Announcements
• Mountain Dew is an herbal supplement
Trusses – Method of Joints
Today’s Objectives
• Define a simple truss
• Determine the forces in members of a
simple truss
• Identify zero-force members
Class Activities
• Applications
• Simple trusses
• Method of joints
• Zero-force members
• Examples
Engr221 Chapter 6
1
Applications
Trusses are commonly used to
support a roof.
For a given truss geometry and
load, how can we determine the
forces in the truss members and
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 we design a light weight
structure that will meet load, safety,
and cost specifications?
Engr221 Chapter 6
2
Defining a Simple Truss
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 triangular element and can be expanded
by adding two members and a joint.
Analysis and 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 or welding.
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.
Engr221 Chapter 6
3
The Method of Joints
In this method of solving 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 support reactions are not given, draw a FBD of the
entire truss and determine all the support reactions using the
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 compressive 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.
Engr221 Chapter 6
4
Example A
Given: P1 = 200 lb, P2 = 500 lb
Find: The forces in each member of
the truss.
Plan: First analyze pin B and then pin C
200 lb
B 500 lb
54
FBA 3
45 º
+ → ∑ FX = 500 + FBC cos 45° – (3 / 5) FBA = 0
+ ↑ ∑ FY = – 200 – FBC sin 45° – (4 / 5) FBA = 0
FBA = 214 lb (T) and FBC = – 525.3 lb (C)
FBC
FBD of pin B
525.3
+ → ∑ FX = – FCA + 525.3 cos 45° = 0
FCA = 371 (T)
FCA45º
C
CY
FBD of pin C
Zero-Force Members
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, CD, 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.
Engr221 Chapter 6
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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 members, as
shown, on the left, for analyzing the truss
further.
Note that zero-force members are used to
increase stability and rigidity of the truss,
and to provide support for various loading
conditions.
Example B
Given: P1 = 240 lb and P2 = 100 lb
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 FBD’s of pins D and B, and then apply the E-of-E at those
pins to solve for the unknowns
Engr221 Chapter 6
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Example B - continued
FBD of pin D
Y
Analyzing pin D:
↑ + ∑FY = – 100 – (5 / 13) FDB = 0
FDC
FDB = – 260 lb (C)
D 240 lb
13
X
5
12
FDB
100 lb
→ + ∑ FX = 240 – FDC – (12 / 13) (– 260) = 0
FDC = 480 lb (T)
FBC Y
260 lb
13
Analyzing pin B:
↑+ ∑ FY = FBC – (5 / 13) 260 = 0
5
12
BX
FBC = 100 lb (T)
X
B
FBD of pin B
Questions
1. Using this FBD, you find that FBC = – 500 N.
Member BC must be in __________
A) Tension
B) Compression
C) Can not be determined
FBC
B
FBD
BY
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
Engr221 Chapter 6
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Questions
1. One of the assumptions used when analyzing a simple truss is
that the members are joined together by __________
A) welding
B) bolting
C) riveting
D) smooth pins E) super glue
2. In the method of joints, typically _________ equations of
equilibrium are applied at every joint.
A) two
B) three
C) four
D) six
Textbook Problem 6-7
Determine the force in each member of the truss and
state if the members are in tension or compression.
FBC = 3 kN (C)
FBA = 8 kN (C)
FAC = 1.46 kN (C)
FAF = 4.17 kN (T)
FCD = 4.17 kN (C)
FCF = 3.12 kN (C)
FEF = 0 kN
FED = 13.1 kN (C)
FDF = 5.21 kN (T)
Engr221 Chapter 6
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Textbook Problem 6-8
Determine the force in each member of the truss in
terms of the external loading and state if the
members are in tension or compression.
FBC
FCD
FAB
FBD
FAD
= 1.89P (C)
= 1.37P (T)
= 0.471P (C)
= 1.67P (T)
= 1.37P (T)
Summary
• Define a simple truss
• Determine the forces in members of a simple truss
• Identify zero-force members
Engr221 Chapter 6
9
Announcements
Trusses – Method of Sections
Today’s Objective
Determine forces in truss members using
the method of sections
Class Activities
• Applications
• Method of sections
• Examples
Engr221 Chapter 6
10
Applications
Long trusses are often used to construct bridges.
The method of joints requires that many joints be analyzed before
we can determine the forces in the middle part of the truss.
There is another method to determine these forces directly.
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 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.
Engr221 Chapter 6
11
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 forces you have to find).
