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quiz_3 - Solution 2.3-8 Bar with a hole d = diameter of...

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Unformatted text preview: Solution 2.3-8 Bar with a hole d = diameter of hole SHORTENING 8 OF 1115 BAR N-L- 3_ _'._’=_‘F:2fi EiAI E A,- P L14 LM LIZ E 1.— + r + 1.— 3w:2 ‘ 0'2) Id? Id; PL 1 1 2 _;E(d§—E+E+E) (Eq. 1) NUMERICAL VALUES (DATA): 3 = maximum allowable shortening of the bar = 8.0 mm P=llOkN L=1.2m E=4.0 GPa dI = 100mm dm = maximum allowable diameter of the hole d2 = 60mm SUBSTITUTE NUMERICAL VALUES INTO E0. (1) FOR 6 AND SOLVE FOR J = dm: UNITS: Newtons and meters 0.003 = (110000103) 614.0 x 10 1 x 1—1— + 1 + 2 1 (0.1)2 — d1 (0.1 )2 {0.06}2 1 1 2 761.598 = —— + —— + 0.01 — d2 0.01 0.0036 1 —— = 761598 — 100 - 555.556 = 106.042 0.01 — d2 d2 = 569.81 x10“ "12 d = 0.0238? In dm,=23.9mm 4-— ...
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