Simplifiedmodel (Equivalent Beam for Truss)

Simplified models

Chapter 12

12

SIMPLIFIED MODELS

Contents 12.1 Bending Action ...........................................................................................1 12.2 Truss Action ...............................................................................................3 12.2.1 Forces in Truss Members.........................................................................4 12.2.2 Solving Statically Determinate Pin-jointed Frameworks ...........................4 12.2.3 Treating a Parallel Chord Trusses as a Beam ............................................4 12.3 Arch Action ................................................................................................7 12.4 Dome Action...............................................................................................8 12.5 Vierendeel Frames.......................................................................................9 12.5.1 Lateral load on vierendeel frame ..............................................................9 12.6 Distribution of lateral load in buildings ......................................................12 12.6.1 Rigid Plate on Springs Model ................................................................12 12.7 Checking Models for Plate Bending...........................................................15 12.7.1 Formal Solutions ...................................................................................15 12.7.2 Uniaxial 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 can be most useful for checking purposes. The constitutive relationship for bending is discussed in Section 4.4 and the implementation of this in frame elements is discussed in Section 5.3 For a checking model, the basic strategy is to treat the system or part of a system as a beam in bending. Table 12.1 gives a selection of formulae for bending moment and shear 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 ratio shear deformation needs to be taken into account as illustrated in Case Study 12.1. Case study 12.1 - Cantilever Bracket Figure 12.1(a) shows a steel cantilever bracket fixed to a column. Figure 12.1(b) shows a finite element model. The web is treated as plane stress elements and the flanges as beam elements. The results give deflections under the load points of 0.5301 and 0.6365mm. The axial stress in the beam element at A is 76.2N/mm2. To check these values, the system is treated as a cantilever with tip load 200kN, length 530mm.

Handbook of Structural Analysis

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Chapter 12

Estimate deflection at centre of load 200 x 530 3 WL3 Bending deflection = = 0.169 mm = 3EI 3 x 200 x 2.93 x 10 8 200 x 530 WL Shear deflection = = = 0.357 mm AG 8.5 x 453.6 x 77 Total = 0.526 mm Average value for load points from finite element model = 0.853mm Note that shear deformation dominates the deflection. Estimate bending stress at A Using the simplified model: WL 200 x 530 σA = = = 81.9 N / mm 2 3 Z 1293 x 10 From the plane stress element model σA = 76.2 N/mm2 The beam checking model gives a difference in tip deflection as compared with the FE model of -24% and difference in maximum stress of +7%. This is adequate accuracy for a checking model.

(b) PLANE STRESS MODEL

Figure 12.1 Plane stress model of a steel bracket


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Chapter 12

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 stress or the strain is constant and unidirectional over a region. This is commonly done in reinforced concrete for shear reinforcement, for the bracing action in infilled frames and for the stiffening effect of cladding of buildings. The unidirectional parts may then be treated as a members of a truss. The converse concept of treating a truss system as a beam can also be useful.


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Chapter 12

12.2.1 Forces in Truss Members Trusses can often be treated as statically determinate by ignoring moment continuity at the connections and by neglecting the effect of compression diagonals in cross braced systems. Selected member forces can then be estimated on the basis of equilibrium. Ignoring the compression diagonals will cause an overestimate of forces in the other truss members and in deflection.

12.2.2 Solving Statically Determinate Pin-jointed Frameworks Basic Process 1. Draw a free body diagram of the truss 2. Calculate the external reactions on the system 3. Select a joint which has 2 or fewer unknowns 4. Solve for unknowns at this joint 5. 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 Strategies Look for joints with only one unknown, they are easier to calculate Look for a symmetrical structure with symmetrical loading, in which case only half of the structure need be analysed Look for members with zero force. This occurs, for example, at a node in a in pinjointed truss which has no applied load and where there are three members, two of which are in line - Figure 12. 2 This member can take no force

Figure 12.2 Truss member with zero force

12.2.3 Treating a Parallel Chord Trusses as a Beam A parallel chord truss has a structural action analogous to that of a beam. The top and bottom chords are equivalent to the flanges while the posts and diagonals are equivalent to the web. Figure 12.3(a) shows a parallel chord bridge truss and Figure 12.3(b) is a beam equivalent. The properties of the beam are: Equivalent Bending Stiffness EI e = EAc b 2 / 2

