Subject
Year
: S0695 / STATICS
: 2008
Analysis of Statically Determinate Trusses
Session 07 - 10
What are Trusses ?
Æ are structures consisting of two or
more straight, slender members
connected to each other at their
endpoints.
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What for are Trusses used ?
Trusses are often used to support roofs,
bridges, power-line towers, and appear
in many other applications.
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3.1. Type of Trusses
3.1.1. Roof Trusses
Are often used as a part of building frame. Trusses
used to support roofs are selected on the basis of
the span, the slope an the roof material. Roof
trusses are supported either by columns of wood,
masonry, concrete, or steel.
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3.1. Type of Trusses
3.1.1. Roof Trusses
Roof
Purlins
C
D
B
E
A
Roof Truss
Span
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Bay
3.1. Type of Trusses
3.1.1. Roof Trusses
There are common types of Roof trusses as follow :
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3.1. Type of Trusses
3.1.2. Bridge Trusses
Are often used as an infrastructure facility.
Trusses used to support load on deck , then floor
beams and finally to the joints of the supporting
trusses. Trusses serves resist to the lateral
forces caused by wind load and the
sideways caused by vehicle moving.
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3.1. Type of Trusses
3.1.2. Bridges Trusses
Typical Bolted Truss Joint
Gusset Plate
Channel Section
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3.1. Type of Trusses
3.1.2. Bridge Trusses
There are common types of Bridge trusses as follow :
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3.2. Trusses Classification
3.2.1. Coplanar Trusses
3.2.1.1. Simple Truss
Simple truss constructed by starting with a basic
tringular element. The simpliet framework that is
rigid and stable is a triangle
A
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C
B
3.2. Trusses Classification
3.2.1. Coplanar Trusses
3.2.1.2. Compound Truss
Æ Is formed by connecting two or more simple trusses
E
F
A
B
C
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D
3.2. Trusses Classification
3.2.1. Coplanar Trusses
3.2.1.3. Complex Truss
Æ That can not classified as
being simple or compund
trusses
F
D
A
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E
C
B
3.2. Trusses Classification
3.2.2. Space Trusses
C
D
A
B
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3.4. Stability
Determinacy Æ
m + r = 2j Æ Statically determinate
m + r > 2j Æ Statically indeterminate
Degree of determinacy Æ
( m + r ) -2j
m
j
r
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= # members
= # joint
= # exteral reaction
3.4. Stability
A truss can be unstable if it is statically
determinate or indeterminate. Stability will
have to be determinated either by inspection or
by forces analysis.
m + r < 2j a truss will be unstable, it will
If
be collapse.
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3.4. Stability
• External Stability
Truss is externally unstable
if
are concurrent or parralel
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all its reactions
3.4. Stability
• External Stability
unstable concurrent reactions
D
E
B
HA
A
C
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VA
HB
3.4. Stability
unstable parallel reactions
• External Stability
D
E
A
B
C
VA
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VB
VC
3.4. Stability
• Internal Stability
If the the structure doesn’t
in a
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hold on its join
fixed position , it will be unstable
3.4. Stability
• Internal Stability
E
F
G
A
B
C
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H
D
3.4. Stability
• Internal Stability
C
F
D
A
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O
E
B
3.4. Stability
Truss is in unstable condition …..
m + r < 2j a truss will be unstable, it will
• If
be collapse.
m + r > 2j
a truss will be unstable, it
• If
becomes if truss support reaction are concurrent
or parallel , or if some components of the truss
form collapsible mechanism.
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3.4. Trusses Analysis
Main concepts
• Forces only act at the pin joints
•
•
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Member forces are always in the direction of the member.
There are no moments in a member (there can be moments on the
full truss or a section of the truss if it has more than one member).
3.4. Trusses Analysis
The three assumptions (or maybe better called idealizations) are :
1. Each joint consists of a single pin to which the respective members
are connected individually.
2. No member extends beyond a joint.
3. Support forces and external loads are only applied at joints.
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3.4. Trusses Analysis
Assume for member analysis
Join
SA-B
SA-B
B
A
SA-B
SA-B
A
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B
TENSILE FORCE Æ
+
COMPRESSIVE FORCE Æ
-
3.4. Trusses Analysis
If the members forces become higher…..
SA-B
SA-B
B
A
SA-B
SA-B
Buckling
B
A
SA-B
SA-B
A
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Separation
B
Crumbling
3.4. Trusses Analysis
3.4.1. Graphics Method
By drawing equilibrium diagrams to find force
members, it called Cremona Methods
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3.4. Trusses Analysis
3.4.1. Graphics Method
Drawing Step :
1. Define the scale for diagram that you will draw
2.Draw all the external reaction
3.Start from join that has max. 2 unknown force members
4. Make a simple complete polygon for that join.
5. Pay attention about the assume for tensile or
compressive force ( + / - )
6. Repeat these steps for another join.
7. Measure the all segmen, and fit with its scale
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3.4. Trusses Analysis
Scale Æ 1 P ~ 4 cm
3.4.1. Graphics Method
D
Example :
Joint A Æ VA – HA – S3 – S1 ; HA = 0
Joint C Æ S1 – S4 – S2 – P
Joint D Æ S4 – S3 – S5
S5
A
3m
S3
S4
S1
HA = 0
S2
Check :
OK !!!
