Page 343 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
15:14
Brown˙C08
Brown.cls
U.S. Customary MACHINE ASSEMBLY SI/Metric 325
solution solution
Step 1. Using Eq. (8.16), calculate the Step 1. Using Eq. (8.16), calculate the
threaded length (L T ) as threaded length (L T ) as
L T = L threaded − L total + L grip L T = L threaded − L total + L grip
= (1.25 in) − (2.5in) + (1.75 in) = (0.03 m) − (0.06 m) + (0.045 m)
= 0.5in = 0.015 m
Step 2. Using Eq. (8.17), calculate the un- Step 2. Using Eq. (8.17), calculate the un-
threaded length (L UT ) as threaded length (L UT ) as
L UT = L grip − L T L UT = L grip − L T
= (1.75 in) − (0.5in) = (0.045 m) − (0.015 m)
= 1.25 in = 0.03 m
Step 3. Using the nominal bolt diameter, Step 3. Using the nominal bolt diameter,
calculate the unthreaded cross-sectional area calculate the unthreaded cross-sectional area
(A UT ) as (A UT ) as
πd 2 πd 2
A UT = bolt A UT = bolt
4 4
π(0.5in) 2 π(0.012 m) 2
= =
4 4
= 1.96 × 10 −1 in 2 = 1.13 × 10 −4 m 2
Step 4. Using the threaded length (L T ) found Step 4. Using the threaded length (L T ) found
in step 1, the given threaded cross-sectional area in step 1, the given threaded cross-sectional area
(A T ), and the modulus of elasticity (E), calcu- (A T ), and the modulus of elasticity (E), calcu-
late the threaded stiffness (k T ) using Eq. (8.6) late the threaded stiffness (k T ) using Eq. (8.6)
as as
A T E A T E
k T = k T =
L T L T
2
2
2
2
9
6
(1.42 × 10 −1 in )(30 × 10 lb/in ) (8.43 × 10 −5 m )(207 × 10 N/m )
= =
(0.5in) (0.015 m)
6
9
= 8.52 × 10 lb/in = 1.16 × 10 N/m
Step 5. Using the unthreaded length (L UT ) Step 5. Using the unthreaded length (L UT )
found in step 2, the unthreaded cross-sectional found in step 2, the unthreaded cross-sectional
area (A UT ) found in step 3, and the modulus of area (A UT ) found in step 3, and the modulus of
elasticity (E), calculate the unthreaded stiffness elasticity (E), calculate the unthreaded stiffness
(k UT ) using Eq. (8.7) as (k UT ) using Eq. (8.7) as
A UT E A UT E
k UT = k UT =
L UT L UT
2
2
2
9
6
2
(1.96 × 10 −1 in )(30 × 10 lb/in ) (1.13 × 10 −4 m )(207 × 10 N/m )
= =
(1.25 in) (0.03 m)
6
8
= 4.70 × 10 lb/in = 7.80 × 10 N/m