Module
4
Analysis of Structure
OVERVIEW
In this module, we will apply the concept of equilibrium of force
systems in structural analysis of rigid structures. Determinacy and stability will
also be discussed. Different types of supports and their corresponding
reactions will be taken up. 2D and 3D framing structures will be given focus.
LEARNING OUTCOMES
After studying this module, you must be able to:
• understand and define the moment of a force,
• determine moments of a force in 2-D
• define a couple, and
• determine the moment of a couple
• Identify support reactions.
• Draw a free body diagram.
• Analyse the equilibrium of rigid bodies in two-dimensional problems.
INTRODUCTION
The moment of a force about a point or an axis provides a measure of
the tendency of the force to cause the body to rotate about that point or axis.
It is also sometimes called a torque.
How is the moment of the force about a point calculated?
1. Draw the line of action of the force:
• a co-linear extension of the force vector.
• the line of action of F and the point O define the plane above.
2. Draw a moment arm from O perpendicular to line of action.
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Moment of the force about O:
where d is the perpendicular distance from the point to the line of action of the
force
Usual sign convention:
• impending rotation is CCW (counter-clockwise) about axis then M is
positive
• impending rotation is CW (clockwise) about axis then M is negative
The force F will tend to rotate the object about an axis perpendicular to the plane
defined by the force and the moment arm. This axis is referred to as the moment
axis. Thus, MO is also referred to as the moment of the force about the axis passing
through O.
If the line of action of the force passes through the point then there is no moment as
d=0
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MOMENTS WITH UNKNOWN LEVER ARM
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Exercises:
1. What is the moment at A in Figure 16
Figure 16
2. What is the moment at A in Figure 17, if F = 20 kN and θ = 250?
Figure 17
3. What is the moment at A in Figure 18?
Figure 18
COUPLES
A couple is defined as two parallel forces with the same magnitude, but opposite in
direction, separated by a perpendicular distance d as shown in Figure 19.
The moment of a couple �� = � ∙ �
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Couples have zero net force. Since the moment of a couple depends only on the distance
between the forces, the moment of a couple is a free vector it can be moved anywhere on
the body and have the same external effect on the body.
We will calculate the moment at a point in Figure 20, using both the moment of a couple
and the sum of the moments from the individual forces.
Figure 20
a) Moment of a couple:
The parallel forces are 10 kN with a perpendicular distance between them.
The moment of the couple is therefore �� = � ∙ � = 10 ∙ 2 = 20 kNm, the
moment is turning in an anti-clockwise direction, so is therefore positive in our
sign convention. As the moment of a couple is a free vector, the moment at A
is also 20 kNm.
b) Individual moments:
The moment of the force 2.5 m away from A is 0��1 = −(10 × 2.5) = −25
kNm (clockwise)
Similarly
�2 = 10 × 4.5 = 45 kNm (anti-clockwise)
The total moment at A is therefore
�� = �1 + �2 = −25 + 45 = 20 kNm
Coplanar Distributed Loading
A simple distributed loading can be represented by its resultant force, which
is equivalent to the area under the loading curve. This resultant has a line of action
that passes through the centroid or geometric center of the area or volume under the
loading diagram.
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EQUILIBRIUM OF A RIGID BODY
In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent
and may cause rotation of the body (due to the moments created by the forces).
Figure 21 A rigid body subject to a system of forces
For a rigid body to be in equilibrium, the net force as well as the net moment about any
arbitrary point O must be equal to zero.
SOLVING RIGID BODY EQUILIBRIUM PROBLEMS
 For analyzing an actual physical system, first we need to create an idealized model.
 Then we need to draw a free-body diagram (FBD) showing all the external (active
and reactive) forces.
 Finally, we need to apply the equations of equilibrium to solve for any unknowns.
Examples:
1. Figure 22 shows a concrete beam of length 10 m spanning the gap
between two masonry walls. The beam is subject to a 20 kN force 4 m
from its left hand edge (for this example the self-weight of the beam will
be ignored). What are the forces from the masonry wall that will be
required to keep the beam in equilibrium?
