COSMOS Chapter 9 Solution 137

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 137. m = ρV = m1 = Then: γ g tA  1 ft  490 lb/ft 3 × 0.08 in. × ( 3.6 in.)(1.2 in.) ×   2 32.2 ft/s  12 in.  3 lb ⋅ s 2 ft = 3.0435 × 10−3  1 ft  490 lb/ft 3 1 m2 = × 0.08 in. × (1.8 in.)(1.2 in.) ×   2 2 32.2 ft/s  12 in.  lb ⋅ s 2 ft = 760.87 × 10−6  1 ft  490 lb/ft 3 π 2 m3 = × 0.08 in. × (1.8 in.) ×   2 2 32.2 ft/s  12 in.  = 3.5855 × 10−3 3 3 lb ⋅ s 2 ft I x = ( I x )1 + ( I x )2 + ( I x )3 Now 2 2 1  2   1 ft  −3 lb ⋅ s  I x =   3.0435 × 10  (1.2 in.)    ft       2 2 2 1   −6 lb ⋅ s  −6 lb ⋅ s  2 2 2 2 2 2  1 ft  +   760.87 × 10  1.8 + 1.2 in +  760.87 × 10  0.6 + 0.4 in    ft  ft  18    12 in.   ( ) ( ) 2 2  2   1 ft  −3 lb ⋅ s   1  +  3.5855 × 10    (1.8 in.)    ft   4     12 in.  ( ) ( ) ( ) =  10.1450 × 10−6 + 1.37379 × 10−6 + 2.7476 × 10−6 + 20.168 × 10−6  lb ⋅ ft ⋅ s 2   = 34.435 × 10−6 lb ⋅ ft ⋅ s 2 I x = 34.4 lb ⋅ ft ⋅ s 2
continued COSMOS: Complete Online Solutions Manual Organization System ( )1 + ( I y )2 + ( I y )3 Iy = Iy 2 1  lb ⋅ s 2  2   1 ft  =   3.0435 × 10−3 3.6 in. )  (  ft   3    12 in.  2 2 1   lb ⋅ s 2  2 −6 lb ⋅ s  2 2 2  1 ft  +   760.87 × 10−6 1.8 in. + 760.87 × 10 1.8 + 0.6 in  )  (   ft  ft  18    12 in.   ( ) 2 2 2    2   1 ft 2  16  2 −3 lb ⋅ s   1 −3 lb ⋅ s   4 × 1.8  2 +  3.5855 × 10  −     (1.8 in.) +  3.5855 × 10  + 1.8  in   ft   2 9π 2  ft   3π      12 in.   ( ) ( ) ( ) =  91.305 × 10−6 + 0.95109 × 10−6 + 19.0218 × 10−6 + 25.805 × 10 −6 + 92.205 × 10−6  lb ⋅ ft ⋅ s 2   = 232 × 10−6 lb ⋅ ft ⋅ s 2 I y = 232 × 10−6 lb ⋅ ft ⋅ s 2
I z = ( I z )1 + ( I z )2 + ( I z )3 2 1   lb ⋅ s 2  2 2 2  1 ft  3.6 1.2 in =   3.0435 × 10−3 +    ft   3    12 in.  ( ) 2 2 1   lb ⋅ s 2  2 −6 lb ⋅ s  2 2 2  1 ft  1.2 in. 760.87 10 1.8 0.4 in +   760.87 × 10−6 + × + )   (   ft  ft  18    12 in.   ( ) 2 2 1   lb ⋅ s 2  2 2   1 ft  −3 lb ⋅ s  1.8 in. 3.5855 10 1.8 in. +   3.5855 × 10−3 + × )  )  ( (  ft  ft   4    12 in.   ( ) ( ) ( ) =  101.45 × 10−6 + 0.42271 × 10−6 + 17.9650 × 10−6 + 100.842 × 10−6  lb ⋅ ft ⋅ s 2   = 221 × 10−6 lb ⋅ ft ⋅ s 2 I z = 221 × 10−6 lb ⋅ ft ⋅ s 2
COSMOS: Complete Online Solutions Manual Organization System To the Instructor: The following formulas for the mass moment of inertia of thin plates and a half cylindrical shell are derived at this time for use in the solutions of Problems 9.137–9.142. Thin rectangular plate ( I x )m ( )m + md 2 = I x′ =  b  2  h  2  1 m b 2 + h 2 + m   +    12  2    2  = 1 m b2 + h2 3 ( ) ( ) ( I y )m = ( I y′ )m + md 2 = 1 b mb 2 + m   12 2 = 1 2 mb 3 2 ( )m + md 2 I z = I z′ 1 h = mh 2 + m   12 2 = 2 1 2 mh 3 Thin triangular plate Have 1  m = ρV = ρ  bht  2   1 3 bh 36 and I z, area = Then I z, mass = ρ tI z, area = ρt × = 1 3 bh 36 1 mh 2 18 continued COSMOS: Complete Online Solutions Manual Organization System I y, mass = Similarly, Now 1 mb 2 18 1 m b2 + h2 18 ( I x, mass = I y, mass + I z, mass = ) Thin semicircular plate Have π  m = ρV = ρ  a 2t  2  And I y, area = I z, area = Then π 8 a4 I y, mass = I z , mass = ρ tI y, area = ρt × π 8 a4 1 2 ma 4 = I x, mass = I y, mass + I z, mass = Now 1 2 ma 2 Also I x, mass = I x′, mass + my 2 or 1 I x′, mass = m  2 And I z, mass = I z′, mass + my 2 or 1 I z′, mass = m  4 Thin quarter-circular plate y = z = Have π  m = ρV = ρ  a 2t  4  and I y, area = I z, area = Then I y, mass = I z, mass = ρ tI y, area π 16 a4 = ρt × = 4a 3π π 16 a4 1 2 ma 4 continued COSMOS: Complete Online Solutions Manual Organization System 1 2 ma 2 Now I x, mass = I y, mass + I z, mass = Also I x, mass = I x′, mass + m y 2 + z 2 ( or 32  2 1 I x′, mass = m  − a 2  2 9π  and I y, mass = I y′, mass + mz 2 or 16  2 1 I y′, mass = m  − a 2  4 9π  )