Analysis of Statically Determinate Trusses Session...
Transcript of Analysis of Statically Determinate Trusses Session...
Analysis of Statically Determinate TrussesSession 07 - 10
Subject : S0695 / STATICS
Year : 2008
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What are Trusses ?
are structures consisting of two or more straight, slender membersconnected 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
Purlins
A
B
C
D
E
Roof Truss
Span
Bay
Roof
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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 TrussesThere 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 B
C
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3.2. Trusses Classification
3.2.1. Coplanar Trusses 3.2.1.2. Compound Truss
Is formed by connecting two or more simple trusses
AC
BD
E F
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3.2. Trusses Classification
3.2.1. Coplanar Trusses
3.2.1.3. Complex Truss
That can not classified as being simple or compund trusses
A C B
D E
F
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3.2. Trusses Classification
3.2.2. Space Trusses
AB
CD
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3.4. Stability
Determinacy
m + r = 2j Statically determinatem + r > 2j Statically indeterminate
Degree of determinacy
( m + r ) -2jm = # membersj = # jointr = # exteral reaction
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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.
If m + r < 2j a truss will be unstable, it will be collapse.
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3.4. Stability
• External Stability
Truss is externally unstable if all its reactions
are concurrent or parralel
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3.4. Stability
• External Stability unstable concurrent reactions
A
C
B
D E
VA
HA
HB
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3.4. Stability
• External Stability unstable parallel reactions
A
D E
VA VB VC
CB
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3.4. Stability
• Internal Stability
If the the structure doesn’t hold on its join
in a fixed position , it will be unstable
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3.4. Stability
• Internal Stability
AC
BD
E F HG
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3.4. Stability
• Internal Stability
A
C
B
D E
F
O
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3.4. Stability
Truss is in unstable condition …..
• If m + r < 2j a truss will be unstable, it will be collapse.
• If m + r > 2j a truss will be unstable, it 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
• 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).
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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
COMPRESSIVE FORCE -A B
SA-BSA-B
Join
SA-B
A BTENSILE FORCE +SA-B
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3.4. Trusses Analysis
If the members forces become higher…..
SA-B
A B
SA-B Separation
A B
SA-BSA-B
A B
SA-BSA-B Buckling
Crumbling
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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 MethodDrawing Step :1. Define the scale for diagram that you will draw2.Draw all the external reaction3.Start from join that has max. 2 unknown force members4. 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
3.4.1. Graphics Method
Example :
Scale 1 P ~ 4 cm
Joint A VA – HA – S3 – S1 ; HA = 0
VA
AB
C
D
S2
S3
S5
S1S4
P8 m
3 m
VA = ½ PVB = ½ P
HA = 0
VB
P
S3
S1
P
( fit with scale )
Tension/Compressive
S5
S4
S3
S2
S1
Members ForceNumber of
Members
….cm ~ ….. P
….cm ~ ….. P
Tension / +
Compressive / -
Joint C S1 – S4 – S2 – P
S4
S2
4 cm ~ 1 PTension / +
….cm ~ ….. PTension / +
Joint D S4 – S3 – S5
S5
….cm ~ ….. PCompressive / -
HA
Check :
Joint B S2 – S5 – VB OK !!!
ΣV = 0 VA + VB = P
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.1. Method of Joint
The method of joints examines each joint as an independent staticstructure. 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 Method3.4.2.1. Method of Joint
AB
C
D
S2
S3
S5
S1S4
P8 m
3 m
VA = ½ PVB = ½ P
HA = 0
VA = ½P
A S3
HA = 0S1
DS5S4S3
CS2S1
S4
P
B
VB = ½P
S2
S5
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.1. Method of Joint
VA = ½P
A S3HA = 0
S1
Y
X
Y
X
VA = ½ P
HA = 0
S3
S1
Σ KY = 0
VA + S3Y = 0
VA + S3. sin α = 0
½ P + S3 . 3/5 = 0
α
S3Y
S3X
S3 = - 5/6 P
Σ KX = 0
HA + S3X + S1 = 0
HA + S3.cos α + S1 = 0
0 + -5/6 . 4/5 + S1 = 0
S1 = + 2/3 P
JOINT “A”
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.1. Method of Joint
Y
X
P
S1
Σ KY = 0
-P + S4 = 0
S4 = + 1 P
Σ KX = 0
- S1 + S2 = 0
-2/3 P + S2 = 0
S2 = + 2/3 P
JOINT “C”
Y
X
CS2S1
S4
P
S2
S4
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.1. Method of Joint
Y
X
Y
S4 = +1 P
Σ KY = 0
- S4 - S3Y - S5Y = 0
- 1 P- (-5/6 P).sin α - S5 sin α = 0
- 1 P- (-5/6 P).3/5 - S5. 3/5 = 0
- ½ P - S5. 3/5 = 0
S5 = - 5/6 P
Σ KX = 0
- S3X + S5X = 0
- (+5/6 P). cos α + S5 cos α = 0
- (+5/6 P). 4/5 + S5.4/5 = 0
- 2/3 P + S5.4/5 = 0
S5 = -5/6 P
JOINT “D”
DS5S4S3
X
S3 = - 5/6 PS5
α α
S3Y
S3X
S5Y
S5X
OK !!!
