Two kinds of products of three vectors are used in engineering mechanics.
The mixed triple product (or scalar product) is used in calculating moments.
It is the dot product of vector A with the vector product of vectors B and C,

The vector triple product (A ? B) ? C = V ? C is easily calculated (for use in
dynamics), but note that

(A x B) x C # A x (B x C)

Moment of a Force about a Line

It is common that a body rotates about an axis. In that case the moment M,
of a force F about the axis, say line ,, is usefully expressed as

where n is a unit vector along the line ,, and r is a position vector from point O
on , to a point on the line of action of F. Note that M, is the projection of
M_O on line ,.

Special Cases

1. The moment about a line , is zero when the line of action of F intersects ,
(the moment arm is zero).
2. The moment about a line , is zero when the line of action of F is parallel to ,
(the projection of M_O on , is zero).

Moment of a Couple

A pair of forces equal in magnitude, parallel in lines of action, and opposite in
direction is called a couple. The magnitude of the moment of a couple is

M = fd

where d is the distance between the lines of action of the forces of magnitude F.
The moment of a couple is a free vector M  that can be applied anywhere to a
rigid body with the same turning effect, as long as the direction and magnitude of
M are the same. In other words, a couple vector can be moved to any other
location on a given rigid body if it remains parallel to its original position
(equivalent couples). Sometimes a curled arrow in the plane of the two forces
is used to denote a couple, instead of the couple vector M, which is perpendicular
to the plane of the two forces.

Force-Couple Transformations

Sometimes it is advantageous to transform a force to a force system acting at
another point, or vice versa. The method is illustrated in Figure 1.2.9.

FIGURE 1.2.9 Force-couple transformations.

1. A force F acting at B on a rigid body can be replaced by the same force F
acting at A and a moment M_A = r ? F about A.
2. A force F and moment M_A acting at A can be replaced by a force F acting
at B for the same total effect on the rigid body.

Simpli?cation of Force Systems

Any force system on a rigid body can be reduced to an equivalent system of a
resultant force R and a resultant moment MR. The equivalent force-couple
system is formally stated as

where MR depends on the chosen reference point.

Common Cases

1. The resultant force is zero, but there is a resultant moment: R = 0, M_R ? 0.
2. Concurrent forces (all forces act at one point): R ? 0, M_R = 0.
3. Coplanar forces: R ? 0, M_R ? 0. M_R is perpendicular to the plane of the forces.
4. Parallel forces: R ? 0, M_R ? 0. M_R is perpendicular to R.

Example 2

The torque wrench in Figure 1.2.10 has an arm of constant length L but a variable
socket length d = OA because of interchangeable tool sizes. Determine how
the moment applied at point O depends on the length d for a constant force F
from the hand.

FIGURE 1.2.10 Model of a torque wrench.

Solution. Using M_O = r ? F with r = Li + dj and F = Fk in Figure 1.2.10,

M_o = (Li + dj) x Fk = Fdi – FLj

Judgment of the Result

According to a visual analysis the wrench should turn clockwise, so the j
component of the moment is justi?ed. Looking at the wrench from the positive x
direction, point  A has a tendency to rotate counterclockwise. Thus, the i component
is correct using the right-hand rule.

Equilibrium of Rigid Bodies

The concept of equilibrium is used for determining unknown forces and
moments of forces that act on or within a rigid body or system of rigid bodies.
The equations of equilibrium are the most useful equations in the area of statics,
and they are also important in dynamics and mechanics of materials.
The drawing of appropriate free-body diagrams is essential for the application
of these equations.

Conditions of Equilibrium

A rigid body is in static equilibrium when the equivalent force-couple system
of the external forces acting on it is zero. In vector notation, this condition is
expressed as

SF = 0
SMo = S(r x F) = 0

where O is an arbitrary point of reference.
In practice it is often most convenient to write Equation 1.2.13 in terms of
rectangular scalar components,

SF_x = 0    SM_x = 0
SF_y = 0    SM_y = 0
SF_z = 0    SM_z = 0

Maximum Number of Independent Equations for One Body

1. One-dimensional problem: ?F = 0
2. Two-dimensional problem:

SF_x = 0    SF_y = 0   SM_A = 0
or  SF_x = 0    SM_A = 0  SM_B = 0          x axis not   AB) ( ?
or  SM_A = 0   SM_B = 0  SM_C = 0    ( AB not  BC)

3. Three-dimensional problem:

SF_x = 0   SF_y = 0   SF_z = 0
SM_x = 0  SM_y = 0  SM_z = 0

where xyz are orthogonal coordinate axes, and A, B, C are particular points
of reference.

