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Framed structures

Framed structures

 

 

Framed structures

A frame structure is an assembly of members and joints (usually called Nodes) which is designed to support a load.  Examples of frame structures include roof trusses, bridges, pylon towers etc.

 

The members in this framed structure can be as ties or struts, depending on the type of force they support.framed structrures

 

 

 

STRUTS AND TIES

 

Strut

Members that are in compression, due to external forces trying to compress them, are known as Struts.


framed structrures

Tie

Members that are in tension, due to external forces trying to pull them apart, are known as Ties.


framed structrures

 


Nodal Analysis

framed structruresNodes

The members are either in compression (strut) or tension (tie).  They can be represented at the node as shown below -

FB is under compression and pushes into the node, FA is under tension and pulls away from the node.

Conditions of Static Equilibrium

 

SUM OF THE MOMENTS = 0                                            S Mo = 0
SUM OF THE VERTICAL FORCES = 0                            S FV = 0
SUM OF THE HORIZONTAL FORCES = 0                      S FH = 0

 

When solving frame structures some analysis of the structure is required to determine the best starting point and which of the conditions of static equilibrium to apply first.

Solving Simple Frame Structures

 

Cantilevered frame structures

For this frame structure it is not necessary to use moments to help find the forces in the members. .
The node with the 900N load acting has only one unknown vertical component.

 

framed structrures

 

As all vertical forces acting on the node must equal zero then FV must equal 900N.

By using trigonometry it is now possible to find the other forces acting on this node.

 

framed structruresWorked Example

 

For the crane shown below we shall find the forces in the members and the reaction forces at the wall.

 

 

The framework is supported at two points.  The hinge support at the top is being pulled away from the wall. R1 will act against this pull and keep the hinge attached to the wall. The roller support is being pushed into the wall. R2 will act against this force and in the opposite direction as shown below.

As member B is acting on a roller then R2 will be at 90° to member C

As R1 is acting at a hinge support there will be a vertical and horizontal component.  At this stage guess the direction of R1.

 

framed structrures

 

If a wrong assumption about the member being in tension or compression or the direction of a force is made then a negative value will be produced from the calculation.  If this is the case the direction of the force is simply reversed

To find the forces in the members and the reactions at the supports, study each of the structure nodes separately.

framed structrures
Node 1
framed structruresThe forces acting on node 1 are shown below.

 

Note - an assumption has been made about the direction of FA and FB.

FA is assumed to be in tension and acting away from the node 1.

framed structruresFB is assumed to be in compression and acting in towards the node 1.

Split FB into its horizontal (FHB) and vertical (FVB) components.

 

Apply a condition of static equilibrium - 

SFV = 0

 

As the sum of Vertical forces is equal to zero then the vertical component of FB (FVB) is 1000N acting up.

From this we can find FB - we can redraw FB and the two components FHB and FVB to form a triangle as shown below.

 

framed structrures

 

framed structrures

framed structruresThe force in member A (FA) will equal and opposite to FHB.

 

 

Find FHB -

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From above FA = 1732N

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This diagram shows all the forces acting at node 1.

 

 

Node 2

framed structruresThere are three forces acting at node 2.

 

 

 

During analysis of node 1, FB was found to be 2000N and was not negative so the assumed direction towards the node (compression) was correct.

In node 1, FB is shown to be acting as a compressive force towards the node with a magnitude of 2000N.
framed structrures

 

FC will be equal and opposite to the Vertical component of FB (FVB).

From analysis of node 1 FVB was equal to 1000 N therefore -
FC = 1000 N
framed structruresAs R2 is a roller support the reaction will be at 90° to the surface.

R2 will be equal to the Horizontal component of FB.

 

 

From above HB was equal to 1732 N therefore -

R2 = 1732N
framed structrures
The forces acting at node 2 are shown below.

 

 

 

 

 

Node 3

framed structruresNode 3 has three forces acting on it.

R1 will be equal and opposite to the resultant of FA and FC.

 

From analysis of node 1 FA was found to be 1732N.
From analysis of node 2 FC was found to be 1000N

 

Find resultant of FA & FC - FR

 

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Find angle of resultant (q) -

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The task was to find the forces in the members and the reactions forces at the supports.  The results can be shown as below -

 

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Source: http://www.earlstonhigh.scotborders.sch.uk/learningzone/technical/Revision%20Notes/HTS%20Revision%20notes/3-2%20Revision.doc

Web site to visit: http://www.earlstonhigh.scotborders.sch.uk

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Framed structures

 

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Framed structures

 

 

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Framed structures