structural analysis 7 th edition in si units russell c. hibbeler chapter 14: truss analysis using...

Structural Analysis 7Structural Analysis 7thth Edition in SI Units Edition in SI UnitsRussell C. HibbelerRussell C. Hibbeler

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

Fundamentals of the stiffness methodFundamentals of the stiffness method

• The stiffness method:The stiffness method:• Is a disp method of analysisIs a disp method of analysis• Can be used to analyse both statically Can be used to analyse both statically

determinate and indeterminate structuresdeterminate and indeterminate structures• Yields the disp & forces directlyYields 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 necessary matrices for the computer using the stiffness methodthe stiffness method

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 14: Truss Analysis Using the Stiffness Method


• Application of the stiffness method requires Application of the stiffness method requires subdividing the structure into a series of subdividing the structure into a series of discrete finite elements & identifying their discrete finite elements & identifying their end points as nodesend points as nodes

• For truss analysis, the finite elements are For truss analysis, the finite elements are represented by each of the members that represented by each of the members that compose the truss & the nodes represent the compose the truss & the nodes represent the jointsjoints

• The force-disp properties of each element are The force-disp properties of each element are determined & then related to one another determined & then related to one another using the force eqm eqn written at the nodesusing the force eqm eqn written at the nodes




• These relationships for the entire structure These relationships for the entire structure are then grouped together into the structure are then grouped together into the structure stiffness matrix, Kstiffness 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 determined for any given loading on the structurestructure

• When these disp are known, the external & When these disp are known, the external & internal forces in the structure can be internal forces in the structure can be calculated using the force-disp relations for calculated using the force-disp relations for each membereach member



Member stiffness matrixMember stiffness matrix

• To establish the stiffness matrix for a single To establish the stiffness matrix for a single truss member using local x’ and y’ truss member using local x’ and y’ coordinates as shown When a +ve disp dcoordinates as shown When a +ve disp dNN is is imposed on the near end of the member imposed on the near end of the member while the far end is held pinnedwhile the far end is held pinned

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



NFNN dL

AEqd

L

AEq ' ;'


• Likewise, a +ve disp dLikewise, a +ve disp dFF at the far end, at the far end, keeping the near end pinned and results in keeping the near end pinned and results in member forcesmember forces

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



FFFN dL

AEqd

L

AEq '' ;''

NFF

FNN

dL

AEd

L

AEq

dL

AEd

L

AEq


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

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



1 1

1 1 '

'

1 1

1 1

L

AEk

dkq

d

d

L

AE

q

q

F

N

F

N




+ =

Displacement & Force Transformation Displacement & Force Transformation matricesmatrices

• Since a truss is composed of many Since a truss is composed of many members, we will develop a method for members, we will develop a method for transforming the member forces q and disp transforming the member forces q and disp d defined in local coordinates to global d defined in local coordinates to global coordinatescoordinates

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

xx and and yy as shown as shown




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

shownshown• The coordinates of N & F The coordinates of N & F

are (xare (xNN, y, yNN ) and (x ) and (xFF, y, yFF ) )



yyxx cos ;cos




22

22

)()(

cos

)()(

cos

NFNF

NF

NFyy

NFNF

NF

NFxx

yyxx

yyL

yy

yyxx

xxL

xx


• Disp Transformation matrixDisp Transformation matrix• In global coordinates each end of the In global coordinates each end of the

member can have 2 degrees of freedom or member can have 2 degrees of freedom or independent disp; namely joint N has Dindependent disp; namely joint N has DNxNx and and DDNyNy

• Joint F has DJoint F has DFxFx and D and DFyFy




• Disp Transformation matrixDisp Transformation matrix• When the far end is held pinned & the near When the far end is held pinned & the near

end is given a global disp, the corresponding end is given a global disp, the corresponding disp along member is Ddisp along member is DNxNxcoscosxx

• A disp DA disp Dny ny will cause the member to be will cause the member to be displaced Ddisplaced DNyNycoscosyy along the x’ axis along the x’ axis



yFxFF

yNxNN

yx

yx

DDd

DDd

coscos

coscos


• Disp Transformation matrixDisp Transformation matrix



0 0

0 0

form,matrix In

;

cos ;cosLet

TDd

D

D

D

D

d

d

DDdDDd

y

x

y

x

yxyx

F

F

N

N

yx

yx

F

N

yFxFFyNxNN

yyxx


• Force Transformation matrixForce Transformation matrix

• If qIf qFF is applied to the bar, is applied to the bar, the global force components at F are:the global force components at F are:

• UsingUsing



yNNxNN qQqQyx

cos ;cos

yyxx cos ;cos yFFxFF qQqQ

yx cos ;cos

yFFxFF

yNNxNN

qQqQqQqQ

yx

yx

; ;


• Force Transformation matrixForce Transformation matrix• In matrix formIn matrix form



qTQ

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


• Force Transformation matrixForce Transformation matrix• In this case, TIn this case, TTT transforms the 2 local forces q transforms the 2 local forces q

acting at the ends of the member into 4 acting at the ends of the member into 4 global force components Qglobal force components Q

