simplifiedmodel (equivalent beam for truss)

Simplified models Chapter 12

Handbook of Structural Analysis 1 22/10/01

12 SIMPLIFIED MODELS

Contents12.1 Bending Action...........................................................................................112.2 Truss Action...............................................................................................3

12.2.1Forces in Truss Members.........................................................................412.2.2Solving Statically Determinate Pin-jointed Frameworks...........................412.2.3Treating a Parallel Chord Trusses as a Beam............................................4

12.3 Arch Action................................................................................................712.4 Dome Action...............................................................................................812.5 Vierendeel Frames.......................................................................................9

12.5.1Lateral load on vierendeel frame..............................................................912.6 Distribution of lateral load in buildings......................................................12

12.6.1Rigid Plate on Springs Model................................................................1212.7 Checking Models for Plate Bending...........................................................15

12.7.1Formal Solutions...................................................................................1512.7.2Uniaxial Bending Models......................................................................16

This chapter describes some models that can be solved with a minimum of calculation.They can be used: In preliminary analysis. For checking models - see Section 2. For design provided they are judged to be valid - see Section 2.

12.1 Bending Action

Bending theory is one of the most widely used models in structural mechanics and canbe most useful for checking purposes. The constitutive relationship for bending isdiscussed in Section 4.4 and the implementation of this in frame elements is discussedin Section 5.3For a checking model, the basic strategy is to treat the system or part of a system as abeam in bending. Table 12.1 gives a selection of formulae for bending moment andshear in single span beams. For other sources see Roark and Young ( )Where the structure (or part of the structure) has a relatively low length-to-depth ratioshear deformation needs to be taken into account as illustrated in Case Study 12.1.

Case study 12.1 - Cantilever BracketFigure 12.1(a) shows a steel cantilever bracket fixed to a column. Figure 12.1(b) showsa finite element model. The web is treated as plane stress elements and the flanges asbeam elements. The results give deflections under the load points of 0.5301 and0.6365mm. The axial stress in the beam element at A is 76.2N/mm2. To check thesevalues, the system is treated as a cantilever with tip load 200kN, length 530mm.



Estimate deflection at centre of load

Bending deflection = 8

33

1093.22003530200

3 xxxx

EIWL

= = 0.169 mm

Shear deflection = 776.4535.8

530200xx

x

GA

WL= = 0.357 mm

Total =0.526 mmAverage value for load points from finite element model = 0.853mm

Note that shear deformation dominates the deflection.

Estimate bending stress at AUsing the simplified model:

23

/9.81101293

530200mmN

x

xZ

WLA ===s

From the plane stress element model sA = 76.2 N/mm2

The beam checking model gives a difference in tip deflection as compared with the FEmodel of -24% and difference in maximum stress of +7%. This is adequate accuracyfor a checking model.

(b) PLANE STRESS MODEL

Figure 12.1 Plane stress model of a steel bracket



Table 12.1 Formulae for single span beams

12.2 Truss Action

The simplest assumption that one can make about structural behaviour is that the stressor the strain is constant and unidirectional over a region. This is commonly done inreinforced concrete for shear reinforcement, for the bracing action in infilled frames andfor the stiffening effect of cladding of buildings. The unidirectional parts may then betreated as a members of a truss. The converse concept of treating a truss system as abeam can also be useful.



12.2.1 Forces in Truss MembersTrusses can often be treated as statically determinate by ignoring moment continuity atthe connections and by neglecting the effect of compression diagonals in cross bracedsystems. Selected member forces can then be estimated on the basis of equilibrium.Ignoring the compression diagonals will cause an overestimate of forces in the othertruss members and in deflection.

