Trusses are lightweight structures capable of carrying heavy loads. In bridge design, the individual members of the truss are connected with rotatable pin joints that permit forces to be transferred from one member of the truss to another. The accompanying figure shows a truss that is held stationary at the lower left endpoint $(1),$ is permitted to move horizontally at the lower right endpoint $(4),$ and has pin joints at $(1),(2),(3),$ and $(4).$ A load of $10,000$ newtons $(N)$ is placed at joint $(3),$ and the resulting forces on the joints are given by $f_1,f_2,f_3,f_4,$ and $f_5,$ as shown in figure $1.$ When positive, these forces indicate tension on the truss elements, and when negative, compression. The stationary support member could have both a horizontal force component $F_1$ and a vertical force component $F_2,$ but the movable support member has only a vertical force component $F_3.$

Figure $1$: Diagrammatic Representation of the Truss

NUMERICAL ANALYSIS

ASSIGNMENT

MATH $351$

If the truss is in static equilibrium, the forces at each joint must add to the zero vector, so the sum of the horizontal and vertical components at each joint must be $0.$ This produces the system of linear equations shown in the accompanying table below:

$\backslash$table$[[$Joint$,$Horizontal Component,Vertical Component$],[(1),-F_1+\frac{\sqrt[\phantom{2}]{2}}{2}{f}_1+f_2=0,\frac{\sqrt[\phantom{2}]{2}}{2}{f}_1-F_2=0$