3. If required, determine the 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. You need to
indicate the unknown forces at the cut members. Initially we
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 your assumption. If the answer
is negative, the member must be in compression. (Please note
that you can also assume forces to be either in tension or
compression by inspection as was done in the figures above.)
Engr221 Chapter 6
12
Steps for Analysis - continued
5. Apply the E-of-E to the selected cut section of the truss to solve
for the unknown member forces. Note that in most cases it is
possible to write one equation to solve for one unknown
directly.
Textbook Example 6.7
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 Σ MD = 0
C) Yes, using Σ ME = 0
D) Yes, using Σ MB = 0
Engr221 Chapter 6
13
Textbook Example 6.7 - continued
2. If you know FED, how will you
determine FEB?
A) By taking section b-b and
using ΣME = 0
B) By taking section b-b, and
using ΣFX = 0 and ΣFY = 0
C) By taking section a-a and
using ΣMB = 0
D) By taking section a-a and
using ΣMD = 0
Textbook Problem 6.36
Given: Loading on the truss as shown
Find:
The force in members BE, EF,
and CB
Plan:
a)
b)
c)
d)
Engr221 Chapter 6
Take a cut through members BE, EF, and CB
Analyze the top section (no support reactions!)
Draw the FBD of the top section
Apply the E-of-E such that every equation yields an
answer to one unknown
14
Textbook Problem 6.36 - continued
+ → ΣFX = 5 + 10 – FBE cos 45º = 0
FBE = 21.2 kN (T)
+ Σ ME = – 5(4) + FCB (4) = 0
FCB = 5 kN (T)
+ Σ MB = – 5 (8) – 10 (4) – 5 (4) – FEF (4) = 0
FEF = – 25 kN or 25 kN (C)
Textbook Problem 6.53
Given: Loads as shown on the
roof truss
Find: The force in members
DE, DL, and ML
Plan:
a) Take a cut through the members DE, DL, and ML.
b) Work with the left part of the cut section. Why?
c) Determine the support reactions at A.
d) Apply the E-of-E to find the forces in DE, DL, and ML.
Engr221 Chapter 6
15
Problem 6.53 - continued
Analyzing the entire truss, we get Σ FX = AX = 0. By symmetry,
the vertical support reactions are
AY = IY = 36 kN
+ MD = – 36 (8) + 6 (8) + 12 (4) + FML (5) = 0
FML = 38.4 kN (T)
Problem 6.53 - continued
+Σ ML = –36 (12) + 6 (12) + 12 (8) + 12 (4) – FDE ( 4/√17)(6) = 0
FDE = –37.11 kN or 37.1 kN (C)
→ + Σ FX = 38.4 + (4/√17) (–37.11) + (4/√41) FDL = 0
FDL = –3.84 kN or 3.84 kN (C)
Engr221 Chapter 6
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Questions
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
Question
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
Engr221 Chapter 6
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Question
When determining the force in
member HG in the previous
question, which one equation of
equilibrium is best to use?
A) ΣMH = 0
B) ΣMG = 0
C) ΣMB = 0
D) ΣMC = 0
Summary
• Determine forces in truss members using the method of
sections
Engr221 Chapter 6
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Announcements
Frames and Machines
Today’s Objectives
• Draw the free body diagram of a frame
or machine and its members
• Determine the forces acting at the joints
and supports of a frame or machine
Class Activities
• Applications
• Frame/machine analysis
• Examples
Engr221 Chapter 6
19
Applications
Frames are commonly used
to support various external
loads.
How is a frame different than
a truss? How can you
determine the forces at the
joints and supports of a
frame?
Applications - continued
Machines, like these 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? It
is required to know these loads when designing the machine
members.
Engr221 Chapter 6
20
Frames and Machines - Definitions
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 the 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
FAB
Engr221 Chapter 6
c) For a joint with more than two members or an
external force, it is advisable to draw a FBD
of the pin
2. Develop a strategy to apply the E-of-E
to solve for the unknowns. Problems are
going to be challenging since there are
usually several unknowns. Practice is
needed to develop good strategies.
21
Textbook Problem 6.72
Given: A frame and loads as shown
Find: The reactions that the pins
exert on the frame at A, B
and C
Plan:
a) Draw a FBD of members AB and BC
b) Apply the E-of-E to each FBD to solve for the six unknowns.
Think about a strategy to easily solve for the unknowns.
