lecture 6 pin jointed structures - complete

8/13/2019 Lecture 6 Pin Jointed Structures - Complete

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Statics Lecture 6

Pin Jointed Structures

andthe Method of Joints

Dr JD Shephard


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Pin Jointed Structures or trusses

What is a Pin Jointed Structure or Truss?

It is an Engineering term used to describe a structurecomposed of pin jointed members.

Only two forces can act on the member (there are onlytwo joints in a given member)

Assumption: The mass of the members can beneglected; and a truss is only loaded or supported at thejoints of its members.

Previously we have considered equilibrium of a single rigid body.

We drew a FBD of this body showing all the external forcesacting on the body and then used force and moment equilibriumconditions.

Here we will consider the internal forces of a structure i.e. forces

of action and reaction between connected bars or struts -members.

Principle of analysis

To analyse we dismember the structure and look at separateFBDs of individual members or combinations of members.

NB: Have to think carefully about Newtons 3rd law!


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Members and Joints (or Pins)

Correct analysis requires that you distinguish between forceson joints and those on members. e.g. If the member shown

below is pinned at A and B and is under Tension, T, orCompression, C, then the following force diagrams can bedrawn:

BA

BA

Tensile force on member

T T

BA

Tensile force onjoints (or pins)

T T

BA

Compressive force on member

C C

BA

Compressive force onjoints (or pins)

C C


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Loading of a simple truss

A

B

C

F

Simplest truss or pin jointed structure is a triangle.

A

B

C

Tension in member

Compression

in member

Compressionin member

BIsolating joint (or pin) B:

F

Compression

in memberBC

Compression

in memberAB


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Building structures from triangles

Structures are built from a basic triangle and are knownas simple trusses.

Some common bridge trusses

PrattHowe

K TypeWarren

Real example of Warren Truss


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3 Key assumptions:

All joints are frictionless pin joints therefore each memberis subject ONLY to tensile or compressive forces. Nobending.

External loads are applied ONLY atjoints.

Members are light compared to applied loads or inducedforces and therefore the weight of the member is negligible.

Formal analysis of pin jointed structures or trusses

Two methodsof analysis

1. Method of Joints (this lecture)

In this method we consider satisfying the conditions forequilibrium for the forces acting on the connecting pin ofeachjoint.

Equilibrium of concurrent forces at pins only twoindependent equilibrium equations are involved (forceequilibrium)

2. Method of Sections (next week)

In this method we consider an entire section of thestructure for the free body under equilibrium.

Takes advantage of the third equilibrium equation(moment equilibrium) as well as the force equilibrium.


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The Method of Joints (example)

A

F E

D

CBF

Structure loaded at B with a force F (= 10 kN).

Step 1: The FBD (for the whole structure)

CB

F =10 kN

We need this initial step to find reactions at A and D

RA RDRoller can only can

support vertical

No other horizontal

components

30o

60o

60o

60o30o

30o 30o

60o

A

F E

D

30o

60o

60o

60o30o

30o 30o

60o

1 m 1 m 1 m

1 m


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We need to use moment equilibrium first to help find reactions.(NB: This is the only use of moments in method of joints.)

MD = 0: RA(3) 10(2) = 0 RA = 6.67 kN

Then use force equilibrium:

FV = 0: RA + RD 10 = 0 RD = 3.33 kN

Now use Method of Joints and start at Joint A:

RA = 6.67 kN

AB

AF

60o

Compressive

Tensile

NB: Take an educated guess at

direction of force (arrows). Aslong as you use same directionfor equilibrium equations thenthe sign (+ or -) will reveal truedirection!

Force equilibrium (Joint A):

FV = 0: RA AFsin60 = 0 AF = 6.67/sin60

AF = 7.70 kN

FH = 0: AB AFcos60 = 0 AB = 7.70cos60

AB = 3.85 kN

Step 2: Find the reactions at supports

Step 3: Equilibrium at Joint A

A


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Force equilibrium (Joint F):

FV = 0: AFcos30 BF = 0 BF = 7.70cos30

BF = 6.67 kN

FH = 0: AFsin30 EF = 0 EF = 7.70sin30

EF = 3.85 kN

Step 4: Equilibrium at Joint F

BFAF = 7.70 kN

EFThink about directions again (but dont worry!)

30o

Step 5: Equilibrium at Joint B

F = 10 kN

BC

Force equilibrium (Joint B):

FV = 0: BF + BEsin60 F = 0 BE = (10 6.67)/sin60

BE = 3.85 kN

FH = 0: BC + BEcos60 AB = 0 BC = 3.85 3.85cos60

BC = 1.93 kN

F

B

BE

BF = 6.67 kN

AB = 3.85 kN

30o

60o


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Step 6: Equilibrium at Joint C

CD

Force equilibrium (Joint C):

FV = 0: CE = 0 !!!!!

So in this load condition the force in member CE = 0. However,

this would not be zero if a vertical load was applied to joint C.Think about a car travelling over a bridge?

FH = 0: BC - CD = 0 CD = 1.93 kN

C

CE

BC = 1.93 kN

Step 7: Equilibrium at Joint E

CE = 0

DE

Force equilibrium (Joint E):

FV = 0: DEcos30 BEcos30 = 0 DE = BE

DE = 3.85 kN

E

EF = 3.85 kN

BE = 3.85 kN

60o

30o30o


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Step 8: Equilibrium at Joint D

We have now found the forces in all members but lets use jointD to check:

RD = 3.33 kN

DE = 3.85 kN

Force equilibrium (Joint D):

FV = 0: RD DEsin60 = 0

3.33 3.85sin60 = 0 3.33 3.33 = 0

Phew, it works!

DCD = 1.93 kN60o


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Next week, the method of sections!

All the internal forces in the structure

RA F RD

A B C D

EF

lecture 6 pin jointed structures - complete

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