(12.1)

where :


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•

• •

Chapter 12

Ac is the area of a chord member assumed to be equal top and bottom. If they are not equal, Ie can be calculated as the second moment of the chord areas about the centroid of the chord areas. That is, a 'neutral axis' within the depth of the truss needs 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)

= EAd f 1 cos θ sin 2 θ

(12.2)

where Ad is the area of a diagonal member, θ 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 sustain compressive load) = 0.5 for a K-braced truss. Ac

Ad b θ

(a) Truss E, Ae, Ie , (AG)e

(b) Equivalent beam

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 12.1 often 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:


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Chapter 12

Consider the bent of a truss shown in Figure 12.4. The deflection δ in the line of the shear force S is: SLd Sb + (12.3) 2 EAd sin θ EAp where Ad and Ap are the areas of the diagonal and post members respectively, a, b, Ld δ are the dimensions as shown in Figure 12.4, is the slope of the lateral displacement a of a truss. Treating the truss as being in the xy plane as shown in Figure 12.4, this slope can be defined as: δ =

δ dν = a dx Substituting this into equation (12.3) and rearranging gives:      dν  a (12.4) S=  Ld b   dx     EA sin 2 + EA   θ p  d  The factor in square brackets in equation (12.4) can be defined as an equivalent shear stiffness. For checking calculations the post flexibility can be ignored and the bracketed term reduces to the equivalent shear stiffness quoted in equation (12.2). Estimation of Member Forces in Parallel Chord Trusses A quick check on the axial member forces in a parallel chord truss can be carried out as follows: 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/(sinθ) where: V is the support reaction θ is the angle between the chord and the diagonal at the support Bracing trusses in buildings The same technique can be used for bracing trusses in buildings. For a multibay truss of this type - Figure 12.5 - the equivalent bending stiffness is based on all the column areas 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 mode behaviour. This effect can be modelled by adding an area Ad cos3θ to the column areas.


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Chapter 12

Figure 12.5 K Braced Truss

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

12.3

Arch Action

Figure 12.6(a) shows the force actions in an arch. The main arch force is assumed to be Compressive, 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 supports i.e. it redistributes vertical load laterally. A main difference is that it requires a horizontal thrust at the supports. This is either compressive from outside the arch or from a tie across the bottom of the arch. Another arch situation is in shear walls which are supported on columns. In such a case the arch forces can be roughly estimated by considering equilibrium of a free body diagram of half of the arch taking the height to the centre-line of the arch as half the span as shown in Figure 12.6(b) (Green 1972). Arch action occurs in lintel beams above openings in walls. The lintel acts more as a tie 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 compression zone slants downward to the reaction area - Figure 12.6(d)


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Chapter 12

Figure 12.6 Arch Action

12.4 Dome Action

Figure 12.7 Dome Action


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Chapter 12

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

12.5

Vierendeel Frames

A vierendeel frame has rectangular panels with no cross bracing. It is not an efficient means of transmitting transverse load but it is sometimes used to resist lateral load in buildings and for architectural reasons when cross bracing is undesirable. 12.5.1 Lateral load on vierendeel frame Figure 12.8(a) shows a typical unbraced rigid jointed building frame of this type. For a simplified lateral load analysis of such a frame the first step is to reduce the frame to a single bay equivalent - see Figure 12.8(b). The properties of the single bay equivalent frame are: Ice = ΣIci/2 (12.5) where Ice is the second moment of area of a column of the equivalent frame Ici is the second moment of column i of the actual frame The summation is over the columns of the frame at the base or at each level at which the properties are different I  I be = Σ bi  (12.6) le  li  where Ibe is the second moment of area of a beam of the equivalent frame l e is the span of a beam of the equivalent frame. This can be given an arbitrary value typically the longest span of the beams of the actual frame Ibi is the second moment of area of a beam of the actual frame l i 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 each level at which the properties are different. Having analysed the equivalent one storey frame either by a computer solution or using the Portal Method (described later in this section) the shears and moments and moments in members of the frame can be calculated using: I M ci = M ce ci (12.7) 2I ce where Mci is the moment in column i of the actual frame Mce is the moment in a column of the equivalent frame I /l (12.8) M bi = bi i I be / l e where Mbi is the moment in beam i of the actual frame


