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W. E. Hudson. ) DYNAMICS (continued) Figure Area & Centroid A = 4ab/3 y xc = 3a/5 b C a yc = 0 x b y C b n y = (h/b )x 31 h x b y = (h/b1/n)x1/n C b 3 I yc = 16a b 175 3 I y = 4a b 7 A = 2ab/3 Ix = 2ab3/15 xc = 3a/5 Iy = 2ab3/7 rx2c = rx2 = b 2 5 ry2c ry2 2 = 12a 175 h x A = bh (n + 1) n +1 xc = b n+2 h n +1 yc = 2 2n + 1 n bh n +1 n +1 xc = b 2n + 1 n +1 yc = h 2(n + 2) A= bh 3 3(3n + 1) hb 3 Iy = n+3 rx2 = b 2 5 ry2 = 3a 2 7 h 2 (n + 1) 3(3n + 1) n +1 2 ry2 = b n+3 Ix = rx2 = n bh 3 3(n + 3) n Iy = b3h 3n + 1 rx2 = Ix = Product of Inertia I xc yc = 0 I xy = 0 = 3a 2 7 yc = 3b/8 n C y I xc = I x = 4ab 3 15 x a y (Radius of Gyration)2 Area Moment of Inertia n +1 2 h 3(n + 1) n +1 2 ry2 = b 3n + 1 Housner, George W.

The solution to this differential equation is x(t) = C1 cos x(t) = (k m ) t + C2 sin (k m ) t , where the displacement in the x-direction, and C1, C2 = constants of integration whose values depend on the initial conditions of the problem. ACθ = rω, and v = BCθ , where The quantity b C = the instantaneous center of rotation, θ = the rotational velocity about C, and k m is called the undamped natural frequency ωn or ωn = k m AC, BC = radii determined by the geometry of the situation. • Timoshenko, S.

27 DYNAMICS (continued) From the static deflection relation The solution to the equation of motion is ( θ0 = The period of vibration is θ 0 = τ = 2π ωn = 2π m k = 2π δ st g If the initial conditions are x(0) = x0 and x (0 ) = v0 , then If the initial conditions are x(0) = x0 and x (0 ) = 0 , then The period of torsional vibration is τ = 2 π ω n = 2π IL GJ which is the equation for simple harmonic motion where the amplitude of vibration is x0. Torsional Free Vibration θ + ω2 θ = 0 , where n ω n = k t I = GJ IL , I = the mass moment of inertia of the body, G = the shear modulus, J = the area polar moment of inertia of the round shaft cross section, and L = the length of the round shaft.