Live Load Forces: Influence Lines Influence Lines For .

Live Load Forces:Influence Lines forDeterminate StructuresIntroductionPrevious developments havebeen limited to structuressubjected to fixed loads.Structures are also subjected tolive loads whose position mayvary on the structure. Thischapter focuses on such loads forstatically determinate structures.Influence LinesConsider the bridge in Fig. 1. Asthe car moves across the bridge,the forces in the truss memberschange with the position of thecar and the maximum force ineach member will be at a differentcar location. The design of eachmember must be based on themaximum probable load eachmember will experience.1Figure 1. Bridge Truss StructureSubjected to a VariablePosition LoadTherefore, the truss analysisfor each member wouldinvolve determining the loadposition that causes thegreatest force or stress ineach member.32If a structure is to be safelydesigned, members must beproportioned such that themaximum force produced by deadand live loads is less than theavailable section capacity.Structural analysis for variableloads consists of two steps:1.Determining the positions ofthe loads at which theresponse functionfti isimaximum; and2.Computing the maximumvalue of the response function.4See pages 49 - 77 in your class notes.1

Influence LineDefinitionsOnce an influence line isconstructed:Response Function supportreaction, axial force, shear force, orbending moment.Influence Line graph of aresponse function of a structure asa function of the position of adownward unit load moving acrossthe structure.structure Determine where to place liveload on a structure to maximizethe drawn response function;and Evaluate the maximummagnitude of the responsefunction based on the loading.NOTE: Influence lines forstatically determinate structuresare always piecewise linear.561Calculating ResponseFunctions(Equilibrium Method)xMBa0 x aVBAy Fy 0 V B A y 1 Ma 0 M B A y a 1(a x)ILD for Ay1Ay0L1ILD for Cy0MBaL7VBa x L Fy 0 V B A y Ma 0 M B A y a82

1 – a/LBeam Example 1VB0a-a/L/LLILD for VBa (1 – a/L)MB0aCalculate and draw the supportreaction response functions.LILD for MB9Beam Example 210Frame ExampleBD: LinkMemberCalculate and draw the responsefunctions for RA, MA, RC and VB.Calculate and draw theresponse functions for Ax, Ay,AB. NOTE: Unit loadand VBtraverses span AC.11123

CAUTION: Principle is only validfor force response functions.Muller-BreslauPrincipleMuller-Breslau Principle Theinfluence line for a responsefunction is given by the deflectedshape of the released structuredue to a unit displacement (orrotation) at the location and in thedirection of the responsefunction.A released structure is obtainedby removing the displacementconstraint corresponding to theresponse function of interest fromthe original structure.13Releases:Support reaction - removetranslational support restraintrestraint.Internal shear - introduce aninternal glide support to allowdifferential displacementmovement.Bending moment - introduce aninternal hinge to allow differentialrotation movement.14Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Influence Line for ShearInfluence Line for Bending Moment15164

Application of MullerBreslau Principle1718y (L – x) (a/L)θ1 θ2 119205

Qualitative InfluenceLinesIn many practical applications, it isnecessary to determine only thegeneral shape of the influencelines but not the numerical valuesof the ordinates. Such aninfluence line diagram is known asa qualitative influence line diagramgram.NOTE: An advantage ofconstructing influence lines usingthe Muller-Breslau Principle isthat the response function ofinterest can be determineddirectly. It does not requiredetermining the influence linesfor other functions, as was thecase with the equilibriummethod.An influence line diagram withnumerical values of its ordinates isknown as a quantitative influence line diagram.2122Influence Lines forTrussesIn a gable-truss frame building,roof loads are usually transmittedto the top chord joints through roofpurlins as shown in Fig. T.1.Similarly, highway and railwaybridge truss-structures transmitfloor or deck loads via stringers tofloor beams to the truss joints asshown schematicallyy in Fig.g T.2.Fig. T.2. Bridge TrussFig. T.1. Gable Roof Truss23246

These load paths to the truss jointsprovide a reasonable assurancethat the primary resistance in thetruss members is in the form ofaxial force. Consequently,influence lines for axial memberforces are developed by placing aunit load on the truss and makingjudicious use of free bodydiagrams and the equations ofstatics.Due to the load transferprocess in truss systems, nodiscontinuity will exist in themember force influence linediagrams. Furthermore, sincewe are restricting our attention tostatically determinate structures, the influence linediagrams will be piecewiselinear.2526Use of Influence LinesExample Truss StructurePoint Response Due to aSingle MovingConcentrated LoadCalculate and draw the responsefunctions for Ax, Ay, FCI and FCD.27Each ordinate of an influenceline gives the value of theresponse function due to asingle concentrated load ofunit magnitude placed on thestructure at the location of thatordinate. Thus,287

