Page 359 - Marks Calculation for Machine Design

P1: Sanjay
15:14
January 4, 2005
Brown.cls
Brown˙C08
U.S. Customary MACHINE ASSEMBLY SI/Metric 341
Example 5. Determine the load factor (n load ), Example 5. Determine the load factor (n load ),
factor-of-safety against separation (n separation ), factor-of-safety against separation (n separation ),
fatigue factor-of-safety (n fatigue ), and the factor- fatigue factor-of-safety (n fatigue ), and the factor-
of-safety against yielding (n yield ) for a per- of-safety against yielding (n yield ) for a per-
manent high-strength cap screw and washer manent high-strength cap screw and washer
assembly like that shown in Fig. 8.6, where assembly like that shown in Fig. 8.6, where
d capscrew = 0.625 in (nominal) d bolt = 16 mm = 0.016 m (nominal)
2
2
A T = 0.226 in = 2.26 × 10 −1 in 2 A T = 157 mm = 1.57 × 10 −4 m 2
6
3
S proof = 85 kpsi = 85 × 10 lb/in 2 S proof = 600 MPa = 600 × 10 N/m 2
6
3
S e = 18.6 kpsi = 18.6 × 10 lb/in 2 S e = 129 MPa = 129 × 10 N/m 2
6
3
S ut = 120 kpsi = 120 × 10 lb/in 2 S ut = 830 MPa = 830 × 10 N/m 2
6
3
S y = 92 kpsi = 92 × 10 lb/in 2 S y = 660 MPa = 660 × 10 N/m 2
D = 1.5 d bolt (washer face) D = 1.5 d bolt (washer face)
α = 30 (cone angle) α = 30 (cone angle)
◦
◦
t washer = 0.125 in (steel) t washer = 3mm = 0.003 m (steel)
t 1 = 0.625 in (steel) t 1 = 16 mm = 0.016 m (steel)
t 2 = 0.75 in (cast iron) t 2 = 20 mm = 0.02 m (cast iron)
9
6
E steel = 30 × 10 lb/in 2 E steel = 207 GPa = 207 × 10 N/m 2
6
9
E cast iron = 16 × 10 lb/in 2 E cast iron = 110 GPa = 110 × 10 N/m 2
P = 4,000 lb P = 18,000 N
solution solution
Step 1. As the thickness (t 2 ) is greater than Step 1. As the thickness (t 2 ) is greater than
the nominal diameter (d), the guidelines in the nominal diameter (d), the guidelines in
Eq. (8.63) give the effective depth (h) as Eq. (8.63) give the effective depth (h) as
d 0.625 in d 0.016 m
h = = = 0.3125 in h = = = 0.008 m
2 2 2 2
Step2. Substitutetheeffectivedepth(h)found Step2. Substitutetheeffectivedepth(h)found
in step 1, the given thicknesses (t washer ) and (t 1 ) in step 1, the given thicknesses (t washer ) and (t 1 )
in Eq. (8.62) to determine the length of the grip in Eq. (8.62) to determine the length of the grip
(L grip ) as (L grip ) as
L grip = t washer + t 1 + h L grip = t washer + t 1 + h
= (0.125 in) + (0.625 in) = (0.003 m) + (0.016 m)
+ (0.3125 in) + (0.008 m)
= 1.0625 in = 0.027 m
Step3. UsingEq.(8.67),theeffectivethreaded Step3. UsingEq.(8.67),theeffectivethreaded
length of the cap screw (L T ) is equal to the grip length of the cap screw (L T ) is equal to the grip
length (L grip ), therefore length (L grip ), therefore
L T = L grip = 1.0625 in L T = L grip = 0.019 m
Step 4. Using the threaded length (L T ) found Step 4. Using the threaded length (L T ) found
in step 3, the given tensile-stress area (A T ), and in step 1, the given tensile-stress area (A T ), and
the modulus of elasticity (E steel ), calculate the the modulus of elasticity (E steel ), calculate the
cap screw stiffness (k T ) using Eq. (8.5) as cap screw stiffness (k T ) using Eq. (8.5) as