How to Conduct a Static Analysis of a Truss?

[littlemac1]

Below is a sketch of a truss in which external support reactions are given as R1 = 0 kN, R2 = 3.5 kN, R3 = 4.5 kN. Members (M) are labeled in boxes, joints labeled in circles, and forces R.

Draw 2 Free body diagrams, one for each joints C and E.
Define what direction represents a positive force, and write equations for all forces acting on that joint.

View attachment 9138

For example, for joint D it's:
("SUM" is used for summation symbol)
SUMF_y = 0
SUMF_y = F4 sin(30) + R3 = 0
SUMF_y = F4(1/2) + 4.5 kN = 0
F4 = -9kN
View attachment 9139

 

[jvicens]

I would like to provide some feedback and additional information on the truss sketch and the requested free body diagrams.

Firstly, it is important to note that trusses are commonly used in structural engineering to support loads and distribute forces. They are composed of interconnected members, which are typically straight and connected at joints. In the sketch provided, the truss is supported by external reactions R1, R2, and R3 at joints A, B, and F respectively.

To accurately draw the free body diagrams for joints C and E, some assumptions need to be made about the truss. For the purpose of this response, I will assume that the truss is in equilibrium and that all members are in tension.

For joint C, the positive force direction will be upwards, as indicated by the arrow in the sketch. The equations for all forces acting on joint C can be written as:

SUMFy = 0
SUMFy = F2 cos(30) + F5 = 0
SUMFy = F2(√3/2) + F5 = 0
F5 = -F2(√3/2)

SUMFx = 0
SUMFx = R1 + F2 sin(30) = 0
SUMFx = 0 + F2(1/2) = 0
F2 = 0

Therefore, the force acting on member F2 is zero and the force acting on member F5 can be calculated as F5 = -√3/2 F2 = 0.

For joint E, the positive force direction will be downwards, as indicated by the arrow in the sketch. The equations for all forces acting on joint E can be written as:

SUMFy = 0
SUMFy = F4 sin(30) + F6 = 0
SUMFy = F4(1/2) + F6 = 0
F6 = -F4(1/2)

SUMFx = 0
SUMFx = R3 + F4 cos(30) = 0
SUMFx = 4.5 + F4(√3/2) = 0
F4 = -4.5/√3

Therefore, the force acting on member F4 can be calculated as F4 = -4.5/√3 kN and the force acting on member F