Transcript ENGR-36_Lec-14_Trusses-1_H13e

Engineering 36
Chp 6:
Trusses-1
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-36: Engineering Mechanics - Statics
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Bruce Mayer, PE
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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
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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.
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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
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Types of Trusses
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http://www.caltruss.com/products.htm
Yet More Trusses
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Simple Trusses
 A rigid truss will not
collapse under the
application of a load.
 A simple truss is
constructed by
successively adding
two members and one
connection to the basic
TRIANGULAR truss
 In a simple truss,
n7
m = 2n − 3 where m is
the total number of
m  11
members and n is the
number of joints
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Bruce Mayer, PE
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Method of Joints for Trusses
 Dismember the truss and create a freebody
diagram for each member and pin.
 The two forces exerted on each member are
equal, have the same line of action, and
opposite sense as the load on the
associated Pin
 Forces exerted by a member on the pins or
joints at its ends are directed along the
member and are equal & opposite.
 Conditions of equilibrium on the pins provide
2n equations for 2n unknowns. For a simple
truss, 2n = m + 3. May solve for m member
forces and 3 reaction forces at the supports.
 Conditions for equilibrium for the entire
truss provide 3 additional equations which
are not independent of the pin equations.
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Zero-Force Members
• Forces in opposite members intersecting
in two straight lines at a joint are equal
(to maintain pin equilibrium).
• The forces in two opposite members
are equal when a load is aligned
with a third member. The third
member force is equal to the load
(including zero load). If P=0, Then AC
Applies NO Force to the pin; AC is
then a ZERO-FORCE MEMBER
• The forces in two members connected
at a joint are equal if the members are
aligned and zero otherwise.
• Recognition of joints under special
loading conditions (ZFMs) simplifies
truss analysis.
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Zero Force Members (ZFMs)
 TWO-Member Version
• When only TWO members form a NON-CoLinear
truss joint and NO external load or support
reaction is applied to the joint then the members
MUST be ZERO-FORCE members.
 THREE-Member Version
• When THREE members form a truss joint for
which two of the members are CoLinear and the
third is forms an angle with the first two, then the
NON-CoLinear member is a ZERO-FORCE
member provided NO external force or support
reaction is applied to the joint
– NOTE that The CoLinear members carry EQUAL loads.
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Zero Force Members (ZFMs)
 Truss members that
canNOT carry load
are called Zero
Force Members.
 Examples of Zero
Force Members
(ZFMs) are the
colored members
(AB, BC, and DG) in
the truss shown at
Right
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F1
B
C
D
E
A
F2
F
G
F3
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Zero Force Members (ZFMs)
 Summing the forces
in the y-direction in
the AB FBD shows
that FAB must be
ZERO since it is
NOT balanced by
another
y-force.
y
 Again Consider
F1
C
B
E
D
A
F2
F
G
F3
 FBD for AB & BC
y
B
x
FBC
FAB
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 With FDAB =0 F
x
Summing forces in
the x-direction shows
that
FBC must also
F
be zero
FDC
DE
DG
Bruce Mayer, PE
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Zero Force Members (ZFMs)
 Summing forces in
the y-direction in the
DG free-bodydiagram, reveals
that FDG must be
zero since it is not
balanced by
another y-force
 Again Consider
F1
B
D
C
E
A
F2
F
G
F3
 The FBD for DG
y
FDC
D
FDE
x
FDG
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Space Trusses
• An elementary space truss consists of 6
members connected at 4 joints to form a
tetrahedron.
• A simple space truss is formed and can be
extended when 3 new members and 1 joint are
added at the same time.
• In a simple space truss, m = 3n − 6 where m
is the number of members and n is the
number of joints.
• Conditions of equilibrium for the Ball-andSocket joints provide 3n equations (no
moments). For a simple truss, 3n = m + 6 and
the equations can be solved for m member
forces and 6 support reactions.
• Equilibrium for the entire truss provides 6
additional equations which are not
independent of the joint equations.
Bruce Mayer, PE
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Example  2D Truss
30°
B
 Solution Plan
3 kN
C
2m
A
E
2m
D
2m
 Using the method of
joints, determine the
force in each member of
the truss
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• Based on a free-body diagram of
the entire truss, solve the 3
equilibrium eqns for the reactions
at A and D
• Joint D is subjected to only two
unknown member forces.
Determine the member forces from
the joint (or pin) equilibrium
requirements.
• In succession, determine unknown
member forces at joints C, B, and A
from pin equilibrium requirements.
• Use Pin-E equilibrium requirement
to check the results
Bruce Mayer, PE
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Black/White Board Work
30°
Solve by
Method of
Joints
B
3 kN
C
2m
A
E
D
2m
2m
y
Weight of Members & Truss is Negligible
Relative to Applied Load(s)
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x
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30°
B
3 kN
C
2m
A
E
D
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2m
2m
Example  2D Truss
 SOLUTION PLAN
 Using the method of
joints, determine the
force in each member
of the truss.
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• Based on a free-body diagram of the
entire truss, solve the 3 equilibrium
equations for the reactions at E and C.
• Joint A is subjected to only two
unknown member forces. Determine
these from the joint (or pin) equilibrium
requirements.
• In succession, determine unknown
member forces at joints D, B, and E
from pin equilibrium requirements.
• All member forces and support
reactions are known at joint C.
However, the joint equilibrium
requirements may be applied
to check the results
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-142_Trusses-1.pptx
Example  2D Truss
M
C
0
 2000lb24 ft  1000lb12 ft  E 6 ft
E  10,000 lb 
 Fx  0  C x
 SOLUTION
 Fy  0  2000 lb - 1000 lb  10,000 lb  C y
• Based on a free-body
diagram of the entire
truss, solve the 3
equilibrium equations for
the reactions at E and C.
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Cx  0
C y  7000 lb 
 So-Far, So-Good
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-142_Trusses-1.pptx
Example  2D Truss
 Joint A is subjected to only
two unknown member
forces. Determine these
from the pin (or joint)
equilibrium requirements.
4
3
FAD
FAD  FAB
5
5
2000lb FAB FAD



