A similar calculaoon for concrete demonstrates

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Unformatted text preview: , BOLT and SLEEVE TEND TO ELONGATE ²་  NOTE NO RIGID WALLS THIS TIME TO CANCEL OUT DISPLACEMENTS (SO ENTIRE ASSEMBLY LENGTHENS) ²་  SLEEVE TENDS TO ELONGATE MORE THAN BOLT (SINCE αs > αb) ²་  CONSTRAINT BY BOLT HEAD, NUT, AND WASHERS FORCES SLEEVE INTO COMPRESSION ²་  BOLT SHAFT FORCED INTO TENSION 1/17/13 M. Mello/Georgia Tech Aerospace steel sleeve (iniOal) T s L T steel sleeve (thermal) L steel sleeve (mechanical) Rs Rs s Rs s steel sleeve (final length) L net displacement T of sleeve: = s Rs s 12 Example 2- 8 (cont.) Displacement of bolt: T s steel sleeve (thermal) BOLT IS IN TENSION L 1/17/13 T b T L Rb Rb b M. Mello/Georgia Tech Aerospace Net elongaOon T of BOLT: = b + Rb b 13 Example 2- 8 (cont.) SOLUTION: ②  EQUILIBRIUM ~ ~ R a = Rb Ra = Rb ③  COMPATIBILITY ²་  NET DISPLACEMENT = SLEAVE THERMAL EXPANSION MINUS SLEEVE MECHANICAL COMPRESSION ²་  OR, ²་  NET DISPLACEMENT = BOLT THERMAL EXPANSION PLUS BOLT MECHANICAL EXTENSION T b + 1/17/13 T T b Rb b L Rs s Rb Rb b T Rs =s s Rb L Rs L = ↵b T L + = ↵s T L ($) E b Ab E s As Rs L Rb L + = ↵s T L ↵b T L E s As E b Ab = T s T s = ↵s T L Rs s = Rs L E b As T b = ↵b T L Rb b = Rb L E b Ab M. Mello/Georgia Tech Aerospace 14 Rs Example 2- 8 (cont.) SOLUTION: ④  SOLVE FOR LOADS Since RA = RB, it follows that: Rs = Rb = ( ↵s ↵b )( T )Es As Eb Ab E s As + E b Ab T s T (2- 19) ④  SOLVE FOR STRESSES: (SIMPLY DIVIDE 2- 19 BY THE APPROPRIATE AREA FACTOR T b Rb b L Rs s Rb Rs (↵s ↵b )( T )Es Eb Ab = = As E s As + E b Ab (2- 20a) T Rs L Rs s = ↵s T L s = E b As Rb (↵s ↵b )( T )Es As Eb Rb RL T = EbbAb = ↵b T L (2- 20b) = = b b b Ab E s As + E b Ab ⑤  INCREASE IN LENGTH OF THE ENTIRE ASSEMBLY: subsOtute either Rs or Rb into the compaObility equaOon ($) s = 1/17/13 (↵s Es As + ↵b Eb Ab )( T )L E s As + E b Ab (2- 21) M. Mello/Georgia Tech Aerospace 15 Rs Example 2- 8 : Special case soluOons Case #1 : ASSUME BOLT IS INFINITELY RIGID AND UNAFFECTED BY TEMPERATURE CHANGES Rs (↵s ↵b )( T )Es Eb Ab R = R = (↵s ↵b )( T )Es As Eb Ab = = s s b As E s As + E b Ab E s As + E b Ab Rb (↵s ↵b )( T )Es As Eb ²་  This implies E b = 1 and ↵b = 0 = = b Ab E s As + E b Ab ²་  Divide numerator and denominator by Eb ²་  Let αb = 0 (↵s Es As + ↵b Eb Ab )( T )L = ²་  Recover equaOons derived in example 2- 7! E s As + E b Ab Bar between rigi...
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This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Institute of Technology.

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