5.2
Trusses
Figure 5 – 6: Undetermined
(Zoe)
Two-Force Members
The truss is a very popular type of structure. It’s used in roof systems, bridges
and towers and it’s an aesthetic feature in many buildings (See Fig. 5 – 6). The
popularity of the truss lies both in its simplicity and its efficiency.
A truss is a structure composed exclusively of two-force members. The
external forces acting on a truss, both the external reactions and the applied forces, act
at the joints of the members. In those cases in which an applied force does not actually
act at either end of a structural member, the applied force can be replaced with an
equivalent force at each end of the member. For example, the weight load is a
distributed force that acts along the length of a structural member. It can be replaced
with an equivalent pair of point forces at each end of the member. This procedure
creates a truss system that can be analyzed using the methods developed in this section.
Trusses are designed for either strength, stiffness, action, or a combination
thereof. Forces need to be determined in trusses designed for strength and
displacements need to be determined in trusses designed for stiffness. The next two
sub-sections describe three ways of finding the forces in a planar truss: the method of
joints, the method of sections, and the sequential method. Finally, it is shown how to
find the displacements in a planar truss.
Truss Connections
Figure 5 – 7: A two-force member
Planar Trusses, Space Trusses
Figure 5 – 8: The ends of a structural
member are often welded or riveted to a
plate, called the gusset plate. The
connection is considered to be a pin
connection if the center lines of the
members intersect the center of the
gusset plate.
The two-force member does not carry moments at either of its ends. The
member remains in equilibrium by an internal reaction that passes through the line
between the two ends of the member, as shown in Fig. 5 – 71. The internal reaction is a
stretching force. The stretching force can be compressive or tensile. Most frequently,
the end of a two-force member is represented by a pin connection or a rolling
connection. The pinned connection is created by welds, bolts, or rivets. They act like a
pinned connection when the centerlines of the structural members intersect at the center
of the joint. By design, the moments they carry are relatively small (See Fig. 5 – 8). In
practice, the rolling connection is frequently created by a roller, a rocker, or an
expansion joint.
In many trusses, the structural members, internal reactions, external reactions
and applied forces all lie in one plane. These are called planar trusses. In general,
though, the trusses are space trusses. The truss shown in Fig. 5 – 6 is an example of a
space truss.
A fully constrained planar truss is fixed by at least three external reactions.
The external reactions constrain the planar truss’s two translational degrees of freedom
and its rotational degree of freedom. When the number of external reactions and the
number of structural members are “minimal” the truss becomes statically determinate.
The statically determinate truss can freely expand or contract when the temperature
changes.
There is a simple relationship between the number of members and the
number of joints in a statically determinate truss. Denote the number of members by m,
the number of joints by n, and the number of external reactions by p. The entire truss is
maintained in static equilibrium if each of the truss’s joints is maintained in static
The two-force member was also discussed in Section 3.4 in the sub-section entitled “Two-Force and Three-Force
Members.”
1
Force Equations
equilibrium. Since the forces acting on a joint are concurrent, static equilibrium of the
truss is maintained by the resultant forces acting on the truss’s joints. The sum of the
moments is not needed. In the case of a planar truss, the resultant forces acting at the
joints in the x and y directions are summed to zero to produce 2n force equations. The
number of unknowns is m + p, so for the statically determinate planar truss
(5 – 1)
2n = m + p.
Equations (5 – 1) is used in the equation step to check a truss’s static determinacy.
Space Trusses
In the case of a space truss, the resultant force acting at the joints in the x, y,
and z directions are summed to zero to produce 3n force equations. For the statically
determinate space truss
(5 – 2)
3n = m + p.
The Method of Joints and the Method of Sections
The unknown forces in a truss problem consist of stretching forces in
members and the force reactions at joints. In some situations the interest lies in finding
all of the unknown forces and in other situations that interest lies in finding selected
forces of interest. The method of joints refers to a problem-solving process that makes
use of all of the 2n force equations described earlier. The method of joints is used in
two situations: (1) when the truss is made up of just two or three members, and (2)
when the problem is solved by computer. When only a few of the stretching forces or
force reactions are sought and the truss consists of more than three members, then it
becomes more convenient to find the stretching forces and the force reactions by
“cutting” the truss into sections, called body sections. A body section is a rigid body in
itself that must be maintained in static equilibrium. Therefore, the resultant forces and
the resultant moment acting on the body section are zero, which produces three
equilibrium equations. Finding stretching forces and member reactions from the
equations governing the equilibrium of one or more body sections is called the method
of sections.
The Method of Joints
The method of joints is a systematic method of determining all of the
reactions and the stretching forces in a truss. In the transition step, a free body diagram
is drawn for each joint. It’s helpful to either a) label the forces as though they’re all in
tension (the compressive members later turn out to be negative) or b) to label the force
in each member depending on whether you expect the member to be in compression or
in tension. Both approaches help you keep track of the senses of the forces. In the
equation step, write down the force equations, namely, Fx = 0 and Fy = 0 for each
joint. In the answer step, a simultaneous set of linear algebraic equations needs to be
solved. In the knowledge step, the signs of the unknown forces are checked. Some of
the members are expected to act in tension and others to act in compression. Also
inspect the magnitudes of the forces. Some truss members will be loaded more than
others. The examples at the end of the section demonstrate the method of joints.
2
The Method of Sections
Figure 5 – 9: The truss below is cut
through line A-A. This creates the body
section shown. Notice that the body
section has three unknown forces.
Body Section
The method of sections is used when the focus is on a few of the structural
members. In the transition step, the truss is cut along a section containing the force that
is sought. (Don’t cut directly through a joint.) The body that remains after the cut is
called the body section. The body section is a rigid body in itself. Draw the free body
diagram of the body section. Remember that only the forces that are external to the
body section need to be included in the free body diagram. It’s important to cut the
truss where the number of unknown reactions is no greater than three (See Fig. 5
– 9). In the equation step, write down for the body section Fx = 0, Fy = 0, and  =
0. The number of equations and the number of unknowns should be three.
Note that you’ve already been using the method of sections in this course even
though the term “method of sections” was not used, per say. In the simplest case, when
external forces and moments acting on a body are found by summing forces and
moments acting on the body, the method of sections is being used. The body section is
the entire body itself. The entire body is being cut away from the external boundary
leaving the external reactions and the applied forces as the only external forces acting
on the body.
The examples at the end of the section demonstrate the method of sections.
The Sequential Method
In many truss problems that are solved by hand, the stretching forces and the
force reactions can be found more easily if the method of sections and the method of
joints are used in combination. The method of sections is used first to find the truss’s
external reactions and the method of joints is then used to find the truss’s stretching
forces. The external reactions are found using the method of sections by letting the
body section be the entire truss. The unknown forces are found by looking at the force
equations sequentially, moving from joint to joint. At each successive joint, at most
two forces are unknown. This way, the unknown forces can be found by summing
forces in the x and y directions at the joint. Then, you move on to the next joint and
find the unknown forces at that joint. These steps are repeated until you’ve covered the
entire truss. The following describes the problem-solving steps.
Set-up
 To begin, check whether or not the structure is a planar truss. It’s a planar
truss if
(a) all of the members and forces lie in a plane, and
(b) the structural members are each two-force members.
The truss is a space truss if (a) does not hold but (b) does hold. It’s not a truss
if (b) does not hold.

Use Eq.(5 – 1) to check whether the planar truss is statically determinate.
Transition
 If the structure is not a truss because an external force(s) does not act at a
joint, then the structure can be reduced to a truss by replacing the external
force(s) with equivalent forces that act at each end of the member. The
3
stretching forces are accurately represented by the equivalent truss throughout,
except in the member in question.

Figure 5 – 10: Situations that eliminate
unknown reactions
Inspect the structure at each of the joints to determine whether there are any
unknown forces that can be eliminated in advance. Unknown forces can be
eliminated in advance in the following three situations (Refer to Fig. 5 – 10):
(a) Members A and B are pinned at a single joint. First assume that member A
is aligned with member B. It follows that the stretching force PA in member A
and the stretching force PB in member B are equal. This eliminates one
unknown. Also, the stretching forces in the two members need to act in
tension. Otherwise, these two members become unstable. Next, assume that
member A is not aligned with member B. Then PA = PB = 0. This eliminates
both unknowns.
