Question
Please clear hand writing and matlab codes
Derive the Taylor series expansions for f(x- 3 Delta x) and f(x + 4 Delta x) with base point x. Include the first four terms. Substitute the expansions obtained in (a) into 9 f (x + 4 Delta x) + 7 f (x) -16 f(x - 3 Delta x) / a Delta x and show that the above expression can be used to approximate the first derivative f'(x) if Delta x is small, AND determine the constant a. A bridge is modeled by a rigid horizontal bar supported by three elastic vertical columns as shown. A force P = 40 kN, which represents a car on the bridge, applied to the rigid bar at a distance d from the end of the bar The forces in columns FAB, FCD, and FFG can be determined from the solution of the following system of equation: FAB + FCD + FFG = -P, 10FAB + 28FCD + 40FFG = -d middot P, 12FAB LAB - 30FCDLCD +18FFGLFG = 0 where L, E, and A, denote the length, elastic modulus, and cross-sectional area, respectively, of the bars. MATLAB Write a MATLAB script file that determines the forces in the three bars for 0 d 50 m. The program displays the three forces as a function of d in one plot. Assume that LAB = 12 m, LCD = 8m, LFG = 10 m, E = 70 GPa and A = 25(10-4)m2.. What is the absolute maximum axial force when the car drives from the left to the right end? In which column?
Explanation / Answer
%Program to analyze trusses with supports and forces specified. %Specified support displacements are not allowed. clear %*** INPUT STARTS HERE *** %Specify structure properties nnodes=4; %specify the number of nodes nmem=5; %specify the number of members %Define nodal coordinates in order starting with node one x(1)=4; y(1)=3; x(2)=4; y(2)=0; x(3)=8; y(3)=0; x(4)=0; y(4)=0; %Define member connectivity start with member one % Start node end node mconn(1,1)=4; mconn(1,2)=1; mconn(2,1)=1; mconn(2,2)=3; mconn(3,1)=2; mconn(3,2)=3; mconn(4,1)=4; mconn(4,2)=2; mconn(5,1)=2; mconn(5,2)=1; %input properties for each member %For now E is constant. E=200*10^6; %A for each member A(1)=0.0015; A(2)=0.0015; A(3)=0.0015; A(4)=0.0015; A(5)=0.0015; %Specify supports nsupp=3; %number of degrees of freedom with supports sup(1)=7;%for the 1 to nsupp indicate which dof are supports sup(2)=8; sup(3)=6; %sup(4)=6; %sup(5)=7; %sup(6)=8; %Specify forces %zero all the forces for i=1:nnodes*2; F(i)=0.0; end; nforce=5; %number of degrees of freedom with forces given F(1)=0; %specify force at each degree of freedom where the external force is known, F(dof) F(2)=0; F(3)=0; F(4)=-10; F(5)=0 %***INPUT ENDS HERE *** %Calculate the truss member lengths, and angles for i=1:nmem; dx=x(mconn(i,2))-x(mconn(i,1)); dy=y(mconn(i,2))-y(mconn(i,1)); L(i)=sqrt(dx^2+dy^2); c(i)=dx/L(i); s(i)=dy/L(i); end; %For each element create the global stiffness matrix and put it into the %global stiffness matrix. %zero the global stiffness matrix for i=1:nnodes*2; for j=1:nnodes*2; kg(i,j)=0.0; end; end; %Create each element global stiffness matrix as transpose[T][k][T] for m=1:nmem; %loop over each element %zero k and T for i=1:4; for j=1:4; k(i,j)=0.0; T(i,j)=0.0; end; end; %create T