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



Member global stiffness matrixMember global stiffness matrix

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

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



TDkq '

k'TTk

TDkTQ

T

T

kDQor

'

Member global stiffness matrixMember global stiffness matrix

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



Truss stiffness matrixTruss stiffness matrix

• Once all the member stiffness matrices are Once all the member stiffness matrices are formed in the global coordinates, it becomes formed in the global coordinates, it becomes necessary to assemble them in the proper necessary to assemble them in the proper order so that the stiffness matrix K for the order so that the stiffness matrix K for the entire truss can be foundentire truss can be found

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



Truss stiffness matrixTruss stiffness matrix

• The structure stiffness matrix will then have The structure stiffness matrix will then have an order that will be equal to the highest an order that will be equal to the highest code number assigned to the truss since this code number assigned to the truss since this rep the total no. of degree of freedom for the rep the total no. of degree of freedom for the structurestructure

• This method of assembling the member This method of assembling the member matrices to form the structure stiffness matrices to form the structure stiffness matrix will now be demonstrated by matrix will now be demonstrated by numerical e.g.numerical e.g.

• This process is somewhat tedious when This process is somewhat tedious when performed by hand but is rather easy to performed by hand but is rather easy to program on computerprogram on computer© 2009 Pearson Education South Asia Pte Ltd


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

Example 14.1Example 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.

SolutionSolution



Member 1We have:

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

SolutionSolution



03

00 ;1

3

03

yx

Member 2Since 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:

SolutionSolution



8.05

04 ;6.0

5

03

yx

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.

SolutionSolution



Application of the stiffness method Application of the stiffness method for truss analysisfor truss analysis

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

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



KDQ


• Expanding yieldsExpanding yields

• Often DOften Dkk = 0 since the supports are not = 0 since the supports are not displaceddisplaced

• Thus becomes Thus becomes



uk DKQ 11

kuu

kuk

DKDKQ

DKDKQ

2221

1211


• Since the elements in the partitioned matrix Since the elements in the partitioned matrix KK1111 represent the total resistance at a truss represent the total resistance at a truss joint to a unit disp in either the x or y joint to a unit disp in either the x or y direction, then the above eqn symbolizes direction, then the above eqn symbolizes the collection of all the force eqm eqn the collection of all the force eqm eqn applied to the joints where the external applied to the joints where the external loads are zero or have a known value Qloads are zero or have a known value Qkk

• Solving for DSolving for Duu, we have:, we have:



ku QKD 111


• With DWith Dkk =0 yields =0 yields

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



uu DKQ 21


• Since with qSince with qNN = -q = -qFF for eqm, for eqm,



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




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 D3=D4=D5=D6=0

Also, the known external loads are Q1=0, Q2=-2kN.

Hence,

SolutionSolution



2

12

0

6543

0000

kk QD

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 K11 and thereby determine Du

By matrix multiplication,

SolutionSolution



DD

. . . .AE

0

01280096009604050

20

2

1

AED

AED

003.19 ;

505.421

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,

SolutionSolution



Expanding & solving for the reactions

The force in each member can be found.Using the data for x and y in example 14.1, we have:

SolutionSolution



kNQkNQ

kNQkNQ

0.25.1

05.1

6

5

4

3

kNqmL

kNqmL

yx

yx

5.25 ,8.0 ,6.0

2,member For

5.13 ,0 ,1

1,member For

2

1

Nodal CoordinatesNodal Coordinates

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

• When this occurs, the constraint of zero When this occurs, the constraint of zero deflection at the support (node) cannot be deflection at the support (node) cannot be directly defined using a single horizontal & directly defined using a single horizontal & vertical global coordinate systemvertical global coordinate system

• Consider the trussConsider 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” axisalong the y” axis




• Because the roller can displace along the x” Because the roller can displace along the x” axis this node will have disp components axis this node will have disp components along both global coordinates axes x & yalong both global coordinates axes x & y

• To solve this problem, so that it can easily To solve this problem, so that it can easily be incorporated into a computer analysis, be incorporated into a computer analysis, we will use a set of nodal coordinates x”, y” we will use a set of nodal coordinates x”, y” located at the inclined supportlocated at the inclined support

• These axes are oriented such that the These axes are oriented such that the reactions & support disp are along each of reactions & support disp are along each of the coordinate axesthe coordinate axes




• To determine the global stiffness eqn for the To determine the global stiffness eqn for the truss, it becomes necessary to develop force truss, it becomes necessary to develop force & disp transformation matrices for each of & disp transformation matrices for each of the connecting members at this support so the connecting members at this support so that the results can be summed within the that the results can be summed within the same global x, y coordinate systemsame global x, y coordinate system




• Consider truss member 1 having a global Consider truss member 1 having a global coordinate system x, y at the near node and coordinate system x, y at the near node and a nodal coordinate system x”, y” at the far a nodal coordinate system x”, y” at the far nodenode




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




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



"" cos cos

cos cos

0 00 0

""

"

"

yFFxFF

yNNxNN

F

F

N

N

yx

yx

F

N

qQqQ

qQqQ

DDDD

dd

yx

yx

y

x

y

x


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

• The disp & force transformation matrices in The disp & force transformation matrices in the above eqn are used to develop the the above eqn are used to develop the member stiffness matrix for this situationmember stiffness matrix for this situation



F

N

y

x

y

x

F

F

N

N

qq

QQQQ

y

x

y

x

0 0

0 0

"

"

"

"


• We haveWe have



TkTk T '


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

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

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



Determine the support reactions for the truss as shown.




