. Applied calculus; principles and applications . Remark. — A full treatment of the subject of moment ofinertia and product of inertia is beyond the scope of thisbook. What has been given has been confined for the mostpart to areas, as that part of the subject has more immediateapplication in engineering. 350 INTEGRAL CALCULUS 186. Deduction of Formulas for Moment of Inertia. — 1. Rectangle of base h and altitude h: L = fy dA = Jbf dy = tV W. §)■=! lab =^h-\-Aa^j^ h¥ + bh hh\ ly = jx^dA = r hx^dx = j\¥h. bh Again, Iab = £bydy = ibh\Ix = tV^^ J = |6^; for square. A Y f X-^-A -- - _i__. X B 2. T
. Applied calculus; principles and applications . Remark. — A full treatment of the subject of moment ofinertia and product of inertia is beyond the scope of thisbook. What has been given has been confined for the mostpart to areas, as that part of the subject has more immediateapplication in engineering. 350 INTEGRAL CALCULUS 186. Deduction of Formulas for Moment of Inertia. — 1. Rectangle of base h and altitude h: L = fy dA = Jbf dy = tV W. §)■=! lab =^h-\-Aa^j^ h¥ + bh hh\ ly = jx^dA = r hx^dx = j\¥h. bh Again, Iab = £bydy = ibh\Ix = tV^^ J = |6^; for square. A Y f X-^-A -- - _i__. X B 2. Triangle about the axes: (a) Through the apex parallel to the base. (5) Through the center of gravity parallel to the (c) Through the base. (1) (2)(3) J = h + h = f^hh + ^Vh = ^{V + h^). (4) (2) DEDUCTION OF FORMULASLet the base be b and the altitude h. ,. J r . , b¥ .bh /h^\ hW 0 y 1 1 / \ 1 /////////a y//m. 1h / V \ - \....... b- / 3. Circle: (a) Polar moment of inertia, axis through center.(6) Moment of inertia about a (a) J = Tr^ dA = J\ tt/ dp = ^ wd^32* 351 (1)(2)(3) (1) 352 INTEGRAL CALCULUS For a sector with angle 6, J = jdpdp = 4 _ ^1 _ 7rr^ (2)(3) (4) /. + /, = 27, = JFor a circular quadrant, 7, = 7For a semicircle, Ix — I (6) Since for the circle the moments of inertia about alldiameters are equal, 16 _ _ _ 7rr*^~2^ ~ 8 * 4. Ellipse: _ 1 ^2 1 (5)(6)(7) 7, = Jx dA = ^/x2 (a2 - a;2)ld(a; = ^. /x + 7, = ^(a2 + 62). (1)(2)(3)
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