Joint B Æ S2 – S5 – VB
B
C
P
S2
8m
VA = ½ P
VB
VB = ½ P
Number
of
Members
Tension/Compressive
P
( fit with scale )
S1
Tension / +
….cm ~ ….. P
S2
Tension / +
….cm ~ ….. P
S3
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Members Force
Compressive / -
S4
Tension / +
S5
Compressive / -
S5
S4
HA
S3
VA
P
S1
….cm ~ ….. P
4 cm ~ 1 P
….cm ~ ….. P
ΣV = 0 Æ VA + VB = P
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.1. Method of Joint
The method of joints examines each joint as an independent static
structure. The summation of all forces acting on the joint must equate to
zero
. Both member forces and external forces are applied to the joint and then the
force equilibrium equations are applied.
For two dimensions, the equations are
ΣFx = ΣFy = 0
For three dimensions each joint must also be in equilibrium. For 3D problems, however,
the vector form is generally easier. The equation becomes
ΣF = 0 (three dimensional vector form )
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.1. Method of Joint
D
A
HA = 0
S3 S4
S3
S1
S5
D
S5
A
S3
VA = ½P
3m
S4
S1
HA = 0
S2
S5
B
C
S2
P
VA = ½ P
8m
VB = ½ P
S4
S1
P
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B
S2
C
VB = ½P
3.4. Trusses Analysis
Y
3.4.2. Analytics Method
3.4.2.1. Method of Joint
JOINT “A”
Y
S3
S3Y
α
HA = 0
HA = 0
A
S3
S1
VA = ½P
S3X
S1
VA = ½ P
X
Σ KY = 0
Σ KX = 0
VA + S3Y = 0
HA + S3X + S1 = 0
VA + S3. sin α = 0
HA + S3.cos α + S1 = 0
½ P + S3 . 3/5 = 0
0 + -5/6 . 4/5 + S1 = 0
S3 = - 5/6 P
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X
S1 = + 2/3 P
3.4. Trusses Analysis
Y
3.4.2. Analytics Method
3.4.2.1. Method of Joint
JOINT “C”
S4
Y
S1
S2
X
P
S4
S1
P
S2
C
X
Σ KY = 0
Σ KX = 0
-P + S4 = 0
- S1 + S2 = 0
-2/3 P + S2 = 0
S4 = + 1 P
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S2 = + 2/3 P
3.4. Trusses Analysis
Y
3.4.2. Analytics Method
3.4.2.1. Method of Joint
JOINT “D”
Y
S3X
S5X
X
X
D
S3 S4
α
S5
α
S4 = +1 P
S3Y
S3 = - 5/6 P
S5Y
S5
Σ KY = 0
Σ KX = 0
- S4 - S3Y - S5Y = 0
- S3X + S5X = 0
- 1 P- (-5/6 P).sin α - S5 sin α = 0
- (+5/6 P). cos α + S5 cos α = 0
- 1 P- (-5/6 P).3/5 - S5. 3/5 = 0
- ½ P - S5. 3/5 = 0
S5 = - 5/6 P
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OK !!!
- (+5/6 P). 4/5 + S5.4/5 = 0
- 2/3 P + S5.4/5 = 0
S5 = -5/6 P
3.4. Trusses Analysis
Y
3.4.2. Analytics Method
3.4.2.1. Method of Joint
Y
JOINT “B”
Check
S5 = - 5/6 P
S5
S2
S5Y
B
X
α
X
S5X
S2 = +2/3 P
VB = ½P
VB = ½ P
Σ KY = 0
Σ KX = 0
VB + S5Y = 0
- S2 - S5X = 0
½ P +S5 sin α = 0
½ P +S5. 3/5 = 0
½ P + (- 5/6 P). 3/5 = 0
½P–½P=0
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0=0
OK !!!