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,

SolutionSolution



707.0 ,707.0 ,0 ,1 "" yxyx

Member 2,

SolutionSolution



707.0 ,707.0 ,1 ,0 "" yxyx

Member 3,

SolutionSolution



6.0 ,8.0 yx

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

SolutionSolution



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.

SolutionSolution



AED

AED

AED

3.127 ,

5.157 ,

5.352321

kNQkNQkNQ 5.22 , 5.7 , 8.31 654

Trusses having thermal changes & Trusses having thermal changes & fabrication errorsfabrication errors

• If some of the members of the truss are If some of the members of the truss are subjected to an increase or decrease in subjected to an increase or decrease in length due to thermal changes or fabrication length due to thermal changes or fabrication errors, then it is necessary to use the errors, then it is necessary to use the method of superposition to obtain the method of superposition to obtain the solutionsolution

• This requires 3 stepsThis requires 3 steps




• First, the fixed end forces necessary to First, the fixed end forces necessary to prevent movement of the nodes as caused prevent movement of the nodes as caused by temperature or fabrication are calculatedby temperature or fabrication are calculated

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

• Third, the actual forces in the members & Third, the actual forces in the members & the reactions on the truss are determined by the reactions on the truss are determined by superposing these 2 resultssuperposing these 2 results




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




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

• If a truss member of length L is subjected to If a truss member of length L is subjected to a temperature increase a temperature increase T, the member will T, the member will undergo an increase in length of undergo an increase in length of L = L = TLTL

• A compressive force qA compressive force q00 applied to the applied to the member will cause a decrease in the member will cause a decrease in the member’s length of member’s length of L’ = qL’ = q00L/AEL/AE

• If we equate these 2 disp qIf we equate these 2 disp q00 = AE = AE TT




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

• If a temperature decrease occurs then If a temperature decrease occurs then T T becomes negative & these forces reverse becomes negative & these forces reverse direction to hold the member in eqmdirection to hold the member in eqm



TAEq

TAEq

F

N

0

0

)(

)(


• If a truss member is made too long by an If a truss member is made too long by an amount amount L before it is fitted into a truss, the L before it is fitted into a truss, the force qforce q00 needed to keep the member at its needed to keep the member at its design length L is qdesign length L is q00 = AE = AEL /LL /L



L

LAEq

L

LAEq

F

N

0

0

)(

)(


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

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




• With the truss subjected to applied forces, With the truss subjected to applied forces, temperature changes and fabrication errors, temperature changes and fabrication errors, the initial force-disp relationship for the the initial force-disp relationship for the truss then becomes:truss then becomes:

• QQoo is the column matrix for the entire truss is the column matrix for the entire truss of the initial fixed-end forces caused by of the initial fixed-end forces caused by temperature changes & fabrication errorstemperature changes & fabrication errors



0QKDQ


• Carrying out the multiplication on the RHS, Carrying out the multiplication on the RHS, we obtain:we 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 determined from the first eqn by subtracting KK1212DDkk and (Q and (Qkk)0 from both sides & then )0 from both sides & then solving for Dsolving for Duu



02221

01211

)(

)(

ukuu

kkuk

QDKDKQ

QDKDKQ


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

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



0' qTDkq

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(103)kN.




Since the member is short, then L = -0.01m.For member 2, we have

SolutionSolution



6.0 8.0 yx

Assembling the stiffness matrix, we have

SolutionSolution



Partitioning the matrices as shown & carrying out the multiplication to obtain the eqn for the unknown disp yields,

Solving simultaneous eqn gives:

SolutionSolution



mD

mD

002084.0

003704.0

2

1

Member 1

Member 2

SolutionSolution



kNq

kNLyx

56.5

)10(8AE ,3m ,1 ,0

1

3

kNq

kNLyx

26.9

)10(8AE ,5m ,6.0 ,8.0

2

3

Space-truss analysisSpace-truss analysis

• The analysis of both statically determinate The analysis of both statically determinate and indeterminate space trusses can be and indeterminate space trusses can be performed by using the same procedure performed by using the same procedure discussed previouslydiscussed previously

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

• Consider the truss memberConsider the truss member




• By inspection the direction cosines bet the By inspection the direction cosines bet the global & local coordinates can be foundglobal & local coordinates can be found



222 )()()(

cos

NFNFNF

NF

NFxx

zzyyxx

xxL

xx




222

222

)()()(

cos

)()()(

cos

NFNFNF

NF

NFzz

NFNFNF

NF

NFyy

zzyyxx

zz

L

zz

zzyyxx

yy

L

yy


• 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 haveBy substitution, we have




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

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



structural analysis 7 th edition in si units russell c. hibbeler chapter 14: truss analysis using...

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