(a) Two members are pinned at the
joint
(b) Three members are pinned at
the joint
(c) Four members are pinned at the
joint
(b) Members A, B, and C are pinned at a single joint. Assume that member A
is aligned with member B. The stretching force PC in member C is then zero.
This eliminates one unknown. But, with PC = 0, the joint now looks like case
(a). Therefore, use case (a) to evaluate the remaining stretching forces in
members A and B.
(c) Members A, B, C, and D are pinned at a single joint. Assume that
members A and B are aligned and that members C and D are aligned. Then,
the stretching force in A is equal to the stretching force in B and the stretching
force in C is equal to the stretching force in D, that is, PA = PB and PC = PD.

Select a body section that has at most three unknown forces (Don’t count the
unknowns that are internal to the body section) and determine the sequence of
joints needed to solve the problem. Remember that at most two forces
(stretching forces and reactions) are unknown at a joint when the equations of
the joint are solved.

Draw the free body diagram of the body section and the free body diagrams of
the truss’s joints in sequential order.
Equation
 Write down Fx = 0, Fy = 0, and  = 0 for the body section. Then, write
down the force equations in sequence, that is Fx = 0 and Fy = 0 for each
joint.

Use Eq.(5 – 1) to check that the planar truss is statically determinate.
Answer
 The equations governing the equilibrium of the truss are linear algebraic
equations. Solve the equations of the body section first. Then solve the
equations for each joint in the order of the sequence.
Knowledge
Check the signs of the unknown forces. Some of the members are
expected to act in tension and others to act in compression. Also inspect the
magnitudes of the forces. Some truss members will be loaded more than
others.
The examples at the end of the section demonstrate the sequential method.
4
Finding Displacements in Planar Trusses
Figure 5 – 11: The length dx of a truss
member stretches an amount du.
Displacement Equations
Displacement Equations =
Member Displacement Equations +
Joint Displacement Equations +
Member-Joint Displacement
Equations
Strain
When designing for stiffness, it becomes important to keep a truss’s joint
displacements within specified tolerances. It can even be important when the joint
displacements are so small that they’re not easy to see. For example, the back truss on
a large microwave antenna dish needs to be extremely stiff to maintain the alignment of
electrical signals transmitted or received by the antenna.
The joint displacements of a structure are caused by the deformations of its
members. A truss member deforms along its axis. The deformation depends on the
stretching force and the material characteristics of the member. Assume that the
stretching force in the member has already been found by one of the methods
developed earlier in this section. As shown below, the unknowns will consist of a
stretch in each member and displacements of each joint. They’ll be found from
displacement equations. The displacement equations are of three types, called
member displacement equations, joint displacement equations and member-joint
displacement equations.
First consider the member displacement equations. The question arises as to
the relationship between the known stretching force and the unknown stretch in the
member. Figure 5 – 11 shows an element of a truss member of length dx acted on by a
stretching force P. The stretch du of the element divided by the original length dx of
the element is called the strain of the element. The stretching force of the member is
linearly proportional to the strain of the element, written
P  ku
(5 – 3)
Stretching Modulus
du
dx
where the proportionality constant ku is called the stretching modulus of the member.
The dimension of the stretching modulus ku is [F]. The stretching modulus depends on
the geometry of the cross-section and the material properties of the member. Assume
that the cross-sectional properties, both the geometry and the material properties, are
the same anywhere along the length of the member. This causes ku to be constant
throughout the member. Both sides of Eq. (5 – 3) can be multiplied by dx and then
integrated to get
L
u
 0 Pdx   0 ku du .
Since P and ku are constant, it follows that
u
(5 – 4)
P
ku L
where u denotes the stretch of the member. Let the truss be composed of m members.
Then, Eq. (5 – 4) is rewritten as
5
us 
(5 – 5)
Ps
, ( s  1, 2, , m),
kus Ls
where the subscript s refers to the member. Equations (5 – 5) are m member
displacement equations. They describe the relationship between the stretch in the
members and the stretching forces in the members.
Figure 5 – 12: The r-th joint displacement
Next, consider the joint displacement equations. Assume that the truss has n
joints numbered from 1 to n. The axes of the coordinate system for the truss are
denoted by X and Y, respectively. The location of a given joint before the load is
applied is designated as point A. After the load is applied, the joint moves to point A’.
The joint displacement vector is U A  [U X UY ], (r  1, 2, , n) as shown in Fig. 5 –
12. There are 2n joint displacements.
The joint displacements are constrained in two ways. They’re constrained by
force reactions and they’re constrained by member displacements. The joint
displacement equations express the constraining of the joint displacements by the force
reactions. For example, if there is a force reaction RX acting at a given external joint,
then the displacement UX at that joint is zero (See Fig. 5 – 13). Since there are p force
reactions, there are p corresponding joint displacement equations.
Figure 5 – 13: When a force reaction
acts at a joint in a given direction, the
displacement is zero at the joint in that
direction.
Figure 5 – 14: The joint displacements
and the member displacements
Next, consider the member-joint displacement equations, which express the
constraining of the joint displacements by the member displacements. Figure 5 – 14
shows a typical member. Before the load is applied, the left end of the member is
located at point A and the right end is located at point B. After the load is applied, the
left end of the member is located at point A’ and the right end is located at point B’.
The joint displacement of the left end is UA and the joint displacement of the right end
is UB. Before the load is applied, the length of the member is L and the member acts in
the n direction where n is the unit vector directed along the length of the unstretched
member. After the load is applied, the member displaces and the length of the member
becomes L + u where u is the stretch of the member. The displaced member acts in the
nS direction. From Fig. 5 – 14
U A  ( L  u)n S  Ln  U B ,
(5 – 6)
Multiply each side of Eq. (5 – 6) by n· to get
(5 – 7)
u(n S  n)  L(n S  n  1)  (U B  U A )  n.
In Eq. (5 – 7) n S  n = cos where is the angle of rotation of the member (See Fig. 5
– 14). Since the angle of rotation is small, cosEquation (5 – 7) reduces to
u  (U B  U A )  n.
(5 – 8)
Given that the truss is composed of m members, Eq. (5 – 8), applied to each member,
becomes
(5 – 9)
us  (U B  U A )  n s , (s  1, 2,  m).
6
Equations (5 – 9) are m member-joint displacement equations.
The stretch in each member is unknown and the two displacements of each
joint are unknown so there are a total of m + 2n unknowns. As seen above the number
of member displacement equations is m, the number of joint displacement equations is
p, and the number of member-displacement equations is m, so for a statically
determinate planar truss
(5 – 10)
2m + p = m + 2n.
An example at the end of this section illustrates step-by-step how to find the
displacements in statically determinate planar trusses.
Space Trusses
In the case of a space truss, there are 3n joint displacements instead of 2n so
the total number of unknowns is m + 3n. For a statically determinate space truss
(5 – 11)
m + p = 3n.
Space trusses are considered further in Section 5.5.
Statically Indeterminate Trusses
Recall that it was pointed out in section 3.6 in the discussion on determinacy
that the forces in a statically determinate system can not be found without considering
the system’s displacements. In the previous sub-section, a procedure was presented for
finding displacements. Does this mean that we can now solve statically indeterminate
truss problems? The answer is yes.
In a statically determinate truss, the force equations were solved first, after
which displacement equations were solved. All of the equations were linear algebraic
equations. In statically indeterminate trusses, it’s not possible to first solve force
equations and then displacement equations. But this doesn’t mean that the equations
can’t be solved. It’s simply not possible to solve the equations in that order because the
equations become coupled, meaning only that they need to be solved together. This
does not present a problem when solving the equations by computer. But when solving
the equations by hand, the number of simultaneous equations becomes large, except in
simple cases, and therefore considerably more difficult to solve by hand.
Let’s look at the number of equations and the number of unknowns in a
statically indeterminate planar truss. The number of equations and the number of
unknowns in the different types of equations considered in this section are tabulated
below.
Equation
Type
Force
Displacement
Total
7
Number of
Equations
2n
2m + p
2m + 2n + p
Number of
Unknowns
m+p
m + 2n
2m + 2n + p
Forces
In general, the truss can be partially constrained or fully constrained. If partially
constrained, the truss has free degrees of freedom creating a system that has no unique
solution. In practice, the truss problem needs to be set up so that the truss is fully
constrained. In the Table we see that the total number of equations is the same as the
total number of unknowns. This table shows that the statically indeterminate truss
problem can be solved.
