SIMPLE TRUSSES, THE METHOD OF JOINTS, &
ZERO-FORCE MEMBERS
Today’s Objectives:
Students will be able to:
In-Class Activities:
a) Define a simple truss.
• Check Homework, if any
b) Determine the forces in members of a
simple truss.
• Reading Quiz
c) Identify zero-force members.
• Simple Trusses
• Applications
• Method of Joints
• Zero-Force Members
• Concept Quiz
• Group Problem Solving
• Attention Quiz
READING QUIZ
1. One of the assumptions used when analyzing a simple truss is that
the members are joined together by __________.
A) Welding
B) Bolting
D) Smooth pins
E) Super glue
C) Riveting
2. When using the method of joints, typically _________ equations of
equilibrium are applied at every joint.
A) Two
B) Three
C) Four
D) Six
APPLICATIONS
Trusses are commonly used to support
roofs.
For a given truss geometry and load, how
can you determine the forces in the truss
members and thus be able to select their
sizes?
A more challenging question is that for a
given load, how can we design the
trusses’ geometry to minimize cost?
APPLICATIONS
(continued)
Trusses are also used in a variety of
structures like cranes and the frames
of aircraft or space stations.
How can you design a light weight
structure that will meet load, safety,
cost specifications, be easy to
manufacture, and allow easy
inspectioin over its lifetime?
SIMPLE TRUSSES
(Section 6.1)
A truss is a structure composed of slender members joined together at
their end points.
If a truss, along with the imposed load, lies in a single plane
(as shown at the top right), then it is called a planar truss.
A simple truss is a planar truss which begins
with a a triangular element and can be expanded
by adding two members and a joint. For these
trusses, the number of members (M) and the
number of joints (J) are related by the equation
M = 2 J – 3.
ANALYSIS & DESIGN ASSUMPTIONS
When designing both the member and the joints of a truss, first it is
necessary to determine the forces in each truss member. This is called
the force analysis of a truss. When doing this, two assumptions are
made:
1. All loads are applied at the joints. The weight of the truss
members is often neglected as the weight is usually small as
compared to the forces supported by the members.
2. The members are joined together by smooth pins. This
assumption is satisfied in most practical cases where the joints
are formed by bolting the ends together.
With these two assumptions, the members act as
two-force members. They are loaded in either
tension or compression. Often compressive
members are made thicker to prevent buckling.
THE METHOD OF JOINTS
(Section 6.2)
A free body diagram of Joint B
When using the method of joints to solve for the forces in truss
members, the equilibrium of a joint (pin) is considered. All
forces acting at the joint are shown in a FBD. This includes all
external forces (including support reactions) as well as the forces
acting in the members. Equations of equilibrium ( FX= 0 and
 FY = 0) are used to solve for the unknown forces acting at the
joints.
STEPS FOR ANALYSIS
1. If the truss’s support reactions are not given, draw a FBD of
the entire truss and determine the support reactions (typically
using scalar equations of equilibrium).
2. Draw the free-body diagram of a joint with one or two
unknowns. Assume that all unknown member forces act in
tension (pulling the pin) unless you can determine by
inspection that the forces are compression loads.
3. Apply the scalar equations of equilibrium,  FX = 0 and
 FY = 0, to determine the unknown(s). If the answer is
positive, then the assumed direction (tension) is correct,
otherwise it is in the opposite direction (compression).
4. Repeat steps 2 and 3 at each joint in succession until all the
required forces are determined.
ZERO-FORCE MEMBERS
(Section 6.3)
If a joint has only two non-collinear
members and there is no external
load or support reaction at that joint,
then those two members are zeroforce members. In this example
members DE, DC, AF, and AB are
zero force members.
You can easily prove these results by
applying the equations of
equilibrium to joints D and A.
Zero-force members can be
removed (as shown in the
figure) when analyzing the
truss.
ZERO – FORCE MEMBERS (continued)
If three members form a truss joint for
which two of the members are collinear
and there is no external load or reaction at
that joint, then the third non-collinear
member is a zero force member.
Again, this can easily be proven. One can
also remove the zero-force member, as
shown, on the left, for analyzing the truss
further.
Please note that zero-force members
are used to increase stability and
rigidity of the truss, and to provide
support for various different loading
conditions.
EXAMPLE
Given: Loads as shown on the truss
Find: The forces in each member
of the truss.
Plan:
1.
2.
3.
4.
Check if there are any zero-force members.
