     # Truss Analysis Using The Stiffness Chapter 14 Truss Analysis Using The Stiffness Method. Stiffness Method

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• Chapter 14

Truss Analysis

Using The Stiffness Method

• Stiffness Method

• Fundamentals of the stiffness method

– There are essentially two ways in which structures can be analyzed using

matrix methods.

• Flexibility method

• Stiffness method

– The stiffness method can be used to analyzed both statically determinate

& indeterminate structures, whereas the flexibility method required a

different procedure for each of these cases.

– The stiffness method yields the displacements & forces directly, whereas

with the flexibility method the displacements are not obtained directly.

• – Application of this method requires subdividing the structure into a series

of discrete finite elements & identifying their end points as nodes.

– The force-displacement properties of each element are determined & then

related to one another using the force equilibrium equations written at the

nodes

– These relationships, for the entire structure, are then grouped together into

what is called the structure stiffness matrix K.

– Once the K is established, the unknown displacements of the nodes can

– When the displacements are known, the external & internal forces in the

structure can be calculated using the force-displacement relations for each

member.

Stiffness Method

• Preliminary Definitions & Concepts

• Member & node identifications – We will specify each member by a number enclosed within a square,

& use a number enclosed within a circle to identify the nodes.

– Also the ‘near’ & ‘far’ ends of the member must be identified, this will

be done using an arrow written along the member with the head of the

arrow directed toward the far end.

• • Global & member coordinates. – We will use two different type of coordinate systems, global or

structure coordinate system and local or member coordinate system.

– Global system x, y, used to specify the sense of each of the external

force & displacement components at the nodes.

– Local system x’,y’ used to specify the sense of direction of members

Preliminary Definitions & Concepts

• • Degrees of freedom – The unconstrained truss has two degree of freedom or two possible

displacements for each joint (node).

– Each degree of freedom will be specified on the truss using a code

number, shown at the joint or node, & referenced to its positive global

coordinate direction using an associated arrow.

• For example – The truss has eight degree of freedom or eight possible displacements.

– 1 through 5 represent unknown or

unconstrained degree of freedom.

– 6 through 8 represent constrained

degree of freedom

Lowest code numbers will always be used

to identify the unknown displacements.

Highest code numbers will be used to

identify the known displacements

Preliminary Definitions & Concepts

• Member Stiffness Matrix

• Case I

– Positive displacement dN on the near end

• Case II

– Positive displacement dF on the far end

• Case I + Case II

– Resultant forces caused by both displacements are

'N N AE

q d L

 'F N AE

q d L

 

''N F AE

q d L

  ''F F AE

q d L

N N F

AE AE q d d

L L  

F N F

AE AE q d d

L L   

• – These load-displacement equations written in matrix form

1 1

1 1

N N

F F

q dAE

q dL

             

'q k d

1 1 '

1 1

AE k

L

     

or

where

The matrix, k` is called the member stiffness matrix.

It is of the same form for each member of the truss.

Member Stiffness Matrix

• Transformation Matrices • Displacement & force transformation

– We will now develop a method for

transforming the member forces q and

displacements d defined in local coordinate

to global coordinates

cos F Nx x x x

L  

  

cos F Ny y y y

L  

  

    2 2

cos F Nx x F N F N

x x

x x y y  

  

  

    2 2

cos F Ny y F N F N

y y

x x y y  

  

  

• Transformation Matrices • Displacement transformation matrix

– In global coordinate each end of the member can have two independent

displacements.

– Joint N has displacements DNx & DNy in global coordinate

cos cosN Nx x Ny yd D D  

N Nx x Ny yd D D  

or

• – Also joint F has displacements DFx & DFy

– Displacements at N & F

cos cosF Fx x Fy yd D D  

F Fx x Fy yd D D  

or

N Nx x Ny yd D D  

F Fx x Fy yd D D  

0. 0.

0. 0.

Nx

x y NyN

x y FxF

Fy

D

Dd

Dd

D

 

 

   

                

d TD or

0. 0.

0. 0.

x y

x y

T  

 

      

where

Transformation Matrices

• • Force transformation matrix

– Force qN applied to the near end of the member

– Force qF applied to the far end of the member

– Rewrite in a matrix form:

cosNx N xQ q  or

cosNy N yQ q 

Nx N xQ q  Ny N yQ q 

cosFx F xQ q  or

cosFy F yQ q 

Fx F xQ q  Fy F yQ q 

0.

0.

0.

0.

Nx x

Ny y N

Fx x F

Fy y

Q

Q q

Q q

Q

       

                   

TQ T q or where

0.

0.

0.

0.

x

yT

x

y

T

           

Transformation Matrices

• Member Global Stiffness Matrix

• Stiffness matrix

– We will determine the stiffness matrix for a member which relates

the member’s global force components Q to its global

displacements D. 'q k d d TDand

'q k T D 

TQ T q 'TQ T k TD  Q kD

'Tk T k T 0.

0. 0. 0.1 1

0. 0. 0.1 1

0.

x

y x y

x x y

y

AE k

L

  

  

   

                

• Member Global Stiffness Matrix

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAE k

FL

F

     

     

     

     

    

           

x y x yN N F F

'Tk T k T

• Example 1  Determine the structure stiffness matrix for the two-member truss

shown. AE is constant

Solution:

Establish the x, y global system

Identify each joint & member numerically.

• Member 1:

Determine x & y, where L = 3ft

3 0 1

3 x

  

0 0 0

3 y

  

2 2

2 2

2 2

2 2

xx x y x x y

yy x y y x y

xx x y x x y

yy x y y x y

N

NAE k

FL

F

     

     

     

     

    

           

x y x yN N F F

1

0.333 0. 0.333 0. 1

0. 0. 0. 0. 2

0.333 0. 0.333 0. 3

0. 0. 0. 0. 4

k AE

           

1 2 3 4

Dividing each element by L = 3ft

  22 1

0.333 3

x

L

    

• Member 2: Determine x & y, where L = 5ft

3 0 0.6

5 x

  

4 0 0.8

5 y

  

2 2

2 2

2 2

2 2

xx x y x x y

yy ##### m4l25 Lesson 25 The Direct Stiffness Method: Truss Analysis (Continued)
Documents ##### Structural Analysis 7 th Edition in SI Units Russell C. Hibbeler Chapter 14: Truss Analysis Using the Stiffness Method
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