Ra p rb 0 equilibrium problem

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Unformatted text preview: (a) (b) (c) (d) NOTE: INVOKE SIMPLE INTUITION HERE i.e., the lower sec>on CB is compressed (σ < 0) while the upper sec>on AC is extended (σ > 0) RA a RB b (tension) C B = (compression) AC = EA EA ²་  separately compute displacement of each “cut” sec>on ²་  RA and RB represent magnitudes of upper and lower reac>on forces con>nued next page M. Mello/Georgia Tech Aerospace 11 RA ²་  EQUILIBRIUM EQUATION RA ²་  COMPATIBILITY CONDITON RA B = AC + A RB RB (b) (a) (c) (d) RECALL, NEED TO DETERMINE: ²་  Reac>on forces RA and RB ²་  Displacement δc of point C = RA b EA = RB b EA ²་  Stresses within the bar (above and below point C) RB RA += = c c A A c+ 1/17/13 c CB = 0 (2A) P + R B = 0 (1) RA a RB b = 0 (2B) EA EA (tensile + compressive) ²་  COMBINE EQS (1) and (2) AND SOLVE FOR INTERNAL FORCES RA , RB. RA P + RB = 0 (1) AND RA a = B b (2) R bP bP RA = a+b ; RA = L INTERNAL force in AC aP aP RB = a+b ; RB = L INTERNAL force in CB ²་  DISPLACEMENT IN AC AND CB P ab displacement ABOVE POINT C (in AC) c+ = L P ab displacement BELOW POINT C (in CB) c = L ²་  STRESS IN AC AND CB P b Pa c= c+ = AL AL M. Mello/Georgia Tech Aerospace (stress in AC) (stress in CB) 12 EXAMPLE 2- 5 GIVEN: ²་  Solid cylinder (S) is encased in hollow circular Copper tube (C) ²་  CYLINDER AND TUBE ARE COMPRESSED BETWEEN RIGID PLATES OF A TESTING MACHINE BY COMPRESSIVE FORCE (P). ²་  AS = Cross- sec>onal area of steel cylinder ²་  Ac = Cross- sec>onal area of Copper tube ²་  Es = steel modulus of elas>city ²་  Ec = Copper modulus of elas>city ²་  L = Length of cylinder and tube DETERMINE: ²་  Compressive forces Ps and PC in cylinder and tube, respec>vely ²་  Corresponding compressive stresses σs and σC ²་  Shortening δ of the en>re assembly SOLUTION: P Fvert = 0 ²་  EQUILIBRIUM EQUATION Ps + Pc P = 0 ²་  COMPATIBILITY CONDITON steel = C opper ²་  FORCE- DISPLACEMENT 1/17/13 s = Ps L E s As c = Pc L E c Ac M. Mello/Georgia Tech Aerospace 13 EXAMPLE 2- 5 (cont.)...
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