First analyze pin D and then pin A
Note that member BD is zero-force member. FBD = 0
Why, for this problem, do you not have to find the external
reactions before solving the problem?
EXAMPLE (continued)
D 450 lb
45 º
FAD
45 º
FCD
FBD of pin D
+   FX = – 450 + FCD cos 45° – FAD cos 45° = 0
+   FY = – FCD sin 45° – FAD sin 45° = 0
FCD = 318 lb (Tension) or (T)
and FAD = – 318 lb (Compression) or (C)
EXAMPLE (continued)
Analyzing pin A:
FAD
45 º
A
FAB
AY
FBD of pin A
+   FX = FAB + (– 318) cos 45° = 0;
FAB = 225 lb (T)
Could you have analyzed Joint C instead of A?
CONCEPT QUIZ
1. Truss ABC is changed by decreasing its
height from H to 0.9 H. Width W and load
P are kept the same. Which one of the
following statements is true for the revised
truss as compared to the original truss?
A) Force in all its members have decreased.
B) Force in all its members have increased.
C) Force in all its members have remained
the same.
D) None of the above.
P
A
H
C
B
W
CONCEPT QUIZ
(continued)
F
F
F
2. For this truss, determine the number of zero-force
members.
A) 0
B) 1
D) 3
E) 4
C) 2
GROUP PROBLEM SOLVING
Given: Loads as shown on the
truss
Find: Determine the force in
all the truss members
(do not forget to
mention whether they
are in T or C).
Plan:
a) Check if there are any zero-force members.
b) Draw FBDs of pins D and E, and then apply EE at those pins to
solve for the unknowns.
c) Note that Member CE is zero-force member so FEC = 0. If you
didn’t see this simplification, could you still solve the problem?
GROUP PROBLEM SOLVING (continued)
FBD of pin D
From geometry, tan-1(1/2)=26.57
Y
600N
D
X
Analyzing pin D:
→ + FX = 600 – FCD sin 26.57 = 0
FDE
26.57
FCD
FCD = 1341 N = 1.34 kN (C) (Note that FCD = FBC!)
+  FY = 1341 cos 26.57 – FDE = 0
FDE = 1200 N = 1.2 kN (T)
GROUP PROBLEM SOLVING (continued)
FBD of pin E
Y
FDE
900 N
E
X
45
FEA
Analyzing pin E:
→ + FX = 900 – FEB sin 45 = 0
FEB = 1273 N = 1.27 kN (C)
+  FY = 1200 + 1273 cos 45 – FEA = 0
FEA = 2100 N = 2.1 kN (T)
FEB
ATTENTION QUIZ
1. Using this FBD, you find that FBC = – 500 N.
Member BC must be in __________.
A) Tension
FBC
B
FBD
B) Compression
C) Cannot be determined
2. For the same magnitude of force to be
carried, truss members in compression
are generally made _______ as compared
to members in tension.
A) Thicker
B) Thinner
C) The same size
BY
THE METHOD OF SECTIONS
Today’s Objectives:
Students will be able to determine:
1. Forces in truss members using the
method of sections.
In-Class Activities:
• Check Homework, if any
• Reading Quiz
• Applications
• Method of Sections
• Concept Quiz
• Group Problem Solving
• Attention Quiz
READING QUIZ
1. In the method of sections, generally a “cut” passes through no more
than _____ members in which the forces are unknown.
A) 1
B) 2
C) 3
D) 4
2. If a simple truss member carries a tensile force of T along its length,
then the internal force in the member is ______ .
A) Tensile with magnitude of T/2
B) Compressive with magnitude of T/2
C) Compressive with magnitude of T
D) Tensile with magnitude of T
APPLICATIONS
Long trusses are often used to construct large cranes and large
electrical transmission towers.
The method of joints requires that many joints be analyzed before we
can determine the forces in the middle part of a large truss.
So we need another method to determine such forces.
THE METHOD OF SECTIONS
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 members will also be
either tensile or compressive with the same magnitude. This result is
based on the equilibrium principle and Newton’s third law.
STEPS FOR ANALYSIS
1. Decide how you need to “cut” the truss. This is based on:
a) where you need to determine forces, and, b) where the total
number of unknowns does not exceed three (in general).
2. Decide which side of the cut truss will be easier to work with
(minimize the number of reactions you have to find).
3. If required, determine any necessary support reactions by drawing
the FBD of the entire truss and applying the E-of-E.
