SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE...
Transcript of SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE...
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SIMPLE TRUSSES, THE METHOD OF JOINTS,
& ZERO-FORCE MEMBERS
In-Class Activities:
• Check Homework, if any
• Reading Quiz
• Applications
• Simple Trusses
• Method of Joints
• Zero-force Members
• Concept Quiz
• Group Problem Solving
• Attention Quiz
Today’s Objectives:
Students will be able to:
a) Define a simple truss.
b) Determine the forces in members of a
simple truss.
c) Identify zero-force members.
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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 C) Riveting
D) Smooth pins E) Super glue
2. When using the method of joints, typically _________ equations of
equilibrium are applied at every joint.
A) Two B) Three
C) Four D) Six
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APPLICATIONS
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?
Trusses are commonly used to support
roofs.
A more challenging question is, that for
a given load, how can we design the
trusses’ geometry to minimize cost?
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APPLICATIONS (continued)
Trusses are also used in a variety of
structures like cranes and the frames
of aircraft or the space station.
How can you design a light weight
structure satisfying load, safety, cost
specifications, is simple to
manufacture, and allows easy
inspection over its lifetime?
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SIMPLE TRUSSES
(Section 6.1)
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 truss is a structure composed of slender members joined together at
their end points.
A simple truss is a planar truss which begins
with 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.
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ANALYSIS & DESIGN ASSUMPTIONS
When designing the members and 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.
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THE METHOD OF JOINTS
(Section 6.2)
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.
A free-body diagram of Joint B
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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 on 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.
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ZERO-FORCE MEMBERS (Section 6.3)
You can easily prove these results
by applying the equations of
equilibrium to joints D and A.
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 zero-
force members. In this example
members DE, DC, AF, and AB are
zero force members.
Zero-force members can be
removed (as shown in the
figure) when analyzing the
truss.
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ZERO – FORCE MEMBERS (continued)
Again, this can easily be proven.
One can also remove the zero-
force member, as shown, on the
left, for analyzing the truss further.
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, e.g., DA.
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.
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EXAMPLE
1. Check if there are any zero-force members.
2. First analyze pin D and then pin A
3. Note that member BD is zero-force member. FBD = 0
4. Why, for this problem, do you not have to find the external
reactions before solving the problem?
Given: Loads as shown on the truss
Find: The forces in each member
of the truss.
Plan:
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EXAMPLE (continued)
+ 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)
45 º
FCD
D 450 lb
FAD
FBD of pin D
45 º
Note BD is a zero force member.
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EXAMPLE (continued)
+ FX = FAB + (– 318) cos 45° = 0; FAB = 225 lb (T)
Could you have analyzed Joint C instead of A?
45 º
FAB
A
FBD of pin A
FAD
AY
Analyzing pin A:
We would analyze C is we wanted the pin reaction.
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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.
H
P
A
BC
W
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CONCEPT QUIZ
(continued)
2. For this truss, determine the number of zero-force
members.
A) 0 B) 1 C) 2
D) 3 E) 4
F F
F
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GROUP PROBLEM SOLVING
a) Check if there are any zero-force members.
Is Member CE zero-force member?
b) Draw FBDs of pins D, C, and E, and then apply E-of-E at those
pins to solve for the unknowns.
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:
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GROUP PROBLEM SOLVING (continued)
Analyzing pin D:
+ FX = FDE (3/5) – 600 = 0
FCD = 1000 N = 1.00 kN (C)
+ FY = 1000 (4/5) – FCD = 0
FDE = 800 N = 0.8 kN (T)
FBD of pin D
FCD
Y
D 600N
X
FDE3
45
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GROUP PROBLEM SOLVING (continued)
Analyzing pin C:
→ + FX = FCE – 900 = 0
FCE = 900 N = 0.90 kN (C)
+ FY = 800 – FBC = 0
FBC = 800 N = 0.80 kN (T)
FBD of pin C
FBC
Y
C 900 N
X
FCE
FCD = 800 N
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GROUP PROBLEM SOLVING (continued)
Analyzing pin E:
→ + FX = FAE (3/5) + FBE (3/5) – 1000 (3/5) – 900 = 0
+ FY = FAE (4/5) – FBE (4/5) – 1000 (4/5) = 0
Solving these two equations, we get
FAE = 1750 N = 1.75 kN (C)
FBE = 750 N = 0.75 kN (T)
FBD of pin E
FAE
Y
E FCE = 900 N
X
FDE = 1000 N
FBE
3
4 5
3
4 5
3
4 5
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ATTENTION QUIZ
1. Using this FBD, you find that FBC = – 500 N.
Member BC must be in __________.
A) Tension
B) Compression
C) Cannot be determined
2. When supporting the same magnitude of
force, truss members in compression are
generally made _______ as compared to
members in tension.
