Chapter 14: Chapter 14: Truss Analysis Using the Stiffness Method Truss Analysis Using the Stiffness Method

Structural Analysis 7

Structural Analysis 7 ^{th th} Edition in SI Units Edition in SI Units

Chapter 14:

Chapter 14:

Truss Analysis Using the Stiffness Method

Truss Analysis Using the Stiffness Method

Fundamentals of the stiffness method Fundamentals of the stiffness method

• The stiffness method: The stiffness method:

• Is a disp method of analysis Is a disp method of analysis

• Can be used to analyse both statically determinate Can be used to analyse both statically determinate and indeterminate structures

and indeterminate structures

• Yields the disp & forces directly Yields the disp & forces directly

• It is generally much easier to formulate the It is generally much easier to formulate the

necessary matrices for the computer using the necessary matrices for the computer using the

stiffness method

stiffness method

Fundamentals of the stiffness method Fundamentals of the stiffness method

• Application of the stiffness method requires Application of the stiffness method requires

subdividing the structure into a series of discrete subdividing the structure into a series of discrete

finite elements & identifying their end points as finite elements & identifying their end points as

nodes nodes

• For truss analysis, the finite elements are For truss analysis, the finite elements are

represented by each of the members that compose represented by each of the members that compose

the truss & the nodes represent the joints the truss & the nodes represent the joints

• The force-disp properties of each element are The force-disp properties of each element are determined & then related to one another using determined & then related to one another using

the force eqm eqn written at the nodes

the force eqm eqn written at the nodes

Fundamentals of the stiffness method Fundamentals of the stiffness method

• These relationships for the entire structure are These relationships for the entire structure are

then grouped together into the structure stiffness then grouped together into the structure stiffness

matrix, K matrix, K

• The unknown disp of the nodes can then be The unknown disp of the nodes can then be

determined for any given loading on the structure determined for any given loading on the structure

• When these disp are known, the external & When these disp are known, the external &

internal forces in the structure can be calculated internal forces in the structure can be calculated

using the force-disp relations for each member

using the force-disp relations for each member

Member stiffness matrix Member stiffness matrix

• To establish the stiffness matrix for a single truss To establish the stiffness matrix for a single truss member using local x’ and y’ coordinates as shown member using local x’ and y’ coordinates as shown

When a +ve disp d

When a +ve disp d _{N N} is imposed on the near end of is imposed on the near end of the member while the far end is held pinned

the member while the far end is held pinned

• The forces developed at the ends of the members The forces developed at the ends of the members are: are:

N F

N

N d

L q AE

L d

q '  AE ; '  

Member stiffness matrix Member stiffness matrix

• Likewise, a +ve disp d Likewise, a +ve disp d _{F F} at the far end, keeping the at the far end, keeping the near end pinned and results in member forces

near end pinned and results in member forces

• By superposition, the resultant By superposition, the resultant forces caused by both disp are forces caused by both disp are

F F

F

N d

L q AE

L d

q ''   AE ; '' 

F N

N

AE d AE d

q

L d d AE

L q AE









Member stiffness matrix Member stiffness matrix

• These load-disp eqn may be written in matrix form These load-disp eqn may be written in matrix form as: as:

• This matrix, k’ is called the member stiffness This matrix, k’ is called the member stiffness matrix

matrix

 

 







 



 

 



 



 







 

 

 





1 1

1 1 '

'

1 1

1 1

L k AE

d k q

d d L

AE q

q

F N F

N

Member stiffness matrix Member stiffness matrix

+ =

Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Since a truss is composed of many members, we Since a truss is composed of many members, we will develop a method for transforming the

will develop a method for transforming the member forces q and disp d defined in local member forces q and disp d defined in local

coordinates to global coordinates coordinates to global coordinates

• Global coordinates convention: +ve x to the right Global coordinates convention: +ve x to the right and +ve y upward

and +ve y upward

   _{x x} and and   _{y y} as shown as shown

Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• The cosines of these angles will be used in the The cosines of these angles will be used in the matrix analysis as follows

matrix analysis as follows

• These will be identified as These will be identified as

• For e.g. consider member NF of the truss as For e.g. consider member NF of the truss as shown

shown

• The coordinates of N & F The coordinates of N & F are (x

are (x _{N N} , y , y _{N N} ) and (x ) and (x _{F F} , y , y _{F F} ) )

y y

x

x   

  cos ;  cos

Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

2 2

2 2

) (

) (

cos

) (

) (

cos

N F

N F

N F

N y F

y

N F

N F

N F

N F

x x

y y

x x

y y

L y y

y y

x x

x x

L x x







 

