
STATICS  CASE STUDY SOLUTION



Support

Free Body Diagram (FBD) of Truss 

When solving any truss, the reaction forces at each support should be determined before trying to calculate individual member loads. Starting with a FBD of the entire truss, the basic moment equilibrium equation gives,
ΣM_{1} = 0
2.5(4/3 + 4) + 5(4/3 + 4) + 7.5(4/3 + 4)  10R = 0
R = 8.0 kips
Due to symmetry, the reaction forces at both the left and right wall
are equal,
R_{L} = R_{R} = R = 8 kips 





Joint 1

Joint 1 

With the reactions known, it is now possible to analyze joint 1. Note
that both F_{31}
and F_{21} are drawn away from the joint, representing tension.
The two unknowns forces, F_{31} and F_{21},
can be found using the two force equilibrium equations,
ΣF_{x}
= 0
 F_{21}  F_{31} cos40
= 0
ΣF_{y}
= 0
F_{31} sin40 + R = 0
Substituting known values and solving gives,
F_{31} = 8/sin40 = 12.45 kips
F_{21} = F_{31} cos40
= 9.534 kips 





Joint 2

Joint 2 

The value of F_{21}, 9.534 kips, was determined at joint 1 and
can now be used at joint 2.
The force in member F_{32} is easy to determine since all members
and loads are perpendicular. Summing the forces in the vertical gives,
ΣF_{y}
= 0
F_{32} = 1.333 kips
Since there are only two horizontal member forces, they must be equal,
ΣF_{x}
= 0
F_{21} = F_{42} = 9.534 kips 





Joint 3

Joint 3 

The value of F_{31}, 12.45 kips, and F_{32},
1.333 kips, were determined from joint 1 and joint 2, respectively. It
is now possible to determine forces, F_{53} and
F_{43},
at joint 3,
ΣF_{x}
= 0 = cos40 [F_{53}  F_{43} + (12.45)]
F_{53} = F_{43}  12.45
ΣF_{y} = 0 = sin40 [F_{53}  F_{43}  (12.45)]  4  1.333
Eliminating F_{53} gives,
0 = 2F_{43}  4/sin40
 1.333/sin40
F_{43} = 4.148 kips
and
F_{53} = 4.148  12.45
= 8.302 kips 





Joint 4

Joint 4 

Now that F_{42}, 9.534 kips, and F_{43},
4.148 kips, are known, joint 4 can be analyzed. Due to symmetry, the
left member forces are the same as the right member forces. There is
only
one unknown force, F_{54}. Using ΣF_{y} =
0 gives,
F_{54} + 2(4.148) sin40  1.333 = 0
F_{54} = 6.666 kips 





Summary

Final Loads (kips) in Truss Members 

As shown by the force diagram, the largest tensile force is 9.534 kips
and the largest compressive force is 12.45 kips.
The best truss angle, θ, that minimizes the member loads can be determined by using the
truss simulation. The best angle is the largest angle possible in the simulator, 65^{o}. Thus
a steep truss reduces the load. However, the cost is higher and buckling
problems will also need to be considered.




