Page 382 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C08
Brown.cls
364
U.S. Customary 15:14 APPLICATION TO MACHINES SI/Metric
Step 4. Substitute the given information in Step 4. Substitute the given information in
Eq. (8.86) to determine the polar moment of Eq. (8.86) to determine the polar moment of
inertia (J group ) as inertia (J group ) as
2
2
(2 L 1 + L 2 ) 3 L (L 1 + L 2 ) 2 (2 L 1 + L 2 ) 3 L (L 1 + L 2 ) 2
1
1
J group = − J group = −
12 2 L 1 + L 2 12 2 L 1 + L 2
(2(5in) + (10 in)) 3 (2(0.13 m) + (0.26 m)) 3
= =
12 12
2
2
(5in) (5in + 10 in) 2 (0.13 m) (0.13 m + 0.26 m) 2
− −
2(5in) + 10 in 2(0.13 m) + 0.26 m
2
2
(20 in) 3 (25 in )(15 in) 2 (0.52 m) 3 (0.0169 m )(0.39 m) 2
= − = −
12 20 in 12 0.52 m
8,000 in 3 5,625 in 4 0.1406 m 3 0.00257 m 4
= − = −
12 20 in 12 0.52 m
= (666.67 − 281.25) in 3 = (0.01172 − 0.00494) m 3
= 385.4in 3 = 0.00678 m 3
Step 5. Substitute the radial distance (r o ) Step 5. Substitute the radial distance (r o )
found in step 3, the polar moment of inertia found in step 3, the polar moment of inertia
(J group ) found in step 4, and the given infor- (J group ) found in step 4, and the given infor-
mation in Eq. (8.73) to determine (τ torsion ) as mation in Eq. (8.73) to determine (τ torsion ) as
PL o r o PL o r o
τ torsion = τ torsion =
J group J group
(18,000 lb)(10 in)(6.25 in) (81,000 N)(0.26 m)(0.1625 m)
= =
385.4in 3 0.00678 m 3
1,125,000 lb · in 2 in 3,422 N · m 2 m
= × = 3 ×
385.4in 3 in 0.00678 m m
2
2
= 2,919 (lb/in ) · in = 504,800 (N/m ) · m
= 2.9 kpsi · in = 0.50 MPa · m
Step 6. Substitute the distance (D o ) from Step 6. Substitute the distance (D o ) from
step 2 and the given information in Eq. (8.87) step 2 and the given information in Eq. (8.87)
to determine the angle (α) as to determine the angle (α) as
α = tan −1 L 1 − D o α = tan −1 L 1 − D o
L 2 L 2
2 2
(5in) − (1.25 in) (0.13 m) − (0.0325 m)
= tan −1 = tan −1
(10 in) (0.26 m)
2 2
3.75 in 0.0975 m
= tan −1 = tan −1 (0.75) = tan −1 = tan −1 (0.75)
5in 0.13 m
= 37 ◦ = 37 ◦
Step 7. Substitute the angle (α) found in step 6 Step 7. Substitute the angle (α) found in step 6
in Eq. (8.88) to determine the angle (β) as in Eq. (8.88) to determine the angle (β) as
◦
◦
◦
◦
◦
◦
β = 180 − α = 180 − 37 = 143 ◦ β = 180 − α = 180 − 37 = 143 ◦