12.2.2 Solving Statically Determinate Pin-jointed Frameworks

Basic Process1. Draw a free body diagram of the truss2. Calculate the external reactions on the system3. Select a joint which has 2 or fewer unknowns4. Solve for unknowns at this joint5. Mark the values and directions of the joint forces in a composite joint force diagram.6. Mark the equal and opposite joint forces at the joints at opposite ends of the members for which forces have been calculated.7. Follow round the joints to find all internal forces - always choose a joint which has 2 or fewer unknowns

StrategiesLook for joints with only one unknown, they are easier to calculateLook for a symmetrical structure with symmetrical loading, in which case only half ofthe structure need be analysedLook for members with zero force. This occurs, for example, at a node in a in pin-jointed truss which has no applied load and where there are three members, two ofwhich are in line - Figure 12. 2

12.2.3 Treating a Parallel Chord Trusses as a BeamA parallel chord truss has a structural action analogous to that of a beam. The top andbottom chords are equivalent to the flanges while the posts and diagonals are equivalentto the web. Figure 12.3(a) shows a parallel chord bridge truss and Figure 12.3(b) is abeam equivalent. The properties of the beam are:

Equivalent Bending Stiffness2/2bEAEI ce = (12.1)

where :

Figure 12.2 Truss member with zero force

This member can take no force



Ac is the area of a chord member assumed to be equal top and bottom. If they arenot equal, Ie can be calculated as the second moment of the chord areas about thecentroid of the chord areas. That is, a 'neutral axis' within the depth of the trussneeds to be established about which the second moments of area are based

b is the depth of the truss E is the E value of the truss material

Equivalent Shear Stiffness(AsG)e (equivalent shear stiffness) qq 21 sincosfEAd= (12.2)

where Adis the area of a diagonal member,q is the angle of a diagonal member to the horizontal,f1 = 1.0 for single bracing = 2.0 for cross bracing (where the diagonals can sustaincompressive load) = 0.5 for a K-braced truss.

AdAc

q

b

(a) Truss

(b) Equivalent beam

E, A , I , (AG)ee e

Figure 12.3 Treatment of a parallel chord truss as a beam

Use of equations (12.1) and (12.2) together with the deflection formulae of Table 12.1often give estimates of truss deflection to a useful degree of accuracy.

Derivation of equation (12.2)

Figure 12.4 Truss bent

Equation (12.2) is derived as follows:



Consider the bent of a truss shown in Figure 12.4. The deflection d in the line of the shear force S is:

pd

d

EASb

EA

SL+=

qd

2sin (12.3)

where Ad and Ap are the areas of the diagonal and post members respectively, a, b, Ld

are the dimensions as shown in Figure 12.4, ad

is the slope of the lateral displacement

of a truss. Treating the truss as being in the xy plane as shown in Figure 12.4, this slope can bedefined as:

dxd

and

=

Substituting this into equation (12.3) and rearranging gives:

dxd

EAb

sinEA

L

aS

p2

d

d

n

+

q

= (12.4)

The factor in square brackets in equation (12.4) can be defined as an equivalent shearstiffness. For checking calculations the post flexibility can be ignored and thebracketed term reduces to the equivalent shear stiffness quoted in equation (12.2).

Estimation of Member Forces in Parallel Chord TrussesA quick check on the axial member forces in a parallel chord truss can be carried out asfollows:1. Axial force in chords = Mmax/d where:

Mmax is the maximum moment in the truss (analysed as a beam)d is the depth of the truss

2. Axial force in diagonal at support = V/(sinq) where:V is the support reactionq is the angle between the chord and the diagonal at the support

Bracing trusses in buildingsThe same technique can be used for bracing trusses in buildings. For a multibay trussof this type - Figure 12.5 - the equivalent bending stiffness is based on all the columnareas and the equivalent shear stiffness is the sum of the values given by equation (12.2)for each braced bay. In a K-braced truss the diagonals contribute to the bending modebehaviour. This effect can be modelled by adding an area Ad cos3q to the columnareas.



Figure 12.5 K Braced Truss

Equivalent beam elementA parallel chord lattice truss in a structural system can be treated as an equivalent beamelement using equations (12.1) and (12.2).The cross-sectional area of the equivalentbeam can be taken as the sum of the areas of the chord members.