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Chapter 12

I Sci = S ci (12.9) I ce where Sci in column i at level j in the frame S is the applied shear at level j of the frame Sbi = Mbi/ l i (12.10) where Sbi is the shear in beam i at level j of the frame

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

Figure 12.8 Vierendeel Frame Models

Validation information The mid-span assumption for points of contraflexure in the beam is not accurate unless the frame is 'proportioned'. A proportioned frame can be divided into a set of equivalent single bay frames each of which has the same column to beam stiffness ratio (λ). Real frames probably seldom conform to this condition but this does not invalidate the use of the portal method for checking.


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Chapter 12

The validity of the mid height position for the column points of contraflexure depends on the value of the ratio: I /h (12.11) λ = ce Ibe / l e where h is the storey height When λ is small i.e. significantly less than 1.0, then the joint rotation is low and the column points of contraflexure will be close to mid-height (sometimes called a shear beam frame). When λ is high, i.e. greater than 5.0, the degree to which the beams transfer vertical shear becomes less significant and the column points of contraflexure will not be close to mid-height. Thus below λ =5, the assumption tends to give an order of magnitude estimate of column moment improving as λ decreases. The error is most noticeable 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 of deformation due to bending of the beams and columns - Figure 12.9(a). There will also be a bending mode deformation due to the axial deformation of the columns - Figure 12.9(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 an equivalent beam with only shear deformation. The equivalent shear area is develop as follows: The deflection δ of the bent of Figure 12.8(d) is given by: Sh 3 (12.12) [1 + 2λ] 12E ∑ I ci where ∑Ici = sum of the I values for all columns of the frame. δ/h is the slope of the lateral displacement of the frame. Treating the frame as being in the xy plane as shown in Figure 12.8(c), this slope can be defined as: δ=


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Chapter 12

δ du = h dy Substituting this into equation (12.12) and rearranging gives: 12 E ∑ I c du S= 2 (12.13) h [1 + 2λ ] dx Equation (12.13) has the same form as for shear deformation of a beam where S = AsG dv/dx Therefore the equivalent shear stiffness for the equivalent beam for a vierendeel frame is: 12E ∑ I c (A s G ) e = (12.14) h 2 [1 + 2λ ] Thus the deflection of a vierendeel frame can be estimated using the beam deflection formulae 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 shear stiffness. The relevant bent is shown in Figure 12.8(e). The corresponding shear stiffness is: 12 E ∑ I e ( AG ) e = 2 (12.15) h (1 − Β D ) 3 + 2λ (1 − Β C ) 3 where βD = D/h, βC = C/l, D = beam depth, C = column width.

[

12.6

]

Distribution of lateral load in buildings

12.6.1 Rigid Plate on Springs Model For this model the building is treated in plan as a rigid plate supported by springs in its own plane - Figure 12.10. A set of bracing elements is defined. A bracing element is a frame, wall or core which is deemed to deemed to provide lateral support to the system. Each bracing element is defined by a single spring stiffness which is the lateral load to cause 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.


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Chapter 12

(a) Plan of 4 storey building

(b) Rigid beam on springs model

Figure 12.10 Rigid beam on springs model

Basic Procedure 1. 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 this stage). Extract or calculate the top lateral deflection ∆top of the bracing element. The stiffness of the bracing element is then ki = W/∆top 4. Treat the system as a rigid beam and springs subject to loading W 5. 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 as W). 6. Analyse each bracing element under the loading pi (e.g. multiply the results of the analysis used to get ki by the factor pi/W).


