structural analysis 7 th edition in si units russell c. hibbeler chapter 15: beam analysis using the...

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

Chapter 15: Chapter 15: Beam Analysis Using the Stiffness MethodBeam Analysis Using the Stiffness Method

Preliminary remarksPreliminary remarks

• Member & node identificationMember & node identification• In general each element must be free from In general each element must be free from

load & have a prismatic cross sectionload & have a prismatic cross section• The nodes of each element are located at a The nodes of each element are located at a

support or at points where members are support or at points where members are connected together or where the vertical or connected together or where the vertical or rotational disp at a point is to be determinedrotational disp at a point is to be determined

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 15: Beam Analysis Using the Stiffness Method


• Global & member coordinatesGlobal & member coordinates• The global coordinate system will be The global coordinate system will be

identified using x, y, z axes that generally identified using x, y, z axes that generally have their origin at a node & are positioned have their origin at a node & are positioned so that the nodes at other points on the beam so that the nodes at other points on the beam are +ve coordinatesare +ve coordinates




• Global & member coordinatesGlobal & member coordinates• The global coordinate system will be The global coordinate system will be

identified using x, y, z axes that generally identified using x, y, z axes that generally have their origin at a node & are positioned have their origin at a node & are positioned so that the nodes at other points on the beam so that the nodes at other points on the beam are +ve coordinatesare +ve coordinates

• The local or member x’, y’, z’ coordinates The local or member x’, y’, z’ coordinates have their origin at the near end of each have their origin at the near end of each elementelement

• The positive x’ axis is directed towards the The positive x’ axis is directed towards the far endfar end




• Kinematic indeterminacyKinematic indeterminacy• The consider the effects of both bending & The consider the effects of both bending &

shearshear• Each node on the beam can have 2 degrees Each node on the beam can have 2 degrees

of freedom, vertical disp & rotationof freedom, vertical disp & rotation• These disp will be identified by code numbersThese disp will be identified by code numbers• The lowest code numbers will be used to The lowest code numbers will be used to

identify the unknown disp & the highest identify the unknown disp & the highest numbers are used to identify the known dispnumbers are used to identify the known disp




• Kinematic indeterminacyKinematic indeterminacy• The beam is kinematically indeterminate to The beam is kinematically indeterminate to

the 4the 4thth degree degree• There are 8 degrees of freedom for which There are 8 degrees of freedom for which

code numbers 1 to 4 rep the unknown disp & code numbers 1 to 4 rep the unknown disp & 5 to 8 rep the known disp5 to 8 rep the known disp

• Development of the stiffness method for Development of the stiffness method for beams follows a similar procedure as that for beams follows a similar procedure as that for trussestrusses




• Kinematic indeterminacyKinematic indeterminacy• First we must establish the stiffness matrix First we must establish the stiffness matrix

for each elementfor each element• These matrices are combined to form the These matrices are combined to form the

beam or structure stiffness matrixbeam or structure stiffness matrix• We can then proceed to determine the We can then proceed to determine the

unknown disp at the nodesunknown disp at the nodes• This will determine the reactions at the beam This will determine the reactions at the beam

& the internal shear & moments at the nodes& the internal shear & moments at the nodes



Beam-member stiffness matrixBeam-member stiffness matrix

• We will develop the stiffness matrix for a We will develop the stiffness matrix for a beam element or member having a constant beam element or member having a constant cross-sectional area & referenced from the cross-sectional area & referenced from the local x’, y’, z’ coordinate systemlocal x’, y’, z’ coordinate system




• Linear & angular disp associated with these Linear & angular disp associated with these loadings also follow the same sign loadings also follow the same sign conventionconvention

• We will impose each of these disp separately We will impose each of these disp separately & then determine the loadings acting on the & then determine the loadings acting on the member caused by each dispmember caused by each disp




• y’ displacementy’ displacement• A positive disp dA positive disp dNy’Ny’ is imposed while other is imposed while other

possible disp are preventedpossible disp are prevented• The resulting shear forces & bending The resulting shear forces & bending

moments that are created are shownmoments that are created are shown




• z’ rotationz’ rotation• A positive rotation dA positive rotation dNz’Nz’ is imposed while other is imposed while other

possible disp are preventedpossible disp are prevented• The required shear forces & bending The required shear forces & bending

moments necessary for the deformation are moments necessary for the deformation are shownshown

• When dWhen dNz’Nz’ is imposed, the resultant loadings is imposed, the resultant loadings are shownare shown