12.3 Arch ActionFigure 12.6(a) shows the force actions in an arch. The main arch force is assumed to beCompressive, giving rise to horizontal and vertical reaction components at each end.The function of an arch is the same as a beam in that it concentrates load to its supportsi.e. it redistributes vertical load laterally. A main difference is that it requires ahorizontal thrust at the supports. This is either compressive from outside the arch orfrom a tie across the bottom of the arch. Another arch situation is in shear walls which are supported on columns. In such acase the arch forces can be roughly estimated by considering equilibrium of a free bodydiagram of half of the arch taking the height to the centre-line of the arch as half thespan as shown in Figure 12.6(b) (Green 1972). Arch action occurs in lintel beams above openings in walls. The lintel acts more as atie to the arch than as a bending element - Figure 12.6(c). The ends of a simply supported beam has arch type action where the compressionzone slants downward to the reaction area - Figure 12.6(d)



Figure 12.6 Arch Action

12.4 Dome Action

Figure 12.7 Dome Action



Dome action is a three dimensional form of arch action. Figure 12.7 is a verticalsection through a dome. The thrust from the compressive forces in the shell of thedome are transferred into a lower ring beam in tension and an upper compressive ringbeam (if present). The hoop forces will therefore vary from being tensile at the topthrough to compressive at the base. A pyramid shape will exhibit the same type ofbehaviour.

12.5 Vierendeel Frames

A vierendeel frame has rectangular panels with no cross bracing. It is not an efficientmeans of transmitting transverse load but it is sometimes used to resist lateral load inbuildings and for architectural reasons when cross bracing is undesirable.

12.5.1 Lateral load on vierendeel frameFigure 12.8(a) shows a typical unbraced rigid jointed building frame of this type. For asimplified lateral load analysis of such a frame the first step is to reduce the frame to asingle bay equivalent - see Figure 12.8(b).

The properties of the single bay equivalent frame are:Ice = SIci/2 (12.5)

where Ice is the second moment of area of a column of the equivalent frameIci is the second moment of column i of the actual frameThe summation is over the columns of the frame at the base or at each level atwhich the properties are different

S=

i

bi

e

be IIll

(12.6)

where Ibe is the second moment of area of a beam of the equivalent frame

el is the span of a beam of the equivalent frame. This can be given an arbitraryvalue typically the longest span of the beams of the actual frameIbi is the second moment of area of a beam of the actual frame

il is the span of a beam of the actual frame.The summation is over the beams of the frame at the first storey level or at eachlevel at which the properties are different.

Having analysed the equivalent one storey frame either by a computer solution or usingthe Portal Method (described later in this section) the shears and moments and momentsin members of the frame can be calculated using:

ce

ciceci I2

IMM = (12.7)

where Mci is the moment in column i of the actual frameMce is the moment in a column of the equivalent frame

ebe

ibibi /I

/IM

l

l= (12.8)

where Mbi is the moment in beam i of the actual frame



ce

cici I

IS S = (12.9)

where Sci in column i at level j in the frameS is the applied shear at level j of the frameSbi = Mbi/ il (12.10)

where Sbi is the shear in beam i at level j of the frame

The Portal MethodThe portal method assumes that there are points of contraflexure at mid-height ofcolumns and mid-span of beams of such a frame. The applied shear is distributedequally to the columns and hence a statically determinate equivalent frame is producedas shown in Figure 12.7(c). Figure 12.7(d) shows a bent from this frame whichillustrates how the internal actions can be calculated. S is the total shear applied to theframe above the beam level being considered.

Figure 12.8 Vierendeel Frame Models

Validation informationThe mid-span assumption for points of contraflexure in the beam is not accurate unlessthe frame is 'proportioned'. A proportioned frame can be divided into a set ofequivalent single bay frames each of which has the same column to beam stiffness ratio(l). Real frames probablyseldom conform to this condition but this does not invalidate the use of the portalmethod for checking.



The validity of the mid height position for the column points of contraflexuredepends on the value of the ratio:

eb

ce/eI

h/I

l=l (12.11)

where h is the storey height When l is small i.e. significantly less than 1.0, then the joint rotation is low and thecolumn points of contraflexure will be close to mid-height (sometimes called a shearbeam frame). When l is high, i.e. greater than 5.0, the degree to which the beamstransfer vertical shear becomes less significant and the column points of contraflexurewill not be close to mid-height. Thus below l =5, the assumption tends to give an orderof magnitude estimate of column moment improving as l decreases. The error is mostnoticeable at the base of the columns where, unfortunately, it is most important.