4
3
5
2000lb 
FAB  1500 lb T
FAD  2500 lb C
 There are now only two
unknown member forces at
joint D.
FDB  FDA
FDE  2
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 FDA
3
5
FDB  2500 lb T
FDE  3000 lb C
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-142_Trusses-1.pptx
Example  2D Truss
 There are now only two
unknown member forces at joint
B. Assume both are in tension.
 Fy  0  1000  54 2500  54 FBE
FBE  3750 lb
FBE  3750 lb C
 Fx  0  FBC  1500  53 2500  53 3750
FBC  5250 lb
FBC  5250 lb T
 There is one unknown member
force at joint E. Assume the
member is in tension.
 Fx  0  53 FEC  3000  53 3750
FEC  8750 lb
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FEC  8750 lb C
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-142_Trusses-1.pptx
Example  2D Truss
 CHECK
• All member forces and support
reactions are known at joint C.
However, the joint equilibrium
requirements may be applied to
check the results.
F
F
x
y
  5250 53 8750  0
 7000 54 8750  0
checks
checks

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A Real Truss Joint Idealized
2 in Tension, 3 in Compression
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WhiteBoard Work
Let’s Work
These Nice
Problems
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Engineering 36
APPENDIX
Truss Analysis
Bruce Mayer, PE
Licensed Electrical & Mechanical
Engineer
r  s  r  s r  s 
[email protected]
2
–
2
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sin
cos
Engineering-36: Engineering Mechanics - Statics
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→2049N←
A
0 (ZFM)
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3 kN
C
→2049N←
B
→2049N←
m
30°
Truss-Member Load Analysis
E
←2049N→
Bruce Mayer, PE
[email protected] • ENGR-36_Lec-142_Trusses-1.pptx
D