => y1^0=0.998, mendekati 1 =1.
GP RA; Btng CBA adlh
=>Y2^0 =>1/926.68*(0.229X2^3 + 3.75X^2)
Masukan nilai x;
x1=2,=> y2^0=0.0181
x2=4,=> y2^0=0.0807
x3=6,=> y2^0=0.199058
Risang R.
(0753010039)
Jawab:
segmen b-a;0>x1>7
y1=1/178.57*(-3/28*x1+30.76x1)
x1=1.75 ---- y1=0.298
x1=3.5 ---- y1=0.577
x1=5.25 ---- y1=1,818
x1=7 ---- y1=1
segmen c-a;0>x2>16
y1=1/178.57*(-8/105*x2+30.66x2)
x2=4 ---- y2=0.659
x2=8 ---- y2=1.55
x2=12---- y2=1.323
x2=16---- y2=1
yayan ahmad irawan
0753010044
Segmen B-A;0 < X < 7
X=0, Y=0
X=3, Y=0.599
X=5, Y=0.724
X=7, Y=1
Segmen C-B;0 < X < 11
X=0, Y=0
X2=3, Y=0.74
X2=5, Y=1.27
Digna eka Putra
(0653010023)
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=0.02
X=2, Y=0.08
X=3, Y=0.18
X=4, Y=0.33
X=5, Y=0.51
X=6, Y=0.73
X=7, Y=1
Segmen B-C;0 < X < 3
X=0, Y=0
X=1, Y=0.19
X=2, Y=0.15
X=3, Y=0
0753010035X1=2, Y1=0.39
X1=4, Y1=0.77
X1=5, Y1=0.96
X1=7, Y1=1.39
X2=2, Y2=0.41
X2=4, Y2=0.81
X2=5, Y2=1
Ganda.apriliansyah
(0653010058)
GP RA; Btng AB/BA adlh
=>Y1^0 =>1/926.68*(1/12X1^3-128.30X^2)
Masukan nilai x;
x1=1.75,=> y1^0=0.24180
x2=3.5, => y1^0=0.4807
x3=5.25,=> y1^0=0.7138
x4=7, => y1^0=0.998, mendekati 1 =1.
GP RA; Btng CBA adlh
=>Y2^0 =>1/926.68*(0.229X2^3 + 3.75X^2)
Masukan nilai x;
x1=2,=> y2^0=0.0181
x2=4,=> y2^0=0.0807
x3=6,=> y2^0=0.199058
Vishe Nurdanatri0753010054Segmen B-A;0 < X < 7
X=0, Y=0
X=3, Y=0.599
X=5, Y=0.724
X=7, Y=1
Segmen C-B;0 < X < 11
X=0, Y=0
X2=3, Y=0.74
X2=5, Y=1.27
ichwan frendi
(0753010030)
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=-0.05
X=2, Y=-0.06
X=3, Y=-0.02
X=4, Y=0.8
X=5, Y=0.28
X=6, Y=0.58
X=7, Y=1
Segmen C-B;0 < X < 1
X=0, Y=0
X=0.5, Y=0.25
X=1, Y=0.5
zakiyul fuad
0753010056
x1=1m,y=0.188
x1=2m,y=0.37
x1=3m,y=0.541
x1=4m,y=0.695
x1=5m,y=0.827
x1=6m,y=0.93
x1=7m,y=1
x2=1m,y=0.2477
x2=2m,y=0.489
x2=2m,y=0.489
x2=3m,y=0.722
x2=4m,y=0.942
x2=5m,y=1.145
x1=6m,y=1.327
m. rochman arif
(0753010060)
1. Perletakan
RB = 4.5 t ( ↑ )
RC = 3.5 t ( ↓ )
2. Segmen B – A → 0 < x1 < 7
Mx1 = 4.5x + 4.5
y’’= - = - = - ( 4.5x +4.5 )
Rotasi : y’ = ( - - + C1 )
Defleksi : y = ( - - + C1X + C2 )
3. Segmen C – B → 0 < x2 < 1
Mx2 = - 7X2
y’’= - = ( - )
