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Engineering 36
Chp 6:
Trusses-2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Introduction: MultiPiece Structures
 For the equilibrium of structures made of several
connected parts, the internal forces as well the
external forces are considered.
 In the interaction between connected parts,
Newton’s 3rd Law states that the forces of action
and reaction between bodies in contact have the
same magnitude, same line of action, and
opposite sense.
 The Major Categories of Engineering Structures
• Frames: contain at least one multi-force member,
i.e., a member acted upon by 3 or more forces
• Trusses: formed from two-force members, i.e.,
straight members with end point connections
• Machines: structures containing moving parts
designed to transmit and modify forces
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Definition of a Truss
 A truss consists of straight members
connected at joints. No member is
continuous through a joint.
 A truss carries ONLY those loads which
act in its plane, allowing the truss to be
treated as a two-dimensional structure.
 Bolted or welded connections are assumed to
be pinned together. Forces acting at the
member ends reduce to a single force and NO
couple. Only two-force members are
considered  LoA CoIncident with Geometry
 When forces tend to pull the member apart, it
is in tension. When the forces tend to push
together the member, it is in compression.
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Truss Defined
 Members of a truss are SLENDER and NOT
capable of supporting large LATERAL loads
• i.e.; IN-Plane, or 2D, loading only
• Members are of NEGLIBLE Weight
 Loads MUST be applied at the JOINTS to
Ensure AXIAL-ONLY Loads on Members.
• Mid-Member Loads Produce BENDING-Loads
which Truss Members are NOT Designed to Support
Beams Apply
RoadBed Load at
JOINTS Only
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Trusses Made of Simple Trusses
• Compound trusses are statically
determinant, rigid, and completely
constrained.
m  2n  3
• Truss contains a redundant
member and is statically
indeterminate. m  2n  3
• Additional reaction forces may be
necessary for a nonrigid truss.
non-rigid
m  2n  3
rigid
m  2n  4
Engineering-36: Engineering Mechanics - Statics
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• Necessary but INsufficient
condition for a compound truss to
be statically determinant, rigid,
and completely constrained,
m  r  2n
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Method of Sections
 When the force in only one
member or the forces in a very
few members are desired, the
method of sections works well.
Engineering-36: Engineering Mechanics - Statics
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 To determine the force in
member BD, pass a section
through the truss as shown and
create a free body diagram for
the left side.
 With only three members cut by
the section, the equations for
static equilibrium may be
applied to determine the
unknown member forces,
including FBD
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
 Given the Truss
with Loading and
Geometry Shown
 Use the Method
of Sections to
Determine the
Force in
Member FD
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
 Take Section to Expose FFD
 Now Take ΣME = 0
0  15kip 10'15kip  20' FFD 10'
FFD
 150  300kip  ft

 45kip
10ft
Engineering-36: Engineering Mechanics - Statics
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FFD  45kip Compressio n
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
 SOLUTION PLAN
 Determine the force in
members just right of
Center:
• Take the entire truss as a free
body. Apply the conditions for
static equilibrium to solve for the
reactions at A and L.
• Pass a section through members
FH, GH, and GI and take the
right-hand section as a free body.
• Apply the conditions for static
equilibrium to determine the
desired member forces.
• FH
• GH
• GI
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
 SOLUTION PLAN
• Take the entire truss as a free
body. Apply the conditions for
static equilibrium to solve for the
reactions at A and L
M
A
 0  5 m 6 kN   10 m 6 kN   15 m 6 kN 
 20 m 1 kN   25 m 1 kN   30 m L
L  7.5 kN 
F
y
 0  20 kN  L  Ay
Ay  12.5 kN 
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
 Pass a section (n-n) through
members FH, GH, and GI
and take the right-hand
section as a free body
 Apply the conditions for static
equilibrium to determine the
desired member forces.
MH  0
7.50 kN 10 m   1 kN 5 m   FGI 5.33 m   0
FGI  13.13 kN
FGI  13.13 kN T
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Example  Method of Sections
tan  
FG 8 m

 0.5333
GL 15 m
0
  28.07
 MG
7.5 kN 15 m   1 kN 10 m   1 kN 5 m 
  FFH cos 8 m   0
FFH  13.82 kN
tan  
M
L
FFH  13.82 kN C
GI
5m
 2
 0.9375
HI 3 8 m 
  43.15
0
1 kN 10 m   1 kN 5 m   FGH cos  15 m   0
FGH  1.371 kN
Engineering-36: Engineering Mechanics - Statics
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FGH  1.371 kN C
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Method of Sections - Summary
1. If needed Determine Support Reactions
2. Decide on How to CUT the Truss into
Sections and draw the Corresponding
Free Body Diagrams
3. Try to Apply the Eqns of Equilibrium to
avoid generation of simultaneous Eqns
•
Moments should be Summed about points
that lie at the intersection of the LoA’s of
2+Forces, making simpler the solution for
the remaining forces
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Pick: Pivot & PoA
 When doing Sections Recall that the
LoA for Truss Members are defined by
the Member Geometry
 Use Force Transmissibility → Forces are
SLIDING Vectors
• Pick a Pivot Point, on or Off the Body,
where the LoA’s of many Force LoA’s Cross
• Apply the Force of interest so that ONE of
its X-Y Components passes Thru the Pivot
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Pick: Pivot & PoA Example
 After Finding support RCNs find force in
Member ED → Use Section a-a
 Pick Pt-B as Pivot to Eliminate from
Moment Calc FAB, FFB, FEB, 1000N
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Pick: Pivot & PoA Example
 Pick Pt-C as the Point of Appliction
(PoA) for FED
 Using Pt-C as the PoA permits using
F•d to find the Moment about Pivot-B
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
WhiteBoard Work
Let’s Work
Some Truss
Problems
Find Forces in
EL & LM
Find Forces in
EL & LM
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Engineering 36
Appendix
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-15_Trusses-2.pptx