PDF

Tugas Analisis Struktur II Kevin Nathaniel / 15012109

SOAL 1

Gunakan program MSA.m untuk untuk menghitung distribusi gaya dalam (geser dan momen).

Lalu, hitung momen serta perpindahan maksimum pada masing-masing bentang pada problem

berikut.

SOAL 2

Lakukan modifikasi pada program MSA.m untuk menyelesaikan problem dengan beda suhu pada

permukaan balok berikut. Lalu, tentukan/gambar distribusi gaya dalam (geser dan momen) dan

hitung momen serta perpindahan maksimum pada masing-masing bentang.

SOAL 3

Lakukan modifikasi pada program MSA.m untuk menyelesaikan problem beban termal pada

rangka batang. Lalu, tentukan perpindahan, gaya batang dan reaksi perletakan dan hitung gayaserta

perpindahan maksimum yang terjadi.


(Anda akan diminta mendemostrasikan bahwa modifikasi program bekerja!)


Jawaban Soal 1 - Tugas 2

Berikut ditampilkan input data beam untuk soal 1 dengan program Matlab :

DataBeam.m

function D=DataBeam

% Units: KN & m

m=3;n=4;

Coord=[0 0;4 0;6 -3;7 (-3-tan(0.389*pi))];Coord(:,3)=0;

Con=[1 2 1 1;2 3 1 1;4 3 1 0];

Re=ones(n,6);Re(:,[1,2,6])=[1 1 1;0 0 0;0 0 0;1 1 0];

Load=zeros(n,6);Load(:,[1,2,6])=[0 0 0;-50 0 -80;0 0 0;0 0 0];

w=[0 -20 0;0 0 0;0 0 0];

E=[200000000 200000000 200000000];

nu=0.3;G=E/(2*(1+nu));

A=[0.06 0.06 1000];

Iz=ones(1,m)*0.0004;

Iy=ones(1,m)*0.0004;

J=ones(1,m)*1;

St=zeros(n,6);

be=zeros(1,m);

D=struct('m',m,'n',n,'Coord',Coord','Con',Con','Re',Re','Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',

J','St',St','be',be');

Clc

Berikut ditampilkan modifikasi program MSA pada Matlab :

MSA.m

function [Q,V,R]=MSA(D)

m=D.m;n=D.n;Ni=zeros(12,12,m);S=zeros(6*n);Pf=S(:,1);Q=zeros(12,m);Qfi=Q;Ei=Q;

for i=1:m

H=D.Con(:,i);C=D.Coord(:,H(2))-D.Coord(:,H(1));e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];c=D.be(i);

[a,b,L]=cart2sph(C(1),C(3),C(2));ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);

r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];T=kron(eye(4),r);

x=D.A(i)/L;g=D.G(i)*D.J(i)/(D.E(i)*L);ez=D.E(i)*D.Iz(i)/(D.A(i)*D.G(i));ey=D.E(i)*D.Iy(i)/(D.A(i)*D.G

(i));

z=D.Iz(i)/(L*(L^2/12+ez))*[1 L/2 (L^2/3+ez) (L^2/6-ez)];

y=D.Iy(i)/(L*(L^2/12+ey))*[1 L/2 (L^2/3+ey) (L^2/6-ey)];

K1=diag([x,z(1),y(1)]);K2=[0 0 0;0 0 z(2);0 -y(2) 0];K3=diag([g,y(3),z(3)]);K4=diag([-g,y(4),z(4)]);

K=D.E(i)*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];

w=D.w(:,i)';Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';Qfs=K*T*D.St(e)';

A=diag([0 -0.5 -0.5]);B(2,3)=1.5/L;B(3,2)=-1.5/L;W=diag([1,0,0]);Z=zeros(3);M=eye(12);p=4:6;q=10:12;

switch 2*H(3)+H(4)

case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];

end

K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);Qfi(:,i)=M*Qf;Pf(e)=Pf(e)+T'*M*(Qf+Qfs);Ei(:,i)=e;

end

V=1-(D.Re|D.St);f=find(V);V(f)=S(f,f)\(D.Load(f)-Pf(f));R=reshape(S*V(:)+Pf,6,n);R(f)=0;V=V+D.St;

for i=1:m

Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i);

end

clc


Berikut adalah output yang dihasilkan dari program tersebut :