Figure 22
To analyse the problem the problem in Figure 22, it must be converted to a
free body diagram of the beam alone.
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Figure 23 Free body diagram
Figure 23 shows the beam in isolation subject to both the applied force of 20 kN and the
reaction forces that the beam would feel from the masonry walls that it sits upon. We can
now apply the equations of equilibrium in order to determine the unknown reactions.
We will first sum the forces in the x direction, and impose the constraint that the sum must
be equal to zero for equilibrium.
In this problem there are no forces acting in the x direction and therefore the horizontal
reaction is zero. We will now sum the forces acting in the y direction.
The above equation leaves us with 2 unknowns, we will derive a 2nd equation by taking
moments about some point along the beam. The choice of where to take moments is
completely arbitrary, as the sum of the moments should be zero everywhere for equilibrium,
however it is sometimes advantageous to choose a point that eliminates one of the
unknowns immediately.
If we were to take moments 4 m along the beam, we would get another equation relating
RAY to RBY, which we could then substitute into the equation above and solve for one of the
reactions. A more direct way would be to take moments about either of the ends, A or B, or
the beam. Taking moments about the left hand end (A), yields
RBY x 10 - 20 x 4 = 0
Hence
RBY = 8 kN
With ��� known we can now substitute this value into our equation for
equilibrium in the � direction to obtain ��y?
RAY + 8 - 20 = 0
RAY = 12 kN
2.
A 10 m beam is now embedded into a wall at its left hand end, as shown
in Figure 24.
Figure 24
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What are the support reactions required at the left hand end, in order for the beam to
remain in equilibrium?
We again begin the solution process by drawing a free body diagram (FBD) of the beam in
isolation from its surroundings.
Figure 25 Free body diagram
As the beam is embedded in the wall, we assume that the wall is able to impose a moment
MA on the left hand end of the beam, in addition to a horizontal reaction RAX and vertical
reaction RAY.
We can now use the three equations of equilibrium in 2D to compute the unknown
reactions.
By inspection, we can immediately see that the horizontal reaction RAX is zero, as no other
horizontal loads are acting on the beam.
Summing the forces in the y direction
Thus
RAX = 0
RAY − 20 = 0
RAY = 20 kN
In order to find the reaction moment, we take moments about the left hand end
MA − 20 × 10 = 0
Thus
MA = 200 kN.m
The moment is given a positive sign as it must act in a counter-clockwise direction.
3.
Given the rigid body in Figure 26, if P1 = 12 kN and P2 = 8 kN and the distances a = 2 m,
b = 3 m and c = 4 m. Find the vertical reactions RAY and RBY required to support the
beam.
Figure 26
Supports for Rigid Bodies Subjected to Two-Dimensional Force Systems
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LESSON 2: CONDITIONS FOR EQUILIBRIUM OF PARALLEL FORCES
The sum of all the forces is zero.
ΣF=0
The sum of moment at any point O is zero.
ΣMO=0
SAMPLE PROBLEMS
1. Determine the reactions for the beam shown.
100 lb/ft
Solution:
��2 = 0
10R1+4(400) =16(300) +9[14(100)]
R1=1580 lb answer
��1 = 0
10R2+6(300)=14(400)+1[14(100)]
R2=520 lb
answer
2. The roof truss shown is supported by a roller at A and a hinge at B. Find
the values of the reactions.
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Replace the 3-20 kN and 2-10 kN forces by a single 80 kN force
ΣMB=0
15RA = 10(60) + 7.5(80) + 5(50)
RA = 96.67 kN
answer
ΣMA=0
15RB = 5(60) + 7.5(80) + 10(50)
RB = 93.33 kN
answer
PROBLEM EXERCISES
1. Determine the reactions for the beam shown.
200 lb
200 lb
350 lb
2. Determine the reactions for the beam loaded as shown.
15 kN
20 kN/m
3. The cantilever beam shown in the Figure is built into a wall 2 ft thick so
that it rests against points A and B. The beam is 15 ft long and weighs 100
lb per ft.