 









 

 











Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Disp Transformation matrix Disp Transformation matrix

• In global coordinates each end of the member can In global coordinates each end of the member can have 2 degrees of freedom or independent disp;

have 2 degrees of freedom or independent disp;

namely joint N has D

namely joint N has D _{Nx Nx} and D and D _{Ny Ny}

• Joint F has D Joint F has D _{Fx Fx} and D and D _{Fy Fy}

Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Disp Transformation matrix Disp Transformation matrix

• When the far end is held pinned & the near end is When the far end is held pinned & the near end is given a global disp, the corresponding disp along given a global disp, the corresponding disp along

member is D

member is D _{Nx Nx} cos cos   _{x x}

• A disp D A disp D _{ny ny} will cause the member to be displaced will cause the member to be displaced D D _{Ny Ny} cos cos   _{y y} along the x’ axis along the x’ axis

y F

x F

F

y N

x N

N

y x

y x

D D

d

D D

d









cos cos

cos cos









Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Disp Transformation matrix Disp Transformation matrix

0 0

0 0 form, matrix

In

;

cos

; cos Let

TD d

D D D D d

d

D D

d D

D d

y x y x

y x

y x

F F N N

y x y x F

N

y F x

F F

y N x

N N

y y

x x



 

 







 

 







 





 



 

 

 









































Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Force Transformation matrix Force Transformation matrix

• If q If q _{F F} is applied to the bar, is applied to the bar,

the global force components at F are:

the global force components at F are:

• Using Using

y N

N x

N

N q Q q

Q

 cos  ;

 cos 

y y

x

x   

  cos ;  cos

y F

F x

F

F q Q q

Q

 cos  ;

 cos 

y F F

x F F

y N N

x N N

q Q

q Q

q Q

q

Q

















;

;

Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Force Transformation matrix Force Transformation matrix

• In matrix form In matrix form

q T Q

q q Q

Q Q Q

T

F N

y x y

x

F F N N

y x y x



 

 





 

 







 

 









 

 







 

 







0

0

0

0









Displacement & Force Transformation Displacement & Force Transformation

matrices matrices

• Force Transformation matrix Force Transformation matrix

• In this case, T In this case, T ^{T T} transforms the 2 local forces q transforms the 2 local forces q

acting at the ends of the member into 4 global force acting at the ends of the member into 4 global force

components Q components Q

• This force transformation matrix is the transpose of This force transformation matrix is the transpose of the disp transformation matrix

the disp transformation matrix

Member global stiffness matrix Member global stiffness matrix

• We can determine the member’s forces q in terms We can determine the member’s forces q in terms of the global disp D at its end points

of the global disp D at its end points

• Substitution yields the final result: Substitution yields the final result:

TD k

q  '

k'T T

k

TD k

T Q

T T







kD Q

or

'

Member global stiffness matrix Member global stiffness matrix

• Performing the matrix operation yields: Performing the matrix operation yields:

Truss stiffness matrix Truss stiffness matrix

• Once all the member stiffness matrices are formed Once all the member stiffness matrices are formed in the global coordinates, it becomes necessary to in the global coordinates, it becomes necessary to

assemble them in the proper order so that the assemble them in the proper order so that the

stiffness matrix K for the entire truss can be found stiffness matrix K for the entire truss can be found

• This is done by designating the rows & columns of This is done by designating the rows & columns of the matrix by the 4 code numbers used to identify the matrix by the 4 code numbers used to identify the 2 global degrees of freedom that can occur at the 2 global degrees of freedom that can occur at

each end of the member

each end of the member

Truss stiffness matrix Truss stiffness matrix

• The structure stiffness matrix will then have an The structure stiffness matrix will then have an order that will be equal to the highest code

order that will be equal to the highest code

number assigned to the truss since this rep the number assigned to the truss since this rep the total no. of degree of freedom for the structure total no. of degree of freedom for the structure

• This method of assembling the member matrices This method of assembling the member matrices to form the structure stiffness matrix will now be to form the structure stiffness matrix will now be

demonstrated by numerical e.g.

demonstrated by numerical e.g.

