CHAPTER 9 - PROBLEM 9.1 Determine by direct integration the...

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PROBLEM 9.1 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. SOLUTION At 5 2 , : x ay b b k a == = or 5 2 b k a = 52 5 2 5 2 or ba yx xy ab ∴= = 3 1 3 y dI x dy = 6 5 6 5 3 1 3 a yd y b = Then 6 5 6 5 3 0 1 3 b y a I y b = 11 5 6 5 3 0 15 311 b a y b = 11 5 6 5 3 5 33 a b b = or 3 5 33 y I = W
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PROBLEM 9.2 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. SOLUTION At 2 0, 0: 0 x yk a c = == + 2 c k a = − ,: x ayb bc = 2 b k a ∴= Then () 2 2 b yx a b a = −− + ( ) 22 2 2 b x ax a b a = + + Now 4 3 2 2 2 2 y bb dI x dA x ydx x x bx bx dx a a  = −+− +   43 2 2 x xd x a a =− + Then 2 0 2 a yy I dI x x dx a a −+ ∫∫ 54 2 0 12 a xx b a a + 33 3 21 1 2 5 aa a =+= 3 3 10 y ab I = W
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PROBLEM 9.3 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. SOLUTION By observation h yh x b =− 1 x h b    Now ( ) 22 y dI x dA x h y dx == 2 1 x x hh d x b 3 hx dx b = Then 34 4 0 0 44 b b yy hx hx hb Id I d x bb b = = ∫∫ 3 4 y bh I = W
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PROBLEM 9.4 Determine by direct integration the moment of inertia of the shaded area with respect to the y axis. SOLUTION Have 2 yk x c = + At ( ) 0, : 0 x ybbk c = == + or cb = At () 2 2, 0 : 0 2 x ay k a b = ==+ or 2 4 b k a =− Then 2 2 4 b yx b a = −+ ( ) 22 2 4 4 b ax a Then ( ) 2 2 ,4 4 y b I xdA dA ydx a x dx a = ( ) 2 2 2 4 4 aa y b I x a a ∫∫ 2 35 2 2 4 4 a a bx x a a ( ) ( ) 33 55 2 83 2 3 20 bb a 73 1 32 0 ab 3 47 60 y I = W
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PROBLEM 9.5 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis. SOLUTION At 3 2 , : x ay b b k a == = or 3 2 b k a = 3 2 3 2 b yx a ∴= 2 x I ydA dA xdy 2 5 2 5 2 0 b a yy d y b = 17 5 17 5 22 55 0 17 17 b aa b y bb = 3 5 or 17 x I ab = W
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PROBLEM 9.6 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis. SOLUTION At 2 0, 0: 0 x yk a c = == + 2 c k a = − , x ay bb c = 2 b k a = − Then () 2 2 b yb xa a =− Now ( ) 22 x dI y dA y xdy From above 2 2 a x ab y b −= Then 2 1 y xa a b and 2 1 y x aa b = −+ Then 2 11 x y dI ay dy b  =+   and 2 0 b xx y I dI a y dy b + ∫∫ 3 2 0 0 1 36 b b yy y d y
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PROBLEM 9.6 CONTINUED For the second integral use substitution 1 1, ( 1 ) y ud u d y y b u bb = −⇒ = = − 01 0 yu ybu = = = = Now 1 2 22 2 00 1( 1 ) y yd y b u u d u b  −=   ∫∫ 35 3 5 7 1 2 2 2 2 0 0 33 1 1 242 2 357 buu u d u b u u u =− + + 3 7 08 43 0 1 6 1 0 5 1 0 5 b −+ =+ + = = Then 3 16 51 31 0 51 0 5 x ba b I aa b = or 3 17 35 x I ab = W
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PROBLEM 9.7 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis. SOLUTION By observation h yh x b =− 1 x h b    or 1 y xb h Now ( ) 22 x dI y dA y b x dy ==− 2 by ybb d y h + 3 by dy h = Then 34 4 0 0 44 h h x by by bh Id y hh h == = 3 or 4 x bh I = W
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PROBLEM 9.8 Determine by direct integration the moment of inertia of the shaded area with respect to the x axis.
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CHAPTER 9 - PROBLEM 9.1 Determine by direct integration the...

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