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5000 lb
15 ft
A
4. The upper beam in Figure shown is supported at D and a roller at C which
separates the upper and lower beams. Determine the values of the
reactions at A, B, C, and D. Neglect the weight of the beams.
80 kN
150 kN
300 kN
5. For the system of pulleys shown below, determine the ratio of W to P to
maintain equilibrium. Neglect axle friction and the weights of the pulleys.
LESSON 3: EQUILIBRIUM OF NON-CONCURRENT FORCE SYSTEM
(COPLANAR FORCE SYSTEMS)
There are three equilibrium conditions that can be used for nonconcurrent, non-parallel force system.
The sum of all forces in the x-direction or horizontal is zero.
ΣFx=0 or ΣFH=0
The sum of all forces in the y-direction or vertical is zero.
ΣFy=0 or ΣFV=0
The sum of moment at any point O is zero.
ΣMO=0
The three equilibrium conditions can solved up to three unknowns in the
system.
SAMPLE PROBLEMS
1. The frame shown below is supported in pivots at A and B. Each member
weighs 5 kN/m. Compute the horizontal reaction at A and the horizontal
and vertical components of the reaction at B.
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Solution:
Length of DF
LDF = 42 + 32 = 5 m
Weights of members
WAB=6(5)=30 kN WCE=6(5)=30 kN
WDF=5(5)=25 kN
ΣMB=0
6AH=3WCE+2WDF+6(200)
6AH=3(30)+2(25)+6(200)
AH=223.33 kN
ΣFH = 0
BH=AH
BH=223.33 kN
answer
ΣFV=0
BV=WAB+WCE+WDF+200
BV=30+30+25+200
BV=285 kN
answer
2.Compute the total reactions at A and B on the truss shown.
3. The beam shown is supported by a hinge at A and a roller on a 1 to 2
slope at B. Determine the resultant reactions at A and B.
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PROBLEM EXERCISES
1. A boom AB is supported in a horizontal position by a
hinge A and a cable which runs from C over a small
pulley at D as shown in the figure. Compute the
tension T in the cable and the horizontal and vertical
components of the reaction at A. Neglect the size of
the pulley at D.
2. Repeat problem #1 if the cable pulls the boom AB into
a position at which it is inclines at 30° above the
horizontal. The loads remain vertical.
3. The truss shown in the figure
is supported on roller at A and
hinge at B. Solve for the
components of the reactions.
7. The uniform rod in Fig. P-357
4. Compute the total reactions at
A and B for the truss shown.
6. The forces acting on a 1-m
length of a dam are shown in
the figure below. The upward
ground
reaction
varies
uniformly from an intensity of p1
kN/m to p2 kN/m at B.
Determine p1 and p2 and also
the horizontal resistance to
sliding.
weighs 420 lb and has its
center of gravity at G.
Determine the tension in the
cable and the reactions at the
smooth surfaces at A and B.
8. A bar AE is in equilibrium under
the action of the five forces
shown in Fig. P-358. Determine
P, R, and T.
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9. A 4-m bar of negligible weight
rests in a horizontal position on
the smooth planes shown.
Compute the distance x at
which load T = 10 kN should be
placed from point B to keep the
bar horizontal.
LESSON 4: SIMPLE TRUSSES
A truss is a structure composed of slender members joined together at
their end points. The members commonly used in construction consist of
wooden struts or metal bars. In particular, planar trusses lie in a single plane
and are often used to support roofs and bridges. The truss shown in Fig. 6-1a
is an example of a typical roof-supporting truss. In this figure, the roof load is
transmitted to the truss at joints by means of a series of purlins. Since this
loading acts in the same plane as the truss, Fig. 6-1b, the analysis of the
forces developed in the truss members will be two-dimensional.
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In the case of a bridge, such as shown in Fig. 6-2a, the load on the
deck is first transmitted to stringers, then to floor beams, and finally to the
joints of the supporting side trusses. Like the roof truss, the bridge truss
loading is also coplanar, Fig. 6-2b.
