18
45 + N 40 NO. REGIS DATA PENAMPANG 5 1 P = 115 L = 20 2 P = 60 L = 40 3 T = 50 A = 35 KOORDINAT CA 80 70 60 48.3333 3 1 2 3 1 2

Tugas 1 Mekanika Bahan

Embed Size (px)

DESCRIPTION

MEKBAN

Citation preview

BESARAN PENAMPANG

20

30 + N

25

45 + N4025

NO. REGISDATA PENAMPANG

51P=115L=202P=60L=403T=50A=35 nKOORDINAT CARTESIUS

807060

48.333333333330

25

033.3350709011557.5LUAS PENAMPANG

F1=115x20=2300CM2F2=60x40=2400CM2F3=1/2x35x50=875CM2+F=5575CM2TITIK KOORDINAT

O1(X1,Y1)X1=0+115/2 =57.5;Y1=60+80/2 =70O1=57.5;70O2(X2,Y2)X2=50+90/2 =70;Y2=0+60/2 =30O2=70;30O3(X3,Y3)X3=2/3x50=33.3333333333;Y3=35x2/3 + 25=48.3333333333O3=33.3333333333;48.3333333333TITIK BERAT PENAMPANG

XS=(F1 x X1) + ( F2 x X2 ) + ( F3 x X3 )=2300x57.5+2400x70+875x33.3333333333F5575=132250+168000+29166.66666666675575=329416.6666666675575=59.0881913303YS=(F1 x Y1) + ( F2 x Y2 ) + ( F3 x Y3 )=2300x70+2400x30+875x48.3333333333F5575=161000+72000+42291.66666666675575=275291.666666667 5575=49.379671151OS(XS,YS)59.0881913303;49.379671151

3

123

12

BESARAN INERSIAMOMEN INERSIA PENAMPANG MAJEMUK

MOMEN INERSIA TERHADAP SUMBU - X ( Ix )Ix1=1/12 x b1 x h13 +ay12F1

=1x115x20+ys - y1x230012=1x115x20+49.4-70x2300=1054622cm412

Ix2=1/12 x b2 x h23 +ay22F2

=1x40x60+ys - y2x240012=1x40x60+49.4-30x2400=1621372cm412

Ix3=1/36 x b3 x h33 +ay32F3

=1x35x50+ys - y3x87536=1x35x50+49.4-48.3333333333x875=122486cm436+2798480cm4

MOMEN INERSIA TERHADAP SUMBU - Y ( Iy )Iy1=1/12 x h1 x b13 +ax12F1

=1x20x115+xs - x1x230012=1x20x115+59.1-57.5x2300=2540593cm412

Iy2=1/12 x h2 x b23 +ax22F2

=1x60x40+xs - x2x240012=1x60x40+59.1-70x2400=605762cm412 Iy3=1/36 x h3 x b33 +ax32F3

=1x50x35+xs - x3x87536=1x50x35+59.1-33.3333333333x875=639947cm436+3786302cm4

MOMEN INERSIA POLAR ( Ip )

Ip=Ix+Iy

=2798480+3786302

=6584782cm4

JARI - JARI INERSIA ( ix ; iy ), JARI2 INERSIA POLAR ( ip )

ix=Ix=2798480=22.405CMF5575

iy=Iy=3786302=26.061CMF5575

ip=Ip=6584782=34.368CMF5575

323323323323323323

Iu,Iv,Zuv,IpMENGHITUNG ( Iu, Iv , Zuv , Ip )

JIKA SUMBU CARTESIUS DIPUTAR SEBESAR=30

Dari data di atas dapat ditentukan=30Zxy=Zoxy+a.b.Fa1=X1-Xs=57.5-59.0881913303=-1.5881913303cma2=X2-Xs=70-59.0881913303=10.9118086697cma3=X3-Xs=33.3333333333-59.0881913303=-25.754857997cmb1=Y1-Ys=70-49.379671151=20.620328849cmb2=Y2-Ys=30-49.379671151=-19.379671151cmb3=Y3-Ys=48.3333333333-49.379671151=-1.046cm

Zxy1=Zoxy1+a1.b1.F1;Zoxy1=0=0+-1.5881913303x20.620328849x2300=-75322.7632657895cm4Zxy2=Zoxy2+a2.b2.F2;Zoxy2=0=0+10.9118086697x-19.380x2400=-507521.43283262cm4Zxy3=Zoxy3+a3.b3.F3;Zoxy3=0=0+-25.754857997x-1.046x875=23579.7466714037cm4Zxy=-559264.449cm4

SEHINGGA:Iu=1Ix + Iy+1Ix - IyCos 2-Zxy.Sin 222=16584782+12798480-3786302Cos2x30--559264.449427005Sin2x3022=3292391.08432985+470407.617-170469.744

=3592328.95749908cm4

Iv=1Ix + Iy-1Ix - IyCos 2+Zxy.Sin 222=16584782-12798480-3786302Cos2x30+-559264.449427005Sin2x3022=3292391.08432985-470407.617+170469.744

=2992453.21116061cm4

Zuv=1Ix - IyCos 2+Zxy.Cos 22=12798480-3786302Cos2x30+-559264.449427005cos2x302=470407.617359474+532650.721119016

=1003058.338cm4

Ip=Iu+Iv

=3592328.957+2992453.211

=6584782cm4

O

M.I SUMBU UTAMA ( ANALITIS )MOMEN INERSIA SUMBU UTAMA ( ANALITIS )

Imax=1Ix + Iy+1Ix - Iy+4xZxy2IX-IY222798480-3786302=-987823

=16584782+1-987823+4x-559264.44942700522=3292391.08432985+746140.190

=4038531cm4

Imin=1Ix + Iy-1Ix - Iy+4xZxy222

=16584782-1-987823+4x-559264.44942700522=3292391.08432985-746140.190

=2546251cm4

=arc tan-2xZxy =-0.424x180Ix-Iy2=arc tan-2x-559264.449427005=-24.288-987822.7772=arc tan1118528.89885401-987822.7772=arc tan-1.132

2=-0.847=-0.424Rad

2

o

222222