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2.2 Governing Equations - Transverse Bending of Planar Beams 57Equilibrium equations Figure 2.4 shows a differential beam element subjected to an external transverseloading, , and restra

Discrete systems are governed by algebraic equations, and the problemreduces to finding the elements of the system stiffness matrix The static caseinvolves solving

(2.1)for , where and are the prescribed displacement and loading vectors.Some novel numerical procedures for solving eqn (2.1) are presented in a latersection

KU* = P*

In the dynamic case, the equilibrium equation for undamped periodicexcitation of the fundamental mode is used:

(2.5)The solution technique for eqn (2.1) also applies for eqn (2.5) Once isknown, the stiffness can be scaled by specifying the frequency, Anappropriate value for is established by converting the system to an equivalentone degree-of-freedom system and using the SDOF design approaches discussed

in the introduction

Continuous systems such as beams are governed by partial differentialequations, and the degree of complexity that can be dealt with analytically islimited The general strategy of working with equilibrium equations is the same,but now one has to determine analytic functions rather than discrete values forthe stiffness Analytical solutions are useful since they allow the keydimensionless parameters to be identified and contain generic informationconcerning the behavior

In what follows, the first topic discussed concerns establishing the stiffnessdistribution for static loading applied to a set of structures consisting ofcontinuous cantilever beams, building type structures modeled as equivalentdiscontinuous beams with lumped masses, and truss-type structures Closedform solutions are generated for the continuous cantilever beam example Thenext topic concerns establishing the stiffness distribution for the case of dynamicloading applied to beam-type structures The process of calibration of the

2.2 Governing Equations - Transverse Bending of Planar Beams 53

fundamental frequency is described and illustrated for both periodic and seismicexcitation The last section of the chapter deals with the situation where thehigher modes cannot be ignored An iterative numerical scheme is presented andapplied to a representative range of beam-type structures

2.2 Governing equations - transverse bending of planar beams

In this section, the governing equations for a specialized form of a beam aredeveloped The beam is considered to have a straight centroidal axis and a cross-section that is symmetrical with respect to a plane containing the centroidal axis.Figure 2.1 shows the notation for the coordinate axes and the displacementmeasures (translations and rotations) that define the motion of the member Thebeam cross-section is assumed to remain a plane under loading This restriction isthe basis for the technical theory of beams, and reduces the number ofdisplacement variables down to three translations and three rotations which are

functions of x and time.

When the loading is constrained to act in the symmetry plane for the section, the behavior involves only those motion measures associated with thisplane In this discussion, the plane is taken as the plane of symmetry, and ,, and are the relevant displacement variables If the loading is further

cross-restricted to act only in the y direction, the axial displacement measure, , isidentically equal to zero The behavior for this case is referred to as transversebending In what follows, the governing equations for transverse bending of acontinuous planar beam are derived The derivation is then extended to deal withdiscontinuous structures such as trusses and frames that are modelled asequivalent beams

u y βz

u x

Fig 2.1: Notation - planar beam.

Planar deformation-displacement relations

Figure 2.2 shows the initial and deformed configurations of a differential beamelement The cross-sectional rotation, , is assumed to be sufficiently small suchthat In this case, linear strain-displacement relations are acceptable.Letting denote the transverse shearing strain and the extensional strain at anarbitrary location from the reference axis, and taking and , thedeformation relations take the form

(2.6)(2.7)

(2.8)where denotes the bending deformation parameter

x y

2.2 Governing Equations - Transverse Bending of Planar Beams 55

Fig 2.2: Initial and deformed elements

Optimal deformation and displacement profiles

Optimal design from a motion perspective corresponds to a state of uniform shearand bending deformation under the design loading This goal is expressed as

(2.9)(2.10)Uniform deformation states are possible only for statically determinate structures.Building type structures can be modeled as cantilever beams, and therefore thegoal of uniform deformation can be achieved for these structures

Consider the vertical cantilever beam shown in Fig 2.3 Integrating eqns(2.7) and (2.8) and enforcing the boundary conditions at leads to

(2.11)

(2.12)The deflection at the end of the beam is given by

(2.13)

βγ

O

A

A y X

Y' Y

-=

u H( ) γ∗H χ∗H2

2

+

-=

where is the contribution from shear deformation and is thecontribution from bending deformation For actual buildings, the ratio of height

to width (i.e aspect ratio) provides an indication of the relative contribution ofshear versus bending deformation Buildings with aspect ratios on the order ofunity tend to display shear beam behavior and On the other hand,buildings with aspect ratios greater than about display bending beam behavior