STEPS FOR ANALYSIS (continued)
4. Draw the FBD of the selected part of the cut truss. We need to
indicate the unknown forces at the cut members. Initially we may
assume all the members are in tension, as we did when using the
method of joints. Upon solving, if the answer is positive, the member
is in tension as per our assumption. If the answer is negative, the
member must be in compression. (Please note that you can also
assume forces to be either tension or compression by inspection as
was done in the figures above.)
STEPS FOR ANALYSIS (continued)
5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut
section of the truss to solve for the unknown member forces. Please
note, in most cases it is possible to write one equation to solve for one
unknown directly. So look for it and take advantage of such a
shortcut!
EXAMPLE
Given: Loads as shown on the
truss.
Find: The force in members
KJ, KD, and CD.
Plan:
a) Take a cut through the members KJ, KD, and CD.
b) Work with the left part of the cut section. Why?
c) Determine the support reactions at A. What are they?
d) Apply the EofE to find the forces in KJ, KD, and CD.
EXAMPLE (continued)
Analyzing the entire truss for the reactions a A, we get
 FX = AX = 0. Then do a moment equation about G to find AY.
∑MG = AY (18) – 20 (15) – 30 (12) – 40 (9) = 0; AY = 45.6 kN
Now take moments about point D. Why do this?
+ MD = – 45.6 (9) + 20 (6) + 30 (3) – FKJ (4) = 0
FKJ = − 50.1 kN or 50.1 kN ( C )
EXAMPLE (continued)
Now use the two force equations of equilibrium.
↑ +  FY = 45.6 – 20 – 30 – (4/5) FKD = 0;
FKD = − 5.5 kN , or 5.5 kN (C)
→ +  FX = (– 50.1) + (3/5) ( –5.5 ) + FCD = 0;
FCD = 53.4 kN (T)
CONCEPT QUIZ
1. Can you determine the force in
member ED by making the cut at
section a-a? Explain your answer.
A) No, there are 4 unknowns.
B) Yes, using  MD = 0 .
C) Yes, using  ME = 0 .
D) Yes, using  MB = 0 .
CONCEPT QUIZ
2. If you know FED, how will you determine FEB ?
A) By taking section b-b and using  ME = 0
B) By taking section b-b, and using  FX = 0 and  FY = 0
C) By taking section a-a and using  MB = 0
D) By taking section a-a and using  MD = 0
GROUP PROBLEM SOLVING
Given: The internal drag truss for
the wing of a airplane is subjected
to
the forces
shown.
Find: The force in members IH,
BH, and BC.
Plan:
a) Take a cut through the members IH, BH, and BC.
b) Analyze the right section (no support reactions!).
c) Draw the FBD of the right section.
d) Apply the equations of equilibrium (if possible try to do it so
that every equation yields an answer to one unknown.
SOLUTION
+ ↑ FY = 80 + 60 + 40 – FBH sin 45º = 0;
FBE = 255 lb (T)
+  MH = – FBC (2) + 60 (2) + 40 (3.5) = 0;
FBC = 130 lb (T)
+ → FX = FIH – 130– 255 cos 45º = 0; FIH = 310 lb (T)
ATTENTION QUIZ
1. As shown, a cut is made through
members GH, BG and BC to
determine the forces in them.
Which section will you choose
for analysis and why?
A) Right, fewer calculations.
B) Left, fewer calculations.
C) Either right or left, same
amount of work.
D) None of the above, too
many unknowns.
ATTENTION QUIZ
2. When determining the force in
member HG in the previous question,
which one equation of equilibrium is
best to use?
A)  MH = 0
B)  MG = 0
C)  MB = 0
D)  MC = 0
FRAMES AND MACHINES
Today’s Objectives:
Students will be able to:
a) Draw the free body diagram of a
frame or machine and its members.
b) Determine the forces acting at the
joints and supports of a frame or
machine.
In-Class Activities:
• Check Homework, if any
• Reading Quiz
• Applications
• Analysis of a Frame/Machine
• Concept Quiz
• Group Problem Solving
• Attention Quiz
READING QUIZ
1. Frames and machines are different as compared to trusses since they
have ___________.
A) Only two-force members
B) Only multiforce members
C) At least one multiforce member D) At least one two-force
member
2. Forces common to any two contacting members act with _______
on the other member.
A) Equal magnitudes but opposite sense
B) Equal magnitudes and the same sense
C) Different magnitudes but opposite sense
D) Different magnitudes but the same sense
APPLICATIONS
Frames are commonly used
to support various external
loads.
How is a frame different than a
truss?
To be able to design a frame,
you need to determine the
forces at the joints and
supports.
