Page 364 - Marks Calculation for Machine Design

P1: Sanjay
January 4, 2005
Brown˙C08
Brown.cls
346
U.S. Customary 15:14 APPLICATION TO MACHINES SI/Metric
Step 13. Use the guidelines in Eq. (8.26) to Step 13. Use the guidelines in Eq. (8.26) to
determine the bolt preload (F preload ) as determine the bolt preload (F preload ) as
F preload = 0.90 F proof F preload = 0.90 F proof
= (0.90)(19,200 lb) = (0.90)(94,200 N)
= 17,280 lb = 84,780 N
Step 14. Substitute the proof load (F proof ) Step 14. Substitute the proof load (F proof )
found in step 12, the bolt preload (F preload ) found in step 12, the bolt preload (F preload )
found in step 13, the joint constant (C) found found in step 13, the joint constant (C) found
in step 11, and the given external load (P) in in step 11, and the given external load (P) in
Eq. (8.45) to determine the load factor (n load ) Eq. (8.45) to determine the load factor (n load )
as as
F proof − F preload F proof − F preload
n load = n load =
CP CP
(19,200 lb) − (17,280 lb) (94,200 N) − (84,780 N)
= =
(0.27)(4,000 lb) (0.28)(18,000 N)
1,920 lb 9,420 N
= = 1.78 < 2 = = 1.87 < 2
1,080 lb 5,040 N
Step 15. Substitute the bolt preload (F preload ) Step 15. Substitute the bolt preload (F preload )
found in step 13, the joint constant (C) found found in step 13, the joint constant (C) found
in step 11, and the given external load (P) in step 11, and the given external load (P)
in Eq. (8.49) to determine the factor-of-safety in Eq. (8.49) to determine the factor-of-safety
against separation (n separation ) as against separation (n separation ) as

F preload F preload
n separation = n separation =
P (1 − C) P (1 − C)
17,280 lb 84,780 N
= =
(4,000 lb)(1 − 0.27) (18,000 N)(1 − 0.28)
17,280 lb 84,780 N
= = 5.9 ∼ 6 = = 6.5
=
2,920 lb 12,960 N
Step 16. Substitute the joint constant (C) Step 16. Substitute the joint constant (C)
found in step 11, and the given external load found in step 11, and the given external load
(P) and tensile-stress area (A T ) in Eq. (8.55) to (P) and tensile-stress area (A T ) in Eq. (8.55) to
determine the alternating stress on the bolt (σ a ) determine the alternating stress on the bolt (σ a )
as as
CP (0.27)(4,000 lb) CP (0.28)(18,000 N)
σ a = = 2 σ a = = −4 2
2 A T (2)(2.26 × 10 −1 in ) 2 A T (2)(1.57 × 10 m )
1,080 lb 5,040 N
= = −4 2
4.52 × 10 −1 in 2 3.14 × 10 m
7
3
= 2.39 × 10 lb/in 2 = 1.61 × 10 N/m 2
= 2.39 kpsi = 16.1MPa