Rotasi : y’ = ( + C3 )
Defleksi : y = ( - + C3X + C4 )
4. Syarat Batas
Titik B : X=0 ; y = 0 → C2 = 0
Titik C : X=0 ; y = 0 → C4 = 0
5. Defleksi di Titik A
7C1 – 2C3 = 185.32
C1 – C3 = 73.21
6. Subtitusi
C1 = 7.78
C3 = - 65.43
7. Faktor Pembagi
AA = -129.3/EI
GPRA
BA = y0 = -1/129.3
( - 4.5/12X3
- 4.5/4 X2+7.78X )
8. Defleksi Masing-masing titik
X1 = 3.5 m → y0 = 0.22
X2 = 7 m → y0 = 1
Mirawati (npm. 0753010037) *ralat tugas*
B-C : x1=0 y1=0
x1=3,5 y1=-0,047
x1=7 y1=0
A-B : x2=0 y2=1
x2=3,5 y2=0,359
x2=7 y2=0
Diana Eka Kartika(npm. 0753010048) *ralat tugas*
B-C : x1=0 y1=0
x1=4 y1=-0,06
x1=8 y1=0
A-B : x2=0 y2=0
x2=3,5 y2=-0,366
x2=7 y2=0
Destriawan Santoso (npm. 075310041)*ralat tugas*
B-C : x1=0 y1=0
x1=0,5 y1=-0,00003
x1=1 y1=0
A-B : x2=0 y2=0
x2=3,5 y2=-0,313
x2=7 y2=0
Anak Agung G. Eroviantara (npm. 0853310086)
A-B : x1=0 y1=1
x1=3,5 y1=0,0625
x1=7 y1=0
B-C : x2=0 y2=0
x2=3 y2=-0,2142
x2=6 y2=0
(posted by : mira, diana, iwan, & anak agung)
Mirawati (npm. 0753010037) *ralat tugas*
B-C :
x1=0 y1=0
x1=3,5 y1=-0,047
x1=7 y1=0
A-B :
x2=0 y2=1
x2=3,5 y2=0,359
x2=7 y2=0
luthfi efendy
npm 0653010044
Segmen B-A;0 < X < 7
X=0, Y=0
X=3, Y=0.599
X=5, Y=0.724
X=7, Y=1
Segmen C-B;0 < X < 11
X=0, Y=0
X2=3, Y=0.74
X2=5, Y=1.27
*kurnia cahya septian
*0653010001
Segmen B-A;0 < X < 7
X=0, Y=0
X=1, Y=-0.05
X=2, Y=-0.06
X=3, Y=-0.02
X=4, Y=0.8
X=5, Y=0.28
X=6, Y=0.58
X=7, Y=1
Segmen C-B;0 < X < 1
X=0, Y=0
X=0.5, Y=0.25
X=1, Y=0.5
wahyu L.P
( 0753210062 )
RB = 4.5 t ( ↑ )
RC = 3.5 t ( ↓ )
2. Segmen B – A → 0 < x1 < 7
Mx1 = 4.5x + 4.5
y’’= - = - = - ( 4.5x +4.5 )
Rotasi : y’ = ( - - + C1 )
Defleksi : y = ( - - + C1X + C2 )
3. Segmen C – B → 0 < x2 < 1
Mx2 = - 7X2
y’’= - = ( - )
Rotasi : y’ = ( + C3 )
Defleksi : y = ( - + C3X + C4 )
4. Syarat Batas
Titik B : X=0 ; y = 0 → C2 = 0
Titik C : X=0 ; y = 0 → C4 = 0
5. Defleksi di Titik A
7C1 – 2C3 = 185.32
C1 – C3 = 73.21
6. Subtitusi
C1 = 7.78
C3 = - 65.43
7. Faktor Pembagi
AA = -129.3/EI
GPRA
BA = y0 = -1/129.3
( - 4.5/12X3
- 4.5/4 X2+7.78X )
8. Defleksi Masing-masing titik
X1 = 3.5 m → y0 = 0.22
X2 = 7 m → y0 = 1
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