Textbook Problem 6.72 - continued
FBD’s of members AB and BC
BY
B
BX
1000N
BX
BY
B
0.4m
500N
C
A X A 45º
0.2m 0.2m
0.2m
0.4m
CY
AY
Writing moment equations at A and C, we get:
+ ∑ MA = BX (0.4) + BY (0.4) – 1000 (0.2) = 0
+ ∑ MC = - BX (0.4) + BY (0.6) + 500 (0.4) = 0
BY = 0 and BX = 500 N
Engr221 Chapter 6
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Textbook Problem 6.72 - continued
FBD’s of members AB and BC
BY
B
BX
1000N
BY
BX B
0.4m
500N
C
A X A 45º
0.2m 0.2m
0.2m
0.4m
CY
Applying E-of-E to bar AB:
→ + ∑ FX = AX – 500 = 0 ;
↑+ ∑ FY = AY – 1000 = 0 ;
Applying E-of-E to bar BC:
→ + ∑ FX = 500 – CX = 0 ;
↑ + ∑ FY = CY – 500 = 0 ;
AY
AX = 500 N
AY = 1000 N
CX = 500 N
CY = 500 N
Textbook Problem 6.83
Given: The wall crane supports an
external load of 700 lb
Find: The force in the cable at the
winch motor W and the
horizontal and vertical
components of the pin
reactions at A, B, C, and D
Plan:
a) Draw FBD’s of the frame’s members and pulleys
b) Apply the E-of-E and solve for the unknowns
Engr221 Chapter 6
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Textbook Problem 6.83 - continued
FBD of the Pulley E
T
T
E
700 lb
Necessary Equations of Equilibrium:
↑+ ∑ FY = 2 T – 700 = 0
T = 350 lb
Textbook Problem 6.83 - continued
350 lb
C
C
Y
CX
→ + ∑ FX = CX – 350 = 0
CX = 350 lb
↑ + ∑ FY = CY – 350 = 0
CY = 350 lb
350 lb
A FBD of pulley C
350 lb
BY
BX
30°
B
→ + ∑ FX = – BX + 350 – 350 sin 30° = 0
BX = 175 lb
↑ + ∑ FY = BY – 350 cos 30° = 0
BY = 303.1 lb
350 lb
A FBD of pulley B
Engr221 Chapter 6
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Textbook Problem 6.83 - continued
Note that member BD is a two-force
member.
TBD
AX
A
AY
45°
4 ft
350 lb
B
175 lb
303.11 lb
4 ft
700 lb
A FBD of member ABC
+ ∑ MA =
TBD sin 45° (4) – 303.1 (4) – 700 (8) = 0
TBD = 2409 lb
↑ + ∑ FY = AY + 2409 sin 45° – 303.1 – 700 = 0
AY = – 700 lb
→ + ∑ FX = AX – 2409 cos 45° + 175 – 350 = 0
AX = 1880 lb
Textbook Problem 6.83 - continued
A FBD of member BD
2409 lb
D
45°
B
2409 lb
At D, the x and y components are:
→ + DX = –2409 cos 45° = –1700 lb
↑ + DY = 2409 sin 45° = 1700 lb
Engr221 Chapter 6
25
Questions
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
Questions
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
directly solve for the first unknown?
A) ∑ MC = 0
B) ∑ MB = 0
C) ∑ MA = 0
D) ∑ FX = 0
Engr221 Chapter 6
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Example Problem
The compound beam is fixed at A and supported
by a rockers at B and C. There are hinges (pins) at
D and E. Determine the reactions at the supports.
Cy = 0N
By = 7.5 KN
Ax = 0N
Ay = 7.5 KN
MA = 45.0 KN-m
Example Problem
The pumping unit is used to recover oil. When the walking beam ABC is
horizontal, the force acting in the wire-line at the well head is 250 lb.
Determine the torque M which must be exerted by the motor in order to
overcome this load. The horse-head C weighs 60 lb and has a center of
gravity at GC. The walking beam ABC has a weight of 130 lb and a center of
gravity at GB, and the counterweight has a weight of 200 lb and a center of
gravity at GW. The pitman, AD, is pin-connected at its ends and has negligible
weight.
FAD = 449 lb
M = 313 ft-lb
Engr221 Chapter 6
27
Summary
• Draw the free body diagram of a frame or machine and its
members
• Determine the forces acting at the joints and supports of a
frame or machine
Engr221 Chapter 6
28