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Chapter 12

No torsion (one degree of freedom system) Figure 12.11 shows a system with springs only in the y direction and the model is therefore a rigid beam on springs. If torsion is neglected the system moves without rotating. The distribution of load to the supports is then in direct proportion to their top stiffnesses ki. The equilibrium condition is: W = Σpi (12.16) where pi is the load in spring i The compatibility condition is δi = ∆ (12.17) where δi is the deflection of spring i and ∆ is the deflection of the system The force-deformation relationships are pi = ki δi (12.18) where ki is the stiffness of spring i Combining (12.16), (12.17) and (12.18) gives the system stiffness relationship: W = ∑ ki ∆ (12.19) from which ∆ can be calculated using: W ∆= (12.20) ∑ ki and the load on a support frame is: k pi = k i ∆ = i W (12.21) ∑ ki The distribution of load to the supports is thus in direct proportion to their top stiffnesses ki. With torsion (two degrees of freedom system) If torsion is considered then an extra freedom (corresponding to a rotation) needs to be added to the rigid beam on springs - Figure 12.11. The system deformations are ∆ and θ. The equilibrium equations are then 1 W  (12.22)   = ∑  p i x i  M  where M is the moment of the applied loads about the origin. (M = Wa for the system of Figure 12.13) xi is the distance from the origin to the position of spring i The compatibility condition is: ∆  (12.23) δ i = [l x i ]  θ  The force-deformation relationship of a spring is pi = ki δi i.e pi = ki(∆ + xi θ) (12.24)


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Chapter 12

Figure 12.11 Rigid beam on springs model Combining equations (12.22), (12.23) and (12.24) gives: ∑ k i x i  ∆  W   ∑ k i (12.25)   2    M  ∑ k i x i ∑ k i x i   θ  Equation (12.25) can be solved to get the system deformations ∆ and θ from which the spring forces are obtained using (12.24). The solution in this case can be conveniently programmed using a spreadsheet. It is helpful to use the 'centre of stiffness' of the spring system as the origin. If this is done, then the off-diagonal terms of the matrix of equation (12.25) are zero and the solution of the equations is simplified. The centre of stiffness is found by taking first moments of spring stiffness about any point to find the position of the resultant of the spring stiffnesses.

12.7 Checking Models for Plate Bending For checking, the plate bending problem being considered can be amended to a form for which a solution is available. This can be achieved by altering the boundary conditions and/or the loading. One should make an estimate of whether the checking model will be stiffer or more flexible than the finite element model. Corresponding results from the two models are then compared with some expectation of the sign of the difference between them. A worthwhile approach is to use two checking models one of which tends to be stiffer than the finite element model and the other tending to be more flexible. These should give results on either side of those being checked. 12.7.1 Formal Solutions The model can be converted to a form for which a formal solution is available - see for example Timoshenko & Woinowsky Krieger (1959), Roark (1965).


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Figure 12.12

Chapter 12

Checking model for a simply supported rectangular plate

12.7.2 Uniaxial Bending Models It can be useful to take a strip of a plate in biaxial bending and treat is as being one-way spanning. This will tend to overestimate deformations and stresses. This is particularly useful if the plate has a dominant span direction. For example, for the simply supported plate with side ratio of 2:1 and uniformly distributed load shown in Figure 12.12(a), one can take a unit width strip as shown in Figure 12.12(b) and treat it as a simply supported one-way span. The maximum deflection and maximum moment for the strip can be calculated using: δ max =

5 qa 4 384 EI

and

M max =

qa 2 8

where ‘a’ is the span of the strip. The coefficients in the above expressions are quoted in Table12.2 in which the corresponding coefficient for Timoshenko & Woinowsky-Krieger (1959) are also given.


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Chapter 12

6.7.3 Grillage Models Results for the simple grillage model shown in Figure 12.13(c) are also shown in Table 12.2. For the grillage one quarter of the total load is applied as a point load at the centre. Member loads could also be used but it is easier to solve the point load case (the grillage of Figure 12.13(c) can be solved by treating it as being singly statically indeterminate). The grillage gives a slightly better estimate of deflection but further overestimates the moments as compared with the one-way strip. Figure 12.13(a) shows a square simply supported uniformly distributed plate. Figure 12.13(b) shows a simple grillage model. Again, one quarter of the total load is applied at the centre. Comparison of the maximum deflection and maximum moment values are quoted in Table 12.2. The checking model results quoted in Table 12.2 do not give close correlation but do establish the likely orders of magnitude. Such information is 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 δmax for Mmax 2a 5 1 a One-way strip 0.01302 = 0.125 = 384 8 grillage 0.01225 0.1366 Timoshenko 0.00922 0.1017 a a

δ max = coefficient x

Grillage

0.00251 =

Timoshenko

0.00406

qa 4 EI


1 192

0.0625 =

1 16

0.0479

M max = coefficient x qa 2

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