Chapter 1 - Equations of Statics & Free Body Diagrams
1.1 BASIC DEFINITIONS
Equilibrium:
A state of no acceleration, in either translational or rotational senses.
Scalar:
A quantity which has only magnitude. Examples include mass and area.
Vector: A quantity which has both magnitude and direction, and satisfies the
parallelogram rule of vector addition. Examples include displacement and
force.
Force:
The interaction between bodies which gives rise to an acceleration or to
the deformation of the body.
Moment:
The product of the magnitude of a force and the perpendicular distance of
its line of action from a particular point. (Also a vector.)
Couple:
It consists of two forces equal in magnitude but opposite in direction whose line
of action are parallel but no collinear.
1.2 PHYSICAL FUNDAMENTALS
Equilibrium:
where everything in a body (or part of it) is in equilibrium.
Compatibility: where all the deformations in the body are smooth. We are not interested
in what happens to a body in the event of a catastrophic occurrence, such
as an explosion, or two objects colliding.
Constitutive laws: where the forces exerted to a body are related to the way the body
deforms. For example if we apply a force to a body, it will deform always
by the same amount if the force is the same.
Energy: The work done in deforming the body is retained by that structure as
internal strain energy which will be released when the applied force is
removed. The classical example is the bow and arrow. You apply a force
to the bow and deflect it by pulling on its string. But when you let go of
the string the energy stored in the bow is enough to propel the arrow for
hundreds of meters.
The types of problems that we will be dealing with are all concerned with the bodies all being
in static equilibrium. In the context of this course the state of Equilibrium is when a body,
although acted upon by many forces or moments experiences no acceleration.
1.3 EQUATIONS OF STATICS (4th SI Ed p.2)
For a 3-D body at rest the coordinate system used is the x-y-z Cartesian system, in which the
definition of positive moments is given by the right hand rule that states that moments are
positive is their sense is counterclockwise as shown in Fig. 1.1.
Chapter 1: Eqns of Statics & FBDs
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Force equilibrium equations are given as:
and Moment equilibrium as:
  Fx

 F y
 F
 z
(1.1)
 M x

 M y
 M
 z
(1.2)
The 2D x & y -axis system looks like Fig 1.2: For a two dimensional body in the xy-axis
system, the 3D equilibrium Eqns. (1.1) and (1.2) simplify to:
  Fx

  Fy
 M
 z
(1.3)
NOTE : Not only a body/structure, but every part of a body/structure must be in equilibrium.
1.4 FREE BODY DIAGRAMS (4th SI Ed p.2)
They are a complete diagram or simplified line sketch of the structure (or body), showing the
position, direction and point of application of all externally applied forces (e.g. P) acting on the
structure, including ground reaction forces (e.g. RDX, RDY and RBX) and/or moments. Figs. 1.3 and
1.4 depict the extraction process from an original structure to a FBD.
Chapter 1: Eqns of Statics & FBDs
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1.5 TYPES OF SUPPORTS
There are 6 commonly used types of supports which prevent a structure, or part of it, from
accelerating when acted upon by external forces. In general, supports are there to keep a
structure in equilibrium. These six types of the supports are given in Table 1.1:
Chapter 1: Eqns of Statics & FBDs
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Example 1.1: Solve for the ground reactions of the structure shown in Fig. 1.3.
Step 1: Free Body Diagram
From Table 1.1: B - External pin, two unknown reactions RDX and RDY
A - Pin joint, there should be two unknown reactions: RBX and RBY.
But there are only two pins connecting bar BC with others. That implies bar BC is a “two-force
member”, i.e. it is acted upon by two equal but opposite forces directed along the bar axis. In
this case, we therefore have vertical component RBY = 0. The free body diagram is given by Fig
1.4.
Step 2: Equilibrium
Totally, there are three unknowns RDX, RDY and RBX. They can be solved based on the three
equilibrium equations as given in eqn (1.3).

  Fx  0  RBX  RDX  P sin   0
(1)
   Fy  0  RDY  P cos   0
_____
 M D  0  RBX  DC  0
Chapter 1: Eqns of Statics & FBDs
RDY  P cos 
(2)
 RBX  0
(3)
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 RDX  P sin 
Substituting (3) into (1) gives
Example 1.2: The beam of Figure 1.5 is subjected to a vertical force, a horizontal force and a
heavy box of 30kN as shown in Fig. 1.5. Determine the reactions at the supports A and B.
Step 1: Free Body Diagram
From Table 1.1:
A - External pin, unknown reactions RAX and RAY
B - Roller support, unknown reaction: RBY
The box can be approximately treated as a “uniform distributed load” (UDL) w on the beam,
which will be further represented by a concentrated resultant force F at the centre of the box in
the equilibrium as shown.
Step 2: Equilibrium:



 Fx  0  R AX  45  0
   Fy  R AY  15  3  10  RBY  0
 ccw M A  0  15  4  3  10  4  5  1.5  RBY  13.5  0
Substituting (3) into (2) gives
 R AX  45kN
 R AY  RBY  45
 RBY  27.78kN
1
2
3
 R AY  17.22kN
Chapter 1: Eqns of Statics & FBDs
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Example 1.3: Determine the ground reactions at A and B of structure as shown.
Step 1 FBD: The reaction force at roller support B should be perpendicular to the surface.
Step 2 Equilibrium:



 Fx  0  R AX  RB cos 45  0
   Fy  R AY RB sin 45  20  0
 ccw M A  0  20  3 RB sin 45  6  0
 RB  10 sin 45kN
1
2
3
Substituting (3) into (1) & (3) into (2) gives
R AX  RB cos 45  10kN
 R AY  10kN
Chapter 1: Eqns of Statics & FBDs
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