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LEC 3
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   Look at equilibrium of Pin A: F  X  = 0 P   cos θ  +  F  3  = 0 F  3  = − P   cos θ Note  θ  = 60 ◦ , so: P F  3  = − 2So:EXAMPLE:4   Q: What are the reactions at the supports?FBD: F  x  = 0 R A x +  R D x = 0 F  y  = 0 R A y +  R D y − P   = 0 M  A  = 0 − 2 dP   +  dR D x = 0Solve. R D x = 2 P    R A x = − 2 P    R A y +  R D y =  P    Note: BD is a 2-Force member.Draw FBD of BD and Pin D.5   Solve for Pin D. F  x  = 02 P   +  F  BD cos θ  = 0 F  y  = 0 R D y +  F  BD sin θ  = 0Solve.2 P F  BD  = − cos θ 2 P R D y −  sin θ  = 0cos θR D y = 2 P   tan θ Note:  θ  = 45 ◦ , so: R D y = 2 P  Substituting: R A y + 2 P   − P   = 0 R A y = − P  Check: Statically Determinate  : A situation in which the equations of equilibrium de-termine the forces and moments that support the structure.EXAMPLE:6   FBD F  x  = 0 R A x = 0 F  y  = 0 R A y +  R B y − P   = 0 M  B  = 0 − aR A y +  M  A − ( l − a ) P   = 0Note: 4 unknowns and 3 equations.This is  statically indeterminate  .7
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