Fig 2.3: Simple cantilever beam

One establishes the values of , based on the performance constraintsimposed on the motion, and selects the stiffness such that these targetdeformations are reached Introducing a dimensionless factor , which is equal tothe ratio of the displacement due to bending and the displacement due to shear atx=H,

(2.14)transforms eqn (2.13) to a form that is more convenient for low rise buildings

2.2 Governing Equations - Transverse Bending of Planar Beams 57

Equilibrium equations

Figure 2.4 shows a differential beam element subjected to an external transverseloading, , and restrained by the internal transverse shear, , and bendingmoment, By definition,

where , J are the mass and rotatory inertia per unit length When the member

is supported only at (see Fig 2.3), the equilibrium equations can beexpressed in the following integral form

Fig 2.4: Forces acting on a differential element.

In the case of static loading, the acceleration terms are equal to 0, and V and M can

be determined by integrating eqns (2.20) and (2.21)

(2.22)(2.23)

where and are defined as the transverse shear and bending rigidities.These equations have to be modified when the deformation varies with time Thisaspect is addressed in Section 2.4 Examples which illustrate how to determine therigidity coefficients for a range of beam cross-sections are presented below

x y

V ∂V

∂x -dx +

b

x -dx +

τ ∂∂τ

x -dx +

στ

V x( ) = DT( )γx ( )x

M x( ) = DB( )χx ( )x

DT DB

2.2 Governing Equations - Transverse Bending of Planar Beams 59

Example 2.1 Composite sandwich beam

Figure 2.5 shows a sandwich beam composed of 2 face plates and a core

Fig 2.5: Composite beam cross section

The face material is usually much stiffer than the core material, and therefore thecore is assumed to carry only shear stress Noting eqn (2.6), the strains in the faceand core are

(2.24)(2.25)

The face thickness is also assumed to be small in comparison to the depth.Considering the material to be linear elastic, the expressions for shear andmoment are:

Example 2.2 Equivalent rigidities for a discrete Truss-beam

The term, truss-beam, refers to a beam type structure composed of a pair of chordmembers and a diagonal bracing system Figure 2.6 illustrates an x-bracingscheme Truss beams are used as girders for long span horizontal systems Trussbeams are also deployed to form rectangular space structures which are theprimary lateral load carrying mechanisms for very tall buildings The typical

“mega-truss” has large columns located at the 4 corners of a rectangular section, and diagonal bracing systems placed on the outside force planes Thesestructures are usually symmetrical, and the behavior in one of the symmetrydirections can be modelled using an equivalent truss beam When the spacing, h,

cross-is small in comparcross-ison to the overall length, one can approximate the dcross-iscretestructure as a continuous beam having equivalent properties In this example,approximate expressions for these equivalent properties are derived for the case

of x-bracing

Fig 2.6: Parameters and internal forces - Truss-beam

The key assumption is that the members carry only axial force Thisapproximation is reasonable when the members are slender, and diagonal orchevron bracing is used Noting Fig 2.6, the cross section force quantities arerelated to the member forces by

Ad

Ac

2.2 Governing Equations - Transverse Bending of Planar Beams 61

(2.30)(2.31)Assuming linear elastic behavior, the member forces are also related to theextensional strains by

(2.32)(2.33)

It remains to express the extensional strains in terms of the bending and sheardeformation measures

Figure 2.7 shows the deformed shapes of a panel of the truss beam Theextensional strain in the diagonals, , due to the relative motion betweenadjacent nodes, , is a function of and

(2.34)

Neglecting the extensional strain in the diagonal due to , and approximatingas

(2.35)one obtains the following approximation for the total extensional strain:

(2.36)

Similarly, the extensional strain in the chord, , is related to the change in angle,, between adjacent sections by

(2.37)Noting that is related to the bending deformation ,

(2.38)the strain can be expressed as

(2.39)

Fig 2.7: Deformed truss-beam section

Substituting for and and combining eqns (2.30)-(2.33) results in

(2.40)

(2.41)Comparing these expressions with the definition equations for the rigidityparameters leads to the following relations for the equivalent continuous beamproperties:

