Bridge Design Assignment #2 Solution 1/6 Spring 2012 Influence Lines for Bridge Deck 2. Maximum +’ve and –‘ve moments due to moving loads. (a) Two 16 k loads, spaced 6 ft apart. Set up two 16 k loads on the “Moving Loads” menu. Make sure both loads are on the bridge for all time increments. • Add a node (N1A) for the starting position of the right wheel load; and add a node (N5A) for the ending position of the left wheel. • Specify starting locations (1 st Joint) and ending locations (2 nd Joint) so that the loads appear at the left end simultaneously, and disappear at the right end simultaneously. Starting Position: . . . Intermediate Position: . . . Ending Position:
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BridgeDesign Assignment#2Solution 1/6Spring2012 InfluenceLinesforBridgeDeck2.Maximum+’veand–‘vemomentsduetomovingloads.(a)Two16kloads,spaced6ftapart.Setuptwo16kloadsonthe“MovingLoads”menu.Makesurebothloadsareonthebridgeforalltimeincrements.
• Addanode(N1A)forthestartingpositionoftherightwheelload;andaddanode(N5A)fortheendingpositionoftheleftwheel.
• Specifystartinglocations(1stJoint)andendinglocations(2ndJoint)sothattheloadsappearattheleftendsimultaneously,anddisappearattherightendsimultaneously.
StartingPosition:
...
IntermediatePosition:
...
EndingPosition:
BridgeDesign Assignment#2Solution 2/6Spring2012 InfluenceLinesforBridgeDeck
• Combinebothmovingloadsintooneloadcombination.SpecifyPDelta=“Y”sothat“DesignResults”canbedisplayedforfindinglocationsofmax.moments.“Solve”theloadcombinationanddisplaytheenvelopeofbendingmomentsfromthemovingloads.The“DesignResults”(CodeChecks)menushowsthatthelocationofthemaximumpositivemomentinSpan1is2.905ftfromtheleftsupport.
Therefore,max.momentsduetotwo16kwheelloads: Max.+’veM:23.7k‐ftat2.91ftfromtheexteriorsupport Max.–‘veM:23.2k‐ftat1stinteriorsupport(b)twopairsof16kwheelloads.Varyspacingbetween2ndand3rdwheelloadfrom4ft(minimum)to14ft.Movingloadmenufor4ftspacingbetween2ndand3rdwheelload:
BridgeDesign Assignment#2Solution 3/6Spring2012 InfluenceLinesforBridgeDeckStartinglocationsofwheelloads:
Results:max+’veM=21.8k‐ftat3.8995ftfromexteriorsupportmax–‘veM=‐24.5k‐ftatfirstinteriorsupport.
Spacingbetween2ndwheel&3rdwheel,ft
max.+’veM,k‐ft
Locationmax+’veM
max.‐’veM,k‐ft
Locationmax‐’veM
4 21.8 3.9ftfromext.sup. ‐24.5 1stint.support6 23.5 3.0ftfromext.support ‐21.2 1stint.support8 24.5 2.9ftfromext.support ‐20.9 1stint.support10 24.5 2.9ftfromext.support ‐21.9 1stint.support12 24.2 2.8ftfromext.support ‐23.1 1stint.support14 23.8 2.8ftfromext.support ‐23.8 1stint.support
3.InfluenceLineformaximum–‘vemomentatfirstinteriorsupport(nodeN10).3.1CalculatedeflectionsusingRISAmodel
• TheRISAmodelismodifiedsothattherearetwonodeswiththesamecoordinatesatthelocationforwhichtheinfluencediagramisdesired:N10LandN10R(seeJointCoordinatesscreenbelow).
• N10LisconnectedtothenodetotheleftviamemberM10,andN10RisconnectedtothenodetotherightviamemberM11.ThereisnomemberconnectingN10LtoN10R.(SeeMemberPrimaryDatascreenbelow).
• TheXandYdegreesoffreedom(DOF)ofnodeN10RareslavedtothecorrespondingDOFofnodeN10L.TherotationabouttheZaxisisnotslavedsothatthetwonodescanrotateindependentlyofoneanother.(SeeJointBoundaryConditionsscreenbelow).
BridgeDesign Assignment#2Solution 4/6Spring2012 InfluenceLinesforBridgeDeck
• ApairofequalandoppositemomentsareappliedtojointsN10LandN10R(seeJointLoadsscreenbelow).Themomentsareappliedtocauseacounter‐clockwiserelativerotation(slopeatN10R–slopeatN10R).TheresultingYdeflectionswillbeusedtoformtheinfluencediagramforbendingmomentatthefirstinteriorsupport.TheZrotationsofN10LandN10Rwillalsobeused.
3.2CalculatetheinfluencelineusingthedeflectionsfromRISA.FromBetti’slaw:
Mi
Px
Px
=!x
!i
!i
,!!!!or !!Mi
Px = Px
!x
!i
!i
Mi
Px = Px!!I,!!!!!where!!I =
!x
!i
!i
TheinfluencevaluesforthemomentatnodeN10isequaltothedeflectionsintheYdirection(Δ)dividedbytherelativerotationofthetwoendsofthebeam(θ).Forexample,Iat5ftfromtheexternalsupport:
I5 ft =!5 ft
!=
.2418"!1 ft
12in
(.009308rad
" (".01075rad))=1.0046 ft
BridgeDesign Assignment#2Solution 5/6Spring2012 InfluenceLinesforBridgeDeckTest:applya10kload5ftfromtheexternalsupportandcalculateMatthefirstinternalsupport(N10).
MN10 = (!10k)(1.0046 ft) = !10.046
k! ft (checks)3.3Adjusttheinfluencevaluesforfive‐spanbeamswithspanlengthsdifferentthan10ft.Sincemomentdueapointloadisdirectlyproportionaltothespanlength,theinfluencevaluesforsayabeamwithfiveequal8‐ftspansis:
I8 ft spans = I10 ft spans
8 ft
10 ft
Test:applya10kloadinthemiddleofthefirstspanoffive‐spanbeamwith8ftspans.
I8 ft spans = I10 ft spans
8 ft
10 ft
I8 ft spans =1.0046 ft
8
10= 0.8037
I4 ft = (!10k)(0.8037) = !8.037
k! ft
BridgeDesign Assignment#2Solution 6/6Spring2012 InfluenceLinesforBridgeDeck4.Testmax.MsusingInfluenceLinewithmax.MsusingmovingloadsinRISA.
Influencelineforfirstinteriorsupport,fromspreadsheet:
Mat1stinteriorsupportduetoonetruck(two1kloadsseparatedby6ft),fromspreadsheet:
ComparisonofRISAMovingLoadAnalysisvs.InfluenceLineMovingLoadAnalysis:
RISAMovingLoadAnalysis
‐‐‐‐‐‐‐‐InfluenceLineMovingLoadAnalysis‐‐‐‐‐‐‐‐
‐‘veM,duetoonetruck ‐23.2k‐ft ‐23.2k‐ft [email protected]’fromextsupport ‐‘veM,duetotwotrucks ‐24.5k‐ft,
4’betweentrucks ‐24.5k‐ft
4’[email protected]’fromext.support
+‘veM,duetoonetruck 23.7k‐ft 23.7k‐ft* [email protected]’fromext.support +‘veM,duetotwotrucks 24.5k‐ft
8’to10’betweentrucks
24.6k‐ft*9’betweentrucks
[email protected]’fromext.support
*Influencelinefor+Mconstructedat0.35L=2.8ftfromext.support
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