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.1. Method of Joint
Y
S2 = +2/3 P
Σ KY = 0
VB + S5Y = 0
½ P +S5 sin α = 0
½ P +S5. 3/5 = 0
½ P + (- 5/6 P). 3/5 = 0
½ P – ½ P = 0
0 = 0
Σ KX = 0
- S2 - S5X = 0
- (+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
JOINT “B”Check
X
S5 = - 5/6 P
α
OK !!!
Y
XB
VB = ½P
S2
S5
VB = ½ P
S5Y
S5X
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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 Method3.4.2.2. Cross Section Method
AB
C
D
S2
S3
S5
S1S4
P8 m
3 m
VA = ½ PVB = ½ P
HA = 0
I
I
II
II
Cutpart of structure at
max. 3members
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.2. Cross Section Method
A
CS2
S3
S1S4
PVA = ½ P
HA = 0
I
I
D
S5
VB = ½ P
I
I
S2
S4
S3
B
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3.4. Trusses Analysis3.4.2. Analytics Method3.4.2.2. Cross Section Method
A
CS2
S3
S1S4
PVA = ½ P
HA = 0
I
I
D
S5
VB = ½ P
I
I
S2
S4
S3
B
CROSS SECTION I-I
D ΣMD=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
ΣMC=0
ΣMC=0S3Y.4m+ VA.4m = 0 S3.sin α. 4m + ½ P.4m = 0S3.3/5.4m + 2 P.m = 0
S3 = - 5/6 P
ΣV=0S3Y + VA + P + S4 = 0 (-5/6).(3/5) + ½ P + P+ S4 = 0-3/6 + ½ P + P + S4 = 0
S4 = + 1 P
CΣMC=0
ΣMD=0
ΣMD=0S2.3m + VB.4m = 0S2.3m + ½ P.4m = 0-S2.3m + 2 P.m = 0
S2 = + 2/3 PΣMC=0- S3X.3m+ VB.4m = 0 - S3.cos α. 3m + ½ P.4m = 0- S3.4/5.3m + 2 P.m = 0
S3 = - 5/6 P
ΣV=0- S3Y + VB + S4 = 0 - (-5/6).(3/5) + ½ P + S4 = 03/6 + ½ P + S4 = 0
S4 = + 1 P
OK !!!
OK !!!
OK !!!
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3.4. Trusses Analysis
3.4.2. Analytics Method3.4.2.2. Cross Section Method
A
D
S3
S5
S1S4
VA = ½ P
HA = 0
II
II
CROSS SECTION II-II
BCS2
S5
S1S4
PVB = ½ P
II
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 trusses,
where there are no join has max. 2 unknown forces.
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
m + r = 2j11 + 3 = 2. 7
14 = 14
OK !!! Stable ..
Statically Determinate Truss
A C B
D
E F
G
S1 S2
S3S4
S5
S6
S7
S9S11 S10
S8
P
VA VB
HA
But… which joint has max.
2Unknown forces ??
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 1st :
A C B
D
E F
G
S1 S2
S3S4
S5
S6
S7
S9
Y
S10S8
P
VA VB
HA
Change the members position & make a new member “Y”
Change Position of member 11 & new member is “Y”
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 2nd :
A C B
D
E F
G
S1 S2
S3S4
S5
S6
S7
S9
Y
S10S8
P
VA VB
HA
Determine all the members forces with external force(by using any methods )
So
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 3rd :
A C B
D
E F
G
S1 S2
S3S4
S5
S6
S7
S9
Y
S10S8
P
VA VB
HA
Put 1 unit forceson the member position that has changed its position before. Assume that 1 unit force is a Tension force
1 unit
1 unit
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3.4. Trusses Analysis
3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 4th :
A C B
D
E F
G
S1 S2
S3S4
S5
S6
S7
S9
Y
S10S8
VA VB
HA
1 unit
1 unitDetermine all the members forces
without external force (by using any methods )
S’
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3.4. Trusses Analysis3.4.2. Analytics Method
3.4.2.3. Henneberg Method Step 5th :
Check with member “Y” that used before the structure never has member “Y”
So…S’Y (x) + SoY = 0 x = - SoY
S’Y
Member forces are :So + S’(x) = 0
Determine x value for that structure
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3.4. Trusses Analysis3.4.2. Analytics Method
3.4.2.3. Henneberg Method
Step 6th :Put all your results in Step 2nd and step 4th on the table below….
i….
So + S’(x) S’. xS’SoNo of Members
Input So from step 2ndInput S’ from step 4th
x from step 5th
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.
If m + r < 3j a truss will be unstable, it will be collapse.
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3.5. Space Truss Analysis
Like a plannar trusses , Stable condition for :
m + r = 2j Statically determinate
m + r > 2j Statically indeterminate For
External stablity Check Reactions
Internal Stability Check Members arrangement
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3.5. Space Truss Analysis
That 3 dimensions there are
3 equations of equilibrium for each joint
Σ Fx = 0Σ Fy = 0Σ Fz = 0
Z
Y
X
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3.5. Space Truss Analysis
AB
CD
Truss ConstructionBracing
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3.5. Space Truss Analysis
AB
CD
E
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3.5. Space Truss Analysis
Joint E Z
Y
X
Σ Fx = 0Σ Fy = 0Σ Fz = 0