2.5-15 a - SECTION 2.5 Thermal Effects 163 COMPATIBILITY...

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Unformatted text preview: SECTION 2.5 Thermal Effects 163 COMPATIBILITY EQUATION ' Reactions must be equai. _ P (SI—firs 0r .'.RA=RB P=2RH REL-7' fl _ BEE : 5 (Eq. 1) Substitute for RB in Eq. (1): BEA EA ‘ - 'ZPL PL PL — ._____ = S or 2 "~— <- Eocuaaauut EQUATION 3 EA BEA 65A Ra 2 I‘CflCliOT! Hi C11d A (10 the left) NOTE: The gap closes when the load reaches tlte vaiue PM. When the load reaches the value P, equal to GEM/L. the reactions are equal (RA = R5 3 PE). Ra “it“ Re When the load is between PM and P. RA is greater titan R”. ll’the load exceeds P. R3 is greater than“ RA. R5- = reaction at end B (to the left) , P ll Problem 2.5-15 Pipe 2 has been inserted snugly into Pipe 1, Pipe 1 (steel) but the holes for a connecting pin do not line up: there is a gap 5. _ _ G‘lP r The user decides to apply either force P. to Pipe 1 or lbrce P3 to / .i i‘ i | Pipe 2. whichever is smaller. Determine the foliowing using the numerical properties in the box. .74....» RA Pipe 2 (brass) {a} If only P] is applied, [ind P1 (kips) required to close gap 5; il’a pin is then inserted and P1 removed. what are reaction ‘3 \ -: forces RA and R3 for this load case? P3 k Pl at L; (bl lt‘only P3 is applied, lind P3 (kips) required to close gap 5:. \\ [.1 il' a pin is inserted and P3 removed, what are reaction P: at 7,: forces RA and R3 for this load case? ___ (c) What is the maximum shear stress in the pipes, for the Numerical properties loads in (a) and (b)? * E El = 30,000 lcsi. £3 = 14.000 ltsi cc] = 6.5 a iii—61°F. a: = ll x 104W?“ Gap 3:005 in. . Ll = 56 in., (1'1 x 6i]1..i'1= 0.5 in.._ A1: 8.64 in.2 L1 = 36 ill., a: z 5 in.. {2 = 0.25 in., A: = 3.73 in.l (d) It's temperature increase AT is to be applied to the entire structure to close gap 5 (instead rgfcrppb'itigflJrres P , and P2), find the ATrcquired to close the gap. Il’a pin is inserted after the gap has closed, what are reaction Forces R,l and RH for this case? (e) Finaliy, if the structure (with pin inserted) then cools to the original ambient temperature, what are reaction forces RA and RB? Soluitan 2.545 (a) find reactions at A (SE B for applied force P1: first W S _ -_ _ ‘ RB—WWWWMWM— IRE—33.2% <— compute P1, requiietl to ciose gap ( Ll L1 ) . ____ + _ EIAI . ' EJAI 1529*: P; = s P. m 23L4§ups <-—- Ll . RA 2 "R13 (— stat—indet analysis with Rg as the redundant {bl lind reactions m A & B [-0]. applied force P3 I-'l I L” EaA-i 5 2 "' (7‘ t = F ( ‘l‘ _ 2 _ 'I 1 z ' {- éw—r Bl s t as \n EIAI‘ 132%) P3 L: s P_ l45.l lips compatibility: 5.3. + 6m : O 3 analysis after removing P2 is same as in (a) so reaction forces are the same ...
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