• This process is somewhat tedious when performed This process is somewhat tedious when performed by hand but is rather easy to program on

by hand but is rather easy to program on computer

computer

Determine the structure stiffness matrix for the 2 member truss as shown. AE is constant.

Example 14.1

Example 14.1

By inspection, member 2 will have 2 unknown disp components whereas joint 1 & 3 are constrained from disp.

Consequently, the disp component at joint 2 are code numbered first, followed by those at joints 3 & 1.

The origin of the global coordinate system can be located at any point.

The members are identified arbitrarily & arrows are written along 2 members to identify the near & far ends of each member.

The direction cosines & the stiffness matrix for each member can now be determined.

Solution

Solution

Member 1 We have:

Using eqn 14.16, dividing each element by L = 3m, we have:

Solution Solution

3 0 0 0

; 3 1

0

3  



 

 _y

x 



Member 2

Since 2 is the near end & 1 is the far end, we have: (amend eqn)

Using eqn 14.16, dividing each element by L = 5m, we have:

Solution Solution

8 . 5 0

0 4

; 6 . 5 0

0

3  



 

 _y

x 



This matrix has an order of 6x6 since there are 6 designated degrees of freedom for the truss.

Corresponding elements of the above 2 matrices are added algebraically to form the structure stiffness matrix.

Solution

Solution

Application of the stiffness method for Application of the stiffness method for

truss analysis truss analysis

• The global force components Q acting on the truss The global force components Q acting on the truss can then be related to its global displacements D can then be related to its global displacements D

using using

• This eqn is referred to as the structure stiffness This eqn is referred to as the structure stiffness eqn eqn

KD

Q 

Application of the stiffness method for Application of the stiffness method for

truss analysis truss analysis

• Expanding yields Expanding yields

• Often D Often D _{k k} = 0 since the supports are not displaced = 0 since the supports are not displaced

• Thus becomes Thus becomes

u

k K D

Q  ₁₁

k u

u

k u

k

D K

D K

Q

D K

D K

Q

22 21

12 11









Application of the stiffness method for Application of the stiffness method for

truss analysis truss analysis

• Since the elements in the partitioned matrix K Since the elements in the partitioned matrix K _{11 11}

represent the total resistance at a truss joint to a represent the total resistance at a truss joint to a

unit disp in either the x or y direction, then the unit disp in either the x or y direction, then the

above eqn symbolizes the collection of all the force above eqn symbolizes the collection of all the force

eqm eqn applied to the joints where the external eqm eqn applied to the joints where the external

loads are zero or have a known value Q loads are zero or have a known value Q _{k k}

• Solving for D Solving for D _{u u} , we have: , we have:

  _k

u K Q

D  ₁₁ ^{ 1}

Application of the stiffness method for Application of the stiffness method for

truss analysis truss analysis

• With D With D _{k k} =0 yields =0 yields

• The member forces can be determined The member forces can be determined

u

u K D

Q  ₂₁

Application of the stiffness method for Application of the stiffness method for

truss analysis truss analysis

• Since with q Since with q _{N N} = -q = -q _{F F} for eqm, for eqm,

Determine the force in each member of the 2-member truss as Determine the force in each member of the 2-member truss as shown. AE is constant.

shown. AE is constant.

Example 14.3

Example 14.3

The origin of x,y and the numbering of the joints & members are shown.

By inspection, it is seen that the known external disp are D

=D

=D

=D

=0

Also, the known external loads are Q

=0, Q

=-2kN.