Note that the conclusions arrived at above also apply to space trusses. For
space trusses
Equation
Type
Force
Displacement
Total
Number of
Equations
3n
2m + p
2m + 3n + p
Number of
Unknowns
m+p
m + 3n
2m + 3n + p
Again, the total number of equations is equal to the total number of unknowns.
Therefore, the forces and displacements in fully constrained space trusses can also be
solved using the equations developed in this section.
Statically indeterminate truss problems are generally solved by computer
because of the relatively large number of equations involved. As the table shows, a
statically indeterminate planar truss consisting of only n = 3 joints, m = 2 members,
and p = 3 reactions has 2m + 2n + p = 13 equations. Examples illustrating how to
solve statically indeterminate truss problems are deferred to section 5.5 entitled
Computer Analysis.
Key Terms
Body Section; Displacement Equations; Force Equations; Gusset Plate; Joint Displacement Equations; Member
Displacement Equations; Member-Joint Displacement Equations; Method of Joints; Method of Sections; Planar Truss;
Sequential Method; Space Truss; Strain; Stretching Modulus, Two-Force Members; Truss;
Review Questions
1. Define a truss structure.
2. The weight load of a structural member acts throughout the length of a truss member, which violates the definition of a truss.
How is this overcome?
3. What is a planar truss? What is a space truss?
4. State the relationship between the number of joints, the number of members, and the number of external reactions in a planar
truss and in a space truss.
5. Describe the method of joints.
6. Describe the method of sections.
7. Describe the method of sequence.
8
8. Which truss method is used to solve space truss problems by computer?
9. Define strain.
10. Name the three types of displacement equations.
11. Name the different types of equations needed to solve statically indeterminate truss problems.
9
Examples
Fig. 1a
Fig. 1b
5–1
The Method of Joints
A wench supports a 450 lb load using the truss structure shown. Find the stretching
force in each of the truss’s members.
Solution
Set-up
The forces lie in a plane and at the ends of each member so this is a planar truss
problem. Since there are only three members, the method of joints is used. The joints
and members were labeled, and a coordinate system was introduced (See Fig. 1b).
Also, notice that 2n = m + p where n = 3 is the number of joints, m = 3 is the number
of members, and p = 3 is the number of external reactions. This is necessary for the
truss to be statically determinate.
Transition
The free body diagrams of the joints are shown in Fig. 1b.
Equation
The resultant forces in the x and y directions are summed to zero to get
joint A : Fx  0  PAB  PAC cos 30  R Ax ,
F y  0  PAC sin 30  R Ay ,
(a)
joint B : Fx  0   PAB  PBC cos 60 ,
F y  0  PBC sin 60  R By ,
joint C : Fx  0   PAC cos 30  PBC cos 60
F y  0   PAC sin 30  PBC sin 60  450
Notice that the member forces were taken to be positive in tension and that the
reactions were taken to be positive in the directions of the axes. The forces are all
measured in lbs.
Answer
Equations (a) are a set of six linear algebraic equations expressed in terms of the six
unknowns PAB, PAC, PBC, RAx, RAy, and RBx. The solution is
PAB  390 lb, PAC  450 lb, PBC  779 lb,
(b)
R Ax  0, R Ay  225 lb, R Bx  675 lb.
Knowledge
Members AB and BC are in compression and member AC is in tension. The reaction at
A in the y direction is negative, so the truss would tip over if point A were not anchored
or weighted to the ground. Also, the compressive force in member BC is twice as large
as the compressive force in member AB.
Fig. 2a
5–2
The Method of Joints
A 200 lb load acts on a 180 lb tapered beam, as shown. The tapered beam leans against
a smooth wall on one end and is supported by wires on each end, as shown. Find the
wall reactions and the tension in the wires. Assume that the force per unit length of
member BC at B is twice as large as the force per unit length at C.
Solution
Set-up
Structural members AB, BC, and AC lie in a plane although the external forces do not
act at joints. The discrete load acts at point D and the weight load is distributed along
the length of BC – neither act at the ends of member BC. However, the discrete load W1
= –200 lb and the distributed load W2 = –180 lb can be replaced with equivalent point
loads that act at points B and C, thereby converting the structure into a truss.
Fig. 2b
10
Fig. 2c
The point forces at points B and C that are equivalent to the discrete load W1 are
4W1
WC1 
 160 lb, WB1  W1  WC1  40 lb.
(a)
5
The force per unit length of the tapered beam is of the form q = q0(1– x/10) lb/ft for
some constant q0. (Notice that q(0) = 2q(5).) Since the weight of the tapered beam is
W2 = –180 lb,
W2 
(b)
5
5
 qdx  q  (1  x / 10)dx  3.75q ,
0
0
0
0
so q0 = W2/3.75 = –48 lb/ft . The point forces at points B and C that are equivalent to q
are (See Fig. 2b)
1 5
48 5
qxdx 
(1  x / 10 ) xdx  64 lb, WB 2  W2  WC 2  116 lb.
(c) WC 2 
5 0
5 0
The point forces at each end of member BC that are equivalent to the discrete load and
the distributed load are
(d)
WC  WC1  WC 2  224 lb, WB  WB1  WB 2  156 lb.
The labeling of the forces is shown in Fig. 2c. Also, notice that 2n = m + p where n =
3 is the number of joints, m = 3 is the number of members, and p = 3 is the number of
external reactions.
Transition
The free body diagrams of the joints are shown in Fig. 2c.
Equation
The resultant forces in the x and y directions at the joints are
joint A : Fx  0  PAC cos  R Ax ,
2


Fy  0   PAB  PAC sin   R Ay ,
(e)
joint B : Fx  0  PBC  R Bx ,
Fy  0  PAB  156 ,
joint C : Fx  0   PBC  PAC cos ,
Fy  0  PAC sin   224
From Fig. 2c,  = tan (2/5) = 21.801○.
Answer
Equations (e) are six linear algebraic equations expressed in terms of the six unknowns
PAB, PAC, PBC, RAx, RAy, and RBx. The solution is
PAB  156 lb, PAC  603 lb, PBC  560 lb,
(f)
R Ax  560 lb, R Ay  380 lb, R Bx  650 lb.
-1
Knowledge
Member BC is in compression and members AB and AC are in tension. The tensile
force in member AB is much smaller than the other forces, which are of the same order
of magnitude.
Fig. 3a
5–3
The Method of Sections
A 400 lb load acts on the end of the truss shown. Find the internal forces in members
CD, CF, and EF.
Solution
Set-up
The structural members lie in a plane and they’re all two-force members so the
structure is a planar truss. Since, only a few internal forces are sought, the method of
sections will be used. First, a body section consisting of the entire body will be selected
in order to find the external reactions. Then, a body section that cuts the truss through
members CD, CF, and EF will be selected to find the internal forces in members CD,
CF, and EF. Also, notice that 2n = m + p where n = 17 is the number of joints, m = 31
11
is the number of members, and p = 3 is the number of external reactions.
Transition/Equation
The free body diagram of the entire truss and the free body diagram of the “cut” truss
are shown in Fig 3b. Considering the free body diagram of the entire truss, the resultant
forces in the x and y directions and the sum of the moments about point A are
Fx  0  R Ax  R Bx ,
Fig. 3b
Fy  0  R Ay  400 ,
(a)
M A  0  2.25 R Bx  6.00 F
The solution to (a) is
(b)
R Ax  1066.7 lb, R Ay  400 lb, RBx  1066.7 lb.
Considering the free body diagram of the cut truss, the resultant forces in the x and y
directions and the sum of the moments about point A are
Fx  0  1066 .7  1066 .7  PCD  PCF cos  PEF ,
(c)
Fy  400  PCF sin  ,
M A  2.25(1066 .7)  2.25 PEF  PCF 12 sin  .
From Fig. 3a, the angle  is 36.870○. The solution to (c) is
(d)
PCF  667 lb, PEF  3200 lb, PCD  3730 lb.
Knowledge
The magnitude of PCF in the cross-member is considerably smaller than the magnitudes
of the forces PCD and PEF in the top and bottom members. The top member is in
compression and the bottom member is in tension, as expected.
Fig. 4a
Fig. 4b
5–4
The Sequential Method
A truss system for the roof of a mountain cottage is acted on by snow loads. The snow
loads are transferred from the roof to the truss through point loads at B, D, and G, as
shown. For simplicity, the loads are equal to each other. Find the truss reactions at
points A and I and the forces in the members of the truss.