Deflection of a vierendeel frame

1

(a) Shear Mode Deformation (b) Bending Mode Deformation

Figure 12.9 Deformation modes for vierendeel frame in a building

The deflection of a vierendeel frame is likely to be dominated by a shear mode type ofdeformation due to bending of the beams and columns - Figure 12.9(a). There will alsobe a bending mode deformation due to the axial deformation of the columns - Figure12.9(b). This latter type of deformation will tend to be prominent only in tall frames.The equivalent single storey frame (Figure 12.8) can therefore normally be treated as anequivalent beam with only shear deformation. The equivalent shear area is develop asfollows:

The deflection d of the bent of Figure 12.8(d) is given by:

[ ]l+

=d 21IE12

Sh

ci

3 (12.12)

where Ici = sum of the I values for all columns of the frame. d/h is the slope of the lateral displacement of the frame. Treating the frame as beingin the xy plane as shown in Figure 12.8(c), this slope can be defined as:



dydu

h=

d

Substituting this into equation (12.12) and rearranging gives:

[ ] dxdu

h

IES c

l2112

2 +

= (12.13)

Equation (12.13) has the same form as for shear deformation of a beam whereS = AsG dv/dx

Therefore the equivalent shear stiffness for the equivalent beam for a vierendeel frameis:

[ ]l+

=21h

IE12)GA(

2c

es (12.14)

Thus the deflection of a vierendeel frame can be estimated using the beam deflectionformulae given in Table 12.1 using the equivalent shear stiffness of equation (12.14).

The finite sizes of the beams and of the columns can be considered in the shearstiffness. The relevant bent is shown in Figure 12.8(e).

The corresponding shear stiffness is:

[ ]332 )1(2)1(12

)(CD

ee h

IEGA

B-+B-

=l

(12.15)

where bD = D/h, bC = C/l, D = beam depth, C = column width.

12.6 Distribution of lateral load in buildings

12.6.1 Rigid Plate on Springs ModelFor this model the building is treated in plan as a rigid plate supported by springs in itsown plane - Figure 12.10. A set of bracing elements is defined. A bracing elementis a frame, wall or core which is deemed to deemed to provide lateral support to thesystem.

Each bracing element is defined by a single spring stiffness which is the lateral load tocause unit top deflection of the bracing element.

Basic Assumptions The stiffness of each bracing element can be modelled as a single spring Normal linear material and geometric assumptions. The floors are assumed to be fully rigid in their own planes. The effect of differences in mode of deformation is neglected, i.e. the non-uniform

interaction between walls and moment resisting frames is not considered.



(b) Rigid beam on springs model

Figure 12.10 Rigid beam on springs model

Basic Procedure1. Identify the bracing elements i.e. those parts of the structure which are assumed to

resist lateral load.2. Establish the position and magnitude of the total lateral load on the building -W.3. Calculate the top stiffness ki for each bracing element. Apply The lateral load W to

the model of the bracing element (the magnitude of this load is not important at thisstage). Extract or calculate the top lateral deflection Dtop of the bracing element. Thestiffness of the bracing element is then ki = W/Dtop

4. Treat the system as a rigid beam and springs subject to loading W5. Solve the beam on springs problem to get the spring forces - pi . These are assumed

to be the loads on the bracing elements (having the same distribution with height asW).

6. Analyse each bracing element under the loading pi (e.g. multiply the results of theanalysis used to get ki by the factor pi/W).

(a) Plan of 4 storey building



No torsion (one degree of freedom system)Figure 12.11 shows a system with springs only in the y direction and the model istherefore a rigid beam on springs.If torsion is neglected the system moves without rotating. The distribution of load tothe supports is then in direct proportion to their top stiffnesses ki.The equilibrium condition is: W = Spi (12.16)where pi is the load in spring iThe compatibility condition is di = D (12.17)where di is the deflection of spring i and D is the deflection of the systemThe force-deformation relationships are pi = ki di (12.18)where ki is the stiffness of spring i

Combining (12.16), (12.17) and (12.18) gives the system stiffness relationship:D= ikW (12.19)

from which D can be calculated using:

ikW

=D (12.20)

and the load on a support frame is:

Wk

kkp

i

iii

=D= (12.21)

The distribution of load to the supports is thus in direct proportion to their topstiffnesses ki.