Tabel Internal Forces

Tabel Defleksi

Elemen 1 Elemen 2 Elemen 3

Fx beginning 65.95341 45.35912 46.68942

Fy beginning 36.12072 -11.0658 -2.22E-15

Fz beginning 0 0 2.74E-22

Mx beginning 0 0 -1.04E-12

My beginning 0 0 -8.01E-22

Mz beginning 24.58457 -39.8983 -8.88E-16

Fx end -65.9534 -45.3591 -46.6894

Fy end 43.87928 11.06581 2.22E-15

Fz end 0 0 -2.74E-22

Mx end 0 0 1.04E-12

My end 0 0 0

Mz end -40.1017 0 0

Support 1 Support 2 Support 3 Support 4

Dx 0 -2.20E-05 0.000917 0

Dy 0 -0.00031 0.000333 0

Dz 0 0 0 0

Rx 0 0 0 0

Ry 0 0 0 0

Rz 0 -0.00028 0.000615 -0.00033


Tabel Reaksi Perletakan

Tabel Output Gaya Dalam


Fx support 65.95341 0 0 -15.9534

Fy support 36.12072 0 0 43.87928

Fz support 0 0 -1.95E-15 1.95E-15

Mx support 0 0 -3.57E-13 3.57E-13

My support 0 0 9.81E-13 -9.81E-13

Mz support 24.58457 0 0 0


Jawaban Soal 2 Tugas 2

Berikut ditampilkan input data beam untuk soal 1 dengan program Matlab :

DataBeam.m

function D=DataBeam

m=2;n=3;

Coord=[0 0;4 0;8 0];Coord(:,3)=0;

Con=[1 2 1 1;2 3 1 1];

Re=ones(n,6);Re(:,[1,2,6])=[1 1 1;0 0 0;1 1 1];

Load=zeros(n,6);

w=zeros(m,3);

E=[400000000 200000000];

nu=0.30;

G=E/(2*(1+nu));

A=ones(1,m)*0.02;

Iz=ones(1,m)*50e-6;

Iy=ones(1,m)*50e-6;

J=ones(1,m)*1000;

St=zeros(n,6);

be=zeros(1,m);

TTop=[40 0];

TBot=[25 0];

Tr=28;

Diameter=[0.182 0];

Alpha=12e-6;

D=struct('Diameter',Diameter','Tr',Tr,'TBot',TBot','TTop',TTop','Alpha',Alpha,'m',m,'n',n,'Coord',Coord','Co

n',Con','Re'Re','Load',Load','w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',J','St',St','be',be');

Clc


MSA.m function [Q,V,R]=MSA(D)

m=D.m;

n=D.n;

Ni=zeros(12,12,m);

S=zeros(6*n);

Pf=S(:,1);

Q=zeros(12,m);

Qfi=Q;

Ei=Q;

for i=1:m

H=D.Con(:,i);

C=D.Coord(:,H(2))-D.Coord(:,H(1));

e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];

c=D.be(i);

[a,b,L]=cart2sph(C(1),C(3),C(2));

ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);

r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];

T=kron(eye(4),r);

co=2*L*[6/L 3 2*L L];

x=D.A(i)*L^2;

y=D.Iy(i)*co;

z=D.Iz(i)*co;

g=D.G(i)*D.J(i)*L^2/D.E(i);

K1=diag([x,z(1),y(1)]);


K2=[0 0 0;0 0 z(2);0 -y(2) 0];

K3=diag([g,y(3),z(3)]);

K4=diag([-g,y(4),z(4)]);

K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];

w=D.w(:,i)';

Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';

Mt=(D.E(i,1)*D.Iz(i,1)*D.Alpha*(D.TTop(i,1)-D.TBot(i,1))/D.Diameter(i,1));

Pt=(D.E(i,1)*D.A(i,1)*D.Alpha*((0.5*(D.TTop(i,1)+D.TBot(i,1)))-D.Tr));

Pft=zeros(1,m);

Mft=zeros(1,m);

Pft(1,1)=Pt;

Mft(1,1)=Mt;

Qft=[Pft(1,i) 0 0 0 0 -Mft(1,i) -Pft(1,i) 0 0 0 0 Mft(1,i)]';

Qfs=K*T*D.St(e)';

A=diag([0 -0.5 -0.5]);

B(2,3)=1.5/L;B(3,2)=-1.5/L;

W=diag([1,0,0]);

Z=zeros(3);

M=eye(12);

p=4:6;

q=10:12;

switch 2*H(3)+H(4)

case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];end

K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);

Qfi(:,i)=M*(Qf+Qfs+Qft);

Pf(e)=Pf(e)+T'*M*(Qf+Qfs+Qft);

Ei(:,i)=e; end


for i=1:m

Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i); end


Berikut adalah output yang dihasilkan dari program tersebut :

