ANALYSIS OF DETERMINATE TRUSS
TRUSSES
A truss is a structure that is made of straight,
slender bars that are joined together to form a
pattern of triangles. Trusses are usually designed
to transmit forces over relatively long spans;
common examples are bridge trusses and roof
trusses. Truss members are connected at their
extremities only; thus no member is continuous
through a joint.
In general, the members of a truss are slender
and can support little lateral load; all loads,
therefore, must be applied to the various joints,
and not to the members themselves.
TRUSSES
The analysis of trusses is based on the following three
assumptions:
1. The weights of the members are negligible. A truss can be classified as
a lightweight structure, meaning that the weights of its members are
generally much smaller than the loads that it is designed to carry.
2. All joints are pins. In practice, the members at each joint are usually
riveted or welded to a plate, called a gusset plate, as shown in Fig. (b).
However, if the members at a joint are aligned so that their centroidal axes
(axes that pass through the centroids of the cross-sectional areas of the
members) intersect at a common point, advanced methods of analysis
indicate that the assumption of pins is justified.
3. The applied forces act at the joints. Because the members of a truss are
slender, they may fail in bending when subjected to loads applied at
locations other than the joints. Therefore, trusses are designed so that the
major applied loads act at the joints.
TRUSSES
Although these assumptions may appear to oversimplify
the real situation, they lead to results that are adequate in
most applications.
Using the assumptions, the free-body diagram for any
member of a truss will contain only two forces—the
forces exerted on the member by the pin at each end.
Therefore, each member of a truss is a two-force body.
When dealing with the internal force in a two-force body,
commonly distinguish between tension and compression.
Figure 4.16 shows the external and internal forces in
tension and compression. Tensile forces elongate (stretch)
the member, whereas compressive forces compress
(shorten) it.
TRUSSES
Typical Trusses:
TRUSSES
SIMPLE TRUSSES
Consider the truss of Fig. a. If a load is applied at B, the truss will greatly deform. In
contrast, the truss of Fig. b, will deform only slightly under a load applied at B. The only
possible deformation for this truss is one involving small changes in the length of its
members. The truss of Fig. b is said to be a rigid truss, the term rigid being used here to
indicate that the truss will not collapse. As shown in Fig. c, a larger rigid truss can be
obtained by adding two members BD and CD to the basic triangular truss of Fig. b. This
procedure can be repeated as many times as desired, and the resulting truss will be
rigid if each time two new members are added, they are attached to two existing
joints and connected at a new joint.† A truss which can be constructed in this manner
is called a simple truss.
ANALYSIS OF DETERMINATE TRUSS
(Method of Joints)
TRUSSES - METHOD OF JOINTS
We have learned that a truss can be considered as a
group of pins and two-force members. The truss can
be dismembered, and a free-body diagram can be
drawn for each pin and each member (See figure).
Each member is acted upon by two forces, one at each
end; these forces have the same magnitude, same line
of action, and opposite sense. Therefore, the forces
exerted by a member on the two pins it connects must
be directed along that member and be equal and
opposite.
TRUSSES - METHOD OF JOINTS
When using the method of joints to calculate the
forces in the members of a truss, the equilibrium
equations are applied to individual joints (or pins) of
the truss. Because the members are two-force bodies,
the forces in the FBD of a joint are concurrent.
Consequently, two independent equilibrium equations
are available for each joint. σ 𝐹𝑥 = 0 𝑎𝑛𝑑 σ 𝐹𝑦 = 0
To illustrate this method of analysis, consider the truss
shown in Fig. a. The supports consist of a pin at A and
a roller at E.
TRUSSES - METHOD OF JOINTS
The FBD of the entire truss in Fig. b contains three
unknown reactions: Ax, Ay, and N, which can be
found from the three available equilibrium equations.
σ 𝑴𝑨 = 𝟎
σ 𝑭𝒙 = 𝟎
40,000 𝟐 + 15,000 𝟒 − 𝑵𝑬 𝟖 = 𝟎
𝑵𝑬 = 𝟏𝟕, 𝟓𝟎𝟎 𝑵
𝑨𝒙 = 𝟎 𝑵
Note that Ax is zero. This result indicates that the truss would be in equilibrium under the
given loading even if the pin at A were replaced by a roller. However, we would then have an
improper constraint, because an incidental horizontal force would cause the truss to move
horizontally. Therefore, a pin support at A (or B) is necessary to properly constrain the truss.
σ 𝑭𝒚 = 𝟎
−𝟒𝟎, 𝟎𝟎𝟎 − 𝟏𝟓, 𝟎𝟎𝟎 + 𝟏𝟕, 𝟓𝟎𝟎 + 𝑨𝒚 = 𝟎
𝑨𝒚 = 𝟑𝟕, 𝟓𝟎𝟎 𝑵
TRUSSES - METHOD OF JOINTS
Determining the forces in the individual members of the truss:
• To determine the force in members AB and AH, we can draw
the FBD of joint A. This FBD, shown in Fig. 4.18(a),
contains the external reactions Ax and Ay and the member
forces PAB and PAH
TRUSSES - METHOD OF JOINTS
• To compute the forces in members BC and BH, we draw the FBD
of joint B. This FBD, shown in Fig. 4.18(b), note that the force PAB
is equal and opposite to the corresponding force in Fig. 4.18(a),
and that we again assumed PBC and PBH to be tensile. Knowing
that PAB =−46,875 N, the equilibrium equations of the joint yield:
• We could continue the procedure, moving from joint to
joint, until the forces in all the members are determined.
METHOD OF JOINTS
Sample Problem:
Using
the
method
of
joints,
determine the force in each
member of the truss shown. State
whether each member is in tension
or compression.
Solution:
M = 0 +
R (15.75 ) − 945 (12 ) = 0
B
C
F
RC = 720 lb 
y
=0 +
RB + 720 − 945 = 0
RB = 225 lb 
RB
RC
METHOD OF JOINTS