- (+2/3 P ) - S5 cos α = 0
- (+2/3 P) - (- 5/6 P).4/5 = 0
- 2/3 P + 2/3 P = 0
0=0
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.2. Cross Section Method
This method uses
free-body-
diagrams of sections of the truss to
obtain unknown forces
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.2. Cross Section Method
Cut
II
I
part of structure at
max. 3
D
members
S5
A
S3
3m
S4
S1
HA = 0
S2
B
C
P
VA = ½ P
8m
VB = ½ P
II
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I
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.2. Cross Section Method
D
I
I
S3
S5
S4
S2
A
S3
HA = 0
S2
C
P
VA = ½ P
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I
S4
S1
B
I
VB = ½ P
3.4.2. Analytics Method
3.4.2.2. Cross Section Method
S3
A
S4
S1
HA = 0
ΣMD=0
P
VA = ½ P
ΣMC=0
ΣMD=0
-S2.3m + VA.4m – HA.3m = 0
-S2.3m + ½ P.4m – 0.3m = 0
-S2.3m + 2 P.m = 0
S2 = + 2/3 P
S3
ΣV=0
- S3Y + VB + S4 = 0
- (-5/6).(3/5) + ½ P + S4 = 0
3/6 + ½ P + S4 = 0
S2
C
I
ΣMD=0
S5
S4
S4 = + 1 P
ΣV=0
S3Y + VA + P + S4 = !0
!!
(-5/6).(3/5) + ½ P
OK+ P+ S4 = 0
-3/6 + ½ P + P + S4 = 0
S4 = + 1 P
S2
C
ΣMC=0
B
I
VB = ½ P
ΣMD=0
S2.3m + VB.4m = 0
S2.3m + ½ P.4m = 0
-S2.3m + 2 P.m = 0
ΣMC=0
ΣMC=0
- S3X.3m+ VB.4m = 0
S3Y.4m+ VA.4m = 0
S3.sin α. 4m + ½ P.4m OK
= 0 !!!- S3.cos α. 3m + ½ P.4m = 0
S3.3/5.4m + 2 P.m = 0
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D
I
I
D
CROSS SECTION I-I
3.4. Trusses Analysis
S3 = - 5/6 P
- S3.4/5.3m + 2 P.m = 0
OK !!!
S3 = - 5/6 P
S2 = + 2/3 P
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.2. Cross Section Method
CROSS SECTION II-II
II
II
D
S5
S5
A
S3
S1
S4
S4
S1
S2
B
C
HA = 0
P
VA = ½ P
II
VB = ½ P
II
Determine the force in members for Cross Section II-II !!
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
We used this method for complex
where there are
forces.
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trusses,
no join has max. 2 unknown
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
G
HA
S10
S4
S3
11 + 3 = 2. 7
S1
A
S2
B
C
P
OK !!! Stable ..
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S9
S8
S11
Unknown
forces ??
Statically Determinate Truss
S7
D
2
14 = 14
F
S5
But… which
joint has max.
m + r = 2j
S6
E
VA
VB
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
S6
E
F
S7
S5
Y
D
Step 1st :
G
S9
S8
S10
Change the
S4
S3
members position &
make a new
member Æ “Y”
Æ Change Position of member
11 & new member is “Y”
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HA
S1
A
S2
B
C
P
VA
VB
3.4. Trusses Analysis
S6
E
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
F
S7
S5
Y
D
Step 2nd :
G
S9
S8
S10
Determine all
S4
S3
the members forces
with external force
(by using any methods )
o
ÆS
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HA
S1
A
S2
B
C
P
VA
VB
3.4. Trusses Analysis
S6
E
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
1 unit
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S7
S5
1 unit Y
D
Step 3rd :
Put
forces
on the member
position that has
HA
changed its position
before. Assume that 1
unit force is a
Tension force
F
G
S9
S8
S10
S3
S4
1 unit
S1
A
S2
B
C
P
VA
VB
3.4. Trusses Analysis
S6
E
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
1 unit Y
D
G
S9
S8
Determine all
S10
S3
the members forces
force (by using any
S7
S5
Step 4th :
without external
F
HA
S4
1 unit
S1
A
S2
B
C
methods )
’
ÆS
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VA
VB
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 5th :
“Y”
Check with member
that used before Æ the
structure never has member “Y”
So…
S’Y (x) + SoY = 0
Determine x value for
that structure
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SoY
x = - S’Y
Member forces are :
So + S’(x) = 0
3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 6th :
Put all your results in Step 2nd and step 4th on the table
below….
Input S’ from step 4th
x from step 5th
Input So from step 2nd
No of Members
So
S’
S’. x
So + S’(x)
i….
Final member
forces
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3.5. Space Truss Analysis
A space truss can be unstable if it is statically
determinate or indeterminate. Stability will
have to be determinated either by inspection or
by forces analysis.
m + r < 3j a truss will be unstable, it will
If
be collapse.
Bina Nusantara
3.5. Space Truss Analysis
Like a plannar trusses , Stable condition for :
m + r = 2j Æ Statically determinate
m + r > 2j Æ Statically indeterminate
For Æ
External stablity Æ Check Reactions
Internal Stability Æ Check Members arrangement
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3.5. Space Truss Analysis
That
3 dimensions there are
3for each
equations of equilibrium
joint Æ
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
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Z
Y
X
3.5. Space Truss Analysis
Bracing
Truss Construction
D
C
A
B
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3.5. Space Truss Analysis
E
D
C
A
B
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3.5. Space Truss Analysis
Joint E
Z
Y
X
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
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