Page 373 - Marks Calculation for Machine Design

P1: Sanjay
15:14
January 4, 2005
Brown˙C08
Brown.cls
U.S. Customary MACHINE ASSEMBLY SI/Metric 355
solution solution
Step 1. Substitute the given information in Step 1. Substitute the given information in
Eq. (8.72) to determine (τ shear ) as Eq. (8.72) to determine (τ shear ) as
P P
τ shear = τ shear =
2 (HL) 2 (HL)
3,000 lb 13,500 N
= =
2 (0.619 in)(4in) 2 (0.014 m)(0.1m)
2
2
= 606 lb/in = 0.61 kpsi = 4,821,000 N/m = 4.82 MPa
Step 2. Substitute the given information in Step 2. Substitute the given information in
Eq. (8.74) to determine the radial distance (r o ) Eq. (8.74) to determine the radial distance (r o )
as as

2 2
L L
r o = + d o 2 r o = + d 2 o
2 2

2 2
4in 0.1m

= + (1.5in) 2 = + (0.04 m) 2
2 2

2
2
= (4 + 2.25) in = (0.0025 + 0.0016) m

2
2
= 6.25 in = 2.5in = 0.0041 m = 0.064 m
Step 3. Substitute the given information in Step 3. Substitute the given information in
Eq. (8.75) to determine the polar moment of Eq. (8.75) to determine the polar moment of
inertia (J group ) as inertia (J group ) as
3 3 3 3
LH HL LH HL
J group = 2 + + LHd o 2 J group = 2 + + LHd o 2
12 12 12 12
(4in)(0.619 in) (0.1m)(0.014 m)
 3   3 
 12   12 
 3   3 
(0.619 in)(4in) (0.014 m)(0.1m)
= 2   = 2  
+ +
   
 12   12 
+(4in)(0.619 in)(1.5in) 2 +(0.1m)(0.014 m)(0.04 m) 2
−2 −8 −6
(7.9 × 10 + 3.301 (2.29 × 10 + 1.167 × 10
= 2 = 2
+5.571 ) in 4 +2.24 × 10 −6 ) m 4
= 8.95 in 4 = 3.43 × 10 −6 m 4
Step 4. Substitute the radial distance (r o ) Step 4. Substitute the radial distance (r o )
found in step 2, the polar moment of inertia found in step 2, the polar moment of inertia
(J group ) found in step 3, and the given infor- (J group ) found in step 3, 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
(3,000 lb)(12 in)(2.5in) (13,500 N)(0.3m)(0.064 m)
= =
8.95 in 4 3.43 × 10 −6 m 4