T(1,1)=c(m) ; T(1,2)=s(m); T(1,3)=0; T(1,4)=0; T(2,1)=-s(m); T(2,2)=c(m); T(2,3)=0; T(2,4)=0; T(3,1)=0; T(3,2)=0; T(3,3)=c(m); T(3,4)=s(m); T(4,1)=0; T(4,2)=0; T(4,3)=-s(m); T(4,4)=c(m); %create k k(1,1)=1.0; k(1,3)=-1.0; k(3,1)=-1.0; k(3,3)=1.0; k=A(m)*E/L(m)*k; %transform k into the element global stiffness matrix k=T'*k*T; %put each element k into the global stiffness matrix kg for i=1:2; for j=1:2; kg(mconn(m,i)*2-1,mconn(m,j)*2-1)=kg(mconn(m,i)*2-1,mconn(m,j)*2-1)+k(i*2-1,j*2-1); kg(mconn(m,i)*2,mconn(m,j)*2)=kg(mconn(m,i)*2,mconn(m,j)*2)+k(i*2,j*2); kg(mconn(m,i)*2-1,mconn(m,j)*2)=kg(mconn(m,i)*2-1,mconn(m,j)*2)+k(i*2-1,j*2); kg(mconn(m,i)*2,mconn(m,j)*2-1)=kg(mconn(m,i)*2,mconn(m,j)*2-1)+k(i*2,j*2-1); end; end; %now kg should be complete end; %end loop over all the elements %Put kg into kgs and modify kgs to solve for the unknown displacements %we save kg so that we can calculate the reactions also. kgs=kg; %modify kgs based on the support conditions. Zero the entire row for each %dof sup(i) except for entry kgs(sup(i),sup(i)) set it equal to 1. When we solve %for the displacements we have essentially set them equal to zero since %they are supports. for i=1:nsupp; %loop through all the specified supports for j=1:nnodes*2; kgs(sup(i),j)=0.0; %zero the whole row end; kgs(sup(i),sup(i))=1.0; %put a 1 in position kgs(sup(i),sup(i)) end; kgs %solve for the global nodal displacements display('The global nodal displacements are:') U=inv(kgs)*F' %Solve for the global external forces, here we use kg that we set aside %above. display('The global external forces are:') F=kg*U %Calculate the forces for each member for m=1:nmem; %loop over each element %zero k and T for i=1:4; for j=1:4; k(i,j)=0.0; T(i,j)=0.0; end; end; %create T T(1,1)=c(m) ; T(1,2)=s(m); T(1,3)=0; T(1,4)=0; T(2,1)=-s(m); T(2,2)=c(m); T(2,3)=0; T(2,4)=0; T(3,1)=0; T(3,2)=0; T(3,3)=c(m); T(3,4)=s(m); T(4,1)=0; T(4,2)=0; T(4,3)=-s(m); T(4,4)=c(m); %create k k(1,1)=1.0; k(1,3)=-1.0; k(3,1)=-1.0; k(3,3)=1.0; k=A(m)*E/L(m)*k; %Extract the global displacements associated with element m u(1)=U(mconn(m,1)*2-1); u(2)=U(mconn(m,1)*2); u(3)=U(mconn(m,2)*2-1); u(4)=U(mconn(m,2)*2); %calculate the local element end forces f=k*T*u'; %save the important end forces fm(m)=f(3); %positive values mean truss member in tension, negative means compression end;%end loop over each element display('The member forces are:') fm' %plot the truss on a graph plot(0,0,'w') hold %loop over all the elements and plot each element, one at a time magnify=2000; %magnification factor for deflections for m=1:nmem; xm(1)=x(mconn(m,1)); ym(1)=y(mconn(m,1)); xm(2)=x(mconn(m,2)); ym(2)=y(mconn(m,2)); xu(1)=U(mconn(m,1)*2-1)*magnify; yu(1)=U(mconn(m,1)*2)*magnify; xu(2)=U(mconn(m,2)*2-1)*magnify; yu(2)=U(mconn(m,2)*2)*magnify; if fm(m) < 0.0; plot(xm,ym,'k'); %undisplaced truss members in compression will be blue else; plot(xm,ym,'b'); %undisplaced truss members in tension will be red end; plot(xm+xu,ym+yu,'r--'); %displaced truss members end; xlabel('xcoordinates') ylabel('ycoordinates') title('Undeflected(solid) and Magnified Deflected(dashed) Truss') hold