Solution
Set-up
The structural members are all two-force members acted on by forces in a plane so the
structure is a planar truss. There are more than three members, so the sequential
method is used to solve the problem. The reactions will be found by letting the entire
truss be a body section and then the stretching forces in the members will be found
sequentially, joint-by-joint. Notice that 2n = m + p where n = 9 is the number of joints,
m = 15 is the number of members, and p = 3 is the number of external reactions.
Transition
The free body diagram of the entire truss and the free body diagrams of the individual
joints are shown in Fig. 4b. Notice in Fig. 4a that joint C connects two parallel
members and a third member so PCB = 0 and PAC = PCF. The three truss members AC,
BC, and CF are equivalent to a single truss member AF. Similarly, joint H connects
two parallel members and a third member. The three truss members FH, HI, and HG
are equivalent to a single truss member FI. Also, joint E connects two parallel
members and a third member so PEF = 0 and PBE = PEG. The three truss members EF,
BE, and EG are equivalent to a single truss member BG. The free body diagrams of the
joints were drawn accordingly in Fig. 4b. The equation 2n = m + p still holds where
now n = 6 is the number of joints, m = 9 is the number of members, and p = 3 is the
number of external reactions. First, the reactions RAx, RAy, and RIy will be found from
the free body diagram of the entire truss. Then the internal forces will be found in a
sequence. The sequence of joints is A, I, F, B, and then G. Notice that the internal
forces of at most two members are unknown at a joint when the joint is being analyzed.
Also, note that there are other sequences that would be suitable.
Equation/Answer
12
Considering the free body diagram of the entire truss, the resultant forces in the x and y
directions and the sum of the moments about point A are
Fx  0  R Ax
(a)
Fy  0  R Ay  RIy  3F0 ,
M A  0  ( L / 2) F0  LF0  (3L / 2) F0  (2 F0 ) RIy
The solution to (a) is
(b)
RAx  0, RAy  1.5F0 , RIy  1.5F0 .
Next, the free body diagrams of the individual joints are considered.
For joint A:
Fx  0  R Ax  PAF  PAB cos 30 ,
(c)
F y  0  R Ay  PAB sin 30 .
The solution is PAB = –3F0, PAF = 2.60F0.
For joint I:
Fx  0   PFI  PGI cos 30 ,
(d)
F y  0  R Iy  PGI sin 30 .
The solution is PFI = 2.60F, PGI = –3F.
For joint F:
Fx  0   PAF  PFI  PBF cos 30  PFG cos 30 ,
(e)
F y  0  PBF sin 30  PFG sin 30 .
The solution is PBF = 0, PFG = 0.
For joint B:
Fx  0   PAB cos 30  PBF cos 30  PBD cos 30  PBG ,
(f)
Fy  0   PAB sin 30  PBF sin 30  PBD sin 30  F0 .
The solution is PBD = –F0, PBG = –1.73F0.
For joint G:
Fx  0  PGI cos 30  PFG cos 30  PGD cos 30  PBG ,
(g)
Fy  0   PGI sin 30  PFG sin 30  PGD sin 30  F0 .
The solution is PGD = –F0.
To summarize, the internal forces and the reactions are
PAB  PIG  3F0 , PBD  PGD   F0 , PBG  1.73 F0 ,
(h)
PBF  PFG  0,
PAF  PIF  2.60 F0 ,
R Ax  0, R Ay  RIy  1.5F0 .
Knowledge
The members on the top of the roof are all in compression in addition to the horizontal
cross member BG. The cross members BF and FG carry none of the snow load (for the
snow load under consideration). Members AF and FI carry the tensile loading in the
truss. Also, you may have noticed that the problem was symmetric about the vertical
axis going through the center of the truss. The symmetry could have been used to
shorten the analysis.
Fig. 5a
5–5
The Sequential Method
A truss supports a 150 lb/ft load, as shown. Find the external reactions acting on the
truss along with the stretching forces in all of the truss’s members. Note that this type
of truss is called a Warren truss.
13
Solution
Fig. 5b
Fig. 5c
Set-up
The structural members act in a plane but not all of the members are two-force
members. The 150 lb/ft distributed load needs to be replaced with point loads to reduce
the structure to a truss. Referring to Fig. 5b, the distributed load is equivalent to the
point loads
(a)
FAy = –750 lb, FCy = –1500 lb, FEy = –750 lb,
Also, since there are more than three members in the structure, this problem will be
solved by the sequential method. Notice that 2n = m + p where n = 5 is the number of
joints, m = 7 is the number of members, and p = 3 is the number of external reactions.
Transition
The free body diagram of the entire truss and the free body diagrams of the individual
joints are shown in Fig. 5c. The reactions RAx, RAy, and REy will be found from the free
body diagram of the entire truss. Then the internal forces will be found in a sequence.
The selected sequence of joints is A, E, B, and then D.
Equation/Answer
Considering the free body diagram of the entire truss:
Fx  0  R Ax
(b)
Fy  0  R Ay  REy  750  1500  750 ,
M A  10  1500  20  750  20 R Ey
The solution to (b) is
(c)
R Ax  0, R Ay  1500 lb, REy  1500 lb.
Next, the free body diagrams of the individual joints are considered.
For joint A:
Fx  0  PAB cos 60  PAC  R Ax ,
(d)
F y  0  PAB sin 60  R Ay  750 .
The solution is PAB = –866 lb, PAC = 433 lb.
For joint E:
Fx  0   PCE  PDE cos 60 ,
(e)
F y  0  PDE sin 60  R Ey  750 .
The solution is PCE = 433 lb, PDE = –866 lb.
For joint B:
Fx  0   PAB cos 60  PBC cos 60  PBD ,
(f)
F y  0   PAB sin 60  PBC sin 60 .
The solution is PBC = 866 lb, PBD = –866 lb.
For joint D:
Fx  0   PBD  PCD cos 60  PDE cos 60 ,
(g)
F y  0   PCD sin 60  PDE sin 60 .
The solution is PCD = 866 lb.
To summarize, the internal forces and the reactions are
PAB  PDE  PBD  866 lb ,
(h)
PBC  PCD  866 lb,
PAC  PCE  433 lb ,
R Ax  0, R Ay  R Ey  1500 lb.
Knowledge
The members on the top of the truss are all in compression. The cross members and the
bottom truss members are all in tension. Of course, the bottom members are equivalent
forces, so the internal forces PAC and PCE are not the actual internal forces in these
members. Also, notice the symmetry of the loads about the vertical axis going through
14
the center of the truss. Finally, it’s interesting that the largest internal forces are in the
top members and the cross members and that the smaller internal forces are in the
bottom members AC and CE even though the external loads act on members AC and
CE. Clearly, the Warren truss does a good job of redistributing the loads away from
members AC and CE.
Fig. 6a
Fig. 6b
5–6
The Sequential Method
A roof truss for a stadium is acted on by a 1000 lb load, as shown. Find the external
reactions acting on the truss along with the stretching forces in all of the truss’s
members. Take advantage of symmetry to simplify the analysis.
Solution
Set-up
The roof truss is a planar truss. Since the structure is made from more than three
members, the sequential method will be used. Notice that 2n = m + p where n = 10 is
the number of joints, m = 17 is the number of members, and p = 3 is the number of
external reactions. Symmetry will be taken advantage of to solve the problem.
Transition
The angles between the different truss members are shown in Fig. 6b. The free body
diagram of the entire truss and the free body diagrams of joints 1, 2, 3, 4, and 6 are
shown in Fig. 6c. The free body diagrams of joints 7, 8, 9, and 10 will not be needed
because of the symmetry in the problem. The reactions RAx, RAy, and REy will be found
from the free body diagram of the entire truss. Then the internal forces will be found in
a sequence. The selected sequence of joints is 1, 2, 4, 3, and 6.
Equation/Answer
Considering the free body diagram of the entire truss:
Fx  0  R1x
(a)
Fig. 6c
F y  0  R1 y  R10 y  1000 ,
M A  20  1000  40 R10 y
The solution to (a) is
(c)
R1x  0, R1y  500 lb, R10 y  500 lb.
Next, the free body diagrams of the individual joints are considered. The angles of the
joints are shown in Fig. 6b.
For joint 1:
Fx  0  P12  P13 cos 67 .5  R1x ,
(d)
Fy  0  P13 sin 67 .5  R1 y .
The solution is P12 = 207.11 lb, P13 = –541.20 lb.
For joint 2:
Fx  0   P12  P24 cos 67 .5,
(e)
F y  0  P23  P24 sin 67 .5.
