135_Mechanics Homework Mechanics of Materials Solution

135_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 2.97 Knowing that the hole has a diameter of 33- -in., determine (a) the radius rf of the ﬁllets for which the same maximum stress oceurs at the hole A and at the ﬁllets, (b) the corresponding maximum allowable-load P if the allowable stress is 15 ksi. PROBLEM 2.97 SOLUTION For Hie circufw he)? {or ‘(J— )(i)= 0.: I875 in cl = ‘1‘— i = 3.625 1n5’13—F”O:L8;;-‘- o 0517 Am = dt= (3 ewxgy 13594 a. From FE: 2.64 0. KW = 2,31 6' = w “um Ana? cm 13-- AME“ = ‘____—"3§f';f“‘ = 7. 25 xx,” .4 (a) For -P.‘Ue+ .D = H a».J a a 2.5’ m \$2: 27—; = [.60 Ass = At = (2.530%) r 0.93.75 A." 6-“, KﬂAf'P KR“ = A3336...” 3 (mfg-Ems) __, mm Fm... Fe: 2.64!» 3E“ are P;%O.l7JI=(O.‘l7‘7l(2-5')=' 0.143 in 4 2.98 For P = 8.5 kips, determine the minimum plate thickness t required if the PROBLEM 2'98 allowable stress is 18 ksi. SOLUTION M H: impel n. = & in clﬁ 2.2—1.0 z L2 in“. 1‘ _ l/z‘._ A "' ’.1 " 0-4,? “(I mm M H»: ﬁllet ‘D= 2-1 }n :93: LG in =39: = 125% = 1.375 res *3 = 0.275“ t. 1‘: - 01.37;“ 0,23“ From F345 2.9+ L K 7' [-10 6;,“ 1' 5&5“ = alga—E- _ KP __ (1.3- OKs-.5) __ h t' 5136” " (1.6)(l3) " 0'50 The J’mer waive is Ht: requtweol minimum Phi-e {4“ places t= O. 8.7 in 4 ...
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