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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 Spring '09
 ZHU
 pH, Force, RL circuit, Propaganda Due

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