Page 405 - Marks Calculation for Machine Design

P1: Naresh
January 4, 2005
15:28
Brown.cls
Brown˙C09
U.S. Customary MACHINE ENERGY SI/Metric 387
solution solution
Step 1. Using Eq. (9.12), calculate the spring Step 1. Using Eq. (9.12), calculate the spring
index (C) as index (C) as
D 0.9in D 0.022 m
C = = = 9 C = = = 11
d 0.1in d 0.002 m
Step 2. Using Eq. (9.14), calculate the shear- Step 2. Using Eq. (9.14), calculate the shear-
stress correction factor (K s ) as stress correction factor (K s ) as
1 1 1 1
K s = + 1 = + 1 K s = + 1 = + 1
2C 2(9) 2C 2(11)
= (0.056) + 1 = 1.056 = (0.045) + 1 = 1.045
Step 3. Using Eq. (9.16), calculate the Step 3. Using Eq. (9.16), calculate the
Bergstr¨asser factor (K B ) as Bergstr¨asser factor (K B ) as
4C + 2 4(9) + 2 38 4C + 2 4(11) + 2 46
K B = = = = 1.152 K B = = = = 1.122
4C − 3 4(9) − 3 33 4C − 3 4(11) − 3 41
Step 4. Use Eqs. (9.47) and (9.48) to find the Step 4. Use Eqs. (9.47) and (9.48) to find the
mean and alternating spring forces. maximum and minimum spring forces.
F max + F min (40 lb) + (10 lb) F max + F min (175 N) + (45 N)
F m = = F m = =
2 2 2 2
50 lb 220 N
= = 25 lb = = 110 N
2 2
F max − F min (40 lb) − (10 lb) F max − F min (175 N) − (45 N)
F a = = F a = =
2 2 2 2
30 lb 130 N
= = 15 lb = = 65 N
2 2
Step 5. Use the shear-stress correction factor Step 5. Use the shear-stress correction factor
(K s ) found in step 2 in Eq. (9.49) to find the (K s ) found in step 2 in Eq. (9.49) to find the
mean shear stress (τ m ). mean shear stress (τ m ).
8F m D 8F m D
τ m = K s τ m = K s
πd 3 πd 3
(8)(25 lb)(0.9in) (8)(110 N)(0.022 m)
= (1.056) = (1.045)
π(0.1in) 3 π(0.002 m) 3
180 lb 19.36 N
= (1.056) = (1.045)
0.00314 in 2 0.000000025 m 2
= (1.056)(57.3 kpsi) = (1.045)(770.3MPa)
= 60.5 kpsi = 805 MPa
Step 6. Use the Bergstrasser factor (K B ) found Step 6. Use the Bergstrasser factor (K B ) found
in step 3 in Eq. (9.50) to find the alternating in step 3 in Eq. (9.50) to find the alternating
shear stress (τ a ). shear stress (τ a ).
8F a D 8F a D
τ a = K B 3 τ a = K B 3
πd πd
(8)(15 lb)(0.9in) (8)(65 N)(0.022 m)
= (1.152) = (1.122)
π(0.1in) 3 π(0.002 m) 3
108 lb 11.44 N
= (1.152) = (1.122)
0.00314 in 2 0.000000025 m 2
= (1.152)(34.4 kpsi) = (1.122)(455.2MPa)
= 39.6 kpsi = 511 MPa