**Example 1:**Before setting up the equilibrium equations to solve for the support reactions, an appropriate free-body diagram of the structure must be drawn to identify all the known and unknown forces acting on the structure. We should keep in mind that if a structure is in the state of equilibrium, then every part of that structure has to be in equilibrium as well. Therefore, if the free-body diagram of the whole or part of the structure is drawn correctly, then it should demonstrate the condition of equilibrium.

For a truss structure, we are normally interested in two separate free-body diagrams. The free-body diagram of the truss as a whole is used for the calculation of reactions, and another set of free-body diagrams is used for the calculation of member forces. We will discuss the latter case in a later section. Let us now develop the free body diagram of a truss for the calculation of its support reactions.

The nine-bar truss shown below is supported at joints 1 and 3 and is loaded at joints 4, 5, and 6.

**Solution:**Since we are interested in calculating the support reactions, it is not necessary to show all the internal details of the truss in the free-body diagram. To demonstrate this fact, the internal members are shown in light gray color in the free-body diagram.

Joint 1 is pinned. Thus, its movements in x and y directions are restrained. Joint 3 is supported by a roller in the horizontal plane, which restrains its movement only in the y direction. Consistent with this fact, the free-body diagram of the truss shows two reactions at joint 1 and one reaction at joint 3.

It is a good practice to show the reactions in the positive x and y directions. With the help of the equilibrium equations, we solve for the unknown reactions. If a reaction force is found to be a negative number, then its true direction is opposite to that shown in the diagram.

The equilibrium of forces in the x direction is expressed as

S F_{x} = 0 |
=> | R_{1-x} + F_{4-x} = 0 |
=> | R_{1-x} = - F_{4-x} |

Thus R_{1-x} is in the negative x direction because of the negative sign in front of its magnitude.

Using the counter-clockwise rotation as positive, the equilibrium of moments about joint 1 is expressed as

S M_{1} = 0 |
=> | R_{3-y} (x_{3} - x_{1}) - F_{4-y}(x_{4} - x_{1}) - F_{5-y}(x_{5} - x_{1}) - F_{6-y}(x_{
6} - x_{1}) = 0 |

=> | R_{3-y} = [F_{4-y}(x_{4} - x_{1}) + F_{5-y}(x_{5} - x_{1}) + F_{6-y}(x_{6} - x_{1})] / (x_{3} ñ x_{1}) |

The positive value of R_{3-y} indicates the assumed direction is correct. The equilibrium of forces in the y direction is written as

S F_{} = 0 |
=> | R_{1-} + R_{3-y} - F_{4-y} - F_{5-y} - F_{6-y} = 0 |

=> | R_{1-y} = - R_{3-y} + F_{4-y} + F_{5-y} + F_{6-y} |

Once again, the positive value of R_{1-y} indicates the assumed direction is correct.