COSMOS Chapter 9 Solution 181

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 181. (a) First compute the moments of inertia using Figure 9.28 and the parallel-axis theorem.  a 2  a 2  13 1 2 2 2 Ix = Iz = m 3a + a + m   +  2   = 12 ma 12 2        1 3 2 I y = ma 2 + m ( a ) = ma 2 2 2 ( ) Next observe that the centroidal products of inertia are zero because of symmetry. Then 0 1  a  a  I xy = I x′y′ + mx y = m  ma 2  − 2  = − 2 2 2     0 1  a  a  I yz = I y′z ′ + my z = m  −  =− ma 2  2 2  2  2  I zx = I 0 ′ ′  a  a  1 2 + mz x = m    = ma  2  2  2 Substituting into Equation (9.56)  13 3 13  2 2 K3 −  + +  ma K  12 2 12  2 2 2  13 3   3 13   13 13   1  1   1  2 +  ×  +  ×  +  ×  −  − − − −    ma    12 2 2 12 12 12 2       2 2     2 2  ( 2 2 2 ) 2 K 3  13 3 13   13   1   3   1   13   1  1  1   1   2 −  × ×  −    − −    −  − − 2− −    ma = 0       2 2   2 2   2   12 2 12   12   2 2   2   2   12   2 2  ( Simplifying and letting K = ma 2ζ yields ζ3 − 11 2 565 95 ζ + ζ − =0 3 144 96 continued ) COSMOS: Complete Online Solutions Manual Organization System Solving yields ζ 1 = 0.363383 ζ2 = 19 12 ζ 3 = 1.71995 The principal moments of inertia are then K1 = 0.363ma 2
K 2 = 1.583ma 2
K3 = 1.720ma 2
(b) To determine the direction cosines λx , λ y , λz of each principal axis, we use two of the equations of Equations (9.54) and Equation (9.57). Thus (9.54a) ( I x − K ) λx − I xyλ y − I zxλz = 0 − I zxλx − I yz λ y + ( I z − K ) λz = 0 (9.54c) λx2 + λ y2 + λz2 = 1 (9.57) Note: Since I xy = I yz , Equations (9.54a) and (9.54c) were chosen to simplify the “elimination” of λ y during the solution process. Substituting for the moments and products of inertia in Equations (9.54a) and (9.54c) 1    13 2  1  ma 2  λ y −  ma 2  λz = 0  ma − K  λx −  −  12  2   2 2  1   1   13  −  ma 2  λx −  − ma 2  λ y +  ma 2 − K  λz = 0 2   12   2 2  or and 1 1  13  − ζ  λx + λ y − λz = 0  2 2 2  12  (i) 1 1  13  − λx + λ y +  − ζ  λz = 0 2 12 2 2   (ii) Observe that these equations will be identical, so that one will need to be replaced, if 13 1 19 ζ = −ζ = − or 12 2 12 Thus, a third independent equation will be needed when the direction cosines associated with K 2 are determined. Then for K1 and K 3 continued COSMOS: Complete Online Solutions Manual Organization System  13  1  13  1   Eq.(i) – Eq.(ii)  − ζ −  −   λx +  − −  − ζ   λz = 0  2   12  2  12 λz = λx or 1 1  13  − ζ  λx + λ y − λx = 0  2 2 2  12  Substituting into Eq.(i) 7  λ y = 2 2  ζ −  λx 12   or Substituting into Equation (9.57) λx2 or 2  7  2  +  2 2  ζ −  λx  + ( λx ) = 1 12     2  7    2 + 8  ζ −   λx2 = 1 12     (iii) K1 : Substituting the value of ζ 1 into Eq.(iii) 2  7  2   2 + 8  0.363383 −   ( λx )1 = 1 12     or and then ( λx )1 = ( λz )1 = 0.647249 ( λ y )1 = 2 7  2  0.363383 −  ( 0.647249 ) 12   = −0.402662 ∴ (θ x )1 = (θ z )1 = 49.7° (θ y )1 = 113.7°
K 3 : Substituting the value of ζ 3 into Eq.(iii) 2  7  2   2 + 8 1.71995 −   ( λx )3 = 1 12     or ( λx )3 = ( λz )3 and then ( λ y )3 = 2 = 0.284726 7  2 1.71995 −  ( 0.284726 ) 12   = 0.915348 ∴ (θ x )3 = (θ z )3 = 73.5° (θ y )3 = 23.7°
continued COSMOS: Complete Online Solutions Manual Organization System K 2 : For this case, the set of equations to be solved consists of Equations (9.54a), (9.54b), and (9.57). Now ( ) − I xy λx + I y − K λ y − I yz λz = 0 (9.54b) Substituting for the moments and products of inertia. 1 1     3  ma 2  λx +  ma 2 − K  λ y −  − ma 2  λz = 0 − − 2 2 2 2 2       or 1 3 1  λx +  − ζ  λ y + λz = 0 2 2 2 2 2  (iv) Substituting the value of ζ 2 into Eqs.(i) and (iv) 1  13 19  −  ( λx )2 + λy  12 12 2 2   ( )2 − 12 ( λz )2 = 0 1 2 2 3 19  −  λy  2 12  ( )2 + 2 1 2 ( λz )2 = 0 ( λx )2 +  or − ( λx )2 + 1 λy 2 ( λx )2 − 2 λy 6 and ( )2 − ( λz )2 = 0 ( ) 2 + ( λz ) 2 = 0 Adding yields ( λ y )2 = 0 and then ( λz )2 = − ( λx ) 2 Substituting into Equation (9.57) 0 ( λx )22 + ( λ y )2 + ( −λx )22 2 or ( λx )2 ∴ = 1 2 (θ x )2 and =1 ( λz )2 =− 1 2 ( )2 = 90.0° (θ z )2 = 135.0°
= 45.0° θ y (c) Principal axes 1 and 3 lie in the vertical plane of symmetry passing through points O and B. Principal axis 2 lies in the xz plane.