The solution is P23 = –500 lb, P24 = 541.20 lb.
For joint 4:
Fx  0   P24 cos 67 .5  P34 cos 45  P46 cos 22 .5,
(f)
Fy  0   P24 sin 67 .5  P34 sin 45  P46 sin 22 .5.
The solution is P34 = 414.22 lb, P46 = 541.20 lb.
For joint 3:
Fx  0   P13 cos 67 .5  P35 cos 22 .5  P36  P34 cos 45,
(g)
Fy  0   P13 sin 67 .5  P35 sin 22 .5  P34 sin 45 .
The solution is P35 = –1847.8 lb, P36 = 1207.1 lb.
For joint 6:
(g)
Fy  0  2P46 sin 22.5  P56 .
The solution is P56 = 414.22 lb.
15
To summarize, the internal forces and the reactions are
P13  541 lb , P35  1850 lb, P23  500 lb ,
(h)
P12  207 lb, P24  P46  541 lb, P34  P56  414 lb, P36  1210 lb,
R1x  0, R1y  R10 y  500 lb.
Knowledge
The members on the outer radius of the truss are all in compression and the members
on the inner radius are all in tension. The cross members are in tension except for the
vertical cross member. Members (3, 5) and (5, 8) carry the largest tensile forces and
members (3, 6) and (6, 8) carry the largest compressive forces. The magnitudes of the
forces vary widely. Can the wide range of the internal forces be viewed as a
disadvantage?
Fig. 7a
(same as 1a)
Fig. 7b
5–7
Finding Displacements in Planar Trusses
Consider Example 5 – 1 once again. Assume that the stretching moduli of the members
are ku = 100,000 lb. Find the joint displacements of each of the truss’s joints.
Solution
Set-up
The system is a statically determinate planar truss. The joints and members were
labeled, and a coordinate system was introduced earlier (See Fig. 1b). The number of
joints is n = 3 and the number of members is m = 3. The stretching forces were also
determined earlier. To solve this problem, m = 3 member displacement equations will
need to be solved first. Then, p joint displacement equations will need to be solved.
Finally, m = 3 member-joint displacement equations will be solved.
Transition
The lengths of the members are found from Fig. 7b. It is found that LBC = 3 ft and LAC
= 5.1962 ft. Also, you can see that the member unit vectors are nAB = i, nBC = cos60i +
sin60j, and nAC = cos30i + sin30j.
Equation
First, consider the three member displacement equations. From Eq. (5 – 5)
P L
389 .71  3
member AB : u AB  AB AB 
 0.011691 ft ,
ku
100 ,000
(a)
member BC : u BC 
PBC L BC  779 .42  3

 0.023383 ft ,
ku
10,000
PAC L AC 450  5.1962

 0.023383 ft ,
ku
10,000
Next consider the joint displacement equations. They are
(b)
U AX  0, U AY  0, U BY  0.
From Eq. (5 – 9), the three member-joint displacement equations are
member AC : u AC 
u AB  (U B  U A )  n AB  ((U BX  U AX )i  (U BY  U AY ) j)  i
 U BX  U AX  0.011691 ,
(c)
u BC  (U C  U B )  n BC  ((U CX  U BX )i  (U CY  U BY ) j)  (cos 60 i  sin 60 j)
 (U CX  U BX ) cos 60  (U CY  U BY ) sin 60  0.023383 ,
u AC  (U C  U A )  n AC  ((U CX  U AX )i  (U CY  U AY ) j)  (cos 30 i  sin 30 j)
 (U CX  U AX ) cos 30  (U CY  U AY ) sin 30  0.023383 .
Answer
Equations (b) and (c) are six linear algebraic equations expressed in terms of the six
unknowns UAX, UAY, UBX, UBY, UCX, and UCY. The solution is
Fig. 7c
16
U AX  0, U AY  0,
U BX  0.0167 ft, U BY  0,
(b)
U CX  0.0697 ft, U CY  0.0740 ft,.
Knowledge
Joint C moves down and to the right. It moves down about the same as it moves to the
right (See Fig. 7c).
17
Problems
Method of Joints
5.2 – 1 (L)
A three-member truss supports a 100 lb load, as shown. Find the truss’s member forces and the external reactions.
Answer: PAB = 64 lb, PAC = 60 lb, PBC = –80 lb, RAx = –48 lb, RAy = 100 lb, and RBx = 48 lb.
5.2– 2 (L)
A three-member truss supports a 100 lb load, as shown. Find the truss’s member forces and the external reactions.
Answer: PAB = –36 lb, PAC = 60 lb, PBC = –80 lb, RAx = –48 lb, RBx = 48 lb, and RBy = 100 lb.
5.2 – 3 (L)
A four-member truss supports a 200 lb load, as shown. Find the truss’s member forces and the external reactions.
Answer: P12 = 173 lb, P13 = 100 lb, P23 = –112 lb, P34 = 50 lb, R2x = –173 lb, R2y = 0, R4x = –50 lb, R4y = 0.
5.2 – 4 (L)
A five-member truss supports two 75 lb loads, as shown. Find the truss’s member forces and the external reactions.
Answer: P12 = 173 lb, P13 = –100 lb, P23 = –62.5 lb, P24 = 188 lb, P34 = 37.5 lb, R3x = 150 lb, R4x = –150 lb, R4y = 150 lb.
5.2 – 5 (L)
A five-member truss supports a 150 lb load, as shown. Find the truss’s member forces and the external reactions.
Answer: P12 = 250 lb, P13 = –200 lb, P23 = 0, P24 = 250 lb, P34 = 0 lb, R3x = 200 lb, R4x = –200 lb, R4y = 150 lb.
5.2 – 6 (L)
A 1 m by 1 m box truss supports a 100 N load at an angle , as shown. The cross-members are wire braces that can only
act in tension. One of them will act in tension while the internal force in the other is zero. Find the truss’s member forces
and external reactions when  = 30○. Over what range of  is brace (1, 4) under tension?
Answer: P13 = P23 = P24 = P34 = 0, P14 = 173 N, P12 = –36.6 N, R2y = 36.6 N, R4x = –86.6 N, R4y = –86.6 N,   90 .
5.2 – 7 (L)
A 1 m by 1 m box truss supports a 50 N load and a 100 N load at an angle , as shown. The cross-members are wire braces
that can only act in tension. One of them will act in tension while the internal force in the other is zero. Find the truss’s
member forces and external reactions when  = 0○. Over what range of  is brace (1, 4) under tension?
Answer: P13 = –50 N, P23 = P24 = P34 = 0, P14 = 212 N, P12 = –150 N, R2y = 150 N, R4x = –150 N, R4y = –150 N,   60 .
5.2 – 8 (L)
A five-member truss supports a 1000 N load, as shown. Find the truss’s member forces and external reactions.
Answer: P12 = P23 = P24 = 0, P34 = 1410 N, P13 = –1000 N, R1x = 1000 N, R4x = –1000 N, R4y = 1000 N.
5.2 – 9 (L)
A five-member truss supports a 1000 N load, as shown. Find the truss’s member forces and external reactions.
Answer: P23 = 0, P12 = P24 = –1000 N, P13 = –1000 N, P34 = 1410 N, R1x = 1000 N, R1y = 1000 N, R4y = –1000 N.
5.2 – 10 (M)
A five-member truss supports a 200 lb load, as shown. The maximum allowable tensile force in the members is 300 lb.
(There’s no limit placed on the compressive force in the members.) Find the minimum angle min of the truss.
Answer: min = 30○.
18
5.2 – 11 (M)
A five-member truss supports a 200 lb load, as shown. The maximum allowable compressive force in the members is 300
lb. (There’s no limit placed on the tensile force in the members.) Find the minimum angle of the truss.
Answer: min = 41.8○.
5.2 – 12 (M)
A three-member structure supports a 100 lb load, as shown. Replace the load with equivalent forces that reduce the
structure to a truss. Then, find the truss’s member forces and external reactions.
Answer: PAB = 75 lb, PAC = –125 lb, PBC = 100 lb, RAx = 100 lb, RBx = –100 lb, RBy = 100 lb.
.
5.2 – 13 (M)
A three-member structure supports a 200 lb load, as shown. Replace the load with equivalent forces that reduce the
structure to a truss. Then, find the truss’s member forces and external reactions.
Answer: PAB = 300 lb, PAC = –500 lb, PBC = 400 lb, RAx = 400 lb, RBx = –400 lb, RBy = 200 lb.