Hence,

Solution Solution

2 1 0 2

6 5 4 3 0 0 0 0

 

 

 

 







 







 _k

k Q

D

Using the same notation as used here, this matrix has been developed in example 14.1.

Writing eqn 14.17, Q = KD for the truss we have

We can now identify K

and thereby determine D

By matrix multiplication,

Solution Solution

D D .

. . AE .

 

 

 

 

 

 

 

 

 

 0 0

128 0

096

0 405 0 096 2 0

0

2

1

By inspection one would expect a rightward and downward disp to occur at joint 2 as indicated by the +ve & -ve signs of the

answers.

Using these results,

Solution

Solution

Expanding & solving for the reactions

The force in each member can be found.

Using the data for 

and 

in example 14.1, we have:

Solution Solution

kN

Q kN

Q kN

Q kN

Q

0 . 2 . 5 1 0 1 . 5

6 5 4 3

   



kN q

m

y

L

x

2, member For

5 . 1 3

, 0

, 1

1, member For

1

 







 



Nodal Coordinates Nodal Coordinates

• A truss can be supported by a roller placed on a A truss can be supported by a roller placed on a incline

incline

• When this occurs, the constraint of zero deflection When this occurs, the constraint of zero deflection at the support (node) cannot be directly defined at the support (node) cannot be directly defined

using a single horizontal & vertical global using a single horizontal & vertical global

coordinate system coordinate system

• Consider the truss Consider the truss

• The condition of zero disp The condition of zero disp at node 1 is defined only at node 1 is defined only

along the y” axis

along the y” axis

Nodal Coordinates Nodal Coordinates

• Because the roller can displace along the x” axis Because the roller can displace along the x” axis this node will have disp components along both this node will have disp components along both

global coordinates axes x & y global coordinates axes x & y

• To solve this problem, so that it can easily be To solve this problem, so that it can easily be

incorporated into a computer analysis, we will use incorporated into a computer analysis, we will use

a set of nodal coordinates x”, y” located at the a set of nodal coordinates x”, y” located at the

inclined support inclined support

• These axes are oriented such that the reactions & These axes are oriented such that the reactions &

support disp are along each of the coordinate axes

support disp are along each of the coordinate axes

Nodal Coordinates Nodal Coordinates

• To determine the global stiffness eqn for the truss, To determine the global stiffness eqn for the truss, it becomes necessary to develop force & disp

it becomes necessary to develop force & disp

transformation matrices for each of the connecting transformation matrices for each of the connecting members at this support so that the results can be members at this support so that the results can be

summed within the same global x, y coordinate summed within the same global x, y coordinate

system

system

Nodal Coordinates Nodal Coordinates

• Consider truss member 1 having a global Consider truss member 1 having a global

coordinate system x, y at the near node and a coordinate system x, y at the near node and a

nodal coordinate system x”, y” at the far node

nodal coordinate system x”, y” at the far node

Nodal Coordinates Nodal Coordinates

• When disp D occur so that they have components When disp D occur so that they have components along each of these axes as shown

along each of these axes as shown

Nodal Coordinates Nodal Coordinates

• This eqn can be written in matrix form as This eqn can be written in matrix form as

"

" cos

cos

cos

cos 0 0

0 0

"

"

"

"

y F

F x

F F

y N

N x

N N

F F N N y

x y x

F N

q Q

q Q

q Q

q Q

D D D D d d

y x

y x

y x y x











 











 

 





 

 





 

 

 

 



Nodal Coordinates Nodal Coordinates

• This can be expressed as: This can be expressed as:

• The disp & force transformation matrices in the The disp & force transformation matrices in the above eqn are used to develop the member

above eqn are used to develop the member stiffness matrix for this situation

stiffness matrix for this situation

 

 

 







 









 

 





 

 





F N y

x y

x

F F N N

q q Q

Q Q Q

y x y x

0 0

0 0

"

"

"

"

 

 

Nodal Coordinates Nodal Coordinates

• We have We have

T k T

k  ^T '

Nodal Coordinates Nodal Coordinates

• Performing the matrix operation yields: Performing the matrix operation yields:

• This stiffness matrix is used for each member that This stiffness matrix is used for each member that is connected to an inclined roller support

is connected to an inclined roller support

• The process of assembling the matrices to form The process of assembling the matrices to form the structure stiffness matrix follows the standard the structure stiffness matrix follows the standard

procedure

procedure

Determine the support reactions for the truss as shown.