5

Fx = 0 +
4 
 + T BC = 0 ; (− 375 )(0.8 ) + T BC = 0
5 
T BC = 300
T BA 
3
4
Force Polygon:
TBC

225
720
225
y
B
TBC
225
F
TBA
Y
x
=0 +
T BC
TBA
3 
 = tan  
4 
 = 36.87 o
−1
@ joint B
T BC
o
225
sin 36.87
= 375
o
T BA =
T BA
T BC = 300 lb Tension
225
=
tan 36.87
= 300
3 
T BA   + 225 = 0
5 
T BA = −375
T BA = 375 lb Compresssi on
METHOD OF JOINTS
@ joint C
TCA
C
13
12
y
B 300
5
F
x
=0 +
5
− TCA 
13

x
720
225
720
Force Polygon:
300
TCA
TCA = 300 2 + 720 2
TCA = 780
720

 − 300 = 0 ; TCA = −780

TCA = 780 lb Compressio n
METHOD OF JOINTS
Sample Problem:
Using the method of joints, determine
the force in each member of the truss
shown. State whether each member is
in tension or compression.
Solution:
M
D
=0 +
R E (21 ) − 693 (16 ) = 0
R E = 528 lb 
F
y
=0 +
R D + 528 − 693 = 0
R D = 165 lb 
RD
RE
METHOD OF JOINTS

Fx = 0 +
T DC cos 36.87
13

5
F
Y
12
3

4
+ T DA cos 67.38
T DC = 125
o
+ T DA sin 67.38
T DC = 125 lb (T
165
528
@ joint D
y
TDA

D
165

 12 

5 
 = tan −1 
TDC
 = 67.38 o
x
3 
 = tan −1  
4 
 = 36.87 o
o
=0
=0 +
T DC sin 36.87
5
o
)
o
+ 165 = 0
T DA = −260
T DA = 260 lb (C )
METHOD OF JOINTS

Fx = 0 +
T AB + T AC cos 36.87
13

5
F
Y
12
3

4
T AC = 400
o
+ 260 sin 67.38
T AC = 400 lb (T
165
528
@ joint A
y
A

260
TAB

x
TAC
 = 36.87
 = 67.38 o
o
+ 260 cos 67.38
o
=0
=0 +
− T AC sin 36.87
5
o
)
o
T AB = −420
=0
T AB = 420 lb (C )
METHOD OF JOINTS

Fx = 0 +
− T EC cos 36.87
13

5
F
Y
12
3

4
T EC = 400
o
+ T EB sin 67.38
T AC = 400 lb (T
165
528
@ joint E
TEB
TEC
y
 = 36.87
 = 67.38 o


E
528
x
o
− T EB cos 67.38
o
=0
=0 +
T EC sin 36.87
5
o
)
o
+ 528 = 0
T EB = −832
T EB = 832 lb (C )
METHOD OF JOINTS