When bridge or roof trusses extend over large distances, a rocker or
roller is commonly used for supporting one end, for example, joint A in Fig. 61a and 6-2a. This type of support allows freedom for expansion or contraction
of the members due to a change in temperature or application of loads.
Assumption for Design
To design both the members and the connections of a truss, it is
necessary first to determine the force developed in each member when the
truss is subjected to a given loading. To do this we will make two important
assumptions:

All loadings are applied at the joints. In most situations, such as for
bridge and roof trusses, this assumption is true. Frequently the weight of
the members is neglected because the force supported by each member is
usually mush larger than its weight. However, if the weight is to be included
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in the analysis, it is generally satisfactory to apply it as a vertical force, with
half of its magnitude applied at each end of the member.

The members are joined together by smooth pins. The joint
connections are usually formed by bolting or welding the ends of the
members to a common plate, called a gusset plate, as shown in Fig. 6-3a,
or by simply passing a large bolt or pin through each of the members, Fig.
6-3b. We can assume these connections act as pins provided the center
lines of the joining members are concurrent, as in Fig. 6-3.
Because of these two assumptions, each truss member will act as a
two force member, and therefore the force acting at each end of the member
will be directed along the axis of the member. If the force tends to elongate the
member, it is a tensile force (T), Fig. 6-4a; whereas if it tends to shorten the
member, it is a compressive force (C), Fig. 6-4b. In the actual design of a truss
it is important to state whether the nature of the force is tensile or compressive.
Often, compression members must be made thicker than tension members
because of the buckling or column effect that occurs when a member is in
compression.
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Simple Truss.
If three members are pin connected at their ends they form a triangular
truss that will be rigid, Fig. 6-5. Attaching two or more members and
connecting these members to a new joint D forms a larger truss, Fig.6-6. This
procedure can be repeated as many times as desired to form an even larger
truss. If a truss can be constructed by expanding the basic triangular truss in
this way, it is called a simple truss.
A
LESSON 5: THE METHOD OF JOINTS
This method is based on the fact that if the entire truss is in equilibrium,
then each of its joints is also in equilibrium. Therefore, if the free-body diagram
of each joint is drawn, the force equilibrium equations can then be used to
obtain the member forces acting on each joint. Since the members of a plane
truss are straight two-force members lying in a single plane, each joint is
subjected to a force system that is coplanar and concurrent. As a result, only
ΣFx = 0 and ΣFy = 0 needed to be satisfied for equilibrium.
For example, consider the pin at joint B of the truss in Fig. 6-7a. Three
forces act on the pin, namely, the 500-N force and the forces exerted by
members BA and BC. The free-body diagram of the pin is shown in Fig. 6-7b.
Here, FBA is “pulling” on the pin, which means that the member BA is in
tension; whereas FBC is “pushing” on the pin, and consequently member BC is
in compression. These effects are clearly demonstrated by isolating the joint
with small segments of the member connected to the pin, Fig. 6-7c. The
pushing or pulling on these small segments indicates the effect of the member
being either in compression or tension
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When using the method of joints, always start at a joint having at least
one known force and at most two unknown forces, as in Fig. 6-7b. In this way,
application of ΣFx = 0 and ΣFy = 0 yields two algebraic equations which can
be solved for the two unknowns. When applying these equations, the correct
sense of an unknown member force can be determined using one of two
possible methods.

The correct sense of direction of an unknown member force can, in many
cases, be determined “by inspection.” For example, FBC in Fig. 6-7b must
push on the pin (compression) since its horizontal component, FBC sin 45°,
must balance the 500-N force (ΣFx = 0). Likewise, FBA is a tensile force
since it balances the vertical component, FBC cos 45° (ΣFy = 0). In more
complicated cases, the sense of an unknown member force can be
assumed; then, after applying the equilibrium equations, the assumed
sense can be verified from the numerical results. A positive answer
indicates that the sense is correct, whereas a negative answer indicates
that the sense shown on the free-body diagram must be reversed.

