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# 2.5-7 - PrDIJIEITI 2.5 A circular steel rodAB{diameter:11 =...

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Unformatted text preview: PrDIJIEITI 2.5-? A circular steel rodAB {diameter :11 = 1.!) in., length Id'1 “'1 L1 = 3J3! ﬂ) has a bronze sleeve (outer diameter d2 = 1.25 in., length A 3 L2 = 1.!) ﬂ) shrunk onto it so that the two parts ane seeuneljur bonded [see ﬁgure}. Calculate the total elongation 5 of the steel bar due to a temperatlme rise 311'" = 500°}? {Material properties are as follows: for steel, E, = 30 X It!"3 psi L1 and as = 6.5 X Ill—SIDE for bronze, Es = 15 X 105 psi and Inll = 11 X Ill—EFF.) Bnlutlnn 2.5-? Steel rod with bronze sleeve 11 d1 {:2 B SUBS'ITTUTE NUMERJCAL VALUES: #E) oJr = 15.5 X Ill—EFT “a = 11 X lﬂ‘GFF . L2 . sissy-clown Eb=15><lﬂﬁpsi L1 a, = 1.0 in. L1 = 35 in- L; = 12 “1- A, = E sf = 0.13540 in.2 ELDNGATICIN OF III-[E TWCI ﬂUTER PARTS OF THE BAR 51 = “llﬁTl'U-q — L2} = {as x Ill—GFFHSEIO‘THEE in. — 12 in.) a, = 1.25 in. 11' A, = Em; — a? = 0.44179 in?- = ﬂﬂTBDEI “1' as" = sno°F L, = 12.0 in. ELoNoMtoN on no; tam-nus Pant" o1= ms EAR '52 = 13.134493 i“- The steel rod and bronze sleeve lengthen the same TOTAL MGMICIN amount, so the}.r are in the same condition as the bolt and sleeve of Example 2—3. Thus, we can calculate ﬂie elongation from Eq. (2—21]: _ [as E: As + [1", Eb Animal-1L2 _ sm+nm s=s, +s2 = 0.123 in. 4— 32 ...
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