Question

Consider the idealized beam cross-section shown in the figure. The simplified piece-wise constant temperature-induced change in Young's modulus, the channel beam cross- section no longer has an axis of symmetry. Within the top flange, E=0.8Eo and AT =2 To. Within the web, E =0.9Eo and AT = To. Within the bottom flange, E = Eo and AT-O. (i) Locate the modulus weighted centroid. (ii) Calculate the area moment of inertia about the z axis (iii) Determine the area product of inertia 20 8t 121 t, typ yo 8t *

Answer 1

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A Z we have Erē & E=QT - 0= EE >> TERET B lat t st Stoen at top Henge, top = 2(0-85)(270) = 1.62EoTo Ž stress at web, web =0 = 2 (0.96)(To) = 1.92 EoTo #Y Stren at bottom flange, bottom = 2(5) (To) = 2E, TO (i) Location of modules of weighted centroid, Στ2 from y-aura (1.6dEsto x lat +0.9E$T)6+ +(5 To) x 0 04.604 Fotot =) ZE 3,5EoTo 1.6dEoTot 0.9.Esto + xeoto (7) e in given by TELE = 20 P ) * [tx&$} +(£xest)eft arby"]x2 +[1729 E ) Izz 12 ZE = 7.03€ | from the y-axers (ii) Area moment of inertia about Z-axels, Tea les élustefjes + Cupertinent litxt)x(8+) -952z= 6043.34* (iii) Area product of prestia about yoz axels 0 +(9.4 xt) (4:34) (194) *[0 +(1+x)(04)64]+[0+ks!4M2800) Iyz - Iyz = 967.5++