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

Documents

PDF