Fx = 0 +
420 − T BC cos 36.87
13

5
o
− 832 cos 67.38
3
@ joint B
420

TBC
T BC = 125 lb (T
Checking:
FY = 0 +


4
− 693 − 125 sin 36.87
165
528
y
693
 = 36.87
B
x

832
=0
T BC = 125
12
5
o
 = 67.38 o
o
o
+ 832 sin 67.38
o
?
=0
− 0.000933  0

)
TRUSSES - METHOD OF JOINTS
Sample Problem:
Using the method of joints, determine the force in each
member of the truss shown.
Steps by Step:
1. Draw a free-body diagram of the entire truss, and use
this diagram to determine the reactions at the supports.
σ 𝑴𝑪 = 𝟎
− 2000 𝑙𝑏 24𝑓𝑡
+ − 1000 𝑙𝑏 12 𝑓𝑡
+ 𝑬 𝟔𝒇𝒕 = 𝟎
𝑬 = 𝟏𝟎, 𝟎𝟎𝟎 𝒍𝒃
σ 𝑭𝒙 = 𝟎
𝑪𝒙 = 𝟎 𝑵
σ 𝑭𝒚 = 𝟎
−2000 𝑙𝑏 − 1000 𝑙𝑏 + 10,000 𝑙𝑏 + 𝐶𝑦 = 0
𝑪𝒚 = 𝟕𝟎𝟎𝟎 𝒍𝒃
TRUSSES - METHOD OF JOINTS
Steps:
• 2. Locate a joint connecting only two members, and draw the freebody diagram of its pin. Use this free-body diagram to determine the
unknown force in each of the two members. A positive answer means
that the member is in tension, a negative answer that the member is
in compression.
FBD @Joint A
σ 𝑭𝒚 = 𝟎
2000
σ 𝑭𝒙 = 𝟎
AB
A
4
5
4
−2000 − 𝑨𝑫 = 0
5
𝑨𝑫 = −𝟐𝟓𝟎𝟎 𝒍𝒃
3
𝑨𝑫 = 0
5
3
𝑨𝑩 +
−2500 = 0
5
𝑨𝑩 +
3
AD
𝑨𝑩 = 𝟏𝟓𝟎𝟎 𝒍𝒃
TRUSSES - METHOD OF JOINTS
Steps:
• 3. Next, locate a joint where the forces in only two of the
connected members are still unknown. Draw the free-body
diagram of the pin and use it as indicated above to determine the two
unknown forces.
FBD @Joint D
AD
σ 𝑭𝒚 = 𝟎
BD
4
5
5
3
3
D
σ 𝑭𝒙 = 𝟎
4
ED
4
4
𝑨𝑫 + 𝑩𝑫 = 0
5
5
𝑩𝑫 = 𝟐𝟓𝟎𝟎 𝒍𝒃
3
3
𝑨𝑫 + 𝑩𝑫 = 0
5
5
3
3
𝑬𝑫 −
−2500 + 2500 = 0
5
5
𝑬𝑫 −
𝑬𝑫 = −𝟑𝟎𝟎𝟎 𝒍𝒃
TRUSSES - METHOD OF JOINTS
Steps:
• 4. Repeat this procedure until the forces in all the members of
the truss have been found.
FBD @Joint B
1000
σ 𝑭𝒚 = 𝟎
AB
−1000 −
−1000 −
BC
5
4
3
BD
B
4
4
4
𝑬𝑩 − 𝑩𝑫 = 0
5
5
4
4
𝑬𝑩 − 2500 = 0
5
5
𝑬𝑩 = −𝟑𝟕𝟓𝟎 𝒍𝒃
5
3
EB
3
3
−𝑨𝑩 + 𝑩𝑪 − 𝑩𝑫 + 𝑬𝑩 = 0
5
5
3
3
−1500 + 𝑩𝑪 − 2500 + (−3750) = 0
5
5
σ 𝑭𝒙 = 𝟎
𝑩𝑪 = 𝟓𝟐𝟓𝟎 𝒍𝒃
TRUSSES - METHOD OF JOINTS
Steps:
• 4. Repeat this procedure until the forces in all the members of
the truss have been found.
FBD @Joint C
Cy
4
𝑪𝑬 = 0
σ 𝑭𝒚 = 𝟎
5
4
−7000 − 𝑪𝑬 = 0
5
−𝐶𝑦 −
BC
C
5
4
3
CE
𝑪𝑬 = −𝟖𝟕𝟓𝟎 𝒍𝒃
TRUSSES - METHOD OF JOINTS
𝑨𝑩 = 𝟏𝟓𝟎𝟎 𝒍𝒃
Tension
𝑨𝑫 = 𝟐𝟓𝟎𝟎 𝒍𝒃
Compression
𝑩𝑫 = 𝟐𝟓𝟎𝟎 𝒍𝒃
Tension
𝑬𝑩 = 𝟑𝟕𝟓𝟎 𝒍𝒃
Compression
𝑬𝑫 = 𝟑𝟎𝟎𝟎 𝒍𝒃
Compression
𝑩𝑪 = 𝟓𝟐𝟓𝟎 𝒍𝒃
Tension
𝑪𝑬 = 𝟖𝟕𝟓𝟎 𝒍𝒃
Compression
A
B
D
C
E