2.2 Governing Equations - Transverse Bending of Planar Beams 63

When the truss beam model is used to represent a tall building, the chordscorrespond to the columns of the building These elements are required to carryboth gravity and lateral loading whereas the diagonals carry only lateral loading.Since the column force required by the gravity loading can be of the same order asthe force generated by the lateral loading, the allowable incremental deformation

in the column due to lateral loading should be less than the corresponding valuefor the diagonal To allow for this reduction, a factor, , which is defined as theratio of the diagonal strain to the chord strain for lateral loading is introduced

Governing equations for buildings modelled as pseudo shear beams

This section considers a class of planar rectangular building frames having aspectratios of order O(1) and moment resisting connections Figure 2.8 shows a typicalcase This type of structure is the exact opposite to the truss beam with respect tothe way the lateral loading is carried In the case of the truss beam, the transverseshear is provided by the axial forces in the braces Here, the shear is produced bybending of the columns The axial deformation of the columns is usually small forlow rise frames, so it is reasonable to assume the “floors” experience only lateraldisplacement and slide with respect to each other Considering the structure as apseudo-beam, there is no rotation of the cross-section, i.e., ; there is onlyone displacement variable per floor; and the transverse shearing strain at a storylocation is equal to the interstory displacement divided by the story height Inwhat follows, the formulation of the governing equations is illustrated using asimple structure and then generalized for more complex structures

Fig 2.8: Low rise rigid frame

The 2-story frame shown in Fig 2.9 is modeled as a 2 DOF system havingmasses concentrated at the floor locations and shear beam segments whichrepresent the action of the columns and beams in resisting lateral displacement.The shear forces in the equivalent beam segments (see Fig 2.10) are expressed interms of shear stiffness factors:

hH

B

2.2 Governing Equations - Transverse Bending of Planar Beams 65

(2.48)Noting the definition of transverse shear strain,

(2.49)

one can relate the ‘s to the equivalent transverse shear rigidity factors:

Fig 2.9: A discrete shear beam model

Fig 2.10: External and internal forces for discrete shear beam model.The equivalent shear stiffness factors are determined by displacing thefloors of the actual frame, determining the shear forces in the columns, summingthese forces for each story, and equating the total shear forces to and as

defined by eqn (2.48) The shear force in the ’th column of story is expressed as:

(2.51)

where depends on the frame geometry and member properties Then,summing the column shears for story and generalizing eqn (2.48) leads to

(2.52)

For this example, and ranges from to

An approximate expression for the column shear stiffness factors can beobtained by assuming the location of the inflection points in the columns andbeams Taking these points at the mid-points, as indicated in Fig 2.9, leads to thefollowing estimates for interior and exterior columns:

2.2 Governing Equations - Transverse Bending of Planar Beams 67

of the applied lateral loading The corresponding equations for this discretesystem are obtained from eqn (2.56) by combining the individual equations:

an DOF system with lumped mass and equivalent shear springs, as shown inFig 2.11 The strategy for determining the equivalent shear stiffness factors is thesame as discussed above The only difference for this case is the form of thesystem matrices They are now of order

Fig 2.11: General shear beam modelThe equilibrium equation for mass is given by

(2.64)Expressing the shear forces in terms of the nodal displacements

(2.65)and substituting in eqn (2.64) results in

m n

=

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading 69

(2.73)

Fig 2.12: Continuous cantilever beam

A cantilever beam having a linear shear rigidity distribution and aquadratic bending rigidity distribution will be in a state of uniform deformationunder uniform transverse loading Taking typical values for , , and the aspectratio for a tall building modelled as a truss beam,

(2.74)and evaluating ,

(2.75)one obtains

(2.76)This result corresponds to the extreme load value for displacement One would

B x

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading 71

use these typical values together with and to establish an appropriate valuefor at As will be seen later, the rigidity distributions need to bemodified near in order to avoid excessive deformation under dynamicload

Example 2.3 Cantilever beam - quasi static seismic loading

The cantilever beam loading shown in Fig 2.13 is used to simulate, in aquasi-static way, seismic excitation for low rise buildings The triangular loading

is related to the inertia forces associated with the fundamental mode response,and the concentrated force is included to represent the effect of the higher modes.Evaluating and applying eqn (2.68) leads to

Example 2.4 Truss-beam revisited

This example extends the treatment of the truss beam discussed in Example 2.2and focuses on comparing the cross-sectional parameters required to satisfy thestrength based versus the stiffness based performance criteria

Considering elastic behavior and given the desired design deformationsand , the corresponding extensional strains and must be less thanthe yield strains for the element materials, and respectively That is