1. The value of a responsefunction due to any singleconcentrated load can beobtained by multiplying themagnitude of the load by theordinate of the responsefunction influence line at theposition of the load.PABCDxyBDABCILD for MB-yD (M B ) max place P at B (M B ) max place P at D29Point Response Due to aUniformly Distributed LiveLoadInfluence lines can also beemployed to determine thevalues of response functions ofstructures due to distributedloads. This follows directly frompoint forces by treating theuniform load over a differentialsegment as a differential pointforce, i.e., dP w l dx. Thus, aresponse function R at a pointcan be expressed as312. Maximum positive value ofthe response function isobtained by multiplying thepoint load by the maximumpositive ordinate. Similarly, themaximum negative value isobtained by multiplying thepoint load by the maximum30negative ordinate.dR dP y w l dx ywhere y is the influence lineordinate at x, which is the point ofapplication of dP.To determine the total responsefunction value at a point for adistributed load between x a to x b, simply integrate:R bbaa wlydx wl ydx328

in which the last integral expression represents the area under thesegment of the influence line,which corresponds to the loadedportion of the beam.SUMMARY1. The value of a responsefunction due to a uniformlydistributed load applied over aportion of the structure can beobtained by multiplying the loadintensity by the net area underthe corresponding portion of theresponse function influence 33line.352. To determine the maximumpositive (or negative) value of aresponse function due to auniformlyuo yddistributedst buted livee load,oad,the load must be placed overthose portions of the structurewhere the ordinates of theresponse function influence lineare positive (or negative).Points 1 and 2 are schematicallydemonstrated on the next slide formoment MB considered in the pointload case.34369

Where should a CLL(Concentrated Live Load), a ULL(Uniform Live Load) and UDL(Uniform Dead Load) be placedon the typical ILD’s shown belowto maximize the responsefunctions?Typical InteriorBeam Shear ILDTypical InteriorBending Moment ILDTypical End Shear(Reaction) ILDPossible Truss Member ILD37Live Loads forHighway andRailroad BridgesLive loads due to vehicular trafficon highway and railway bridgesare represented by a series ofmoving concentrated loads withspecified spacing between theloads. In this section, we discussthe use of influence lines todetermine: ((1)) the value of theresponse function for a givenposition of a series of concentratedloads and (2) the maximum valueof the response function due to aseries of moving concentrated 39loads.38To calculate the responsefunction for a given position ofthe concentrated load series,simply multiply the value of eachseries load Pi by the magnitudeoff theh influencei flliline didiagramyordinate i at the position of Pi ,i.e.R Pi yiiThe ordinate magnitude yi can becalculated from the slope of theinfluence line diagram (m) viayi m x i4010

where x i is the distance to point imeasured from the zero y-axisintercept, as shown in theschematic ILD below.mxiFor example,consider the ILDshown on the next slidesubjected to the given wheelloading:L dPLoadPositionii 11:1ybyaVB1 8( 1 20) 10( 1 16) 15( 1 13) 5( 1 8)x a m Pi xi 18.5k18 5kbya y b similar trianglesabyy ya b a ; m bbb303030301( )(8(20) 10(16) 15(13) 5(8))30i41422/310 ft.20 ftft.Position 1-1/3ILD for Internal Shear SBPosition 2Wheel Loads434411

Load Position 2:VB2 1 ( 8(6) 10(20) 15(17) 5(12))30 15.6kThus,, load positionp1 results in themaximum shear at point B.NOTE: If the arrangement ofloads is such that all or most of theheavier loads are located near oneof the ends of the series, then theanalysis can be expedited byselecting a direction of movementfor the series so that the heavierloads will reach the maximuminfluence line ordinate before thelighter loads in the series. In sucha case, it may not be necessary toexamine all the loading positions.Instead the analysis can beInstead,ended when the value of theresponse function begins todecrease; i.e., when the value ofthe response function is less thanthe preceding load position. Thisprocess is known as thep“Increase-Decrease Method”.45CAUTION: This criterion is notvalid for any general series ofloads. In general, depending onthe load magnitudes, spacing,and shape of the influence line,theh valuel off theh responsefunction, after declining for someloading positions, may startincreasing again for subsequentloading positions and may attain ahigher maximum.46Zero Ordinate LocationLinear Influence Lineb 1m xm1x b-L474812

x b b b; m m Lx b b b; m m LExample Truss Problem:Application of Loads toMaximize ResponseNOTE: Both of these solutionsare obtained fromy mx bwith y 0.ML49Also see page 69 in your class notes.CM50A larger version is on page 69a in yourclass notes.Force and MomentEnvelopesPlaceUDL 1.0 k/ft;ULL 4.0 k/ft;CLL 20 kipsto maximize the tension andcompression axial forces inmembers CM and ML.Calculate the magnitudes of thetension and compression forces.51A plot of the maximumresponse function as afunction of the location of theresponse function is referredto as the envelope of themaximum values of aresponse function for theparticular load case beingconsidered.id d5213

For a single concentratedforce for a simply supported beam:For a uniformly distributedload on for a simply supported beam:((V)) max a (V) max P 1 L wl( L a )22La(V) max P(V) max a M max P a 1 L M maxma LPlot is obtained by treating “a”as a variable.wl a22Lwl a(L a )2Plot is obtained by treating “a”as a variable.5354555614

bending moment. Influence Line graph of a response function of a structure as a function of the position of a downward unit load moving across the structure. 5. the structure. NOTE: Influence lines for statically determinate structures are always piecewise linear. Once an influ