With torsion (two degrees of freedom system)If torsion is considered then an extra freedom (corresponding to a rotation) needs to beadded to the rigid beam on springs - Figure 12.11. The system deformations areD and q. The equilibrium equations are then

ii

px

1

M

W

=

(12.22)

where M is the moment of the applied loads about the origin. (M = Wa for the systemof Figure 12.13)

xi is the distance from the origin to the position of spring i

The compatibility condition is:

[ ]

qD

=d ii xl (12.23)

The force-deformation relationship of a spring ispi = ki di

i.e pi = ki(D + xi q) (12.24)



Figure 12.11 Rigid beam on springs model

Combining equations (12.22), (12.23) and (12.24) gives:

qD

2iiii

iii

xkxk

xkk

M

W (12.25)

Equation (12.25) can be solved to get the system deformations D and q from which thespring forces are obtained using (12.24). The solution in this case can be convenientlyprogrammed using a spreadsheet.It is helpful to use the 'centre of stiffness' of the spring system as the origin. If this isdone, then the off-diagonal terms of the matrix of equation (12.25) are zero and thesolution of the equations is simplified. The centre of stiffness is found by taking firstmoments of spring stiffness about any point to find the position of the resultant of thespring stiffnesses.

12.7 Checking Models for Plate Bending

For checking, the plate bending problem being considered can be amended to a form forwhich a solution is available. This can be achieved by altering the boundary conditionsand/or the loading. One should make an estimate of whether the checking model willbe stiffer or more flexible than the finite element model. Corresponding results fromthe two models are then compared with some expectation of the sign of the differencebetween them. A worthwhile approach is to use two checking models one of which tends to bestiffer than the finite element model and the other tending to be more flexible. Theseshould give results on either side of those being checked.

12.7.1 Formal SolutionsThe model can be converted to a form for which a formal solution is available - see forexample Timoshenko & Woinowsky Krieger (1959), Roark (1965).



Figure 12.12 Checking model for a simply supported rectangular plate

12.7.2 Uniaxial Bending ModelsIt can be useful to take a strip of a plate in biaxial bending and treat is as being one-wayspanning. This will tend to overestimate deformations and stresses. This is particularlyuseful if the plate has a dominant span direction. For example, for the simply supportedplate with side ratio of 2:1 and uniformly distributed load shown in Figure 12.12(a), onecan take a unit width strip as shown in Figure 12.12(b) and treat it as a simply supportedone-way span. The maximum deflection and maximum moment for the strip can be calculatedusing:

83845 2

max

4

max

qaMand

EIqa

==d

where a is the span of the strip.

The coefficients in the above expressions are quoted in Table12.2 in which thecorresponding coefficient for Timoshenko & Woinowsky-Krieger (1959) are alsogiven.



6.7.3 Grillage ModelsResults for the simple grillage model shown in Figure 12.13(c) are also shown in Table12.2. For the grillage one quarter of the total load is applied as a point load at thecentre. Member loads could also be used but it is easier to solve the point load case (thegrillage of Figure 12.13(c) can be solved by treating it as being singly staticallyindeterminate). The grillage gives a slightly better estimate of deflection but furtheroverestimates the moments as compared with the one-way strip. Figure 12.13(a) shows a square simply supported uniformly distributed plate. Figure12.13(b) shows a simple grillage model. Again, one quarter of the total load is appliedat the centre. Comparison of the maximum deflection and maximum moment valuesare quoted in Table 12.2. The checking model results quoted in Table 12.2 do notgive close correlation but do establish the likely orders of magnitude. Such informationis valuable in checking.

Figure 12.16 Grillage checking model for a plate

Table 12.2 Checking models for uniformly loaded simply supported plates Problem Model Coefficient Coefficient for dmax for Mmax 2a

a One-way strip 3845

01302.0 = 81

125.0 =

grillage 0.01225 0.1366 Timoshenko 0.00922 0.1017

a

a Grillage 1921

00251.0 = 161

0625.0 =

Timoshenko 0.00406 0.0479

2max

4

max qaxtcoefficienMEIqa

xtcoefficien ==d

simplifiedmodel (equivalent beam for truss)

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