Page 362 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C08
Brown.cls
344
U.S. Customary 15:14 APPLICATION TO MACHINES SI/Metric
C 2 = 2 (1.0625 in/2) tan 30 ◦ C 2 = 2 (0.027 m/2) tan 30 ◦
+ 2.5 (0.625 in) +2.5 (.016 m)
= 2.18 in = 0.056 m
Therefore, Therefore,

lb N
(1.814) 30 × 10 6 (0.625 in) (1.814) 2.07 × 10 11 (.016 m)
in 2 m 2
k 1 = k 1 =
0.93 in 0.024 m
ln 5 ln 5
2.18 in 0.056 m
9
7
3.40 × 10 lb/in 6.01 × 10 N/m
= =
0.758 0.762
7
9
= 4.49 × 10 lb/in = 7.88 × 10 N/m
Step 9. Substitute the effective depth (h), the Step 9. Substitute the effective depth (h), the
bolt diameter (d bolt ) and the modulus of elastic- bolt diameter (d bolt ), and the modulus of elastic-
ity (E cast iron ) in Eq. (8.22) to find the stiffness ity (E cast iron ) in Eq. (8.22) to find the stiffness
(k 2 ) as (k 2 ) as
◦
◦
(π tan 30 ) Ed (π tan 30 ) Ed
k 2 = k 2 =
◦
◦
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 cast iron d bolt (π tan 30 ) E cast iron d bolt
= =
C 1 C 1
ln 5 ln 5
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 h tan 30 + 0.5 d bolt = 2 h tan 30 + 0.5 d bolt
= 2 (0.3125 in) tan 30 ◦ = 2 (0.008 m) tan 30 ◦
+0.5 (0.625 in) +0.5 (.016 m)
= 0.67 in = 0.017 m
and and
◦
◦
C 2 = 2 t tan 30 + 2.5 d C 2 = 2 t tan 30 + 2.5 d
◦ ◦
= 2 h tan 30 + 2.5 d bolt = 2 h tan 30 + 2.5 d bolt
= 2 (0.3125 in) tan 30 ◦ = 2 (0.008 m) tan 30 ◦
+ 2.5 (0.625 in) + 2.5 (.016 m)
= 1.92 in = 0.049 m
Therefore, Therefore,
lb N
(1.814) 16 × 10 6 (0.625 in) (1.814) 1.1 × 10 11 (.016 m)
in 2 m 2
k 2 = k 2 =
0.67 in 0.017 m
ln 5 ln 5
1.92 in 0.049 m
9
7
1.81 × 10 lb/in 3.19 × 10 N/m
= =
0.557 0.551
7
9
= 3.26 × 10 lb/in = 5.80 × 10 N/m