.
5.2 – 14 (M)
A three-member structure is composed of two very light members and a heavy member that weighs 200 lb. Replace the
weight load of the heavy member with equivalent forces that reduce the structure to a truss. Then, find the truss’s member
forces and external reactions.
Answer: PAB = –200 lb, PAC = 320 lb, PBC = –250 lb, RAx = –250 lb, RAy = 0, RBx = 250 lb, RBy = 200 lb.
.
5.2 – 15 (M)
A three-member structure is acted on by two loads, as shown. Reduce the structure to a truss, and find the truss’s member
forces and external reactions.
Answer: PAB = 75 lb, PAC = 150 lb, PBC = –168 lb, RAx = –150 lb, RAy = 100 lb, RBx = 150 lb.
.
5.2 – 16 (M)
A three-member structure is composed of two very light members and a heavy member that weighs 150 N. Replace the
weight load of the heavy member with equivalent forces that reduce the structure to a truss. Then, find the truss’s member
forces and external reactions.
Answer: PAB = 0, PAC = 150 N, PBC = –168 N, RAx = –150 N, RAy = 0, RBx = 150 N, RBy = 150 N.
.
5.2 – 17 (M)
A three-member structure is composed of two very light members and a heavy member that weighs lb. The heavy
member’s weight per unit length decreases linearly from point B to zero at point C. Replace the weight load of the heavy
member with equivalent forces that reduce the structure to a truss. Then, find the truss’s member forces and external
reactions.
Answer: PAB = 600 lb, PAC = PBC = 0, RAx = 0, RAy = 600 lb, RBx = 130 N, RCx = 0, RCy = 300 lb.
.
5.2 – 18 (M)
A three-member truss is acted on by a 100 lb load, as shown. Find the truss’s member forces and external reactions.
Answer: PAB = 80 lb, PAC = –48 lb, PBC = 60 lb, RAx = –51.2 lb, RAy = 38.4 lb, RCx = –28.8 lb, RCy = 21.6 lb.
.
5.2 – 19 (M)
A truss supports a 300 lb sign, as shown. Find the truss’s member forces and the external reactions.
Answer: PAB = 0, PAC = 0, PCD = -300 lb, PAD = 424 lb, PCD = –300 lb, PBD,TOP = 0, PBD,BOTTOM = –300 lb, RAx = –300 lb, RAy
= 300 lb, RCx = 300 lb.
Method of Sections
5.2 – 20 (M)
A Pratt type roof truss supports three 200 lb loads from above, as shown. Find the forces in members (3, 8) and (4, 8).
Answer: P3, 8 = 283 lb, P4, 8 = –200 lb.
5.2 – 21 (M)
A Pratt type roof truss supports three 200 lb loads from below, as shown. Find the forces in members (3, 8) and (4, 8).
Answer: P3, 8 = 283 lb, P4, 8 = 0.
19
5.2 – 22 (M)
A Pratt type roof truss supports five 200 lb loads from above, as shown. Find the forces in members (3, 8) and (4, 9).
Answer: P3, 8 = 361 lb, P4, 9 = 424 lb.
5.2 – 23 (M)
A Pratt type roof truss supports is subjected to a wind load that is represented as three 75 shear loads, as shown. Find the
forces in members (3, 8) and (5, 12).
Answer: P3, 8 = P5, 12 = 0.
5.2 – 24 (M)
A Pratt type roof truss supports is subjected to a 300 lb load from below, as shown. Find the forces in members (10, 11),
(11, 12), (4, 11), and (5, 11).
Answer: P10, 11 = 300 lb, P11, 12 = 600 lb, P4, 11 = 424 lb, P5, 11 = 0.
5.2 – 25 (M)
A Howe type roof truss supports a 400 lb load from below, as shown. Find the forces in members (4, 8), (5, 8), and (7, 8).
Answer: P4, 8 = 400 lb, P5, 8 = 600 lb, P7, 11 = 600 lb.
5.2 – 26 (M)
A Howe type roof truss supports a 400 lb load from below, as shown. Find the forces in members (2, 7), (3, 7), and (6, 7).
Answer: P2, 7 = 0, P3, 7 = 400 lb, P6, 7 = 400 lb.
5.2 – 27 (M)
A Howe type roof truss supports five 100 lb loads from above, as shown. Find the forces in members (2, 8), (3, 9), and
(4, 10).
Answer: P2, 8 = 0, P3, 9 = 50 lb, P4, 10 = 800 lb.
5.2 – 28 (M)
A Howe type roof truss is subjected to a wind load that is represented as six 75 lb shear loads, as shown. Find the forces in
members (10, 11), (11, 12), and (12, 7) and the external reactions.
Answer: P10, 11 = 213 lb, P11, 12 = 213 lb, P12, 7 = 213 lb, R1x = –427 lb, R1y = –71.2 lb, R7y = 71.2 lb.
5.2 – 29 (M)
A Fink type roof truss is acted on by a 1000 N load, as shown. Find the forces in members (5, 6), (5, 15), and (12, 15). Let
 = 20○.
Answer: P5, 6 = –1460 lb, P5, 15 = P12,15 = 0.
5.2 – 30 (M)
A Fink type roof truss is acted on by four 250 lb loads from below, as shown. Find the forces in members (5, 6), (5, 13),
(6, 13), (7, 13), (7, 14), (7, 15) and (8, 15).
Answer: P5, 6 = –1100 lb, P5, 13 = 583 lb, P6, 13 = P7, 13 = 0, P7, 14 = –266 lb, P7, 15 = 778 lb, P8, 15 = 0.
5.2 – 31 (M)
A bridge truss is acted on by a distributed 10,000 lb load between points 3 and 5, as shown. Find the forces in member (2,
6).
Answer: P2, 6 = –2500 lb.
5.2 – 32 (M)
A bridge truss is acted on by a distributed 10,000 lb load, between points 4 and 5, as shown. Find the forces in member
(2,6).
Answer: P2, 6 = –2500 lb.
5.2 – 33 (M)
A bridge truss is acted on by five 2000 lb loads, as shown. Find the forces in members (2, 3), (2, 9), and (8, 9).
Answer: P2, 3 = –2500 lb, P2, 9 = 0, P8, 9 = 8000 lb.
20
5.2 – 34 (M)
A bridge truss is acted on by a 9000 lb load, as shown. Find the forces in member (3, 4).
Answer: P3, 4 = –9000 lb.
5.2 – 35 (M)
A bridge truss is acted on by a 9000 lb load, as shown. Find the forces in member (3, 4).
Answer: P3, 4 = –4500 lb.
5.2 – 36 (M)
A bridge truss is acted on by a 9000 lb load, as shown. Find the forces in member (3, 4).
Answer: P3, 4 = 13,500 lb.
5.2 – 37 (M)
A Warren type bridge truss is acted on by two 500 lb loads, as shown. Find the forces in member (1, 2).
Answer: P1, 2 = –500 lb.
5.2 – 38 (M)
A Warren type bridge truss is acted on by two 1000 lb loads, as shown. Find the forces in member (1, 2).
Answer: P1, 2 = –1000 lb.
5.2 – 39 (M)
A Warren type tower truss is acted on by a 500 lb load, as shown. Find the forces in members (1, 2) and (2, 3).
Answer: P1, 2 = 1500 lb, P2, 3 = 1000 lb.
5.2 – 40 (M)
A K-type bridge truss is acted on by three 1000 lb loads, as shown. Find the forces in member (5, 8).
Answer: P5, 8 = 1060 lb.
5.2 – 41 (M)
A K-type bridge truss is acted on by three 1000 lb loads, as shown. Find the forces in member (5, 2).
Answer: P5, 8 = 1060 lb.
5.2 – 42 (M)
A K-type bridge truss is acted on by five 1000 lb loads, as shown. Find the forces in member (4, 9).
Answer: P4, 9 = –1060 lb.
5.2 – 43 (M)
A K-type bridge truss is acted on by five 1000 lb loads, as shown. Find the forces in member (9, 14).
Answer: P9, 14 = 1060 lb.
The Sequential Method
5.2 – 44 (M)
Find all of the member forces and the external reactions in Problem 5 – 20.
Answer: P1, 2 = P2, 3 = P3, 4 = P4, 5 = –671 lb, P1, 6 = P8, 5 = 600 lb, P6, 7 = P7, 8 = 400 lb, P2, 6 = P4, 8 = –200 lb, P3, 7 = 0, P3, 6 =
P3, 8 = 283 lb, R1x = 0, R1y = R2y = 300 lb.