(2.78)

(2.79)

Once the dimensions and the design deformations are specified, the structuralmaterial can be chosen to satisfy the motion design constraints defined by eqns(2.78) and (2.79) When the column strain is constrained to be related to thediagonal strain by

(2.80)eqn (2.79) can be written as

(2.81)

To provide more options in satisfying the design requirements, different materialsmay be used One must also insure that the stresses due to the design forces,and , are less than the yield stresses

The axial forces in the columns, , and diagonals, , are related to thetransverse shear and moment by

(2.82)(2.83)

V = 2Fdcosθ

M = BFc

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading 73

The cross-sectional areas required to provide the strength capacity follow fromeqns (2.82) and (2.83)

(2.84)

(2.85)

where denotes the allowable stresses based on strength considerations

The rigidity terms for this model are

(2.86)

(2.87)Substituting eqns (2.86) and (2.87) in the motion based design criteria,

Table 2.1: values for various steel strengths.

2.4 Stiffness Distribution for a Discrete Cantilever Shear Beam - Static Loading 75

2.4 Stiffness distribution for a discrete cantilever shear beam - static loading

Consider the set of equilibrium equations relating the nodal forces andstory displacements for an ‘th order discrete shear beam:

(2.95)

These equations are derived in section 2.2 In the normal analysis problem, onespecifies and , and solves for The problem is statically determinate sincethere are equations for the unknown displacements In this problem, onespecifies and , and attempts to determine the stiffness factors Since thereare linear algebraic equations, it should be possible to solve for the stiffnesscoefficients by rearranging the equations such that the ‘s are the unknowns Thevector containing the stiffness coefficients is denoted by

(2.96)

With this definition, eqn (2.95) is written as

(2.97)where the elements of are linear combinations of the prescribed displacementcomponents, , and contains the prescribed loads The entries in the ‘th

row of follow from eqn (2.66).

(2.98)for

These entries define to be an “upper triangular” matrix The diagonal entriesfor are interstory displacements Normally, one would not specify an interstorydisplacement to be since it would require an infinite shear stiffness It followsthat will be non-singular, and there will be a unique solution for

Example 2.5 3 DOF shear beam

Consider a 3 DOF cantilever beam subjected to uniform nodal loading andrequired to have a linear displacement profile The design values are:

(2.99)

(2.100)

Applying eqn (2.98) results in

(2.101)

Since the coefficient matrix is upper triangular, one solves for , , and then

by backsubstitution The solution is

0

P*

19.619.619.6

kN

=

U*

0.0250.0500.075

=

k3 k2 k1

2.5 Stiffness Distribution - Truss under Static Loading 77

(2.102)

As illustrated with the above example, solving eqn (2.97) is relatively simple, andone can easily handle arbitrary loading and permissible (non-zero interstory drift)displacement profiles

2.5 Stiffness distribution - truss under static loading

The approach discussed in the previous sections can be extended to deal withother types of structures such as trusses and frames Since trusses involve simpleequations, the approach is illustrated using a planar truss A more generaldiscussion is contained in Gallegos (1998)

Fig 2.14: Planar truss

The truss shown in Fig 2.14 has members and displacement variables;

the supports are assumed to be rigid so there are no support movements Each

member has an extension, , and corresponding force, Assuming linear elasticbehavior, the force-displacement relation for member is

(2.103)

k

23521558784

where is the member stiffness factor,

2.5 Stiffness Distribution - Truss under Static Loading 79

(2.108)

Since there are unknown stiffness factors and only equations, eqn (2.108)represents an undetermined system of equations and one cannot obtain a uniquesolution for This situation is typical of indeterminate structures A solution ofeqn (2.108) can be generated by selecting 2 stiffness factors as the primaryvariables, such as and , and solving for these variables in terms of , ,and the third stiffness factor, The result is written as

(2.109)

where , depend on the forces, and Equation (2.109) represent aconstraint on the stiffness parameters An additional condition must beintroduced in order to determine This additional condition is usuallyestablished by formulating an optimization problem

One possibility is to work with a weighted volume measure that is related

to the member stiffness factor, and seek the solution that corresponds to theminimum value of the sum of the weighted volumes For example, one can take

Another possibility is to use a least square approach, i.e., to select suchthat the sum of the squares of the weighted member volumes is a minimum Theobjective function for this optimization problem is

(2.115)