Tabel Internal Forces

Tabel Defleksi


Elemen 1 Elemen 2

Fx beginning 144 144

Fy beginning -3.5964 -3.5964

Fz beginning 0 0

Mx beginning 0 0

My beginning 0 0

Mz beginning -23.3766 -8.99101

Fx end -144 -144

Fy end 3.596404 3.596404

Fz end 0 0

Mx end 0 0

My end 0 0

Mz end 8.991009 -5.39461

Support 1 Support 2 Support 3

Dx 0 0.000144 0

Dy 0 -0.00048 0

Dz 0 0 0

Rx 0 0 0

Ry 0 0 0

Rz 0 -0.00072 0

Support 1 Support 2 Support 3

Fx support 144 0 -144

Fy support -3.5964 0 3.596404

Fz support 0 0 0

Mx support 0 0 0

My support 0 0 0

Mz support -23.3766 0 -5.39461


Tabel Output Gaya Dalam


Jawaban Soal 2 Tugas 3

Berikut ditampilkan input data truss untuk soal 1 dengan program Matlab :

DataF2D.m

function D=DataF2D

% Satuan: kN & m - Beban Thermal

m=3;n=4;

Coord=[0 0;4 0;4 3;0 3];

Coord(:,3)=0;

Con=[1 3 1 0;2 3 1 0;3 4 0 1];

Re=ones(n,6);

Re(:,[1,2,6])=[1 1 0;1 1 0;0 0 0;1 1 0];

Load=zeros(n,6);Load(:,[1,2,6])=[0 0 0;0 0 0;-4 -8 0;0 0 0];

w=zeros(m,3);

E=ones(1,m)*200000000;

nu=0.3;G=E/(2*(1+nu));

A=[4e-5 4e-5 4e-5];

Iz=ones(1,m)*0.0004;

Iy=ones(1,m)*0.0004;

J=ones(1,m)*1000;

St=zeros(n,6);

St(1,2)=-0.0025;

be=zeros(1,m);

T1=zeros(1,m);T1(1,1)=48; %T1 pada member 1

T2=zeros(1,m);T2(1,1)=48; % T2 pada member 1

d=ones(1,m);d(1,1)=1; % d pada member 1

Tr=28; % suhu ruangan

alfa=12e-6; % koefisien muai material

D=struct('Tr',Tr,'d',d','T2',T2','T1',T1','alfa',alfa,'m',m,'n',n,'Coord',Coord','Con',Con','Re',Re','Load',Load',

'w',w','E',E','G',G','A',A','Iz',Iz','Iy',Iy','J',J','St',St','be',be');


MSA.m function [Q,V,R]=MSA(D)

m=D.m;

n=D.n;

Ni=zeros(12,12,m);

S=zeros(6*n);

Pf=S(:,1);

Q=zeros(12,m);

Qfi=Q;

Ei=Q;

for i=1:m

H=D.Con(:,i);

C=D.Coord(:,H(2))-D.Coord(:,H(1));

e=[6*H(1)-5:6*H(1),6*H(2)-5:6*H(2)];

c=D.be(i);

[a,b,L]=cart2sph(C(1),C(3),C(2));

ca=cos(a);sa=sin(a);cb=cos(b);sb=sin(b);cc=cos(c);sc=sin(c);

r=[1 0 0;0 cc sc;0 -sc cc]*[cb sb 0;-sb cb 0;0 0 1]*[ca 0 sa;0 1 0;-sa 0 ca];

T=kron(eye(4),r);

co=2*L*[6/L 3 2*L L];

x=D.A(i)*L^2;

y=D.Iy(i)*co;

z=D.Iz(i)*co;

g=D.G(i)*D.J(i)*L^2/D.E(i);


K1=diag([x,z(1),y(1)]);

K2=[0 0 0;0 0 z(2);0 -y(2) 0];

K3=diag([g,y(3),z(3)]);

K4=diag([-g,y(4),z(4)]);

K=D.E(i)/L^3*[K1 K2 -K1 K2;K2' K3 -K2' K4;-K1 -K2 K1 -K2;K2' K4 -K2' K3];

w=D.w(:,i)';

Qf=-L^2/12*[6*w/L 0 -w(3) w(2) 6*w/L 0 w(3) -w(2)]';

Mt=(D.E(i,1)*D.Iz(i,1)*D.Alpha*(D.TTop(i,1)-D.TBot(i,1))/D.Diameter(i,1));

Pt=(D.E(i,1)*D.A(i,1)*D.Alpha*((0.5*(D.TTop(i,1)+D.TBot(i,1)))-D.Tr));

Pft=zeros(1,m);

Mft=zeros(1,m);

