Analysis of Statically Determinate Structures

1 1!Idealized Structure!Principle of Superposition!Equations of Equilibrium!Determinacy and Stability!Beams!Frames!Gable Frames!Application of the Equations of Equilibrium! Analysis of Simple Diaphragm and ShearWall Systems ProblemsAnalysis of Statically Determinate Structures2 Classification of Structures Support Connectionstypical roller-supported connection (concrete)typical fixed-supported connection (concrete)typical pin-supported connection (metal)stiffenersweldweldtypical fixed-supported connection (metal)3fixed-connected jointpin-connected jointfixed supportABPactual beamL/2L/2torsional spring jointpin supporttorsional spring supportidealized beamABL/2L/2P4 One unknown.

2 The reaction is aforce that acts perpendicular to the surface at the point of unknown. The reaction is aforce that acts perpendicular to the surface at the point of unknown. The reaction is aforce that acts in the direction of the cable or unknown. The reaction is aforce that acts perpendicular to the surface at the point of of ConnectionIdealized SymbolNumber of UnknownsReactionTable 2-1 Supports for Coplanar Structures (3)F(4)F(1) Light cable rockers(2)rollers5fixed-connected collarTwo unknowns. The reactions are two force of ConnectionIdealized SymbolNumber of UnknownsReactionFMTwo unknowns. The reactions are a force and unknowns.

3 The reactions are the moment and the two (7)fixed support(5)Smooth pin or hinge(6)slider6 Idealized m4 mABactual structureidealized structure3 m4 mABF7idealized framing planABCD joistslabcolumngirderfixed-connected beamidealize beamfixed-connected overhanging beamIdealized beam8idealized framing planidealized framing plan9 Tributary beamgirderslabdeck girderpierveihicleslabstringerfloor beamgirder10wall footingslab on gradecolumnlandingstairsfoundationwallba sement1st floorsupported slab2nd floorbeambeamjoist slabjoistspread footingspandrelbeam11 ABCDEF idealized framing plan4 m4 m2 m2 kN/m2 One-Way midealized beam1 m1 m1 m1 m1 kN/m2 kN2 kNidealized girderFB2 m2 m2 kN1 kN1 kN12

4 ABCDEFI dealized framing planfor one-way slab actionrequires2/12 LLL2L1L1L1/2L1/2concrete slab isreinforced in two directions, poured on plane formsAgirderbeamcolumn136 m4 midealized framing planABCDL2/L1 = < 24 m4 midealized framing planABCDL2/L1 = 1 Two-Way m4 m4 kN/m245o45o2 m45o45o2 m2 mACidealized beam1 kN/m2 m2 mAB1kN/m2 m2 m 2 m2 m2 mABidealized beam, all1 kN/m2 m2 m1 kN/m14 Principle of Superposition P1d+Two requirements must be imposed for the principleof superposition to apply : 1. The material must behave in a linear-elastic manner, so that Hooke s law is valid, and therefore the load will be proportional to displacement.

5 = P/A = PL/AE 2. The geometry of the structure must not undergo significant change when the loads are applied, , small displacement theory applies. Large displacements will significantly changeand orientation of the loads. An example wouldbe a cantilevered thin rod subjected to a force at its P = P1 + P2d P2d15 Equations of Equilibrium Fx = 0 Fy = 0 Fz = 0 Mx = 0 My = 0 Mz = 0 Vinternal loadingsNMMNV16 Determinacy and Stabilityr = 3n, Statically determinater > 3n, Statically indeterminaten = the total parts of structure = the total number of unknown reactive force and moment components Determinacy17 Example 2-1 Classify each of the beams shown below as Statically Determinate or staticallyindeterminate.

6 If Statically indeterminate , report the number of degrees ofindeterminacy. The beams are subjected to external loadings that are assumed tobe known and can act anywhere on the = 3, n = 1, 3 = 3(1) Statically determinater = 5, n = 1, 5 - 3(1) = 2 Statically indeterminate to the second degreer = 6, n = 2, 6 = 3(2) Statically determinater = 10, n = 3, 10 - 3(3) = 1 Statically indeterminate to the first degreehinge19 Example 2-2 Classify each of the pin-connected Structures shown in figure below as staticallydeterminate or Statically indeterminate . If Statically are subjected to arbitraryexternal loadings that are assumed to be known and can act anywhere on = 7, n = 2, 7 - 3(2) = 1 Statically indeterminate to the first degreer = 9, n = 3, 9 = 3(3) Statically determinate21r = 10, n = 2, 10 - 6 = 4 Statically indeterminate to the fourthdegreer = 9, n = 3, 9 = 3(3) Statically determinate22 Example 2-3 Classify each of the frames shown in figure below as Statically Determinate orstatically indeterminate .