Always assume the unknown member forces acting on the joint’s free-body
diagram to be in tension; i.e., the forces “pull” on the pin. If this is done,
then numerical solution of the equilibrium equations will yield positive
scalars for members in tension and negative scalars for members in
compression, once an unknown member force is found, use its correct
magnitude and sense (T or C) on subsequent joint free-body diagrams.
Procedure for Analysis
The following procedure provides a means for analyzing a truss using
the method of joints.

Draw the free-body diagram of a joint having at least one known force and
at most two unknown forces. (If this joint is at one of the supports, then it
may be necessary first to calculate the external reactions at the support.)
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
Use one of the two methods described above for establishing the sense of
an unknown force.

Orient the x and y axes such that the forces on the free-body diagram can
be easily resolved into their x and y components and then apply the two
force equilibrium equations ΣFx = 0 and ΣFy = 0. Solve for the two
unknown member forces and verify their correct sense.

Using the calculated results, continue to analyze each of the other joints.
Remember that a member in compression “pushes” on the joint and a
member in tension “pulls” on the joint. Also, be sure to choose a joint
having at most two unknowns and at least one known force.
SAMPLE PROBLEM
1. Determine the force in each member of the truss shown in Fig. 6-8a and
indicate whether the members are in tension or compression.
Fig.
6-8a
Solution:
Since we should have no more than two unknown forces at the joint
and at least one known force acting there, we will begin our analysis at joint B.
FBD of Joint B
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Fig. 6-8b
Joint B. The free body diagram of the joint at B is shown in Fig. 6-8b,
Applying the equations of equilibrium, we have
Since the force in member BC has been calculated, we can proceed to
analyze joint C to determine the force in member CA and the support reaction
at the rocker.
FBD of Joint C
Fig. 6-8c
Joint C. From free-body diagram of joint C, Fig. 6-8c, we have
FBD of Joint A
Fig. 6-8d
Joint A. Although it is not necessary, we can determine the components of
the support reactions at joint A using the results of FCA and FBA. From the
free-body diagram, Fig. 6-8d, we have
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Therefore, we have:
Note: The results of the analysis are summarized in Fig. 6-8e. Note
that the free-body diagram of each joint (or pin) shows the effects of all
the connected members and external forces applied to the joint, whereas
the free-body diagram of each member shows only the effects of the end
joints on the member.
2. Determine the force in each member of
the truss shown in Fig. 6-9a and indicate
whether the members are in tension or
compression.
Solution:
Since joint C has one unknown and only two
unknown forces acting on it, it is possible to start at
this joint, then analyze joint D, and finally joint A. This
way the support reactions will not have to be
determined prior to starting the analysis.
FBD of Joint C
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Fig. 6-9b
Joint C. By inspection of the force equilibrium, Fig 6-9b, it can be seen that
both members BC and CD must be in compression.
FBD of Joint D
Fig. 6-9c
Joint D. Using the result FCD = 400N (C), the force in members BD and
AD can be found by analyzing the equilibrium on joint D. We will assume
FAD and FBD are both tensile forces or you can use inspection to
determine the correct direction of these forces, Fig 6-9c.
ΣFy = 0;
ΣFx = 0;
- FBD 2 25 - FAD cos 45° = 0
FBD 24 5 + FAD sin 45° - 400 = 0
FBD 24 5 + FAD sin 45° = 400
Using your calculator, input the coefficients of the two (2) equations.
FBD = 894.43 N (T)
FAD = -565.69 N = 565.69 (C)
FBD of Joint A
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Fig. 6-9d
Joint A.
The force in member AB can be found by analyzing the
equilibrium of joint A. Fig 6-9d. We have
ΣFx = 0;
(565.69) cos 45° - FAB = 0
FAB = 400 N (C)
Zero-Force Members
Truss analysis using the method of joints is greatly simplified if we can
first identify those members which support no loading. These zero-force
members are used to increase the stability of the truss during construction
and to provide added support if the loading is changed.
The zero-force members of the truss can generally be found by
inspection of each of the joints. For example, consider the truss shown in Fig.
