Activity 2.1.6 Step-by-Step Truss mechanism Answer KeyIntroductionTruss systems are essential contents within structural systems varying from residential building and construction to large scale civil design projects such as bridges. Regardless of the device application, trusses are designed come utilize material strength, mitigate costs, and support a determined load. Designers must be able to understand how loads plot on a truss structure and within the framework to ensure style feasibility and also safety. Activity 3.1.7 will guide you with the step-by-step process of calculating reaction forces and also member forces within a truss system.EquipmentStraight edgeCalculator PencilProcedureIn this task you will certainly calculate reaction and member pressures for the truss system depicted below. It is crucial to monitor each action within the procedure come ensure suitable calculations and complimentary body diagrams. Calculate external Reaction Forcesx and y Reaction pressure at pin A and also y Reaction force at Roller CDraw a freebody diagram for the entire truss structure portrayed above. Make certain to incorporate all known and unknown angles, forces, and also distances.Calculate and also determine all angles using trigonometry and geometry.(1 box = .5 Units)Algebra help hints: sin θ = O/H cos θ = A/Htan θ = O/Aa2 + b2 = c2Calculate reaction pressures at the roller and also pin connections.List revolution equilibrium equations – Hint: They all involve summations.ΣFx=0ΣFy=0ΣM=0List every know and also unknown forces acting and also reacting ~ above the truss framework Label direction of pressure with one arrow.Forces in the x-directionRAx→Forces in the y-directionRAy↑RCy↑1000lb↓500lb↓250lb↓Moment pressures – determined from pen AFormula review: M = Fd1000 x 3 250 x 5500 x 7RFCy x 10Solve because that RCY by utilizing the minute static equilibrium equation exhilaration upon pin A. ΣM=0-3000 - 1250 - 3500 + (10RCy)= 0EquationSubstitution10RCy = 7750RCy= 775 lbNote: The price is positive, so the presume direction is correct.SimplificationSolutionSolve for unknown reaction pressure in the x-direction (RAx). Usage the sum of forces in the x-direction equilibrium equation.ΣFx=0RAx=0RAx= 0EquationSubstitutionSolutionSolve for unknown reaction pressures in the y-direction. Usage the sum of forces in the y-direction equilibrium equation.ΣFy=0RAy + RCy +(-1000lb) + (-500lb) + (-250lb) = 0EquationSubstitutionRAy + 775lb = 1750lbRAy = 975lbSolutionDraw a freebody diagram for the entire truss system portrayed on web page 1.Make sure to include your calculated assistance reactions (1 box = .5 Units).Calculate individual Truss Member ForcesCalculate member forces ad and ABDraw the freebody diagram because that joint A.Make sure to incorporate all known and also unknown angles and forces (including x and y vector components). Execute not encompass lengths.Updated drawing Use SOH CAH TOA to express ADX and ADY in terms of AD.Calculate ADXADX = ADsin45°EquationSubstitutionSolutionCalculate ADYADY = ADcos45°EquationSubstitutionSolutionList all know and also unknown forces.Label direction of pressure with an arrow.Forces in the x-directionRAx = 0lb →AB →(ADsin45°) →Forces in the y-direction975 lb ↑(ADcos45°) ↑Use revolution equilibrium equations to solve for advertisement and AB.Solve for ad by calculating y-direction static equilibrium.ΣFy=0975 lb + (AD x cos45°)AD x cos45° = -975 lbEquationSubstitutionSimplificationAD = -1378.86 lbNote: The answer is negative, so adjust the assumed direction.SimplificationSolutionSolve for abdominal by calculating x-direction revolution equilibrium.ΣFx=0RAx + abdominal muscle + (ADsin45o) = 00 lb + abdominal + (AD x Sin45°) = 0AD x Sin45° = -ABEquationSubstitutionSimplification-1378.86 lb x sin45° = -ABAB = 975 lbAB is under tensionSubstitution – Insert calculated FAD valueSolutionUpdate the joint A freebody diagram through calculated forces for advertisement and AB.Calculate CB and CE.Draw the freebody diagram for jointC.Make certain to incorporate all known and also unknown angles and also forces (including x and y vector components). Execute not incorporate lengths.Updated illustration Use SOH CAH TOA come express CEx and CEy in regards to CE.Calculate CExCEx = CEsin45°EquationSubstitutionSolutionCalculate FCEYCEy = CEcos45°EquationSubstitutionSolutionList all know and unknown forces.Label direction of force with an arrow.Forces in the x-directionCB ←CEsin45°←Forces in the y-direction775 lb ↑CEcos45° ↑Use static equilibrium equations to settle for ad and AB.Solve because that CE through calculating y-direction revolution equilibrium.ΣFy=0775 lb + (CEcos45°) = 0CEcos45° = -775 lbEquationSubstitutionSimplificationCE = -1096.02 lbNote: The prize is negative, so change the assumed direction.SolutionSolve because that CB through calculating x-direction static equilibrium.ΣFx=0CB + (CEsin45°) = 0-CB = CESin45°EquationSubstitutionSimplification-CB = -1096.02 lb Sin45° CB = 775 lbSubstitution – Insert calculation CE valueSolutionUpdate joint C free-body diagram v calculated pressures for CE and also CB.Calculate EB and EDDraw the free-body diagram for jointE.Make sure to incorporate all known and also unknown angles and also forces (including x and also y vector components).


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Do not incorporate lengths.Updated illustration Use SOH CAH TOA come express EBx and EBy in terms of EB.Calculate EByEBy = EB sin56°EquationSubstitutionSolutionCalculate FEBXEBx = EB cos56°EquationSubstitutionSolutionList every know and unknown forces.Label direction of pressure with one arrow.Forces in the x-directionED ←775 lb ←EBcos56° ←Forces in the y-direction500 lb↓775 lb↑EB sin56° ↓Use static equilibrium equations to resolve for EB.Calculate y-direction static equilibrium.ΣFy=0-500lb + 775 lb + (-EB sin56°) = 0275lb - EB sin56° = 0EquationSubstitutionSimplification-EB sin56° = -275 lbEB = 331.71 lbSubstitutionSimplificationSolutionCalculate x-direction revolution equilibrium.ΣFX=0-ED - 775lb - (EB cos56°)=0-ED - (331.71 cos56°) = 775EquationSubstitutionSimplification-ED - (185.27lb) = 775lb-ED = 185.27lb + 775lbED = -960.27lbNote: The answer is negative, so adjust the presume direction.SubstitutionSimplificationSolutionUpdate joint E freebody diagram v calculated pressures for EB and ED.Calculate DBDraw the freebody diagram for jointD.Make sure to encompass all known and also unknown angles and forces (including x and also y vector components). Do not include lengths.Updated illustration Use SOH CAH TOA to express DBx and also DBy in terms of DB.Calculate DByDBY = DB sin56°EquationSubstitutionSolutionCalculate FDBXDBX = DB cos56°EquationSubstitutionSolutionList every know and unknown forces.Label direction of pressure with one arrow.Forces in the x-direction975 lb→960.27 lb ← DB cos56°→Forces in the y-direction 975 lb ↑1000 lb ↓DB sin56°↓ use static equilibrium equations to resolve for DB.Solve for DB by Calculatingy-direction static equilibrium.ΣFy=0975lb + (-1000lb) - (DB sin56°)=0-25lb = DB sin56°EquationSubstitutionSimplificationDB = -30.05 lbSimplificationSolutionUpdate share D freebody diagram through calculated forces for DB and DE.Draw Completed cost-free Body DiagramDraw a completed freebody diagram for whole truss framework using all calculated reaction and also member forces.................................Online PreviewDownload