Page 348 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C08
Brown.cls
330
U.S. Customary 15:14 APPLICATION TO MACHINES SI/Metric
where where
◦
◦
C 1 = 2 t tan 30 + 0.5 d C 1 = 2 t tan 30 + 0.5 d
◦ ◦
= 2 (L grip /2) tan 30 + 0.5 d bolt = 2 (L grip /2) tan 30 + 0.5 d bolt
◦
◦
= 2 (1.75 in/2) tan 30 + 0.5 (0.5in) = 2 (0.045 m/2) tan 30 + 0.5 (.012 m)
= 1.26 in = 0.032 m
and and
◦
C 2 = 2 t tan 30 + 2.5 d C 2 = 2 t tan 30 + 0.5 d
◦
◦ ◦
= 2 (L grip /2) tan 30 + 2.5 d bolt = 2 (L grip /2) tan 30 + 2.5 d bolt
◦
◦
= 2 (1.75 in/2) tan 30 + 2.5 (0.5in) = 2 (0.045 m/2) tan 30 + 2.5 (.012 m)
= 2.26 in = 0.056 m
Therefore, Therefore,
lb N
(1.814) 16 × 10 6 (0.5in) (1.814) 1.1 × 10 11 (.012 m)
in 2 m 2
k 2 = k 2 =
1.26 in 0.032 m
ln 5 ln 5
2.26 in 0.056 m
7
9
1.45 × 10 lb/in 2.39 × 10 N/m
= =
1.025 1.05
9
7
= 1.42 × 10 lb/in = 2.28 × 10 N/m
Step 5. Substitute the thickness (t 1 ), the bolt Step 5. Substitute the thickness (t 1 ), the bolt
diameter (d bolt ), and the modulus of elasticity diameter (d bolt ), and the modulus of elasticity
(E 1 ) in Eq. (8.22) to find the stiffness (k 1 ) as (E 1 ) in Eq. (8.22) to find the stiffness (k 1 ) as
◦
◦
(π tan 30 ) Ed (π tan 30 ) Ed
k 1 = k 1 =
◦
2 t tan 30 + 0.5 d 2 t tan 30 + 0.5 d
◦
ln 5 ln 5
◦
◦
2 t tan 30 + 2.5 d 2 t tan 30 + 2.5 d
◦ ◦
(π tan 30 ) E 1 d bolt (π tan 30 ) E 1 d bolt
= =
ln 5 C 1 ln 5 C 1
C 2 C 2
where where
◦
◦
C 1 = 2 t tan 30 + 0.5 d C 1 = 2 t tan 30 + 0.5 d
◦ ◦
= 2 t 1 tan 30 + 0.5 d bolt = 2 t 1 tan 30 + 0.5 d bolt
◦
◦
= 2 (0.75 in) tan 30 + 0.5 (0.5in) = 2 (0.02 m) tan 30 + 0.5 (.012 m)
= 1.17 in = 0.029 m
and and
◦
◦
C 2 = 2 t tan 30 + 2.5 d C 2 = 2 t tan 30 + 2.5 d
◦ ◦
= 2 t 1 tan 30 + 2.5 d bolt = 2 t 1 tan 30 + 2.5 d bolt
◦
= 2 (0.75 in) tan 30 + 2.5 (0.5in) = 2 (0.02 m) tan 30 + 2.5 (.012 m)
◦
= 2.17 in = 0.053 m