A) Thicker
B) Thinner
C) The same size
FBD
FBC
B
BY
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GROUP PROBLEM SOLVING
Rollers exert upward normal at a and b.
9 unknown, two normals and seven members.
Draw FBDs of pins a, b, c, d, and e, and then apply E-of-E at those
pins to solve for the unknowns.
Given: Load is P = 1 kN as shown
on the equilateral truss
Find: Determine the force in all the
truss members (do not forget
to mention whether they are
in T or C).
Plan:
Pa
b
c
d
e
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a
Na
Tab
Tac
60Na + Tab sin60 = 0
Tabcos60 + Tac = 0
b
PTab
Tbd6060
Tbc
P + Tab sin60 + Tbc sin60 = 0
Tbd + Tbccos60 Tabcos60 = 0
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Tbc sin60 + Tcd sin60 = 0
Tcdcos60 + Tce
Tbccos60 Tac = 0
d
Tcd
Tbd
6060
Tde
Tcd sin60 + Tde sin60 = 0
Tdecos60 Tbd Tcdcos60 = 0
c
Tcd
Tce
6060
Tac
Tbc
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e
Ne
Tde
Tce
60Ne + Tde sin60 = 0
Tdecos60 + Tce = 0
10 equations but 9 unknowns. One equation
is redundant (not needed)
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TAB TAC TBC TBD TCD TCE TDE Na Ne
sin60 0 0 0 0 0 0 1 0 0
0.5 1 0 0 0 0 0 0 0 0
sin60 0 sin60 0 0 0 0 0 0 -1000
-0.5 0 0.5 1 0 0 0 0 0 0
0 0 sin60 0 sin60 0 0 0 0 0
0 -1 -0.5 0 0.5 1 0 0 0 0
0 0 0 0 sin60 0 sin60 0 0 0
0 0 0 -1 -0.5 0 0.5 0 0 0
0 0 0 0 0 0 sin60 0 1 0
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TAB -866.03 C
TAC 433.01 T
TBC -288.68 C
TBD -288.68 C
TCD 288.68 T
TCE 144.34 T
TDE -288.68 C
Na 750.00
Nb 250.00
Solution
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Reducing the computation load!
Consider the trusses as a
single rigid body and
apply Eof E.P
a
b
c
d
e
Na Ne
Na + Ne = P
P (L/4) + Ne L = 0
We find:
Ne = P/4 = 250 N and Na = 750 N
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Corners have a known force and equations simplify
a
750
Tab
Tac
60
750 + Tab sin60 = 0
Tabcos60 + Tac = 0
Tab = 866.03 N & Tac = 433.01 N
e
250
Tde
Tce
60
250 + Tde sin60 = 0
Tdecos60 + Tce = 0
Tde = 288.68 N & Tce = 144.34 N
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b
P866.03
Tbd6060
Tbc
1000 + 866.03sin60 +
Tbc sin60 = 0
Tbd + Tbccos60
866.03cos60 = 0
Tbc = 288.68 N & Tbd = 288.68 N
d
Tcd
Tbd
6060
Tde
Tcd sin60 + Tde sin60 = 0
Tdecos60 Tbd Tcdcos60 = 0
Tcd = 144.34 N
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What if the Load is not at a Joint?
Given: Load is P = 1 kN as shown
on the equilateral truss
Find: Determine the force in all the
truss members (do not forget
to mention whether they are
in T or C).
a
b
c
d
e
P
The normals are still the same since the macroscopic picture is
still the same.
Text suggests that if a
member has non-negligible
weight, just split the load at
the joints.a
b
c
d
e
P/2 P/2
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Finding the solution
a
750
Tab
Tac
60
250 + Tab sin60 = 0
Tabcos60 + Tac = 0
Tab = 288.67 N & Tac = 144.34 N
e
250
Tde
Tce
60
250 + Tde sin60 = 0
Tdecos60 + Tce = 0
Tde = 288.68 N & Tce = 144.34 N
500
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b
288.68
Tbd6060
Tbc
288.68sin60 + Tbc sin60 = 0
Tbd + Tbccos60
288.68cos60 = 0
Tbc = 288.68 N & Tbd = 288.68 N
d
Tcd
Tbd
6060
Tde
Tcd sin60 + Tde sin60 = 0
Tdecos60 Tbd Tcdcos60 = 0
Tcd = 144.34 N
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TAB -288.67 C
TAC 144.371 T
TBC -288.68 C
TBD -288.68 C
TCD 288.68 T
TCE 144.34 T
TDE -288.68 C
Na 750.00
Nb 250.00
Solution - distributed load
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What if the Load is not centered?
a
b
c
d
e
P
d
d
P
L
P1 P2
P1 + P2 = P and d*P = L*P2
Share the load according to above equations