6-11a. If a free-body diagram of the pin at joint A is drawn, Fig. 6-11b, it is
seen that members AB and AF are zero-force members. (We could not have
come to this conclusion if we have considered the free-body diagrams of
joints F and B simply because there are five unknowns at each joint.) In
similar manner, consider the free-body diagram of joint D, Fig 6-11c. Here
again it is seen that DC and DE are zero-force members. From these
observations, we can conclude that if only two members form a truss joint
and no external load or support reaction is applied to the joint, the two
members must be zero-force members. The load in the in Fig 6-11a is
therefore supported by only five members as shown in Fig. 6-11d.
Now consider the truss shown in Fig. 6-12a. The free-body diagram of
the pin at joint D is shown in Fig. 6-12b. By orienting the y axis along
members DC and DE and the x axis along member DA, it is seen that DA is a
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zero-force member. Note that this is also the case for member CA, Fig. 6-12c.
In general, if three members form a truss joint for which two of the
members are collinear, the third member is a zero-force member
provided no external force or support reaction is applied to the joint.
The truss shown in Fig. 6-12d is therefore suitable for supporting the load P.
PROBLEM EXERCISES
1. Determine the force in each member of the truss shown and state if the
members are in tension or compression.
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2. Determine the force in each member of the truss, and state if the members
are in tension or compression. Set P = 800lb.
3. Determine the force in each member of the truss and state if the members
are in tension or compression. The load has a mass of 40 kg.
4. Determine the force in each member of the truss in terms of the load P,
and indicate whether the members are in tension or compression.
LESSON 6: THE METHOD OF SECTIONS
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When we need to find the force in only a few members of a truss, we
can analyze the truss using the method of section. It is based on principle that
if the truss is in equilibrium then any segment of the truss is also in equilibrium.
In the method of sections, a truss is divided into two parts by taking an
imaginary “cut” (shown here as a-a) through the truss.
Since truss members are subjected to only tensile or compressive
forces along their length, the internal forces at the cut member will also be
either tensile or compressive with the same magnitude. This result is based
on the equilibrium principle and Newton’s third law.
Procedure for Analysis:
The process used in the method of sections is outlined below:
1.
In the beginning it is usually useful to label the members in your
truss. This will help you keep everything organized and consistent in
later analysis. In this book, the members will be labeled with letters.
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2. Treating the entire truss structure as a rigid body, draw a free body
diagram, write out the equilibrium equations, and solve for the external
reacting forces acting on the truss structure. This analysis should not
differ from the analysis of a single rigid body.
3. Next you will imagine cutting your truss into two separate sections. The
cut should travel through the member that you are trying to solve for the
forces in, and should cut through as few members as possible (The cut
does not need to be a straight line).
4. Next you will draw a free body diagram for either one, or both sections that
you created. Be sure to include all the forces acting on each section.



Any external reaction or load forces that may be acting at the
section.
An internal force in each member that was cut when splitting the
truss into sections. Remember that for a two force member, the
force will be acting along the line between the two connection
points on the member. We will also need to guess if it will be a
tensile or a compressive force. An incorrect guess now though
will simply lead to a negative solution later on. A common
strategy then is to assume all forces are tensile, then later in the
solution any positive forces will be tensile forces and any
negative forces will be compressive forces.
Label each force in the diagram. Include any known magnitudes
and directions and provide variable names for each unknown.
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5. Write out the equilibrium equations for each section you drew a
free body diagram of. These will be extended bodies, so you will
need to write out the force and the moment equations.

For 2D problems you will have three possible equations for
each section, two force equations and one moment equation.

For 3D problems you will have six possible equations for each
section, three force equations and three moment equations.
6. Finally, solve the equilibrium equations for the unknowns. You
can do this algebraically, solving for one variable at a time, or you
can use matrix equations to solve for everything at once. If you
assumed that all forces were tensile earlier, remember that
negative answers indicate compressive forces in the members.