Pft(1,1)=Pt;

Mft(1,1)=Mt;

Qft=[Pft(1,i) 0 0 0 0 -Mft(1,i) -Pft(1,i) 0 0 0 0 Mft(1,i)]';

Qfs=K*T*D.St(e)';

A=diag([0 -0.5 -0.5]);

B(2,3)=1.5/L;B(3,2)=-1.5/L;

W=diag([1,0,0]);

Z=zeros(3);

M=eye(12);

p=4:6;

q=10:12;

switch 2*H(3)+H(4)

case 0;B=2*B/3;M(:,[p,q])=[-B -B;W Z;B B;Z W];case 1;M(:,p)=[-B;W;B;A];case 2;M(:,q)=[-B;A;B;W];end

K=M*K;Ni(:,:,i)=K*T;S(e,e)=S(e,e)+T'*Ni(:,:,i);

Qfi(:,i)=M*(Qf+Qfs+Qft);

Pf(e)=Pf(e)+T'*M*(Qf+Qfs+Qft);

Ei(:,i)=e; end


for i=1:m

Q(:,i)=Ni(:,:,i)*V(Ei(:,i))+Qfi(:,i); end


Berikut adalah output yang ditampilkan dari program tersebut :

Tabel Internal Forces and Moments

Tabel Defleksi


Elemen 1 Elemen 2 Elemen 3

Fx beginning 0.47037 6.277778 1.703704

Fy beginning -3.84 3.93E-16 1.78E-15

Fz beginning 0 0 3.91E-16

Mx beginning 0 0 8.11E-10

My beginning 0 0 0

Mz beginning -19.2 2.91E-16 0

Fx end -0.47037 -6.27778 -1.7037

Fy end 3.84 -3.93E-16 -1.78E-15

Fz end 0 0 -3.91E-16

Mx end 0 0 -8.11E-10

My end 0 0 -1.56E-15

Mz end 0 0 0


Dx 0 0 -0.00085 0

Dy -0.0025 0 -0.00235 0

Dz 0 0 0 0

Rx 0 0 0 0

Ry 0 0 0 0

Rz 0.000126 0.000284 -0.00024 -0.00059


Fx support 2.296296 -8.76E-18 0 1.703704

Fy support 1.722222 6.277778 0 -1.78E-15

Fz support 0 0 -1.83E-16 1.83E-16

Mx support 0 0 -8.11E-10 8.11E-10

My support 0 0 0 -1.56E-15

Mz support 0 0 0 0

PDF

Documents

PDF 007.pdf

2357. kozhasznu02_3.pdf /14700/14714/pdf/kozhasznu02_3.pdf

7199. erdely3.pdf /06800/06860/pdf/erdely3.pdf

275. erkel2_02.pdf /08600/08689/pdf/erkel2_02.pdf

Pdf joni pdf

H20youryou[2] · 2020. 9. 1. · 65 pdf pdf xml xsd jpgis pdf ( ) pdf ( ) txt pdf jmp2.0 pdf xml xsd jpgis pdf ( ) pdf pdf ( ) pdf ( ) txt pdf pdf jmp2.0 jmp2.0 pdf xml xsd

ELEKTRICNI_STROJI_1.pdf .pdf

Biodecodificaciòn pdf..pdf

IMG 0085.pdf IMG 0084.pdf IMG 0083.pdf IMG 0082.pdf IMG

book1631.pdf pdf

Pdf ed176 pdf

GRANTEE - Waupaca County, WisconsinGTE_A_1851-1867.pdf GTE_A_1957.pdf GTE_B_1917-1930.pdf ... GTE_A_1873-1876.pdf GTE_A_1965-1971.pdf GTE_B_1943-1948.pdf GTE_A_1876-1885.pdf GTE_A_1971.pdf

skripsi PDF 4.pdf

Foral1.pdf · PDF file

Pdf ed177 pdf

Pdf ed168 pdf

VAIO - OoCities...DSC-P9.pdf DSC-P92.pdf DSC-S30.PDF DSC-S40.pdf DSC-S50.PDF DSC-S60.pdf DSC-S70.pdf DSC-S75.pdf DSC-S80.pdf DSC-S85.pdf DSC-S90.pdf DSC-T1.pdf DSC-T3.pdf DSC-T33.pdf

PDF in Theorie und Praxis - qucosa.de · Inhalt Quellen Was ist PDF? PDF-Struktur PDF erstellen PDF analysieren PDF bearbeiten PDF drucken und Troubleshooting Rechtliches Spezialthemen

Pdf ed166 pdf

Pdf ed169 pdf