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2.5-10 a

# 2.5-10 a - 158 CHAPTER 2 Axiaily Loaded Members COppBl“ E...

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Unformatted text preview: 158 CHAPTER 2 Axiaily Loaded Members COppBl“ E. ___ l8 ‘Dlllil liql a,“ z 9 5 X love/BF SUBSTITUTE NUMERICAL VALUES: Aluminum: a1, = 10,000 ksi _ (3.5 >< 10“)“ F_)(IUO°F)(18.000 more in?) PU m PC I 8' p, 0 a", = 13 >< lD‘ﬁloF , ' l + (”Xi") IO 2.0 Use the results of Example 2-9. = 4500113 Find the forces Pr, and P‘. in the aluminum bar and FREE-BODY DIAGRAM or pm A? THE LEW END copper bar, respectively, from Eq. (2-19). PL. Replace the subscript “"S in that equation by “n" (for 7 aluminum) and replace the subscript “B" by “c" (for p“ copper). because a for aluminum is larger than a for P. copper. ”:7. p = P A; W V = shear force in pin [1 C EUAH + EL'AL' = PL]: Note that P” is the compressive force in. the aluminum _ . . . . . __ 2,250 lb bar and PL. ts the combmed tensne force in the two copper bars. 7' = average shear stress on cross section of pin P _ P _ (at, — ac._){AT)Ei.AU ' _3_ l _ 2.2501b ” w " W 1 Jr Ei-Ac ' A7: 0150331119 Ear-l” r m 15.0 ksi <— Problem 2.5-“) A rigid bar ABCD is pinned at end A and supported by two cables atpoints B and C (see ﬁgure). The cable at B has nominal diameter (1,; w—« 12 mm and the cable at C has nominaldiameter dc 2 20 mm. A load P acts at end 1) of the bar. What is the allowable load P if the temperature rises by 6(PC and each cable is required to have a factor of safety of at least 5 against its ultimate load? ' (Note: The cables have effective modulus ot’elastieity E = 140 GPa and coefﬁcient of thermal expansion a r 12 K lOWﬁ/“C. Other properties of the cables can be found in Table 2-1, Section 2.2.) ‘ Solution 2.5-10 Rigid har supported by two eahies FREE~BODY omen-m 0F BAR ABCD From Table 2-1: An = 76.7 mm2 E = 140 GPA or : 600C AC : 173 mm2 a = [2 x 10‘“/°c EQUATION 0F EQUILIBRIUM EMA = (.l m (A THtIZb) + TC-(4b) - P(5b) : (l or 2TB + 4TC = 51’ (Eq. I) ' TRAV T3 = force in cable B TC = force in cable C :11, =12 mm 71.; = 20 mm ...
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