SAMPLE PROBLEMS
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1. Determine the force in member CF of the truss show in Fig. 6-17a. Indicate
whether the member is in tension or compression. Assume each member
is pin connected.
Solution:
Fig 6-17a
Free-Body Diagram. Section aa in Fig. 6-17a will be used since this
section will “expose” the internal force in member CF as “external” on
the free-body diagram of either the right or left portion of the truss. It is
first necessary, however, to determine the support reactions on either
the left or right side. Verify the results shown on the free-body diagram
in Fig. 6-17b
Fig. 6-17b
The free-body diagram of the right portion of the truss, which is the
easiest to analyze, is shown in Fig. 6-17c. There are three unknowns,
FFG, FCF, and FCD.
Fig. 6-17c
Equations of Equilibrium. We will apply the moment equation about
point O in order to eliminate the two unknowns FFG and FCD. The location of
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point O measured from E can be determined from proportional triangles, i.e.,
4/(4 + x) = 6/8 + x), x = 4m. or stated in another manner, the slope of member
GF has a drop of 2 m to a horizontal distance of 4 m, Fig. 6-17c, then from D
to O the distance must be 8 m.
An easy way to determine the moment of FCF about point O is to use
the principle of transmissibility and slide FCF to point C, and then resolve FCF
into its two rectangular components, We have
2. Determine the force in member EB of the roof truss show in Fig. 6-18a.
Indicate whether the member is in tension or compression.
Fig. 6-18a
Solution:
Free-Body Diagram. By the method of sections, any imaginary
section that cuts through EB, Fig. 6-18a, will also have to cut through three
other members for which the forces are unknown. For example, section aa
cuts through ED, EB, FB, and AB. If a free-body diagram of the left side of this
section is considered, Fig. 6-18b, it is possible to obtain FED by summing
moments about B to eliminate the other three unknowns; however, FEB cannot
be determined from the remaining two equilibrium equations. One possible
way of obtaining FEB is first to determine FEB is first to determine FED from
section aa, then use this result on section bb, Fig 6-18a, which is shown in Fig
6-18a, which is shown in Fig. 6-18c. Here the force system is concurrent and
our sectioned free-body diagram is the same as the free-body diagram for the
joint at E.
Fig. 6-18b
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Fig. 6-18c
Equations of Equilibrium. In order to determine the moment of FED
about point B, Fig 6-18b, we will use the principle of transmissibility and slide
the force to point C and then resolve it into its rectangular components as
shown. Therefore,
Considering now the free-body diagram of section bb, Fig 6-18c, we have
PROBLEM EXERCISES
1. Determine the force in each members KJ, KD, and CD of the Pratt
truss. State if the members are in tension or compression.
2. Determine the force in members EF, CF, and BC of the truss. State
if the members are in tension or compression.
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3. Determine the force in members DC, HI, and JI of the truss. State if
members are in tension or compression.
4. Determine the force in members JK, CJ, and CD of the truss, and
state if the members are in tension or compression.
5. Determine the force in members HI, FI, and EF of the truss, and
state if the members are in tension.
Problem 4 & 5
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LESSON 7: FRAMES AND MACHINES
Frames and machines are two types of structure which are often
composed of pin-connected multiforce members, i.e., members that are
subjected to more than two forces. Frames are used to support loads,
whereas machines contain moving parts and are designed to transmit and
alter effect of forces. Provided a frame or machine contains no more supports
or members than are necessary to prevent its collapse, the forces acting at the
joints and supports can be determined by applying the equations of equilibrium
to each of its members. Once these forces are obtained, it is then possible to
design the size of the members, connections, and supports using the theory of
mechanics of materials and an appropriate engineering design code.
Free-Body Diagrams. In order to determine the forces acting at the
joints and supports of a frame or machine, the structure must be disassembled
and the free-body diagrams of its parts must be drawn. The following
important points must be observed:



Isolate each part by drawing its outlined shape. Then show all the
forces and/or couple moments that act on the part. Make sure to
label or identify each known and unknown force and couple moment
with reference to an established x, y coordinate system. Also,
indicate any dimensions used for taking moments. Most often the
equations of equilibrium are easier to apply if the forces are
represented by their rectangular components. As usual, the sense of
an unknown force or couple moment can be assumed.
