Activity 2.1.6 Step by Step Truss System



Activity 2.1.6 Step-by-Step Truss SystemIntroductionTruss systems are essential components within structural systems ranging from residential construction to large scale civil engineering projects such as bridges. Regardless of the system application, trusses are designed to utilize material strength, reduce costs, and support a determined load. Engineers must be able to understand how loads act on a truss structure and within the structure to ensure design feasibility and safety. Activity 2.1.6 will guide you through the step-by-step process of calculating reaction forces and member forces within a truss system. EquipmentStraight edgeCalculator PencilProcedureIn this activity you will calculate reaction and member forces for the truss system illustrated below. To ensure proper calculations and free body diagrams, it is essential that you follow each step in the procedure. Calculate External Reaction Forces x and y Reaction Force at Pin A and Y Reaction Force at Roller CDraw a free body diagram for the entire truss structure illustrated above. Make sure to include all known and unknown angles, forces, and distances. Calculate and determine all angles using trigonometry and geometry.(1 Box = .5 Units)Algebra hints: sin θ = O/H cos θ = A/H tan θ = O/A a2 + b2 = c2Calculate reaction forces at the roller and pin connections.List static equilibrium equations. Hint: (They all involve summations)∑ FX = 0 ∑ FY = 0∑ MA = 0List all known and unknown forces acting and reacting on the truss structure. Label direction of force with an arrow.Forces in the x direction AXForces in the y directionAYBYCYDYEYMoment Forces – Determined from Pin AFormula review: M = FdCY * 10-1000 * 3-500 * 7-250 *5 Use the moment static equilibrium equation acting upon pin A to solve for RCy. ∑MA = 0 0 = CY (10) -1000(3) -500(7) -250(5) EquationSubstitution0 = CY (10) -77507750 = CY (10)RCy = 775 lb SimplificationSolutionSolve for unknown reaction force in the x direction (RAx). Use the sum of forces in the x direction equilibrium equation.∑ FX = 0 AX = 0RAx = 0 lbEquationSubstitutionSolutionSolve for unknown reaction forces in the y direction. Use the sum of forces in the y direction equilibrium equation.∑ FY = 0 0 = AY + 775 -250 -500 -10000 = AY + 775 -17500 = AY -975975 = AYEquationSubstitutionRAy = 975 lb .SolutionDraw a free body diagram for the entire truss system illustrated on page 1.Make sure to include your calculated support reactions (1 Box = .5 units).Calculate Individual Truss Member ForcesCalculate member forces AD and AB.Draw the free body diagram for joint A. Make sure to include all known and unknown angles and forces (including x and y vector components). Do not include lengths.Use SOH CAH TOA to express ADx and ADy in terms of AD.Calculate ADxSin ? = O / HSin 45° = FADx / FAD ADx = FAD * Sin 45°EquationSubstitutionSolutionCalculate ADyCos ? = A / HCos 45° = FADY / FADADy = FAD * Cos 45°EquationSubstitutionSolutionList all known and unknown forces. Label direction of force with an arrow.Forces in the x direction0 lbFAB ( FAD * Sin 45° )Forces in the y direction975 lb( FAD * Cos 45° )Use static equilibrium equations to solve for AD and AB.Solve for AD by calculating y direction static equilibrium.ΣFY = 0 975 + ( FAD * Cos 45°) = 0FAD * Cos 45° = -975EquationSubstitutionSimplificationFAD = (-975) / (Cos 45°)AD = 1378.56 lbSimplificationSolutionSolve for AB by calculating x direction static equilibrium.