APPLICATIONS (continued)
“Machines,” like those above, are used in a variety of
applications. How are they different from trusses and frames?
How can you determine the loads at the joints and supports?
These forces and moments are required when designing the
machine members.
FRAMES AND MACHINES: DEFINITIONS
Frame
Machine
Frames and machines are two common types of structures that
have at least one multi-force member. (Recall that trusses have
nothing but two-force members).
Frames are generally stationary and support external loads.
Machines contain moving parts and are designed to alter the
effect of forces.
STEPS FOR ANALYZING A FRAME OR MACHINE
1. Draw a FBD of the frame or machine and its
members, as necessary.
Hints:
a) Identify any two-force members, b) forces
on contacting surfaces (usually between a pin
and a member) are equal and opposite, and,
c) for a joint with more than two members or an
external force, it is advisable to draw a FBD of
the pin.
FAB
2. Develop a strategy to apply the equations of
equilibrium to solve for the unknowns.
Problems are going to be challenging since there
are usually several unknowns. A lot of practice
is needed to develop good strategies.
EXAMPLE
Given: The frame supports an
external load and moment
as shown.
Find: The horizontal and vertical
components of the pin
reactions at C and the
magnitude of reaction at B.
Plan:
a) Draw FBDs of the frame member BC. Why pick this
part of the frame?
b) Apply the equations of equilibrium and solve for the
unknowns at C and B.
EXAMPLE
800 N m
400 N
CX
CY
1m
1m
2m
B
45°
FBD of member BC
FAB
Please note that member AB is a two-force member.
Equations of Equilibrium:
+  MC = FAB sin45° (1) – FAB cos45° (3) + 800 N m + 400 (2) = 0
FAB = 1131 N
EXAMPLE
800 N m
400 N
CX
CY
1m
1m
2m
B
45°
FBD of member BC
FAB
Now use the x and y direction Equations of Equilibrium:
 +  FX = – CX + 1131 sin 45° = 0
CX = 800 N
 +  FY = – CY + 1131 cos 45° – 400 = 0
CY = 400 N
CONCEPT QUIZ
1. The figures show a frame and its FBDs. If an additional couple
moment is applied at C, then how will you change the FBD of
member BC at B?
A)
B)
C)
D)
No change, still just one force (FAB) at B.
Will have two forces, BX and BY, at B.
Will have two forces and a moment at B.
Will add one moment at B.
CONCEPT QUIZ (continued)
D
2. The figures show a frame and its FBDs. If an additional force is
applied at D, then how will you change the FBD of member BC
at B?
A)
B)
C)
D)
No change, still just one force (FAB) at B.
Will have two forces, BX and BY, at B.
Will have two forces and a moment at B.
Will add one moment at B.
GROUP PROBLEM SOLVING
Given: A frame supports
a 50-kg cylinder.
Find: The reactions that
the pins exert on the
frame at A and D.
Plan:
a) Draw a FBD of member ABC and another one for CD.
b) Apply the equations of equilibrium to each FBD to solve for
the six unknowns. Think about a strategy to easily solve for
the unknowns.
GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD
CY
CX
50(9.81) N
0.7 m
1.2 m
1.6 m
DX
Applying E-of-E to member ABC:
DY
+  MA = CY (1.6) – 50 (9.81) (0.7) – 50 (9.81) (1.7) = 0 ;
CY = 735.8 N
+  FY = AY – 735.8 – 50 (9.81) – 50 (9.81) = 0 ; AY = 245 N
+  FX = CX – AX = 0 ; CX = AX
GROUP PROBLEM SOLVING (continued)
FBDs of members ABC and CD
CY
CX
50(9.81) N
0.7 m
1.2 m
1.6 m
DX
Applying E-of-E to member CD:
DY
+  MD = CX (1.2) + 50 (9.81) (0.7) – 735.8(1.6) = 0 ; CX = 695 N
+  FY = DY – 735.8 + 50 (9.81) = 0 ; DY = 245 N
+  FX = DX – 695 = 0 ; DX = 695 N
AX = CX = 695 N
ATTENTION QUIZ
1. When determining the reactions
at joints A, B, and C, what is the
minimum number of unknowns
for solving this problem?
A) 3
B) 4
C) 5
D) 6
2. For the above problem, imagine that you have drawn a FBD
of member AB. What will be the easiest way to write an
equation involving unknowns at B?
A)  MC = 0
B)  MB = 0
C)  MA = 0
D)  FX = 0