• By superposition, the resulting four load-disp By superposition, the resulting four load-disp relations for the member can be expressed in relations for the member can be expressed in matrix form asmatrix form as

• These eqn can be written asThese eqn can be written as



kdq

Beam-structure stiffness matrixBeam-structure stiffness matrix

• Once all the member stiffness matrices have Once all the member stiffness matrices have been found, we must assemble them into been found, we must assemble them into the structure stiffness matrix, Kthe structure stiffness matrix, K

• The rows & columns of each k matrix, the The rows & columns of each k matrix, the eqns are identified by the 2 code numbers at eqns are identified by the 2 code numbers at the near end (Nthe near end (Ny’y’, N, Nz’z’) of the member ) of the member followed by those at the far end (Ffollowed by those at the far end (Fy’y’, F, Fz’z’) )

• When assembling the matrices, each When assembling the matrices, each element must be placed in the same element must be placed in the same location of the K matrixlocation of the K matrix



Beam-structure stiffness matrixBeam-structure stiffness matrix

• K will have an order that will be equal to the K will have an order that will be equal to the highest code number assigned to the beamhighest code number assigned to the beam

• Where several members are connected at a Where several members are connected at a node, their member stiffness influence node, their member stiffness influence coefficients will have the same position in coefficients will have the same position in the K matrix & therefore must be the K matrix & therefore must be algebraically added to determine the nodal algebraically added to determine the nodal stiffness influence coefficient for the stiffness influence coefficient for the structurestructure



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

• Once the stiffness matrix is determined, the Once the stiffness matrix is determined, the loads at the nodes of the beam can be loads at the nodes of the beam can be related to the disp using the structure related to the disp using the structure stiffness eqnstiffness eqn

• Partitioning the stiffness matrix into the Partitioning the stiffness matrix into the known & unknown elements of load & disp, known & unknown elements of load & disp, we havewe have



KDQ


• This expands into 2 eqn:This expands into 2 eqn:

• The unknown disp DThe unknown disp Duu are determined from are determined from the first of these eqnthe first of these eqn

• Using these values, the support reactions QUsing these values, the support reactions Quu are computed for the second eqnare computed for the second eqn



kuu

kuk

DKDKQ

DKDKQ

2221

1211


• Intermediate loadingsIntermediate loadings• It is important that the elements of the beam It is important that the elements of the beam

be free of loading along its lengthbe free of loading along its length• This is necessary as the stiffness matrix for This is necessary as the stiffness matrix for

each element was developed for loadings each element was developed for loadings applied only at its endsapplied only at its ends

• Consider the beam element of length L which Consider the beam element of length L which is subjected to uniform distributed load, wis subjected to uniform distributed load, w




• Intermediate loadingsIntermediate loadings• First, we will apply FEM & reactions to the First, we will apply FEM & reactions to the

element which will be used in the stiffness element which will be used in the stiffness methodmethod

• We will refer to these loadings as a column We will refer to these loadings as a column matrix qmatrix qoo

• The distributed loading within the beam is The distributed loading within the beam is determined by adding these 2 resultsdetermined by adding these 2 results




• Member forcesMember forces• The shear & moment at the ends of each The shear & moment at the ends of each

beam element can be determined by adding beam element can be determined by adding on any fixed end reactions qon any fixed end reactions qoo if the element is if the element is subjected to an intermediate loadingsubjected to an intermediate loading

• We have:We have:



0qkdq

Determine the reactions at the supports of the beam as shown. EI is constant.

Example 15.1Example 15.1



The beam has 2 elements & 3 nodes identified.The known load & disp matrices are:

Each of the 2 member stiffness matrices can be determined.

SolutionSolution



6

500

4321

0 0 50

kk DQ

We can now assemble these elements into the structure stiffness matrix.The matrices are partitioned as shown,

SolutionSolution



Carrying out the multiplication for the first 4 rows, we have

SolutionSolution



43

4321

321

321

2000

45.10

05.15.15.15

05.120

DD

DDDD

DDDEI

DDD

Solving, we have:

Using these results, we get:

SolutionSolution



EID

EID

EID

EID

33.3 ,

67.6

67.26 ,

67.16

43

21

kNQ

kNQ

5

10

6

5

structural analysis 7 th edition in si units russell c. hibbeler chapter 15: beam analysis using the...

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unknown disp

structure stiffness

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local x

z coordinate system