ΣFX = 0 0 + FAB + (FAD * Sin 45°)=0FAD * Sin 45° = - FABEquationSubstitutionSimplification-1378.56 * Sin 45° = -FABAB = 975 lbSubstitution – Insert calculated AD valueSolutionUpdate the joint A free body diagram with calculated forces for AD and AB.Calculate CB and CE.Draw the free body diagram for joint C.Make sure to include all known and unknown angles and forces (including x and y vector components). Do not include lengths.Use SOH CAH TOA to express CEx and CEy in terms of CE.Calculate CExSin ? = O / HSin 45° = FCEX / FCE CEX = FCE * Sin 45°EquationSubstitutionSolutionCalculate CEyCos ? = A / HCos 45° = FCEY / FCE CEy = FCE * Cos 45°EquationSubstitutionSolutionList all known and unknown forces. Label direction of force with an arrow.Forces in the x directionFCBFCE * Sin 45°Forces in the y direction775 lbFCE * Cos 45°Use static equilibrium equations to solve for AD and AB.Solve for CE by calculating y direction static equilibrium. ΣFY = 0 775 + (FCE * Cos 45°) = 0 FCE * Cos 45° = -775EquationSubstitutionSimplificationCE = 1096.02 lbSolutionSolve for CB by calculating x direction static equilibrium.ΣFX = 0 FCB + (FCE * Sin 45°) = 0FCB = FCE * Sin 45°EquationSubstitutionSimplificationFCB = Sin 45° * 1096.02CB = 7750 lbSubstitution – Insert calculated CE valueSolutionUpdate joint C free-body diagram with calculated forces for CE and CB.Calculate EB and EDDraw the free-body diagram for joint E.Make sure to include all known and unknown angles and forces (including x and y vector components). Do not include lengths.Use SOH CAH TOA to express EBx and EBy in terms of EB.Calculate EBySin ? = O / HSin 56° = FEBY / FEB EBy = Sin 56° * FEBEquationSubstitutionSolutionCalculate EBxCos ? = A / HCos 56° = FEBX / FEB EBx = Cos 56° * FEBEquationSubstitutionSolutionList all known and unknown forces. Label direction of force with an arrow.Forces in the x directionFED775 lbCos 56° * FEBForces in the y direction-500 lb775 lbSin 56° * FEBUse static equilibrium equations to solve for EB.Calculate y direction static equilibrium. ΣFY = 0(Sin 56° * FEB) – 500 + 775275 + (Sin 56° * FEB) = 0EquationSubstitutionSimplification-275 = -Sin 56° * FEBFEB = -275 / -Sin 56° * FEBEB = -331.325 lbSubstitutionSimplificationSolutionCalculate x direction static equilibrium.ΣFX = 0(Cos 56° * FEB) + FED +775 = 0(Cos 56° * FEB) + FED = -775EquationSubstitutionSimplificationFED + 185.27 = -775 FED = -775 -185.27 ED =-960.775 lbSubstitutionSimplificationSolutionUpdate joint E free body diagram with calculated forces for EB and ED.Calculate DBDraw the free body diagram for joint D.Make sure to include all known and unknown angles and forces (including x and y vector components). Do not include lengths.Use SOH CAH TOA to express DBx and DBy in terms of DB.Calculate DBySin ? = O / HSin 56° = FDBY / FDB DBy = Sin 56° * FDBEquationSubstitutionSolutionCalculate DBxCos ? = A / HCos 56° = FDBX / FDB DBx = Cos 56° * FDBEquationSubstitutionSolutionList all known and unknown forces. Label direction of force with an arrow.Forces in the x direction-975-FDESin 56° * FDBForces in the y direction-9751000Cos 56° * FDBUse static equilibrium equations to solve for DB.Solve for DB by calculating y-direction static equilibrium. ΣFY = 0-975 +1000 + (Sin 56° * FDB) = 0-25 = (Sin 56° * FDB)EquationSubstitutionSimplification-25 / Sin 56° = FDBDB = 30.05 lbSimplificationSolutionUpdate joint D free body diagram with calculated forces for DB and DE.Draw Completed Free Body DiagramDraw a completed free body diagram for the entire truss structure using all calculated reaction and member forces. ................
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