5.2 – 45 (M)
Find all of the member forces and the external reactions in Problem 5 – 21.
Answer: P1, 2 = P2, 3 = P3, 4 = P4, 5 = –671 lb, P1, 6 = P8, 5 = 600 lb, P6, 7 = P7, 8 = 400 lb, P2, 6 = P4, 8 = 0 lb, P3, 7 = 200 lb, P3, 6
= P3, 8 = 283 lb, R1x = 0, R1y = R2y = 300 lb.
21
5.2 – 46 (M)
Find all of the member forces and the external reactions in Problem 5 – 22.
Answer: P1, 2 = P2, 3 = P5, 6 = P6, 7 = –1580 lb, P3, 4 = P4, 5 = –1260 lb, P1, 8 = P12, 7 = 1500 lb, P8, 9 = P11, 12 = 1200 lb,
P9, 10 = P10, 11 = 900 lb, P2, 8 = P6, 12 = –200 lb, P3, 9 = P5, 11 = –300 lb, P4, 10 = 0, P3, 8 = P5, 12 = 361 lb, P4, 9 = P4, 11 = 424
lb, R1x = 0, R1y = R7y = 500 lb.
5.2 – 47 (M)
Find all of the member forces and the external reactions in Problem 5 – 23.
Answer: P1, 2 = 225 lb, P2, 3 = 150 lb, P3, 4 = 75 lb, P4, 5 = P5, 6 = P6, 7 = P1, 8 = P8, 9 = P9, 10 = P10, 11 = P11, 12 = P12, 7 = P2, 8
= P3, 9 = P4, 10 = P5, 11 = P6, 12 = P3, 8 = P4, 9 = P4, 11 = P5, 12 = 0, R1x = –213 lb, R1y = –71 lb, R7y = 0.
5.2 – 48 (M)
Find all of the member forces and the external reactions in Problem 5 – 24.
Answer: P1, 2 = P2, 3 = P3, 4 = –316 lb, P4, 5 = P5, 6 = P6, 7 = –632 lb, P1, 8 = P8, 9 = P9, 10 = P10, 11 =300 lb, P11, 12 = P12, 7
=600 lb, P2, 8 = P3, 9 = P4, 10 = P5, 11 = P6, 12 = P3, 8 = P4, 9 = 0, P4, 11 = 424 lb, P5, 12 = 0, R1x = 0, R1y = 100 lb, R7y = 200 lb.
5.2 – 49 (M)
Find all of the member forces and the external reactions in Problem 5 – 25.
Answer: P1, 2 = P2, 3 = P3, 4 = –224 lb, P4, 5 = –671 lb, P1, 6 = P6, 7 = P3, 7 = 200 lb, P2, 6 = P2, 7 = 0, P4, 8 = 400 lb,
P4, 7 = –447 lb, R1x = 0, R1y = 100 lb, R5y = 300 lb.
5.2 – 50 (M)
Find all of the member forces and the external reactions in Problem 5 – 26.
Answer: P1, 2 = P2, 3 = P3, 4 = P4, 5 = –447 lb, P1, 6 = P6, 7 = P7, 8 = P8, 5 = 400 lb, P2, 6 = P2, 7 = P4, 7 = P4, 8 = 0, P3, 7 = 400
lb, R1x = 0, R1y = R5y = 200 lb.
5.2 – 51 (M)
Find all of the member forces and the external reactions in Problem 5 – 27.
Answer: P1, 2 = P6, 7 –791 lb, P2, 3 =P5, 6 = –632 lb, P3, 4 = P4, 5 = –474 lb, P1, 8 = P8, 9 =P11, 12 = P12, 7 = 750 lb, P9, 10 =
P10, 11 = 600 lb, P2, 8 = P6, 12 = 0, P3, 9 = P5, 11 = 50 lb, P4, 10 = 200 lb, P2, 9 = P6, 11 = –158 lb, P3, 10 = P5, 10 = –180 lb,
R1x = 0, R1y = R5y = 250 lb.
5.2 – 52 (M)
Find all of the member forces and the external reactions in Problem 5 – 28.
Answer: P1, 2 = 150 lb, P2, 3 = 75 lb, P3, 4 = 0, P4, 5 = –75 lb, P5, 6 = –150 lb, P6, 7 = –225 lb, P1, 8 = P8, 9 = P9, 10 = P10, 11 =
P11, 12 = P12, 7 = 213 lb, P2, 8 = P3, 9 = P4, 10 = P5, 11 = P6, 12 = P2, 9 = P3, 10 = P5, 10 = P6, 11 = 0, R1x = –427 lb,
R1y = R7y = –71.2 lb.
5.2 – 53 (M)
Find all of the member forces and the external reactions in Problem 5 – 29.
Answer: P1, 2 = P2, 3 = P3, 4 = P4, 5 = P5, 6 = P6, 7 = P7, 8 = P8, 9 = –686 lb, P1, 10 = P10, 11 = P11, 12 = P12, 13 = P13, 9 = 470 lb,
P3, 10 = P7, 13 = P3, 14 = P15, 7 = P2, 10 = P3, 11 = P4, 14 = P6, 15 = P7, 12 = P8, 13 = P11, 14 = P5, 14 = P5, 15 = P15, 12 = 0, R1x = 0,
R1y = R9y = 500 lb.
5.2 – 54 (M)
Find all of the member forces and the external reactions in Problem 5 – 30.
Answer: P1, 2 = P2, 3 = –686 lb, P3, 4 = P4, 5 = P5, 6 = P6, 7 = –515 lb, P7, 8 = P8, 9 = –686 lb, P1, 10 =P13, 9 = 470 lb, P10, 11 =
P12, 13 = 368 lb, P11, 12 = 200 lb, P3, 10 = P7, 13 = 270 lb, P3, 14 = P15, 7 = P2, 10 = P4, 14 = P6, 15 = P7, 12 = P8, 13 = P11, 14 = P5, 14
= P5, 15 = P15, 12 = 0, R1x = 0, R1y = R9y = 500 lb.
22
5.2 – 55 (M)
Find all of the member forces and the external reactions in Problem 5 – 31.
Answer: PAC =PBE =–2800 lb, PAF =PFB = –2500 lb, PCD =PDE = 1200 lb, PFD = 0, PAD = PBD = 2800 lb, RCx = 0, RCy = REy
= 5000 lb.
5.2 – 56 (M)
Find all of the member forces and the external reactions in Problem 5 – 32.
Answer: PAC =PBE =–2800 lb, PAF =PFB = –2500 lb, PCD =PDE = 1200 lb, PFD = 0, PAD = PBD = 2800 lb, RCx = 0, RCy =
2500 lb, REy = 7500 lb.
5.2 – 57 (M)
Find all of the member forces and the external reactions in Problem 5 – 33.
Answer: P1, 2 = –8000 lb, P2, 3 = P3, 4 = –9000 lb, P4, 5 = –8000 lb, P1, 6 = –7070 lb, P6, 7 =P7, 8 = 5000 lb, P8, 9 = P9, 10 =
8000 lb, P10, 11 = P11, 12 = 5000 lb, P5, 12 = –7070 lb, P1, 7 = 0, P2, 8 = –3000 lb, P3, 9 = –2000 lb, P4, 10 = –3000 lb, P5, 11 = 0,
P1, 8 = 4240 lb, P2, 9 =P4, 9 = 1410 lb, P5, 10 = 4240, R6x = 0, R6y = R12y = 5000 lb.
5.2 – 58 (M)
Find all of the member forces and the external reactions in Problem 5 – 34.
Answer: P1, 2 = –6000 lb, P2, 3 = P3, 4 = –9000 lb, P4, 5 = –12000 lb, P1, 6 = –4240 lb, P6, 7 =P7, 8 = 3000 lb, P8, 9 = 6000 lb,
P9, 10 = 12000 lb, P10, 11 = P11, 12 = 6000 lb, P5, 12 = –8490 lb, P1, 7 = 0, P2, 8 = –3000 lb, P3, 9 = 0, P4, 10 = 3000 lb, P5, 11 = 0,
P1, 8 = P2, 9 = 4240 lb, P4, 9 = –4240 lb, P5, 10 = 8490, R6x = 0, R6y = 3000 lb, R12y = 6000 lb.