Substituting for and transforms to

(2.116)Requiring to be stationary with respect to a change in ,

(2.117)

leads to

(2.118)The problem with this approach is that it tends to eliminate redundant members,

-=

2.5 Stiffness Distribution - Truss under Static Loading 81

such as member 3 for this example This tendency is evidenced by “optimal”volume values which are low, and may even be negative A negative value for thevolume indicates the member should be deleted

A third approach partially overcomes the limitation of the least squareapproach by working with the deviation from the mean value as the membervariable Let denote the mean value,

(2.119)and represent the deviation from the mean value for member i,

(2.120)The optimization problem is stated as

Example 2.6 Application of least square approaches

The procedures described above are applied to Fig 2.14 specialized for the datacontained in (1)

2.5 Stiffness Distribution - Truss under Static Loading 83

(4)

Equation (4) represents the constraint on and Substituting for k in terms of

weighted volume measure, V=AEL, using eqn (2.111) transforms eqn (4) to

(5)

Least square procedure

The objective function is

(6)Differentiating with respect to ,

(7)

substituting for and , and then setting the resulting expression equal tozero results in

(8)The “optimal” least square values are

The low value for indicates that this member is “redundant”, and could beremoved.

Mean value least square procedure

One starts by establishing the mean value,

(10)and the deviations

2.5 Stiffness Distribution - Truss under Static Loading 85

(15)

Note that now is the same order of magnitude as and

The solution is sensitive to the prescribed loading and displacementquantities To illustrate this point, the computation is repeated taking the sameimposed displacements but different sets of nodal forces The results are listedbelow:

Mean value least square:

The case 2 result indicates that members 2 and 3 should be deleted This isthe correct solution for the specified loading and displacement Taking

corresponds to a load at 45o, which coincides with the direction of member 1.Furthermore, an extension of member 1 corresponds to , which is thespecified displacement condition

The procedures described above can be generalized to deal with anarbitrary truss Suppose there are degrees of freedom and members Itfollows that there are equilibrium equations relating the stiffness factorsand the prescribed nodal forces These equations are written as

(2.123)

where is of order When , the structure is indeterminate, and eqn(2.123) does not have a unique solution for In this case the problem isundetermined, and an optimization statement has to be formulated One canwork with either stiffness or weighted volume measures as the variables To allowfor different choices, the form of eqn (2.123) is generalized to

(2.124)

where is an th order vector containing the selected variables, and is oforder If one selects V defined by eqn (2.111) as the measure, is obtained

by multiplying the i’th column of by , and taking i from 1 to

Background material on computational techniques for solving linearalgebraic equations is contained in Strang,1993 In what follows, the use of one ofthese techniques to solve eqn (2.124) is described in general terms The sameprocedure is also applied in quasi-static active control problems in Chapter 7

2.5 Stiffness Distribution - Truss under Static Loading 87

The least square approach works with a scalar quantity, f, that is equal to

the sum of the squares of the elements of Using matrix notation, f is given by

(2.125)

The elements of X are constrained by eqn (2.124) It follows that elements can

be expressed in terms of elements, which can be interpreted as theunknowns that need to be selected such that is a minimal value Assuming thefirst columns of A are independent, eqn (2.124) can be expressed in partitioned

solution corresponds to an absolute minimum value of (see Strang, 1993) TheMATLAB statement, , generates this least square solution.

The mean value least square approach works with the deviation from themean,

(2.131)For this case, there are members, and is given by

(2.132)Using matrix notation, the deviation vector, , can be expressed as,

2.5 Stiffness Distribution - Truss under Static Loading 89

(2.137)

where is a vector containing the n Lagrangian multipliers Requiring J to be

stationary with respect to both X and leads to the following set of equations:

Example 2.7 Comparison of strength vs displacement based design

Consider the 3 member truss shown in the figure Suppose members 1 and

3 to have the same properties It follows that and The memberelongations are:

u1

P2,u2

u1 = 0 e3≡e1

Noting eqn (1), the elongations are constrained by

(2)The member flexibility factor, , is defined as

(3)Then, the member force-deformation relation can be expressed as

(4)Using (4), the geometric compatibility equation expressed in terms of memberforces has the form

(5)The solution for and is

(6)

• Strength based design

Ideally, one would want the member stresses to be equal to an allowablestress, However, a state of uniform stress is not possible because of theindeterminate nature of the structure Noting that

(7)for a member, eqn (1) can be written as