09/11/95 1EquilibriumCEIG - 125 Introduction to Civil EngineeringFall 2005Lecture 3Outline• Free Body Diagrams• Supports and Reaction Forces• Equilibrium (Newton’s Laws)• Applying Equilibrium Equations• Statically Determinate Reactions• ExamplesFree Body Diagrams (FBDs)• Sketch representing a body and forces acting on it.• Isolates things of interest.• FBDs are a must for every statics problem being solved• First thing you do is draw an FBDFBDs for Supports1. Body on an inclined planeWR2. Weight on a chain or ropeWRFBDB CAAAARABRACWAABRBCRBXRBYACRCXRCYRBCRAB, RAC and WAare concurrentWCWBR1R1R2R1R2MSupports and Reaction Forces• Pin• Roller• Fixed09/11/95 2PinRollerComplex FBD ExampleComplex FBD ExampleCar wheel loadssupport reactionfixed end reactionsLoads from hanging pipeLoad from blowerDead load from deckDead load from beamEquilibrium• Newton’s Laws:• First Law: A particle continues to move in a straight line or remains at rest if there is no unbalanced force• Second Law: The acceleration of a particle is proportional to the force acting on it.• Third Law: To every action there is an equal and opposite reaction.Equilibrium• Law 1 - Defines static equilibriumAt rest => no unbalanced forceΣ M = 0 (moments)Σ F = 0 (forces)• Law 2 - States that F ∝ a• Law 3 - Defines interaction between bodies and the rest of the world - reactionsUsing Equilibrium• To determine the magnitude of the reaction forces provided by structure supports:• Draw a free body diagram to isolate the structure from its environment and replace all supports with reaction forces• Confirm that reactions are statically determinate and structure is stable• Generate and solve equilbrium equations09/11/95 3Statically Determinate Reactions• In two dimensions, we have 3 equilibrium equations (in 3D, 6 equilibrium eqns)• Statically determinate reactions in 2D means:• we have 3 unknown forces • we have 3 equilibrium equationsStatically Indeterminate Reactions and Unstable Structures• If > 3 reactions (6 in 3D), then we have a statically indeterminate set of reactions• If < 3 reactions (6 in 3D), then we have an unstable structureStatically Determinate?staticallydeterminatestaticallydeterminatestaticallyindeterminateunstablein x directionunstable inrotation aboutAAExample 1 -Simply Supported Beam• Consider a 10 m long simply supported beam with a mass of 8 kg/m. The beam supports a mass of 50 kg, 2m from the left end. • Find the reactions.Distributed Loads• Extend over an area and along a line.• Distributed loads can be replaced by a single resultant force acting through a centroid.w N/mwLLL/2Distributed Load Examples• Dead load of beam09/11/95 4• Drifting snow on a bridge• Wind on a buildingx N/mhxh Nh/2w N/ms N/mwLsL/2LL/2 L/3Distributed Load Examples, con’tStep 1: Draw FBD for Example 18.00 kg/m50.0 kg2 m 8 mA B491 N = 50.0 kg * 9.81 N/m^278.4 N/m = 8 kg/m * 9.81 m/s^2Problem DrawingFBD DrawingRBYRAYRAXDistributed dead load:Step 2: Statically Determinate?• We have 3 unknown reactions• We have 3 equilibrium equations• Structure’s reactions are statically determinateStep 3: Determine ReactionsΣ FX= 0 = RAX, RAX= 0Σ FY= 0 = +RA+ RB - 491 N - (78.4 N/m * 10 m )RAY+ RBY= 491 N + 784 = 1275 NΣ MA= 0 = +(RBY* 10 m) - (491 N * 2m) - (78.4 N/m * 10 m * 5 m)RBY= 490 N , RAY= 784 N+++45o5 kNGH8m 7m 15mABAnother Harder ExampleGiven the force system shown below, determine the reactions at G & H. The uniform beam has a mass of 15.0 kg/m.6mhingeStep 1: FBD of BH and HA45o5 kNGH8m 7m 15m45o5 kNHHVHVG147 N/mFBDVHHHHAVAAMAB09/11/95 5Step 2: Statically Determinate?• For section BH, we have 3 unknown reactions: VH , HH, and VG• We have 3 equilibrium equations• Section BH’s reactions are statically determinate• Note: after solving section BH, section HA has only 3 unknowns and its reactions can be found using equilibriumStep 2: Apply Equilibrium to BHΣ FX= 0 = (5 kN cos 45 ) - HHHH= 3.54 kNΣ FY = 0 = VH+ VG- (5 kN cos 45) - 0.147 kN/m * 30 mVH+ VG= 7.95 kNΣ MH = 0 = +(5 kN cos 45) * 22 m -(VG* 15 m) +(0.147 kN/m*30m*15m)VG = 9.60 kN , VH= 1.65 kN+++Results for Section BH45o 5 kNHH = 3.54 kN VH = 1.65 kNVG = 9.6 kN 147 N/mFBDApply Equilibrium to Section HAHAVAMAHH = 3.54 kN VH = 1.65 kNΣ FX= 0 = 5.54 - HAHA= 3.54 kNΣ FY = 0 = 1.65 kN + VA- 0.147 kN/m * 6 mVA= 0.77 kNΣ MA= 0 = -(1.65 kN * 6 m) + MAMA= 9.9 kNm = 9.9 kNm +++VHand HHare applied in equallyand oppositely to the way they were applied to section BH.Results for Section HAHA = 3.54 kN VA = 2.53 kN MA = 9.9 kNmHH = 3.54 kN VH = 1.65 kNSummary• Equilibrium means that no unresisted forces exist• If we know a structure is in equilibrium, we can use equilibrium equations to find reaction forces on the structure• What happens if structure is statically indeterminate? What can we do?09/11/95 6Next Week• Homework 3 is due at the beginning of class• Bridges• Gunn and Vesilind: Chapter 3, Groups 5 and