Temperature Distribution-Material and Structures-Assignment Solution, Exercises of Structures and Materials

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7. . E=U4GPa k= 4.2 pefy 2 oun = 7 ye 08 Tam = O'C >, ery chess be {Com - X Hess bem cca 4 V7 dew cao Let's find the tewpevatuce distribubion first Aguouing 0 quabyadic ditithedion of the form, Tle) = at bl? + o.08u)” oud plygivg iv dee giver dempentuves ot tla cum cide ond the shade © Gide, we get Tl-0.054) = a= - 40% Te+0.05m) = -400 + blotny = 8D. > b= |ooo "the Tle) = - 40°C + 1) 000 Cfut (2+ 0.08)” AT = Te)- Tee TR) - 0° > AT =- 40°C t Move ‘Cfa*(2t 0-05u) —- of AT = [N000*Che 2* + M00'Ym2~ ¥)SE & docsity.com