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

2022
Views:

255

Pages:

2
Author:

garciavelascogr16

COSMOS: Complete Online Solutions Manual Organization System Chapter 9, Solution 161. First note m = ρV = γ g V = γ g AL Specific weight of aluminium = 0.10 lb/in 3 = 172.8 lb/ft 3 Then m= 2 172.8 lb/ft 3  π  0.075   ft    L 32.2 ft/s 2  4  12   ( ) = 0.16464 × 10−3 L lb ⋅ s 2/ft 2 Now L1 = L4 = 12.5 in. = 1.04167 ft m1 = m4 = 0.17150 × 10−3 lb ⋅ s 2 /ft L2 = L5 = 9 in. = 0.75 ft m2 = m5 = 0.12348 × 10−3 lb ⋅ s 2 /ft L3 = L6 = 15 in. = 1.25 ft m3 = m6 = 0.20580 × 10−3 lb ⋅ s 2 /ft and I x′y′ = I y′z ′ = I z ′x′ = 0 continued COSMOS: Complete Online Solutions Manual Organization System m, lb ⋅ s 2 /ft x , ft y , ft z , ft mx y, lb⋅ft ⋅s2 my z , lb ⋅ ft ⋅ s 2 m z x , lb ⋅ ft ⋅ s 2 1 0.17150 × 10−3 0.75 0.5208 0 0.06699 × 10−3 0 0 2 0.12348 × 10−3 0.375 1.04167 0 0.04823 × 10−3 0 0 3 0.20580 × 10−3 0 1.04167 0.625 0 0.13398 × 10−3 0 4 0.17150 × 10−3 0 0.5208 1.25 0 0.111646 × 10−3 0 5 0.12348 × 10−3 0.375 0 1.25 0 0 0.05788 × 10−3 6 0.20580 × 10−3 0.75 0 0.625 0 0 0.09647 × 10−3 Σ 0.11522 × 10−3 0.24563 × 10−3 0.15435 × 10−3 0 I xy = Σ ( I x′y′ + mx y ) = 0.115222 × 10−3 lb ⋅ ft ⋅ s 2 I xy = 0.1152 × 10−3 lb ⋅ ft ⋅ s 2 or 0 my z = 0.24563 × 10−3 lb ⋅ ft ⋅ s2 I yz = Σ ( I y′z ′ + ) or I yz = 0.246 × 10−3 lb ⋅ ft ⋅ s 2 I zx = Σ ( I z ′x′ +0 mz x ) = 0.15435 × 10−3 lb ⋅ ft ⋅ s 2 or I zx = 0.1543 × 10−3 lb ⋅ ft ⋅ s 2