Analysis of Statically Determinate Trusses Session …repository.binus.ac.id/content/s0695/s069577921.pdfA truss can be unstable if it is statically determinate or indeterminate. Stability

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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3.1.1. Roof Trusses

Purlins

A

B

C

D

E

Roof Truss

Span

Bay

Roof

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3.1.1. Roof Trusses

There are common types of Roof trusses as follow :

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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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3.1.2. Bridges Trusses

Typical Bolted Truss Joint

Gusset Plate

Channel Section

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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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3.2.1. 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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3.2.1. 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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3.2.2. 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


AC

BD

E F HG

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3.4. 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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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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Assume for member analysis

COMPRESSIVE FORCE -A B

SA-BSA-B

Join

SA-B

A BTENSILE FORCE +SA-B

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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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3.4.1. Graphics Method

By drawing equilibrium diagrams to find force

members, it called Cremona Methods

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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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3.4.1. 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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3.4.2. 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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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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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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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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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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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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3.4.2. 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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3.4.2. 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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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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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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3.4.2.3. Henneberg Method

We used this method for complex trusses,

where there are no join has max. 2 unknown forces.

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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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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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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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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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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


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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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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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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AB

CD

Truss ConstructionBracing

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AB

CD

E

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Joint E Z

Y

X

Σ Fx = 0Σ Fy = 0Σ Fz = 0

Analysis of Statically Determinate Trusses Session …repository.binus.ac.id/content/s0695/s069577921.pdfA truss can be unstable if it is statically determinate or indeterminate. Stability

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