Analysis summery
The truss that we decided to use was a Warren Truss. Warren trusses are symmetrical and can be identified by their composition of equilateral or isosceles triangles. Our truss was made entirely of equilateral triangles with sides of 15 inches. We chose this design because from our analysis of the equilateral Warren truss we found that the top and side members would fail simultaneously. After further analysis we could not find any other way of maximizing our Warren truss without adding further welds.
Another reason that we chose the Warren truss was because of the small number of welds. A big percentage of our performance index depended on the number of welds that your truss required. We referred to this as the “cost” of the truss because the more welds that the truss had, the more time and materials it would take to make the truss. Another benefit from the Warren truss was its simplicity. It was much easier to construct an equilateral triangle than it would have been to be cutting members of all different lengths, and then assemble them perfectly to try and achieve the desired angles.
Analysis code
restart;eq1:=0=Ay+By-F;
> eq2:=0=Ax;
> eq3:=0=By*30-F*15;
> eq4:=0=-Tad*sin(60/180*Pi)-Tac*sin(60/180*Pi);
> eq5:=0=-Tac*cos(60/180*Pi)+Tad*cos(60/180*Pi)+Tab;
> eq6:=0=Ay+Tac*sin(60/180*Pi);
> eq7:=0=Ax+Tcd+Tac*cos(60/180*Pi);
This is an arbitrary force that we chose just to see which member(s) will be most likely to fail
>
> eq8:=F=372;
> solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8},{Ay,By,F,Ax,Tad,Tac,Tab,Tcd});
eq1 := 0 = Ay + By - F
eq2 := 0 = Ax
eq3 := 0 = 30 By - 15 F
1/2 1/2
Tad 3 Tac 3
eq4 := 0 = - -------- - --------
2 2
Tac Tad
eq5 := 0 = - --- + --- + Tab
2 2
1/2
Tac 3
eq6 := 0 = Ay + --------
2
Tac
eq7 := 0 = Ax + Tcd + ---
2
eq8 := F = 372
1/2 1/2 1/2
{Ax = 0, Ay = 186, Tcd = 62 3 , Tab = -124 3 , Tad = 124 3 ,
1/2
Tac = -124 3 , F = 372, By = 186}
The piece most likely to fail will be AC, AB or BE.
From the Compression failure equation we determine Tca to have a max compression of about 37.2 pounds before failing. We'll plug this back into the previous series of equations to find the max F.
> Cf:=(Pi^2*29*10^6*(1/4)*Pi*((5/32)/2)^4)/(15^2);
> evalf(%);
3
11328125 Pi
Cf := ------------
9437184
37.21904522
> restart;
> eq1:=0=Ay+By-F;
> eq2:=0=Ax;
> eq3:=0=By*30-F*15;
> eq4:=0=-Tad*sin(60/180*Pi)-Tac*sin(60/180*Pi);
> eq5:=0=-Tac*cos(60/180*Pi)+Tad*cos(60/180*Pi)+Tab;
> eq6:=0=Ay+Tac*sin(60/180*Pi);
> eq7:=0=Ax+Tcd+Tac*cos(60/180*Pi);
Now insert the force that member AC will fail at.
>
> eq8:=Tac=-37.21904522;
> solve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8},{Ay,By,F,Ax,Tad,Tac,Tab,Tcd});
eq1 := 0 = Ay + By - F
eq2 := 0 = Ax
eq3 := 0 = 30 By - 15 F
1/2 1/2
Tad 3 Tac 3
eq4 := 0 = - -------- - --------
2 2
Tac Tad
eq5 := 0 = - --- + --- + Tab
2 2
1/2
Tac 3
eq6 := 0 = Ay + --------
2
Tac
eq7 := 0 = Ax + Tcd + ---
2
eq8 := Tac = -37.21904522
{Tac = -37.21904522, Ax = 0., Tad = 37.21904522, Tcd = 18.60952261,
Ay = 32.23263867, By = 32.23263867, F = 64.46527733,
Tab = -37.21904522}
> Low:=64.446527733*2;
> High:=Low*4;
>
>
Low := 128.8930555
High := 515.5722220
The range in which the member CA, AB, or BE will buckle is from 128.89 lbs to 515.57 lbs.
For 5 welds (the number on one side)
At worst our performance index would be
>
> ((128.8930555/500)+(0))/((10/2)+(270.89/288));
0.04339402297
At best our performance index would be
>
> ((515.5722220/500)+(1))/((10/2)+(270.89/288));
0.3419095324
This page by Micheal Minor