5.2 – 59 (M)
Find all of the member forces and the external reactions in Problem 5 – 35.
Answer: P1, 2 = –3000 lb, P2, 3 = P3, 4 = –4500 lb, P4, 5 = –6000 lb, P1, 6 = –2120 lb, P6, 7 =P7, 8 = 1500 lb, P8, 9 = 3000 lb,
P9, 10 = 6000 lb, P10, 11 = P11, 12 = 7500 lb, P5, 12 = –10600 lb, P1, 7 = 0, P2, 8 = –1500 lb, P3, 9 = 0, P4, 10 = 1500 lb, P5, 11 =
9000 lb, P1, 8 = P2, 9 = 2120 lb, P4, 9 = P5, 10 = –2120, R6x = 0, R6y = 1500 lb, R12y = 7500 lb.
5.2 – 60 (M)
Find all of the member forces and the external reactions in Problem 5 – 36.
Answer: P1, 2 = –9000 lb, P2, 3 = P3, 4 = –13500 lb, P4, 5 = –9000 lb, P1, 6 = –6360 lb, P6, 7 =P7, 8 = 4500 lb, P8, 9 = P9, 10 =
9000 lb, P10, 11 = P11, 12 = 4500 lb, P5, 12 = –6360 lb, P1, 7 = 0, P2, 8 = –4500 lb, P3, 9 = 0, P4, 10 = –4500 lb, P5, 11 = 0 lb, P1, 8
= P2, 9 = P4, 9 = P5, 10 = 6360, R6x = 0, R6y = R12y = 4500 lb.
5.2 – 61 (M)
Find all of the member forces and the external reactions in Problem 5 – 37.
Answer: P1, 2 = –500 lb, P1, 3 = –707 lb, P3, 4 = 500 lb, P1, 4 = P2, 4 = 0, P4, 5 = 500 lb, P2, 5 = –707 lb,R3x = 0, R3y = R5y =
500 lb.
5.2 – 62 (M)
Find all of the member forces and the external reactions in Problem 5 – 38.
Answer: P1, 2 = –1000 lb, P1, 3 = –707 lb, P3, 4 = 500 lb, P1, 4 = P2, 4 = 707 lb, P4, 5 = 500 lb, P2, 5 = –707 lb,R3x = 0, R3y =
R5y = 500 lb.
5.2 – 63 (M)
Find all of the member forces and the external reactions in Problem 5 – 39.
Answer: P1, 2 = 1500 lb, P2, 3 =1000 lb, P3, 4 = 500 lb, P5, 6 = –1250 lb, P6, 7 = –750 lb, P7, 8 = –250 lb, P1, 5 = 500 lb, P2, 5 =
P3, 6 = P4, 7 = = –559 lb, P2, 6 = P3, 7 = P4, 8 = 559 lb, R1x = –500 lb, R1y = –1500 lb, R5y = 1500 lb.
23
5.2 – 64 (M)
Find all of the member forces and the external reactions in Problem 5 – 40.
Answer: P1, 2 = P2, 3 = –750 lb, P1, 6 = –1680 lb, P6, 7 =P7, 8 = P8, 9 = P9, 10 = 750 lb, P3, 10 = –1680 lb, P1, 4 = 1500 lb, P4, 7
= 1000 lb, P3, 5 = 1500 lb, P5, 9 = 1000 lb, P2, 4 = P2, 5 = –354 lb, P4, 8 = P5, 8 = 354 lb, P2, 8 = 500 lb, R6x = 0, R6y = R10y =
1500 lb.
5.2 – 65 (M)
Find all of the member forces and the external reactions in Problem 5 – 42.
Answer: P1, 2 = –1250 lb, P2, 3 = P3, 4 = –2000 lb, P4, 5 = –1250 lb, P1, 10 = –2800 lb, P10, 11 = P11, 12 = 1250 lb, P12, 13 =
2000 lb, P14, 15 = P15, 16 = 1250 lb, P5, 16 = –2800 lb, P1, 6 = 1500 lb, P6, 11 = 0, P2, 7 = –250 lb, P7, 12 = –750 lb, P4, 8 = –250
lb, P8, 14 = –750 lb, P5, 9 = 1500 lb, P9, 15 =0, P2, 6 = P6, 12 = –1060 lb, P3, 7 = P3, 8 = –354 lb, P7, 13 = P8, 13 = 354 lb, P4, 9 = –
1060 lb, P9, 14 = 1060 lb, P3, 13 = –500 lb, R10x = 0, R10y = R16y = 2500 lb.
Finding Displacements in Planar Trusses
5.2 – 66 (M)
A three-member truss supports a 100 lb load, as shown. Find the joint displacements of the truss. The stretching modulus
of each member is ks = 5000 lb.
Answer: UAX = UAY = 0, UBX = 0, UBY = -0.064 ft, UCX = -0.0403 ft, UCY = -0.114 ft.
5 – 67 (M)
A three-member truss supports a 100 lb load, as shown. Find the joint displacements of the truss. The stretching modulus
of each member is ks = 5000 lb.
Answer: UAX = 0, UAY = -0.036 ft, UBX = UBY = 0, UCX = -0.00768 ft, UCY = -0.0858 ft.
5 – 68 (M)
A three-member structure supports a 100 lb load, as shown. Replace the load with equivalent forces that reduce the
structure to a truss. Find the joint displacements of the truss. The stretching modulus of each member is ks = 5000 lb.
Answer: UAX = 0, UAY = -0.0450 ft, UBX = UBY = 0, UCX = 0.0800 ft, UCY = -0.360 ft.
5 – 69 (M)
A three-member structure supports a 200 lb load, as shown. Replace the load with equivalent forces that reduce the
structure to a truss. Find the joint displacements of the truss. The stretching modulus of each member is ks = 25,000 lb.
Answer: UAX = 0, UAY = -0.0450 ft, UBX = UBY = 0, UCX = 0.0800 ft, UCY = -0.360 ft.
5 – 70 (M)
A three-member structure is composed of two very light members and a heavy member that weighs 150 N. Replace the
weight load of the heavy member with equivalent forces that reduce the structure to a truss. Find the joint displacements of
the truss. The stretching modulus of each member is ks = 25,000 lb.
Answer: UAX = UAY = 0, UBX = UBY = 0, UCX = 0.00480 ft, UCY = -0.0230 ft.
5 – 71 (M)
A three-member structure is composed of two very light members and a heavy member that weighs lb. The heavy
member’s weight per unit length decreases linearly from point B to zero at point C. Replace the weight load of the heavy
member with equivalent forces that reduce the structure to a truss. Find the joint displacements of the truss. The stretching
modulus of each member is ks = 5,000 lb.
Answer: UAX = UAY = 0, UBX = 0.0900 ft, UBY = -0.180 ft, UCX = UCY = 0
5 – 72 (M)
A three-member truss is acted on by a 100 lb load, as shown. Find the joint displacements of the truss. The stretching
modulus of each member is ks = 2,500 lb.
Answer: UAX = UAY = 0, UBX = 0.0960 ft, UBY = -0.172 ft, UCX = -0.0576 ft, UCY = -0.0768 ft.
24
5.2 – 73 (H)
A truss supports a 300 lb sign, as shown. Find the joint displacements of the truss. The stretching modulus of each member
is ks = 10,000 lb.
Answer: UAX = UAY = 0, UBX = 0.398 ft, UBY = -0.230 ft, UCX = UCY = 0, UDX = -0.0600 ft, UDY = -0.230 ft.
5.2 – 74 (H)
A bridge truss is acted on by a distributed 10,000 lb load between points 3 and 5, as shown. Find the joint displacements of
the truss. The stretching modulus of each member is ks = 50,000 lb.
Answer: U1X = 0.300 ft, U1Y = -0.569 ft, U2X = 0, U2Y = -0.569 ft, U3X = U3Y = 0, U4X = 0.150 ft, U4Y = -1.06 ft, U5X =
0.233 ft, U5Y = 0, U6X = 0.150 ft, U6Y = -1.06 ft.
5.2 – 75 (H)
A Warren type bridge truss is acted on by two 1000 lb loads, as shown Find the joint displacements of the truss. The
stretching modulus of each member is ks = 50,000 lb.
Answer: U1X = 0.200 ft, U1Y = -0.341 ft, U2X = 0, U2Y = -0.341 ft, U3X = U3Y = 0, U4X = 0.1 ft, U4Y = -0.583 ft, U5X = 0.2
ft, U5Y = 0.
25