Answers
To find the moment of inertia with respect to x' axis I have used parallel axis theorem.
a a=2.2 in Ga 0 اع a 2a x! е regions ①, ② and ③ as shown in Let consider, three figure. Region Ai faxa=6² yu AcYE a 2 a x60² = 3a3 za 3a23a - gap (2) (a +20) a = za atza 2 3a 3 [at2a) a = 30² atza 2 39 2 39² x 39 / 4 The centnoid 7 is given by the formula, ΣΑ:Je ya ΣΑ: 993 2 =a (323 + 1293 7 162? + 342 +3a2) 12a2 : The centro id ŷ of the chanell's enoss sectional area is ý za = 2.2 in Ans,
To find moment of inertia :- By the parallel aris theonem, the moment of inentia with respect to the x axis is, Isel [ Izę + Az dye whene is the moment of inentia of each element about itself about a homizontal axis and A į is ainea of each element 9 - Ye and dyi Ize dy:=Y-Yi Ai dy? 2 1 1 3 xGaxa = hat baxa (a-as) = a a4 12 3 12 iz xa xCQ +20) (2-39) = -a Ba?sa=2a1 Ta xa x27Q? qa ® tę xa x ca+2439 =as 30² xa 3 (a-30) old k4 shat 2 14 In' Iri + Az dy 12 a - + gas + 3 4 12 ha + + 24 24294) + 2 LŽ a4) ai (4+9+3) = sat Ini =804 - 8(2-2) iny " The moment of inertia, Izel = 187.4 in! 187.4 in 4 Ans.
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Solution in the uploaded image
Solution: a = 2.2 in 6 a Taking 2D view, Consider the given figure is made up of three regular In I Ba c A 3= shapes 0 2 2 6. A Area each shapes: A = A= ax29=2a² (a+bataxa 8a2 each & shape will be Hence, 2a g, = Y 2 = 9+ 22 Y 3= a/ A3 centraid of along their geometrie centru using super possition, ý = Alyit Asyz +A343 _ 20x (29) + 29 (829) + Bara 292 + 2a²+ & 2 A tA ₂ +A3 2.2 in q 9 3 12a 12a2 9 4 a (2013 + I = I 2 = 12 t using parallel anis Theorem and superposition principle, Jinding so I geach each part about (22x9) (a)?? opf 89 (6) (Baxa)(%)' = Ij = F + 1 2 + Iz = 2 x 20 + 8y = 8 at = 2(2-2) = 187.4004 (you I3= + 12 3 3 # hit like understood the solution!!
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