User:EML4507.s13.team4ever.Tobin/R.1.6



BC is a zero force member D= 2 in σ=F/A=F/(πr^2 )

Summing the moments about joint A and solving for the reaction force at joint F $$ \sum M_a= -F_{fy} (36)+400(9)-1200(24)=0 $$ $$ F_{fy} (36)=32400$$ $$ F_{fy}=900 lb$$

Now we can solve for the reaction forces at joint A $$ \sum F_y= F_{ay}-1200+900=0$$ $$ F_{ay}=300 lb $$ $$ \sum F_x= F_{ax}-400=0$$ $$ F_{ax}=400 lb$$

Making a FBD of joint A,F,D,E and summing forces in the x and y direction we can solve for the axial forces in the two force members. Using this force and the known cross sectional area we can solve for the stiffness.

Joint A FBD: $$ \sum F_x = -400 + F_{ac} - F_{ab} \frac{12}{15} = 0$$ $$ \sum F_y = 300 - F_{ab} \frac{9}{15} = 0$$ $$ F_{ab} =300 \frac{9}{15} =500 lb$$ $$ F_{ac} = 500 \frac{12}{15} +400 = 800$$ $$ \sigma_{ab} = \frac{500lb}{\pi(1in)^2}=159.1549 \frac{lb}{in^2}= 159.2psi$$ $$ \sigma_{ac} =\frac{800}{\pi(1in)^2} =254.6479psi$$

Joint F FBD: $$ \Sigma F_x = -F_{ef} +F_{df} \frac{12}{15} = 0$$ $$ \Sigma F_y = 900 - F_{df}\frac{9}{15} = 0$$ $$ F_{df} =900\frac{15}{9} =1500lb$$ $$ F_{ef}=1500\frac{12}{15} =1200lb$$ $$ \sigma_{df} =\frac{1500lb}{\pi(1in)^2}=477.5psi$$ $$ \sigma_{ef}=\frac{1200}{\pi(1in)^2}=382psi$$

Joint D FBD: $$ \Sigma F_x = 400-F_{bd} -1500\frac{12}{15} = 0$$ $$ \Sigma F_y = -F_{de}+1500\frac{9}{15} = 0$$ $$ F_{bd} =1500\frac{12}{15}-400 =800lb$$ $$ F_{de} =1500\frac{9}{15} =900lb$$ $$ \sigma_{bd}=\frac{800lb}{\pi(1in)^2}=254.6psi $$ $$ \sigma_{de} =\frac{900lb}{\pi(1in)^2}=286.7psi$$

Joint E FBD: $$ \Sigma F_x = 1200 -F_{ce} -F_{be}\frac{12}{15} = 0$$ $$ \Sigma F_y =900 -1200 +F_{be}\frac{9}{15}=0$$ $$ F_{be} =(-900+1200)\frac{15}{9} =500lb$$ $$ F_{ce} =-500\frac{12}{15} +1200 =800lb$$ $$ \sigma_{be} = \frac{500lb}{\pi(1in)^2} = 159.2psi$$ $$ \sigma_{ce} = \frac{800}{\pi(1in)^2}=254.6psi$$

Array showing all finalized answers $$ \begin{bmatrix} Member & \sigma (psi) & \\ AB &159.2 &Compression \\AC &254.6 &Tension \\BD &254.6 &Tension \\BE &159.2 &Tension \\BC &0 &n/a \\ CE &254.6  &Tension \\DF &477.5 &Compression \\DE &286.7 &Tension \\EF &382.0 &Tension \end{bmatrix}$$

The following MATLAB code automatically solves for the same values as above. This code verifies our hand calculations were correct.

Coord=[0 0;12 0;24 0;36 0;12 9;24 9]; Dof=[1 2;3 4;5 6;7 8;9 10;11 12]; Edof=[1 1 2 3 4;2 3 4 5 6;3 5 6 7 8;4 1 2 9 10;5 9 10 3 4;6 9 10 5 6;7 9 10 11 12;8 11 12 5 6;9 11 12 7 8]; [Ex,Ey]=coordxtr(Edof,Coord,Dof,2); K=zeros(12); F=zeros(12,1); F(6)=-1200;F(11)=400; ep=[10000 pi]; for i=1:9 Ke=bar2e(Ex(i,:),Ey(i,:),ep); K=assem(Edof(i,:),K,Ke); end bc=[1 0;2 0;8 0]; Q=solveq(K,F,bc); Ed=extract(Edof,Q); for i=1:9 N(i)=bar2s(Ex(i,:),Ey(i,:),ep,Ed(i,:)); end eldraw2(Ex,Ey); plotpar=[1 3 0];scale=1; eldisp2(Ex,Ey,Ed,plotpar,scale); r=1; A=pi*r^2; stressAB=N(4)/A stressAC=N(1)/A stressBD=N(7)/A stressBE=N(6)/A stressBC=0 stressCE=N(2)/A stressDF=N(9)/A stressDE=N(8)/A stressEF=N(3)/A The negative and positive terms denote compression and tension respectively. stressAB = -159.1549 stressAC = 254.6479 stressBD = 254.6479 stressBE = 159.1549 stressBC =    0 stressCE = 254.6479 stressDF = -477.4648 stressDE = 286.4789 stressEF = 381.9719