7 If Statically indeterminate , report the number of degreesof indeterminacy. The frames are subjected to external loadings that are assumedto be known and can act anywhere on the = 9, n = 2, 9 - 6 = 3 Statically indeterminate to the third degreer = 15, n = 3, 15 - 9 = 6 Statically indeterminate to the sixth degree24 StabilityPartial ConstrainsAPPAMAFAr < 3n, unstable>r 3n, unstable if member reactions are concurrent or parallel or some of the components form a collapsible mechanism25 Improper ConstraintsPABCdOOPABCdFAFBFCABCPABCPFAF BFC26 Example 2-4 Classify each of the Structures in the figure below as stable or unstable.

8 Thestructures are subjected to arbitrary external loads that are assumed to be member is stable since the reactions are non-concurrent and is also Statically compound beam is stable. It is also indeterminate to the second compound beam is unstable since the three reactions are all member is unstable since the three reactions are concurrent at structure is unstable since r = 7, n = 3, so that, r < 3n, 7 < 9. Also, this canbe seen by inspection, since AB can move horizontally without of the Equations of EquilibriumABCDEP1P2 DxDyDxBxByBxByEyExExP2P1 AyAxCxr = 9, n = 3, 9 = 3(3); Statically determinate30 ABCP2P1BP2P1 AyAxCyCxP1P2 AyAxCyCxBxBxByr = 6, n = 2, 6 = 3(2).

9 Statically determinate31 Example 2-5 Determine the reactions on the beam m1 m2 m70 kN m150 mAB32 SOLUTION+ MA = 0:By(4) - ( )(3) + ( )( ) -70 = 0 By = kN, Fy = 0:+Ay - + = 0 Ay = kN , Fx = 0:+Ax - = 0: Ax = kN , 3 m1 m70 kN mAyAxBy265 cos 60o = kN265 sin 60o = kN3 m1 m2 m70 kN m265 mAB33 Example 2-6 Determine the reactions on the beam kN/m5 kN/m12 m34 SOLUTIONA15 kN/m5 kN/m12 m Fx = 0:+Ax = 0 Fy = 0:+Ay - 60 - 60 = 0 Ay = 120 kN , + MA = 0:MA - (60)(4) - (60)(6) = 0 MA = 600 kN m12 m10 kN/m(1/2)(12)(10) = 60 kN4 m6 m(5)(12) = 60 kN5 kN/mAxAyMA35 Example 2-7 Determine the reactions on the beam shown.

10 Assume A is a pin and the support atB is a roller (smooth surface).A3 m2 m4 m7 kN/mB36 SOLUTION+ MA = 0:-28(2) + NBsin (6) + NBcos (3) = 0 NB = kN Fx = 0:+Ax - NBcos = 0; Ax = = 8 kN , Fy = 0:+Ay - 28 + = 0 Ay = kN , tan-1(3/2) = = kN2 m6 m3 mA3 m2 m4 m7 kN/mB37 Example 2-8 The compound beam in figure below is fixed at A. Determine the reactions at A,B, and C. Assume that the connection at pin and C is a m4 m6 kN/m8 kN m388 kN mCyBxBy+ MA = 0:Member AB Fx = 0:+ Fy = 0:+MA - 36(3) + 2(6) = 0 MA = 96 kN mAx - B = 0 ; Ax = Bx = 0Ay - 36 + 2 = 0 Ay = 34 kN , BxAxAyMA36 kN3 m6 mByCy - By = 0; By = Cy = 2 kN , + MB = 0:Member BC Fx = 0:+ Fy = 0:+Cy(4) - 8 = 0 Cy = 2 kN , Bx = 0 SOLUTION hingeABC6 m4 m6 kN/m8 kN m39 Example 2-9 The side girder shown in the photo supports the boat and deck.