Identify all two-force members in the structure and represent their
free-body diagrams as having two equal but opposite collinear
forces acting at their points of application. By recognizing the twoforce members, we can avoid solving an unnecessary number of
equilibrium equations.
Forces common to any two contacting members act with equal
magnitudes but opposite sense on the representative members. If
the two members are treated as a “system” of connected members,
then these forces are “internal” and are shown on the free-body
diagram of the system; however, if the free-body diagram of each
member is drawn, the forces are “external” and must be shown on
each of the free-body diagrams.
The following examples graphically illustrate how to draw the free-body
diagrams of a disassembled frame or machine. In all cases, the weight of the
members is neglected.
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Sample Problems
1. For the frame shown in Fig. 6-21a, draw the free-body diagram of (a) each
member, (b) the pin at B, and (c) the two members connected together.
Fig. 6-21a
Solution:
Part (a). By inspection, members BA and BC are not two-force
members. Instead, as shown on the free-body diagram, Fig. 6-21b BC
is subjected to a force from the pins at B and C and the external couple
moment M. The pin forces are represented by their x and y components.
Fig. 6-12b
Part (b).
The pin at B is subjected to two forces, i.e., the force of
member BC and the force of member AB. For equilibrium these forces
or their respective components must be equal but opposite, Fig. 6-21c.
Realize that Newton’s third law is applied between the pin and its
connected members, i.e., the effect of the pin on the two members, Fig.
6-21b, and the equal but opposite effect of the two members on the pin.
Fig. 6-21c.
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Fig. 6-12c
Part (c).
The free-body diagram of both members connected
together, yet removed from the supporting pins at A and C, is shown in
Fig. 6-12d. The force components Bx and By are not shown on this
diagram since they are internal forces (Fig. 6-21b) and therefore cancel
out. Also, to be consistent when later applying the equilibrium equations,
the unknown force components at A and C must act in the same sense
as those shown in Fig 6-21b.
Fig. 6-21d
2. Determine the horizontal and vertical components of force which the
pin at C exerts on members BC of the frame in Fig. 6-26a.
Fig. 6-26a
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Solution I:
FBD:
By inspection it can be seen that AB is a two-force member. The freebody diagrams are shown in Fig. 6-26b.
Fig 6-26b
Equations of Equilibrium:
The three unknowns can be determined by applying the three equations
of equilibrium to member CB.
Solution II:
FBD:
If one does not recognize the AB is a two-force member, then more
work is involved in solving this problem. The free-body diagrams are
shown in Fig 6-26c.
Fig. 6-26c
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Equations of Equilibrium. The six unknowns are determined by
applying the three equations of equilibrium to each member.
Member AB
Member BC
The result for Cx and Cy can be determined by solving these equations
in following sequence: 4, 1, 5, then 6. The results are
By = 1000 N
Bx = 577 N
Cx = 577 N
Cy = 1000 N
By comparison, Solution I is simpler since the requirement that FAB in
Fig. 6-26b be equal, opposite, and collinear at the ends of member AB
automatically satisfies Eqs. 1, 2, and 3 above and therefore eliminates
the need to write these equations. As a result, save yourself some time
and effort by always identifying the two-force members before starting
the analysis.
Assigned Readings:
Read your reference books in Engineering Mechanics Statics,
Chapter 6. Solve more similar problems from the exercises given
above.
Suggested Readings:
You can also try to access the internet, aside from your
reference books. You can access and watch you tube, just search for
our topic and you will see videos of professors discussing about these
topics that we have undertaken in this module.
References:
 Hibbeler, R.C., Engineering Mechanics, Statics, 14th Edition, 2016
 Khurmi, R.S., A Textbook of Engineering Mechanics
 Singer, Ferdinand L., Engineering Mechanics, Statics and
Dynamics, 3rd Edition
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