Example 14.6

Example 14.6

Since the roller support at 2 is on an incline, we must use nodal coordinates at this node.

The stiffness matrices for members 1 and 2 must be developed.

Member 1,

Solution Solution

707 .

0 ,

707 .

0

, 0

,

1  _"  _"  

 _{y x y}

x   



Member 2,

Solution Solution

707 .

0 ,

707 .

0

, 1

,

0   _"   _"  

 _{y x y}

x   



Member 3,

Solution Solution

6 . 0

, 8 .

0 

 _y

x 



Assembling these matrices to determine the structure stiffness matrix, we have:

Solution

Solution

Carrying out the matrix multiplication of the upper partitioned matrices, the three unknown disp D are determined from solving the resulting simultaneous eqn.

The unknown reactions Q are obtained from the multiplication of the lower partitioned matrices.

Solution Solution

D AE D AE

D AE 127 . 3

, 5 . 157

, 5 . 352

3 2

1

 

 



kN Q

kN Q

kN

Q ₄  31 . 8 , ₅   7 . 5 , ₆   22 . 5

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• If some of the members of the truss are subjected If some of the members of the truss are subjected to an increase or decrease in length due to

to an increase or decrease in length due to

thermal changes or fabrication errors, then it is thermal changes or fabrication errors, then it is

necessary to use the method of superposition to necessary to use the method of superposition to

obtain the solution obtain the solution

• This requires 3 steps This requires 3 steps

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• First, the fixed end forces necessary to prevent First, the fixed end forces necessary to prevent

movement of the nodes as caused by temperature movement of the nodes as caused by temperature

or fabrication are calculated or fabrication are calculated

• Second, equal but opposite forces are placed on Second, equal but opposite forces are placed on the truss at the nodes & the disp of the nodes are the truss at the nodes & the disp of the nodes are

calculated using the matrix analysis calculated using the matrix analysis

• Third, the actual forces in the members & the Third, the actual forces in the members & the reactions on the truss are determined by

reactions on the truss are determined by superposing these 2 results

superposing these 2 results

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• This force will hold the nodes of the member fixed This force will hold the nodes of the member fixed as shown

as shown

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• This procedure is only necessary if the truss is This procedure is only necessary if the truss is statically indeterminate

statically indeterminate

• If a truss member of length L is subjected to a If a truss member of length L is subjected to a temperature increase

temperature increase   T, the member will undergo T, the member will undergo an increase in length of

an increase in length of   L = L =     TL TL

• A compressive force q A compressive force q _{0 0} applied to the member will applied to the member will cause a decrease in the member’s length of

cause a decrease in the member’s length of   L’ = L’ = q q _{0 0} L/AE L/AE

• If we equate these 2 disp q If we equate these 2 disp q _{0 0} = AE = AE     T T

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• This force will hold the nodes of the member fixed This force will hold the nodes of the member fixed as shown in the figure

as shown in the figure

• If a temperature decrease occurs then If a temperature decrease occurs then   T T

becomes negative & these forces reverse direction becomes negative & these forces reverse direction

to hold the member in eqm to hold the member in eqm

T AE

q

T AE

q

F N















0 0

) (

)

(

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• If a truss member is made too long by an amount If a truss member is made too long by an amount

  L before it is fitted into a truss, the force q L before it is fitted into a truss, the force q _{0 0}

needed to keep the member at its design length L needed to keep the member at its design length L is q is q _{0 0} = AE = AE   L /L L /L

L L q AE

L L q AE

F N

 



 

0 0

) (

)

