Medidas Arancelarias y no Arancelarias

Equilibrium equations are of fundamental importance and necessary as a basic tool for structural analysis. Equilibrium equations relate three basic static quantities, i.e. applied loads, support reactions, and the internal force, by means of the conservation of the linear momentum and the angular momentum of the structure that is in equilibrium.

The necessary and sufficient condition for the structure to be in equilibrium is that the resultant of all forces and moments acting on the entire structure and any part of the structure vanishes. For three-dimensional structures, this condition generates six independent equilibrium equations for each part of the structure considered: three equations associated with the vanishing of force resultants in each coordinate direction and the other three equations corresponding to the vanishing of moment resultants in each coordinate direction. These six equilibrium equations can be expressed in a mathematical form as

0 ΣF ; 0 ΣF ; 0 ΣF_X  _Y  _Z  (1.15) 0 ΣM ; 0 ΣM ; 0 ΣM_AX  _AY  _AZ  (1.16) where {O; X, Y, Z} denotes the reference Cartesian coordinate system with origin at a point O and A denotes a reference point used for computing the moment resultants.

  DISP LACEMENT   METHO D  

Applied Loads & Support Reactions

(Known and unknown)

Internal Forces

(Unknown)

Deformation

(Unknown)

Displacement & Rotation

(Known and unknown)

Static Equilibrium Constitutive Law Kinematics   FORCE     ME TH O D  

Copyright © 2011 J. Rungamornrat

For two-dimensional or plane structures (which are the main focus of this text), there are only three independent equilibrium equations: two equations associated with the vanishing of force resultants in two directions defining the plane of the structure and one associated with the vanishing of moment resultants in the direction normal to the plane of the structure. The other three equilibrium equations are satisfied automatically. If the X-Y plane is the plane of the structure, such three equilibrium equations can be expressed as

0 ΣM ; 0 ΣF ; 0 ΣF_X  _Y  _AZ  (1.17)

It is important to emphasize that the reference point A can be chosen arbitrarily and it can be either within or outside the structure. According to this aspect, it seems that moment equilibrium equations can be generated as many as we need by changing only the reference point A. But the fact is these generated equilibrium equations are not independent of (1.15) and (1.16) and they can in fact be expressed in terms of a linear combination of (1.15) and (1.16). As a result, this set of additional moment equilibrium equations cannot be considered as a new set of equations and the number of independent equilibrium equations is still six and three for three-dimensional and two-dimensional cases, respectively. It can be noted, however, that selection of a suitable reference point A can significantly be useful in several situation; for instance, it can offer an alternative form of equilibrium equations that is well-suited for mathematical operations or simplify the solution procedures.

To clearly demonstrate the above argument, let consider a plane frame under external loads as shown in Figure 1.30. For this particular structure, there are three unknown support reactions {RA, RBX, RBY}, as indicated in the figure, and three independent equilibrium equations (1.17) that provide a sufficient set of equations to solve for all unknown reactions. It is evident that if a point A is used as the reference point, all three equations FX = 0, FY = 0 and MAZ = 0 must be solved simultaneously in order to obtain {RA, RBX, RBY}. To avoid solving such a system of linear equations, a better choice of the reference point may be used. For instance, by using point B as the reference point, the moment equilibrium equation MBZ = 0 contains only one unknown RA and it can then be solved. Next, by taking moment about a point C, the reaction RBX can be obtained from MCZ = 0. Finally the reaction RBY can be obtained from equilibrium of forces in Y-direction, i.e. FY = 0. It can be noted, for this particular example, that the three equilibrium equations MBZ = 0, MCZ = 0 and FY = 0 are all independent and are alternative equilibrium equations to be used instead of (1.17). Note in addition that an alternative set of equilibrium equations is not unique and such a choice is a matter of taste and preference; for instance, {MBZ = 0,FX = 0,FY = 0}, {MBZ = 0,FY = 0,MDZ = 0}, {MBZ = 0,MCZ = 0,MDZ = 0} are also valid sets.

Figure 1.30: Schematic of a plane frame indicating both applied loads and support reactions

A

B

R

R

R

X

Y

C

D

Copyright © 2011 J. Rungamornrat

The number of independent equilibrium equations can further be reduced for certain types of structures. This is due primarily to that some equilibrium equations are satisfied automatically as a result of the nature of applied loads. Here, we summarize certain special systems of applied loads that often encounter in the analysis of plane structures.

1.7.1 A system of forces with the same line of action

Consider a body subjected to a special set of forces that have the same line of action as shown schematically in Figure 1.31. For this particular case, there is only one independent equilibrium equation, i.e. equilibrium of forces in the direction parallel to the line of action. The other two equilibrium equations are satisfied automatically since there is no component of forces normal to the line of action and the moment about any point located on the line of action identically vanishes. Truss members and axial members are examples of structures that are subjected to this type of loadings.

Figure 1.31: Schematic of a body subjected to a system of forces with the same line of action

1.7.2 A system of concurrent forces

Consider the body subjected to a system of forces that pass through the same point as shown in Figure 1.32. For this particular case, there are only two independent equilibrium equations (equilibrium of forces in two directions defining the plane containing the body, i.e. FX = 0 and FY = 0). The moment equilibrium equation is satisfied automatically when the two force equilibrium equations are satisfied; this can readily be verified by simply taking the concurrent point as the reference point for computing the moment resultant. An example of structures or theirs part that are subjected to this type of loading is the joint of the truss when it is considered separately from the structure.

Figure 1.32: Schematic of a body subjected to a system of concurrent forces

1.7.3 A system of transverse loads

Consider the body subjected to a system of transverse loads (loads consisting of forces where their lines of action are parallel and moments that direct perpendicular to the plane containing the body)

Line of action

F

F

F

F

F

F

F

X

Y

Copyright © 2011 J. Rungamornrat

as shown schematically in Figure 1.33. For this particular case, there are only two independent equilibrium equations (equilibrium of forces in the direction parallel to any line of actions and equilibrium of moment in the direction normal to the plane containing the body, i.e. FY = 0 and MAZ = 0). It is evident that equilibrium of forces in the direction perpendicular to the line of action is satisfied automatically since there is no component of forces in that direction. Examples of structures that are subjected to this type of loading are beams.

Figure 1.33: Schematic of a body subjected to a system of transverse loads

An initial step that is important and significantly useful for establishing the correct equilibrium equations for the entire structure or any part of the structure (resulting from the sectioning) is to sketch the free body diagram (FBD). The free body diagram simply means the diagram showing the configuration of the structure or part of the structure under consideration and all forces and moments acting on it. If the supports are involved, they must be removed and replaced by corresponding support reactions, likewise, if the part of the structure resulting from the sectioning is considered, all the internal forces appearing along the cut must be included in the FBD. Figure 1.34(b) shows the FBD of the entire structure shown in Figure 1.34(a) and Figure 1.34(c) shows the FBD of two parts of the same structure resulting from the sectioning at a point B. In particular, the fixed support at A and the roller support at C are removed and then replaced by the support reactions {RAX, RAY, RAM, RCY}. For the FBD shown in Figure 1.34(c), the internal forces {FB, VB, MB} are included at the point B of both the FBDs.