2.5-15 a

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern