u8191E3Y6s14_1210016433_jwk572

# u8191E3Y6s14_1210016433_jwk572 - homework 13 – KIM JI –...

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Unformatted text preview: homework 13 – KIM, JI – Due: Nov 28 2007, 4:00 am 1 Question 1, chap 20, sect 2. part 1 of 1 10 points The concrete sections of a certain super- highway are designed to have a length of 28 m. The sections are poured and cured at 12 ◦ C. What minimum spacing should the engi- neer leave between the sections to eliminate buckling if the concrete is to reach a temper- ature of 28 ◦ C? Correct answer: 0 . 005376 m (tolerance ± 1 %). Explanation: Let : L = 28 m , T 1 = 12 ◦ C , T 2 = 28 ◦ C , and α = 1 . 2 × 10 − 5 ( ◦ C) − 1 . The extension for this temperature interval will be Δ L = Lα Δ T = (28 m) bracketleftbig 1 . 2 × 10 − 5 ( ◦ C) − 1 bracketrightbig (16 ◦ C) = . 005376 m . Question 2, chap 20, sect 2. part 1 of 2 10 points A copper rod and a steel rod are heated. At 0 ◦ C the copper rod has a length of L C , the steel one has a length of L S . When the rods are being heated or cooled, a difference of 4 . 6 cm is maintained between their lengths. Find the value of L C . Correct answer: 8 . 43333 cm (tolerance ± 1 %). Explanation: Let : d = 4 . 6 cm , α C = 1 . 7 × 10 − 5 ( ◦ C) − 1 , and α S = 1 . 1 × 10 − 5 ( ◦ C) − 1 . For d = L S- L C = 4 . 6 cm to be constant, the rods must expand by equal amounts: α C L C Δ T = α S L S Δ T L S = α C L C α S d = α C L C α S- L C α S d = α C L C- α S L C , so L C = α S d α C- α S = bracketleftbig 1 . 1 × 10 − 5 ( ◦ C) − 1 bracketrightbig (4 . 6 cm) 1 . 7 × 10 − 5 ( ◦ C) − 1- 1 . 1 × 10 − 5 ( ◦ C) − 1 = 8 . 43333 cm . Question 3, chap 20, sect 2. part 2 of 2 10 points Find the value of L S . Correct answer: 13 . 0333 cm (tolerance ± 1 %). Explanation: L s = L c + d = 8 . 43333 cm + 4 . 6 cm = 13 . 0333 cm ....
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## This note was uploaded on 05/05/2008 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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u8191E3Y6s14_1210016433_jwk572 - homework 13 – KIM JI –...

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