(

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• If the member is too short, then If the member is too short, then   L becomes L becomes negative & these forces will reverse

negative & these forces will reverse

• In global coordinates, these forces are: In global coordinates, these forces are:

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• With the truss subjected to applied forces, With the truss subjected to applied forces,

temperature changes and fabrication errors, the temperature changes and fabrication errors, the

initial force-disp relationship for the truss then initial force-disp relationship for the truss then

becomes:

becomes:

• Q Q _{o o} is the column matrix for the entire truss of the is the column matrix for the entire truss of the initial fixed-end forces caused by temperature

initial fixed-end forces caused by temperature changes & fabrication errors

changes & fabrication errors

Q 0

KD

Q  

Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• Carrying out the multiplication on the RHS, we Carrying out the multiplication on the RHS, we obtain:

obtain:

• According to the superposition procedure According to the superposition procedure described above, the unknown disp are described above, the unknown disp are

determined from the first eqn by subtracting K

determined from the first eqn by subtracting K _{12 12} D D _{k k} and (Q

and (Q )0 from both sides & then solving for D )0 from both sides & then solving for D

0 22

21

0 12

11

) (

) (

u k

u u

k k

u k

Q D

K D

K Q

Q D

K D

K Q













Trusses having thermal changes &

Trusses having thermal changes &

fabrication errors fabrication errors

• Once these nodal disp are obtained, the member Once these nodal disp are obtained, the member forces are determined by superposition:

forces are determined by superposition:

• If this eqn is expanded to determine the force at If this eqn is expanded to determine the force at the far end of the member, we obtain:

the far end of the member, we obtain:

' TD q 0

k

q  

Determine the force in member 1 & 2 of the pin-connected

assembly if member 2 was made 0.01 m too short before it was fitted into place. Take AE = 8(10

)kN.

Example 14.7

Example 14.7

Since the member is short, then L = -0.01m.

For member 2, we have

Solution Solution

6 . 0

8 .

0  



 _y

x 



Assembling the stiffness matrix, we have

Solution

Solution

Partitioning the matrices as shown & carrying out the

multiplication to obtain the eqn for the unknown disp yields,

Solving simultaneous eqn gives:

Solution Solution

m D

m D

002084 .

0

003704 .

0

2 1









Member 1

Member 2

Solution Solution

kN q

kN

y L

x

56 . 5

) 10 ( 8 AE

, 3m

, 1

, 0

1

3











 



kN q

kN

y L

x

26 . 9

) 10 ( 8 AE

, 5m

, 6 . 0

, 8 . 0

2

3













 



Space-truss analysis Space-truss analysis

• The analysis of both statically determinate and The analysis of both statically determinate and

indeterminate space trusses can be performed by indeterminate space trusses can be performed by

using the same procedure discussed previously using the same procedure discussed previously

• To account for the 3-D aspects of the problem, To account for the 3-D aspects of the problem, additional elements must be included in the

additional elements must be included in the transformation matrix T

transformation matrix T

• Consider the truss member Consider the truss member

Space-truss analysis Space-truss analysis

• By inspection the direction cosines bet the global & By inspection the direction cosines bet the global &

local coordinates can be found local coordinates can be found

2 2

2 ( ) ( )

) (

cos

N F

N F

N F

N F

N x F

x

z z

y y

x x

x x

L x x











 

 

 



Space-truss analysis Space-truss analysis

2 2

2

2 2

2

) (

) (

) (

cos

) (

) (

) (

cos

N F

N z F

z

N F

N F

N F

N F

N F

y y

z z

y y

x x

z z

L z z

z z

y y

x x

y y

L y y











 

 













 

 











Space-truss analysis Space-truss analysis

• As a result of the third dimension, the As a result of the third dimension, the transformation matrix becomes:

transformation matrix becomes:

• By substitution, we have By substitution, we have

Space-truss analysis Space-truss analysis

• Carrying out the matrix multiplication yields the Carrying out the matrix multiplication yields the symmetric matrix

symmetric matrix

• This eqn rep the member stiffness matrix This eqn rep the member stiffness matrix expressed in global coordinates

expressed in global coordinates