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The critical locations in a beam are determined from
shear force and bending moment diagrams for the whole length of the beam.
The construction of these diagrams is  facilitated by following the step sillustrated
for a cantilever beam in Figure 1.2.20.

FIGURE 1.2.20 Construction of shear force and bending moment diagrams.

1. Draw the free-body diagram of the whole beam and determine all reactions
at the supports.
2. Draw the coordinate axes for the shear force (V) and bending moment
(M) diagrams directly below the free-body diagram.
3. Immediately plot those values of V and M that can be determined by inspection
(especially where they are zero), observing the sign conventions.
4. Calculate and plot as many additional values of V and M as are necessary for
drawing reasonably accurate curves through the plotted points, or do it all
by computer.

Example 5

A construction crane is modeled as a rigid bar AC which supports the boom
by a pin at B and wire CD. The dimensions are AB = 10,, BC = 2,, BD = DE = 4,.
Draw the shear force and bending moment diagrams for bar AC
(Figure 1.2.21). Solution. From the free-body diagram of the entire crane,

SF_x = 0      SF_y = 0     SM_A = 0
A_x = 0     -P+A_y = 0   -P(8) + M_A = 0
A_y = P    M_A = 8P

FIGURE 1.2.21 Shear force and bending moment diagrams of a
component in a structure.

Now separate bar AC and determine the forces at B and C.

From (a) and (c), B_x = 4P and T_CDx = 4P. From (b) and (c), By = P 2P = P and
T_CDy = 2P.
Draw the free-body diagram of bar AC horizontally, with the
shear force and bending moment diagram axes below it. Measure x from
end C for convenience and analyze sections 0 ? x ? 2, and 2, ? x ? 12,
(Figures 1.2.21b to 1.2.21f).

At point B, x = 2, M_K1= 4P(2) = 8P = M_K2= M_A. The results for section
AB, 2 ? x ? 12, show that the combined effect of the forces at B and C is to
produce a couple of magnitude 8P, on the beam. Hence, the shear force is zero
and the moment is constant in this section. These results are plotted on the axes
below the free-body diagram of bar A-B-C.

Simple Structures and Machines
Ryan Rolof and Bela I. Sandor

Equilibrium equations are used to determine forces and moments acting on
statically determinate simple structures and machines. A simple structure is
composed solely of two-force members. A machine is composed of multiforce
members. The method of joints and the method of sections are commonly
used in such analysis.

Trusses

Trusses consist of straight, slender members whose ends are connected at joints.
Two-dimensional plane trusses carry loads acting in their planes and are
often connected to form three-dimensional space trusses. Two typical trusses
are shown in Figure 1.2.22.

FIGURE 1.2.22 Schematic examples of trusses.

To simplify the analysis of trusses, assume frictionless pin connections at the
joints. Thus, all members are two-force members with forces (and no moments)
acting at the joints.

Members may be assumed

weightless or may have their weights evenly divided to the joints. Method
of Joints Equilibrium equations based on the entire truss and its joints allow
for determination of all internal forces and external reactions at the joints
using the following procedure.
1. Determine the support reactions of the truss. This is done using
force and moment equilibrium equations and a free-body diagram
of the entire truss.
2. Select any arbitrary joint where only one or two unknown forces act.
Draw the free-body diagram of the joint assuming unknown forces are
tensions (arrows directed away from the joint).
3. Draw free-body diagrams for the other joints to be analyzed, using
Newtons third law consistently with respect to the ?rst diagram.
4. Write the equations of equilibrium, ?F_x = 0 and ?F_y = 0, for the forces
acting at the joints and solve them. To simplify calculations, attempt to progress
from joint to joint in such a way that each equation contains only one unknown.
Positive answers indicate that the assumed directions of unknown forces were
correct, and vice versa.

