University:
Michigan State University
Course:
ERG 160 | Success in Science, Technology, Engineering, and Mathematics
Academic year:

2021
Views:

368

Pages:

3
Author:

Athena Harrison

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 164. (a) First divide the tetrahedron into a series of thin vertical slices of thickness dz as shown. a z  z + a = a 1 −  c c   Now x=− and b z  y = − z + b = b 1 −  c c  The mass dm of the slab is 2 z 1  1  dm = ρ dV = ρ  xydz  = ρ ab 1 −  dz c 2  2  c 2 3  c  1 z 1 z   m = ∫ dm = ∫ ρ ab 1 −  dz = ρ ab  − 1 −   c 2 c    02  3  0 c Then = 1 ρ abc 6 dI zx = dI z′x′ + z EL xEL dm Now dI z′x′ = 0 where z EL = z and Then xEL = ( symmetry ) 1 1  z x = a 1 −  3 3  c 2 c  1  z   1 z  I zx = ∫ dI zx = ∫ z  a 1 −    ρ ab  1 −  dz  c    2 c   0 3  = 1 2 c z2 z3 z 4  ρ a b∫ 0  z − 3 + 3 2 − 3  dz 6 c c c   c m 1 z3 3 z 4 1 z5  = a  z2 − + −  c 2 c 4 c 2 5 c3  0 or I zx = 1 mac 20 continued COSMOS: Complete Online Solutions Manual Organization System (b) Because of the symmetry of the body, I xy and I yz can be deduced by considering the circular permutation of ( x, y, z ) and ( a, b, c ) . Thus 1 I xy = mab 20 I yz = 1 mbc 20 Alternative solution for part a First divide the tetrahedron into a series of thin horizontal slices of thickness dy as shown. Now x=− a y  y + a = a 1 −  b b   and z =− c y  y + c = c 1 −  b b  The mass dm of the slab is 2 y 1  1  dm = ρ dV = ρ  xzdy  = ρ ac 1 −  dy b 2  2  Now dI zx = ρ tdI zx, Area where t = dy and dI zx, Area = Then 1 2 2 x z from the results of Sample Problem 9.6 24 2 2  1   y    y    dIzx = ρ ( dy )   a 1 −   c 1 −    b    b    24    4 4 1 y 1m  y  ρ a 2c 2 1 −  dy = = ac 1 −  dy 24 b 4 b  b  Finally I zx = ∫ dI zx 4 1m  y =∫ ac 1 −  dy 4 b  b b 0 b 5 1 m  b  y  = ac  −  1 −   4 b  5  b  0 or I zx = 1 mac 20 continued COSMOS: Complete Online Solutions Manual Organization System Alternative solution for part a The equation of the included face of the tetrahedron is x y z + + =1 a b c x z  y = b 1 − −  a c  so that For an infinitesimal element of sides dx, dy, and dz dm = ρ dV = ρ dydxdz z  x = a 1 −  c  From part a Now I zx = ∫ zxdm = ∫ 0 ∫ 0 ( c = ρ∫0 ∫0 ( c a 1− cz = ρ b∫ c 1 z 0  a 1− cz ) b(1− ax − cz ) zx ( ρ dydxdz ) ∫0 ) zx b (1 −  x a − z c ) dxdz ( a 1− cz 1 x3 1 z 2  x − x  − 3 a 2 c 0 2 2 ) dz 2 3 2 z 1 3 z 1 z 2 z  c 1  a 1 −  − a 1 −   dz = ρ b∫ 0 z  a 2 1 −  − c 3a  c 2c  c    2  c = ρ b∫ 0 = 3 1 2  z a z 1 −  dz 6 c  1 2 c z2 z3 z 4  ρ a b∫ 0  z − 3 + 3 2 − 3  dz 6 c c c   c m 1 z3 3 z 4 1 z5  = a  z2 − + −  c 2 c 4 c 2 5 c3  0 or I zx = 1 mac 20