Example 6

Use the method of joints to determine the forces acting at A, B, C, H, and I of
the truss in Figure 1.2.23a.
The angles are ? = 56.3?, ? = 38.7?, ? = 39.8?, and ? = 36.9?.

FIGURE 1.2.23 Method of joints in analyzing a truss.

Solution. First the reactions at the supports are determined and are shown in
Figure 1.2.23b. A joint at which only two unknown forces act is the best
starting point for the solution. Choosing joint A, the solution is progressively
developed, always seeking the next joint with only two unknowns. In each diagram
circles indicate the quantities that are known from the preceding analysis.
Sample calculations show the approach and some of the results.

Method of Sections

The method of sections is useful when only a few forces in truss members
need to be determined regardless of the size and complexity of the entire
truss structure. This method employs any section of the truss as a free body
in equilibrium. The chosen section may have any number of joints and members
in it, but the number of unknown forces should not exceed three in most cases.
Only three equations of equilibrium can be written for each section of a plane
truss. The following procedure is recommended.
1. Determine the support reactions if the section used in the analysis includes
the joints supported.
2. Section the truss by making an imaginary cut through the members of interest,
preferably through only three members in which the forces are unknowns
(assume tensions). The cut need not be a straight line. The sectioning is
illustrated by lines l-l, m-m, and n-n in Figure 1.2.24.
3. Write equations of equilibrium. Choose a convenient point of reference for
moments to simplify calculations such as the point of intersection of the lines
of action for two or more of the unknown forces. If two unknown forces are
parallel, sum the forces perpendicular to their lines of action.
4. Solve the equations. If necessary, use more than one cut in the vicinity of
interest to allow writing more equilibrium equations. Positive answers indicate
assumed directions of unknown forces were correct, and vice versa.

FIGURE 1.2.24 Method of sections in analyzing a truss.

Space Trusses

A space truss can be analyzed with the method of joints or with the method
of sections. For each joint, there are three scalar equilibrium equations,
?F_x = 0, ?F_y = 0, and ?F_z = 0. The analysis must begin at a joint where there
are at least one known force and no more than three unknown forces.
The solution must progress to other joints in a similar fashion.
There are six scalar equilibrium equations available when the method of sections
is used: ?F_x = 0, ?F_y = 0, ?F_z = 0, ?M_x = 0, ?M_y = 0, and ?M_z = 0.

Frames and Machines

Multiforce members (with three or more forces acting on each member) are
common in structures. In these cases the forces are not directed along the members,
so they are a little more complex to analyze than the two-force members in
simple trusses. Multiforce members are used in two kinds of structure. Frames are
usually stationary and fully constrained. Machines have moving parts, so the
forces acting on a member depend on the location and orientation of the
member. The analysis of multiforce members is based on the consistent use
of related free-body diagrams. The solution is often facilitated by
representing forces by their rectangular components.
Scalar equilibrium equations are the most convenient for two-dimensional
problems, and vector notation is advantageous in three-dimensional situations.
Often, an applied force acts at a pin joining two or more members,
or a support or connection may exist at a joint between two or more members.
In these cases, a choice should be made of a single member at the joint on
which to assume the external force to be acting. This decision should be
stated in the analysis. The following comprehensive procedure is
recommended. Three independent equations of equilibrium are available
for each member or combination of members in two-dimensional loading;
for example, ?F_x = 0, ?F_y = 0, ?M_A = 0, where A is an arbitrary point
of reference.

1. Determine the support reactions if necessary.
2. Determine all two-force members.
3. Draw the free-body diagram of the ?rst member on which the unknown
forces act assuming that the unknown forces are tensions.
4. Draw the free-body diagrams of the other members or groups of members
using Newtons third law (action and reaction) consistently with respect to the
?rst diagram. Proceed until the number of equilibrium equations available is no
longer exceeded by the total number of unknowns.
5. Write the equilibrium equations for the members or combinations of members
and solve them. Positive answers indicate that the assumed directions for unknown
forces were correct, and vice versa.
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